Chemistry Malaysian Matriculation Full Notes & Slides for Semester 1 and 2

WELCOME TO SKO16 CHEMISTRY CHEMISTRY CHEMISTRY

CHEMISTRY SK016 Chapter

1.0 2.0 3.0 4.0 5.0 6.0 7.0

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Topics

Hours

Matter Atomic Structure Periodic Table Chemical Bonding State of Matter Chemical Equilibrium Ionic Equilibria Total

7 7 4 2 7 5 12 54

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Topic

8.0 9.0 10.0 11.0 12.0 13.0 14.0 15.0

Thermochemistry Electrochemistry Reaction Kinetics Intro To Organic Chemistry Hydrocarbons Aromatic Compounds Haloalkanes (Alkyl halides) Hydroxy compounds

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Hours

4 6 7 4 8 3 4 3 3


16.0 17.0 18.0 19.0 20.0

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Topic

Carbonyl Carboxylic acids & Derivatives Amines Amino acids and Proteins Polymers

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Hour

4 4 5 2 1

4

ASSESSMENT 1. COURSEWORK (20%)  Continuous evaluation (tutorial/test/quiz) - 10%  Practical work - 10% 2. MID-SEMESTER EXAMINATION

- 10%

3. FINAL EXAMINATION (70%)  Paper 1 (30 multiple choice questions)- 30%  Paper 2 (Part Astructured) (Part B-long structured) -100% 08/16/11

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REFERENCE BOOKS 

CHEMISTRY ,9th Ed. – Raymond Chang, McGraw-Hill



CHEMISTRY –The Molecular Nature of Matter and Change, 3rd Ed.– Martin Silberberg, McGraw Hill



CHEMISTRY – The Central Science, 9th Ed. Theodore L.Brown, H.Eugene LeMay,Jr, Bruce E Bursten, Pearson Education



GENERAL CHEMISTRY – Principle & Structure, 6th Ed. James E Brady, John Wiley and Sons.

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GENERAL CHEMISTRY – Principle and Modern Applications, 8th Ed. Ralph H. Petrucci, William S. Harwood, Prentice-Hall

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ORGANIC CHEMISTRY, 7th Ed – T.W.Graham Solomon,Craig B.Fryhle, John Wiley and Sons



ORGANIC CHEMISTRY, 4th Ed – L.G. Wade, Jr, Prentice Hall



ORGANIC CHEMISTRY, 6th Ed – John McMurry, Thompson – Brooks/Cole

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Chapter 1 : MATTER 1.1 Atoms and Molecules 1.2 Mole Concept 1.3 Stoichiometry

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1.1 Atoms and Molecules

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Learning Outcome At the end of this topic, students should be able to: (a) Describe proton, electron and neutron in terms of the relative mass and relative charge. (b) Define proton number, Z, nucleon number, A and isotope. (c) Write isotope notation. 08/16/11

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Introduction 

Matter  Anything that occupies space and has mass. e.g: air, water, animals, trees, atoms, etc



Matter may consists of atoms, molecules or ions.

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Classifying Matter

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A substance is a form of matter that has a definite or constant composition and distinct properties. Example: H2O, NH3, O2 

A mixture is a combination of two or more substances in which the substances retain their identity. Example : air, milk, cement 

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An element is a substance that cannot be separated into simpler substances by chemical means. Example : Na, K, Al,Fe A compound is a substance composed of atoms of two or more elements chemically united in fixed proportion. Example : CO2, H2O, CuO

Three States of Matter

SOLID 08/16/11

LIQUID matter

GAS 15

1.1 Atoms and Molecules a) Atoms 



An atom is the smallest unit of a chemical element/compound. In an atom, there are 3 subatomic particles: - Proton (p) - Neutron (n) - Electron (e)

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Modern Model of the Atom



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Electrons move around the region of the atom. matter

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

All neutral atoms can be identified by the number of protons and neutrons they contain.



Proton number (Z)  is the number of protons in the nucleus of the atom of an element (which is equal to the number of electrons). Protons number is also known as atomic number.



Nucleon number (A)  is the total number of protons and neutrons present in the nucleus of the atom of an element. Also known as mass number.

Subatomic Particles Particle

Mass

Charge

Charge

(gram)

(Coulomb)

(units)

Electron (e)

9.1 x 10-28

-1.6 x 10-19

-1

Proton (p)

1.67 x 10-24

+1.6 x 10-19

+1

Neutron (n)

1.67 x 10-24

0

0

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Isotope 

Isotopes are two or more atoms of the same element that have the same proton number in their nucleus but different nucleon number.

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Examples:

1 1

H

235 92

U

2 1

H (D)

238 92

U

3 1

H (T)

Isotope Notation An atom can be represented by an isotope notation ( atomic symbol )

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X=

Element symbol

Z =

Proton number of X (p)

A= =

Nucleon number of X p+n

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Nucleon number of mercury, A = 202

Total charge on the ion

The number of neutrons =A–Z = 202 – 80 = 122

Proton number of mercury, Z = 80

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

In a neutral atom:  number of protons equals number of electrons



In a positive ion:  number of protons is more than number of electrons



In a negative ion:  number of protons is less than number of electrons

Exercise 1 Give the number of protons, neutrons, electrons and charge in each of the following species: Symbol

Number of : Proton

200 80

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Neutron Electron

Hg

63 29

Cu

17 8

O 2

59 27

Charge

Co 3  matter

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Exercise 2 Write the appropriate notation for each of the following nuclide : Species

A B C D 08/16/11

Number of : Notation Proton Neutron Electron for nuclide 2 1 1 7

2 2 1 7 matter

2 0 1 10 26

b) Molecules A molecule consists of a small number of atoms joined together by bonds.

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A diatomic molecule  Contains only two atoms Ex : H2, N2, O2, Br2, HCl, CO A polyatomic molecule  Contains more than two atoms Ex : O3, H2O, NH3, CH4

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Learning Outcomes At the end of this topic, student should be able to : (a) Define relative atomic mass, Ar and relative molecular mass, Mr based on the C-12 scale. (b) Calculate the average atomic mass of an element given the relative abundance of isotopes or a mass spectrum. 08/16/11

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Relative Mass i.

Relative Atomic Mass, Ar A mass of one atom of an element compared to 1/12 mass of one atom of 12C with the mass 12.000 amu

Mass of one atom of element Re lative atomic mass, A r  1 X Mass of one atom of 12 C 12

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

Mass of an atom is often expressed in atomic mass unit, amu (or u).



Atomic mass unit, amu is defined to be one twelfth of the mass of 12C atom



Mass of a 12C atom is given a value of exactly 12 amu 1 u = 1.660538710-24 g



The relative isotopic mass is the mass of an atom, scaled with 12C.

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Example 1 Determine the relative atomic mass of an element Y if the ratio of the atomic mass of Y to carbon-12 atom is 0.45

ANSWER:

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ii) Relative Molecular Mass, Mr A mass of one molecule of a compound compared to 1/12 mass of one atom of 12C with the mass 12.000amu Relative 

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1 12

molecular mass, Mr Mass of one molecule x Mass of one atom of

matter

12

C

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The relative molecular mass of a compound is the summation of the relative atomic masses of all atoms in a molecular formula.

Example 2 Calculate the relative molecular mass of C5H5N, Ar C = 12.01 Ar H = 1.01 Ar N = 14.01

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MASS SPECTROMETER 



An atom is very light and its mass cannot be measured directly A mass spectrometer is an instrument used to measure the precise masses and relative quantity of atoms and molecules

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Mass Spectrum of Monoatomic Elements 

Modern mass spectrum converts the abundance into percent abundance

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Relative abundance

Mass Spectrum of Magnesium 

The mass spectrum of Mg shows that Mg consists of 3 isotopes: 24Mg, 25Mg and 26Mg.



The height of each line is proportional to the abundance of each isotope.

63 8.1 9.1 24

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25 26

m/e (amu) 

matter

24Mg

is the most abundant of the 3 isotopes

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Learning Outcomes At the end of this topic, student should be able to : (a) Calculate the average atomic mass of an element given the relative abundances of isotopes or a mass spectrum.

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How to calculate the relative atomic mass, Ar from mass spectrum? 

Ar is calculated using data from the mass spectrum.



The average of atomic masses of the entire element’s isotope as found in a particular environment is the relative atomic mass, Ar of the atom. (isotopicmass abundance )  Averageatomicmass abundance

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Example 1: Calculate the relative atomic mass of neon from the mass spectrum.

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Solution: Average atomic mass of Ne =

 (% abundance  isotopic mass) =  % abundanc e

(90.5  20 u)  (0.3  21 u)  (9.2  22 u) (90.5  0.3  9.2)

= 20.2 u Relative atomic mass Ne = 20.2

Example 2: Copper occurs naturally as mixture of 69.09% of 63Cu and 30.91% of 65Cu. The isotopic masses of 63Cu and 65Cu are 62.93 u and 64.93 u respectively. Calculate the relative atomic mass of copper.

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Solution: Average atomic mass of Cu = =

(%abundance isotopicmass)  = % abundance

(69.09  62.93 u)  (30.91 64.93 u) (69.09  30.91)

63.55 u

Relative atomic mass Cu = 63.55

Example 3: Naturally occurring iridium, Ir is composed of two isotopes, 191Ir and 193Ir in the ratio of 5:8. The relative isotopic mass of 191Ir and 193Ir are 191.021 u and 193.025 u respectively. Calculate the relative atomic mass of Iridium

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Solution:

Average atomic =  abundance  isotopic mass  abundance  mass of Ir = =

(5  191 .021 u)  (8  193 .025 u) (5  8 )

192.254 u

Relative atomic mass Ir = 192.254 08/16/11

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Mass Spectrum of Molecular Elements A sample of chlorine which contains 2 isotopes with nucleon number 35 and 37 is analyzed in a mass spectrometer. How many peaks would be expected in the mass spectrum of chlorine?

35Cl-35Cl 35Cl-37Cl

MASS SPECTROMETER

37Cl-37Cl

+

_

_

Cl2

35Cl-35Cl+

Cl2 + e 

Cl2+

+ 2e

Cl2 + e  2Cl+ + 2e

35Cl-37Cl+ 37Cl-37Cl+ 35Cl+ 37Cl+

Mass Spectrum of Diatomic Elements

Exercise: How many peaks would be expected in a mass spectrum of X2 which consists of 3 isotopes?

MATTER 1.2 Mole Concept

Learning Outcome At the end of this topic, students should be able to:

a) Define mole in terms of mass of carbon-12 and Avogadro’s constant, NA

Avogadro’s Number, NA 







Atoms and molecules are so small – impossible to count A unit called mole (abbreviated mol) is devised to count chemical substances by weighing them A mole is the amount of matter that contains as many objects as the number of atoms in exactly 12.00 g of carbon-12 isotope The number of atoms in 12 g of 12C is called Avogadro’s number, NA = 6.02 x 1023

Example: 1 mol of Cu contains 6.02  1023

Cu atoms

1 mol of O2 contains 6.02  1023 O2 molecules O atoms 2  6.02  1023 1 mol of NH3 contains 6.02  1023 NH3 molecules 23 atoms 6.02  10N 23 atoms 3  6.02  10H

1 mol of CuCl2 contains 6.02  1023 Cu2+ ions 23- ions 2  6.02  10Cl

Mole and Mass Example: Relative atomic mass for carbon, C = 12.01 Mass of 1 C atom = 12.01 amu Mass of 1 mol C atoms = 12.01 g Mass of 1 mol C atoms consists of 6.02 x 1023 C atoms = 12.01 g

12.01 g  Mass of 1 C atom = 6.02 x 10 23 = 1.995 x 10-23 g

12.01 amu = 1.995 x 10-23 g 1.995 x 10 23 g 1 amu = 12.01 amu

= 1.66 x 10-23 g

Example: From the periodic table, Ar of nitrogen, N is 14.01 The mass of 1 N atom =

14.01 amu

The mass of 1 mol of N atoms = The molar mass of N atom =

14.01 g

14.01 g mol1

The molar mass of nitrogen gas = 28.02 g mol1 

The nucleon number of N = 14

Mr of CH4 is 16.05 The mass of 1 CH4 molecule =

16.05 amu

The mass of 1 mol of CH4 molecules = 16.05 g The molar mass of CH4 molecule = 16.05 g mol1

Learning Outcome At the end of this topic, students should be able to: (a) Interconvert between moles, mass, number of particles, molar volume of gas at STP and room temperature. (b) Define the terms empirical & molecular formulae (c) Determine empirical and molecular formulae from mass composition or combustion data. 08/16/11

MATTER

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Example 1: Calculate the number of moles of molecules for 3.011 x 1023 molecules of oxygen gas. Solution: 6.02 x 1023 molecules of O2 

3.011 x 1023

1 mol of O2 molecules

3.01110 23 molecules  1 mol molecules of O2  6.02  10 23 molecules

= 0.5000 mol of O2 molecules

Example 2: Calculate the number of moles of atoms for 1.204 x 1023 molecules of nitrogen gas. Solution: 6.02 x 1023 molecules of N2 

1.204 x

1023

1mol of N2 molecules 2 mol of N atoms

molecules of N2

23 1 . 204  10 molecules  2 mol  6.02 10 23 molecules

= 0.4000 mol of N atoms

Example 3: Calculate the mass of 0.25 mol of chlorine gas. Solution: 1 mol Cl2 

2  35.45 g

0.25 mol Cl2

 2  35.45 g  0.25 mol 1 mol

18 g

 or

mass = mol x molar mass = 0.25 mol x (2 x 35.45 g mol-1) = 18 g

Example 4: Calculate the mass of 7.528 x 1023 molecules of methane, CH4 Solution: 6.02 x 1023 CH4 molecules  (12.01 + 4(1.01)) g

7.528 x

1023

23 16 . 05 g  7.528  10 CH4 molecules  6.02  10 23

= 20.06 g

Molar Volume of Gases 





Avogadro (1811) stated that equal volumes of gases at the same temperature and pressure contain equal number of molecules Molar volume is a volume occupied by 1 mol of gas At standard temperature and pressure (STP), the molar volume of an ideal gas is 22.4 L mol 1 Standard

Temperature

273.15 K 0 C

and Pressure 1 atm

101325 N m-2 101325 Pa

760 mmHg

Standard Molar Volume

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

At room conditions (1 atm, 25 C), the molar volume of a gas = 24 L mol-1

Example 1: Calculate the volume occupied by 1.60 mol of Cl2 gas at STP. Solution: At STP, 1 mol Cl2 occupies22.4 L 1.60 mol Cl2 occupies

1.60 mol  22.4 L 1 mol = 35.8 L

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Example 2: Calculate the volume occupied by 19.61 g of N2 at STP Solution: 1 mol of N2 occupies 22.4 L 19.61 g 2(14.01) g mol 1

 19.61    mol  22.4 L  2(14.01)  of N2 occupies 1 mol

= 15.7 L 08/16/11

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Example 3: 0.50 mol methane, CH4 gas is kept in a cylinder at STP. Calculate: (a) The mass of the gas (b) The volume of the cylinder (c) The number of hydrogen atoms in the cylinder Solution: 16.05 g (a) Mass of 1 mol CH4 = Mass of 0.50 mol CH4 16=.05 g  0.50 mol

1 mol

= 8.0 g 08/16/11

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(b) At STP; 1 mol CH4 gas occupies 22.4 L 0.50 mol CH4 gas occupies 22.4 L  0.50 mol 1 mol = 11 L (c) 1 mol of CH4 molecules  4 mol of H atoms 0.50 mol of CH4 molecules  2 mol of H atoms 1 mol of H atoms  2 mol of H atoms   08/16/11

MATTER

6.02 x 1023 atoms 2 x 6.02 x 1023 atoms 1.2 x 1024 atoms 72

Exercise A sample of CO2 has a volume of 56 cm3 at STP. Calculate: a) The number of moles of gas molecules (0.0025 mol) a) The number of CO2 molecules (1.506 x 1021 molecules) a) The number of oxygen atoms in the sample (3.011x1021atoms) Notes: 08/16/11

1 dm3 1 dm3

= 1000 cm3 =1L MATTER

73

Empirical And Molecular Formulae -

Empirical formula => chemical formula that shows the simplest ratio of all elements in a molecule.

-

Molecular formula => formula that show the actual number of atoms of each element in a molecule. 08/16/11

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-

The relationship between empirical formula and molecular formula is : Molecular formula = n ( empirical formula )

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Example: A sample of hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass. Its molar mass is 56. Determine the empirical formula and molecular formula of the compound.

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Solution: C

H

Mass

85.7

14.3

Number of moles

85.7 12.01

14.3 1.01

Simplest ratio

= 7.1357

= 14.1584

1

1.984 2

Empirical formula = CH2

56 n 14.03  3.99 4

Molecular formula  n(CH2 )  Molecular formula  C 4H8

Exercise: A combustion of 0.202 g of an organic sample that contains carbon, hydrogen and oxygen produce 0.361g carbon dioxide and 0.147 g water. If the relative molecular mass of the sample is 148, what is the molecular formula of the sample? Answer : C6H12O4

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79

Learning Outcome At the end of this topic, students should be able to: (a) Define and perform calculation for each of the following concentration measurements : i) molarity (M) ii) molality(m) iii) mole fraction, X iv) percentage by mass, % w/w v) percentage by volume, %v/v 08/16/11

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Concentration of Solutions 

A solution is a homogeneous mixture of two or more substances: solvent + solute(s) e.g: sugar + water – solution sugar – solute water –

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solvent 81

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Concentration of a solution can be expressed in various ways : a) molarity b) molality c) mole fraction d) percentage by mass e) percentage by volume

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a) Molarity 

Molarity is the number of moles of solute in 1 litre of solution moles of solute (mol) molarity, M  volume of solution (L)



Units of molarity: mol L-1 mol dm-3 M

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84

Example 1: Determine the molarity of a solution containing 29.22 g of sodium chloride, NaCl in a 2.00 L solution.

Solution:

MNaCl



 =

nNaCl Vsolution 29.22   mol    (22.99  35.45)  2.00 L 0.250 mol L-1

Example 2: How many grams of calcium chloride, CaCl2 should be used to prepare 250.00 mL solution with a concentration of 0.500 M

Solution: nCaCl  M CaCl x Vsolution 2

2

= 0.500 mol L1  250.00  103 L

mass of CaCl2  nCaCl x molar mass 2

= (0.500  250.00  103) mol  (40.08 + 2(35.45)) g mol1

= 13.9 g

b) Molality 

Molality is the number of moles of solute dissolved in 1 kg of solvent moles of solute (mol) molality, m  mass of solvent (kg)



Units of molality:mol kg-1 molal m

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89

Example: What is the molality of a solution prepared by dissolving 32.0 g of CaCl2 in 271 g of water?

Solution:

nCaCl 2

32.0 g  40.08  2(35.45) g mol-1

 32.0 mol    Molality of CaCl2   110.98 3  271 10 kg

 1.06 mol kg1

Exercise: Calculate the molality of a solution prepared by dissolving 24.52 g of sulphuric acid in 200.00 mL of distilled water. (Density of water = 1 g mL-1) Ans = 1.250 mol kg-1

c) Mole Fraction (X) Mole fraction is the ratio of number of moles of one component to the total number of moles of all component present. For a solution containing A, B and C: nA Mol fraction of A, X A  nA  nB  nC nA  nT 08/16/11

MATTER

93







Mol fraction is always smaller than 1 The total mol fraction in a mixture (solution) is equal to one. XA + XB + XC + X….. = 1 Mole fraction has no unit (dimensionless) since it is a ratio of two similar quantities.

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94

Example: A sample of ethanol, C2H5OH contains 200.0 g of ethanol and 150.0 g of water. Calculate the mole fraction of (a) ethanol (b) water in the solution.

Solution: nethanol

nwater

Xethanol

=

200.0 g (2(12.01)  5(1.01)  16.00) g mol1

=

150.0 g 1 (2(1.01)  16.00) g mol

=

 200.0 mol     45.07   200.0 mol    150.0 mol       45.07   18.02 

=

0.3477

Xwater

= 1  0.3477 = 0.6523

d) Percentage by Mass (%w/w) Percentage by mass is defined as the percentage of the mass of solute per mass of solution.

mass of solute %ww  x100 mass of solution Note: Mass of solution = mass of solute + mass of solvent 08/16/11

MATTER

98

Example: A sample of 0.892 g of potassium chloride, KCl is dissolved in 54.362 g of water. What is the percent by mass of KCl in the solution? Solution: 0.892 g % mass   100% 0.892 g  54.362 g = 1.61%

Exercise: A solution is made by dissolving 4.2 g of sodium chloride, NaCl in 100.00 mL of water. Calculate the mass percent of sodium chloride in the solution. Answer = 4.0%

e) Percentage by Volume (%V / V) Percentage by volume is defined as the percentage of volume of solute in milliliter per volume of solution in milliliter.

%v

volume of solute  x 100 v volume of solution

Note:

mass of solution Density of solution  volume of solution 08/16/11

MATTER

101

Example 1: 25 mL of benzene is mixed with 125 mL of acetone. Calculate the volume percent of benzene solution. Solution: 25 mL % volume   100% 25 mL  125 mL

= 17%

Example 2: A sample of 250.00 mL ethanol is labeled as 35.5% (v/v) ethanol. How many milliliters of ethanol does the solution contain?

Solution:

Vethanol % volume of ethanol   100 % Vsolution Vethanol

35.5%  250.00 mL  100%

=

88.8 mL

Example 3: A 6.25 m of sodium hydroxide, NaOH solution has has a density of 1.33 g mL-1 at 20 ºC. Calculate the concentration NaOH in: (a) molarity (b) mole fraction (c) percent by mass

Solution: nNaOH (a) M = Vsolution 6.25 m of NaOH  there is 6.25 mol of NaOH in 1 kg of water

for a solution consists of 6.25 mol of NaOH and 1 kg of water; Vsolution

mass solution = solution

masssolution = massNaOH + masswater massNaOH = nNaOH  molar mass of NaOH = 6.25 mol  (22.99 + 16.00 + 1.01) g mol1 = 250 g

masssolution = 250 g + 1000 g = 1250 g Vsolution =

1250 g 1

1.33 g mL

MNaOH

6.25 mol = 3   1250  10 L    1.33  = 6.65 mol L1

(b) XNaOH =

nNaOH nNaOH  n water

1 kg of water contains 6.25 mol of NaOH

nwater

mass water = molar mass of water 1000 g

=

(2(1.01)  16.00) g mol XNaOH =

6.25 mol 1000   6 . 25 mol  mol   18.02  

= 0.101

1

massNaOH (c) %(w/w) of NaOH =  100% massNaOH  mass water

250 g =  100% 250 g  1000 g = 20.0%

Exercise: An 8.00%(w/w) aqueous solution of ammonia has a density of 0.9651 g mL-1. Calculate the (a) molality (b) molarity (c) mole fraction of the NH3 solution Answer:

a) 5.10 mol kg-1 b) 4.53 mol L-1 c) 0.0842

MATTER 1.3 Stoichiometry

Learning Outcome At the end of the lesson, students should be able to: a) Determine the oxidation number of an element in a chemical formula. b) Write and balance : i) Chemical equation by inspection method ii) redox equation by ion-electron method

Balancing Chemical Equation 



A chemical equation shows a chemical reaction using symbols for the reactants and products. The formulae of the reactants are written on the left side of the equation while the products are on the right.

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MATTER

114

Example:

xA +

yB

zC +

Reactants

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wD

Products

MATTER

115



A chemical equation must have an equal number of atoms of each element on each side of the arrow



The number x, y, z and w, showing the relative number of molecules reacting, are called the stoichiometric coefficients.



A balanced equation should contain the smallest possible whole-number coefficients



The methods to balance an equation: a) Inspection method b) Ion-electron method

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MATTER

116

Inspection Method 1. Write down the unbalanced equation. Write the correct formulae for the reactants and products.

1. Balance the metallic atom, followed by nonmetallic atoms.

1. Balance the hydrogen and oxygen atoms. 1. Check to ensure that the total number of atoms of each element is the same on both sides of equation.

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117

Example: Balance the chemical equation by applying the inspection method. NH3 + CuO → Cu + N2 + H2O

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118

Exercise Balance the chemical equation below by applying inspection method. 1. Fe(OH)3 + H2SO4 → Fe2(SO4)3 + H2O 2. C6H6 + O2 → CO2 + H2O 3. N2H4 + H2O2 → HNO3 + H2O 4. ClO2 + H2O → HClO3 + HCl

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Redox Reaction 

Mainly for redox (reduction-oxidation) reaction

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 

Oxidation is defined as a process of electron loss. The substance undergoes oxidation  loses one or more electrons. 

increase in oxidation number



act as an reducing agent (electron donor)

Half equation representing oxidation: Mg  Mg2+ 2e Fe2+  Fe3+ + e 2Cl-  Cl2 + 2e

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121

 

Reduction is defined as a process of electron gain. The substance undergoes reduction 

gains one or more electrons.



decrease in oxidation number



act as an oxidizing agent (electron acceptor)

Half equation representing reduction: Br2 + 2e → BrSn4+ + 2e → Sn2+ Al3+ + 3e → Al

Oxidation numbers of any atoms can be determined by applying the following rules: 1. For monoatomic ions, oxidation number = the charge on the ion e.g: ion oxidation number Na+ +1 Cl-1 Al3+ +3 S2-2

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123

2. For free elements, e.g: Na, Fe, O2, Br2, P4, S8 oxidation number on each atom = 0 1. For most cases, oxidation number for O = -2 H = +1 Halogens = -1

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Exception: 1. H bonded to metal (e.g: NaH, MgH2) oxidation number for H = -1 1. Halogen bonded to oxygen (e.g: Cl2O7) number for halogen = +ve

oxidation

1. In a neutral compound (e.g: H2O, KMnO4) the total of oxidation number of every atoms that made up the molecule = 0 1. In a polyatomic ion (e.g: MnO4-, NO3-) the total oxidation number of every atoms that made up the molecule = net charge on the ion

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Exercise 1.

Assign the oxidation number of Mn in the following chemical compounds. i. MnO2 ii. MnO4-

1.

Assign the oxidation number of Cl in the following chemical compounds. i. KClO3 ii. Cl2O72-

1.

Assign the oxidation number of following: i. Cr in K2Cr2O7 ii. U in UO22+ iii. C in C2O42-

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Balancing Redox Reaction 



Redox reaction may occur in acidic and basic solutions. Follow the steps systematically so that equations become easier to balance.

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127

Balancing Redox Reaction In Acidic Solution Fe2+ + MnO4- → Fe3+ + Mn2+ 1. Separate the equation into two halfreactions: reduction reaction and oxidation reaction i. Fe2+ → Fe3+ ii. MnO4- → Mn2+ 08/16/11

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128

1. Balance atoms other than O and H in each half-reaction separately

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i.

Fe2+ → Fe3+

ii.

MnO4- → Mn2+

MATTER

129

3. Add H2O to balance the O atoms Add H+ to balance the H atoms i. Fe2+ → Fe3+ ii. MnO4- + → Mn2+ + 8H+

4H2O

4. Add electrons to balance the charges i. Fe2+ → Fe3+ + ii. MnO4- + 8H+ + 1 e → Mn2+ + 4H2O 5e 130 08/16/11

MATTER

3.Multiply each half-reaction by an integer, so that number of electron lost in one half-reaction equals the number gained in the other.

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i.

5 x (Fe2+ → Fe3+ + 1e) 5Fe2+ → 5Fe3+ + 5e

ii.

MnO4- + 8H+ + 5e → Mn2+ + 4H2O

MATTER

131

1. Add the two half-reactions and simplify where possible by canceling the species appearing on both sides of the equation. i. 5Fe2+ → 5Fe3+ + 5e ii. MnO4- + 8H+ + 5e → Mn2+ + 4H2O ___________________________________ 5Fe2+ + MnO4- + 8H+ → 5Fe3+ + Mn2+ + 4H2O ___________________________________

5. Check the equation to make sure that there are the same number of atoms of each kind and the same total charge on both sides.

5Fe2+ + MnO4- + 8H+ → Total charge reactant = 5(+2) + (-1) + 8(+1) = + 10 - 1 + 8 = +17

5Fe3+ + Mn2+ + 4H2O Total charge product = 5(+3) + (+2) + 4(0) = + 15 + (+2) = +17

Exercise: In Acidic Solution C2O42- + MnO4- + H+ → CO2 + Mn2+ + H2O Solution :

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Balancing Redox Reaction In Basic Solution 1. Firstly balance the equation as in acidic solution. 1. Then, add OH- to both sides of the equation so that it can be combined with H+ to form H2O. 1. The number of OH- added is equal to the number of H+ in the equation.

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135

Example: In Basic Solution Cr(OH)3 + IO3- + OH- → CrO32- + I- + H2O

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Exercise: 1. 2. 3. 4.

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H2O2 + MnO4- + H+  O2 + Mn2+ + H2O (acidic medium) Zn + SO42- + H2O  Zn2+ + SO2 + 4OH(basic medium) MnO4- + C2O42- + H+  Mn2+ + CO2 + H2O (acidic medium) Cl2  ClO3- + Cl(basic medium)

MATTER

137

Stoichiometry 



Stoichiometry is the quantitative study of reactants and products in a chemical reaction. A chemical equation can be interpreted in terms of molecules, moles, mass or even volume.

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138

C3H8 + 5O2  3CO2 + 4H2O 



1 molecule of C3H8 reacts with 5 molecules of O2 to produce 3 molecules of CO2 and 4 molecules of H2O 6.02 x 1023 molecules of C3H8 reacts with 5(6.02 x 1023) molecules of O2 to produce 3(6.02 x 1023) molecules of CO2 and 4(6.02 x 1023) molecules of H2O

C3H8 + 5O2  3CO2 + 4H2O 





1 mol of C3H8 reacts with 5 moles of O2 to produce 3 moles of CO2 and 4 moles of H2O 44.09 g of C3H8 reacts with 160.00 g of O2 to produce 132.03 g of CO2 and 72.06 g of H2O 5 moles of C3H8 reacts with 25 moles of O2 to produce 15 moles of CO2 and 20 moles of H2O



At room condition, 25 ºC and 1 atm pressure; 22.4 dm3 of C3H8 reacts with 5(22.4 dm3) of O2 to produce 3(22.4 dm3) of CO2

Example 1: How many grams of water are produced in the oxidation of 0.125 mol of glucose? C6H12O6(s) +

O2(g) 

CO2(g) +

H2O(l)

Solution: From the balanced equation; 1 mol C6H12O6 produce 6 mol H2O 0.125 mol

C6H12O6 produce

H2O 0.125 mol  6 mol 1 mol

mass of H2O = (0.125 x 6) mol x (2.02 + 16.00) g mol-1 = 13.5 g

Example 2: Ethene, C2H4 burns in excess oxygen to form carbon dioxide gas and water vapour. (a) Write a balance equation of the reaction (b) If 20.0 dm3 of carbon dioxide gas is produced in the reaction at STP, how many grams of ethene are used?

Solution: (a) C2H4 + (b)

O2 

CO2 +

H2O

22.4 dm3 is the volume of 1 mol CO2

20.0 dm3 is the volume of

20.0 dm  1 mol

2 mol CO2 produced by 1 mol C2H422.4 dm mol CO2 produced by C2H4

20.0 22.4

CO2

3

3

 20.0  mol    1 mol  22.4  2 mol

massethane

 20.0      22.4  mol x [2(12.01)  4(1.01)] g mol-1 2

= 12.5 g

Learning Outcome At the end of this topic, students should be able to: a) Define the limiting reactant and percentage yield b) Perfome stoichiometric calculations using mole concept including limiting reactant and percentage yield.

Limiting Reactant/Reagent 



Limiting reactant is the reactant that is completely consumed in a reaction and limits the amount of product formed Excess reactant is the reactant present in quantity greater than necessary to react with the quantity of limiting reactant

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148

Example: 3H2 + N2  2NH3 If 6 moles of hydrogen is mixed with 6 moles of nitrogen, how many moles of ammonia will be produced? Solution: 3 mol H2 reacts with

1 mol N2

6 mol H2 reacts with

6 mol  1 mol 3 mol = 2 mol N2

N2 is the excess reactant H2 is the limiting reactant  limits the amount of products formed 3 mol H2

produce

6 mol H2 produce

2 mol NH3

6 mol  2 mol 3 mol

= 4 mol NH3

or 1 mol N2 6 mol N2

react with 3 mol H2 react with

6 mol  3 mol molNH3 1 mol = 18 mol H2

H2 is not enough

 limiting reactant  H2 limits the amount of products formed

3 mol H2

produce

6 mol N2

produce

2 mol NH3

6 mol  2 molmol NH3 3 mol

= 4 mol NH3

Exercise: Consider the reaction: 2 Al(s) + 3Cl2(g)  2 AlCl3(s) A mixture of 2.75 moles of Al and 5.00 moles of Cl2 are allowed to react. (a) What is the limiting reactant? (b) How many moles of AlCl3 are formed? (c) How many moles of the reactant remain at the end of the reaction?

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PERCENTAGE YIELD 



The amount of product predicted by a balanced equation is the theoretical yield The theoretical yield is never obtain because: 1. The reaction may undergo side reaction 2. Many reaction are reversible 3. There may be impurities in the reactants

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4. The product formed may react further to form other product 5. It may be difficult to recover all of the product from the reaction medium 

The amount product actually obtained in a reaction is the actual yield

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Percentage yield is the percent of the actual yield of a product to its theoretical yield

actual yield % yield  x 100 theoretica l yield

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Example 1: Benzene, C6H6 and bromine undergo reaction as follows: C6H6 + Br2  C6H5Br + HBr In an experiment, 15.0 g of benzene are mixed with excess bromine (a) Calculate the mass of bromobenzene, C6H5Br that would be produced in the reaction. (b) What is the percent yield if only 28.5 g of bromobenzene obtain from the experiment?

CHAPTER 2 ATOMIC STRUCTURE

2.1 Bohr’s Atomic Model

Learning Outcomes At the end of this topic students should be able to:a) Describe the Bohr’s atomic model. b) Explain the existence of electron energy levels in an atom. c) Calculate the energy of electron at− 18 its level 1

E n= − R H

(orbit) using.

n

2

, R H = 2.18x10

J

Learning Outcomes d) Describe the formation of line spectrum of hydrogen atom. e) Calculate the energy change of an electron 1 1 RH − 2 duringE =transition. 2 n1 n2

f)

where RH = 2.18× 10

− 18

J

Calculate the photon of energy emitted by an E= h where = c / electron that produces a particular wavelength during transition.

Learning Outcomes g) Perform calculation involving the Rydberg equation for Lyman, Balmer, Paschen, 1 1 Pfund 1 7 −1 Brackett and series: = RH − where R = 1.097× 10 m H 2 2 n1

n2

n1 n2

g) Calculate the ionisation energy of hydrogen atom from the Lyman series.

Learning Outcomes i)

State the weaknesses of Bohr’s atomic model.

j)

State the dual nature of electron using de Broglie’s postulate and Heisenberg’s uncertainty principle.

Bohr’s Atomic Model In 1913, a young Dutch physicist, Niels Böhr proposed a theory of atom that shook the scientific world. The atomic model he described had electrons circling a central nucleus that contains positively change proton.

Bohr’s Atomic Model Böhr also proposed that these orbits can only occur at specifically “permitted” levels according to the energy levels of the electron and explain successfully the lines in the hydrogen spectrum.

Bohr’s Atomic Model First Postulates Electron moves in circular orbits about the nucleus. While moving in the orbit, the electron does not radiate or absorb any energy.

[orbit = energy level=shell] Orbit is a pathway where the electron is move around the nucleus. Orbit

Bohr’s Atomic Model Second Postulate The moving electron has a specific amount of energy; its energy is quantised.

The energy of an electron in its level is given by:

1 En= − RH 2 n RH (Rydberg constant) = 2.18 x 10-18J. n (principal quantum number) = 1, 2, 3 …. ∞ (integer) Note: n identifies the orbit of electron Energy is zero if electron is located infinitely far from nucleus

Bohr’s Atomic Model Third postulate At ordinary conditions, the electron is at the ground state (lowest level). If energy is supplied, electron absorbed the energy and is promoted from a lower energy level to a higher ones. (Electron is excited)

Bohr’s Atomic Model Fourth Postulate Electron at its excited states is unstable. It will fall back to lower energy level and released a specific amount of energy in the form of light (photon). The energy of the photon equals the energy difference between levels.

Electron is excited from lower to higher energy level. A specific amount of energy is absorbed ∆E = hν = E3-E1

n =1

n=2 n=3 n=4

Electron falls from higher to lower energy level . A photon is emitted. ∆E = hν = E1-E3

Radiant energy emitted when the electron moves from higher-energy state to lower-energy state is given by:

E= Ef − E i

Where:

Ef = − RH

1 2 nf

 1    1  ∆E = −R H  2  −  − R H  2     n n  f    i 

E i= − R H

1 2 ni

 1 1  Thus, ∆E = R H  2 − 2  n  n f   i

The amount of energy released by the electron is called a photon of energy. A photon of energy is emitted in the form of radiation with appropriate frequency and wavelength.

E = hv Where : h (Planck's constant) = 6.63 x 10-34 J s v = frequency (s-1)

c = v Where : c (speed of light) = 3.00 x 108 ms-1 Thus :

∆E =

hc

λ

Rydberg Equation Wavelength emitted by the transition of electron between two energy levels is calculated using Rydberg equation:

1

= RH

1 1 − 2 2 n1 n2

7

−1

where R H = 1.097× 10 m n1 n2

Example 1 Calculate the wavelength, in nanometers of the spectrum of hydrogen corresponding to n = 2 and n = 4 in the Rydberg equation.

Exercises: 1.

Calculate the energy of hydrogen electron in the:

(a) 1st orbit (b) 3rd orbit (c) 8th orbit 1.

Calculate the energy change (J), that occurs when an electron falls from n = 5 to n = 3 energy level in a hydrogen atom.

2.

Calculate the frequency and wavelength (nm) of the radiation emitted in question 2.

Emission Spectra Emission Spectra

Continuous Spectra

Line Spectra

Continuous Spectrum A spectrum consists of radiation distributed over all wavelength without any blank spot. Example : electromagnetic spectrum, rainbow It is produced by white light (sunlight or incandescent lamp) that passed through a prism

Formation of Continuous Spectrum

Regions of the Electromagnetic Spectrum

Line Spectrum (atomic spectrum) A spectrum consists of discontinuous & discrete lines with specific wavelength. It is composed when the light from a gas discharge tube containing a particular element is passed through a prism.

Formation of Atomic / Line Spectrum The emitted light (photons) is then separated into its components by a prism. Each component is focused at a definite position, according to its wavelength and forms as an image on the photographic plate. The images are called spectral lines.

Formation of Atomic / Line Spectrum

Formation of Atomic / Line Spectrum Example : The line emission spectrum of hydrogen atom Line spectrum are composed a few wavelengths giving a series of discrete line separated by blank areas It means each line corresponds to a specific wavelength or frequency.

Formation of Line Spectrum When electron absorbed radiant energy, they will move from lower energy level to higher energy level (excited state). This excited electron is unstable and it will fall back to lower energy level. During the transition, electron will release energy in the form of light with specific wavelength and can detected as a line spectrum.

Differences Between Line & Continuous Spectra Continuous Spectrum A spectrum that contains all wavelength without any blank spots. Example: Rainbow.

Line Spectrum A spectrum that contain only specific wavelengths. A spectrum of discrete lines with certain wavelengths. Example: Emission spectrum an element.

Formation of Line Spectrum (Lyman Series) n=∞ n=5 n=4 Energy

n=3 n=2

n=1

Emission of photon

Line spectrum

E λ Lyman Series

Formation of Line Spectrum (Balmer Series) n=∞ n=5 n=4 Energy

n=3 n=2

n=1

Emission of photon

Line spectrum

E λ Lyman Series

Balmer Series

Energy Level in Hydrogen Atom

Example Complete the following table Spectrum region

Series

nf

ni

Lyman

1

2,3,4,…

ultraviolet

Balmer

2

3,4,5,…

Visible/uv

Paschen

3

4,5,6,…

Infrared

Brackett

4

5,6,7,…

Infrared

Pfund

5

6,7,8,…

Infrared

Example The following diagram is the line spectrum of hydrogen atom. Line A is the first line of the Lyman series. A

B

C D E

Line spectrum

E λ v

Specify the increasing order of the radiant energy, frequency and wavelength of the emitted photon. Which of the line that corresponds to i) the shortest wavelength? ii) the lowest frequency?

Example The line spectrum of Balmer is given as below: W

Y

Describe the transitions of electrons that lead to the lines W, and Y, respectively. Solution For W: transition of electron is from n =4 to n = 2 For Y: electron shifts from n = 7 to n = 2

Example E

D

C B A

Line spectrum

(a)

Paschen series

Which of the line in the Paschen series corresponds to the longest wavelength of photon?

(b)

Describe the transition that gives rise to the line.

Solution Line A. The electron moves from n=4 to n=3.

Example With refer to the second line in the Balmer series of the hydrogen spectrum, Calculate; a)the wavelength in nm b)the frequency c)the energy

Example Refer to last line of hydrogen spectrum in Lyman series, Calculate: a) Wavelength b) Frequency c) Wave number; where wave number = For Lyman series; n1 = 1 & n2 = ∞ Ans: i. 9.116 x10-8 m ii. 3.29 x1015 s-1 iii. 1.0970 X 107 m-1

1

Ionization Energy Defination : Ionization energy is the minimum energy required to remove an electron in its ground state from an atom (or an ion) in gaseous state. M (g) → M+ (g) + e

∆H = +ve

Ionization Energy The hydrogen atom is ionised when electron is removed from its ground state (n = 1) to n = ∞. At n = ∞, the potential energy of electron is zero, here the nucleus attractive force has no effect on the electron (electron is free from nucleus)

Example n1 = 1, n2 = ∞ ∆E =

=

RH (1/n12 – 1/n22) =

2.18 x 10-18 (1/12 – 1/∞2)

=

2.18 x 10-18 (1 – 0)

2.18 x 10-18 J

Ionisation energy = =

2.18 x 10-18 x 6.02 x 1023 J mol-1

1.312 x 106 J mol-1 =

1312 kJ mol-1

Example Calculate the energy to ionized : (a) a hydrogen atom. (b) 1 mol of hydrogen atom.

Solution (a) E= − R H

1 1 − 2 2 n1 n2 1 1 − 2 2 1 ∞

− 18

= − 2.18× 10

=

− 18

2.18× 10

J

Solution (b) 1 H atoms need 2.18 x 1018 J 1 mol H atom = 2.18 x 1018 x 6.02 x 1023 = 1.31 x 106 J The energy to ionized 1 mol of hydrogen atom is 1.31 X 106 J

Example 10.97 10.66

10.52

10.27

9.74

8.22

The Lyman series of the spectrum of hydrogen is shown above. Calculate the ionisation energy of hydrogen from the spectrum.

Solution ∆E

= h X c/λ

= h x c x wave no.

= (6.626x10-34 Js)(3x108 ms-1)(10.97x106 m-1) = 218.06 x 10-20 J = 2.18 x 10-18J Ionisation energy = (2.18 x 10-18) (6.02x1023 J mol-1) = 1.312 x 106 J mol-1 = 1312 kJ mol-1

The weaknesses of Bohr’s Theory It can only explain the hydrogen spectrum or any spectrum of ions contain one electron. example: He+, Li2+.Therefore, it did not account for the emission spectrum of atom containing more than 1 electron. Electron are wavelike, we can’t define the precise location of a wave because a wave extends in space.

de Broglie’s Postulate In 1924 Louis de Broglie proposed that not only light but all matter has a dual nature and possesses both wave and particle properties. Electron is both particle and wave. Tiny particle such as electron does have wave properties. De Broglie deduced that the particle and wave properties are related by the expression:

Example Electron has dual nature properties. Why don't we see the wave properties of a Baseball?

de Broglie’s Postulate h = mµ h = Planck constant (J s) m = particle mass (kg) µ = velocity (m/s) λ = wavelength of a matter wave

Heisenberg’s Uncertainty Principle It is impossible to know simultaneously both the momentum p (defined as mass times velocity) and the position of a particle with certain. Stated mathematically,

x where

h p≥ 4π

∆x = uncertainty in measuring the position ∆p = uncertainty in measuring the momentum

=

∆mv h = Planck constant

2.2 Quantum Mechanical Model

Learning Outcomes At the end of this topic students should be able to:Define the term orbital. State the four quantum numbers in an orbitals. sketch the 3-D shape of s, p and d orbitals.

Atomic Orbital Definition An orbital is a threedimensional region in space around the nucleus where there is a high probability of finding an electron.

Quantum Numbers Each of the electrons in an atom is described and characterised by a set of four quantum numbers, namely principal quantum number, n angular momentum quantum number, l magnetic quantum number, m electron spin quantum number, s.

Principal Quantum Number, n n determines the energy level (electron shell) and size of an orbital. The principal quantum number n, may have +ve value starting from n =1, 2, 3, …, ∞. As n increase : i) the orbital become larger ii) electron has higher energy

Principal Quantum Number, n n

1

2

3

Orbital size

Energy

increases

4

Angular Momentum Quantum Number, l Alternative name: - Subsidiary Quantum Number - Azimuthal Quantum Number - Orbital Quantum Number The value of l indicates the shape of the atomic orbital. The allowed values of l are 0, 1, 2,…, ( n - 1)

Angular Momentum Quantum Number, l Letters are assigned to different numerical values of Value of l 0 1 2 3

Symbol Orbital shape s spherical p dumbbell d cloverleaf f

Angular Momentum Quantum Number, l value is depend on n. (i.e., 0 ≤ l < n). One subshell If n = 1, l = 0 (s-orbital) (s orbital) If n = 2, l = 0 (s-orbital) two subshells (s and p orbitals) = 1 (p-orbital) If n = 3, l = 0 (s-orbital) = 1 (p-orbital) = 2 (d-orbital)

three subshells (s, p, and dorbitals)

Magnetic Quantum Number, m Describe the orientation of orbitals in space. Possible values of m depend on the value of l. For a given l, m can be : -l, …, 0, …, +l Example: If l = 0, m = 0 » 1 orientation of s orbital If l = 1, m = -1,0,+1 » 3 orientation of p orbital (px, py, pz) If l = 2, m = -2,-1, 0,+1,+2 » 5 orientation of d orbital ( dxy,dxz,dyz,dx2-y2,dz2)

Electron Spin Quantum Number, s The value of s represent the direction of an electron rotation on its own axis. either clockwise or anticlockwise It has 2 value : +½ and -½

Shape of Atomic Orbital s orbitals Spherical shape with the nucleus at the centre. When l = 0 , m = 0 , only 1 orientation of s orbital. The larger value of n, the size of s orbital gets larger.

Shape of s orbital with different n

Z

Z

Y

Y

Y

X

X

1s

Z

2s

X

3s

Shape of Atomic Orbitals p orbitals Can be represent as a pair of dumb-bell shaped When l = 1, m = -1, 0, +1 3 orientation of p-orbitals px, py, and pz. As n increases, the p orbitals get larger.

Shape of p orbital

px

py

pz

Shape of Atomic Orbitals d orbitals All the d orbitals do not look alike. When l = 2 , m = -2, -1, 0, +1, +2. There are five orientation of d orbitals.

Shape of d orbital

dx2-y2

dxy

dz2

dxz

dyz

Set of Four Quantum Numbers 4 quantum number n,l,m and s enable us to label completely an electron in any orbital of an atom. Example: 4 quantum numbers of 2s orbital electron are n = 2 , l = 0 , m = 0 and s = +½ and -½ Can be simplified as (2,0,0,+½) or (2,0,0,-½) n, l , m, s n, l , m, s

Exercise

Shell n 1 2

1 2

l

Orbital notation

m

No. of orbitals

0

1s

0

1

0

2s

0

1

1

2p

-1,0,+1

3

Exercise Predict the following quantum numbers whether they are allowed or not (a) (1,0,0,-½) (b) (2,0,1,1) (c) (0,1,1,+½) (d) (4,1,0,-½)

2.3 Electronic Configuration

Learning Outcomes At the end of this topic students should be able to:State and apply Aufbau principle, Hund's rule and Pauli exclusion principle in filling of electrons in orbitals of an atom. Write the electronic configuration of atoms and monoatomic ions. (a) Orbital diagram (b) spdf notation

Learning Outcomes Explain the anomalous electronic configurations of chromium and copper.

Introduction The electronic configuration of an atom show how electron are filled in the orbital. Electronic configuration describes the arrangement of electron in an atom.

Electronic Configuration Method 1: Orbital diagram Example: 8O Box 1s

2s

2p

Electronic Configuration Method 2: s,p,d,f notation Example: 8O 2 1s

Principal quantum number, n

2 2s

Number of electrons in the subshells 4 2p

Azimuthal quantum number, l

Electronic Configuration To enable us to do electronic configuration, we have to obey the following rules: a) The Aufbau Principle b) The Pauli Exclusion Principle c) The Hund rule

Aufbau Principle State that electrons are filled in the orbitals in order of increasing energy. Electrons should occupy the orbital with the lowest energy first before enters the one with higher energy.

Relative Energy Level of Atomic Orbitals 4d n=5

5s 4p 3d

energy

Energy

n=4 4s n=3

3p 3s

n=2

2p 2s

n=1 1s

Orbital energy levels in a many-electron atom

Order of orbitals (filling) in multi-electron atom

Order of orbitals (filling) in multi-electron atom The order of filling orbitals is: 1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s Start with the 1s orbital and move downward, following the arrows. Example: (a) 4Be

(b)

10Ne

Electronic configuration 4Be : 1s2 2s2 Electronic configuration 10Ne : 1s2 2s2 2p6

Pauli Exclusion Principle No two electrons in an atom can have the same four quantum numbers (n, l, m, s). Eg : Li (3 electrons)

Hund’s Rule States that when electrons are added to the orbital of equivalent energy (degenerate orbitals), each orbital are filled singly with electron of the same spin first before it is paired. The electron in half-filled orbitals have the same spins, that is, parallel spins.

Example 1) Carbon (6 electron)

1s

2s

2p

2) Oxygen (8 electron)

1s

2s

2p

Exercise Write the electronic configuration of the following atom or ion: (a)C (b)Ne (c)Al (d)Al3+ (e)Cl (f)

Cl-

The Anomalous Electronic Configurations of Cr and Cu Cr and Cu have electron configurations which are inconsistent with the Aufbau principle. The anomalous are explained on the basis that a filled or half-filled orbital is more stable.

Element

Expected (Aufbau Principle)

Observed/actual

Cr (Z=24)

1s22s22p63s23p6 4s2 3d4

1s22s22p63s23p6 4s1 3d5

Cu (Z=29)

1s22s22p63s23p6 3d9 4s2

1s22s22p63s23p6 3d10 4s1

Orbital diagram 24Cr

:

18[Ar] 3d

4s

3d orbital with a half filled orbital arrangement are more stable. 24Cr

:

Actual

18[Ar] 3d

4s

2 2s2 2p6 3s2 3p6 Ar : 1s 18

*Half filled orbital arrangement increase stability of Cr atom

Copper expected orbital notation (Aufbau Principle)

Cu :

18[Ar]

3d

4s

3d orbital with fully filled orbital arrangement is more stable. Copper actual orbital notation

Cu :

18[Ar]

3d 2 2s2 2p6 3s2 3p6 Ar : 1s 18

4s

3.0 PERIODIC TABLE

1

LEARNING OUTCOMES

• At the end of the lesson the students should be able to : • (i) Indicate period, group and block (s, p, d, f). • (ii) Specify the position of metals, metalloids and non-metals in the periodic table. • (iii) Deduce the position of elements in the periodic table from its electronic configuration. 2

3.1 Classification of elements • The periodic table is a table that arranges all the known elements in order of increasing proton number. • This order generally coincides with increasing atomic mass.

3

• A vertical column of elements is called a group and a horizontal row is known as a period. • Elements in the same group have the same number of valence electrons. Group number = number of valence electrons (if the element is in block s and d) Group number = number of valence electrons + 10 (if the element is in block p) 4

Transition metals

For example, oxygen and sulphur are both found in group 16 which means that they both have 6 valence electrons. 5

GROUP • Main Groups in Periodic Table – Group 1 – Group 2 – Group 3-11 – Group 17 – Group 18 – Group 12

: : : : : :

alkali metals (except H) alkaline earth metals transition metals halogens inert/ noble gases Zn, Cd, Hg

6

• The periods in the Periodic Table are numbered from 1 to 7 • For example, hydrogen and helium are in Row 1 or Period 1 because their principal quantum number, n, of the main electron shell is 1. (H:1s1 ;He: 1s2) Period number = Principle quantum number

7

Blocks • All the elements in the Periodic Table can be classified into 4 main blocks according to their valence electrons configuration. • These main blocks are s, p, d and f block.

8

s- block • Group 1 and 2 • The filling of valence electrons involve the s orbital • Configuration of the valence electrons : ns1 to ns2 • Eg: 2 2s2 2p6 3s1 Na : 1s 11 2 2s2 2p6 3s2 3p6 4s2 Ca : 1s 20 9

p-block • Group 13 to 18 • The filling of valence electrons involve s and p orbital. • The configuration of valence electrons: ns2 np1 to ns2 np6

• Eg. 2 2s2 2p6 3s2 3p1 Al : 1s 13 2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 Te : 1s 52 5p4 10

d-block • Groups 3 to 12 • The filling of valence electrons involve s and d orbitals. • Group 3 to 11 known as Transition metal. • Configuration of valence electron : • Eg. (n-1) d1 ns2 to (n-1) d10 ns2 2 2 6 2 6 3 2 or 23V : 1s 2s 2p 3s 3p 3d 4s [Ar] 3d3 4s2 where [Ar] = 18 electrons 11

f-block • Involve the elements in the series of lanthanides (Ce to Lu) and actinides (Th to Lr). • The filling of valence electrons happen in the subshell of 4f and 5f.

12

elements

block

period

group

2 2s2 K : 1s 19 2p6 3s2 3p6 2 1 Mg : 1s 12 4s 2 2p62 3s22 2s P 15 : 1s 2s 2p6 3s2 3p3 2 2s2 Bv : 1s 35 2p6 3s2 3p6 2 2s52 2 10 Ni : 1s 4s 4p 28 3d 2p6 3s2 3p6 2 3d 8 2 2s2 Zr : 1s 4s 40 2p6 3s2 3p6 4s2 3d10 4p6 5s2 3d2

s

4

1

s p

3 3

2 15

Numbe r of velenc 1 e electro 2 n 5

p

4

17

7

d

4

10

10

d

5

4

4

Example

• Classify the following elements into its appropriate group, period and block. A ……1s2 2s2 2p6 3s2 3p6

B …….1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p5 C …….1s2 2s2 2p6 3s2 3p6 4s2 D …….1s2 2s2 2p6 3s2 3p6 3d3 4s2 E …….1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 15

Element

Group

Period

Class/block

A

18

3

Inert gas / block p Block p

B C D

17

4

2

4

5

4

Block s Transition element/ block d Inert gas / block p

E

18

4

16

3.2 Periodicity 4.2.1 Periodic trends in the size of atom (atomic radii) • The size /radius of atom is difficult to be defined exactly because the electron cloud has no clear boundary. • Therefore, the atomic radius is taken as half of the distance between the nuclei of two adjacent identical atom. 17

Radius, r = a/2 (Å) Size



volume

V = 4/3 πr3 , V  r

a

• Down the group, atomic radii increases.  Across period, atomic radii decreases. • Across the period of d-block (transition elements) the change in atomic radii is small as valence electrons are filled in the 3d degenerate orbitals. 18

Two factors that influence the changes of atomic radii in the Periodic Table are: i. Effective nuclear charge experienced by the valence electrons ii. The principal quantum number, n, of the valence electrons 19

i. Effective nuclear charge (Zeff ) • Electrons around the nucleus experience different nucleus attraction. • Those electrons closer to the nucleus experience a greater attraction than those that are farther away. • The actual nuclear charge experienced by an electron is called the effective nuclear charge, Zeff 20

• Effective nuclear charge increase, nucleus attraction stronger, atomic radii decrease • Across the period, the effective nuclear charge increases as proton number increase. • As a result, the attraction between the nucleus and valence electrons become stronger, causing the atomic radius to decrease.

21

ii. The principal quantum number of the valence electrons

• As we move down a group, the number of shells increases, more inner electrons are present to shield the valence electrons from the nucleus. • The valence electrons are farther from the nucleus. • Thus, the attraction between the nucleus and valence electrons decreases, therefore, the atomic radius increase. 22

• Down a group, the atomic radius increases because of the increasing principal quantum number (n) of the valence electron.

23

Across period 3 Across Period 2

24

The graph shows that : •Atomic radius decreases when : * Across a period (from left to right) * Moving up a group in the periodic table. •Atomic radius increases when * Going down the group • The greater the nucleus attraction, the smaller the atomic radius. 25

Example Arrange the following atoms in order of increasing radius P,Si,N.

Solution • N and P are in the same group and N is above P. • Atomic radius increases as we go down the group. • Therefore, the radius of N is smaller than that of P • Both Si and P are in the third period and Si is to the left of P. • Atomic radius decreases as we move from left to right. • Therefore, the radius of P is smaller than Si. • Thus the order of increasing radius :

Group 14

Group 15 N

Si

P

N
3.2.2 Trends in the ionic radii • When electrons are added to an atom, the mutual repulsions between them increase. • This enlarge the domain of electron cloud. • Therefore, negative ions (anions) are larger than the atoms from which they are formed.

27

• When electrons are removed from the valence shell, the electron-electron repulsions decrease but the nuclear charge remains the same. • So the remaining electrons are to be pulled closer together around the nucleus. • Therefore, cations are smaller than the atoms from which they are formed. 28

29

Isoelectronic species • Isoelectronic species are groups of atoms and ions which have the same electronic configuration. Within isoelectronic species: a) the more positive the charge, the smaller the species E.g : • Na+, Mg2+, Al3+and Si4+ ions are isoelectronic (10 e) with the electron configurations as 1s2 2s2 2p6. 30

Isoelectronic species with electronic configuration 1s2 2s2 2p6 (10 electrons) species Number of Na+ 11 proton Mg2+ 12 Al3+ 13 Si4+ 14 •When proton number increase, effective nuclear charge increase. •The attraction between nucleus and remaining electron increase. •Therefore, the ionic radii decrease. •The ionic radii of Na+ > Mg2+ > Al3+ > Si4+

Isoelectronic species with electronic 1s2 2s2 2p6 3s2 3p6 (18 electrons) species Number of P315 proton S216 Cl17 •When proton number increase, effective nuclear charge increase. •The attraction between nucleus and remaining electron increase. •Therefore, the ionic radii decrease. •The ionic radii of Cl- < S2- < P3-

Exercise Na+,Si4+ ,Mg2+, N3- ,O2- ,Al3+ and F- are isoelectronic with the electronic configuration as 1s2 2s2 2p6. Arrange in an descending order the size of those isoelectronic species.

Answer : N3- > O2- > F- > Na+ > Mg2+ > Al3+> Si4+

33

34

3.2.3 Trends in the ionization energies • The ionization energy (IE) is the minimum energy required to remove an electron from a gaseous atom in its ground state.  The first ionization energy (IE ) is the 1 minimum energy required to remove the first electron from the atom in its ground state. E.g: energy + X(g) → X+(g) + e- ΔH = IE1

35

i) Ionization energy across a period :  The effective nuclear charge increases, the atomic size decreases.  Electrons are held tightly to the nucleus thus it is difficult to remove the first electron.  Therefore the first ionisation energy is high. It can be said that the first ionization energy increases from left to right. However, there are some irregularities in the trend. 36

‫٭‬Be ♦B

■N ▲O

■ ▲

‫٭‬

♦

37

Anomalous cases in Period 2 a) Between group 2 and 13 •

2 2s2 2p1 in group 13 has a lower IE B : 1s 5 1 than 4Be: 1s2 2s2 in group 2.

• Be loses a 2s electron while B loses a 2p electron. • Less energy is needed to remove an electron from partially-filled 2p orbital in B than to remove an electron from fully/completely filled 2s orbital in Be. 38

b) Between group 15 and 16 • O (group16) has lower IE1 than N (group 15) • 7N :1s2 2s2 2p3 (the half-filled 2p orbital ) 2 2s2 2p4( the partially-filled 2p orbital) O :1s 8

• When N loses an electron it must come from the half-filled 2p orbital which is more stable than that of electron of the partially-filled orbital in O. • As a result, the first ionization energy of N is higher than of O.

39

ii) Ionization energy going down the group • Going down the group, the atomic size increases as the energy level, n increases. • Therefore the outer electrons are farther from the nucleus and are held less tightly (weaker attraction) by the nucleus. • Thus, it is easy to remove the first electron. • Hence the Ionization Energy decreases down the group. 40

• Second ionization energy (IE2) is the minimum energy required to remove an electron from a positive gaseous ion. X+(g) → X2+(g) + e• When an electron is removed from a neutral atom, the mutual repulsion among the remaining electrons decrease. • Since the nuclear charge remain constant, the electron are held tightly to the nucleus. • Therefore more energy is needed to remove another electron from the positively charged ion. 41

• Thus, ionization energies always increase in the following order : IE1< IE2< IE3< IE4<….. • Although the removal of a subsequent electron from an atom requires an increment amount of energy but it may not be consistence. • We can determine the electronic configuration of the valence electron for an element by using the ionization energy.

42

Example 1 • 4Be • The ionization energies (kJmol-1) of Beryllium are shown below.

IE1 899

IE2 1757

IE3 14850

IE4 21005

43

• The ratio between the ionization energies are: IE 2 IE 1 IE 3 IE 2 IE 4 IE 3

1757   1.95 899 14850   8.45 1757 21005   1.41 14850

44

• A sharp increase in ionization energy occurs when an inner-orbital electron is removed. • The sharp increase is in IE3. It means the 3rd electron occupies the inner shell. • Therefore, there are 2 valence electrons. Hence, Be is in group 2 with valence configuration ns2.

45

Example 2 • Five successive ionization energies (kJmol-1) of atom M is shown below: IE1 800

IE2 1580

IE3 3230

IE4 IE5 4360 16000

• Determine i) the electron configuration of the valence electron. ii) the group number in the periodic table. 46

• The sharp increase is in IE5, this means the 5th electron occupies the inner shell. • Therefore, there are 4 valence electrons. • Hence, the valence electronic configuration for M is ns2 np2 in group 14 of the periodic table.

47

Trends in the electronegativity •

Electronegativity is the relative tendency of an atom to attract electrons to itself when bonded with another atom.

•

Electronegativity increases up a group and across a period. This follows the trends for ionization energy and electron affinity.

48

a) Across period • The nuclear charge increase • The atomic size decrease • Hence, the nucleus attraction stronger • Therefore, the electronegativity increase

49

b) Down a group • • • •

The principle quantum number increase The atomic size increase Hence, weaker nuclear attraction Therefore electronegativity decrease

50

Trends in the melting or boiling point •

The melting or boiling point depends on the types of intermolecular forces that exist between the molecules.

51

Across the period • Melting and boiling point of the 3rd period elements Element Melting point (oC) Boiling point (oC)

Na 98

Mg Al Si P S Cl Ar 650 660 1410 44 120 -101 -189

890 1120 2450 2360 280 445 -34 -186

52

• The variation of melting and boiling point of elements in the 3rd period can be discussed as: (a) Metallic structure (Na to Al) (b) Gigantic covalent structure (Si) (c) Simple molecular structure (P to Ar)

53

a)Metallic structure (Na to Al) Metal has positive metal ions attracted to the electrons sea which form the metallic bonding. Strength of metallic bonding is proportional to the number of valence electrons.

The more valence electrons, the stronger the metallic bond and the higher the melting / boiling point

e

e

Na+ Na+

Na+ e

Na+ Na+

e

Na+ e

e

Na+ Na+

Na+ e

e

e

e

Na+ Na+

e

e

Na+

54

b) Gigantic covalent structure (Si) • Silicon has a gigantic covalent structure. • Melting and boiling point of Si is very high because high energy is needed to break the infinity amount of the strong covalent bond.

55

c) Simple molecular structure ( P to Ar) • The non-metal that exist as molecules of P4, S8, Cl2 and Ar (monoatom). • The covalent bond between the atoms is very strong but the intermolecular force (Van der Waals), is very weak. • The strength of Van der Waals force is proportional to molecular size (relative molecular weight) – Molecular size: Ar < Cl2 < P4 < S8 – therefore melting / boiling point : Ar < Cl < P < S

56

Melting and boiling points down the group

Group 1 • The size increase, the attraction between nucleus and electron sea become weaker. • Therefore, less energy is needed to overcome the attraction. • Thus, melting and boiling point decrease. 57

Group 17 • The size of molecules increase, the intermolecular forces (Van der Waals) become stronger. • Therefore, more energy is needed to overcome the attraction - Thus, melting and boiling point increase.

58

Metallic character • Metallic properties are: - shiny with various colours and most are silvery. - malleable and ductile - good thermal and electric conductor • Generally, metallic character : - Increases down a group - Decreases across the period

59

• The easier to the electrons to be removed from an atom, the more metallic the element. • Therefore metallic character increases down a group and decreases across a period.

metal

Metalloids (semimetal)

nonmetal

60

Acid-base behavior of oxides of Period 3 For Period 3: Na Mg Al

Si

P

S Cl

• When react with oxygen : (a) Na & Mg form basic oxide (b) Al form amphoteric (both acidic and basic) oxide. (c) Si, P, S & Cl form acidic oxide 61

• Na reacts with oxygen to form a basic oxide. 4Na (s) + O2 (g)→2Na2O (s) • The oxide will produce base solution when react with water. Na2O (s) + H2O (l) → 2NaOH (aq) • Mg burns in oxygen to form a basic oxide, MgO. 2Mg (s) + O2 (g) →2MgO (s)

MgO (s) + 2HCl (aq) →MgCl2 (aq) + H2O (l) base acid 62

• Al forms amphoteric oxide, can react either with an acid or a base. Al2O3(s) + 6HCl(aq) → 2AlCl3(aq) + 3 H2O (l) base

acid

Al2O3(s) + 2NaOH(aq) + 3H2O(l) → 2NaAl(OH)4 (aq) acid base sodium aluminate

63

Si, P, S & Cl burn in oxygen to form acidic oxide. Si

:

Si (s) + O2 (g) → SiO2 (s) SiO2 (s) + NaOH (aq) → Na2SiO3(aq)+H2O (l) acid

base

64

P: P4 (s) + 3O2 (g) → P4O6 (s) P4O6 (s) + 6H2O (l) → 4H3PO3 (aq) phosphorus acid

65

S: S (s) + O2 (g) →SO2 (g) SO2 (g) + H2O (l) → H2SO3 (aq) sulfurous acid Cl : Cl2O7 (g) + H2O (l) → 2HClO4 (aq) hypochloric acid

66

CHAPTER 4 CHEMICAL BONDING





Chemical bond is the force that holds two atoms together in a molecule or compound Valence electrons play an important role in the formation of chemical bonds

CHAPTER 4 CHEMICAL BONDING 4.1 Lewis Structure 4.2 Molecular Shape and Polarity 4.3 Orbital Overlap and Hybridization 4.4 Intermolecular Forces 4.5 Metallic Bond

4.1 Lewis Structure 4.1.1

Lewis Symbol

A Lewis symbol consists of:  the symbol of an element  dots or cross is used to represent the valence electrons in an atom of the element.

Example The Lewis symbol of atom

Group

1

2

13

14

15

16

17

18

Valence electron

1

2

3

4

5

6

7

8

Lewis dot symbol



Elements in the same group have the same valence electronic configurations  similar Lewis symbols.

4.1.2 



Octet Rule

Octet rule states that atoms tend to form bonds to obtain 8 electrons in the valence shell Atoms combine to achieve stablility  to have the same electronic configuration as a noble gas





Atoms achieve noble gas configuration through: i) transferring electrons ii) sharing electron Bond formation involve transferring or sharing of only valence electrons

Electronic Configuration of Cations and Anions 1)Noble gas configuration Group 1, 2 and 13 elements donate valence electrons to form cations with noble gas configurations Example: Na : 1s22s22p63s1 Na+ : 1s22s22p6 (isoelectronic with Ne) Ca : 1s22s22p63s23p64s2 Ca2+ : 1s22s22p63s23p6 (isoelectronic with Ar)



Group 15, 16 and 17 elements accept electrons to form anions with noble gas configurations Example: O : 1s22s22p4 O2 : 1s22s22p6 (isoelectronic with neon) Cl : 1s22s22p63s23p5 Cl : 1s22s22p63s23p6 (isoelectronic with Ar)

2) 

Pseudonoble gas configuration

d block elements donate electrons from 4s orbitals to form cations with pseudonoble gas configuration. Example: Zn : 1s22s22p63s23p64s23d10 Zn2+ : 1s22s22p63s23p63d10 (pseudonoble gas configuration )

3) Stability of the half-filled orbitals 

d block element can also donate electrons to achieve the stability of half-filled orbitals Example: Mn : 1s22s22p63s23p64s23d5 Mn2+ : 1s22s22p63s23p63d5 (stability of half-filled 3d orbital ) Fe : 1s22s22p63s23p64s23d6 Fe3+ : 1s22s22p63s23p63d5 (stability of half-filled 3d orbital)

4.1.3 Formation of the bonds using Lewis Symbols i.Ionic (electrovalent) bond ii.Covalent bond iii.Dative (coordinate) bond

4.1.3.1

Ionic bond (Electrovalent bond)

Ionic bond (electrovalent bond) is an electrostatic attraction between positively and negatively charged ions.  Ionic compounds are formed when electrons are transferred between atoms (metal to nonmetal) to give electrically charged particles that attract each other . 

Example 1: NaCl

Sodium, an electropositive metal, tends to remove its valence electron to obtain noble gas electronic configuration (Ne)

Chlorine, an electronegative element, tend to accept electron from Na to obtain noble gas electronic configuration (Ar)





The electrostatic forces between Na+ and Cl- produce ionic bond These two processes occur simultaneously

+

Example 2: CaCl2 Ca: 1s2 2s2 2p6 3s2 3p6 4s2 (Has two electrons in its outer shell) Cl: 1s2 2s2 2p6 3s2 3p5 (Has seven outer electrons)

Calcium Chloride If Ca atom transfer 2 electrons, one to each chlorine atom, it become a Ca2+ ion with the stable configuration of noble gas. 

At the same time each chlorine atom to achieve noble gas configuration gained one electron becomes a Cl- ion to achieve noble gas configuration. 

The electrostatic attraction formed ionic bond between the ions. 

Ionic bond

(Formed by transfer of electrons) Calcium Chloride +

+

2

Example 3: LiF +

Lewis structure and formation of ionic compounds

1) CaCl2 +

+

2) MgO +

2

3) CaBr2

+

+



Ionic bond is very strong, therefore ionic compounds: 1. Have very high melting and boiling points 2. Hard and brittle 3. Can conduct electricity when they are in molten form or aqueous solution because of the mobile ions

Exercises:

By using Lewis structure, show how the ionic bond is formed in the compounds below. ( a ) KF ( b ) BaO ( c ) Na2O

4.1.3.2 Covalent Bond Definition of covalent bond

i. Chemical bond in which two or more electrons are shared by two atoms. ii. The electrostatic force between the electrons being shared the nuclei of the atoms. Why should two atoms share electrons?

To gain stability by having noble gas configuration (octet)

Example

F

+

7e-

F

F F

7e-

8e- 8e-

Lewis structure of F2 single covalent bond

lone pairs

F

F

lone pairs

single covalent bond lone pairs

F F

lone pairs 9.4

Covalent compounds: 

Compounds may have these covalent bonds: i. Single bond ii. Double bond iii. Triple bond.

Lewis structure of water

H + O+ H

H O H 2e- 8e- 2e-

or single covalent bonds

H

O

H

Double bond – two atoms share two pairs of electrons

O 8e-

C

O

8e-

8e-

double bonds

or O

C

O

double bonds

Triple bond – two atoms share + three pairs of electrons

N 8e-

N

or

N

N

8e-

triple bond

triple bond

4.1.3.3 Coordinate Covalent Bond (Dative Bond) 



Dative bond is a bond in which the pair of shared electrons is supplied by one of the two bonded atoms Involve overlapping of a full orbital and an empty orbital

Requirement for dative bonds: i. Donor atoms should have at least one lone pair electrons ii. The atoms that accepts these electrons should have empty orbitals.

i.Single bond i.Double bond i.Triple bond

Steps in Writing Lewis Structures 1. Count total number of valence e- of atoms involved. 2. Add 1 for each negative charge. Subtract 1 for each positive charge. 3. Draw skeletal structure of the compound. Put least electronegative element in the center. 4. Complete an octet for all atoms except hydrogen 5. If structure contains too many electrons, form double and triple bonds on central atom as needed.

Example Draw the Lewis structure for each of the following compounds: i. HF ii. CH4 iii. CHCl3 iv. NH3 v. H2O

Total no. of valence electrons H: 1e F: 7e Total : 8e

Number of electrons C : 4e 4H: 4e Total : 8e

Center atom: N 

Count electrons: C : 4e H : 1e 3Cl: 21e Total: 26 e

Count number of electrons N : 3H : Total :

5e 3e 8e

4.1.5 Bond Length Compare the bond length between single, double and triple bond Bond length :

The distance between nuclei of the atoms involves in the bond C

C

1.54 Å

C

1.34 Å

C

C C

1.20 Å

As the number of bonds between the carbon increase, the bond length decreases because C are held more closely and tightly together As the number of bonds between two atoms increases, the bond grows shorter and stronger

The sum of formal charge on each atom should equal: i.zero for a molecule ii.the charge on the ion for a polyatomic ion Formal charge is used to find the most stable Lewis structure

EXAMPLE 1) Draw all the possible Lewis structure of COCl2. 2) Predict the most plausible structure.

SOLUTION 1)

2)

The most plausible structure is (2) Formal charge is determined before completing a Lewis structure to predict the most stable structure because formal charge closest to zero.

EXERCISE 1

Draw the possible Lewis structures for HNO2.  Determine the most plausible Lewis structures for HNO2. 

EXERCISE 2 

Suggest the possible Lewis structure for H2SO4. Explain your answer.

EXERCISE 3 1) HCN 1) CO2 1) SCN



Three conditions:

1) Incomplete octet 2) Expanded octet 3) Odd no. electron







Occurs when central atom has less than 8 electrons. Elements that can form incomplete octet are: Boron,B , Beryllium, Be & Aluminium, Al This is due to elements being relatively small in size but having high nuclear charge.







Occurs when central atom has more than 8 electrons. Formed by non-metals that have d orbitals OR Non-metals of the 3rd, 4th, 5th….rows in the periodic table



Nitrogen may form compounds that contain odd number electrons. Example: Nitric oxide, NO Nitrogen dioxide, NO2

The use of two or more Lewis structures to represent a particular molecule.  Requirement: Molecules/ions must have multiple bonds and lone pairs electrons at the terminal atoms. 

RESONANCE STRUCTURE FOR NO3-

EXERCISE: Write Lewis structures of the following compounds/ ions: CCl4

CO32-

HCN

PCl3

HNO3

PO43-

C2H4

C2H2

CH2Cl2

ICl

NH4+

NF3

H2S

N2H4

PH3

CS2

NO2-

XeF4

NH3

HCOOH

SO42-

ICl4-

SF6

O3

NO2

4.2 MOLECULAR SHAPE AND POLARITY

LEARNING OUTCOMES At the end of the lesson, students should be able to; 1. Explain valence shell electron repulsion theory. (VSEPR) 2. Draw the basic molecular shapes : linear, trigonal planar, tetrahedral, trigonal bipyramidal and octahedral. 3. Predict and explain the shapes of molecule and bond angles in a given species. 4. Explain bond polarity and dipole moment. 5. Deduce the polarity of molecules based on the shapes and the resultant dipole moment 63

4.2 MOLECULAR SHAPE AND POLARITY i. ii. iii.

64

VSEPR theory 5 basic shapes polarity

Molecular shape:

Introduction

shows the 3-dimensional arrangement of atoms in a molecule Predicted by using Valence Shell Electron Pair Repulsion (VSEPR) theory

65

4.2.1 VSEPR The Valence-Shell Electron Repulsion theory states that: The valence electron pairs around the central atom are oriented as far apart as possible to minimize the repulsion between them.

66

The repulsion may occur either between: a) bonding pair & another bonding pair b) bonding pair & lone pairs or c) between lone pair & another lone pairs 67

The strength of repulsion: The order of repulsive force is: Lone pairlone pair repulsion

> Lone pair-bonding > Bonding pair-bonding pair repulsion pair repulsion

Decrease of the repulsion force Note: The electron pairs repulsion will determine the orientation of atoms in space 68

4.2.2 Shape of a molecule  Basic shapes are based on the repulsion between the bonding pairs.

69

 Tips to determine the molecular shape : Step 1 Draw Lewis structure of the molecule Step 2 Consider the number of bonding pairs Step 3 Place bonding pairs as far as possible to minimize repulsion.

A.

Molecules with 2 bonding pairs  shape

Example: BeCl2 Lewis structure Be : 2e

180°

2Cl :14e Total : 16 e .. .. Cl: Cl Be ..

..

: 70

Linear

B.

Molecules with 3 bonding-pairs

 Example: BCl3  Repulsive forces Lewis structure between pairs are the B: 3e same 3Cl : 21e Total: 24e :

120°

..

B

.. Cl : ..

..

: 71

.. Cl

..

Cl

Trigonal planar

C. Molecules with 4 bonding pairs  Example: CH4 Lewis structure

 Equal repulsion between bonding pairs – equal angle

H H

C

H

109.5°

H

Tetrahedral 72

D. Molecules with 5 bonding pairs  Example: PCl5  Lewis structure :

90°

.. ..

:

..

..

Cl

:

73

.. Cl :

120°

..

P

.. Cl : ..

.. Cl

..

Cl

 Shape:

Trigonal bipyramidal

E. Molecules with 6 bonding pairs  Example: SF6 Lewis structure S : 6e 6F : 42e Total : 48e

F F F

S F F

74

F

Octahedral 90o 90o

2 electron pairs in the valence shell of central atom: Class of molecules

Number of bonding pairs

Number of lone pairs

AB2

2

0

Shape

180°

Linear 75




AB3

3

0

Shape

120°

76

trigonal planar




AB4

4

0

Shape

109.5o

77

Tetrahedral




AB5

5

0

Shape

90°

120°

78

Trigonal pyramidal




AB6

6

0

Shape

90°

90°

79

Octahedral

4.2.3 Effect of lone pairs on molecular shape The geometries of molecules and polyatomic ions, ions with one or more lone pairs around the central atom can be predicted using VSEPR. The molecular geometry is determined by the repulsions of electron pairs in the valence shell of the central atoms. atoms 80

 Repulsion between electron pairs decreases in the order of: Lone pairlone pair repulsion

> Lone pairbonding pair repulsion

> Bonding pairbonding pair repulsion

Stronger to weaker repulsion

81

 Electrons in a bond are held by the attractive forces exerted by the nuclei of the two bonded atoms therefore, they take less space of repulsion.  Lone- pair electrons in a molecule occupy more space; therefore they experience greater repulsion from neighboring lone pairs and bonding pairs

82

Number of electron pair : 3 Example : SO2  Class of molecules : AB2E  Molecular shape : Bent / V-shaped

83

Number of electron pair : 4 Example : NH3  Class of molecules : AB3E  Molecular shape : Trigonal pyramidal

84

Number of electron pair : 4 Example : H2O  Class of molecules : AB2E2  Molecular shape : Bent / V-shaped

85

Number of electron pair : 5 Example : SF4  Class of molecules : AB4E  Molecular shape : Distorted tetrahedron / seesaw

86

Number of electron pair : 5 Example : ClF3  Class of molecules : AB3E2  Molecular shape : T-shaped

87

Number of electron pair : 5 Example : I3 Class of molecules : AB2E4  Molecular shape : Linear

88

Number of electron pair : 6 Example : BrF5  Class of molecules : AB5E  Molecular shape : Square pyramidal

89

Number of electron pair : 6 Example : XeF4  Class of molecules : AB4E2  Molecular shape : Square planar

90

Shape of molecules which the central atom has one or more lone pairs Class of molecules



AB2E

2

1

Shape

Bent / V-shaped 91

Bond angle : < 120o

4 electron pairs in the valence shell of central atom:

92

Class of molecules



AB3E

3

1

Shape

Trigonal pyramidal Bond angle : < 109.5o




AB2E2

2

2

Shape

Bent / V-shaped Bond angle : < 109.5o 93


94

Class of molecules



AB4E

4

1

Shape

Distorted tetrahedral (see-saw) Bond angle : < 90o


95

Class of molecules



AB3E2

3

2

Shape

T-shaped Bond angle : < 90o


96

Class of molecules



AB2E3

2

3

Shape

Linear Bond angle : 180o


97

Class of molecules



AB5E

5

1

Shape

Square pyramidal Bond angle :90o and 180o


98

Class of molecules



AB4E2

4

2

Shape

Square planar Bond angle : 90o

99

COMPARISON OF BOND ANGLE IN CH4, NH3 AND H2O

109.5o

100

107.3o

104.5o

a) CH4  Has 4 bonding pairs electrons.  The repulsion between the bonding pairs electrons are equal. equal  The bond angles are all 109.5o

101

b) NH3

 has 3 bonding pairs electron and 1 lone pair electron.  according to VSEPR, lone pair - bonding pair > bonding pair - bonding pair repulsion. repulsion  Lone- pair repels the bonding-pair more strongly, strongly the three NH bonding-pair are pushed closer together, thus HNH angle in ammonia become smaller, 107.3o. 102

c) H2O

 Has 2 bonding pairs electrons and 2 lone pair electrons.  According to VSEPR, lone pair – lone pair > lone pair – bonding pair > bonding pair – bonding pair repulsion. repulsion  Lone-pair tend to be as far from each other as possible.  Therefore, the two OH bonding-pairs are pushed toward each other.  Thus, the HOH angle is 104.5o. 103

4.2.4 POLAR AND NONPOLAR MOLECULES  A quantitative measure of the polarity of a bond is

its dipole moment ( µ ). µ µ = Qr

µ

Where : µ = dipole moment Q = the product of the charge from electronegativity r = distance between the charges. Dipole moments are usually expressed in debye units(D) 104

E.g : Polarity of HF  Hydrogen fluoride is a covalent molecule with a polar bond.  F atom is more electronegative than H atom, so the electron density will shift from H to F.  The symbol of the shifted electron can be represented by a crossed arrow to indicate the direction of the shift. H 105

F

The consequent charge separation can be represented by : δ + : partial positive charge δ - : partial negative charge

106

 Diatomic molecules containing atoms of different elements (e.g. : HCl, NO and CO) have dipole moments and are called polar molecules.  Diatomic molecules containing atoms of the same element (e.g. : H2, N2 and Cl2) do not have dipole moments and are called nonpolar molecules.

107

For polyatomic molecules, the polarity of the bond and the molecular geometry determine whether there is a dipole moment. Even if polar bond are present, the molecules will not necessarily have a dipole moment.

108

Example Predict the polarity of the following molecules:

– – – –

109

Carbon dioxide, CO2 Carbon tetrachloride, CCl4 Chloromethane, CH3Cl Ammonia, NH3

(a) Carbon dioxide, CO2

110

molecular geometry : linear oxygen is more electronegative than carbon, Dipole moment can cancell each other has no net dipole moment (µ = 0) therefore CCl4 is a nonpolar molecule.

(b) Carbon tetrachloride, CCl4

111

molecular geometry : tetrahedral Chlorine is more electronegative than carbon, Dipole moment can cancell each other has no net dipole moment (µ = 0) therefore CCl4 is a nonpolar molecule.

( c) Chloromethane, CH3Cl

- molecular geometry : tetrahedral - Cl is more electronegative than C, C is more electronegative than H - Dipole moment cannot cancell each other - has a net dipole moment (µ ≠ 0) - therefore CH3Cl is a polar molecule. 112

(d ) Ammonia, NH3

- molecular geometry : tetrahedral - N is more electronegative than H, - Dipole moment cannot cancell each other - has a net dipole moment (µ ≠ 0) - therefore NH3 is a polar molecule. 113

Factors that affected the polarity of molecules molecular geometry electronegativity of the bonded atoms.

114

BOND NON-POLAR POLAR NON-POLAR MOLECULES

NON-POLAR MOLECULES Symetrical molecules - basic molecular shape with the same terminal atom - molecules with lone pairs linear (from trigonal bipyramidal) and square planar with the same terminal atom 115

POLAR MOLECULES Non-symetrical molecules - basic molecules with different terminal atom - molecules with lone pairs except linear and square planar

Exercises : Predict the polarity of the following molecules:



116

SO2 ; HBr ; SO3 ; CH2Cl2 ; ClF3 ; CF4 ; H2O ; XeF4 ; NF3 ; Cis-C2H2Cl2 ; trans-C2H2Cl2

4.3 ORBITAL OVERLAP AND HYBRIDIZATION

1. Formation Covalent Bond 2. Formation Hybrid orbitals 3. Orbital Overlapping 117

Objectives At the end of this subtopic, students should be able to: 1. Draw and describe the formation of sigma( and pi() bonds from overlapping of orbitals. 2. Draw and explain the formation of hybrid orbitals of a central atom: sp, sp2, sp3, sp3d, sp3d2 using appropriate examples. 3. Draw orbitals overlap and label sigma( and pi() bonds of a molecule. 118

4.3.1

Valence Bond theory

• explains the formation of covalent bonds and the molecular geometry outlined by the VSEPR. • States that a covalent bond is formed when the neighboring atomic orbitals overlap. • Overlapping may occur between: a) orbitals with unpaired electrons b) an orbital with paired electrons and another empty orbitals (dative bond) 119

Example: H

H

The s-orbital of the Hydrogen atom

Change in electron density as two hydrogen atoms approach each other. High electron density as the orbitals overlap (covalent bond formed) 120

10.3

FORMATION OF COVALENT BOND • Valence bond theory - Covalent bond is formed when two neighbouring atomic half-filled orbitals overlap. • Two types of covalent bonds are

a) sigma bond (σ) b) pi bond ()

121

a)  bond • formed when orbitals overlap along its internuclear axis (end to end overlapping) Example: i. overlapping s orbitals

H

+

H

H

H

 bond 122

ii.

Overlapping of s and p orbitals

Px orbital H

+

x

H

 bond

123

x

• iii. Overlapping of p orbitals

x

+

x

x

 bond

124

b) bond • Formed when two p-orbitals of the same orientation overlap sideways y

y

y

y

+

125

 bond

y

y

y

+

 bond 126

y

Formation of bonds in a molecule • Covalent bonds may form by: a) overlapping of pure orbitals b) overlapping of hybrid orbitals

127

Overlapping of pure orbitals • Example : i. O2 ii. N2

128

O2 Consider the ground state configuration: O : 1s2 2s2 2p4

Two unpaired electrons to be used in bonding.

y 1s

2s

2p

Overlapping occurs between the p-orbitals of each atom

129

O

y



σ



O

x

y

y



O

O σ



130

x

N2

131

4.3.2

Formation Hybrid orbitals

•

Overlapping of hybrid orbitals and the pure orbitals occur when different type of atoms are involved in the bonding.

•

Hybridization of orbitals: mixing of two or more atomic orbitals to form a new set of hybrid orbitals

•

The purpose of hybridisation is to produce new orbitals which have equivalent energy

•

Number of hybrid orbitals is equal to number of pure atomic orbitals used in the hybridization process.

132

Hybridization • Hybrid orbitals have different shapes from original atomic orbitals • Types of hybridisation reflects the shape/geometry of a molecule • Only the central atoms will be involved in hybridisation

133

10.4

Hybridization of orbitals i. sp ii. sp2 iii. sp3 iv. sp3d v. sp3d2

134

sp3 hybridization • one s orbital and three p orbitals are mixed to form four sp3 hybrid orbitals • the geometry of the four hybrid orbitals is tetrahedral with the angle of 109.5o .

135

sp3 hybrid • Mixing of s and three p orbitals sp3

sp3

sp3 sp 3

136

Example: 1)

CH4

• Lewis structure : • Valence orbital diagram ; H: C ground state : C excited : C hybrid : • Orbital Overlap : • Molecular Geometry : 137

Example : Methane, CH4 Ground state : C : 1s2 2s2 2p2 1s

2s

Lewis Structure

H H C H H

2p

Excitation: to have 4 unpaired electrons

H sp3

Excited state : 1s

2s

2p

C sp3

sp3 hybrid shape: tetrahedral 138

H

sp3

H

sp3 H

Fig. 10.8

sp3-Hybridized C atom in CH4

sp3 1s

sp3

sp3 sp3

1s 139

1s

Example 2 :

NH3 Lewis structure : Valence orbital diagram ; H: N ground state : N excited : N hybrid : Orbital Overlap : Molecular Geometry : 140

Fig. 10.9

sp3

1s

sp3

sp3 sp3

1s 141

1s

Example: 3) H2O • Lewis structure

:

• Valence orbital diagram; O ground state : O hybrid : • Orbitals overlap:

142

2 sp

hybridization

• one s orbital and two p orbitals are mixed to form three sp2 hybrid orbitals • the geometry of the three hybrid orbitals is trigonal planar with the angle of 120o .

143

Fig. 10.12

s

px

sp2

py

144 one s orbital + two p orbitals

sp2

sp2

three sp2 orbitals

• simplified drawing of sp2 orbitals:

Shown together (large lobes only)

145

Example: 1) BF3 • Lewis structure

:

• Valence orbital diagram; F: B ground state : B excited : B hybrid : • Orbital overlap: 146

Example: BF3

Pure p orbital



 sp2

sp2

F : 1s22s22p5 sp2

 Shape : trigonal planar

147

Example: 2) C2H4 • Lewis structure

:

• Valence orbital diagram; C ground state : C excited : C hybrid : • Orbital overlap:

148

Fig. 10.16a-c

 bonds

 bond 149

150

10.5

sp hybridization • one s orbital and one p orbital are mixed to form two sp hybrid orbitals • the geometry of the two hybrid orbitals is linear with the angle of 180o

151

Types of hybrid orbitals

Formation of sp Hybrid Orbitals

sp

sp

Produces linear shape

152

10.4

Example: 1) BeCl2 • Lewis structure

:

• Valence orbital diagram; Cl : Be ground state : Be excited : Be hybrid : • Orbital overlap: 153

Fig. 10.11

154

Example: 2) C2H2 • Lewis structure

:

• Valence orbital diagram; C ground state : C excited : C hybrid : • Orbital overlap:

155

Fig. 10.19a-c

156

Example: 3) CO2 • Lewis structure

:

• Valence orbital diagram; O: C ground state : C excited : C hybrid : • Orbital overlap: 157

3 sp d

hybridization

• one s orbital, three p orbitals and one d orbital are mixed to form five sp3d hybrid orbitals. • the geometry of the five hybrid orbitals is trigonal bipyramidal with the angle of 120o and 90o

158

• simplified drawing of sp3d orbitals:

159

Example: 1) PCl5 • Lewis structure

:

• Valence orbital diagram; Cl : P ground state : P excited : P hybrid : • Orbital overlap: 160

Example: 2) ClF3 • Lewis structure

:

• Valence orbital diagram; F: Cl ground state : Cl excited : Cl hybrid : • Orbital overlap: 161

3 2 sp d

hybridization

• one s orbital, three p orbitals and two d orbitals are mixed to form six sp3d2 hybrid orbitals • the geometry of the six hybrid orbitals is octahedral with the angle of 90o

162

• Simplified drawing of sp3d2 orbitals:

163

Example: 1) SF6 • Lewis structure

:

• Valence orbital diagram; F: S ground state : S excited : S hybrid : • Orbital overlap: 164

Example: 2) ICl5 • Lewis structure

:

• Valence orbital diagram; Cl : I ground state : I excited : I hybrid : • Orbital overlap: 165

How do I predict the hybridization of the central atom? Count the number of lone pairs AND the number of atoms bonded to the central atom No of Lone Pairs + No of Bonded Atoms

Hybridization

Examples

2

sp

BeCl2

3

sp2

BF3

4

sp3

CH4, NH3, H2O

5

sp3d

PCl5

6

sp3d2

SF6

166

10.4

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

167

Exercise: • For each of the following, draw the orbital overlap to show the formation of covalent bond a) XeF2 b) O3 c) ICl4 d) OF2 168

4.4 Intermolecular forces

LEARNING OUTCOMES At the end of the lesson, students should be able to; 1.

2. 3.

1.

Describe intermolecular forces i. van der Waals forces : - dipole-dipole interactions or permanent dipole - London forces or dispersion forces ii. Hydrogen bonding Explain factors that influence the strength of van der Waals forces Explain the effects of hydrogen bonding on i. boiling point ii. Solubility iii. Density of water compared to ice Explain the relationship between : i. intermolecular forces and vapour pressure ii. Vapour pressure and boiling point

4.4 Intermolecular forces 4.4.1

Types of intermolecular forces 4.4.2The effect of intermolecular forces on the physical properties.

171

Intermolecular Forces 

Intermolecular forces are the attractive forces between molecules

172

Effects of intermolecular forces on physical properties 

Have effects on these physical properties: a) boiling point b) melting point c) solubility d) density e) electrical conductivity 173

Intermolecular Forces Van der Waal Forces

Hydrogen Bond

Between covalent molecules

Between covalent molecules with H covalently bonded to F, O or N 174

4.4.1.1

van der Waal Forces

Forces that act between covalent molecules  Three types of interaction: i. Dipole-dipole attractive forces - act between polar molecules ii. London Dispersion forces - act between non-polar molecules 

175

Dipole-dipole forces (permanent dipole forces) Exist in polar covalent compounds  Polar molecules have permanent dipole due to the uneven electron distributions Example: 

-

+

H

Cl

Chlorine is more electronegative, thus it has higher electron density



-

+

H

Cl

Dipole-dipole forces; the partially positive end attracts the partially negative end

176

4.4.1.2. London Dispersion Forces 





attractive forces that exist between non-polar molecules result from the temporary (instantaneous) polarization of molecules The temporary dipole molecules will be attracted to each other and these attractions is known as the London Forces or London Dispersion forces 177

The formation of London forces At any instant, electron distributions in one molecule may be unsymmetrical.  The end having higher electron density is partially negative and the other is partially positive.  An instant dipole moment that exists in a molecule induces the neighboring molecule to be polar. 

178

Example: London forces in Br2 Electrons in a molecule move randomly about the nucleus Br

At any instant, the electron density might be higher on one side +

Br

Temporary dipole molecule

Br

-

Br

+

Br

-

Br

The temporary dipole molecule induce the neighboring atom to be partially polar

London forces

179

Factors that influence the strength of the van der Waals forces. 



The molecular size/molecular mass Molecules with higher molar mass have stronger van der Waals forces as they tend to have more electrons involved in the London forces. Example: CH4 has lower boiling point than C2H6 Note: However if two molecules have similar molecular mass, the dipole-dipole interaction will be more dominant. Example: H2S has higher boiling point than CH3CH3

180

4.4.1.3 Hydrogen intermolecular bond Dipole-dipole interaction that acts between a Hydrogen atom that is covalently bonded to a highly electronegative atom ; F, O ,N in one molecule and F,O or N of another molecule. Example: 



-

+

H

F



-

+

H

F

Hydrogen intermolecular bond 181

Other examples: NH3 liquid

H2O

.. H Hydrogen intermolecular bond

.. N

H

H

H

O

.. H

H H

..

H

O

..

..

H

H

..

N

Hydrogen intermolecular bond

Covalent bond

..

H

H

O

..

O

.. H

Hydrogen intermolecular bond

H

182

Consider ethanol, CH3OH CH3OH

CH3OH and methane

Hydrogen bond .. .. H O O

H

H

.. O

H

H

..

H

H

in

C

H

Not a hydrogen bond

C

..

..

H

C

H

H

H H

H is not bonded to either F, O or N

C H

H H

183

Example: H2O

___ covalent bond ----- hydrogen bond

H O H

H

H O H

O H

H

O

H

H O H

184

185

Properties of compounds with Hydrogen intermolecular forces

Boiling point 

Have relatively high boiling point than compounds having dipole-dipole forces or London forces - the Hydrogen bond is the strongest attraction force compared to the dipoledipole or the London forces. 186

Solubility A. Dissolve in polar solvent  The molecules that posses Hydrogen bonds are highly polar.  They may form interaction with any polar molecules that act as solvent. B. Dissolve in any solvent that can form Hydrogen bonds 187

Example NH3 dissolves in water because it can form Hydrogen intermolecular bond with water.

..



H

O

..

Hydrogen bond

H

..

..

N

N H

H H

H

H H

188

Problem: 

Explain the trend of boiling point given by the graph below: T/oC HF

HI HBr HCl

Molecular mass

189

Answer 

HF can form hydrogen bonds between molecules while HCl, HBr and HI have van der Waals forces acting between molecules. Hydrogen intermolecular bond is stronger that the van der Waals forces. More energy is required to break the Hydrogen bond.



Boiling point increases from HCl to HI. The strength of van der Waals forces increases with molecular mass. Since molecular mass increases from HCl to HI, thus the boiling point will also increase in the same pattern. 190

The effect of Hydrogen bond on water molecules 

The density of water is relatively high compared to other molecules with similar molar mass. Reason: Hydrogen intermolecular bonds are stronger than the dipole-dipole or the London forces. Thus the water molecules are drawn closer to one another and occupy a smaller volume. 191

Density Ice (solid H2O) has lower density compared to its liquid. Refer to the structure of ice

Ice form tetrahedral arrangement

Hydrogen bond takes one of the tetrahedral orientation and occupy some space

193



H2O(l) is denser than H2O(s) because  the hydrogen bond in ice arrange the H2O molecules in open hexagonal crystal  H2O molecules in water have higher kinetic energy and can overcome the hydrogen bond  V-shaped water molecules slide between each other.

194

195

196

Fig. 11.13

197

Boiling points of substance with Hydrogen intermolecular bonds 

The boiling points of these substances are affected by: a) the number of hydrogen bonds per molecule b) the strength of H intermolecular forces which directly depends on the polarity of the hydrogen bond Example: Explain the trend of boiling points given below: The order of the increase in boiling point is: H2O > HF > NH3 > CH4 198

Answer: 







by looking at the polarity of the bond, we have (Order of polarity: HF > H2O > NH3) but H2O has the highest boiling point. For H2O, the number of hydrogen bonds per molecule affects the boiling point. Each water molecule can form 4 hydrogen bonds with other water molecules. More energy is required to break the 4 Hydrogen bonds. HF has higher boiling point than NH3 because F is more electronegative than Nitrogen. CH4 is the lowest - it is a non polar compound and has weak van der Waals forces acting between molecules.

199

Effects of intermolecular forces on physical properties 1)Boiling point  For molecules with similar size, the order of intermolecular strength: Hydrogen bond > dipole-dipole forces > London dispersion forces  Strength of intermolecular forces ↑  boiling point ↑

200

201

Why boiling point H2O > HF and HF > NH3? 

Fluorine is more electronegative than oxygen, therefore stronger hydrogen bonding is expected to exist in HF liquid than in H2O.



However, the boiling point of H2O is higher than HF because each H2O molecules has 4 hydrogen bonds.

202





On the other hand, H-F has only 2 hydrogen bonds. Therefore the hydrogen bonds are stronger in H2O rather than in H-F.

203

Boiling point HF > NH3 

Fluorine is more electronegative than nitrogen ,thus the hydrogen bonding in H-F is stronger than H-N.

204

Vapour Pressure 

Molecules can escape from the surface of liquid at any temperature by evaporation



in a closed system : 



vapour molecules which leaves the surface cannot escape from the system the molecules strike the container wall and exert some pressure

Fig. 11.34



The pressure exerted by those molecules is called vapour pressure (or maximum vapour pressure)

Vapour pressure is the pressure exerted by a vapour in equilibrium with its liquid phase.

In a close system …. Liquid molecules vapourise

Molecules have enough energy to overcome intermolecular forces

Vapour molecules are trapped in the close container Volume of liquid becomes less

Rate of vaporisation is faster than the rate of condensation

Some of the vapour molecules may collide and lose their energy. They re-enter the liquid surface

System reaches equilibrium dynamic equilibrium Volume of liquid remains constant

Rate of vapourisation is equal to the rate of condensation Pressure exerted by the vapour molecules is known as the vapour pressure

Dynamic equilibrium and vapour pressure Number of liquid molecules leaving the surface is the same as the number of vapour molecules entering the liquid surface.

The vapour pressure at this stage is constant and known as the equilibrium vapour pressure.

Dynamic equilibrium is reached when: Rate of evaporation = rate of condensation Note: Equilibrium vapour pressure = saturated vapour pressure = vapour pressure

Factors that affects vapour pressure i. Intermolecular forces Molecules with weak intermolecular forces can easily vapourise. More vapour molecules will be present and exert higher pressure.  the weaker intermolecular forces the higher is the vapour pressure. ii. Temperature Heating causes more molecules to have high kinetic energies that are higher than their intermolecular forces. More liquid molecules will form vapour.  vapour pressure increases with temperature.

Fig. 11.35

Boiling – the process 

Increasing the temperature will increase in the vapour pressure.



As heat is applied, the vapour pressure of a system will increase until it reaches a point whereby the vapour pressure of the liquid system is equal to the atmospheric pressure.



Boiling occurs and the temperature taken at this point is known as the boiling point.



At this point, the change of state from liquid to gas occurs not only at the surface of the liquid but also in the inner part of the liquid.



Bubbles form within the liquid.



Boiling Point: the temperature at which the vapour pressure of a liquid is equal to the external atmospheric pressure.



Normal Boiling Point: the temperature at which a liquid boils when the external pressure is 1 atm (that is the vapour pressure is 760 mmHg)

Factors affecting the boiling point: 1.

Intermolecular forces

A substance with weak intermolecular forces can easily vapourise and the system requires less heat to achieve atmospheric pressure, thus it boils at a lower temperature. 2. Atmospheric pressure When the external atmospheric pressure is low, liquid will boil at a lower temperature.

4.5 Metallic bond

LEARNING OUTCOMES At the end of the lesson, students should be able to; 1. 2.

3. 4.

Explain the formation of metallic bond by using electron sea model. Relate metallic bond to the properties of metal: i. malleability ii. Ductility iii. Electrical conductivity iv. Thermal conductivity Explain the factors that affect the strength of metallic bond Relate the strength of metallic bond to boiling point

Metallic bond An electrostatic force between positive charge metallic ions and the sea of electrons.  Bonding electrons are delocalized over the entire crystal which can be imagined as an array of the ions immersed in a sea of delocalized valence electron. 

217

Metallic bonds Positive ions are immersed in the sea of electrons

e

e

Free moving electrons e

e

e

e

e

e

e

e

e

e

218

Electrostatic force in a metal Metallic Bond (Electron-sea Model)  Metals form giant metallic structure  Each positive ion is attracted to the ‘sea of electrons’.  These atoms are closely held by the strong electrostatic forces acting between the positive ions and the ‘sea of electrons’.  These free moving electrons are responsible for the high melting point of metals and the electrical conductivity. 219

Physical properties of metals  

metals have high melting point high energy is required to overcome these strong electrostatic forces between the positive ions and the electron sea in the metallic bond + e + e + e e + e + e e + e + e + e + e + e + e + e + e e

+ e + e

Metallic bonds formed by the electrostatic forces exist between positive ions and the free moving electrons

220

The strength of the metallic bonds 

The strength of the metallic bond increases with the number of valence electrons and the size of ions.



The smaller the size of positive ions the greater is the attractive force acting between the ions and the valence electrons

221

Boiling points in metals Mg

Na +1

+1 e

e +1

e +1

+1 e

e +1

+1

+1

e e +1 +1

+1 e

+1

+1 e

+1

+1

e e e +1 +1 +1

Has one valence electron

the electrostatic force acting between positive ions and free moving electrons form metallic bonds

ee ee ee ee ee +2 +2 +2 +2 +2 ee ee ee ee ee +2 +2 +2 +2 +2 ee ee ee ee ee ee +2 +2 +2 +2 +2

ee ee ee +2 +2 +2 ee ee ee +2 +2 +2 ee ee ee +2 +2 +2

Has 2 valence electrons Stronger metallic bond due to the size of Mg being smaller than Na and the strong electrostatic force between +2 ions and the two valence electrons,

Mg has higher boiling point than Na 222

Example: Explain the difference in the boiling point of the two metals given: Magnesium – 11300 oC Aluminum – 24500 oC

223

Answer 

 





The cationic size of Al is smaller compared to magnesium and its charge is higher (+3). Mg has two valence electrons Al has three valence electrons involved in the metallic bonding. The strength of metallic bond in Aluminium is greater than that of Magnesium Al has higher boiling point 224

The strength of metallic bond is directly proportional to the boiling point.  The stronger metallic bond,the higher the boiling point. 

225

CHAPTER 5.0 STATES OF MATTER 5.1 5.2 5.3

Gas Liquid Solid

At the end of the lesson, student should be able to : (a) Explain the general properties of gas in terms of arrangement of

particle, density and compressibility. (b) Explain qualitatively the basic assumptions of the kinetic molecular theory of gases for an ideal gas. (c) Define gas laws (i) Boyle’s Law (ii) Charles’s Law (iii) Avogadro’s Law (d) Sketch and interpret the graphs of Boyle’s and Charles’s laws (e) Perform calculations involving gas laws. (f) derive ideal gas equation based on the gas laws

(g) perform calculations using the ideal gas equation (h) determine the molar mass and density of a gas using ideal gas equation (i) Define:  Partial pressure  Dalton's law (j) perform calculation using Dalton's law (k) compare the ideal and non-ideal behaviours of gases in terms of intermolecular forces and molecular volume (l) explain the conditions at which real gases approach the ideal behaviour. (m) explain qualitatively van der Waals equation and relate the values of a and b to intermolecular forces and molecular volume of a gas

General Properties of Gas 

Particles of gas are far apart and fill the available space.



Gases assume the volume and shape of their containers.





can be compressed – due to the particles being so small and are relatively far apart from one another. Gases have relatively low densities.

Kinetic – Molecular Theory of Gases Describes the behavior of an ideal gas  Ideal gas : gases which obey the ideal gas equation (PV = nRT) 

The theory is based on the following assumptions: 1) Gas molecules are very tiny that their size are negligible compared to the volume of the container. (having mass but no volume) 2) Gas molecules move in straight lines and are at a constant motion unless they collide. 3) Molecular collisions are elastic – no energy is lost during collisions. 4) Attractive and repulsive forces between gas particles are negligible. 5) The average kinetic energy of the particles is proportional to the absolute temperature.

The Gas Laws a) Boyle’s Law : The volume of fixed amount of gas at constant temperature is inversely proportional to the gas pressure 1

V  P PV = k

(no of mole and temperature are constant) Where: k = constant V = volume P = pressure T = temperature n = number of moles

 at different pressure and volume : P1 V1 = P2 V2 Where P1 = initial pressure V1 = initial volume P2 = final pressure V2 = final volume

Graph of P versus 1

Graph of P versus V

V

P P

1 V

V

pressure is inversely proportional to volume

pressure is directly 1 proportional to

volume

Graph of PV versus P

PV

PV = constant

P

Example 1 A sample of chlorine gas occupies a volume of 2 L at a pressure of 1 atm. Calculate the pressure of the gas if the volume is increased to 5 L at constant temperature.

0.4 atm

Example 2 The pressure of a sample of hydrogen gas in a 50.0 mL container is 765 mmHg. The sample is then transferred into another container and the measured pressure is 825 mmHg. What is the volume of the second container?

46.36 mL

a) Charles’s Law : The volume of a fixed amount of gas at constant pressure is directly proportional to the absolute temperature of the gas (in Kelvin). V  T (no of mole and pressure constant) V =k T

Where : k = constant T = absolute temperature (K) V = volume

At different volume and temperature:

V1 V2  T1 T2 Where V1 = initial volume T1 = initial temperature V2 = final volume T2 = final temperature

T = absolute temperature in Kelvin (K) T(K) = T°C + 273.15

Graph of volume versus temperature : V

V

0

T(K)

-273.15

T(0C)

Example 1 A sample of carbon monoxide gas occupies 3.2 L At 125 °C. The sample is then cooled at constant pressure until it contracts to 1.54 L. Calculate the final temperature in degree Celsius.

-81.54 °C

Example 2 A sample of gas trapped in a capillary tube by a plug of mercury at 22 oC has a volume of 4.5 mL. Calculate the volume of the gas when the capillary tube is heated to 60 oC.

5.08 mL

The Combination of Boyle’s and Charles’s Law 1 Boyle’s law : P Charles’s law : VT V

T P T V= k P PV =k T V

P1V1 T1

=

P2V2 T2

Example 1

A sample of methane gas occupies 25.5 L at 298.15 K and 153.3 kPa. Find its volume at STP.

Ans : 35.32 L

Example 2 2 moles of chlorine gas kept in a cylinder with piston occupies a volume of 49 L. When another 3 moles of chlorine gas is pumped into the cylinder at constant temperature and pressure the piston moves

upwards

to

accommodate

the

gas.

Calculate the final volume of the gas. Ans : 73.5 L

C) Avogadro’s Law At constant pressure and temperature, temperature the volume of a gas is directly proportional to the number of moles of the gas present V  n (P and T are constant) V = k n V =k n

V1 V2  n1 n2

where : n = number of moles k= constant

Combination of Boyle's law, Charles's law and Avogadro's law : 1 Boyle's Law: V P Charles' Law: VT

Avogadro's Law: V V=R

Vn

nT P nT P

PV = nRT Ideal gas equation

Where : R = gas constant T = Temperature(K) n = number of moles V = volume P = Pressure

Value of R depend on the unit of pressure and volume used in the equation. unit of pressure atm Nm2 or

Pa

unit of volume

value of R

unit of R

L or dm3 0.08206 L atm mol1 K1 m3

8.314

Nm mol1 K1 J mol1 K1

Example 1

A steel gas tank has a volume of 275 L and is filled with 0.485 kg of O2. Calculate the pressure of O2 if the temperature is 29 oC. Ans : 1.36 atm

Example 2

A sample of chlorine gas is kept in a 5.0 L container at 228 torr and 27 °C. How many moles of gas are present in the sample?

Ans : 0.06 mol

MOLAR MASS and DENSITY CALCULATION  Molar mass and density of a gas can be calculated by rearranging Equation:

the

Ideal

Gas

PV  nRT  m  PV=   RT  Mr  mRT Mr = PV

m where: n = Mr Mr = molar mass of a gas

Density of a gas

PV  nRT m PV=   RT  Mr  mRT P= VM r dRT P= Mr PM r d= RT

where, d = density of a gas

Example 1 Calculate the density of ammonia (NH3) in grams per litre (g/L) at 752 mmHg and 55 C. Ans : 0.625 gL-1

Example 2 A chemist has synthesized a greenish-yellow compound of chlorine and oxygen and finds that its density is 7.71 g L-1 at 36 °C and 2.88 atm. Calculate the molar mass of the compound.

Ans : 67.9 gmol-1

Example 3

What mass of KClO3 is required to produce 2.40 L O2 gas that measured at a pressure of 1 atm and a temperature of 26 oC. The reaction equation is 2KClO3(s) 2KCl(s) + 3O2(g)

Ans : 8.0 g

Dalton’s Law of Partial Pressure The total pressure of mixture of non reacting gases is the sum of the partial pressures exerted by each of the gas in the mixture (Partial pressure is the pressure of individual gas component in a mixture).

- For a mixture of 3 gases, A,B and C : PT = PA + PB + PC

According to ideal gas equation :

nA RT pressure exerted by gas A = PA = V

pressure exerted by gas B = PB

nBRT = V

 in the mixture of gases A and B : PT = PA + PB n A RT nBRT  = V V

= (nA + nB )RT V when nA + nB = nT

n TRT  PT = V

Mole fraction and pressures Daltons theory allows us to form a relationship between Mole fractions, partial pressure and a total pressure. Consider the following condition at constant T and V If P  n RT A

A

 

Ptotal  ntotal(RT ) Thus:

PA n A ( RT )  Ptotal ntotal ( RT )

PA  X A Ptotal



PA nA  PTotal nTotal

XA = mole fraction of gas A

Example 1 A gaseous mixture of 7.00 g N2 and 3.21 g CH4 is placed in a 12.0 L cylinder at 25 oC.

a) What is the partial pressure of each gas? b) What is the total pressure in the cylinder? Ans :a) 0.501 atm , 0.41 atm b) 0.92atm

Example 2 A mixture of gases contains 4.53 moles of neon, 0.82 moles of argon and 2.25 moles of xenon. Calculate the partial pressure of the gases if the total pressure is 2.15 atm at a certain temperature.

Ans : P Ne = 1.28 atm,, P Ar = 0.079

atm, P

Xe = 0.63 atm,

Example 3 A sample of gas at 5.88 atm contains 1.2 g CH4, 0.4 g H2 and 0.1 g He. Calculate :

a)

The partial pressure of CH4, H2 and He in the mixture.

b)

What is the partial pressure of CH4 and H2 if He is removed? Ans : a) P CH4= 1.74 atm,P H2 = 3.92 atm, P He= 0.49 atm b) P CH4= 1.6 atm,P H2 = 4.28 atm

 One of the applications of Dalton’s Law is to calculate the pressure of a gas collected over water ( for gases that not soluble in water).

gas

gas + water vapour

The gas collected is actually a mixture of the gas and water vapour.

PT  Pgas  PH 2O Vapour pressure of water, Pwater = 23.8 torr

Example 1 Consider the reaction below :

2KClO3

2KCl + 3O2

A sample of 5.45 L of oxygen is collected over water at a total pressure of 735.5 torr at 25 °C. How many grams of oxygen have been collected? (at 25°C, Pwater = 23.8 torr) Ans : 7.011 g

Example 2 Excess amount of hydrochloric acid is added to 2.5 g of pure zinc. The gas produced is collected over water in a gas cylinder at 28 oC and 100.0 kNm-2. Calculate :

a) the number of mole of gas produced in the reaction. b) the volume of gas collected in the cylinder. Ans : a) 0.038 mol b) 0.95 L

Ideal gas any gas that obeys the ideal gas equation and has the properties as outlined by the Kinetic Molecular Theory

Deviation from Ideal Behavior real gas (non-ideal gas) : gases which do not obey

ideal gas properties Real gases do not behave ideally because: i) gas molecules do have its own volume and they occupy some space. ii) gas molecules do have intermolecular forces acting between them

1. Gas molecules have volumes

•

The free volume where the molecules move about is smaller than the volume of the container.

2. There are intermolecular forces between molecules

• Collision between molecules and the container wall are less frequent • The actual pressure exerted by the gas is lower than calculated by ideal gas equation

Real gases behave almost ideally at a low pressure and high temperature.

Reason: At low pressure: • To achieve a low pressure, the volume of a container is increased. • Thus, the molecules will be far apart from another, hence the intermolecular forces can be neglected. • At a low pressure the size of a container is extremely large compared to the size of molecules, thus the size of molecules can be neglected.

Reason : At low pressure • At low pressure, the volume of a container is very large • Thus the molecules will be more far apart from one another. Hence the intermolecular forces can be neglected. • At low pressure, the volume of the container is extremely large compared to the size of the molecules, thus the volume can be neglected

At high temperature The gas molecules have high kinetic energies and move at high speed The molecules are able to free themselves from the intermolecular forces that act between them. The intermolecular forces can be neglected, thus they behave almost ideally.

Graph PV/RT against P for N2 at different temperatures PV 273K

RT

600K 1000K

The lines approach the ideal line when T increases

The value approaches 1.0 at a very low pressure P

9

Van der Waal’s Equation  Since real gas does not exhibit ideal gas behavior at high pressure and low temperature :  the ideal gas equation (PV=nRT) needs to be adjusted  adjusting the equation, two parameters need to be reconsidered : √ attractive forces between the gas molecules √ volume of the gas molecules

a) Attractive Forces Between Molecules 

Attractive forces which act between the gas molecules will :



make the molecules move slower



give less impact to the wall







pressure exerted by the real gas is less compared to the ideal gas since Preal < Pideal 2 n the term pressure need to be corrected by adding coefficient 2a V

n=number of moles

P ideal = P real + n2a V2

a=correcting factor for pressure V = volume of container

(b) Volume of the gas molecules

• Since the gas occupy a sizeable portion of a container, the space in which the molecules are able to move are less than the volume of the container. Vreal < V container • The correction factor done to the volume is: nb

V = volume

V = Vcontainer - nb

n = number of moles b = size factor

v

• Referring to the ideal gas equation: PV = nRT By replacing these into the equation P ideal = P real + n2a V2

& V = V container - nb

(Preal + n2a ) ( V – nb) = nRT V2

The van der Waals equation

Value of a : • a is a positive constant which depends on the strength of the attractive forces between molecules • Molecules with a higher value of a have stronger attractive forces.

Value of b : • b is a constant to represent the volume occupied by the molecules. • Larger molecules have bigger value of b

5.2 Liquids

At the end of this topic, students should be able to : a) Explain the properties of liquids : shape, volume, surface tension, viscosity, compressibility and diffusion. b)Explain vaporization process and condensation process based on kinetic molecular theory and intermolecular forces. c) Define vapour pressure and boiling point d) Relate : - intermolecular forces to vapour pressure - vapour pressure to boiling point

The properties of Liquids 1.

Volume and Shape



has a definite volume but not a definite shape √

the particles are arranged closely but not rigidly

√

held together by a strong intermolecular forces but not strong enough to hold the particles firmly in place

√ 

particles are able to move freely

thus, a liquid flows to fit the shape of its container and is confined to a certain volume

2. Surface tension 

•

Surface tension is the amount of energy required to stretch or increase the surface of a liquid by a unit area

Surface tension is caused by the attractive forces and the direction of the force acting between molecules √

Molecules within a liquid experienced intermolecular attractive forces in all directions by their neighbouring molecules

√

however, molecules at the surface are pulled inwards and sideways from the neighbouring molecules

Fig. 11.8

molecule at the surface

molecule within the liquid

√

these intermolecular attractive forces will pull the molecules into the liquid

√

Thus, cause the surface to stretch and tighten

√

the stronger the intermolecular attractive forces, the higher the surface tension

3. Viscosity 

is a measure of the resistance of a liquid to flow

the greater the viscosity, viscosity the more slowly the liquid flows.





Viscosity may be explained in terms of the intermolecular forces present in a liquid

The factors affecting the viscosity : i) The strength of intermolecular forces Liquids that have strong intermolecular forces have higher viscosity ii) The size of the molecules Liquids with larger size (higher molar mass) is more viscous because it has stronger intermolecular forces

iii) The temperature of the liquid At higher temperature, molecules have higher kinetic energy,  the molecules move faster  molecules can overcome the intermolecular attractive forces more easily  resistance to flow decrease  Viscosity decrease

4. Compressibility 

in liquid, the particles are packed closely together



thus, there is very little empty space between the molecules



 liquids are much more difficult to compress than gas

5. Diffusion 

Diffusion is the movement of liquid from one fluid through another.



Diffusion can occur in a liquid because molecules are not tightly packed and can move randomly around one another.

Vaporisation Process 

a process in which molecules escape from liquid into gaseous state through a surface √ molecules in a liquid moves freely, collide with each other and posses different magnitudes of kinetic energy √

when the kinetic energy is sufficient enough to overcome the intermolecular forces acting on them, the molecules will escape as vapour.



Factors affecting the rate of vaporization : √

i) Surface area

~ the surface area , the chances for the molecules to escape from the surface   surface area , evaporation rate 

√

ii) Temperature ~ temperature , the kinetic energy   more molecules have enough energy to

overcome the attractive intermolecular forces and escape from the surface of the liquid ~ thus evaporation rate 

Condensation Process

a process whereby gaseous molecules turn to liquid √ some of the vapour molecules may lose their kinetic energy during the collision √ they do not have enough energy to remain as vapour molecules √ they reached the surface of the liquid and trapped by the attractive forces √ if they cannot overcome the attractive forces, these vapour molecules return as liquid molecules √

the process is known as condensation

Vapour Pressure  molecules escape from the surface of a liquid : referred as vapour molecules Pressure exerted by the vapour molecules on the surface of a liquid is known as the vapour pressure

√

If evaporation occurs in an open container, vapour molecules will diffused away until the liquid dries up



in a closed system : √ √ √ √

√

vapour molecules which leaves the surface are trapped in the close container these vapour molecules are in constant random motion the molecules strike the wall of container and exert some pressure As the quantity of molecules in the vapour phase increase, some molecules may lose energy and condense Eventually, the rate of evaporation = the rate of condensation. (The system achieved dynamic equilibrium) At equilibrium, the number of vapour molecules above liquid are constant.

Factor that affects the vapour pressure • Temperature – Applying the heat causes more molecules have high kinetic energy – More molecules will be able to overcome the intermolecular forces and escape from the liquid to form vapour – Temperature , vapour pressure  • Intermolecular forces – Molecules with weak intermolecular forces can easily vapourised – More vapour molecules will exert pressure on the liquid surface – Intermolecular forces , vapour pressure 

Boiling • A process in which liquid molecules change to vapour at a particular temperature and at an atmospheric pressure. • Happens throughout the liquid. Evaporation • liquid molecules change to vapour at any temperature and pressure. • Occurs only at the surface of a liquid

Boiling Point: the temperature at which the vapour pressure of a liquid is equal to the external atmospheric pressure. Normal Boiling Point: the temperature at which a liquid boils when the external pressure is 1 atm (that is the vapour pressure is 760 mmHg)

Boiling Point - process • Increasing the temperature will increase in the vapour pressure. • As heat is applied, the vapour pressure of a system will increase until it reaches a point whereby the vapour pressure of the liquid system is equal to the atmospheric pressure. • Boiling occurs and the temperature taken at this point is known as the boiling point. • At this point, the change of state from liquid to gas occurs not only at the surface of the liquid but also in the inner part of the liquid. • Bubbles form within the liquid.

Factors affecting the boiling point: 1. Atmospheric pressure W hen the external atmospheric pressure is low, liquid will boil at a lower temperature 2. Intermolecular forces A substance with weak intermolecular forces can easily vapourise and the system requires less heat to achieve atmospheric pressure, thus it boils at a lower temperature.

5.3 Solid

OBJECTIVE… (a) State the properties of solid. (b) Explain the process of: - freezing

- sublimation

- melting

- deposition

(c) Differentiate between amorphous and crystalline solids. (d) Describe the types of bonding and the interparticle/intermolecular forces involved in the following crystalline solids using appropriate examples. i. metallic

iii. molecular covalent

ii. ionic

iv. giant covalent

Properties of solid 1.Particles are closely arranged and regularly in order 2. Rigid arrangement- particles can vibrate, rotate about fixed position and cannot move freely without disrupting the whole structure. 3. Has definite shape and volume. 4.Has strong forces between the particles. 5. Has high densities. 6. Incompressible. 7.Diffusion within the solid is extremely slow.

In principle, solid, liquid and gas states are interconvertible

liquid

sublimation deposition solid

gas

5

Melting (Fusion) Process  Solid is changing into a liquid  When a solid substance is heated : - its particles gain energy - therefore able to vibrate more rapidly - at certain temperature, the kinetic energy is higher enough to overcome the intermolecular forces of attraction between solid particles. – the particles are free to move and the solid start to melt • Melting point – the temperature at which solid and liquid coexist in equilibrium

Freezing (Solidification) Process  Liquid is changing into a solid  when the temperature of a liquid is lowered, the kinetic energy of the liquid particles decreases

 the liquid particles vibrate at a slower rate  when the intermolecular forces are strong enough to hold the particles together in a fixed and orderly arrangement, the liquid freezes. Freezing point – a temperature at which the liquid and solid phases of a substances coexist at equilibrium.

Sublimation Process •The process by which a substance changes directly from solid to the gaseous state without passing through the liquid state. •Occurs on solid with weak intermolecular forces of attraction

Deposition Process •The process where molecules from vapour state change to the solid state. •The opposite process of sublimation

Types of Solid

crystalline solid  A solid that has highly ordered

amorphous solid  Solid that does not have a

structure where atoms, ions or

regular three dimensional

molecules show a regular

arrangement of atoms or

repetition in three dimensional

molecules

arrangement  Formed when a saturated

liquid is cooled slowly  Its atoms, molecules or ions

occupy specific position  Example : ice, sugar, salt

 Formed when a saturated liquid is cooled rapidly  Example :  glass  plastic material  charcoal

Types of Crystalline Solid 1. Metallic solid 

Close packed structure



Composed of atoms of the same metal linked together by metallic bond

– – –

The properties of metal: High electrical and thermal conductivity Lustre Ductile and malleable



Examples : all metallic elements : Na, Mg, Fe



2.

Ionic solids 

consist of ions (cation & anion) held together by ionic bonds



Physical properties of ionic solid: - High melting point - Hard but brittle - Does not conduct electricity in the solid state but does conduct electricity in molten state or in aqueous state.



Example: NaCl, CsCl

3. Molecular covalent solid 

composed of molecules held together by intermolecular

forces

and/or hydrogen bonds) 

Example : iodine [I2 (s) ]

(van

der

Waals

4.

Giant covalent solids  very large molecules / gigantic structure  composed of atoms linked together by

covalent bond  examples : diamond, graphite, SiO2

5.4 Phase Diagram

Learning Outcomes : At the end of this lesson students should be able to: (a)

Define phase .

(b)

Sketch the phase diagram of H2O and CO2.

(c)

Define triple point and critical point.

(d)

Explain the anomalous behaviour of H2O.

(e)

Describe the changes in phase with respect to i.

temperature (at constant pressure)

ii. pressure (at constant temperature).

Phase is a homogeneous part of a system in contact with other parts of the system but separated from them by a well-defined boundary Phase Diagram is a diagram that describes the stable phases and phase changes of a substance as a function of temperature and pressure • used to predict the phase that exist under a certain temperature and pressure

(i)

Phase Diagram of H2O

C

B

T A

8/16/11

Regions of the diagram - The diagram has three main regions : Solid, liquid and gas - Each region corresponds to one phase of the substance. - A particular phase is stable for any combination of pressure and temperature. - Any point along a line shows the pressure and temperature at which the two phases exist in equilibrium.

8/16/11

Important Points 1. Point T : - Known as the triple point - Triple point is the point at which the vapour, liquid and solid states of a substance are in equilibrium. - Triple point for water is 0.01°C and 0.06 atm.

2. Point C : - Known as the critical point - Critical point is the point on a phase diagram at which the vapour cannot be condensed to a liquid. - The liquid – gas line ends at the critical point. - Above the critical point, the liquid cannot be distinguished from its vapour form. 8/16/11

(ii)

Phase Diagram of CO2

T

A

8/16/11

B

C

TRIPLE POINT At point T, the triple point; solid, liquid and vapour are in equilibrium CO2(s) ⇌ CO2(l) ⇌ CO2(g)



Triple point for CO2 is at 5.2 atm pressure and temperature 57 OC.





for most compounds, the TB curve slant to the right because solid is denser than liquid

8/16/11

ANOMALOUS BEHAVIOR OF H2O The phase diagram for water is not typical. The melting temperature line , TB , slopes slant to the left (negative slope) i.e. the melting temperature decreases with pressure. • This is because ice is less dense than water, the solid occupies more space than the liquid. An increase in pressure, favours the formation of liquid H2O.( the phase that occupy less space) Thus, the higher the pressure, the lower the temperature at which solid water melts.

Exercise: a) Describe the phase change when carbon dioxide undergoes isobaric heating at 5.2 atm pressure

Exercises: b) Describe the phase change when pressure is applied to water isothermally at 0.01 C

Chemical Equilibrium Chapter 6

6.0 Chemical Reaction Chemical reaction may occur in two ways:a) a non-reversible reaction - occurs in one direction b) a reversible reaction - occurs in both direction i.e (forward & reverse reaction)

6.1.1 A non-reversible reaction Example 1 NaCl + AgNO3

AgCl + NO3

• One/both reactants will be completely consumed and amount of products formed depends on the limiting reactant. • A single arrow (→) is used to represent reaction.

6.1.2

Reversible Reaction

Example 2 H2(g) + I2(g)

⇌

2HI(g)

• Occurs in both direction. • ⇌ shows that reactants react to form product, and product reacts to form reactants. • Amount of products formed does not depend on the amount of reactant used. • Reaction will reach equilibrium when the concentration of products & reactants remained constant.

6.2 Equilibrium System • A system is at equilibrium when there is no observable change occurs. • Equilibrium system can be observed in: a. Physical Equilibrium b. Chemical Equilibrium

6.2.1 Physical Equilibrium Example 3 • vaporisation in a closed container. H2O (l)

⇌

H2O (g)

• Involves physical change of substance. • Level of H2O in the container remains constant at equilibrium because the rate of evaporation equals the rate of condensation.

6.2.2 Chemical Equilibrium Example 4 2NO2(g) ⇌N2O4(g) • Initially NO2 molecules combine to form N2O4 (brown gas appears) • As soon as N2O4 is formed, it undergoes reverse reaction to form NO2. • At equilibrium, the system contains a constant amount of NO2 & N2O4.

6.3 Dynamic Equilibrium A system reaches a dynamic equilibrium when • the rate of forward reaction equals the rate of reverse reaction. • the concentrations of both reactants and products remain constant. • no observable change occurs but the conversions of reactants to products and products to reactants continue.

After time t1, both concentrations remain constant.

concentration

[N2O4] at equilibrium

[NO2 ]at equilibrium The reaction does not stop but the rate of forward reaction equals the rate of reverse reaction. t1

time

6.4 Equilibrium constant, K • Since concentrations at equilibrium remain constant, the equilibrium can be expressed by a constant. • Consider: aA (g) + bB (g) Kc =

⇌

cC (g) + dD (g)

[C]c[D]d [A]a[B]b

• Kc is known as equilibrium constant Concentrations of species are expressed in molar.

Table 6.1

N2O4 (g) K=

⇌

[NO2]2 [N2O4]

2NO2 (g) = 4.63 x 10-3

constant

6.5 Equilibrium Law: Law of Mass Action • When a system has reached equilibrium, the ratio of multiplied concentrations of products to the multiplied concentrations of reactants (each raised to the power of the respective stoichiometric coefficient) is a constant at constant temperature. OR • Value of equilibrium constant, Kc is a constant at constant temperature.

6.6 Types of System in Chemical Equilibrium a) Homogeneous equilibrium Homogenous equilibrium applies to reactions in which all reacting species are in the same phase. b) Heterogeneous equilibrium A system in which the reactants and products are not in the same phase.

6.7 Homogeneous Equilibrium System 6.7.1 Liquid phase • Important variable: concentration • Equilibrium constant: Kc Example 5 CH3COOH (l) + CH3OH (l) ⇌ CH3COOCH3 (l) + H2O (l)

Kc=

Example 6 CH3COOH (aq) ⇌ CH3COO- (aq) + H+ (aq)

Kc=

6.7.2 Gas phase • The quantitative aspects to be considered are the concentration and pressure. • We can use Kcand Kp to represent the equilibrium constant. (PC)c (PD)d Kp =

(PA)a (PB)b

the pressure of gases are expressed in atm or other units of pressure

Example 7 Write the equilibrium law in the form of Kp and Kc for the following reactions: a) N2(g) + 3H2(g) NH3(g) b) 2NO(g) + O2(g) N2O4(g) Answer: Kp = Kc =

Expression of Kc Expression of Kc depends on the equilibrium equation given Example 8 2SO3 (g) ⇌ 2SO2 (g) + O2 (g)

Kc = [SO2]2 [O2] [SO3]2

1. SO3 (g) ⇌ SO2 (g) + ½ O2 (g)

Kc1 = [SO2] [O2]1/2 [SO3]

2. 2SO2 (g) + O2 (g) ⇌ 2SO3 (g)

Kc2 =

[SO3]2 [SO2]2 [O2]

Example 9 Find the relationship between Kc1 and Kc2 for the following equilibrium equations. 1. SO3 (g) ⇌ SO2 (g) + ½ O2 (g)

Kc1 = [SO2] [O2]1/2 [SO3]

2. 2SO2 (g) + O2 (g) ⇌ 2SO3 (g)

Kc2 =

[SO3]2 [SO2]2 [O2]

Example 10 Consider the following reaction: A(g) + B(g) ⇌ 2C(g) If 5 moles of A are allowed to mix and react with 3 moles of B in 1dm3 container, 2 moles of C are produced at equilibrium. What is the value of Kc for this reaction? (Ans: 0.5 )

Example 11 Consider the reaction: CO(g) + 2H2(g) ⇌ CH3OH(g) The following equilibrium concentrations are achieved at 400K: [CO] = 1.03M [H2] = 0.332M [CH3OH] = 1.56M Determine the equilibrium constant at 400K (Ans: 13.74 M-2)

Example 12 Nitrogen gas reacts with oxygen gas at high temperature to form nitrogen monoxide gas, NO. The equilibrium constant for the reaction is 4.1 x 10-4 at 200oC. If the concentration for both N2(g) and O2(g) at equilibrium are 0.40 moldm-3 and 1.3 moldm-3 respectively, calculate the concentration of NO gas at equilibrium? (Ans: 0.0146 M)

Example 13 The equilibrium constant Kp for the reaction; 2NO2 (g) ⇌ 2NO (g) + O2 (g) is 158 at 1000K. What is the equilibrium pressure of O2 if the P NO2 = 0.400 atm and PNO = 0.270 atm?

Answer; 2 PNO PO

Kp =

PO = 2

2

2 PNO 2 2 PNO 2 Kp 2 PNO

PO2 = 158 x (0.400)2 (0.270)2 = 347 atm

6.7.3 Relationship between Kp and Kc Consider aA (g) + bB (g) Kc =

⇌

cC (g) + dD (g)

[C]c[D]d [A]a[B]b

Thus, Kc =

{PC /(RT)}c {PD / (RT)}d {PA /(RT)}a {PB /(RT)}b

Since P = nRT V [ ]=

P RT

For the reaction aA (g) + bB (g) Kc =

⇌

cC (g) + dD (g)

PCc PDd (RT) (a + b) – (c+d) PAa PBb

Kc = Kp (RT) (a + b) – (c+d) Kp = Kc (RT) (c + d) – (a+b)

Kp = Kc (RT)Δn

1. The equilibrium concentrations for the reaction between carbon monoxide and chlorine to form COCl2 (g) at 740C are [CO] = 0.012 M, [Cl2] = 0.054 M, and [COCl2] = 0.14 M. Calculate the equilibrium constants Kc and Kp. CO (g) + Cl2 (g)

⇌

COCl2 (g)

Answer ; 0.14 [COCl2] = Kc = 0.012 x 0.054 [CO][Cl2] = 220 M -1 Kp = Kc(RT)n

n = 1 – 2 = -1

Kp = 220 x (0.08206 x 347)–1 = 7.7 atm–1

6.8 Heterogeneous Equilibrium system • Heterogenous equilibrium applies to reactions in which reactants and products are in different phases. • Kc and Kp can be used to represent the equilibrium constant. • The concentration of solids and pure liquids are not included in the expression for the equilibrium constant.

1. CH3COOH (aq) + H2O (l)

⇌

CH3COO- (aq) + H3O+ (aq)

[CH3COO-][H3O+] Kc = [CH3COOH][H2O] H2O a pure liquid, has a constant concentration and thus is not included in the expression. Therefore, the equilibrium constant is written as, [CH3COO-][H3O+] Kc = [CH3COOH] Unit : M

2.

CH3COOH (aq) + C2H5OH (aq)⇌ CH3COOC2H5 (aq) + H2O (l) Alcohol

Carboxylic

Ester

acid

Kc =

[CH3COOC2H5] [CH3COOH] [C2H5OH]

3.

CaCO3 (s)

⇌

CaO (s) + CO2 (g)

[CaO][CO2] K‘c = [CaCO3] [CaCO3] = constant [CaO] = constant Kc = [CO2] Kp = PCO2

CaCO3 (s)

⇌

CaO (s) + CO2 (g)

PCO2 = Kp PCO2 does not depend on the amount of CaCO3 or CaO

5.

Consider the following equilibrium at 295 K: NH4HS (s)

⇌

NH3 (g) + H2S (g)

The partial pressure of each gas is 0.265 atm. Calculate Kp and Kc for the reaction?

Answer ; Kp = PNH PH S = 0.265 atm x 0.265 atm 2 3 = 0.0702 atm2 Kp = Kc(RT)n

n = 2 – 0 = 2 T = 295 K

Kc =

0.0702 atm2 (0.08206 atm L mol–1 K–1 x 295 K )2

=

1.2 x 10–4 mol2 L–2

=

1.2 x 10–4 M2

6. The value of Kc for the following equation is 1.0 x 10–3 M at 200°C. 2NOBr (g) ⇌ 2NO (g) + Br2 (g) Calculate Kc’ for the following equation at 200°C. NO (g) + ½ Br2 (g) ⇌ NOBr (g)

Answer; Kc = [NO]2 [Br2]

Kc ’ =

[NOBr]2 1 = Kc

1 [NO]2 [Br2]

=

[NOBr] [Br2]1/2 [NO]

[NOBr]2 [NO]2 [Br2]

[NOBr]2 1 Kc

½

=

[NOBr] = Kc ’ [NO] [Br2] ½

½

Kc ’ =

1 Kc

Kc ’ =

1 1.0 x 10–3 M

= 31.6 M-½ = 31.6 dm3/2 mol–½

½

7. At 440 C, the equilibrium constant Kc for reaction, H2(g) + I2(g) ⇌ 2HI(g) has a value of 49.5. If 0.200 mole of H2 and 0.200 mole of I2 are placed into a 10.0 L vessel and permitted to react at this temperature, what will be the concentration of each substance at equilibrium?

Answer; [H2]0 = 0.200 /10.0 = 0.0200 M [I2]0 = 0.200 /10.0 = 0.0200 M H2 (g)

[ ]0 / M Δ[ ] / M [ ]⇌ / M

0.0200 -x 0.0200 - x

+ I2 (g)

⇌

2HI (g)

0.0200

0

-x 0.0200 - x

+2x +2x

Kc =

[HI]2 [H2] [I2]

49.5 =

7.036 =

(2x)2 (0.0200 – x) (0.0200 – x) 2x 0.0200 - x

x = 0.0156

H2 (g)

+

I2 (g)

⇌

2HI (g)

[ ]0 / M

0.0200

Δ[ ] / M

-x

[ ]⇌ / M

0.0200 - x

0.0200 - x

+2x

= 0.0044

= 0.0044

= 0.0312

0.0200 -x

0 +2x

Therefore concentration of each substance at equilibrium: [H2] = 0.0044 M [I2] = 0.0044 M [HI] = 0.0312 M

8. A 1.000-L flask is filled with 1.000 mol of H2 and 2.000 mol of I2 at 448°C. The value of the equilibrium constant, Kp for the reaction, H2 (g) + I2 (g) ⇌ 2HI (g) at 448°C is 50.5. What are the partial pressures of H2 , I2 and HI in the flask at equilibrium.

Answer; Initial pressures of H2 and I2. PH = nH RT = (1.000)(0.08206)(721) 2 2 V

1.000 = 59.19 atm

PI = nI2 RT 2

V

= (2.000)(0.08206)(721) 1.000 = 118.4 atm

H2 (g) P0 ΔP P⇌

Kp

2

= PHI PH2 P I2

59.19 -x 59.19 - x ⇒ 50.5 =

+ I2 (g)

⇌

2HI (g)

118.4

0

-x 118.4 - x

+2x +2x

(2x)2 (59.19 – x) (118.4 – x)

46.5x2 – 8969x + 353904 = 0 Partial pressure of gases at equilibrium: PH2 = 3.89 atm ; P = 63.1 atm ; PHI = 110.6 atm I2

6.2.6

Degree of Dissociation, α

The fraction of a molecule dissociated. ΔC α = C0

ΔC = changes in concentration C0 = initial concentration

A complete dissociation occurs if α = 1

AB

C0 / M ΔC / M C⇌ / M

⇌

A+

+

B–

a

0

0

-x

+x

+x

a-x

+x

+x

x α = a

9. The concentration of H+ ion measured for HCOOH 0.5 M is 0.0089 M. What is the degree of dissociation of the compound? x = 0.0089 a = 0.5 ⇌

H+ 0

C0 / M

HCOOH 0.5

+

ΔC / M

-0.0089

+0.0089

C⇌ / M

0.5-0.0089

0.0089

HCOO– 0

+0.0089 0.0089

= 0.4911

α =

x a

0.0089 = = 0.0178 = 1.78% 0.5

6.2.7 Predicting the direction of reaction 





For any reversible reaction, we can determine the direction of a reaction (whether is moving forward or reverse) by comparing the value of Q with Kp or Kc. Q is the reaction quotient Q has the same expression of Kc and Kp but the numerical value gained is NOT at equilibrium.

Predicting the direction of reaction: aA (g) + bB (g) Qc =

⇌

cC (g) + dD (g)

[C]c[D]d [A]a[B]b

If Q = K

The system is at equilibrium

If Q < K

Reaction is moving forward - more reactants are present in the mixture at t

If Q > K

Reaction is reversed - more products are present in the mixture at t

10. At 12800C the equilibrium constant (Kc) for the

reaction

Br2 (g)

⇌

2Br (g)

is 1.1 x 10-3. If the initial concentrations are [Br2] = 0.063 M and [Br] = 0.012 M, calculate the concentrations of these species at equilibrium.

Br2 (g) C0/M [Br]2 Qc = [Br2]

Qc > Kc

0.063 =

⇌

2Br (g) 0.012

(0.012)2 = 2.29 x 10–3 0.063

Equilibrium position shifts from right to the left.

Br2 (g)

C0/M

0.063

ΔC/M

+x

C⇌/M

⇌

0.063 + x

2Br (g)

0.012 - 2x

0.012 - 2x

[Br]2 Kc = [Br2] 1.1x10-3

(0.012 – 2x)2 = 0.063 + x

4x2 – 0.048x + 1.44x10-4 = 6.93x10-5 + 1.1x10-3x 4x2 – 0.0491x + 7.47x10-5 = 0 x = 0.0105

C⇌

x = 1.78x10-3

[Br] = 0.012 – 2x

[Br2] = 0.063 + x

If x = 0.0105

= -0.009 M

= 0.0735 M

If x = 1.78x10-3

= 0.00844 M

= 0.0648 M

11.

The equilibrium constant Kc for the reaction I2 (g) + H2(g) ⇌ 2HI (g) is 54.0 at certain temperature. At equilibrium, the conc. of I2, H2 and HI are 0.200 M, 0.200 M and 1.47 M respectively. If 0.500M of HI is added, what are the conc. when equilibrium is re-established?

Qc = =

[HI]2 [H2] [I2] (1.47 + 0.5)2 (0.200) (0.200)

= 97.0

Since Qc > Kc, the system is not at equilibrium The net reaction will proceed from RIGHT to LEFT until equilibrium is reestablished.

H2 (g)

+ I2 (g)

⇌

2HI (g)

C0/M

0.200

0.200

1.47 + 0.500

ΔC/M C⇌/M

+x 0.200 + x

+x 0.200 + x

-2x 1.97 – 2x

= 0.2535

= 0.2535

= 1.863

Kc = 54.0 =

Kc = 54 Qc = 97

[HI]2 [H2] [I2] (1.97 – 2x)2 (0.200 + x) (0.200 + x)

7.35 =

1.97 – 2x 0.200 + x

x = 0.0535M

6.3 Factors affecting the equilibrium position The position of equilibrium in a system is affected by: a) concentration b) pressure c) temperature The effects can be predicted & explained by the : Le Chatelier’s Principal

6.3.1

Le Châtelier’s Principle

If an external stress is applied to a

system at equilibrium, the system adjusts in such a way that the stress is partially offset as the system reaches a new equilibrium position.

A) Effect of Concentration: • Changing the concentration of the reactants or products causes the equilibrium to shift to the direction that will reduce the effect . Example: • Increasing the concentration of reactant/s causes the equilibrium position to shift forward in order to reduce the disturbance (to eliminate the additional concentration of reactant/s) • more products will be formed until the equilibrium is reached again (i.e: composition of equilibrium reaches Kc value which remains the same at a fixed temperature).

Example ; N2 (g) + 3H2 (g) ⇌

2NH3 (g)

Change

Equilibrium Shift

Increase concentration of product(s)

To the left

Decrease concentration of product(s)

To the right

Increase concentration of reactant(s)

To the right

Decrease concentration of reactant(s)

To the left

Example : Consider the reaction below: Fe 3+(aq) (Yellowish)

+ SCN-(aq)

Fe(SCN)2+(aq) (Blood red)

Discuss the effect of adding the following substances to the equilibrium mixture. a) Iron (III) or Fe+3 b) Sodium hydroxide, NaOH

Answer; Consider the reaction below: Fe 3+(aq) + SCN-(aq) (Yellowish)


a) When Fe3+ is added to the equilibrium, the concentration of Fe3+ in the system increases. • the position of equilibrium shifts to right in order to eliminate the effect of the increasing Fe3+ ions. • more products will be formed and the solution turns blood red.

b) When NaOH is added to the system:

Consider the reaction below: Fe 3+(aq) + SCN-(aq) (Yellowish)


• It reacts with Fe3+ to form Fe(OH)3; thus the concentration of Fe3+ decreases. • In order to achieve equilibrium, the position of equilibrium moves to the left to replace the Fe3+ that has reacted. • More Fe3+ ions will be formed and the solution becomes yellowish brown.

B) Effect of Temperature: The effect of temperature on an equilibrium depends on the type of reactions, whether it is: i) an endothermic reaction or ii) an exothermic reaction Temperature is the only factor that affects the value of Kc

i) Endothermic reaction • An endothermic reaction is a reaction that absorbs heat to form products. Example: • Consider the following system: N2O4(g)

2NO2(g)

ΔH = +57 kJ

– ‘+’ sign indicates that for the forward reaction absorbs heat – If forward reaction absorbs heat, then the reverse reaction releases heat.

N2O4(g)

2NO2(g)

ΔH = +57 kJ

Increasing the temperature: • Means that heat is added to the system • Reaction will move forward (forward reaction absorbs heat) so as to reduce the added heat. • More NO2 gas will be released. • Since the amount of product increases, value of Kc increases

N2O4(g)

2NO2(g)

ΔH = +57 kJ

Lowering the temperature: • Heat is removed from the system • The position of equilibrium will move from right to left to replace the heat lost. (since reverse reaction is exothermic) • More N2O4 will be formed. • Since more product is formed, value of Kc decreases [ NO2 ]2 Kc  [ N 2O4 ]

Concentration of N2O4 increases, thus Kc value decreases

Example: Consider the reaction: Co(H2O)62+(aq)+ 4HCl(aq) pink To overcome the low temperature reaction should release heat

Disturbance: Low temperature

CoCl42+(aq) + H2O ΔH = +ve blue

System at equilibrium

High temperature

C) Effect of Pressure The pressure of a system may be changed by: i) changing the moles of reactant or product ii) changing the volume iii) adding inert gas to the system Affects only the reversible reaction involving gaseous reactants/products

i) Changing moles of reactant or product Adding or removing the gaseous reactant or product at constant volume has the same effect as changing the concentration Example: Consider the following system at equlibrium 2SO2(g) + O2(g)

2SO3(g)

When SO2 gas is added to the system, the partial pressure of SO2 increases (which means [SO2] increases) •

• Equilibrium shifts to the right to reduce the effect of high concentration of SO2.

ii. Changing the volume of the container: Decreasing the volume: • Causes the pressure of the system to increase • The position of equilibrium will shift in a direction that will lower down the pressure • The pressure can be brought down by having reaction that produces less number of molecules

Example

N2(g) + 3H2(g)

2NH3(g)

Consider the above system at equilibrium:  When the Vol. of the container decreases, the P of the system increases.  position of equilibrium will shift in the direction that brings the pressure down by reducing the number of moles present in the system  The equilibrium shifts to the right to produce less number of moles of gases in the system  More ammonia is formed  Value of Kc remains the same

ii. Changing the volume of the container Increasing the volume: • Causes the pressure in the system to decrease • The equilibrium position will shift in the direction that will raise the pressure up. • It moves to the direction that produces more number of molecules (higher number of moles)

Example Consider the reaction: 2SO2(g) + O2(g)

2SO3(g)

• when the volume of the container increases, the pressure of the system decreases • Equilibrium position will shift to the direction that produces more moles • The equilibrium shifts to the left.

iii) Changing pressure by adding inert gas a) Adding inert gas at constant volume: • Adding inert gas causes the total pressure in the container to increase. • However the partial pressure of each gas reacting in the reversible reaction remains the same. • Since the partial pressure of the gases remain the same & Kp is constant at constant temperature, the equilibrium position does not change.

b) Adding inert gas at constant pressure: • Adding inert gas to the system increases the total pressure. • In order to keep pressure constant, the volume of the container has to expand. • This causes the partial pressures of the gases to drop. • System will move in the direction that produces higher number of moles

Example: For the following chemical equation, N2 (g) + O2 (g) ⇌ 2NO (g)

The value of Kc for the reaction is 1x10–3 at high temperature and 4.8x10–31 at 25°C. Is the forward reaction exothermic or endothermic? Kc =

[NO]2 [N2] [O2]

T↑

Kc↑

[NO]↑

Equilibrium position shifts from LEFT to the RIGHT

Forward reaction is ENDOTHERMIC.

D)

The effect of Catalyst

Catalysts are substances that increase the rate of a chemical reaction but are not used up in the reaction. The rate of forward reaction and reverse reaction increase. A catalysts allows a system to reach equilibrium faster but does not shift the position of an equilibrium system.

Le Châtelier’s Principle Change

Equilibrium shift

Change Equilibrium Constant, K

Concentration

yes

no

Pressure

yes

no

Volume

yes

no

Temperature

yes

yes

Catalyst

no

no

Example : Consider the chemical reaction at equilibrium given below CaCO3 (s) ⇌ CaO (s) + CO2 (g) ΔH =+ve (a)

(b)

Calculate the values of Kp and Kc for the system at 525oC, with the equilibrium pressure of CO2 is 0.220 atm. Predict the equilibrium position when the following changes are made to the system (i) some of the CaO is removed from the system (ii) the pressure of the system is increased (iii) the temperature of the system is raised (iv) Helium gas is added at constant volume (v) Helium gas is added at constant pressure

Kp = PCO

2

= 0.220 atm

Kp = Kc(RT)Δn

Δn = 1 – 0 = 1

Kp= Kc(RT) Kc =

Kp (RT)

=

0.220 atm 0.08206 atm L mol –1 K –1 x 798 K

= 3.36 x 10–3 M

(b)(i)

(ii)

Does not change (the partial pressure of CO2 is not affected by the removal of SOLID CaO from the system). Shifts to the LEFT. (P↑, ; to the side with less number of moles of gas).

(iii)

Shifts to the RIGHT. (Endothermic, T↑, shifts to the right to reduce heat).

(iv)

Does not change (Addition of inert gas at constant V does not change the partial pressure of gas).

(v)

Shifts to the RIGHT. (Addition of inert gas at constant P, V increase, therefore equilibrium position shifts to the side with greater number of moles of gas).

HABER PROCESS A process that produces ammonia from H2 and N2 The process is introduced by German scientists Fritz Haber and Karl Bosch. (sometimes called Haber-Bosch) in 1912. Main objectives of the industry are to obtain a high yield of the product while keeping the cost down

Consider N2(g) + 3H(g)

2NH3(g)

ΔH = -92.6 kJ

A higher yield of NH3 can be obtained by carrying out reaction under high pressures and at the lowest temperature possible. However the rate of production is low at a lower temperature. Commercially, faster production is preferable. So combination of high-pressure, high temperature and proper catalyst is the most efficient way to produce ammonia on a large scale. Temperature used = 500oC Pressure = 500 – 1000 atm Catalyst = Osmium and Ruthenium

7.0 IONIC EQUILIBRIA

1

7.1

ACIDS AND BASES

Theory of Acids and Bases Three important theories 1. Arrhenius theory 2. Bronsted-Lowry’s theory 3. Lewis theory

2

Theory of Acids and Bases 1. Arrhenius Theory Acid: any substance that produces hydrogen ion (H+) or hydronium ion (H3O+) when dissociating in water. Example: HCl (aq)

 H+ (aq) + Cl (aq)

HCl (aq) + H2O (l)  H3O+(aq) + Cl (aq)

3

Base: any substance that produces hydroxide ions (OH-) when dissociating in water. Example: NaOH (aq)

 Na+ (aq) + OH (aq)

4

2. Bronsted-Lowry Theory Acid: any substance that can donate a proton, (H+) to other substance Example: HNO3 (aq) + H2O (l)  NO3 (aq) + H3O+ (aq) acid

NH4+ (aq) + H2O (l)

NH3 (aq) + H3O+ (aq)

acid

5

Base: any substance that can accept a proton from other substance Example: NH3 (aq) + H2O (l)

NH4+ (aq) + OH (aq)

base

CO32- (aq) + H2O (l)

HCO3- (aq) + OH- (aq)

base

H2O is able to act as an acid or a base : AMPHOTERIC

6

Conjugate Acid-base Pairs 

Conjugate base ◦ A species that remains when one proton has been removed from the Bronsted acid.



Conjugate acid ◦ A species that remains when one proton has been added to the Bronsted base.

7

Example HCO3 (aq) + H2O (l) acid



base

CO32 (aq) + H3O+(aq) conjugate base

conjugate acid

The stronger the acid, the weaker is its conjugate base and the stronger the base, the weaker is its conjugate acid.

8

Exercise 1. In the following reactions, identify the acid-base conjugate pairs. a. ClO (aq) + H2O (l) base

HOCl (aq) +

acid

b. CH3NH2 (aq) + H2PO4- (aq) base

acid

OH (aq)

conjugate acid

conjugate base

CH3NH3+(aq) + HPO42 (aq) conjugate acid

conjugate base

9

NH4+(aq) + Br (aq)

c. HBr(aq) + NH3(aq) acid

conjugate acid

base

d. PO43(aq) + H2O(l) base

HPO42(aq) + OH(aq) conjugate acid

acid

e. CH3COOH(aq) + H2O(l) acid

conjugate base

base

conjugate base

CH3COO(aq) + H3O+(aq) conjugate base

conjugate acid

10

2. Write the conjugate base formula for the following acids. a) HS d) H3PO4 b) HCN e) H3O+ c) N2H5+ f) CH3NH3+ Answer : acid

conjugate base

a)

HS

S2-

+

H+

a)

HCN

CN-

+

H+

a)

N2H5+

N2H4

+

H+

a)

H3PO4

H2PO4-

+

H+

a)

H3O+

H2O

+

H+

a)

CH3NH3+

CH3NH2

+

H+

11

3. Write the conjugate acid formula for the following bases. a) HSd) NH3 b) C6H5COO e) H2O c) OH f) H2PO4

12

3. Lewis Theory Acid: a substance that can accept a pair of electrons to form a covalent bond. The species that can be Lewis acid are  Cations such as H+, Fe2+, Al3+  molecules with incomplete octet central atom such as BF3, BeCl2  Molecules with central atom that can expand octet such as PCl3

13

Base: a substance that can donate a pair of electrons to form a covalent bond. The species that can be Lewis base are  anions such as OH-, CN-, Cl molecules with lone pairs electrons at the central atom such as H2O, NH3, ROH  molecules with nonpolar multiple bond such as alkenes

14

Example 1) H3N:

+

Lewis base

2) H+ Lewis acid

BF3

H3N BF3



H2O

Lewis acid

+

OHLewis base

3) 2Cl-(aq) + SnCl4 (aq) Lewis base





SnCl62-(aq)

Lewis acid

15

Exercise Classify each of the following species as Lewis acid or Lewis base. Lewis base

i. NH3 ii.

AlCl3

iii.

BCl3

Lewis acid Lewis acid

iv.

Ag+

v.

I-

Lewis acid Lewis base 16

Strength of Acids and Bases 

Strength of acids and bases depend on the extent of dissociation

% dissociation =

[ ]dissociated  100% [ ]initial

Degree of dissociation,  =

[ ]dissociate d [ ]initial

17

Strong Acid 

Strong acid is an acid that ionizes completely in water.



3 ways to write the equation for acid dissociation: a) HCl (aq) a) HCl (g)



H+ (aq) + Cl (aq)

2 O (l) H 

H+ (aq) + Cl (aq)

a) HCl (g) + H2O (l)  H3O+ (aq) + Cl (aq)

18



Example : for 0.10 M HCl HCl (aq) + H2O (l)  Cl (aq) + H3O+ (aq) [ ]i 0.10

0

0

Δ - 0.10

+ 0.10

+ 0.10

[ ]f 0.00

0.10

0.10

19

[HCl]dissociated % dissociation = [HCl]initial

 100%

0.10=  100% 0.10 = 100 %

[HCl]dissociated Degree of dissociation,  = [HCl]initial 0.10 = 0.10

= 1.00

20



The strong acids are ... ◦ The hydrohalic acids HCl, HBr and HI ◦ Oxoacids in which the number of O atoms exceeds the number of ionizable protons by two or more e.g: HNO3 and HClO4

21

Strong Base 

Strong base is a base that ionizes completely in water.

Example : for 0.50 M NaOH NaOH (aq)

H2O

  Na+ (aq) + OH (aq)

[ ]i

0.50

0

0

Δ

- 0.50

+ 0.50

+ 0.50

[ ]f

0.00

0.50

0.50

22

[ NaOH ]dissociated % dissociation = [ NaOH ]initial

 100%

0.50= 100% 0.50 = 100 % Degree of dissociation,  =

[ NaOH ]dissociated [ NaOH ]initial

0 .50= 0 .50 = 1.00

23



The strong bases... ◦ M2O or MOH, where M is group 1 element : Li, Na, K, Rb, Cs ◦ RO or R(OH)2, where R is Ca, Sr, Ba

24

Weak Acid 

Weak acid is an acid that only ionizes partially in water Example: CH3COOH, HCOOH



Generally for any weak acid, HA the dissociation reaction is: HA (aq) + H2O (l)



% dissociation < 100%



<1

A (aq) + H3O+(aq)

25



Acid dissociation constant expression, Ka

Ka =   

[A  ][H3O  ] [HA]

Ka value depends on temperature Normally Ka is measured at 25 C Value of Ka ↓ ;  ↓ strength of acid ↓

26



The weak acids are .... ◦ The hydrohalic acids HF ◦ Acids in which H is not bonded to O or to halogen e.g: HCN and H2S ◦ Oxoacids in which the number of O atoms equals or exceeds by one the number of ionizable protons e.g: HNO2 and HClO ◦ Organic acids (general formula RCOOH) e.g: CH3COOH

27

Weak Base 

Weak base is an base that only ionizes partially in water Example: NH3, N2H4



Generally for any weak base, B the dissociation reaction is: B (aq) + H2O (l)



% dissociation < 100%



<1

BH+ (aq) + OH- (aq)

28



Base dissociation constant expression, Kb

Kb =   

[BH ][OH ] [B]

Kb value depends on temperature Normally Kb is measured at 25 C Value of Kb ↓ ;  ↓ strength of base ↓

29



The weak bases are ... ◦ Ammonia (NH3) ◦ Amines with general formula RNH2, R2NH, R3N e.g. : CH3CH2NH2; (CH3)2NH; (CH3)3N

30

Example: At 25 C, 4.2% of 0.10 M formic acid, HCOOH dissociated in aqueous solution. Calculate the acid dissociation constant, Ka for formic acid.

31

Solution:

[4.2] [HCOOH] dissociated   0.10M [100]  0.0042 M HCOOH + H2O

H3O+ + HCOO-

[ ]i

0.10

0

0

Δ

-0.0042

+0.0042

+0.0042

[ ]

0.0958

0.0042

0.0042

32

[HCOO - ][H 3O  ] Ka  [HCOOH] [0.0042M]2  [0.0958M ]  1.8  10 -4 M Ka for HCOOH is 1.8 x 10-4 M 

Ka or Kb for many weak acids or weak bases < 10,  ↓↓,



To simplify the calculation for ionic equilibrium the following assumption can be made

[ ]initial  [ ]dissociated  [ ]initial 33



The assumption will give correct answer to at least 2 significant figures if:  value of Ka or Kb < 10 or  % dissociation < 5% Ka =

[H 3O  ] [HCOO  ] [HCOOH]

[ ]initial  [ ]dissociated  [ ]initial 2 (0.0042 M) = 0.1M = 1.8  104 M

34

Try this… 1.

Acid dissociation constant, Ka for hydrofluoric acid HF at 25 C is 6.8104 M. For a solution of 0.20 M HF, calculate: a. the concentration of hydronium ion at equilibrium b. degree of dissociation

2.

Base dissociation constant, Kb for ammonia solution, NH3 at 25 C is 1.8105 M. For a solution of 0.50 M NH3, calculate: a. the concentration of hydroxide ion at equilibrium b. % dissociation 35

Water and the pH Scale a.

The Water Ionization Constant, Kw ◦ When water molecules ionized, transferring a proton from one water molecule to another producing a hydroxonium and hydroxide ion. 2H2O (l)

H3O+ (aq) + OH (aq)

36

2H2O (l)

H3O+ (aq) + OH (aq)

The equilibrium constant expression can be written as follows: Kw = H3O+ OH- Kw is the ionization constant for water at 25C

37

For pure water, at 25C, H3O+ = OH = 1 x 10-7 M Kw

 Kw

= H3O+ OH = (1 x 10-7) (1 x 10-7) =

1 x 10-14 M2

Note: Temperature ↑, Kw ↑ because the dissociation of water is an endothermic process

38

The pH Scale

a.

◦

The pH scale is used to express acidity.

◦

The pH of a solution is defined as the negative logarithm (log) of the hydroxonium ion (hydrogen ion) concentration.

pH =  log H3O+

39



In similar way, pOH is the negative logarithm of the hydroxide ion concentration. pOH =  log OH

40

In pure water, Kw = H3O+ OH = 1 x 10-14

M2

H3O+ = OH = 1 x 10-7 M pH

= pOH =  log (1 x

10-7 ) pH

= pOH = 7 pH + pOH = 14 41

At 25 oC, ◦ Acidic solution : pH < 7.0 ; pOH > 7.0 ◦ Basic solution : pH > 7.0 ; pOH < 7.0 ◦ Neutral solution : pH = 7.0 ; pOH = 7.0

42

Example 1 In a NaOH solution, [OH-] is 2.9 x 10-4 M. Calculate the pH of the solution at 25 oC. (10.46)

43

Example 2 The pH of rainwater in a certain region was 5.68. Calculate the H+ ion concentration of the rainwater. (2.089 x 10-6)

44

Example 3 Calculate the pH of a 0.15 M acetic acid (CH3COOH) solution, Ka = 1.8 x 10-5. (2.79)

45

Example 4 What is the pH of a 0.25 M ammonia solution? Kb = 1.8 x 10-5. (11.33)

46

Example 5 The pH of 0.06 M solution of formic acid (HCOOH) is 3.44. Calculate the Ka of the acid. (2.211 x 10-6)

47

Exercises 1.

The Ka for benzoic acid, (C6H5COOH) is 6.5x10-5. Calculate the pH of a 0.25 M benzoic acid solution. (2.39)

1.

The pH of an acid solution is 6.20. Calculate the Ka for the acid. The acid concentration is 0.01M. (3.98 x 10-11)

3. Calculate the pH for 0.5M C5H5N. Kb = 1.7x10-9 (9.47)

48

1.

Calculate pH for the following solution.

a. b. c. d. e.

Answer 0.55 M CH3COOH (Ka = 1.8x10-5) 2.5 0.23 M NH3 (Kb = 1.8x10-5) 11.31 0.15 M HCl 0.82 0.20 M KOH 13.3 0.45 M HCN (Ka = 4.9 x 10-10) 4.8

49

SALTS 



Produced when acid reacts with base.

3 types of salts are : 1) neutral salts 2) basic salts 3) acidic salts

50

1) Neutral salts 

Produced when strong acid reacts with strong base. e.g HCl + NaOH NaCl + H2O neutral salt - Na+ comes from strong base - Na+ does not react with water (does not hydrolyzed) - Cl- comes from strong acid - Cl- does not react with water (does not hydrolyzed)

51



So, pH of a solution depends on the ionization of water H2O (l)

H+ (aq) + OH- (aq)

Kw = H3O+ OH = 1 x 10-14 M2 [H+ ] = [ OH- ] = 1.0 x 10-7 M pH = 7

52

SALT HYDROLYSIS 

Salt hydrolysis is a chemical reaction between anion or cation of a salt with water molecules.



Acidic salts and basic salts have ions that undergo hydrolysis in aqueous solution.



The pH value of a solution depends on whether OH- or H3O+ is produced during hydrolysis.

53

2) BASIC SALTS 

Normally is produced from the reaction of weak acid and strong base.



produce OH- when hydrolyzed



eg :

CH3COONa(aq)

Na+ (aq) + CH3COO- (aq)

Na+

: comes from strong base (does not hydrolyzed)

CH3COO-

: comes from weak acid and undergoes hydrolysis

54

Hydrolysis of CH3COO- : CH3COO-(aq) + H2O (l)

CH3COOH(aq) + OH-(aq)



The solution is basic because the hydrolysis of CH3COO- produces OH-



pH of the basic salt solution is > 7.0

55

Example Sodium cyanide, NaCN is a salt formed when a strong base, NaOH is reacted with a weak acid, HCN. a)

Write a balanced equation to show the reaction between NaOH and HCN. Classify the salt formed.

b) What would be expected pH of the NaCN solution? Explain the answer using equation(s).

56

3) ACIDIC SALTS 

Normally is produced from the reaction of strong acid and weak base.



produce H3O+ when hydrolysed.



Eg. NH4Cl(s) Cl-



NH4+(aq) + Cl(aq)

: comes from a strong acid (does not hydrolyzed)

NH4+ : comes from weak base and undergoes hydrolysis

57

Hydrolysis of NH4+ : NH4+(aq) + H2O(l)

NH3(aq) + H3O+(aq)



The solution is acidic because hydrolysis of NH4+ produces H3O+



pH of the acidic salt solution is < 7.0

58

Example CH3NH3Cl is a salt formed when a weak base,CH3NH2 is reacted with a strong acid, HCl. a)

Write a balanced equation to show the reaction between CH3NH2 and HCl. Classify the salt formed.

b)

What would be expected pH of the CH3NH3Cl solution ? Explain the answer using equation(s).

59

BUFFER SOLUTION A solution that maintains its pH when a small amount of a strong acid or a strong base is added to it. It contains weak acid or weak base with salt that has its conjugate pair. 2 types of buffer solution: a. acidic buffer solution b. basic buffer solution

60

ACIDIC BUFFER SOLUTION 

Has pH < 7.



An acidic buffer solution is made up of a weak acid and its salt (containing its conjugate base)

Example: A solution of CH3COOH and CH3COONa

61



The dissociation reactions are: CH3COOH (aq)

H+(aq) + CH3COO-(aq)

CH3COONa(aq)

CH3COO-(aq) + Na+(aq)



CH3COONa dissociates completely and produces high concentration of CH3COO-.



The high concentration of CH3COO- will disturb the equilibrium of the dissociation of ethanoic acid, CH3COOH.

62



Equilibrium of acid shifts backward, less CH3COOH dissociates.



Solution now has high concentrations of CH3COOH (from weak base) and its conjugate base ion CH3COO- (from salt)

63

BUFFER’S ACTION 

Since buffer solution contains CH3COOH that acts as acid and the conjugate ion CH3COO- that acts as base, buffer solution will maintain its pH by performing reactions as follows:

(i) Adding a small amount of acid to the solution: ◦ CH3COO- (conjugate base) will neutralize it. CH3COO-(aq) + H3O+ (aq)

CH3COOH(aq) + H2O(l)

◦ The amount of weak acid, CH3COOH increased a little but since the dissociation of acid is small, the pH of the solution is not much affected. 64

(ii) Adding a small amount of base to the solution : ◦ the acid, CH3COOH will neutralize it. CH3COOH(aq) + OH-(aq)

CH3COO-(aq) + H2O(l)

◦ The amount of weak acid, CH3COOH decreased a little but addition amount CH3COOwill cause the equilibrium position to move to the left and replace the amount of acid used ◦ So,the pH of the solution is not much affected.

65

CALCULATION OF pH OF BUFFER SOLUTION 

The pH is obtained by referring to the equilibrium dissociation of a weak acid, HA.



Consider buffer solution containing HA and its conjugate, AHA

Comes from weak acids

H+

+

A-

Comes from the ionisation of a salt

66

We can write the acidic concentration constant, Ka

[ H  ][ A  ]  [ HA ]

or

[H+]

Ka[HA] = [A  ]

By applying –log on both sides, we have -log

[H+]

pH

[ HA] = -log Ka + ( - log [ A ]

)

 [ A ] = pKa + log [HA]

Henderson-Hasselbalch equation 67

Example a.

Calculate the pH of a 1.00 L solution containing 0.30 M CH3OOH and 0.1 M CH3COONa. (Ka CH3COOH = 1.8 x 10-5)

b.

What is the pH when 10 mL of 0.01 M HCl is added to the buffer solution in (a)?

c.

What is the pH when 1 mL of 0.1 M NaOH is added to the buffer solution in (a)?

68

BASIC BUFFER SOLUTION 

Has pH > 7



Basic buffer solution is made up of a weak base and its salt (containing its conjugate)



Example: A basic buffer solution of NH3 and NH4Cl



The dissociation reactions are: NH3(aq) + H2O(l) NH4Cl(aq)

NH4+(aq) + OH-(aq) NH4+(aq) + Cl-(aq) 69



NH4Cl dissociates completely and produce high concentration of NH4+ ions



The high concentrations of NH4+ disturb the equilibrium of the dissociation of NH3.



Equilibrium of base shift backwards, less NH3 dissociate.



Solution now has high concentrations of NH3 and its conjugate acid ion NH4+, originated from salt.

70

BUFFER’S ACTION 

Buffer solution maintains its pH by performing reactions as follow :



NH3 acts as base and the conjugate ion, NH4+ acts as an acid.

71

(i)

Adding a small amount of acid to the solution ◦ NH3 (base) will neutralize it.

NH3(aq) + H3O+(aq)

NH4+(aq) + H2O

◦ The amount of NH4+ will increase a little but this will cause the equilibrium to shift to the left and replace the NH3 used, the pH of solution is not much affected

72

(ii) Adding a small amount of base to the solution ◦ NH4+ (conjugate acid) will neutralize it. NH4+(aq) + OH-(aq)

NH3 (aq) + H2O(l)

◦ The amount of NH3 will increase a little but since the dissociation of NH3 is small, the pH of solution is not much affected

73

CALCULATION OF pH OF BUFFER SOLUTION



The pOH and pH can be calculated by using the Henderson-Hasselbalch equation.



Consider the following base dissociation reaction: B + H2O

BH+ + OH-

74



The base dissociation constant, Kb [BH ][OH ] Kb = [B]

[OH-] = Kb

or

[ B] [ BH  ]

By applying – log on both sides: -log [OH] = -log Kb + ( -log

pOH

[ B] )  [ BH ]

 [ BH ] = pKb + log [B ]

Henderson-Hasselbalch equation

75

Example A buffer solution is prepared by mixing 400mL 1.5 M NH4Cl solution with 600 mL of 0.10 M NH3 solution. [ Kb = 1.8 x 10-5 ] a) Calculate the pH of buffer b) Calculate the new pH of the buffer after the addition of 2.0 mL of 0.10 M HCl c) Calculate the new pH of the buffer after the addition of 2.0 mL of 0.15 M NaOH

76

Exercises a) Calculate the pH of a 1 L buffer solution containing 1.5 M CH3COOH and 0.20 M of CH3COONa. pH = 3.87 b) Calculate the new pH of the buffer after the addition of 2.0 mL of 0.1 M HCl. PH = 3.87 a) Calculate the new pH of the buffer after the addition of 2.0 mL of 0.2 M NaOH. pH = 3.87 [ Ka = 1.8 x 10-5 ]

77

a) Calculate the pH of the solution prepared by mixing 500 mL 0.1 M hydrazinium chloride, N2H5Cl with 500 mL 0.2 M hydrazine, N2H4.

pH = 8.53

a) Calculate the new pH of the buffer after the addition of 2.0 mL of 0.1 M HCl. pH = 8.53 c) Calculate the new pH of the buffer after the addition of 2.0 mL of 0.2 M NaOH pH = 8.54 [ Kb N2H4 = 1.7 x 10-6 ]

78

7.2 ACID-BASE TITRATIONS Titration



◦

a method for determining the concentration of a solution using another solution (known concentration), called standard solution Titration Curve



◦

a graph of pH versus volume of titrant

79

Titration apparatus

Titrant

Analyte 80



The equivalence point ◦ the point at which the number of moles of OH ions added to a solution is equal to the number of moles of H+ ions originally present.



The end point ◦ is the point when the indicator changes colour.



Indicators ◦ is a weak organic acids or bases that change colour over a range of pH values. 81

Table 1: Some Common Acid-Base Indicators Colour Indicator

In Acid

In Base

pH Range *

Red

Yellow

1.2-2.8

Yellow

Bluish Purple

3.0-4.6

Orange

Yellow

3.1-4.4

Red

Yellow

4.2-6.3

Chlorophenol blue

Yellow

Red

4.8-6.4

Bromothymol blue

Yellow

Blue

6.0-7.6

Cresol Red

Yellow

Red

7.2-8.8

Colourless

Reddish pink

8.3-10.0

Thymol blue Bromophenol blue Methyl orange Methyl red

Phenolphthalein

82

TYPES OF TITRATIONS 1.

Strong acid - strong base titrations

2.

Strong acid – weak base titrations

3.

Weak acid – strong base titrations

83

Strong acid-strong base titration curves

* The pH starts

out

low, reflecting the high [H3O+] of the strong acid, and increases gradually as acid is being neutralised by the added base.

84

* The pH rises sharply when the mole of OHthat have been added nearly equal the mole of H3O+ .

85

* An additional drop or two of base neutralises the final tiny excess of acid and introduces a tiny excess of base, so pH jumps 6 to 8 unit.

* Beyond this sharp portion, the pH increases slowly again as more base is added.

86

Weak acid-strong base titration curves * The initial pH is higher than strong acid-strong base titration curve because the weak acid dissociates only slightly, less H3O+ is present.

87

* A gradually rising portion of the curve, called the buffer region, appears before the sharp rise to the equivalence point.

88

* The pH at the equivalence point > 7.0. The solution contains the basic salt.

89

Weak base-strong acid titration curves * The pH starts above 7.0 (~11) because the weak base dissociates only slightly.

* The pH decreases gradually in the buffer region.

90

* After the buffer region, the curve drops vertically to the equivalence point.

* Beyond the equivalence point, the pH decreases slowly as excess H3O+ is added.

91

How to sketch a titration curve Steps : 1. Calculate the initial pH of the solution - identify the analyte (solution in the conical flask). Whether strong acid, weak acid, strong base or weak base.

2. Determine the equivalence point : - Volume and pH

92

3. pH jump (steep portion / sharp portion) - depends on the type of the titration Type

pH jump

Strong acid-strong base

3-11

Strong acid-weak base

3-7

Weak acid-strong base

7-11

4. Identify the final pH - depends on the [ ] of the titrant (solution in the burette). 93

Example 1 : Sketch the titration curve of 25 mL 0.1 M HCl with 0.1 M NaOH. Step 1 : Analyte is a strong acid, HCl – dissociates completely HCl(aq) + H2O(l)  Cl (aq) + H3O+(aq) [H3O+] = [HCl] = 0.1 M pH = - log [H3O+] = 1.0 94

Step 2 : at equivalence point - pH equivalence: - the solution is NaCl (aq) - both Na+ and Cl- do not hydrolyse - therefore pH = 7.0 - Volume equivalence: n HCl MHCl VHCl

= n NaOH = MNaOH VnaOH

VNaOH = 0.1 x 25 0.1

= 25 mL

95

Step 3 : Type of titration : strong acid-strong base pH jump : 3 – 11

Step 4 : Titrant is a strong base, NaOH. NaOH (aq) 

Na+ (aq) + OH (aq)

[OH-] = [NaOH] = 0.1 M pOH = - log 0.1 = 1 pH = 14 – 1 = 13 Final point approaching pH < 13. 96

Example 2 : Sketch the titration curve of 25 mL 0.2 M HCl and 0.1 M NH3. Step 1 : Analyte is a strong acid, HCl HCl (aq) + H2O (l)  Cl (aq) + H3O+ (aq) [H3O+] = [HCl] = 0.2 pH = - log [H3O+] = 0.70

97

Step 2: at equivalence point HCl(aq) + NH3 (aq)

NH4Cl (aq)

- pH equivalence - the solution is NH4Cl (aq) - only NH4+ hydrolyses to form H3O+ - therefore pH < 7.0 - Volume equivalence n NH = n HCl MNH VNH = MHClVHCl VNH = 0.2 x 25 = 50 mL 0.1 3

3

3

3

Step 3 : Type of titration : strong acid-weak base pH jump : 3 – 7

Step 4 : Titrant is a weak base, NH3. Final point approaching pH < 11

99

How to determine the suitable indicator for titrations 

Choose an indicator which the endpoint pH range lies on the steep portion of the titration curve.



This choice ensures that the pH at the equivalent point will fall within the range over which indicator changes color. Types of Titrations

pH jump

Suitable Indicators

Strong Acid-Strong Base Weak Acid-Strong Base Strong Acid-weak Base

3 – 10

Any Indicator except thymol blue

7 – 11

Phenolphthalein, cresol red

3 – 7

Methyl orange, methyl red, chlorophenol blue, bromophenol blue

Exercise 1 What is the colour of the solution when 3 drops of the below indicators are added separately to water (pH = 7) ?

Indicator

pH range

Phenolphthalein

8.2 – 10.0

Colour change Colourless

reddish pink

Methyl orange

3.2 – 4.2

Red

Yellow

Bromothymol blue

6.0 – 7.6

Yellow

Blue

phenol red

6.8 – 8.4

Yellow

Red

Exercise 2 The pH range and the colour change for 3 indicator X, Y and Z is shown in the table below. Determine X, Y and Z. Indicator

pH range

colour change

X

1.2 – 2.8

Red

Y

6.0 – 7.6

Yellow

Z

8.3 – 10.5

Colourless

Yellow Blue Yellow

SOLUBILITY EQUILIBRIUM 

Some salts are soluble but most are insoluble or slightly soluble in water.



A saturated solution is a solution that contains the maximum amount of solute that can dissolve in a solvent.

103



The solubility of a salt is the amount of solid that dissolved in a known value of saturated solution.



The unit of solubility used may be g L-1 or mol L-1



Molar solubility is the maximum number of moles of solute that dissolves in a certain quantity of solvent at a specific temperature.

104

THE SOLUBILITY PRODUCT CONSTANT, Ksp 

Ksp is the product of the molar concentrations of the ions involved in the equilibrium, each raised to the power of its stoichiometric coefficient in the equilibrium equation.



Ksp is called the solubility product constant.



The degree of solubility of a salt is shown by the value of Ksp.

105



Consider the equilibrium system below : MX (s)

M+ (aq) + X- (aq) Kc = [M+] [X-] [MX] Kc [MX] = [M+] [X-] *since [MX] is a constant ; Ksp = [M+] [X-]

106



Soluble salt such NaCl and KNO3 has an extremely high value of Ksp .



The smaller the value of Ksp the less soluble the compound in water.



Temperature ↑ , solubility ↑, Ksp ↑

107

Example Write the solubility product expression and state the units of Ksp for each of the following ionic compounds.

a) Ca3(PO4)2

b) Ag2CO3

108

Example The solubility product of silver chromate(VI), Ag2CrO4 is 2.4 x10-12 mol3dm-9. Calculate the concentration of Ag+(aq) and CrO42-(aq) in the saturated solution. Ans: [Ag+] = 1.68 x 10-5 M [CrO42-] = 8.43 x 10-5 M

109

Example The solubility of silver sulphide, Ag2S is 5.0x10-17. Calculate the solubility product of Ag2S. Ans: 5.0 x 10-49

110

Example Calculate the solubility of copper (II) hydroxide, Cu(OH)2, in g L-1. Ksp Cu(OH)2 = 2 x 10-20 M3, Mr Cu(OH)2 = 97.57g mol-1

111

Predicting the formation of a precipitate 

Precipitate is an insoluble solid formed in and

separates from the solution.



A mixture of two solutions will produce a precipitate

or not depending on the ion product, Q present.



Q has the same form as Ksp but the concentrations of ions are taken at any given time.

112



The solubility equilibrium equation for a slightly

soluble salt, MA : MA (s)

M+ (aq)

+

A- (aq)

Ksp = [M+] [A-]



If we mix a solution containing M+ ions with one

containing A- ions, the ion product, Q is given by : Q = [M+] [A-]

113

Three possible situations: Q < Ksp; Solution is not saturated. Solid will dissolve and no precipitate formed. Q = Ksp; Saturated solution formed. System is in equilibrium. Q > Ksp; Solution is supersaturated; Ions will form precipitate until the ionic concentration product of the system equals the Ksp (until the system reaches equilibrium). 114

Example Will precipitate form when 50 mL 0.001 M NaOH is added to 150 mL of 0.01 M MgCl2. (Ksp Mg(OH)2 = 2 x 10-11)

115

Example Exactly 200 mL of 0.0040 M AgNO3 are added to 800 mL of 0.008 M K2SO4. Will a precipitate form? (Ksp Ag2SO4 = 1.4x10-5 M3)

116

COMMON ION EFFECT 

Common ion is an ion that is common to two or more components in a mixture of an ionic solution.



Common ion effect is the shift in equilibrium caused by the addition of a compound having an ion in common with the dissolved substances.

117

Example 

Consider a saturated solution of silver chloride, AgCl in water.

(Ksp AgCl = 1.6 x 10-10)

118



The equilibrium system is: Ag+ (aq) + Cl- (aq)

AgCl (s) 

If AgNO3 is added to the saturated AgCl solution : AgNO3 (aq)

Ag+ (aq)

+ NO3- (aq)

(common ion) - [Ag+] will increase. - Equilibrium position will shift backward (according to the Le Chatelier's Principle). - Solubility of AgCl will decrease. - So, the addition of a common ion will reduce the solubility of a slightly soluble salt. 119

Example Calculate the solubility of AgCl (mol L-1) in : a.

liquid water

b.

0.05 M of silver nitrate solution. (Ksp AgCl = 1.6 x 10-10)

120

9.0 THERMOCHEMISTRY

Concept of Enthalpy

Important Terms 

Heat is energy transferred between two bodies of different temperatures



System is any specific part of the universe



Surroundings is everything that lies outside the system



Open system is a system that can exchange mass and energy with its surroundings



Closed system is a system that allows the exchange of energy with its surroundings



Isolated system is a system that does not allow the exchange of either mass or energy with its surroundings



Energy is the ability to do work   

SI unit of energy is kg m2 s-2 or Joule (J) Non SI unit of energy is calorie (Cal) 1 Cal = 4.184 J

Thermochemistry  

A study of heat change in chemical reactions. Two types of chemical reactions:  

Exothermic Endothermic

Exothermic reactions 

Enthalpy of products < Enthalpy of reactants, ΔH is negative.



Energy is released from the system to the surroundings.



Consider the following reaction: A (g) + B (g) → C (g) (reactants) (product)

ΔH = ve

reactants enthalpy H = -ve products reaction pathway

Energy profile diagram for exothermic reaction

Endothermic Reactions 

Enthalpy of products > enthalpy of reactants, ΔH is positive



Energy is absorbed by the system from the surrounding



Consider the following reaction A (g) + B (g) → C(g) ΔH = + ve (reactants) (product)

Energy profile of diagram endothermic reactions

Enthalpy, H 

The heat content of a system or total energy in the system



Enthalpy, H of a system cannot be measured when there is a change in the system.



Example: system undergoes combustion or ionisation.

Enthalpy of Reaction, ∆H and Standard Condition 

Enthalpy of reaction:  The enthalpy change associated with a chemical reaction.



Standard enthalpy, ∆Hº  The enthalpy change for a particular reaction that occurs at 298K and 1 atm (standard state)

Thermochemical Equation 

The thermochemical equation shows the enthalpy changes. Example : H2O(s) → H2O(l)

ΔH = +6.01 kJ



1 mole of H2O(l) is formed from 1 mole of H2O(s) at 0°C, ΔH = +6.01 kJ



However, when 1 mole of H2O(s) is formed from 1 mole of H2O(l), the magnitude of ΔH remains the same with the opposite sign of it. H2O(l) → H2O(s) ΔH = 6.01 kJ

Types of Enthalpies 

There are many kind of enthalpies such as:  Enthalpy of formation  Enthalpy of combustion  Enthalpy of atomisation  Enthalpy neutralisation  Enthalpy hydration  Enthalpy solution

Enthalpy of Formation, ∆Hf 

The change of heat when 1 mole of a compound is formed from its elements at their standard states. H2 (g) + ½ O2(g) → H2O (l)



∆Hf = 286 kJ mol1

The standard enthalpy of formation of any element in its most stable state form is ZERO. ∆H (O2 ) = 0 ∆H (Cl2) = 0

Enthalpy of Combustion, ∆Hc 

The heat released when 1 mole of substance is burned completely in excess oxygen.

C(s) + O2(g) → CO2(g)

∆Hc = 393 kJ mol1

Enthalpy of Atomisation, Ha 

The heat change when 1 mole of gaseous atoms is formed from its element



Ha is always positive because it involves only breaking of bonds e.g:



Na(s)  Na(g)

Ha = +109 kJ mol-1

½Cl2(g)  Cl(g)

Ha = +123 kJ mol-1

Enthalpy of Neutralization, ∆Hn 

The heat change when 1 mole of water, H2O is formed from the neutralization of acid and base .



HCl(aq)+ NaOH(aq) → NaCl(aq) +H2O(aq) ΔHn = 58 kJ mol1

Enthalpy of Hydration, Hhyd 



The heat change when 1 mole of gaseous ions is hydrated in water. e.g: Na+(g)  Na+(aq) Hhyd = 406 kJ mol-1 Cl-(g)  Cl-(aq) Hhyd = 363 kJ mol-1

Enthalpy of Solution, Hsoln  The

heat change when 1 mole of a substance is dissolves in water.  e.g: KCl(s)  K+(aq) + Cl(aq) Hsoln = +690 kJ mol-1

Enthalpy of Sublimation, Hsubl The heat change when one mole of a substance sublimes (solid into gas). H subl

I2 (s) →

I2(g)

Hsubl =

+106 kJ mol1

Calorimetry 

 

A method used in the laboratory to measure the heat change of a reaction. Apparatus used is known as the calorimeter Examples of calorimeter  

Simple calorimeter Bomb calorimeter

Simple calorimeter 



The outer Styrofoam cup insulate the reaction mixture from the surroundings (it is assumed that no heat is lost to the surroundings) Heat release by the reaction is absorbed by solution and the calorimeter

A bomb calorimeter

Important Terms in Calorimeter 



Specific heat capacity, c  Specific heat capacity, c of a substance is the amount of heat required to raise the temperature of one gram of the substance by one degree Celsius (Jg 1C1). Heat capacity, C  Heat capacity,C is the amount of heat required to raise the temperature of a given quantity of the substance by one degree Celsius (JC1)

Heat released by substance

=

Heat absorbed by calorimeter

q = mc∆T q = heat released by substance m= mass of substance C= specific heat capacity ∆T = temperature change

Basic Principle in Calorimeter Heat released by a reaction

=

Heat absorbed by surroundings

• Surroundings may refer to the: i. Calorimeter itself or; ii. The water and calorimeter • qreaction= mcΔT or CΔT

Example 1 In an experiment, 0.100 g of H2 and excess of O2 were compressed into a 1.00 L bomb and placed into a calorimeter with heat capacity of 9.08 x 104 J0C1. The initial temperature of the calorimeter was 25.0000C and finally it increased to 25.155 0C. Calculate the amount of heat released in the reaction to form H2O, expressed in kJ per mole.

Solution Heat released = Heat absorbed by the calorimeter q

= C∆T = (9.08 X 104 J0C-1) X (0.1550C) = 1.41 X 104 J = 14.1 kJ H2(g) + ½O2(g) → H2O(c)

mole of H2 = 0.100 2.016 = 0.0496 mol

moles of H2O

=

mole of H2

0.0496 mol of H2O released 14.1 kJ energy 1 mol H2O released

14.1 = 0.0496

∆Heat of reaction, ∆H

= 284 kJ = - 284 kJ mol1

kJ

Example 2 1. Calculate the amount of heat released in a reaction in an aluminum calorimeter with a mass of 3087.0 g and contains 1700.0 mL of water. The initial temperature of the calorimeter is 25.0°C and it increased to 27.8°C.

Given: Specific heat capacity of aluminum = 0.553Jg-1 °C-1 Specific heat capacity of water = 4.18 Jg-1 °C-1 Water density = 1.0 g mL-1 ΔT = (27.8 -25.0 )°C = 2.8°C Solution Heat released

=

Heat absorbed by aluminium calorimeter

q = mwcwΔT + mcccΔT = (1700.0 g)(4.18 Jg-1 °C-1)(2.8 °C) + (3087.0 g)(0.553 Jg-1 °C-1)(2.8°C) = 24676.71 J = 24.7 kJ

+

Heat absorbed by water

HESS’S LAW

Hess Law 



Hess’s Law states that when reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in the series of steps. The enthalpy change depends only on the nature of the reactants and products and is independent of the route taken.  H 1

A  H 2

C

B  H 3

H1  H2  H3

Algebraic Method Step 1 i. List all the thermochemical equations involved i.C

O  CO (S) 2( g ) 2( g )

1 O H O 2 2( g ) 2( g ) 2 ( g) iii.C H 7 O  2CO  3H O 2 2 6( g ) 2( g ) 2( g ) 2 ( g) ii.H

H  - 393kJmol- 1 H  -286kJmol- 1 H  -1560kJmol

-1

Algebraic Method Step 1 i. List all the thermochemical equations involved i.C

O  CO (S) 2( g ) 2( g )

1 O H O 2 2( g ) 2( g ) 2 ( g) iii.C H 7 O  2CO  3H O 2 2 6( g ) 2( g ) 2( g ) 2 ( g) ii.H

H  - 393kJmol- 1 H  -286kJmol- 1 H  -1560kJmol

-1

ii. Write the enthalpy of formation reaction for C2H6

H  ? f C  3H    C H (s) 2( g ) 2 6( g )

iii. Add the given reactions so that the result is the desired reaction. (i )  2

2C ( S )  2O2( g )  2CO2( g )

H1  2  -393 kJ

( ii )  3

3 H 2( g )  3 O2( g )  3 H 2O( g ) 2

H 2  3  -286kJ

reverse (iii) 2CO2(g)  3 H 2O( g )  C 2 H 6( g )  7 O2( g ) H3  1560kJ 2 _______________________________________________________________ H f  ? 2C ( s)  3 H 2( g )    C 2 H 6( g ) - 84kJ Hf  H1  H 2  H3  -84kJ

Energy Cycle Method Draw the energy cycle and apply Hess’s Law to calculate the unknown value.

HOf 2C(s)

+

3H2 (g)

C2H6 (g)

HO2 = 3(-286) HO1 = 2(-393)

2O2(g)

3/2 O2(g)

2CO2 (g) + 3H2O(g)

7/2 O2(g) HO3 = - (-1560)

° ° ° ° ΔH f = 2( H1 ) + 3(H 2 ) + ΔH3  -786 - 858  1560 1  -84 kJmol

Example 1 The thermochemical equation of combustion of carbon monoxide is shown as below. C(s) + ½ O2(g)  CO(g) = ? H given :   H C(s) + O2(g)  CO2(g) ∆H= -394 kJ mol-1  CO(s) + ½ O2(g)  CO2(g) ∆H=H-283 kJ mol-1 Calculate the enthalpy change of the combustion of carbon to carbon monoxide.

Example 2 Calculate the standard enthalpy of formation of methane if the enthalpy of combustion of carbon, hydrogen and methane are as follows: -1  ∆H [C ] = -393 kJ mol (s) Hc ∆H [H2(s)] = -293 kJ mol-1  Hc -1 ∆H [CH ] = -753 kJ mol 4(s)  Hc

Example 3  Standard enthalpy of formation of ammonia, hydrogen chloride and ammonium chloride is -46.1 kJ mol-1, -92.3 kJ mol-1, 314.4 kJ mol-1 respectively. Write the thermochemical equation for the formation of each substance and calculate the enthalpy change for the following reaction. NH3(g) + HCl (g)  NH4Cl(s)

Exercise 1.Calculate the enthalpy of formation of benzene if : ∆H (CO2(g) ) = -393.3 kJ mol-1 ∆H (H2O(l) ) = -285.5 kJ mol-1 -1 ∆H (C H ) = -3265.3 kJ mol  6 6(l) H f H f

H f

Born-Haber Cycle

Lattice Energy, Hlattice 

is the energy required to completely separate one mole of a solid (ionic compound) into gaseous ions

e.g: NaCl(s)  Na+(g) + Cl-(g) 

Na+(g) + Cl-(g)  NaCl(s) 

Hlattice = +771 kJ mol-1 (lattice dissociation)

Hlattice = -771 kJ mol-1 (lattice formation) The magnitude of lattice energy increases as  the ionic charges increase  the ionic radii decrease



There is a strong attraction between small ions and highly charged ions so the H is more negative.



H for MgO is more negative than H for Na2O because Mg2+ is smaller in size and has bigger charge than Na+, therefore Hºlattice (MgO) > Hºlattice (Na2O)

Hydration Process of Ionic Solid 

Na+ and Cl- ions in the solid crystal are separated from each other and converted to the gaseous state (Hlattice)



The electrostatic forces between gaseous ions and polar water molecules cause the ions to be surrounded by water molecules (Hhydr)

Hsoln = Hlattice + Hhdyr

Na+ and Cl- ion in the gaseous state

ne

La

He a

y rg

E e ttic

to fH yd ra

tio

n

Heat of Solution Na+ and Cl- ion in the solid state

Hydrated Na+ and Cl- ion

Born-Haber Cycle   



The process of ionic bond formation occurs in a few stages. At each stage the enthalpy changes are considered. The Born Haber cycle is often used to calculate the lattice energy of an ionic compound. In the Born-Haber cycle energy diagram, by convention, positive values are denoted as going upwards, negative values as going downwards. Consider the enthalpy changes in the formation of sodium chloride.

Example : Na  i. ii. iii. iv. v. vi.

(s)



1

2

Cl

2(g)

 NaCl

Given; Enthalpy of formation NaCl Enthalpy of sublimation of Na First ionization energy of Na Enthalpy of atomization of Cl Electron affinity of Cl Lattice energy of NaCl

= = = = = =

(s)

-411 kJmol-1 +108 kJmol-1 +500 kJmol-1 +122 kJmol-1 -364 kJmol-1 ?

Example: A Born-Haber cycle for NaCl energy Na+(g) + e

+ Cl(g)

Ionisation Energy of Na Na(g)

Electron Affinity of Cl Na+(g) + Cl- (g)

+ Cl(g)

HaCl +ve

Na(g) + ½ Cl2(g)

Lattice energy

HaNa E=0 -ve

Na(s) + ½ Cl2(g)

From Hess’s Law: Hf

Hf NaCl NaCl(s)

NaCl

= HaNa + HaCl +IENa + EACl + Lattice Energy

Calculation:  H 0f   H S  IE   H a ( Cl )  EA   H lattice  H lattice   H 0f   H S  IE   H a ( Cl )  EA   H lattice   411 kJ   108 kJ  500 kJ  122 kJ   364 kJ   H lattice   777 kJ

Exercise: 

i. ii. iii. iv. v. vi.

Construct a Born-Haber cycle to explain why ionic compound NaCl2 cannot form under standard conditions. Use the data below: Enthalpy of sublimation of sodium = +108 kJmol-1 First ionization energy of sodium = +500 kJmol-1 Second ionization energy of sodium = +4562 kJmol-1 Enthalpy of atomization of chlorine = +121kJmol-1 Electron affinity of chlorine = -364 kJmol-1 Lattice energy of NaCl2 = -2489 kJmol-1

Electrochemistry

Electrochemistry Is the study of the relationship between electricity and chemical reaction Chemical reactions involved in electrochemistry are :

Reduction

REDOX REACTION

Oxidation One type of reaction cannot occur without the other.

REDOX Reaction REDUCTION

OXIDATION

gain of electron

loss of electron

Oxidation no. decrease

Oxidation no. increase

Reaction at cathode

Reaction at anode

Remember… RED CAT

Example:

Mg  Mg2+ + 2e-

= REDuction at CAThode RED CAT Oxidation no. 

Example:

Cu2+ + 2e-  Cu Oxidation no.



Electrochemical reaction consists of reduction and oxidation. These two reactions are called ‘half-cell reactions’ The combination of 2 half reactions are called ‘cell reaction’

Example Reduction : Cu2+(aq) + 2e-  Cu(s) Oxidation :

Zn(s)  Zn2+(aq) + 2e-

Overall cell 2+ 2+ Cu + Zn  Cu + Zn (aq) (s) (s) (aq) reaction :

Half-cell reaction

Cells There are 2 type of cells

Electrochemical Cells where chemical reaction produces electricity

Electrolytic Cells Uses electricity to produce chemical reaction

Also called; Galvanic cell or Voltaic cell Chemical Energy

Electrical Energy

Electrical Energy

Chemical Energy

Component and Operation of Galvanic cell Consists of : 1) Zn metal in an aqueous solution of Zn2+ 2) Cu metal in an aqueous solution of Cu2+ - The 2 metals are connected by a wire - The 2 containers are connected by a salt bridge. - A voltmeter is used to detect voltage generated.

Galvanic cell Voltmeter

Cu electrode

Zn electrode Zn2+ ZnSO4(aq) solution

Cu2+

Salt bridge

CuSO4(aq) solution

What happens at the zinc electrode ?  Zinc is more electropositive than copper.  Tendency to release electrons: Zn > Cu.

Zn (s)  Zn2+ (aq) + 2e   

Zinc dissolves. Oxidation occurs at the Zn electrode. Zn2+ ions enter ZnSO4 solution. Zn is the –ve electrode since it is a source of electrons  anode.

What happens at the copper electrode ?  The electron from the Zn metal moves out through the wire enter the Cu metal  Cu2+ ions from the solution accept electrons.

Cu2+ (aq) + 2e-  Cu (s)  Copper is deposited.  Reduction occurs at the Cu electrode.  Cu is the +ve electrode  cathode

Reactions Involved:

Anode : Cathode : Overall cell reaction :

Zn (s)  Zn2+ (aq) + 2eCu2+ (aq) + 2e-  Cu (s)

Zn (s) + Cu2+ (aq)  Zn2+ (aq) + Cu (s)

Salt bridge An inverted U tube containing a gel permeated with solution of an inert electrolyte such as KCl, Na2SO4, NH4NO3.

Functions  Salt bridge helps to maintain electrical neutrality  Completes the circuit by allowing ions carrying charge to move from one half-cell to the other.

What happened if there is no salt bridge? V Cu

Zn

ZnSO4(aq)

Zn2+

e e e e

e e

Cu2+

CuSO4(aq)

 As the zinc rod dissolves, the concentration of Zn2+ in the left beaker increase.  The reaction stops because the nett increase in positive charge is not neutralized. This excess charge build-up can be reduced by adding a salt bridge

Zn (s)  Zn2+ (aq) + 2e-

Cu2+ (aq) + 2e-  Cu (s) CATHODE (+)

ANODE (-) E = +1.10 V

-

Zn

ZnSO4(aq)

Zn2+

Cu

+

e e

Cu2+ Salt bridge (KCl)

CuSO4(aq)

How does the cell maintains its electrical neutrality? Left Cell Zn (s)  Zn2+ (aq) + 2eZn2+ ions enter the solution. Causing an overall excess of tve charge. Cl- ions from salt bridge move into Zn half cell

Right Cell Cu2+ (aq) + 2e-  Cu (s) Cu 2+ ions leave the solution. Causing an overall excess of -ve charge. K+ ions from salt bridge move into Cu half cell

Electrical neutrality is maintained

Electrochemical Cells

anode oxidation

cathode reduction

spontaneous redox reaction

half

19.2

Cell notation Zn (s) + Cu2+ (aq)

Cu (s) + Zn2+ (aq)

Also can be represented as: Salt bridge

Phase boundary

Zn (s) | Zn2+ (aq) || Cu2+ (aq) | Cu (s) anode

cathode Cell notation

Exercise For the cell below, write the reaction at anode and cathode and also the overall cell reaction. Cell notation

Zn (s) | Zn2+(aq) || Cr3+ (aq) | Cr (s) Anode : Cathode :

2e3 Zn(s) → 3Zn2+(aq) + 6e 3e- → 2Cr(s) 2 Cr3+(aq) + 6e

X3 X2

Overall cell 3Zn(s) + 2Cr3+(aq) + → 3Zn2+ (aq) +2Cr(s) reaction:

The difference in electrical potential between the anode and cathode is called: • cell voltage • electromotive force (emf) • cell potential

measured by a voltmeter

Acts as ‘electrical pressure’ that pushes electron through the wire.

A measure of the ability of a half-cell to attract electrons towards it.

Cu2+(aq) + 2e → Cu(s)

Eored = +0.34 V

Zn2+(aq) + 2e → Zn(s)

Eored = -0.76 V Standard reduction potential

•The more positive the half-cell’s electrode potential, the stronger the attraction for electrons. Tendency for reduction ↑ (cathode)

Standard reduction potential of copper half-cell is more positive compared to zinc. Zinc half-cell becomes anode.

Cu2+(aq) + 2e → Cu(s)

Eored = +0.34 V

Zn2+(aq) + 2e → Zn(s)

Eored = -0.76 V

Cell Potential (Eocell)= Eocatode — Eoanode = +0.34 – (-0.76) = +1.1 V

Zn2+ (aq) + 2e-  Zn (s)

E0 = -0.76V

Cu2+ (aq) + 2e-  Cu (s)

E0 = +0.34V

E0cell = E0cathode - E0anode

or

= +0.34 – (-0.76) = +1.10 V

E0cell = E0red + E0ox

Change the sign

= +0.34 + (+0.76) = +1.10 V Half-cell equation at:

Anode :

Zn (s)  Zn2+ (aq) + 2e-

Cathode :

Cu2+ (aq) + 2e-  Cu (s)

Standard Electrode Potentials (Eo) A measure of the ability of half-cell to attract electrons towards it at 25oC, the pressure is 1 atm (for gases), and the concentration of electrolyte is 1M.

The sign of E0 changes when the reaction is reversed •Changing the stoichiometric coefficients of a half-cell reaction does not change the value of E0

For example: Cl2(g) + 2e-  2Cl-(aq)

E0 = +1.36 V

½Cl2(g) + e-  Cl-(aq)

E0 = +1.36 V

Cl-(aq)  ½Cl2(g) + e-

E0 = -1.36 V

Standard Hydrogen Electrode (SHE) Made up of a platinum electrode, immersed in an aqueous solution of H+ (1 M) and bubbled with hydrogen gas at 1 atm pressure, and temperature at 25oC

H2 gas at 1 atm

H+ (aq) 1 M

Pt electrode

The standard reduction of SHE is 0 V

Standard reduction potential of Zinc half cell is measured by setting up the electrochemical cell as below. E0 = +0.76 Zn

e e e 2+ Zn e

ZnSO4(aq) 1M

V

E0 = 0

+ +

H2 (g), 25oC,1

H+(aq),1 M Pt

atm.

Standard Electrode Potentials Zn (s) | Zn2+ (1 M) || H+ (1 M) | H2 (1 atm) | Pt (s) Anode (oxidation): Cathode (reduction):

Zn2+ (1 M) + 2e-

Zn (s) 2e- + 2H+ (aq,1 M)

Cell reaction Zn (s) + 2H+ (aq,1 M)

H2 (g,1 atm) Zn2+(aq) + H2 (g,1 atm)

0 = E 0+ - E 0 2+ Ecell H /H2 Zn /Zn 0 2+ 0.76 V = 0 - EZn /Zn 0 2+ EZn /Zn = -0.76 V

Zn2+ + 2e- Zn

E0 = -0.76 V

Standard reduction potential of Copper half cell is measured by setting up the electrochemical cell as below.

+ + Cu

E0 = 0

V-

H2 (g) 25oC 1 atm.

H+(aq) 1M

Cu2 CuSO4(aq) 1M

Pt

Pt (s) | H2 (1 atm) | H+ (1 M) || Cu2+ (1 M) | Cu (s) Anode (oxidation):

H2 (1 atm)

Cathode (reduction): 2e- + Cu2+ (1 M) H2 + Cu2+

0 0 = E0 Ecell cathode - Eanode 0 = E 0 2+ 0 Ecell Cu /Cu – EH +/H2 0 2+ 0.34 = ECu /Cu - 0 0 2+ ECu /Cu = 0.34 V

2H+ + 2eCu (s) Cu (s) + 2H+

The direction of half-reaction of SHE depends on the other half-cell connected on it. The cell notation for SHE is either:

 H+(aq) | H2(g) | Pt(s) when it is cathode  Pt(s) | H2(g) | H+ (aq) when it is anode

In either case, E0 of SHE remains 0

Zn2+ (aq) + 2e-  Zn (s)

E0 = -0.76V

Cu2+ (aq) + 2e-  Cu (s)

E0 = +0.34V


or

= +0.34 – (-0.76) = +1.10 V

E0cell = E0red + E0ox

Change the sign

= +0.34 + (+0.76) = +1.10 V Half-cell equation at:

Anode :

Zn (s)  Zn2+ (aq) + 2e-

Cathode :

Cu2+ (aq) + 2e-  Cu (s)

At standard-state condition

E0cell = E0cathode - E0anode or E0cell = E0red + E0ox

Exercise Calculate the standard cell potential of the following electrochemical cell. Co(s) | Co2+(aq) || Ag+(aq) | Ag(s) Ag+(aq) + e-  Ag(aq)

Answer

Co2+(aq) + 2e-  Co(s)

+ Cathode (Red) : Ag (aq) + e  Ag(aq)

Anode (Ox) : Co(s)  Co2+(aq) + 2e-

E0cell = E0cathode - E0anode = +0.80 – (-0.28) = +1.08 V

E0 = +0.80V E0 = -0.28V E0 = +0.80V E0ox = +0.28V

Refer to the list of Standard Reduction Potential: Oxidation agent → left of the half cell equation Reduction agent → right of the half cell equation Example : Ni2+ (aq) + 2e- → Ni (s) E0 = -0.25 V Cu2+ (aq) + 2e- → Cu (s) E0 = +0.34 V Ag+ (aq) + e- → Ag (s) E0 = +0.80 V Reducing Oxidation agent agent

Increase strength as reducing agent

The more -ve the value of E0 → the stronger the reducing agent The more +ve the value of E0 → the stronger the oxidizing agent

Exercise Arrange the 3 elements in order of increasing strength of reducing agents X3+ + 3e-  X

E0 = -1.66 V

Y2+ + 2e-  Y

E0 = -2.87 V

L2+ + 2e-  L

E0 = +0.85 V

Answer : L < X < Y

Example Calculate the E0 cell for the reaction :: Mg(p) | Mg2+(ak) || Sn4+(ak),Sn2+(ak) | Pt(p) Given : Mg2+(ak) + 2e→ Sn4+(ak)

Eθ = -2.38 V

Mg(p)

+ 2e → Sn2+(ak) Eθ= +0.15 V

Oxidation

: Mg(p) →

Mg2+(ak)

Reduction

: Sn4+(ak)

+ 2e →

+

= +2.38 + 0.15 = +2.53 V

E o ox

=+2.53V

+ 2e

Eoox = +2.38 V

Sn2+(ak)

Eo = +0.15 V

Mg(p) + Sn4+(ak) → Mg2+(ak) + Sn2+(ak)

Eθcell = E o red

E0cell = Ecathode - Eanode =+0.15- (-2.38)

Ecell = +2.53 V

Exercise A cell is set up between a chlorine electrode and a hydrogen electrode Pt | H2(g, 1 atm) | H+(aq, 1M) || Cl2(g, 1atm) | Cl-(aq, 1M) | Pt

E0cell = +1.36 V (a) Draw a diagram to show the apparatus and chemicals used. (b) Discuss the chemical reactions occurring in the electrochemical cell.

Answer E0cell =1.36V -

H2 (g), 1 atm.

-

V

+ +

Pt

Cl2 (g), 1 atm.

Pt H+(aq), 1M

Cl-(aq), 1M

1. Show the process occur at anode and cathode - Half-cell reaction 2. Overall reaction

Answer  Reduction (cathode) Cl2 (g) + 2e-  2Cl- (aq)  Oxidation(anode) H2 (g)  2H+ (aq) + 2e-

E0 = 0

 Eocell =+1.36 V Eocell = Eocathode - E0anode +1.36 = Eocathode – 0 E0cathode = +1.36 V So the standard reduction potential for Cl2 is: Eo = +1.36 V

Spontaneous & Non-Spontaneous reactions - Redox reaction is spontaneous when Ecell is +ve. - Non spontaneous is when Ecell is –ve. 

Eθ cell = 0

The reaction is at equilibrium

Predict whether the following reactions occur spontaneously or non-spontaneously under standard condition. EoSn4+/Sn2+=+0.15V → Zn + Sn4+ Sn2+ + Zn2+ EoZn/Zn2+ = - 0.76V. The two half-cells involved are:Anod : Zn → Zn2+ + 2e Eoox = +0.76 V Cathode: Sn4+ + 2e → Sn2+ Eo = +0.15 V

Zn + Sn4+

Zn2+ + Sn2+

→

Eocell = Eo Sn4+/Sn2+ → Eo Zn/Zn2+ = +0.15 – (-0.76 ) = +0.91 V spontaneous

Or

Eocell= Eored + Eoox = (+0.15) + (0.76) = +0.91 V

Pb2+(aq) + 2Cl-(aq) → Pb(s) + Cl2(g)

Reduction Pb2+(aq) + 2Cl-(aq)

→ Pb(s) + Cl (g)

→Pb(s) Pb(s) Cl2 (g) Pb2+(aq) + 2Cl-(aq) → + Cl+2(g) Oxidation

cathode: Pb2+(aq) + 2e → Pb(s) anode: 2Cl-(aq) → Cl2(g) + 2e

Eo = -0.13 V Eoox = -1.36

Pb2+(aq) + 2Cl-(aq) → Pb(s) + Cl2(g) Eocell= Eored + Eoox = (-1.36) + (-0.13) = -1.48 V Non-spontaneous No Reaction

Example : Predict whether the following reactions occur spontaneously : 2Ag+(aq)

2Ag(s) + Br2(aq) 0 + EAg

Br2(aq)

= +0.8 V

0 EBr - = +1.07 V 2 /Br

Answer : 2Ag(s)

/Ag

2Ag+(aq) + 2e

2Ag(s) + Br2(aq)

+

+ 2Br-(aq) standard reduction potential

2e

Eθox = - 0.80 V

2Br -(aq)

Eθ = +1.07 V

2Ag+(aq) + 2Br-(aq)

Esel = + 0.27 V

The reaction is spontaneous

Exercise A cell consists of silver and tin in a solution of 1 M silver ions and tin (II) ions. Determine the spontaneity of the reaction and calculate the cell voltage of this reaction.

Ag+ (aq) + e- → Ag (s)

E0 = +0.80 V Sn2+ (aq) + 2e- → Sn (s) E0 = -0.14 V

(cathode) (anode)


E0cell

= +0.80 – (-0.14) = +0.94 V = +ve ( reaction is spontaneous)

Nernst equation Nernst equation can be used to calculate the E for any chosen concentration : Ecell = Eocell –

RT nF

ln

[ product ]x [ reactant]y

At 298 K and R = 8.314 J K-1 mol-1 , 1 F = 96500 C Ecell = Eocell –

0.0257 n

2.303 log


cell

Ecell = Eocell – 0.0592

log


n

[ product ]x [

reactant]y

Ecell = Eocell –

=

0.0592

Q

log Q

n

n = no of e- that are involved Q = reaction quotient

Example 1 Calculate the Ecell for the following cell Zn(s) / Zn2+ (aq, 0.02M) // Cu2+(aq, 0.40 M) / Cu(s)

Answer Zn(s)

+ Cu2+(aq)

Eocell = Eored + Eoox

@

Zn2+(aq)

+ Cu(s)

Eocell = Eocathode – Eoanode

= +0.34 V + 0.76 V

= +0.34 V - (- 0.76 V)

= +1.10 V

= +1.10 V

E = Eo – 0.0592

log [ Zn2+] [ Cu2+]

n

E = +1.10 V – 0.0592 log (0.02 ) 2

= +1.10 V – (-0.0385) = +1.139 V

( 0.40)

At equilibrium: ~ No net reaction occur (Q=K) ~ Ecell = 0


log K

n 0

= Eocell – 0.0592 n

Eθcell

=

0.0592 n

log K

log K

Example 2 Calculate the equilibrium constant (K) for the following reaction. Cu(s) +

2Ag+(ak)

Cu2+(ak)

Answer At equilibrium, E cell = 0 Eocell = Eo cathode -

Eo anode

= +0.80 – ( +0.34) = +0.46 V

+

2Ag(s)


log K

2 0

= 0.46 – 0.0592

log K

2 0.0592 log K = 0.46 2 log K = 15.54 K = 3.467

x 1015

Electrolysis Electrolysis is a chemical process that uses electricity for a non-spontaneous redox reaction to occur. Such reactions take place in electrolytic cells.

Electrolytic Cell  It is made up of 2 electrodes immersed in an electrolyte.  A direct current is passed through the electrolyte from an external source.  Molten salt and aqueous ionic solution are commonly used as electrolytes.

Electrolytic Cell +

-

Oxidation

Reduction Electrolyte (M+X-) X-,OH-

Anion

M+,H+

Cation

A n o d e

 Positive electrode  The electrode which is connected to the positive terminal of the battery  Oxidation takes place Electrons flow from anode to cathode



C a t h o d e

 Negative electrode  The electrode which is connected to the negative terminal of the battery  Reduction takes place

Electrode  as circuit connectors  as sites for the precipitation of insoluble products  example: Platinum , Graphite (inert electrode) Electrolyte  a liquid that conducts electricity due to the presence of +ve and –ve ions  must be in molten state or in aqueous solution so that the ions can move freely example: KCl(l), HCl(aq), CH3COOH(aq)

Comparison between an electrochemical cell and an electrolytic cell Electrolytic Cell +

-

e-

e+

Anode

Electrochemical Cell

-

Cathode

e-

-

Anode

+

e-

Cathode

Electrolytic Cell

Electrochemical Cell

 Cathode = negative

 Cathode = positive

 Anode = positive

 Anode = negative

 Non-spontaneous redox  Spontaneous redox reaction requires energy reaction releases energy to drive it Similarities:  Oxidation occurs at anode, reduction occurs at cathode  Anions move towards anode, cations move towards cathode.  Electrons flow from anode to cathode in an external circuit.

Electrolysis of molten salt  Electrolysis of molten salt requires high temp.  Electrolysis of molten NaCl Cation : Na+ Anode :

Anion : Cl-

Cl- (l) → Cl2(g) + 2e-

Cathode : Na+ (l) + e- → Na (s) 2Na+ (l) + 2e- → 2Na (s) Overall :

2Na+ (l) + 2Cl-(l) → Cl2(g) + 2Na(s)

 Electrolysis of molten NaCl gives sodium metal deposited at cathode and chlorine gas evolved at anode. Electrolysis of molten NaCl is industrially important. The industrial cell is called ‘Downs Cell’

Electrolysis of Aqueous Salt  Electrolysis of aqueous salt is more complex because the presence of water.  Aqueous salt solutions contains anion, cation and water.  Water is an electro-active substance that may be oxidised or reduced in the process depending on the condition of electrolysis. Reduction : 2H2O (l) + 2e-

H2 (g) + 2OH- (aq)

E0 = -0.83 V

Oxidation : 2H2O (l)

4H+ (aq) + O2 (g) + 4e-

E0 = -1.23 V

Predicting the products of electrolysis

Factors influencing the products : 1. Reduction/oxidation potential of the species in electrolyte 2. Concentrations of ions 3. Types of electrodes used – active or inert

Electrolysis of Aqueous NaCl NaCl aqueous solution contains Na+ cation, Cl- anion and water molecules On electrolysis,  the cathode attracts Na+ ion and H2O molecules  the anode attracts Cl- ion and H2O molecules The electrolysis of aqueous NaCl depends on the concentration of electrolyte.

Electrolysis of diluted NaCl solution Cathode Na+ (aq) + e-

Na (s)

E0 = -2.71 V

2H2O (l) + 2e-

H2 (g) + 2OH- (aq)

E0 = -0.83 V

E0 for water molecules is more positive. H2O easier to reduce.

Anode Cl2 (g) + 2eO2 (g) + 4H+ (aq) + 4e-

2Cl- (aq) 2H2O (l)

E0 = +1.36 V E0 = +1.23 V

In dilute solution, water will be selected for oxidation because of its lower Eo.

Reactions involved 2H O (l) + 2e 2 Cathode:

4H2O (l) + 4e-

H2 (g) + 2OH- (aq) E0 = -0.83 V 2H2 (g) + 4OH- (aq) E0 = -1.23 V

Anode: 2H2O (l)

O2 (g) + 4H+ (aq) + 4e-

Cell 6H2O(l) reaction:

O2(g) + 2H2(g) + 4OH-(aq) + 4H+(aq)

2H2O(l)

4 H2O O2(g) + 2H2(g)

E0cell = -2.06 V

Electrolysis of Concentrated NaCl solution Cathode Na+ (aq) + e-

Na (s)

E0 = -2.71 V

2H2O (l) + 2e-

H2 (g) + 2OH- (aq)

E0 = -0.83 V

 E0 for water molecules is more positive  H2O easier to be reduce

Anode Cl2 (g) + 2eO2 (g) + 4H+ (aq) + 4e-

2Cl- (aq) 2H2O (l)

E0 = +1.36 V E0 = +1.23 V

In concentrated solution, chloride ions will be oxidised because of its high concentration.

Reactions involved Cathode: 2H2O (l) + 2eAnode:

2Cl- (aq)

Cell 2H2O(l) + 2Clreaction:

H2 (g) + 2OH- (aq) E0 = -0.83 V Cl2 (g) + 2e-

E0 = -1.36 V

Cl2(g) + H2(g) + 2OH-(aq) E0cell = -2.19 V

Exercise Predict the electrolysis reaction when Na2SO4 solution is electrolysed using platinum electrodes.

Solution  Na2SO4 aqueous solution contains Na+ ion, SO42- ion and water molecules  On electrolysis,  the cathode attracts Na+ ion and H2O molecules  the anode attracts SO42- ion and H2O molecules

Cathode Na+ (aq) + e-

Na (s)

E0 = -2.71 V

2H2O (l) + 2e-

H2 (g) + 2OH- (aq)

E0 = -0.83 V

 E0 for water molecules is more positive H2O easier to reduce

Anode S2O82- (aq) + 2eO2 (g) + 4H+ (aq) + 4e-

2SO42- (aq) 2H2O (l)

E0 = +2.01 V E0 = +1.23 V

 E0 for water molecules is less positive H2O easier to oxidise

Equation Cathode: 2H2O (l) + 2eAnode:

2H2O (l)

Cell 2H2O(l) Reaction:

H2 (g) + 2OH- (aq) E0 = -0.83 V O2 (g) + 4H+ (aq) + 4eO2(g) + 2H2(g)

E0 = -1.23 V

E0cell = -2.06 V

 Cathode = H2 gas is produced and solution become basic at cathode because OH- ions are formed

 Anode = O2 gas is produced and solution become acidic at anode because H+ ions are formed

Faraday’s Law of Electrolysis Describes the relationship between the amount of electricity passed through an electrolytic cell and the amount of substances produced at electrode.

Faraday’s First Law States that the quantity of substance formed at an electrode is directly proportional to the quantity of electric charge supplied.

Faraday’s 1st Law

mαQ Q = electric charge in coulombs (C) m = mass of substance discharged

Q = It Q = electric charge in coulombs (C) I = current in amperes (A) t = time in second (s)

Faraday constant (F) is the charge on 1 mole of electron

1 F = 96 500 C

Example An aqueous solution of CuSO4 is electrolysed using a current of 0.150 A for 5 hours. Calculate the mass of copper deposited at the cathode.

Answer Electric charge, Q = Current (I) x time (t) Q = (0.150 A) x ( 5 x 60 x 60 )s Q = 2700 C 1 mole of electron Ξ 1 F Ξ 96 500 C No. of

e-

passed through =

2700 96 500

= 0.028 mol

Cu2+ (aq) + 2e-  Cu (s) From equation: 2 mol electrons  1 mol Cu 0.028 mol electrons  0.014 mol Cu Mr for Cu = 63.5 Mass of Copper deposited = 0.014 x 63.5 = 0.889 g

11.0 REACTION KINETICS Objectives: 1. 2.

Define reaction rate, average rate, instantaneous rate and initial rate. Determine the reaction rate based on a differential equation.

REACTION KINETICS Chemical kinetics is the study of the rates of chemical reactions, the factors that affect these rates, and the reaction mechanisms by which reactions occur. Important

industrial process -Time -Optimum yield -Optimum conditions control over reaction, obtain products economically, using optimum conditions

Rate of reaction

• Reaction rate is the change in the concentration of a reactant or a product with time. • Unit of rate (mol L-1 s-1)

1 • rate  Time Example;

A

 B

d[A] rate = dt

d[A] = change in concentration of A

d[B] rate = dt

d[B] = change in concentration of B

dt = period of time

Because [A] decreases with time, d[A] is negative.

A

B

time

d[A] rate = dt rate =

d[B] dt

[B] ↑

[A] ↓

Rate of reaction • The average rate is the rate over a period of time. • The rate of reaction at a given time is called an instantaneous rate of reaction. • The instantaneous rate at the beginning of a reaction is called the initial rate of reaction. • Instantaneous rate is determined from a graph of concentration vs time by drawing a line tangent to the curve at that particular time.

Rate of reaction Reaction: H2O2(aq)  H2O(l) + ½ O2(g) Reaction rates are obtained from the slopes of the straight lines; purple

An average rate from the purple line. The instantaneous rate at t =300 s from the red line.

blue

red

The initial rate from the blue line.

Br2 (aq) + HCOOH (aq)

2Br- (aq) + 2H+ (aq) + CO2 (g)

instantaneous rate = rate at a specific time [Br2]final – [Br2]initial d[Br2] average rate = =dt tfinal - tinitial

The differential rate equation A differential rate equation enables the relationship between the rate of disappearance of reactants and the formation of products. Consider the reaction, aA + bB

Rate =



cC + dD

1 d[A] 1 d[B] 1 d[C] 1 d[D]     a dt b dt c dt d dt

a,b,c and d are the stoichiometric coefficients

The differential rate equation Example: The formation of NH3, N2(g) + 3H2(g)  2NH3(g) The differential rate equation is;

d[N 2 ] 1 d[H 2 ] 1 d[NH 3 ]   Rate =  dt 3 dt 2 dt The equation means that the rate of disappearance of N2 is 1/3 the rate of disappearance of H2 and 1/2 the rate of formation of NH3.

Example 1: Consider the reaction, 2HI  H2 + I2, determine the rate of disappearance of HI when the rate of I2 formation is 1.8 x 10-6 M s-1. Solution:

1 d[HI] d[H 2 ] d[I 2 ]   Rate =  2 dt dt dt d[I 2 ] = 1.8  10-6 dt 1 d[HI] d[I 2 ]  Rate =  2 dt dt

d[HI] = 2  1.8  10-6 = 3.6  10-6 M s-1 dt

EXERCISE 1: Hydrogen gas produced nonpolluting product is water vapour when react in O2 due to this reaction has been used for fuel aboard the space shuttle, and may be used by Earth-bound engines in the near future. 2H2(g) + O2(g) 2H2O(g) • Express the rate in terms of changes in [H2], [O2] and [H2O] with time. • When [O2] is decreasing at 0.23 mol L-1 s-1, at what rate is [H2O] increasing? (0.46 mol L-1 s-1)

Exercise 2: Consider the reaction, NO(g) + O2(g) 2NO2(g). Suppose that at a particular time during the reaction nitric oxide (NO) is reacting at the rate of 0.066 M s-1 a) At what rate is NO2 being formed? b) At what rate is molecular oxygen reacting?

Exercise 3: Consider the reaction, N2(g) + 3H2(g)  2NH3(g) Suppose that at a particular moment during the reaction molecular hydrogen is reacting at the rate of 0.074 M s-1 a) At what rate is ammonia being formed? b) At what rate is molecular nitrogen reacting?

11.1 RATE LAW

Objectives: At the end of the lesson the students should be able to: 1. define rate law and write the rate equation 2. define the order of reaction and the rate constant 3. calculate the order with respect to a certain reactant from experimental data 4. determine the overall order of a reaction from experimental data 5. calculate the value and determine the units of rate constants

The Rate Law The rate law expresses the relationship of the rate of a reaction to the rate constant and the concentrations of the reactants raised to some powers. aA + bB

cC + dD

Rate α [reactant]

Rate = k [A]x[B]y reaction order is x with respect to A reaction order is y with respect to B Overall reaction order is (x + y) The exponents x, y, … can be integers, fractions or decimal or negative values. k is called rate constant

Rate Law •

The values of x and y can only be experimentally.

•

Reaction order is usually defined in terms of reactant concentrations.

•

The order of a reactant is not related to the stoichiometric coefficients of the reactants in the balanced chemical equation. F2 (g) + 2ClO2 (g)

2FClO2 (g)

rate = k [F2][ClO2] 1

determined

The units of rate constant, k A

Products

Rate, r = k [A]x i) The reaction is zero order Rate = k [A]0 Rate = k unit k = unit rate = mol L-1 s-1 or M s-1

ii) First order

rate

Rate = k [A]1

k= [A] Unit k =

M s-1 M

= s-1 iii) Second order

Rate = k [A]2 rate k=

[A]2

M s-1 Unit k = M2

= M-1 s-1

Example :

S2O82- + 3I-

2SO42- + I3-

The above reaction is first order with respect to iodide ions and to thiosulphate ions. a) Write the rate of equation for the reaction. b) What is the unit of rate constant, k? Solution : a) Rate = k [S2O82-]1[I-]1 b) Rate = k [S2O82-]1[I-]1 rate k= [S2O82-]1[I-]1

Unit k =

Ms-1 M2

=

M-1 s-1

The order of reaction For reaction

A

Products

Rate = k [A]x i) If x = 0 Rate = k [A]0 Rate = k Rate is not dependent on [A] Therefore this reaction is zero order with respect to A

ii) If x= 1

Rate = k [A]1

Assume [A]i = 1.0M Rate = k (1.0M) If the [A] is doubled from 1.0M to 2.0M, Rate = k (2.0M) = 2k(1.0M) hence Rate = 2k[A] Doubling the [A] will double the rate of reaction. Therefore this reaction is first order with respect to A

iii) If x = 2

Rate = k[A]2

Assume [A]i = 1.0 M Rate = k (1.0 M)2 If the [A] is doubled from 1.0 M to 2.0 M, Rate = k (2.0 M)2 = 4k(1.0 M) hence Rate = 4k[A] Doubling [A], the rate will increase by a factor of 4. Therefore the reaction is second order with respect to A

Example Determining Reaction Order from Rate Law For each of the following reactions, determine the reaction order with respect to each reactant and the overall order from the given rate law. (a)

2NO(g) + O2(g)

2NO2(g); rate = k[NO]2[O2]

The reaction order respect to NO : 2 The reaction order respect to O2 : 1 overall reaction order = 3

(b) CH3CHO (g)

CH4(g) + CO(g);

rate = k[CH3CHO]3/2 Solution: The reaction order with respect to CH3CHO : 3/2 The reaction order (overall) : 3/2 (c) H2O2(aq) + 3I-(aq) + 2H+(aq)

I3-(aq) + 2H2O(l);

rate = k[H2O2][I-] Solution: The reaction of order with respect to H2O2 : 1 The reaction of order with respect to I- : 1 and zero order in H+, while overall order is 2.

Determination of the orders of reaction rate; O2(g) + 2NO(g)

2NO2(g)

Initial Reactant Concentrations (molL-1) Experiment

O2

NO

Initial Rate (M s-1)

1

1.10x10-2

1.30x10-2

3.21x10-3

2

2.20x10-2

1.30x10-2

6.40x10-3

3

1.10x10-2

2.60x10-2

12.8x10-3

4

3.30x10-2

1.30x10-2

9.60x10-3

5

1.10x10-2

3.90x10-2

28.8x10-3

Solution: O2(g) + 2NO(g)

2NO2(g)

rate = k [O2]m[NO]n Compare 2 experiments in which the concentration of one reactant varies and the concentration of the other reactant(s) remains constant.

rate2 rate1 =

k [O2]2m[NO]2n k [O2]1m[NO]1n

6.40x10-3Ms-1

=

3.21x10-3Ms-1

2=

=

[O2]2m

=

[O2]1m

2.20x10-2mol/L

[O2]2 [O2]1

m

1.10x10-2mol/L 2m , m = 1 ;

The reaction is first order with respect to O2 Do a similar calculation for the other reactant(s).

m

To find the order with respect to NO, we compare experiment 3 and 1, in which [O2] is held constant and [NO] is doubled: n m [NO] n k [O ] [NO] Rate 3 23 3 3 = = [NO]1 Rate 1 k [O2]1m [NO]1n n -3 -1 12.8 x 10 Ms 2.60 x 10-2mol/L = -3 -1 3.21 x 10 Ms 1.30 x 10-2mol/L 4 = 2n ; n = 2 The reaction is second order with respect to NO Thus the rate law is : Rate = k[O2][NO]2

Exercise: ClO2(aq) + 2OH- (aq)  products The results of the kinetic studies are given below.

1

[ClO2] M 0.0421

[OH-] M 0.0185

8.21 1 0-3

2

0.0522

0.0185

1.26 1 0-2

3

0.0421

0.0285

1.26 1 0-2

exp

Initial rate, Ms-1

a) Explain what is meant by the order of reaction. b) Reffering to the data determine (i) rate law /rate equation (ii) rate constant, k (iii) the reaction rate if the concentration of both ClO2 and OH- = 0.05 M

Exercise: Write rate law for this equation, A + B  C i)

When [A] is doubled, rate also doubles. But doubling the [B] has no effect on rate. ii) When [A] is increased 3x, rate increases 3x, and increasing of [B] 3x causes the rate to increase 9x. iii) Reducing [A] by half has no effect on the rate, but reducing [B] by half causes the rate to be half the value of the initial rate.

Exercise: Many gaseous reactions occur in a car engine and exhaust system. One of the gas reaction is given below. NO2(g) + CO(g) NO(g) + CO2(g) Rate = k [NO2]m[CO]n Use the following data to determine the individual and overall reaction orders: Experiment

Initial Rate(Ms-1)

Initial [NO2](M)

Initial [CO](M)

1

0.0050

0.10

0.10

2

0.080

0.40

0.10

3

0.0050

0.10

0.20

Integrated Rate Law Objectives: 1.Write the rate law for zero order, 1st order and 2nd order reaction 2. Define half-life. 3. Draw the respective graphs for the different order reactions 4. Solve quantitative problems.

Integrated rate equations Zero Order Reaction A zero order reaction is a reaction independent of the concentration of reactant. A product The rate law is given by rate = k[A]0 rate = k

rate

[A] M

- d[A] = k dt Using calculus, - d[A] = kdt - ∫d [A] = k ∫dt - [A] = kt + c substituting t=0, [A] = [A]0 - [A]0 = k(0)+c c = - [A]0 [A]0 -[A] = kt Unit of k for zero order reaction M s-1

Graphs for zero order reaction

[A]o – [A] = kt

[A]o – [A] y = mx + c

t [A] [A]o

rate

[A] = -k t + [A]o y=m x+c

t

[A]

Half-life (t½) Half life (t½) is the time required for the concentration of a reactant to decrease to half of its initial value. zero order reaction Substituting t = t1/2 , and [A] = [A]0 into the zero order reaction, gives 2

[A]0 - [A] = kt [A]0 – [A]0 = kt1/2 2 Solving for t1/2 gives t1/2 = [A]0 2k

First Order Reactions A first order reaction is a reaction whereby its rate depends on the concentration of reactant raised to the first power.

From the rate law, rate = k[A] To obtain the units of k k = rate [A] unit k = M s-1 M = s-1

Rate = k[A] y = mx + c

Rate Ms-1

[A] ,M

For first order reaction, Rate = k[A] - d[A] =

k[A]

dt - d[A]

t = 0, [A]=[A]0 =

k dt - ln [A]0 = k(0) + c c =  ln[A]0

[A] - d[A] ∫

substituting

= ∫ k dt

[A]

-ln [A] = kt – ln[A]0

- ln [A] = kt + c ln

[A]0 [A]

=

kt

Characteristic graphs of 1st order reaction ln[A]o – ln[A] =kt

ln[A] = - kt +ln[A]o

[A]

[A]0 ln -----[A]

t

=

kt

ln[A] ln[A]o [A]

ln[A]o

t

t

Example The reaction 2A B is first order with respect A with a rate constant of 2.8 x 10-2 s-1 at 800C. How long will it take for A to decrease from 0.88 M to 0.14 M ? [A]0 = 0.88 M ln[A]o – ln[A] = kt [A] = 0.14 M kt = ln[A]0 – ln[A] ln[A]0 – ln[A] = t= k

ln

[A]0 [A] k

ln =

0.88 M 0.14 M

2.8 x

10-2

s-1

= 66 s

Example Decomposition of H2O2 (aq) is first order, given that k = 3.66 x 10-3 s-1 and [H2O2 ]o = 0.882 M, determine a) the time at which [H2O2] = 0.600 M b) the [H2O2 ] after 225 s.

Solution : a)

ln

[H2O2]0

kt

=

[H2O2] ln

0.882 0.600

=

3.66 x 10-3 s-1 x t

ln 1.47 = 3.66 x 10-3 s-1 x t ln 1.47 t= 3.66 x 10-3

= 105.26 s b) ln

[H2O2]0

=

kt

=

3.66 x 10-3s-1x 225 s

[H2O2] 0.882 ln [H2O2] [H2O2] = 0.387 M

Exercise, The conversion of cyclopropane to propene in the gas phase is a first order reaction with a rate constant of 6.7 X 10-4 s-1 at 500°C.

CH2 CH2

CH2

CH3-CH=CH2

a) If the initial concentration of cyclopropane was 0.25 M, what is the concentration after 8.8 minute. (0.18 M) b) How long will it take for the concentration of cyclopropane to decrease from 0.25 M to 0.15 M? (13 min) c) How long will it take to convert 74 percent of the starting material? (33 min)

Half-life of a first-order reaction

The half-life, t½, is the time required for the concentration of a reactant to decrease to half of its initial concentration. t=t½ when [A] = [A]0 2 [A]0 ln [A]0/2 ln2 0.693 t½ = = = k k k What is the half-life of N2O5 if it decomposes with a rate constant of 5.7 x 10-4 s-1? 0.693 t½ = ln2 = = 1200 s = 20 minutes -4 -1 k 5.7 x 10 s How do you know decomposition is first order? units of k (s-1)

First-order reaction A

product

No. of half-lives 1

[A]o = 8 M 4

2

2

3

1

4

1/2 t½ = ln2 k

Example The decomposition of ethane C2H6 to methyl radicals is a first order reaction with a rate constant of 5.36 x 10-4 s-1 at 700o C. C2H6 (g) → 2 CH3 (g) Calculate the half life of the reaction in minutes.

Solution

ln 2 t1/2

= k 0.693 = 5.36 x 10-4 = 1.29 x 103 s =

21.5 min

Problem 2 What is the half-life of a compound if 75% of a given sample of the compound decomposes in 60 min? Assume it is first-order reactions kinetics. (t1/2=30 min)

Problem 3 The decomposition of SO2Cl2 is a first-order reaction. SO2Cl2(g) SO2 (g) + Cl2 (g) i) Write the rate differential equation for the reaction. ii) Calculate the value of rate constant, k at 500 K if 5.00 % SO2Cl2 decomposed in 6.75 min. Ans (7.6X10-3) iii) Specify the half-life for the decomposition reaction. Ans (91.20 )

Second Order Reactions A second order reaction is a reaction which rate depends on the concentration of one reactant raised to the second power or on the concentration of two different reactants each raised to the first power.

Example A product Where

d[A] Rate = dt

= k[A]2

To obtain the units of k rate k

= [A]2

Unit k

=

M/s M2 = M-1 s-1

Using calculus, the following expression can be obtained 1 = [A]

1 [A]0

+ kt

Characteristic graphs for second order reaction

Rate = k [A]2 rate

[A] rate

[A]2

Graphs for second order reaction

1

[A]

[A]

=

1

+ kt

[A]o

t 1/[A] – 1/[A]o

1/[A] M-1

1/[A]o t

t

2nd –order reaction, r = k[A]2 If [A] doubles, r2 = k (2[A])2 = k ( 4 [A]2 ) = 4 k [A]2 =4r R will increase by 4 times if [A] doubles

Half life of a second order reaction 1

1

+ kt

= [A]

[A]0 [A]= [A]o

Substituting t = t1/2 1

2 1

+ kt1/2

= [A]0

[A]0

2 1 t1/2=

k[A]0

Detemination of half-life using graph for second order reaction

[A]0

1 t1/2=

k[A]0

[A]0/2

[A]0/4 [A]0/8

x

2x

4x

t

Example Iodine atoms combine to form molecular iodine in the gaseous phase I (g) + I (g)  I2(g) This reaction is a second order reaction , with the rate constant of 7.0 x 109 M-1 s1

If the initial concentration of iodine was 0.086 M, i) calculate it’s concentration after 2 min. ii) calculate the half life of the reaction if the initial concentration of iodine is 0.06 M and 0.42 M respectively.

Solution : 1 i)

1 =

[A]

+ kt

[A]0

1

1

+ (7.0 x 109 x 2 x 60 )

= [A]

[0.086] = 8.4 x 1011

[A] = 1.190 x 10-12 M ii)

[I2] = 0.06 M 1 t1/2=

1 = 7.0 x 109 x 0.060

k[A]0 =

2.4 x 10-10 s

[I2] = 0.42 M 1 t1/2=

k[A]0 1 = 7.0 x 109 x 0.042

= 3.4 x 10-10 s

Using graph Example, The following results were obtained from an experimental investigation on dissociation of dinitrogen pentoxide at 45oC

N2O5(g)  2 NO2(g) + ½ O2(g)

time, t/min

0

10

20

30

40

50

60

[N2O5] x 10-4 M

176

124

93

71

53

39

29

Plot graph of [N2O5] vs time, determine i)

The order of the reaction

ii)

the rate constant k

Solution :

180 160 140 120

[N2O5] x 100 10-4 /M

80 60 40 20 10

20

30

40 50 60 70

80

Time ( min)

i) Based on the above graph, Time taken for concentration of N2O5 to change from 176 x 10-4 M to 88 x 10-4 M is 20 min Time taken for concentration of N2O5 to change from 88 x 10-4 M to 44 x 10-4 M is also 20 min The half life for the reaction is a constant and does not depend on the initial concentration of N2O5

Thus, the above reaction is first order ii)

ln2 k=

= 0.03 min-1 20 min

Summary of the Kinetics of Zero-Order, First-Order and Second-Order Reactions

Order 0

Rate Law rate = k

1

rate = k [A]

2

[A]2

rate = k

Concentration-Time Equation [A] = [A]0 - kt ln[A] = ln[A]0 - kt 1 1 = + kt [A] [A]0

Half-Life t½ =

[A]0 2k

t½ = ln2 k 1 t½ = k[A]0

1st order A  product r = k [A]1 Unit k = s-1

Zero order A  product r = k [A]0 Unit k = M s-1

r

r

r [A]

[A]

Integrated rate law

[A]0

[A]2

Integrated rate law

ln([A]0 / [A]) = kt

1/[A] – 1/[A]0 = kt

ln[A]

t1/2 = [A]0/2k [A]

r

[A]

Integrated rate law

[A]0 – [A] = kt [A]

2nd order A  product r = k [A]2 Unit k = M-1 s-1

[A]

1/[A]

ln[A]0 1/[A]0

[A]0 - [A]

t

t

t

ln([A]0 / [A])

t

t1/2 = ln2/k

t

t 1/[A] – 1/[A]0

t

t1/2 = 1/k[A]0 t

11.2 Collision Theory Objectives At the end of the lesson the students should be able to: 1. explain reaction rates in terms of collision theory. 2. identify factors affecting the effectiveness of collision. 3. define activation energy. 4. Define and state the characteristics of an activated complex

Collision Theory •

Collision Theory is the theory to explain the rate of chemical reactions. It is based on; Rate 

Number of effective collisions time

1- molecule must collide to react 2- molecules must possess a certain minimum kinetic energy (activation energy) to initiate the chemical reaction.

•

3- molecule must collide in the right orientation in order for the reaction to occur.

•

Only effective collisions cause formation of product; collisions of molecules with Ea and at correct orientation. The activation energy (Ea) is the minimum energy that must be supplied or required by collisions for a reaction to occur.

•

The activation energy (Ea) …..is the minimum energy is required to initiate the chemical reaction.

Importance of Orientation

Orientation is unimportant

Orientation is important

Importance of Orientation

Orientation is important

Transition State Theory • The configuration of the atoms of the colliding species at the time of the collision is called the transition state. • Species formed at transition state is called activated complex.

Characteristics of Activated Complex •

Very unstable i.e. It has a short half-life.

•

Its potential energy is greater than reactants or products.

•

The activated complex and the reactants are in chemical equilibrium.

•

It decomposes to form products or reactants.

Potential energy

Ea Reactant

H

product Progress of reaction

- A reaction profile shows potential energy plotted as a function of the progress of the reaction. - The difference in potential energies between the products and the reactants is H for the reaction. - Reactant molecules must have enough energy to overcome an energy barrier separating products from reactants, Ea.

A Reaction Profile: exothermic reaction Activated complex Transition state

Ea (Forward reaction) Ea (reverse reaction)

H

CO(g) + NO2(g)

CO2(g) + NO(g)

A Reaction Profile for endothermic process activated complex. Ea (reverse reaction) Ea (forward reaction)

Product H Reactant

A Reaction profile: endothermic reaction

Ea H

2NOCl

→ 2NO + Cl2

Example: 1. For the reaction A + B C + D , the enthalphy change of the forward reaction is + 21 kJ/mol. The activation energy of the forward reaction is 84 kJ/mol. a) What is activation energy of the reverse reaction? b) Sketch the reaction profile of this reaction 2. Draw a potential energy diagram for an exorthermic reaction. Indicate on the drawing: a) Potential energy of the reactants and the products b) The activation energy for the forward and the reverse reaction c) The heat of the reaction

Factors affecting rate of reaction • CONCENTRATIONS OF REACTANTS: Reaction rates generally increase as the concentrations of the reactants are increased.

• TEMPERATURE: Reaction rates generally increase rapidly as the temperature is increased. • CATALYSTS: Catalysts speed up reactions. • PARTICLE SIZE: The rate increases as the smaller the size of reacting particles .

A) CONCENTRATIONS OF REACTANTS Reaction rate



collision time

• The frequency of collision increases increases with the concentration 4 particle system (2 and 2)  4 collision

A) CONCENTRATIONS OF REACTANTS • A concentration of reactants increases, the frequency of collision increases. • This would also result in the increase in the quantity of effective collision. Thus the reaction rate increases. 5 particle system (3 and 2)  6 collision

A) CONCENTRATIONS OF REACTANTS • This observation correlates with the RATE LAW that has been previously discussed… x

y

Reaction rate = k [ A ] [ B ] … (A & B = reactants) (x & y = rate order)

• Based on this equation, Reaction rate  concentration of reactants (depending on its rate order) • REMINDER! Only in zero order reactions, the rate of reaction is not dependant upon the concentration of the reactants. Reaction rate = k [ A ]0 = k (constant)

B) TEMPERATURE

• As temperature increases, kinetic energy, of molecules increases • So, more collisions occur in a given time • Furthermore, the higher the kinetic energy, the higher the energy of the effective collisions. • So more molecules will have energy greater than activation energy, Ea • Thus, the rate of reaction increases

Distribution of Kinetic Energies of Molecules

Represent total number of molecules with kinetic energy greater than Ea

B) TEMPERATURE

ARRHENIUS EQUATION • In 1889, Svante Arrhenius proposed the following mathematical expression for the effect of temperature on the rate constant, k:

k = A

-Ea∕ e RT

Where… k = rate constant A = constant known as the collision frequency factor e = natural log exponent Ea = activation energy for the reaction R = universal gas constant (8.314 J mol-1 K-1) T = absolute temperature

B) TEMPERATURE

ARRHENIUS EQUATION • The relation ship between the rate constant, k and temperature can be seen in the k vs T graph:

k = A

-Ea∕ e RT

1/T (K-1)

B) TEMPERATURE

ARRHENIUS EQUATION - DERIVATION • The relationship between k and T is clearer when we further derive the Arrhenius Equation… k  Ae Natural log both ends…

 Ea

ln k  ln( A .e

RT

 Ea

RT

)

 Ea

RT ln k  ln A  ln( e )  Ea ln k  ln e  ln A RT

Thus… See the linear relationship…?

 Ea 1 ln k  ( )  ln A R T y = m x + C

(But ln e = 1)

B) TEMPERATURE

Graph Representation Of The Arrhenius Equation • Plotting a ln k vs 1/T graph would show a clearer relationship between k (Rate constant) and temperature  Ea 1 ln k  ( )  ln A R T

Where, Ea = Activation Energy R = 8.314 Jmol-1K-1 T = Absolute Temp A = Collision freq. factor

B) TEMPERATURE • If the value of A (collision frequency factor) is not known and the same reaction conducted at two different temperatures.The Arrhenius equation at each temperature can be written and combined to formed the equation shown in the box.  Ea 1 ln k1  ( )  ln A R T1

and

ln k 2 

 Ea 1 ( )  ln A R T2

Rearranging the equations would give…

Ea 1 ln k 1  ( )  ln A R T1 Since “A” is a constant… E 1 E 1 ln k 1  a ( )  ln k 2  a ( ) R T1 R T2

Ea 1 Ea 1 ln k1  ln k 2  ( ) ( ) R T2 R T1

Ea 1 ln k 2  ( )  ln A R T2

k1 Ea 1 1 ln  (  ) k2 R T2 T1

Exercise: the Activation energy The decomposition of hydrogen iodide,

2 HI (g)  H2(g) + I2(g) has rate constants of 9.51 x 10-9 L mol-1 s-1 at 500 K and 1.10x10-5 L mol-1 s-1 at 600 K. Find Ea. DATA: k1 = 9.51 x 10-9 L mol-1 s-1

k2 = 1.10 x 10-5 L mol-1 s-1

T1 = 500K T1 = 600K

SOLUTION:

k1 Ea 1 1 ln  (  ) k2 R T2 T1 E a   R ln(

k1 1 1 )(  ) 1 k 2 T 2 T1

1.10  10 5 1 1 1 Ea  (8.314 ) ln( )(  ) 9 9.51 10 600 500 Ea = 1.76 x 105 J/mol = 176 kJ/mol

C) CATALYST

• A catalyst is a substance that increases the rate of a chemical reaction without itself being consumed. • Addition of a catalyst increases the reaction rate by increasing the frequency of effective collision. That is by – Decreasing the Ea, and – Correct orientation

C) CATALYST

• Addition of a catalyst changes the value of k (rate constant) . x

Reaction rate = k [ A ] [ B ]

y

(A & B = reactants) (x & y = rate order)

• The catalyst reacts by reducing the Ea and increasing A, thus increasing the k.

 Ea 1 ln k  ( )  ln A R T

C) CATALYST

• When Ea decreases, k increases, REACTION RATE increases

rateuncatalyzed < ratecatalyzed

uncatalyzed

catalyzed

Ea > E’a

Reaction pathway

D) PARTICLE SIZE The smaller the size of reacting particles, the greater is the total surface area exposed for reaction and consequently the faster the reaction. In the case of heterogeneous systems, in which the reactants are in different phases, the area of contact between the reacting substances will influence the reaction rate considerably.

Introduction To Organic Chemistry

Lecture 1 12.1 Introduction Learning Outcomes: At the end of the lesson the students should be able to : 1. List the elements that made up organic compounds C, H, O, N, P, S and halogens. 2. State the ability of carbon to form 4 covalent bonds with other carbons or elements. 3. Differentiate between saturated and unsaturated organic compounds. 4. Give examples of organic compounds used in medicine, engineering, biotechnology and agriculture.

WHAT IS ORGANIC CHEMISTRY? Organic chemistry is the chemistry of carbon compounds. Organic compounds contain H as well as C, while other common elements are O, N, the halogens, S and P. There are many varieties of organic compounds ( more than 10 millions!!!) They may exist as simple or complex molecules; as gases, liquids or solid and coloured or colourless.

Examples :CH4 methane (a component of natural gas)

OCOCH3 COOH

methyl salicylic acid (aspirin-a drug)

O CH2

C

S

NH O

N COOH

penicillin (an antibiotic)

Cl

CH

Cl

CCl3

dichlorodiphenyltrichloroetane (DDT- a pesticide component)

All organic compounds consist of carbon atom. Properties of carbon atom: -has 4 valence electrons. -can form 4 covalent bonds. C C

Single bond

C

C

Double bond

C

C

Triple bond

Hydrocarbons

saturated Contains only single bonds ( -C-C- ) Examples: alkanes, cycloalkanes

unsaturated Contains at least one carbon-carbon double bond (-C=C-) or triple bond (-C C-). Examples: alkenes, alkynes.

Uses of organic compounds Medicine

Engineering

Biotechnology Agriculture

Antibiotics are used to fight bacterial and fungal infections Gasoline-as a fuel for internal combustion engines. Genetic information like DNA DDT-as insectisides to kill harmful insects.

Lecture 2: 12.2 Molecular and Structural Formulae Learning Outcomes: At the end of the lesson the students should be able to : Define structural formula. Draw structural formula in the form of expanded, condensed and skeletal structures based on the molecular formula. Explain primary (1°), secondary (2°), tertiary (3°) and quaternary (4°) carbon.

Structural formula shows how the atoms in a molecule are bonded to each other. 3 types of structural formula: • condensed structure • expanded structure • skeletal structure

2- Dimensional formula

Condensed Structure Does not show single bonds between carbon and hydrogen atoms, but double and triple bonds are shown. All atoms that are attached to a carbon are written immediately after that carbon. C4H9Cl

CH3CHCH2CH3 (Condensed structure)

Examples: ii) Cyclohexane, C6H12

H2C

H2 C CH2

H2C C H2

CH2

iii) Aldehyde, CH3CHO O CH3CH

Expanded Structure Expanded structures indicate how atoms are attached to each other but are not representations of the actual shapes of the molecules. C4H9Cl Molecular Formula

H

H

H

H H

C

C

C

C

H

Cl

H

H

H

Expanded structure

Examples: i) Alcohol (C2H6O) H

H

H

C

C

H

H

OH

ii) Carboxylic acid (C3H6O2 ) H

H

H

O

C

C

C

H

H

OH

Skeletal Structure Shows only the carbon skeleton. Hydrogen atoms are not written. Other atoms such as O, Cl, N etc. are shown. i)

CH3CH(Cl)CH2CH3

= Cl

ii)

H2C

CH2

H2C

CH2

=

3- Dimensional formula ( wedge – dashed wedge – line formula )

Describes how the atoms of a molecule are arranged in space.

Example : Bromoethane Br H

H

C H

H

Br

H

H

C

C

C

H

H

or

H

H

Br

or

H

Indication ::bonds that lie in the plane :bonds that lie behind the plane :bonds that project out of the plane

Br

H

Classification of C atoms: A carbon atom can be classified as primary carbon(1o) →bonded to 1 C secondary carbon(2o) → bonded to 2 C tertiary carbon(3o) → bonded to 3 C quarternary carbon(4o) → bonded to 4 C

H H

C

H

CH3 10 carbon

10 carbon

CH3 H

C CH3

30 carbon

CH3

1

H H CH3

1

CH3 1

1

H C C C CH2 C CH3 H H H

CH3 1

H H CH3

CH3

2

H C C C CH2 C CH3 2

H H H

CH3

H H CH3 3

CH3 4

H C C C CH2 C CH3 H H H

CH3

Question Expanded Structure

Condensed Structure

Skeletal Structure

CH3(CH2)CCl(CH3)2 O

H

H H

CH3 C C CH H H

CH3

12.3 FUNCTIONAL GROUPS AND HOMOLOGOUS SERIES

Lecture 3 Functional Group and Homologous Series Learning Outcomes: At the end of the lesson the students should be able to : • Define functional group. • Name functional groups and classify organic compounds according to their functional groups. • Define homologous series and explain general characteristics of its members.

Functional group is an atom or group of atoms in an organic molecule which characterised the molecule and enables it to react in specific ways which determines its chemical properties.

Functional groups are important for three reasons: i.

ii. iii.

A basic by which organic compounds are divided into different classes. A basic for naming organic compounds A particular functional group will always undergo similar types of chemical reactions.

Homologous Series is series of compounds where each member differs from the next member by a constant – CH2 unit Members of the same homologous series are called homologs.

Homologs Features 1.

Obey a general formula:

Examples: • Alkane: CnH2n+2 • Alkene: CnH2n • Alcohol : CnH2n+1OH 2. 3.

Differ from the successive homolog by a CH2 unit Show a gradual change in the physical properties

4. 5. 6.

Have same functional group Have similar chemical properties Can be prepared by similar general methods

Classification of organic compound

Class of Compound

Functional Group Example Structure

Alkane

Alkene

Alkyne

Name CH3-CH3

-C=C-

carbon-carbon double bond CH3CH=CH2

-C C-

carbon-carbon triple bond CH3C CCH3

Aromatic

Haloalkane

Benzene ring

X (F, Cl, Br, I) Halogen

Alcohol

-OH

Hydroxyl

Phenol

-OH

Hydroxyl

-C-O-C-

Alkoxide

Ether

-CH3

CH3Cl

CH3-OH

-OH

CH3-O-CH3

Aldehyde

-C=O H

Ketone

R-C=O R

Carboxylic acid

-C=O OH

Ester

-C-O-CO

Acyl chloride

-C=O Cl

Carbonyl

CH3-C=O H

Carbonyl

CH3-C=O CH3

Carboxyl

CH3-C=O OH

Carboalkoxy

CH3-C=O OCH3 CH3-C=O Cl

Anhydride

O O -C-O-C-

Amide

-C=O N-

Amine

-NH2

Nitrile

-C N

O O CH3C-O-CCH3 Carboxamide CH3-C=O NH2 Amino Cyano group

CH3-NH2 CH3C

N

Exercises: 1. Identify the functional group in the following molecules a)(CH3)3CCH2CH=CH2 b)(CH3)3CCH=CHCH2-OH

c)

O

O C

OH C O C

CH3 CH2 OH CH2 CH3

C

C

O

C

NH2 CH C CH2 O

CH3

O

CH

C

O

CH3

NH2

12.4 Isomerism Learning Outcomes: At the end of the lesson the students should be able to : Define isomerism. Explain constitutional isomerism. chain isomers positional isomers functional group isomer

Isomerism

Structural/ Constitutional Isomerism

Chain Isomerism

Positional Isomerism

Stereoisomerism

Functional Group Isomerism

diastreomer enantiomer

cis-trans isomerism

other diastereomers

Isomerism is the existence of different compounds with the same molecular formula but different structural formulae. Different structural formula that have the same molecular formula are called isomers.

1) Constitutional isomers (Structural isomers) • are isomers with the same molecular formula but differ in the order of attachment of atoms. 2) Stereoisomers • are isomers with the same molecular formula but different arrangement of atoms in space

Constitutional isomerism Isomerism resulting from different order of attachment of atoms. Three types a) Chain/skeletal isomerism b) Positional isomerism c) Functional group isomerism •

a) Chain/skeletal isomerism The isomers differ in the carbon skeleton (different carbon chain). They possess the same functional group and belong to the same homologous series. Example:

C5H12 CH3CH2CH2CH2CH3 CH3 CH3CHCH2CH3 CH3

CH3CCH3 CH3

2)Positional isomerism These isomers have a substituent group/ functional group in different positions. Examples • C3H7Cl

CH3CH2CH2Cl 1-chloropropane

CH3CHCH3 Cl 2-chloropropane

C4H8

CH2=CHCH2CH3

CH3CH=CHCH3

1-butene

2-butene

C8H10 CH3

CH3

CH3 CH3

CH3

1,2-dimethylbenzene 1,3-dimethylbenzene

CH3 1,4-dimethylbenzene

C6H13N NH2 CH3

CH3 H2N

CH3 CH2NH2

NH2

3)Functional group isomerism These isomers have different functional groups and belong to different homologous series with the same general formula. Different classes of compounds that exhibit functional group isomerism :General formula Classes of compounds CnH2n+2O ; n > 1

alcohol and ether

CnH2nO ; n ≥ 3

aldehyde and ketone

CnH2nO2 ; n ≥ 2

carboxylic acid and ester

CnH2n ; n ≥ 3

alkene and cycloalkane

Examples CH3CH2OH C2H6O ethanol

C3H6O

CH3CCH3 O

propanone

C3H6O2

CH3CH2COH O propanoic acid

CH3OCH3 dimethyl ether

CH3CH2CH O

propanal

CH3COCH3 O methyl ethanoate

Exercise: 1. State how many are isomers with the following molecular formulae, identify the type of isomerism and draw the structural formula of the isomers. a) C5H10 b) C5H10O2 c) CH3CH=C(Cl)CH3 d) C4H6Cl2 e) CH3CH2CH(OH)CH(Br)CH2CH3

Lecture 5 12.4 Isomerism Learning Outcomes: At the end of the lesson the students should be able to : • Define stereoisomerism. • Describe cis-trans isomerism due to restricted rotation about C=C bond and CC bond in cyclic compounds • Identify cis-trans isomerism of a given structural formula.

Stereoisomerism / optical isomerism : Isomerism that resulting from different spatial arrangement of atoms in molecules. Two subdivisions of stereoisomers: i) Enantiomers (mirror image) ii) Diastereomers (non-mirror image)

Diastereomer Cis-Trans Isomerism  The requirements for geometric isomerism : i) restricted rotation about a C=C,double bond in alkenes, or a C-C single bond in cyclic compounds. ii) each carbon atom of a site of restricted rotation has two different groups attached to it.

Examples H

CH3 C

C

H3 C

H

H

C

H

H

cis-2-butene

CH3 H H

CH 3 CH 3

CH3 C

trans-2-butene

H

H3C

CH3

cis-1,2-dimethylcyclohexane trans-1,2-dimethylcyclohexane

If one of the doubly bonded carbons has 2 identical groups, geometric isomerism is not possible. Example

H3C

CH3 C

H3C

C

H No cis – trans isomer

Lecture 6 12.4 Isomerism Learning Outcomes: • • • • • • •

At the end of the lesson the students should be able to : Identify cis-trans isomerism of a given structural formula. Define chirality centre and enantiomers. Identify chirality centre in a molecule. Explain optical activity of a compound. Draw a pair of enantiomers using 3-dimensional formula. Define racemate. State the applications of chiral compounds in daily life.

Enantiomer Optical Isomerism • Optically active compounds have the ability to rotate plane-polarized light to the right (dextrorotary) and to the left (levorotary) • The angle of rotation can be measured with an instrument called polarimeter.

Polarimeter

The requirements for optical isomerism :i) molecule contains a chiral carbon or chirality centre or stereogenic centre (a sp3-hybridized carbon atom with 4 different groups attached to it) P Q

C*

S

PQRS *designates chiral centre

R ii) molecule is not superimposable with its mirror image.



Enantiomers a pair of mirror-image that are not superimposable. Example:i) 2-butanol ,

CH3CHCH2CH3 OH

H3C

CH2CH3

CH2CH3

C*

C

H OH

H

enantiomers

CH3 OH

ii) 2-hydroxypropanoic acid,

CH3CHCOOH OH

COOH H

OH

COOH HO

H CH3

CH3

enantiomers

12.4.9 Racemate A racemic mixture or racemate is an equimolar mixture of enantiomers which is optically inactive because the two components rotate plane-polarized light equally (same degree of rotation) but in opposite directions. Hence it does not give a net rotation of planepolarized light.

Applications of chiral compounds in daily life.

e.g: 



() Dopa is used for treatment of Parkinson’s disease but (+) dopa is toxic to human. (S)-Ibuprofen the popular analgesic(the active ingredient in motrin, advil, and many other nonaspirin analgesics)

REACTIONS OF ORGANIC COMPOUNDS

Lecture 7 12.5 Reactions of Organic Compounds Learning Outcomes: At the end of the lesson the students should be able to : Explain covalent bond cleavage: homolytic heterolytic

Types of Covalent Bond Cleavage/Fission  All chemical reactions involved bond breaking and bond making.  Two types of covalent bond cleavage : Homolytic cleavage  Heterolytic cleavage

a)

Homolytic Cleavage Occurs in a non-polar bond involving two atoms of similar electronegativity. A single bond breaks symmetrically into two equal parts, leaving each atom with one unpaired electron. Formed free radicals.

Example:

X : X

X + X free radicals

X X

b) Heterolytic cleavage •

Occurs in a polar bond involving unequal sharing of electron pair between two atoms of different electronegativities.

•

A single bond breaks unsymmetrically.

•

Both the bonding electrons are transferred to the more electronegative atom.

•

Formed cation and anion.

A:B

A:+ B+ anion cation A is more electronegative. A+ + B:cation anion B is more electronegative.

Reaction Intermediates a) Carbocation • b) Carbanion • c) Free Radical They are unstable and highly reactive. •

a)

Carbocation Also called carbonium ion. A very reactive species with a positive charge on a carbon atom. Carbocation is formed in heterolytic cleavage.

Example :   (CH3)3C — Cl

(CH3)3C+ + Clcarbocation anion

Chlorine is more electronegative than carbon and the C—Cl bond is polar. The C—Cl bond breaks heterolitically and both the bonding electrons are transferred to chlorine atom to form anion and carbocation.

b) Carbanion

 is

an anion counterpart  a species with a negative charge on a carbon atom.  Carbanion is formed in heterolytic cleavage.

example:   • (CH3)3C — Li Li+ kation

(CH3)3C- + carbanion

b)

Free Radical

A very reactive species with an unpaired electron. Formed in homolytic cleavage. Examples: i)

free radicals

Cl – Cl

uv

Cl● + Cl ●

ii)

C

●C

C

+

●C

iii)

H3C

H

H3C ● + ●H

Lecture 7 12.5 Reactions of Organic Compounds Learning Outcomes: At the end of the lesson the students should be able to:  State the relative stabilities of primary, secondary and tertiary free radicals, carbocations and carbanions.  Explain the inductive effect of alkyl group towards the stability of carbocations and carbanions.  Define electrophile and nucleophile.

Relative Stabilities of Carbocations, Carbanions and Free Radicals

Carbocation, carbanion and free radical can be classified into:  Primary  Secondary  Tertiary

Carbocation Stability The alkyl groups (electron-releasing group) stabilise the positive charge on the carbocation. The stability of carbocation increases with the number of alkyl groups present.

Carbocation Stability: H H R R H C H < H C R
Increasing stability

Carbanion Stability o

Alkyl group and other electron-donating groups destabilise carbanions.

o

Electron withdrawing group (e.g: halogen) stabilise carbanions through the inductive withdrawal of electron density

Carbanion Stability: H H R R < R C R H C H < H C < H R R C Methyl anion

Primary 10

Secondary 20


Tertiary 30

Free-radical stability The stability of free radical increases as more alkyl groups are attached to the carbon atom with unpaired electron.

Free Radical Stability : R H

H H R H C H C H < < . . methyl radical

Primary 10

C .

R R
Secondary 20


Tertiary 30

Reagents and sites of organic reactions A) Electrophile  Means ‘electron loving’.  An electron-deficient species and accepting electron from an attacking nucleophile.  Can be either neutral or positively charged

Examples of electrophiles : • cations such as H+, H3O+, NO2+ etc. • carbocations. • Lewis acids such as AlCl3, FeCl3, BF3 etc. • oxidizing agents such as Cl2, Br2 and etc.

electrophilic sites are molecules with low electron density around a polar bond Examples: ii) i) + + -C = O (carbonyl) ; -C – X (haloalkanes) iii)

+ -C – OH (hydroxy compounds)

b) Nucleophile means ‘nucleus loving’ An electron-rich species and electron-pair donor. A nucleophile can be either neutral or negatively charged.

Examples of nucleophiles :  anions

such as OH-, RO-, Cl-, CN- etc.

 carbanions

( species with a negative charge on carbon atoms ).

Lecture 7 12.5 Reactions of Organic Compounds Learning Outcomes: At the end of the lesson the students should be able to explain the main types of organic reactions: • addition: electrophilic and nucleophilic • substitution: electrophilic, nucleophilic and free radical • elimination • rearrangement

4 Types of Organic Reactions    

Addition Substitution Elimination Rearrangement

I) Addition Reaction  A reaction in which atoms or groups add to adjacent atoms of a multiple bond.  Two types of addition :a) b)

Electrophilic Addition Nucleophilic Addition

a)Electrophilic Addition • Initiated by an electrophile accepting electron from an attacking nucleophile. • Typical reaction of unsaturated compounds such as alkenes and alkynes. Example :

CH3CH=CH2 + Br2

CCl4

CH3CHBrCH2Br

Room temperature

electrophile

b)

Nucleophilic Addition

Initiated by a nucleophile, which attacks an electrophilic site of a molecule. • Typical reaction of carbonyl compounds.

O CH3

CN 

C CH3 + HCN 

CH3 C OH

CH3

II) Substitution Reaction  A reaction in which an atom or group in a molecule is replaced by another atom or group.  Three types of substitution :a) free radical substitution. b) electrophilic substitution. c) nuclephilic substitution.

a) Free-radical Substitution  Substitution which involves free radicals as intermediate species. Example :

CH3CH3 + Cl2

uv light

CH3CH2Cl + HCl

b)

Electrophilic Substitution Typical reaction of aromatic compounds.

• The aromatic nucleus has high electron density, thus it is nucleophilic and is prone to electrophilic attack. Example:

+ Br2 electrophile

Fe catalyst

Br + HBr

c)

Nucleophilic Substitution Typical reaction of saturated organic compounds bearing polar bond as functional group, such as haloalkane with alcohol.

Example :   CH3CH2Br + OH-(aq) nucleophile

Δ

CH3CH2OH + Br-(aq)

•

III) Elimination Reaction An atoms or groups are removed from adjacent

carbon atoms of a molecule to form a multiple bond (double or triple bond). • Results in the formation of unsaturated molecules. • Example : CH3CH2OH

conc. H2SO4 CH2= CH2 + H2O 

IV)

Rearrangement Reaction

 A reaction in which atoms or groups in a molecule change position.  Occurs when a single reactant reorganizes the bonds and atoms. Example :

H

H

C

tautomerisme C R H C

H

OH

H

C R O

Exercises 1.

Explain how the free radicals are formed in homolytic cleavage.

2. Write an equation for the brominebromine bond cleavage in the bromination of methane. State the type of bond cleavage.

3.

Which would you expect to be the most stable free radical ?

CH2CH3 , (CH3)2 CH , CH3 ,

 CH3

Chapter 12.0 HYDROCARBONS

12.1 : ALKANES Hydrocarbon are compounds which contain only carbon and hydrogen atoms.

HIDROCARBONS

AROMATIC (contain one or more benzene ring)

ALIPHATIC

ALKANES (saturated) contain only single bond

ALKENES (unsaturated) contain C=C

CYCLO-ALKANES (saturated) Alkanes which C atoms are join in rings

CYCLO-ALKENES (unsaturated)

ALKYNES (unsaturated) contain C≡C

• Saturated hydrocarbons - compound with C-C - Example : alkanes and cycloalkanes – Contain the maximum number of hydrogen atoms that the carbon compound can possess.

• Unsaturated hydrocarbons -compounds with multiple bonds -Example : alkenes, cycloalkenes, alkynes and aromatic hydrocarbons – They posses fewer than the maximum number of hydrogen atoms.

ALKANES

Alkanes are known as saturated hydrocarbon which contain only single covalent bonds. General formula for straight chain of alkanes is CnH2n+2 where n ≥ 1 General formula for cycloalkanes is CnH2n where n ≥ 3

Each carbon atom in alkanes is - sp3 hybridised - tetrahedral with four sigma bond (formed by the four sp3 hybrid orbitals.) - all bond angles are close to 109.5o Alkanes IUPAC names have the –ane suffix.

The First Ten Unbranched Alkanes Molecular formula CH4 C2H6 C3H8 C4H10 C5H12 C6H14 C7H16 C8H18 C9H20 C10H22

Structural formula CH4 CH3—CH3 CH3—CH2—CH3 CH3—(CH2)2—CH3 CH3—(CH2)3—CH3 CH3—(CH2)4—CH3 CH3—(CH2)5—CH3 CH3—(CH2)6—CH3 CH3—(CH2)7—CH3 CH3—(CH2)8—CH3

No .of C atoms 1 2 3 4 5 6 7 8 9 10

Name Methane Ethane Propane Butane Pentane Hexane Heptane Octane Nonane Decane

• Starting from C4H10 onwards, the alkanes show the phenomenon of chain isomerism. • They can exist as linear or branched alkanes.

Example C4H10 2-

isobutane

CH3(CH2)2CH3

CH3CH(CH3)2

C5H12

IUPAC NOMENCLATURE  IUPAC – International Union of Pure and Applied Chemistry  Branched - chain alkanes are named according to the following rules:

Choose the longest continuous chain of carbon atoms; this chain determines the parent name for alkanes. Examples: CH3CH2CH2CH2CHCH3 | CH3 Parent name: hexane

CH3CH2CH2CH2CHCH3 CH2 CH3 Parent name : heptane

Number the longest chain beginning with the end of the chain nearer the substituent. Example: 6

5

4

3

2

1

CH3CH2CH2CH2CHCH3 CH3

substituent

7

6

5

4

3

CH3CH2CH2CH2CH CH3 | 2 CH2 | 1 CH3

substituent

Use rule number 2 to locate the position of the substituent. 

The position and the name of the substituent must be written in front of the parent chain.

Examples: 6

5

4

3

2

1

CH3CH2CH2CH2CHCH3 CH3 2-methylhexane

Substituent -methyl at C-2

Substituent 7

6

5

4

3

-methyl at C-3

CH3CH2CH2CH2CH CH3 | 2 CH 2 | 1 CH 3 3-methylheptane

Some Common Substituent Groups Alkane methane ethane propane

name methyl ethyl propyl isopropyl

butane

butyl

substituent

CH3 CH2CH3 CH2CH2CH3 CHCH3 CH3 CH2CH2CH2CH3

isobutyl

CH2CHCH3 CH3

sec-butyl

tert-butyl

CHCH2CH3 CH3 CH3 CCH3

neopentyl

CH3 CH3 CH2CCH3 CH3

cyclopropyl cyclobutyl

phenyl C6H5 or benzyl CH2

name

substituent

Bromo

-Br

Chloro

-Cl

Flouro

-F

Iodo

-I

Hydroxyl

-OH

Amino

-NH2

Cyano

-CN

Nitro

-NO2

If two or more substituents are present, give each substituent a number corresponding to its location on the longest chain. 



the substituent should be listed alphabetically. In alphabetizing, the prefixes di, tri, tetra, sec-, tert- are ignored except iso and neo.

Example: 1

2

3

4

5

6

CH3CHCH2CHCH2CH3 | | CH3 CH2 | CH3 4-ethyl-2-methylhexane

If two substituents are present on the same carbon atom, use that number twice Example: CH3 |3 4 5 6 CH3CH2CCH2CH2CH3 | 2 CH 2 | 1 CH 3 3-ethyl-3-methylhexane

If two or more identical substituents are present, use prefixes di-(2 identical substituents),tri-(3 identical substituents), tetra-(4 identical substituents). Commas are used to separate numbers from each other. Example: CH3CH―CH CH3 | | CH3 CH3

2,3-dimethylbutane

If there are two chains of equal length as the parent chain, choose the chain with the greater number of substituents.

7

6

5

4 4

3 3

2 2

1 1

CH3CH2-CH CH CH CHCH3 | | | | CH3 5 CH2 CH3 CH3 | 6 CH2 | 7 CH3 2,3,5-trimethyl-4-propylheptane (four substituents) 4-sec-butyl-2,3-dimethylheptane (three substituents)

If branching occurs at an equal distance from either end of the longest chain, choose the name that gives the lower number at the first point of difference. 6

5

4

3

2

1

CH3CHCH2CH CHCH3 1 2 3 4 5 6 | | | CH3 CH3 CH3 2,3,5-trimethylhexane (NOT 2,4,5-trimethylhexane)

CYCLOALKANES Cycloalkanes – alkanes which carbon atoms are joined in rings. Cycloalkanes are known as saturated hydrocarbon, because it has the maximum number of bonded hydrogen ( only has single bonds). General formula: CnH2n where n = 3, 4, 5, ……

NOMENCLATURE OF CYCLOALKANES Cycloalkanes with only one ring are named with the prefix cyclo- to the names of the alkanes (contain the same number of carbon atoms)

Examples: Monocyclic compounds C3H6

cyclopropane

C4H8

cyclobutane

C5H10

cyclopentane

If only one substituent is present, it is not necessary to designate its position. Examples: Chlorocyclopropane Cl

CH3

Methylcyclohexane

If two substituents are present, number carbon in the ring beginning with the substituent according to the alphabetical order

and number in the direction that gives the next substituent the lowest number possible.

Examples: CH3 3

2

4

1 5

CH2CH3

6

1-ethyl-2-methylcyclohexane NOT 1-ethyl-6-methylcyclohexane

2

Cl

3

1

4

Cl

5

1,3-dichlorocyclopentane (NOT 1,5-dichlorocyclopentane)

When three or more substituents are present, begin at the carbon with substituent that leads to the lowest set of locants.

Locants

Example: 2

Cl

1

3

3 2

4

4 1

5 6

CH2CH3

6

CH3

chloro 1

4

ethyl 3

2

methyl 4

1

5

1-chloro-3-ethyl-4-methylcyclohexane 4- chloro-2-ethyl-1-methylcyclohexane

H3C

CH2CH3

CH3 1-ethyl-1,3-dimethylcyclopentane (NOT 3-ethyl-1,3-dimethylcyclopentane)

When a single ring system is attached to a single chain with a greater number of carbon atoms

or when more than one ring system is attached to a single chain, then it is appropriate to name the compounds as cycloalkylalkane.

Number of C at linear chain

Number of C at ring

1

3 2

1,3-dicyclohexylpropane

Examples:

CH2CH2CH2CH2CH3

1-cyclobutylpentane

Physical properties of alkanes & cycloalkanes

Boiling Point

Physical state

Solubility



Physical state At room temperature (25oC) and atmospheric pressure (1 atm), for unbranched alkanes, C1 – C4 : gases C5 – C17 : liquids C18 - more : solids





Boiling points The boiling points of the straight alkanes show a regular increase with increasing molecular weight. Branching of the alkanes chain, lower the boiling point.

Mr ↑ boiling point ↑





C-H is non polar bond  Intermolecular forces exist – London dispersion The London dispersion forces increase :  As molecular weight increases,  Molecular size increases  Molecular surface area increase  Therefore,

more energy is required to separate molecules from one another  Result - a higher boiling point.

Mr ↑ boiling point ↑



Chain branching :  makes a molecule more compact  Surface area reduces  The strength of the London dispersion forces reduce  Lower boiling points.

Isomeric Alkanes Have different boiling point due to branching Pentane

2-methylbutane

2,2dimethylpropane

CH3CH2CH2CH2CH3

CH3CHCH2CH CH3

CH3 CH3CCH3 CH3 9oC

37oC

28.5oC

For example, butane and 2-methylpropane both have a molecular formula C4H10, but the atoms are arranged differently CH3CH2CH2CH3

butane

CH3CH2CH2CH3

CH3CHCH3

CH3CHCH3

2-methylpropane

CH3 CH3 In butane the C atoms are arranged in a single chain, but 2-methylpropane is a shorter chain with a branch

Name

Molecular structure

Boiling point

Butane has a higher boiling point because the dispersion forces are greater. The molecules are longer (and so set up bigger temporary dipoles) and can lie closer together than the shorter, fatter 2-methylpropane molecules.

Cycloalkanes The boiling points of cycloalkanes are 10oC to 15oC higher than the corresponding straight chain alkanes. Cycloalkane Boiling point

alkane

Boiling point

Cyclobutane

13oC

Butane

-0.5oC

Cyclopentane

49oC

Pentane

36..3oC

Solubility 

Alkanes – less dense than water



Alkanes and cycloalkanes are almost totally insoluble in water  

They are non-polar molecule Unable to form hydrogen bond with H2O



Liquid alkanes and cycloalkanes are soluble in one another, and they generally dissolve in nonpolar solvents.



Good solvents for alkanes are benzene, carbon tetrachloride, chloroform, and other hydrocarbons.

Nature sources of alkanes; Natural gas oil Natural gas contains primarily methane and ethane, with some propane and butane Oil is a mixture of liquid alkanes and other hydrocarbons

CHEMICAL REACTION OF ALKANES Non-reactivity of alkanes 

Alkanes are generally inert towards many chemical reagents (bases, acids, dehydrating agents and aqueous oxidizing agents)



C and H has nearly the same electronegativity, the C-H bonds of alkanes are only slightly polarised (non polar)

Reaction of alkanes  

Combustion Halogenation

Combustion of alkanes 

Alkanes burnt in air ( oxygen ) to give carbon dioxide gas, water and heat.

y y CxHy+ x  O2 → x CO2 + H2O 4 2 + Heat

Examples: Combustion of alkane in excess oxygen

C4H8+ 3O2 → 4CO2 + + Heat

4H2O

Examples: Combustion of alkane in limited oxygen

C4H8+ 4O2 → 4CO + 4H2O + Heat



Alkanes are unreactive towards polar or ionic reagents but can react with non-polar reagents such as oxygen and bromine.

 The

low reactivity of alkanes toward many reagents explain why alkanes were originally called paraffins. (Latin : parum affinis= low affinity).

Halogenation: a free radical substitution reaction 

Alkanes react with halogen (chlorine & bromine) to produce haloalkanes in the presence of light or temperature >> 100 oC. R–H + X2



hv R–X + HX 

With methane, the reaction produces a mixture of halomethane and a hydrogen halide.

Examples: i.

CH4 + Cl2

hv 

CH3Cl + CH2Cl2 + CHCl3 + CCl4 + HCl

ii.

hv CH3CH3 + Cl2  CH3CH2Cl + HCl

CH3

CH3

| iii. CH3 CCH3 + Cl2 | CH3

| hv CH CCH Cl + HCl  3 2 | CH3

(ii) & (iii): 1 product only because all the hydrogen atoms are identical

CH3CH2CH2Br iv.

CH3CH2CH3

hv + Br2 

( minor) + CH3CHCH3 | Br ( major) + HBr

Reaction mechanism CH4 + Cl2 i.

hv CH3Cl + HCl 

Chain initiation step

Cl – Cl

hv 

2Cl•

ii. Chain propagation steps H H-C

H + •Cl  •CH3 + HCl

H 3HC• + Cl Cl_  CH3Cl + Cl•

iii. Chain termination step. •Cl + •Cl

 Cl2

•Cl + •CH3  CH3Cl •CH3 + •CH3  CH3CH3

Example 

The bromination of 2-methylbutane yields a mixture of isomers. CH2Br

CH3

C H

CH2

CH3

CH3

CH3

CH3

C

CHBr

H

Increasing % yield

CH3

CH3

C Br

CH2

CH3

Exercise 

Chlorination reaction of certain alkanes can be used for laboratory preparations, for example in the preparation of chlorocyclopentane from cyclopentane. Give the mechanism for the reaction.

12.2

ALKENES



General formula CnH2n , n  2.



Functional group  double bond C=C



Unsaturated hydrocarbon



Each carbon atom ( C=C ) is sp2 hybridized.



Restricted rotation of carbon-carbon double bond causes cis-trans isomerism

IUPAC Nomenclature Step 1 Determine the parent name by selecting the longest chain that contains the double bond and change the ending ‘-ane’ in alkane to ‘-ene’.

Step 2 When the chain contains more than three carbon atoms, a number is need to indicate the location of the double bond. The chain is numbered starting from the end closest to the double bond..

1

2

3

CH3CH2CH

4

5

6

7

8

CH CH2CH2CH2CH3

3-octene (not 5-octene)

Step 3 Indicate the position of the substituent by the number of the carbon atoms to which they are attached.

CH3 H3 C

C

CH CH3

1 2 3 4 2-methyl-2-butene

(not 3-methyl-2-butene)

1

2

3

H3C

C

CH CH2 CH CH3

CH3

4

5

6

CH3

2,5-dimethyl-2-hexene (not 2,5-dimethyl-4-hexene)

Step 4

The ending of the alkenes with more than one double bond should be change from ene to  

diene – if there are two double bonds triene – if there are three double bonds

1

2

H2C

3

4

CH CH

CH2

1,3-butadiene 1

H2C

2

3

CHCH

4

5

6

7

CHCH CHCH3

1,3,5-heptatriene MORE

Step 5 In cycloalkenes : Number the carbon atoms with a double bond as 1 and 2, in the direction that gives the substituent encountered first with a small number.

CH3 1 5

2 4

3

1-methylcyclopentene (not 2-methylcyclopentene)

1 6

2

5

3

CH3

4

CH3

3,5-dimethylcyclohexene (not 4,6-dimethylcyclohexene) MORE

Step 6 Two frequently encountered alkenyl groups are vinyl group and allyl group. RCH=CHR alkene CH2=CHvinyl group

-H

RCH=CHalkenyl CH2=CHCH2allyl group

Step 7

When two identical groups are attached: a) on the same side of the double bond, the compound is cis b) on the opposite sides of double bond, it is trans.

cis / trans

Cl C H

Cl C H

cis-1,2-dichloroethene

Cl C H

H C Cl

trans-1,2-dichloroethene

Give IUPAC names for the following alkenes 1)

2)

CH3 CH3

3)

4)

CH2CH3

CH3

5)

CH3CH=CHCH2C(CH3)2CH3

Preparation of alkenes

a) Dehydration of alcohols b) Dehydrohalogenation of alkyl halide



C

C

H

OH

Dehydration of alcohols Alcohols react with strong acids in the presence of heat to form alkenes and water.

H2SO4 (conc.) 

C

C

+

H2 O



Concentrated sulphuric acid (H2SO4) or phosphoric acid (H3PO4) : as acidic catalysts and dehydrating agents.



The major product is the most stable alkene, which has greater number of alkyl group attach to C=C is follow Saytzeff’s rule.

Saytzeff’s Rule An elimination usually gives the most stable alkene product, commonly the most highly substituted alkene.

Examples (1) CH3CH2OH

(2) H3C CH CH3 OH

H2SO4 (conc.) 

H2SO4 (conc.) 

H2C CH2

+

H2C

+

H2O

CH CH3 H2O

Mechanism for the dehydration of alcohol H3C CH CH2CH3 OH

H SO (conc.) 2

4



CH 3 CH=CHCH 3 (major product)

+

CH 2 =CHCH 2 CH 3 (minor product)

+

H 2O

Step 1: Protonation of alcohol. H H3C C CH2CH3 :OH ..

+

H + :O H H

H

H H3C

C +

: OH H

CH 2 CH 3

+

:O ..

H

Step 2: Formation of carbocation H H 3 C C CH 2 CH 3 +:OH H

H3 C

H + C CH2 CH3

carbocation H +

:O ..

H

Step 3: Formation of alkenes H

H H H + H C C C CH3 b

H

+

H

Root a : major product Root b : minor product

a

:O H ..

H

H C

H3C

H +

C CH 3

+ : O

H

H

stable alkenes +

CH 2 =CHCH 2 CH 3 (minor product)

Rearrangement During Dehydration of Alcohol Example 1: CH3 H H3C

C

C

CH3OH

CH3

H2 SO4 (conc.) 

CH3 H 3C

C

C

CH3

CH3 (major product)

+

CH 3 H C C CH CH 2 3 CH 3 (minor product)

Step 1 and Step 2 are similar to the previous example.

Step 3:  Now the rearrangement occurs.  The less stable, secondary carbocation rearrange to form more stable tertiary carbocation.

CH H C 3

CH

3

C

C

+

CH

CH H 3

H C

3

3

rearrangement

C

+

C

3

CH

CH H 3

2o carbocation

3o carbocation

(less stable)

(more stable)

3



The rearrangement occurs through the migration of an alkyl group (methyl) from the carbon atom adjacent to the one with the positive charge.



Because a group migrates from the one carbon to the next, this kind of rearrangement is often called a 1,2 shift.

CH

(a) HC 2

(a)

C

3 CH CH

3

CH

H + CH3 H C C C CH3 H CH H 3

3 (minor product)

H

+

:O H ..

less stable alkene CH HC 3

(b)

(b)

C

3 C CH

3

CH 3 (major product) more stable alkene

The The final final step step can can occur occur in in two two ways: ways:

Path (a): leads to less stable, disubstituted alkene and produces the minor product of the reaction. Path (b): leads to highly stable tetrasubstituted alkene and produces the major product according to the Saytzeff’s rule.

Try this! CH H 3 H C C C CH 3 3 H OH

CH 3 H C C CH CH 3 3 (major product)

H2SO4 (conc) 

+

CH 3 H C C CH CH 2 2 3 (minor product)

(2) Dehydrohalogenation of Alkyl Halides  The elimination of a hydrogen and a halogen from an alkyl halide to form an alkene.

C

C

+

alcohol KOH

reflux

C C

H X 

+

HX

Saytzeff’s rule is used to determine the major product

Examples: (1)

Cl H3C

CH CH2CH2CH3

H3C CH

+

CH CH2CH3

(major product)

KOH

alcohol reflux

+

H2C CH CH2CH2CH3 (minor product)

(2)

H3C CH CH3 Br

+

KOH

alcohol reflux

H2C

CH CH3

(3)

Br + CH CH 3

KOH

alcohol reflux

CH CH 3 (minor product)

+

CH2 CH3 (major product)

(4) CH3

CH3 H C C CH3 CH3 Cl

CH2

CH3 H C C CH3 CH3

2,3-dimethyl-1-butene (minor product) 2 alkyl groups

+ KOH

CH3

alcohol reflux

CH3 C C CH3 CH3

2,3-dimethyl-2-butene ( major product) 4 alkyl groups

Chemical Reaction of Alkenes Comparison of The Reactivity Between Alkanes and Alkenes  





Alkenes are more reactive compared to alkanes. Alkanes have carbon-carbon single bonds (σ bonds) while alkenes have carbon-carbon double bonds (π bonds). The double bond is a site of high electron density (nucleophilic). Therefore most alkenes reactions are electrophilic additions.

Electrophilic Additions Mechanism H

H C

H

H

+

C H

X

Y

slow

H

C X

+

+

C

H

H

Y

-

H H

C X

+

C H

H

+

Y

-

fast

H

H

H

C

C H

X

Y

Example: i.

H3C

CH3 C

H3C

+

C

H

Cl

slow

CH3 CH3CH3 H3C

C H

C

+

CH3

+

-

Cl

CH3CH3 H3C

C H

C

+

CH3

+

Cl-

fast

CH3CH3 H3C

C

C

H Cl

CH3

Addition Reaction of Alkenes (1) Hydrogenation  The reaction of an alkene with hydrogen in the presence of catalyst such as platinum, nickel and palladium to form alkane.

C C

+

H2

Pt or Ni or Pd

C C H H

Examples: (1)

H3C

C

CH2

+

H2

Pt /Ni / Pd

CH3

CH3

(2)

H3C CH CH3

+

H2

Pt /Ni / Pd

(2) Halogenation of Alkenes (i) In inert solvent (CH2Cl2)  Alkenes react rapidly with chlorine or bromine in CH2Cl2 at room temperature to form vicinal dihalides.

C

C

+

X2

CH2Cl2

C

C

X

X

Example: CH3 H3C C C CH3

+

CH2Cl2

Cl 2

H Cl Cl H3C

C

C

CH3H

CH3



When bromine is used for this reaction, it can serve as a test for the presence of carbon-carbon double bonds.



If bromine is added to alkene, the reddish brown color of the bromine disappears almost instantly as long as the alkene is present in excess.

Unsaturation test C C

+

Br2

CH2Cl2 room temperature C C Br Br

Observation: The reddish brown bromine decolourised

(3) Hydrohalogenation : Markovnikov’s 

rule

Hydrogen halides (HI, HBr, HCl and HF) add to the double bond of alkenes to form haloalkanes.

C

C

+

HX

C C H X



The addition of HX to an unsymmetrical alkenes, follows Markovnikov’s rule.

 Markovnikov’s Rule:

 In

the addition of HX to an alkenes, the hydrogen atom adds to the carbon atom of the double bond that already has the greater number of hydrogen atoms.

i.

The addition of HBr to propene, could conceivably lead to either 1-bromopropane or 2-bromopropane. The main product, however, is 2bromopropane

H H H2C CH CH3

+

HBr

H C C CH3 H Br

ii.

When 2-methylpropane reacts with HBr, the main product is tert-butyl bromide, not isobutyl bromide.

Br H3C C CH2 CH3 

+

HBr

H3C C CH3 CH3

The addition of HX to an unsymmetrical alkenes, yield the main product according to the Markovnikov’s rule.

Mechanism CH 3 CH 3

C

CH

CH 3

+

slow H

Br

CH3 CH3

C +

CH

30 Carbocation

CH3

+ Br -

CH3 CH3

C +

CH

CH3

+

Br

fast

-

CH3 CH3

C

CH

Br

H

CH3

(4) Addition of HBr to Alkenes in The Presence of Peroxides (Anti Markovnikov’s rule)  When alkenes are treated with HBr in the presence of peroxides, ROOR (eg: H2O2) the addition occurs in an anti-Markovnikov manner 

The hydrogen atom of HBr attached to the doubly bonded carbon with fewer hydrogen atoms.

Example: H2C CH CH3

+

HBr

ROOR Br CH2CH2CH3

Complete the following reactions:

(1)

H3C

CH2C

+

CH2

HBr

H2 O 2

CH3 (2)

CH2

+

HBr

H2 O2

(ii) In aqueous (Halohydrin Formation) 

If the halogenation of an alkene is carried out in aqueous solution, the major product of the overall reaction is a haloalcohol called a halohydrin.

C C

+

X2

+

H2O

C C X OH

X2 = Cl or Br

halohydrin

Example: H

+ Br2 + H2O

C C H

H H

H H

Unsaturated test

H C C H Br OH 2-bromobutanol

Observation : the reddish brown bromine decolourised 

If the alkenes is unsymmetrical, the halogen ends up on the carbon atom with the greater number of hydrogen atoms.

Example: H 3C

C

CH2

+

Br 2

+

H2O

CH3 CH3H H3 C

C

C

H

OH Br 1-bromo-2-methyl-2-propane

Complete the following reaction:

+ CH3

Br2

+

H2O

(5) Hydration of Alkene 

The acid-catalyzed addition of water to the double bond of an alkene (hydration of an alkene) is a method for the preparation of low molecular weight alcohols.



The acid most commonly used to catalyse the hydration of alkenes are dilute solution of sulphuric acid and phosphoric acid.



The addition of water to the double bond follows Markovnikov’s rule.

+

C C

+

H2O

H3O

C C H OH

Example: H3C

C

CH2

+

H3 O

H2O

+

CH3 OH H3C

C CH3

CH3

Complete the following reactions:

(1)

CH2

+

H2 O

H3 O

+

(2)

H2C

CH CH3

+

H2O

H3O

+

Mechanism STEP 1 CH3 CH3

C

H

.. CH

CH3

+

H

O+

H

CH3 CH3

C +

CH H

CH3

STEP 2 CH3 C +

CH

CH3

+

H2O

..

CH3

..

H

CH3 CH3

CH

O+

H

..

H

C

H

CH3

STEP 3

CH3 CH3

CH

O+

H

..

H

C

H

CH3 CH3

C

CH

OH

H

..

H2O

CH3

..

CH3

+

H3O+

..

(6) Addition of Sulphuric Acid to Alkenes 





Alkenes dissolve in concentrated sulphuric acid to form alkyl hydrogen sulphates. Alkyl hydrogen sulphates can be easily hydrolysed to alcohols by heating them with water. The overall result of the addition of sulphuric acid to alkenes followed by hydrolysis is the Markovnikov addition of -H and -OH.

H3 C

CH3 C

H3 C

C

CH3CH3 conc. H2 SO 4

H3 C

C

CH3

H OSO 3 H alkyl hydrogen sulphate

CH3

CH3CH3 H2O heat

C

H3 C

C

C

H

OH

CH3



Example:

H2C CH CH 3

OSO3H

conc.H2SO4

H3C CH CH2 H

H2O heat

OH H3 C

CH CH2 H

7 ) Oxidation of Alkenes 

Alkenes undergo a number of reactions in which the C=C is oxidized

KMnO4  basic,cold,dilute  acidic,hot,concentrated

ozonolysis

i)

With cold and dilute potassium permanganate, KMnO4



Potassium permanganate in alkaline solution can be used to oxidise alkenes to 1,2-diols (glycols).

-

C

C

+

KMnO4 (purple)

C

C

OH OH

OH ,cold

+

MnO2 (brown precipitate)

Observation: Purple colour of KMnO4 decolourised and brown precipitate formed.



This reaction is called Baeyer’s test.



It is a test for the presence of C=C where the purple colour of the KMnO4 decolourised, and brown precipitate of MnO2 is formed.

Example: -

H2C

CH2

+ H

KMnO4 H

H

C

C

OH OH

OH ,H2 O cold

H

+

MnO2

ii) With hot potassium permanganate solutions to alkenes 

When oxidation of the alkene is carried out in acidic solution of KMnO4, cleavage of the double bond occurs and carbonyl-containing products are obtained.

If the double bond is tetrasubstituted, tetrasubstituted the two carbonyl-containing products are ketones CH3 H3C

C

C

-

CH3

+

(i) OH ,heat

KMnO4

(ii) H3O

CH3 O H3C

C

+

O CH3

+

H3C

C

CH3



If a hydrogen is present at double bond, one of the carbonyl-containing products is a carboxylic acid;



If two hydrogens are present on one carbon, CO2 is formed.

2) -

H2C

CH CH3

+

KMnO4

(i) OH ,heat (ii) H3O

+

O HO

C

CH3

+

CO2

+

H2 O



The oxidative cleavage of alkenes can be used to establish the location of the double bond in an unknown alkene.

Example:  An unknown alkene with the formula C7H14 undergoes oxidation with hot basic potasium permanganate solution to form propanoic acid and butanoic acid. What is the structure of this alkene?

C7 H14

+

KMnO4

O H3C

C

CH2

(i) OH-,heat (ii) H3 O

+

O OH

+

CH2 C H3C CH2 OH

Answer: O H3C

C

CH2

O CH2 C H3C CH2 OH

OH

propanoic acid

butanoic acid

H H H3C

CH2C

C

CH2CH2CH3

3-heptene

Example 

An unknown alkene undergoes oxidation in hot basic KMnO4 followed by acidific to give the following product: O

O

CH3CCH2CH2CH2CH2C OH Deduce the structural formula for the unknown alkene.

iii)

Ozonolysis of Alkenes



A more widely used method for locating the double bond of an alkene is the use of ozone (O3). Ozone reacts vigorously with alkenes to form unstable compounds called molozonides, which rearrange spontaneously to form compounds called ozonides.



O

C

C

+

C

C

O3 O

O

ozonide



Ozonides:  very unstable compounds 

can easily explode violently



they are not usually isolated but are reduced directly by treatment with water and in the presence of zinc and acid (normally acetic acid) to give carbonyl compounds (either aldehydes or ketones).

O

H

C

C

+

Zn

H2 O,H

+

R

O

O

ozonide H

C=O

+

O=C R

Example: CH3 H3C

C

(i) O3 CH CH3

(ii) Zn,H2O/H

CH3 H3C

C

O

+

CH3

+

H

C

O

Exercise 1.

Write the structure of alkene that would produce the following products when treated with ozone followed by water, zinc and acid CH3COCH3 and CH3CH(CH3)CHO

Example 

Deduce the structural formula of an alkene that gives the following compound when it reacts with ozone in the presence of Zn / H+. O=CH-CH2-CH2-CH(CH3)CH=O

2. Acid-catalyzed dehydration of neopentyl alcohol, (CH3)3CCH2OH, yields 2-methyl2-butene as the major product. Outline a mechanism showing all steps in its formation.

Exercise 

Compounds A, B, C and D are isomers with the molecular formula C4H8. A and B give a positive Baeyer test, while C and D do not. A exists as cis- and trans- isomers, while B does not have geometrical isomers. C has only secondary hydrogen, while D has primary, secondary and tertiary hydrogen. Give the IUPAC names for A, B, C, and D.

The origin of August Kekulé’s view of benzene structure: “There I sat and wrote my textbook, but things did not go well; my mind was occupied with other matters, I turned the chair towards the fireplace and began to doze.

“Once again the atoms danced before my eyes. This time smaller groups modestly remained in the background. My mental eye, sharpened by repeated apparitions of similar kind, now distinguished larger units of various shapes.” “Long rows, frequently joined more densely; everything in motion, twisting and turning like snakes. And behold, what was that? One of the snakes caught hold of its own tail and mockingly whirled around my eyes.

“1 awoke, as if by lightening; this time, too, I spent the rest of the night working out the consequences of this hypothesis”

“Let us learn to dream, gentlemen, then perhaps we shall find the truth” AUGUST KEKULÉ

Edison, Einstein and many others have used the subconscious mind to give them the insight and the “know-how” to bring about their great achievements.

August Kekulé

“In variably, my device works as I imagined it should. In twenty years there has not been single exception.” NIKOLA TESLA

“Imagination is more important than knowledge.” ALBERT EINSTEIN

AROMATIC COMPOUNDS

14.1

Introduction to aromatic Compounds

Arenes = aromatic compounds: the word aromatic has nothing to do with odour.

KEYWORDS aromatic

resonance structure

Kekulé structure ortho meta para Friedel-Craft acylation

electrophilic substitution activating and deactivating group

ortho-para and meta director

oxidation of alkylbenzene

substitution of toluene

DISCOVERY OF BENZENE 1825

Michael Faraday (British)

Isolated a pure compound of boiling point 80oC Empirical formula = CH Named “bicarburet of hydrogen”

1834

Eilhard Mitscherlich (German)

Prepared benzene from benzoic acid C6H5CO2H + CaO

heat



C6H6 + CaCO3

Molar mass = 78, molecular formula = C6H6 named “benzin” benzene

gum benzoin

benzin benjoin (French)

benzoin

Luban Jawi (Arabic)

1866

Friedrich August Kekulé (German) Proposed a cyclic structure for benzene.

AROMATIC COMPOUNDS In earlier time, compounds are called aromatic because of their pleasant odours.

Benzene has strong pleasant odour.

Today, we use the word aromatic to refer to benzene and its structural relatives.

AROMATIC COMPOUNDS Aromatic compound is a cyclic conjugated molecule or ion that is stabilized by ∏ electron delocalisation. it is characterised by substitution reactions.

KEKULÉ’S STRUCTURE  Kekule was the first to formulate a reasonable representation of benzene H H H C C C H

C

C

C

or

H

H The Kekule structure suggests alternating double and single carbon-carbon bonds

RESONANCE STRUCTURE Benzene is actually a resonance hybrid of the two Kekulé structures.

equivalent to

resonance hybrid

All C–C bond length equal = 139 pm Shorter than typical C–C (148 pm) Longer than typical C=C (134 pm)

The six  electrons completely delocalized around the ring

The circle represents the six  electrons, distributed over the six atoms of the ring!

All six C atoms and six p orbitals are equivalent

THE CRITERIA FOR AROMATICITY 4 structural criteria must be satisfied for compound to be aromatic Cyclic

ly e t e l p Com ted a g u j n co

Planar

Conta in par ticula numb r e r of ∏ electr on  obe ys HÜ Rule ckel’s

THE CRITERIA FOR AROMATICITY HÜckel’s Rule * cyclic, planar and completely conjugated compounds that contain [4n+2] ∏ electron

(n=0,1,2…..) are said to be aromatic Erich HÜckel (1896-1980)  planar monocyclic rings with 2,6,10,14 and so forth ∏ electrons are aromatic

THE CRITERIA FOR AROMATICITY EXAMPLE OF AROMATIC COMPOUNDS

1. Aromatic compounds with a single ring

Benzene

aromatic Benzene is aromatic because:  contains 6∏ electrons (obeys HÜckels Rule)

[4n+2] ∏ = [4(1) + 2]∏ = 6 ∏ electrons

 cyclic, planar and has double bond in the ring

THE CRITERIA FOR AROMATICITY 2. Aromatic compounds with more than one ring EXAMPLE Aromatic naphthalene 4n+2= 4(2)+ 2 10 ∏ electrons * Two benzene rings joined together forms naphthalene

AROMATIC COMPOUNDS

14.2

Nomenclature of benzene and its derivatives


NAMING BENZENE DERIVATIVES Many organic molecules contain a benzene ring with one or more substituents. Many common name are recognized by the IUPAC system

EXAMPLE:

CH3 Common: toluene IUPAC: methylbenzene

MONO SUBSTITUTED BENZENE Benzene is the parent name and the substituent is indicated by a prefix. Cl

F

fluorobenzene

Br

chlorobenzene

NO2

nitrobenzene

bromobenzene CH2CH3

ethylbenzene

IUPAC rules allow some common names to be retained. CH3

toluene COOH

benzoic acid

OH

NH2

phenol

aniline CHO

benzaldehyde

DISUBSTITUTED BENZENE Two Same Substituents

Relative position of subsituents are indicated by prefixes ortho, meta, and para ( o–, m–, and p–) or by the use of number. Br Br 1 2 1 Br 2 Br 1

2 3

3 4

Br

Br

1,2–dibromobenzene 1,4–dibromobenzene or or 1,3–dibromobenzene o–dibromobenzene p–dibromobenzene or m–dibromobenzene

NO2 1

2

NO2 NO2

1 2 3

NO2 1,2–dinitrobenzene or o–dinitrobenzene NO2 1

1,3–dinitrobenzene or m–dinitrobenzene

2 3 4

NO2 1,4–dinitrobenzene or p–dinitrobenzene

DISUBSTITUTED BENZENE Two Different Substituents

Select one of the substituent that give new parent name and numbered as C1. COOH 1 2 NO2

COOH 1

2

COOH 1

3

2 4

3

NO2

NO2

2–nitrobenzoic acid 4–nitrobenzoic acid or or 3–nitrobenzoic acid o–nitrobenzoic acid p–nitrobenzoic acid or m–nitrobenzoic acid

THREE OR MORE SUBSTITUENTS Position of substituents must be indicated by numbers. The substituents are listed alphabetically when writing the name. Cl

Br 1

2

Br

3 4

Br 1,2,4–tribromobenzene

Br I 2–bromo–1–chloro–3–iodobenzene

C atom bearing the subtituent that define the new parent name is numbered as C1. OH 1

COOH 2

NO2

6

3

1

3

5

4

HO

NO2 2,4–dinitrophenol

Br 4

2

4

OH

3,5–dihydroxybenzoic acid 3

2

CH3

1

CH3

4–bromo–1,2–dimethylbenzene

Br 4

3

2

CH3

1

CH3

4–bromo–1,2–dimethylbenzene

correct

4–bromo–o–dimethylbenzene o–, m– and p– naming system is used for arenes with 2 substituents only!

14.2-11

PHENYL GROUP Benzene ring as substituent. If alkyl substituent is larger than the ring (more than 6 C), the compound is named as phenyl-substituted alkane. 1 CH2 3 2 CH–CH

4

5

6

7

2–CH2–CH2–CH2–CH3

2–phenylheptane

Phenyl = C6H5– = Ph

14.2-12

If the chain is unsaturated (have C═C or C≡C) or contains important functional group, the benzene ring is considered as phenyl substituent.

1

2

CH2–C

3 4

C–CH3

1–phenyl–2–butene

2

1

CH2–CH2–OH

2–phenylethanol

14.2-13

BENZYL GROUP CH2—

phenyl group

CH2Br

benzyl bromide

benzyl group

CH2OH

benzyl alcohol

AROMATIC COMPOUNDS

14.3

Chemical properties of benzene and its derivatives


14.3-01

UNUSUAL REACTIONS OF BENZENE Br2 / CCl4

BENZENE

KMnO4 / H2O H2 / Ni

no addition of Br2 (no decolorization) no oxidation (no decolorization) slow addition at high temperature and pressure

Alkenes readily undergo addition reaction, benzene does not!

14.3-03

REACTION OF ARENES Involves the benzene ring itself Electrophilic aromatic substitution + Br2 benzene

Br

FeBr3

bromobenzene

Involves substituents attached to the ring CH3CH2CH2CH3 KMnO 4 H2O butylbenzene

COOH

benzoic acid

14.3-04

ELECTROPHILIC SUBSTITUTION Most characteristic reaction of benzene. A H atom is replaced by an electrophile.

H

E +

E+ electrophile

+

H+

14.3-05

X2, FeX3 (X = Cl Br)

X + HX

HONO2

NO2 + H2O

H2SO4

RCl , AlCl3 (R can rearrange)

O RCCl , AlCl3

halogenation

nitration

R + HCl

Friedel-Crafts Alkylation

O C–R + HCl

Friedel-Crafts Acylation

14.3-07

GENERAL MECHANISM Formation of arenium ion

STEP 1

H H

H

H

H

E+

H H H

+

H

H

E H H

H

H H

H

E H

H

H

+

+

H H H arenium ion

E

H H

14.3-08

ARENIUM ION + E+

benzene ring

E H arenium ion

Electrophile takes two electrons of six–electrons  system to form  bond. This interrupts of cyclic system of  electrons. Benzene ring  arenium ion (aromatic) (nonaromatic) The four  electrons delocalized through these the five C atom (p orbitals)

Loss of

STEP 2 H H

H

E H

+

H

14.3-09

H+

H

H

E

H

H H

H

Substitution reaction allow aromatic six  electrons to be regenerated.

E+ 6  electrons

E H 4  electrons

+

– H+

E 6  electrons

14.3-10

Kekulé structures are more appropriate for writing mechanisms. For simplicity, however, we can show the mechanism in the following way: E

STEP 1 + E+

H

+

arenium ion E

STEP 2

+

E H

+ H+

14.3-11

HALOGENATION Reactants: benzene and halogens (Cl2 or Br2). Conditions: Lewis acid such as FeCl3 and FeBr3 no reaction (decolorization not observed)

+ Br2

+ Br2

FeBr3

Br + HBr bromobenzene

14.2-13

MECHANISM STEP 1

Formation of Br+ +

Br–Br

STEP 2

–

Br—Br–FeBr3 complex

+ FeBr3

Br+

+ FeBr4–

Br

Formation of arenium ion

+

H

+ Br+ STEP 3

Loss of H+ Br FeBr4–

+

H

Br + HBr + FeBr3

14.3-14

FUNCTION OF LEWIS ACIDS

Increase polarity of halogen molecules. Produce positive halogen ions (Br+ or Cl+).

14.3-15

NITRATION Reactants: benzene and concd. HNO3. Conditions: Concd. H2SO4 NO2

HNO3 H2SO4

nitrobenzene + H+

+ HSO4–

12.5-49

MECHANISM STEP 1

.. ..

H-O-NO2 +

Formation of nitronium ion (NO2+)

H

H-OSO3H

+-NO – H-O + HSO .. 2 4

H2O + NO2+ nitronium ion

STEP 2

Formation of arenium ion NO2 NO2+

+

STEP 3

Loss of H+ HSO4– NO2

+

H

+

H

NO2 + H2SO4

12.5-51

FRIEDEL–CRAFT ALKYLATION Reactants: benzene and haloalkane. Conditions: Catalyst: Lewis acid such as AlCl3.

+ R–X

AlCl3

R + HCl alkylbenzene

Alkylation = transfer an alkyl group to benzene

12.5-52

12.5-53

EXAMPLE: Cl

CH(CH3)2

AlCl3

+ CH3CHCH3 2–chloropropane

+ HCl isopropylbenzene

+ (CH3)3C–Cl

AlCl3

C(CH3)3 + HCl

2–chloro–2–methylpropane tert–butylbenzene

12.5-54

MECHANISM Formation of carbocation

(CH3)2CHCl: +

AlCl3

+ (CH3)2CH–Cl–AlCl3

: :

: :

STEP 1

(CH3)2CH+

+ AlCl4–

carbocation


STEP 2

CH(CH3)2 +

+CH(CH ) 3 2

+

H

Loss of H+

STEP 3

AlCl4– CH(CH3)2

+

H

CH(CH3)2 + HCl

12.5-55 OTHER FACTS ABOUT FRIEDEL–CRAFT ALKYLATION

Rearrangement can occur, especially when 1o haloalkanes are used. CH2CH2CH2CH3 EXAMPLE:

35%

CH3CH2CH2CH2Cl AlCl3

+

butylbenzene CH3 CHCH2CH3

sec–butylbenzene

Rearrangement: H

65%

+

CH3—CH2—CH—CH2 1o carbocation

+

CH3—CH2—CH—CH3 2o carbocation

12.5-56

FRIEDEL–CRAFT ACYLATION Reactants: benzene and acid chloride. Product: ketone. Conditions: Catalyst: Lewis acid such as AlCl3. EXAMPLE: O

O +

CH3C—Cl

AlCl3

CCH3

acetyl chloride acetophenone

+ HCl

12.5-57

ACYL GROUP O RC— acyl group EXAMPLE: O

O CH3C— acetyl group

–C— benzoyl group

Acylation = transfer an acyl group to benzene

12.5-59

MECHANISM Formation of acylium ion

STEP 1 :O:

+

AlCl3

R–C–Cl–AlCl3

: :

R–C–Cl

:O:

R–C+═O

R–C≡O+

acylium ion

+ AlCl4–

STEP 2

R


C=O + R–C+═O STEP 3

Loss of H+ R C=O

+

H

AlCl4–

H

+

R C═O + HCl

12.5-61

SUBSTITUENT EFFECT But, what would happen if we were to carry out a reaction on aromatic ring that already has a substituent?

12.5-62

EFFECT ON REACTIVITY Activating groups: Substituents that activate the ring, making it more reactive than benzene. Deactivating groups: Substituents that deactivate the ring, making it less reactive than benzene. EXAMPLE:

relative rate of nitration

NO2 6 x 10–8

Cl 0.033 reactivity

H 1

OH 1000

12.5-63

EFFECT ON ORIENTATION Ortho–para directors Tend to direct the incoming group into ortho and para positions. Meta directors Tend to direct the incoming group into meta position. EXAMPLE: CH3 HNO3 H2SO4

ortho-para director CH3 CH3 NO2 + +

CH3

NO2 (59%)

(37%)

NO2

(4%)

12.5-64

CLASSIFICATION OF SUBSTITUENTS ortho– , para– directing activators

ortho– , para– directing deactivators

meta– directing deactivators

ORTHO– , PARA– DIRECTING ACTIVATORS

12.5-65

Increasing activation ●●

—NH2

●●

●●

—NHR —NR2

●●

—OH ●● ●●

—OR ●●

General structure: —R or —Z● ●

●●

—NHCOR —R

Alkyl groups or have nonbonded electron pair on the atom bonded to benzene ring

12.5-66

EXAMPLE:

CH2CH3

CH2CH3

CH2CH3

Br Br2

+

FeBr3 ethylbenzene

(38%)

(62%)

Br

12.5-67 ORTHO– , PARA– DIRECTING DEACTIVATORS

●● —F ● ● ●● ●●

—Br● ● ●●

●●

—Cl● ● ●●

●● — I● ● ●●

General structure ●● —X● ● ●●

(halogens)

12.5-68

EXAMPLE:

Cl

Cl

Cl NO2

HNO3

+

H2SO4 chlorobenzene

(35%)

(64%)

NO2

12.5-69

META– DIRECTING ACTIVATORS —CHO —COR —COOR —COOH —CN —SO3H —NO2 + —NR3 Increasing deactivation

General structure: —Y (+ or –)

Have a full or partial positive charge on the atom bonded to benzene ring

12.5-69

EXAMPLE:

NO2

NO2 HNO3 H2SO4 NO2

nitrobenzene (93%)

12.5-70

INDUCTIVE EFFECT Due to: Electronegativity of the atoms in the substituent. Polarisability of the substituent. EXAMPLE:

Cl

Cl— electron–withdrawing

CH3

CH3— electron–donating

12.5-71

Activating Groups Release electrons to the ring Stabilise arenium ion

CH3

NO2

+

Form faster

Deactivating Groups Withdraw electrons from the ring Destabilise arenium ion Form slower

Cl

+

NO2

12.5-72

EXAMPLE: CF3

CH3

(trifluoromethyl)benzene

benzene

toluene

increasing rate of nitration CF3

NO2

NO2

+ CF3 withdraws e-,  arenium ion less stable  ring less reactive

+

CH3

NO2

+ CH3 releases e-,  arenium ion more stable  ring more reactive

OXIDATION OF SIDE CHAIN

12.5-75

Reactants: arene with benzylic H. Conditions: Strong oxidizing agent such as KMnO4 and Na2Cr2O7. Heat. benzylic H CH3 CH3

CH(CH3)2

12.5-76

EXAMPLE: CH3

COOH KMnO4 heat

toluene

O2N

benzoic acid CH3

Na2Cr2O7 heat

p–nitrotoluene

CH3

CH(CH3)2

isopropyl toluene

O2N

COOH

p–nitrobenzoic acid

KMnO4 heat

HOOC

COOH

terepththalic acid

12.5-77

CH3

CH(CH3)2

KMnO4 heat

HOOC

COOH

Alkyl group, regardless their chain length are converted to –COOH. Compounds without a benzylic H are inert to oxidation.

CH3

C(CH3)3

KMnO4 heat

HOOC

C(CH3)3

12.5-78

HALOGENATION OF TOLUENE Free radical substitution reaction Take place at high temperature or in the presence of uv light. Mechanism: free–radical substitution Cl or Br replaces H atom of alkyl group EXAMPLE: CH3

toluene

(dichloromethyl)benzene CH2Cl

CHCl2

Cl2

Cl2

Cl2

heat or light

heat or light

heat or light

benzyl chloride

CCl3

(trichloromethyl)benzene

HALOGENATION OF TOLUENE Electrophilic aromatic substitution reaction CH3

CH3

CH3 Br2

Br2

+

FeBr3

toluene

Br2

●

CH2

12.5-79

benzylic radical

Benzylic radicals are even more stable than 3o radicals!

CARCINOGENIC EFFECT

12.5-34

CH3

benzene

toluene

Many aromatic compounds are carcinorgenic and toxic. Example: benzene, benzo[a]pyrene.

12.5-35

At one time, benzene was widely used as solvent. Studies revealed benzene is carcinorgenic (can cause cancer). Replaced by toluene.

12.5-36

benzo[a]pyrene Benzo[a]pyrene is found in cigarette smoke, automobile exhaust, and the fumes from charcoal grills. When ingested or inhaled, it oxidised to carcinogenic products.

12.5-21

12.5-20

Benzoic acid, the simplest organic acid, prevent the growth of many organism.

12.5-22

widely used as a food preservative.

12.5-19

Fresh wild berries

END OF SLIDE SHOW

CHAPTER 15 HALOALKANES FJ / Chemistry Unit, KMPk / Mac 2006

1

15.1 : Introduction  Haloalkanes or alkyl halides

compounds that contains halogen atom to an sp3 hybridized carbon atom.

 General formula :

bonded

R-X or CnH2n+1X (acyclic) or CnH2n-1X (cyclic)

where X : halogen atom (F, Cl, Br or I) 2

15.1.1 : Classification of Haloalkanes  Haloalkanes are classified according to the nature of

carbon atom bonded to the halogen. General Formula CH 3 X

R CH 2 X

R R CH X

Classification methyl halide - halogen is bonded methyl group

to

Primary (10) halide - halogen is bonded to 10 carbon atom Secondary (20) halide - halogen is bonded to 20 3 carbon atom

General Formula

Classification

R

Tertiary (30) halide - halogen is bonded to 30 carbon atom

R C X R

X

Aryl halide - halogen is bonded to aromatic ring

CH 2 X

** Not a aryl halide 4

Example :  Classify the following haloalkanes :

No.

Haloalkanes

Classification

i.

CH 3 CH 2 Br

10

ii.

CH 3 CH (Cl)CH 3

20

iii.

(CH 3 ) 3 C(Br)

30

H3C

iv.

Cl

30 5

15.1.2 : IUPAC Nomenclature  Haloalkanes are named as alkanes with halogen as

substituents.  Locate and number the parent chain from the direction

that gives the substituent encountered first the lower number.  Show halogen substituents by the prefixes flouro-,

chloro-, bromo- and iodo-, and list them in alphabetical order along with other substituents. 6

Example : i.

CH 3 CHCH 2 CH 3 Br 2-bromobutane

ii.

Cl BrCH 2 CH 2 CHCHCH 2 CH 3 CH 3

1-bromo-3-chloro-4-methylhexane 7

Example : iii.

CH 2 CH 2 F CH 3 CH 2 CH 2 CHCH 2 CH 2 CH 3 4-(2-flouroethyl)heptane

iv.

H3C

CH 3 Cl

2-chloro-1,1-dimethylcyclopentane

8

Example : v.

Br

4-bromocyclohexene

vi.

CH 2 Cl

vii.

CH 3 Cl

(chloromethyl)benzene

2-chlorotoluene 9

15.1.3 : Structure of Haloalkane  The carbon – halogen bond in haloalkene is polar

because halogens is more electronegative than carbon. δ+ δC X

electrophilic site

 The polar C – X bond causes the carbon bearing the

halogen is susceptible to nucleophilic attack.  Haloalkanes are reactive and undergo nucleophilic

substitution and elimination reaction. 10

15.2 : Chemical Properties 15.2.1 : Nucleophilic Substitution Reaction  Haloalkanes

undergo nucleophilic substitution reactions in which the halogen atom is replaced by a nucleophile.

 In this reaction, the nucleophile attacks the partially

positive charge (δ+) carbon atom bonded to the halogen (δ-).  General reaction : _ _ R X + N u:

R _N u

+

_ X: 11

(a) : Hydrolysis of Haloalkane with Aqueous Solution of NaOH (H2O/NaOH)  Alkaline hydrolysis is carried out by boiling R-X with

NaOH(aq) to form alcohol.

R_ X + NaOH

H2 O

R_ OH + NaX

 Example :

CH3 H2 O _ _ CH3 C Br + NaOH CH3

CH3 CH3 _ C _OH + NaBr CH3 12

(b) : Reaction of Haloalkane with Potassium Cyanide (KCN)  When R-X is refluxed with KCN in alcohol, the halogen

atom is substituted by the CN- to produce a nitrile compound.

_ _ R X + CN

alcohol reflux

R_ CN

+ X

_

 Example : CH 3 CH 2 Br + KCN

alcohol reflux

CH 3 CH 2 CN

+ KBr 13

(c) : Reaction of Haloalkane with Ammonia (NH3)  When R-X is heated with excess concentrated NH3, the

halogen atom is replaced by the amino group, NH2-.

R_ X

NH 3

_ + RNH 3 X

NH 3

_ _ + R N H2 + NH 4 X (amine)

 Example :

CH 3 CH 2 Cl + excess NH 3

_ + CH 3 CH 2 NH 2 + NH 4 Cl 14

15.2.2 : Mechanisms of Nucleophilic Substitution Reaction  They are 2 important mechanisms for the substitution

reaction: (A). Unimolecular Nucleophilic Substitution (SN1)

Reaction

(B). Bimolecular Nucleophilic Substitution (SN2)

Reaction

15

(A) : Unimolecular Nucleophilic Substitution (SN1)

Reaction

 The term unimolecular means there is only one

molecule involved in the transition state of the ratelimiting step.  SN1 reactions are governed mainly by the relative

stabilities of carbocations.  Relative reactivities of haloalkanes in an SN1 reaction :

R-X < R-X < R-X 1o 2o 3o increasing reactivity 16

 The rate of SN1 reaction does not depend on the

concentration of nucleophile.  The rate depends only on the concentration of the

substrate, alkyl halide.  rate = k [R3C-X]

* SN1 is a first order reaction

17

 The mechanism of SN1 reaction involves 2 steps.

Step 1 : Formation of a carbocation (rate determining step)

R R_ C _ X

s low

R 3o alkyl halide

R _ _ R C+ + X R carbocation

halide ion

Step 2 : Nucleophilic attack on the carbocation

R _ _ R C + + N u: R

fas t

R R_ C _ N u R

18

 Example 1 :

Reaction of 2-bromo-2-methylpropane with H2O. CH 3 CH 3 _ C _ Br + H 2 O CH 3

CH 3 CH 3 _ C _ OH + HBr CH 3

 SN1 mechanism :

Step 1 : Formation of a carbocation CH 3 CH 3 _ C _ Br CH 3

s low

CH 3 CH 3 _ C + CH 3

+ Br

_ 19

 Step 2 : Nucleophilic attack on the carbocation

CH 3 CH 3 _ C +

+

fa s t H2O

CH 3

CH 3 H _ _ CH 3 C O + H CH 3

Loss of proton, H+ to solvent

CH 3 H _ _ CH 3 C O + H CH 3

+

H2O

CH 3 CH 3 _ C _ OH + H 3 O + CH 3 20

 Example 2 :

Write the mechanism for the following reaction. CH3 CH3 _C _CH2Br + NaOH(aq) CH3

CH3 CH3 _ C _CH2CH3 + NaBr OH

 SN1 Mechanism :

Step 1 : Formation of carbocation CH 3 CH 3 _ C _ CH 2 _Br CH 3

s low

CH 3 _ _ _ CH 3 C CH 2 + Br + CH 3

21

Rearrangement : CH 3 CH 3 _ C _ CH 2 + CH 3

1,2-meth yl s hift

CH 3 CH 3 _ C _ CH 2 + CH 3

Step 2 : Nucleophilic attack on the carbocation CH 3 CH 3 _ C _ CH2 + CH 3

+ OH

_

fas t

CH 3 CH 3 _ C _ CH 2 CH 3 OH 22

Exercise 1 : (Feb 2003) Write a reasonable structures of products formed when 1iodobutane reacts with i. ii. iii.

KCN NaOH/H2O excess NH3

Write the mechanism for the reaction in (ii).

23

Exercise 2 : (Mac 2002) The structure of compound A is as follows: CH3 CH3 C Br CH3

i. Give IUPAC name for A ii. Compound A react with OH- forming an alcohol. Write the mechanism for the formation of this alcohol and name the reaction. 24

(B) : Bimolecular Nucleophilic Reaction (SN2)

Substitution

 The term bimolecular means that the transition state of

the rate limiting step involves the collision of two molecules.  SN2 reactions are governed mainly by steric factors

(steric effect).

 Steric effect

is an effect on relative rates caused by the spacefilling properties of those parts of a molecule attached at or near to the reacting site. 25

 The reactivity on SN2 reaction depends on the size of

atoms or groups attached to a C – X.  The presence of bulky alkyl groups will prevent the

nucleophilic attack and slow the reaction rate.  Relative reactivities of haloalkanes in an SN2 reaction :

R-X < R-X < CH3-X 2o 1o increasing reactivity 26

 The rate of reaction depends on the concentration of

the haloalkane and the concentration of nucleophile.  rate = k [R-X] [Nu:-]

* SN2 is a second order reaction.  The mechanism of SN2 occurs in a single step.

General Mechanism : R Nu:-

H

C X H

R slow

Nu C X H H

transition state

fast Nu C

R

H

H

+ X27

 In SN2 reaction, the nucleophile attacks from the back

side of the electrophilic carbon, that is, from the side directly opposite bonded to the halogen.

 The transition state involves partial bonding between

the attacking nucleophile and the haloalkane.  Back-side attack causes the product formed has inverse

configuration from the original configuration. * turns the tetrahedron of the carbon atom inside out, like umbrella caught by the wind.

28

 Example 3 :

Reaction of ethyl bromide with aqueous sodium hydroxide.

CH3 CH2 Br + NaOH(aq)

CH3 CH2 OH + NaBr

 SN2 Mechanism : CH3

CH3 :OHH

C Br H

slow

OH C Br H H

fast

CH3 OH C H

H

+ Br-

transition state 29

Comparison of SN1 and SN2 Reactions SN1

SN2

A two-step mechanism

A one-step mechanism

A unimolecular ratedetermining step Second order : rate = k [RX] [Nu]

A bimolecular ratedetermining step First order : rate = k [RX]

Strong nucleophile

Weak nucleophile

Carbocation rearrangement Reactivity order : 3o > 2o > 1o

No carbocation rearrangement Reactivity order : methyl > 1o > 2o

30

15.2.3 : Elimination Reaction (dehydrohalogenation of haloalkanes)  Halogen can be removed from one carbon of a

haloalkane and hydrogen from an adjacent carbon to form a carbon-carbon double bond in the presence of a strong base.  General reaction :

H _C_C_

+

_

_ C C

base

alkene

:B

+

X

_

X haloalkane

31

 Example : Br

i.

CH 3 CHCHCH 3

CH 3 CH 2 ON a CH 3 CH 2 OH

CH 3 C CHCH 3 CH 3

CH 3

major + CH 3 CHC CH 2 CH 3

minor ii. Br CH 3

CH 3 CH 2 ON a

CH 3

CH 2 +

CH 3 CH 2 OH

major

minor

32

15.2.4 : Synthesis of Organomagnesium Compound ( Grignard Reagent )  Prepared

by the reaction of haloalkanes with magnesium metal in anhydrous ether as a solvent. R-X

+

Mg

ether

R-MgX Grignard Reagent ( alkylmagnesium halide)

 Example :

i. CH 3 CH 2 CH 2 Br + Mg ether CH 3 CH 2 CH 2 MgBr ii.

Cl + Mg

ether

MgCl 33

15.2.4.1 : Synthesis of Alkanes, Alcohols and Carboxylic Acids from Grignard Reagents.  The Grignard reagents undergo many reactions that

make them useful as a starting materials in the synthesis of other organic compounds. (i).

Synthesis of alkane The Grignard reagent is hydrolyzed to an alkane when warmed with H2O. RMgX + H 2 O

H+

R-H + Mg(OH)X 34

Example : + H i. CH 3 CH 2 MgBr + H 2 O

CH 3 CH 3

+ Mg(OH)Br

ii. CH 3 CH-MgBr + H 2 O

H+

CH 3 CH 2 CH 3

+ Mg(OH)Br

CH 3

iii. CH 2 MgCl

H 2 O/H +

CH 3 + Mg(OH)Cl 35

(ii).

Synthesis of 1o alcohol Methanal reacts with the Grignard reagent, followed by the hydrolysis produces primary alcohol. H

O R-MgX + H-C-H

R-C-OMgX H H 2 O,H +

H R-C-OH H

+

Mg(OH)X 36

 Example :

i. O CH 3 MgBr + H-C-H

H 3 O+

H CH 3 -C-OH

+ Mg(OH)Br

H

ii. MgBr

O + H -C-H

H 3 O+

CH 2 OH + Mg(OH)Br

37

(iii).

Synthesis of 2o alcohol Grignard reagent reacts produce secondary alcohol.

with

aldehydes

to

R'

O R-MgX + H-C-R'

R-C-OMgX H H 2 O,H +

R' R-C-OH H

+ Mg(OH)X 38

 Example :

O CH 3 CH 2 MgBr + CH 3 CH 2 -C-H

i.

H 2 O/H + H CH 3 CH 2 -C-CH 2 CH 3 + Mg(OH)Br OH

ii. MgCl

O + CH 3- C-H

CH 3 H 2 O/H +

C-OH H

+ Mg(OH )Cl 39

(iv).

Synthesis of 3o alcohol

Grignard reagent reacts with ketons to the tertiary alcohol.

produce

R'

O R-MgX + R'-C-R"

R-C-OMgX R"

H 2 O,H + R' R-C-OH R"

+ Mg(OH)X 40

 Example :

O

i.

CH 3 CH 2 MgBr + CH 3 -C-CH 3 H 2 O/H + CH 3 CH 3 CH 2 -C-CH 3 + Mg(OH)Br OH

ii. MgCl

CH 3

O + CH 3- C-CH 3

H 3 O+

C-OH CH 3

+ Mg(OH )Cl 41

(v).

Synthesis of carboxylic acid Grignard reagent reacts with carbon dioxide (CO2) followed by hydrolysis to form carboxylic acid.

O RMgX + O C O O R-C-O-MgX + H 2 O

H+

R-C-O-MgX O R-C-OH + Mg(OH)X 42

 Example : O CH 3 CH 2Mg I + CO 2

O CH 3 CH 2- C-O-MgI + H 2 O

CH 3 CH 2- C-O-MgI

H+

O CH 3 CH 2 COH + Mg(OH)I

43

15.2.5 : Wurtz Reaction  Reaction of haloalkane (RX) with an alkali metal

(usually Na) to synthesise longer alkane. i.

To prepare an even number of carbon atoms alkane 2RX + 2Na

dry ether



RR + 2NaX

Example: 2CH3CH2Br + 2Na  CH3CH2CH2CH3 + 2NaBr 44

ii.

To prepare a odd number of carbon atoms alkane RX + R’X + 6Na  RR + RR’ + R’R’ + 6NaX

Example: CH3CH2Br + CH3Br + 6Na  CH3CH2CH2CH3 + CH3CH2CH3 + CH3CH3 + 6NaBr

45

15.2.6 : Importance of Haloalkanes as Inert Substance Haloalkanes

Uses

CCl4 (carbon tetrachloride)

solvent for dry cleaning, spot removing

CHCl3 (chloroform) CF2Cl2 , Freon-12 (dichlorodifluoromethane) CFC (chloroflourocarbons) DDT (DichloroDiphenylTrichloroethane)

solvent for cleaning and degreasing work propellants in aerosol sprays refrigerant gas 46 insecticide protects

HYDROXY COMPOUNDS (ALCOHOLS) Click me for intro!

1

INTRODUCTION  



Alkane with hydroxyl group (–OH) General formula for aliphatic alcohol, CnH2n+1OH Aromatic alcohol – phenol (-OH attached to the benzene rings)

2

INTRODUCTION OH

CH2-OH

CH3CH2-OH

ALIPHATIC ALCOHOL (–OH group attached directly to the alkyl group)

3

INTRODUCTION OH

OH

OH

CH3

Br

AROMATIC ALCOHOL (–OH group attached directly to the aryl group)

4

CLASSIFICATION 

Methyl alcohol CH3-OH



10 Alcohols

R-CH2-OH

20 Alcohols

R R-CH-OH





30 Alcohols

R R-C-OH R 5

NOMENCLATURE Structure

General name

IUPAC name

CH3-OH

methyl alcohol

methanol

CH3CH2-OH

ethyl alcohol

ethanol

CH3CH(OH)CH3

isopropyl alcohol

2-propanol

OH

cyclohexyl alcohol

cyclohexanol

6

Naming the alcohol 1. The suffix ‘–e’ in the alkane parent name is replaced by ‘–ol’ 2. The location of the –OH on the longest chain is given by the smallest possible number

7

IUPAC rules for naming alcohols 1. Determine parent name 2. Give the lowest number to the carbon bearing the –OH group 3. Identify substituent groups and their position 4. Name the alcohol

8

Naming Alcohols, R-O-H: CH3 CH3CHCH2CHCH3 OH 4-methyl-2-pentanol

CH3 CH3CCH2CH2CH3 OH 2-methyl-2-pentanol

2o

3o

CH3 HO-CHCH2CH3

CH3CH2CH2-OH

sec-butyl alcohol 2-butanol

n-propyl alcohol 1-propanol

2o

1o

9

Naming Alcohols

10

For phenol, C attach to the –OH group is C 1

OH

OH

OH CH3

CH2CH3 3-ethylphenol

Br 3-bromophenol

2-methylphenol

11

Alcohol with 2 hydroxy groups Structure CH2(OH)CH2-OH CH3CH(OH)CH2-OH HO-CH2CH2CH2-OH OH-CH2CHCH2-OH OH

General name

IUPAC name

ethylene glycol

1,2-ethanediol

propylene glycol

1,2-propanediol

trimethylene glycol glycerol

1,3-propanediol 1,2,3-propanetriol

12

Benzene derivatives with 2 –OH groups are name as benzenediol

OH

OH

OH OH

OH 1,4-benzenediol (hydroquinone)

OH 1,3-benzenediol (resorcinol)

1,2-benzenediol (catechol)

13

EXCERCISE 1. Name the following compounds using IUPAC nomenclature (a)

(b)

-CH2CH(CH3)CH2-OH

-OH NO2

(b)

-OH Br 14

PHYSICAL PROPERTIES  

 

Most simple alcohols are liquids MeOH & EtOH are volatile liquids, b.p. 65oC & 78oC respectively Mr , b.p. ROH with C > 12 are solids

15

PHYSICAL PROPERTIES  

 

No. of –OH , b.p. b.p. ROH > alkane & RX because of Hbonding b.p. ROH straight chain > ROH branch b.p. 1o ROH > 2o ROH > 3o ROH

16

PHYSICAL PROPERTIES (ii) Boiling point  As molecular weight increase, van der Waals forces increase, boiling point will increase too.

   

For isomeric alcohol (same molecular formula) increase in boiling point with increasing carbon number. decrease in boiling point with branching. boiling point in descending order:1° alcohol  2° alcohol  3° alcohol 17

PHYSICAL PROPERTIES (Solubility) Alcohols and water have the ability to form inter-molecular H-bonding Methanol, ethanol & propanol are completely soluble in water Solubility decreases with relative Mr because of the bigger hydrophobic area Solubility increases with the no. of -OH groups 18

PHYSICAL PROPERTIES (Solubility) δ+

R

δ-

O δ-

H R

H

O δ-

O

H

δ+

H

O

Hydrophilic area

R

OH

δ+

H H

Hydrogen bonding

Hydrophobic area

δ+

δ-

19

PHYSICAL PROPERTIES (Solubility) Phenol is not completely soluble in water below 66oC, but completely soluble above this temp. Phenol is a colourless crystalline solid which is hygroscopic 4-nitrophenol is more soluble in water than 2nitrophenol because can form H-bonding with water molecules

20

PHYSICAL PROPERTIES (Solubility) 4-nitrophenol

2-nitrophenol

δ-

O

δ-

δ+

H δ- O

δ-

O

O δ+ O N

H

H

δ+

N δ+

O

Intramolecular H-bonding

O δ+ δ- H

δ-

O H

δ+

H

Intermolecular H-bonding 21

Acidity of phenol & alcohol



phenol > water > alcohol Ethanol, very weak acid & only can react with strong base such as Na or K to produce a conjugate base (ethoxide ion) CH3CH2-OH CH3CH2-O- + H+

22

Acidity of phenol & alcohol 





CH3CH2O- is e- richer than OH-, thus H2O > acidic than CH3CH2OH Phenol > acidic than H2O & CH3CH2OH because phenoxide ion formed is more stable The phenoxide ion formed is stabilised by resonance effect

23

Acidity of phenol & alcohol

O

O -

H release proton

+ H+ Phenoxide ion

24

Phenoxide ion stability O -

O

O

-

O -

Oδδ-

δδ25

Example: OH

CH2CH2-OH

CH2-OH

OH

acidity increase

26

Example: COOH

OH

>

> CH CH -OH 3

2

Na2CO3

KOH

K, Na

NaHCO3

NaOH

NaH

Weak bases

Strong bases

ACIDITY DECREASE

Very strong bases 27

EXCERCISE 1. Arrange the following compounds in the order of increasing boiling point. Explain your answer. 1,2-ethanediol , n-butane , 1-propanol , 1,3-propanediol 2. Predict which of the compound is more soluble in water. Briefly explain your answer. (a) propane & 1-propanol (b) 1-propanol & 1,2-propanediol (c) 2-propanol & 2-pentanol 28

PREPARATION OF ALCOHOLS 1. Fermentation of carbohydrate

-fermentation of sugars by yeast to form ethanol C6H12O6

2CH3CH2OH + 2CO2

29

PREPARATION OF ALCOHOLS 2.Hydration of Alkenes follows Markovnikov’s rule Example:CH3CH=CH2

H2O / H2SO4

CH3CH-CH3 OH

30

PREPARATION OF ALCOHOLS Mechanism CH3CH=CH2 + H+

CH3CH-CH3 2o carbonium +

●● ● ● OH

CH3CH-CH3 ● ● OH ●●

31

PREPARATION OF ALCOHOLS 3. Hydrolysis of Haloalkanes RX + OH- (aq)

CH2Cl

ROH + X-

CH2OH

+ NaOH(aq)

32

PREPARATION OF ALCOHOLS 4. Addition of Grignard reagent to

carbonyl compounds

10, 20 & 30 alcohols can be prepared by addition of Grignard reagent with carbonyl compounds followed by hydrolysis

33

PREPARATION OF ALCOHOLS (a) Formation of 1o alcohols (Grignard reagent + methanal)

CH3CH2-MgCl + H2C=O

CH3CH2CH2-OMgCl

methanal

H3O+ 1o Alcohol

CH3CH2CH2-OH + Mg(OH)Cl

34

PREPARATION OF ALCOHOLS (b) Formation of 2o alcohols (Grignard reagent + aldehyde) CH3

CH3

CH3CH2-MgCl + HC=O

CH3CH2CH-OMgCl

ethanal

H3O+ CH3 2o Alcohol

CH3CH2CH-OH + Mg(OH)Cl 35

PREPARATION OF ALCOHOLS (c) Formation of 3o alcohol (Grignard reagent + ketone) O-MgCl

O CH3-MgCl + CH3-C-CH3 propanone

CH3-C-CH3 CH3 H3O+ O-H

3o Alcohol

CH3-C-CH3 + Mg(OH)Cl CH3 36

EXCERCISE 1. Write the chemical equation for the following reaction: i.

tert-butyl alcohol from an alkene

ii. isopropyl alcohol from an alkyl halide 2. Draw the structure of A, B and C: i. CH3 O CH3-C-MgCl + ii.

-C-H

A

CH3 -CH=CH-CH3

H2O / H+

B + C 37

ALCOHOL REACTIONS (a) As an acid In an aqueous solution alcohol reacts as a very weak acid ROH + H2O

RO- + H3O+

Reacts with Na or NaH & form a very strong base alkoxide ion CH3CH2OH + Na

CH3CH2O-Na+ + 1/2H2

38

ALCOHOL REACTIONS (a) As a base With the presence of lone pair of e- at the O atom, alcohol can accept proton from an acid

●●

CH3CH2-O-H + HA ●●

●● CH3CH2O+-H

+ A-

H

39

ALCOHOL REACTIONS (1) Esterification Alcohol reacts with carboxylic acid under reflux & is catalysed by mineral acids (H2SO4 or HCl) General reaction:O

H+,

R-C-OH + R’-OH acid

alcohol

O

R-C-OR’ + H-O-H ester

water 40

ALCOHOL REACTIONS (1) Esterification Example:O

H+,

CH3C-OH + CH3-OH ethanoic acid

methanol

O

CH3C-OCH3 + H-O-H methyl ethanoate

41

ALCOHOL REACTIONS (1) Esterification Alcohol reacts with acyl chloride to form ester O

base

CH3C-Cl + CH3-OH ethanoil chloride

methanol

O

CH3C-OCH3 + HCl methyl ethanoate

42

ALCOHOL REACTIONS (2) Dehydration of alcohol •Involve formation of carbocation •Possible of rearrangement

CH3CH2CH2OH propanol

H2SO4(con.) 1800C

CH3CH=CH2 propene

43

ALCOHOL REACTIONS (3) Formation of haloalkane •ROH reacts with HX to form RX •The reaction rates are in the order HI > HBr > HCl (HF non-reactive) •Reactivity of alcohols 3o ROH > 2o ROH > 1o ROH > phenol

44

ALCOHOL REACTIONS Example:-

•

ROH + HX

• CH3CH2CH2OH + HBr

RX + H2O CH3CH2CH2Br + H2O

45

ALCOHOL REACTIONS (3) Formation of haloalkane •ROH reacts with PX3, PX5 , SOCl2 General reaction: (i) 3ROH + PX3

3RX + H3PO3

(ii) ROH + PCl5

RCl + POCl3 + HCl

(iii)ROH + SOCl2

RCl + SO2 + HCl

46

ALCOHOL REACTIONS (4) Oxidation of alcohols

• Common oxidising agents are potassium permanganate (KMnO4/H+) and potassium dichromate (K2Cr2O7/H+) • Oxidation product depends on the class of alcohol used

47

ALCOHOL REACTIONS (4) Oxidation of 1o alcohol • 1o R-OH can be oxidised to give an aldehyde. Further oxidation would give a carboxylic acid

RCH2OH 1o alcohol

[O]

O RC-H aldehyde

[O]

O RC-OH carboxylic acid

48

ALCOHOL REACTIONS (4) Oxidation of 1o alcohol Example:-

CH3CH2OH

K2Cr2O7

/H+

O CH3C-H

K2Cr2O7/H+

O CH3C-OH + Cr3+

green

orange

49

ALCOHOL REACTIONS (4) Oxidation of 1o alcohol • Sarett-Collins reagent is a special oxidation agent to change a 1o alcohol to an aldehyde • CrO3 is added to excess pyridine to form pyridinium chlorochromate (PCC)

CH3CH2CH2OH + PCC propanol

O CH3CH2C-H propanal 50

ALCOHOL REACTIONS (4) Oxidation of 2o alcohol • form ketone which cannot be oxidised further under normal conditions R R’CH-OH 2o alcohol

[O]

R R’C=O + H2O ketone

CH3 CH3 + KMnO4/H CH3CH-OH CH3C=O + H2O 2-propanol

propanone 51

ALCOHOL REACTIONS (4) Oxidation of 3o alcohol • Does not undergo oxidation under normal condition

R R’-C-OH

[O]

no reaction

R’’ 3o alcohol

52

Chemical test for R-OH 

Lucas test

• Lucas reagent : HCl(con.) + ZnCl2 • 3o R-OH + Lucas reagent – reacts • •

immediately to form cloudy haloalkane 2o R-OH + Lucas reagent –takes about 5-10 minutes 1o R-OH + Lucas reagent – no visible changes unless heated

53

OBSERVATIONS :

Example:

CH2OH HCl / ZnCl2

Does not turn cloudy

HCl / ZnCl2

Turn cloudy within 5 minute

OH

54

CH3 OH HCl / ZnCl2

OBSERVATIONS : Turn cloudy immediately

55

Haloform Reaction 

Methyl carbinol cleavage to give Carboxylic acids and Haloform H CH 3 C H OH

X2 , H+ OH-

HCX 3 + H

C OH O

H CH 3 C R OH

X2 , H+ OH-

HCX 3 + R

C OH O 56

IODOFORM TEST OH

CH3CH2CCH3 + 3I2 base (OH-) 2-butanol CH3CH2COO- + CHI3↓ yellow precipitate (iodoform)

57

PHENOL Manufacture of phenol by Cumene process (a) Friedel-Crafts reaction CH3-CH-CH3 + CH3CH=CH2

AlCl3

Isopropylbenzene (cumene)

58

PHENOL (b) Oxidation of Cumene CH3-CH-CH3 + O2

O -OH CH3-C-CH3

1200C

Cumene hydroperoxide

(c) Decomposition & rearrangement CH3 -C-O-OH

H2SO4

O -OH + CH3CCH3

CH3 59

Phenol reactions 1. Reaction at functional group -OH (a) Reaction with Na

O-Na+

OH + Na

+ ½ H2

60

Phenol reactions 1. Reaction at functional group -OH (b) Reaction with NaOH O-Na+

OH + NaOH

CH3CH2OH + NaOH

+ H2O

no reaction

61

Phenol reactions 2. Electrophilic substitution reactions (a) Halogenation (Bromination) -OH is ortho-para director in Br2, aqueous solution, white precipitate 2,4,6tribromophenol will form

OH

OH Br

Br

+ 3Br2 / H2O

+ 3HBr Br 62

Phenol reactions 2. Electrophilic substitution reactions (a) Bromination to get monobromination (catalyst: FeBr3 or CS2) OH

OH + Br2

OH Br

CS2 50C

+

+ HBr

Br 63

Phenol reactions 2. Electrophilic substitution reactions (b) Nitration OH

OH

NO2

Dilute HNO3

+

Room temp.

OH NO2 OH O2N

NO2

Concentrated HNO3

Yellow precipitate

NO2 64

Phenol reactions 3. Esterification phenol is a weaker nucleophile than alcohol since the lone pair e- of O atom are delocalised to the benzene ring reacts with an acid derivative such as acid chloride to form an ester CH3 OH O-C=O Cl + CH3-C=O

+ Cl65

Phenol reactions 4. Identification test for phenols phenol reacts with aqueous solution of iron (III) chloride giving a light purple complex OH

OH + FeCl3 (aq) FeCl3 Purple complex

66

EXERCISE 1 Suggest a chemical test to differentiate the following pairs of compounds. State the reagents, observations and write chemical equations involved. (a) methanol from ethanol (b) 2-methyl-1-propanol from 2-methyl-2-propanol (c) cyclohexanol from cyclohexane (d) phenol from benzyl alcohol (e) o-methylphenol from 2-methylcyclohexanol

67

EXERCISE 2 Compound A (C4H10O) undergoes oxidation to form compound B (C4H8O). Oxidation of B gives compound C (C4H8O2). Reaction of A with hot concentrated sulphuric acid yields compound D (C4H8). Hydration of D gives compound E (C4H10O) which cannot be oxidised. A and E are isomeric. Suggest the structure of A, B, C, D and E. Rasionalise your answer.

68

Common used of alcohol and phenol 



Ethyl alcohol is a poison. LD50 = ~10g/Kg orally in mice.- Nausea, vomiting, flushing, mental excitement or depression, drowsiness, impaired perception, loss of coordination, stupor, coma, death may occur. (intoxication) Phenol is an important industrial chemical. Major use is in phenolic resins for adhesives and plastics. 69

Phenols in Medicine 

Many phenols are used as antiseptics and disinfectants. OH OH OH

OH

OH CH2CH2CH2CH2CH2CH3

Phenol

Resorcinol (antiseptic)

4-Hexylresorcinal (antiseptic) 70

Derivatives of Phenol 

Compounds of phenol are the active ingredients in the essential oils of cloves, vanilla, nutmeg, and mint.

71

17.0

CARBONYL COMPOUNDS: ALDEHYDES AND KETONES

17.1 Introduction  Aldehydes and ketones are carbonyl compounds O

 The functional group :

Aldehyde O R C H

C

ketone O R C

R

1

17.2 Nomenclature  The longest continuous carbon chain with -CHO group gives the parent name.  Aldehydes : the -e ending is replaced by –al  Ketones : the –e ending is replaced by –one.  The chain is numbered in the direction that provides the lower number for the functional group.

2

Example

butanal

O CH3CH2CH2C H CH3

2-ethyl-4-methylpentanal

O

CH3CH CH2CH C H CH2 CH3 CH3

3,5-dimethyl-2-hexanone

O

CH3CH CH2CH C CH3 CH3

3

IUPAC nomenclature and common names of some carbonyl compounds. Compound

Structure formula

Aldehyde

H2C=O

Ketone

IUPAC nomenclature

Common name

Methanal

Formaldehyde

CH3CH=O

Ethanal

Acetaldehyde

CH3CH2CH=O

Propanal

Propionaldehyde

CH3CH2CH2CH=O

Butanal

Butyraldehyde

CH3COCH3

Propanone

Acetone

CH3CH2COCH3

2-butanone

Ethyl methyl ketone

CH3CH2CH2COCH3

2-pentanone

n–propyl methyl ketone

CH3CH2COCH2CH3

3-pentanone

Diethyl ketone 4

Example i)

2- methylpentanal O CH3CH2CH2CH C H CH3

ii) 3- hexanone O CH3CH2C CH2CH2CH3

iii) 5-methyl-3- heptanone O CH3CH CH2 C CH2CH3 CH2CH3

5

iv) 3-ethylcyclopentanone CH3 CH2

O

v) 3-cyclohexylpropanal CH2 CH2 C H O

6

17.3 Synthesis of Aldehydes and Ketones

a) Oxidation Alcohols  When 1o alcohols are oxidised using pyridinium chlorochromate (PCC) in CH2Cl2, aldehydes are formed.

 2o alcohols are oxidised to ketones using oxidation agent such as chromic acid (H2CrO4), acidified potassium dichromate (K2Cr2O7, H+), acidified potassium permanganate (KMnO4,H+).

7

Example 1. H

O

CH3 C

OH

PCC CH2Cl2

CH3 C

H

H ethanol

ethanal

(alcohol 10)

(aldehyde)

CH3

O

2.

CH3 C

OH

H 2-propanol

KMnO4, H+ heat

CH3 C

CH3

propanone

(alcohol 20)

8

3. CH3 CH3 C

OH

no reaction

CH3 2-methyl-2-propanol (alcohol 30)

9

b) Ozonolysis of Alkenes

R

R1 C

C

H

H

R

R1 C

C

R2

O

O i. O3 ii. H2O/Zn

R C

H + R1

ii. H2O/Zn

R C

H

O

O i. O3

C

R2 + R1

C

R3

R3

Example: 1. CH2 CH3

H 3C C H

C

i. O3 ii. H2O/Zn

H

2-pentene 10

2. CH2CH3 CH3 C

i. O3

CH CH2 CH3

ii. H2O/Zn

3-methyl-3-hexene

3. CH3 i. O3 ii. H2O/Zn

4-methylcyclopentene

11

c) Hydration of Alkynes (keto-enol tautomerism)  Alkynes react with water in the presence of mercury sulphate and sulphuric acid. O

OH R C

C R 1 + H 2O

H2SO4 HgSO4

R C

CH R1

Enol

R C CH2 R1 Keto

Example: H C

C H + H 2O

H2SO4 HgSO4

ethyne

H 3C C

C H + H 2O

propyne

H2SO4 HgSO4

12

d) Friedel- Crafts Acylation  Acyl chloride or anhydride acid reacts with benzene in the presence of Lewis acid to form ketone. Example : O AlCl3

+ CH3 C Cl

O + R

O

C O C

R'

AlCl3

13

e) Hydrolysis Geminal Dihalides  Hydrolysis of aromatic gem- dihalides to produce aromatic aldehyde. CH3

CHO

CHCl2

2Cl2

H2O

uv

 Hydrolysis can be carried out by adding one of the following reagents: a) aqueous calcium carbonate b) aqueous sodium carbonate c) 70 % H2SO4 Example :

14

17.4 Chemical Properties A) Nucleophilic addition  Aldehydes and ketones undergo nucleophilic addition reaction. R

+

C

-

O

R1 aldehyde or ketone (R or R1 may be H)

 The positive charge on the carbonyl carbon atom is exposed to attack by a nucleophile. General reaction :

R H

OH C

O + Nu

R C

Nu

H

15

General Mechanism  The nucleophile attacks the partially positive carbonyl group and breaking the carbonyl  bond.

16

a)

Addition of Hydrogen Cyanide  Hydrogen cyanide adds to the carbonyl groups of aldehydes and ketones to form compounds called cyanohydrins.

R

OH O + HCN

C

H

R C CN H cyanohydrins

R

OH O + HCN

C

R1

R C CN R1

 HCN can be obtained from the reaction of NaCN or KCN with diluted sulphuric acid.

NaCN

+

H2SO 4

HCN

+

NaHSO 4

17

Example: CN H C OH

CHO NaCN H2SO 4

Mechanism:

18

Exercises. 1. O H3C

NaCN

C H H2SO 4

2. O H3C

NaCN

C CH3 H2SO 4

19

b) Addition of Grignard Reagent General mechanism :

20

 Grignard

reagents

react

with

carbonyl

compounds to yield alcohol.

 Reaction of Grignard reagent with methanal gives primary alcohol

H CH3CH2CH2CH2MgX

+

+ HCOH

21

 Reaction of Grignard reagent with aldehyde other than methanal gives secondary alcohol

CH2CH2CH3 CH3 CH2MgX + HC O

H+

 Reaction of Grignard reagent with ketone gives tertiary alcohol

CH2CH3

CH3CH2CH2MgX + C O

H+

CH3

22

c) The Hydration Reaction 

The gem-diol results from a nucleophilic addition of water to the carbonyl group of the aldehyde.

R

O H

C

R + H2O

H+

HO

C

H

OH R

O R'

C

R + H2O

H+

HO

C

R'

OH

23

Example: 1. O H

H+

CH3 + H2O

C

ethanal

2. O H

C

H+

H

+ H2O

methanal

3. O CH3 CH2C

CH3 + H2O

H+

2-butanone 24

Mechanism : O H 3C

C

H+

CH3 + H2O

propanone

25

d) The Addition Of Alcohols: Hemiacetals and Hemiketals i.

Hemiacetals and hemiketals  Aldehyde reacts with alcohol to form hemiacetal.

O R

C

OH H

+

R'OH

R

C OR' H

hemiacetal

 A hemiacetal or hemiketal has both – OR and –OH groups attached to the same carbon atom. 26

 Ketone undergoes similar reaction to give hemiketal. O R

C

OH R'

+

R''OH

R

C

OR''

R'

hemiketal  A hemiacetal or hemiketal in acidic medium can reacts further to produce an acetal or ketal. OH R

C OR' H

hemiacetal

+

R''OH

H

OR''

+

R

C OR' H

acetal

27

OH R

C

OR'''

+

OR''

+

H R'''OH

R'

hemiketal

R

C

OR''

R'

ketal

 An acetal has two – OR groups attached to the same carbon atom.

28

Example: 1. H CH3 CH2 CH2 C

O + CH3CH2CH2OH

HCl

2. OCH2CH2CH3 CH3 CH2 CH2 C

OH + CH3CH2CH2OH

HCl

H

29

3. CH3 CH3 CH2 C

O

+ CH3CH2CH2OH

HCl

4. OCH2CH2CH3 CH3 CH2 C

OH + CH3CH2CH2OH

HCl

CH3

30

e) Addition of Sodium bisulfite  Addition of sodium bisulfite (NaHSO3) to aldehyde or ketone give bisulfite salt. _ O C

O + NaHSO3

C

+ Na SO3H

OH C

_ SO3

+ Na

Example: 1.

CH3 C

O + NaHSO3

31

2. H CH3 CH2 C O + NaHSO3

32

Addition of Ammonia and its derivatives.  General reaction

C

O + H2N-Z

aldehyde or ketone

H3O+

C

N-Z + H2O

Z = -H, -OH, -NH2 ,etc.

Example:

33

R NH2 amine 10 H2N NH2 hidrazina C O

C NR C NNH2

imine hydrazone

H2N OH hidroxylamine

C NOH

H 2N

NH

C NNHC6H5

phenylhydrazine

phenylhydrazone O

O H2N NH C NH2 semicarbazide H2N

NH O2N

NO2

2,4-dinitrophenylhydrazine

oxime

C NNH C NH2 semicarbazone C N

NH O2 N

NO2

2,4-dinitrophenylhydrazone 34

(Brady’s reagent)

35

 When react with hydrazine, phenylhydrazine or 2,4-dinitrophenylhydrazine gives yellow precipitate which is used as a test to determine the carbonyl compounds.

a) Oxidation  Aldehydes are oxidized to carboxylic acid while ketone are hardly oxidized. Example : O CH3 (CH2)4 C H + K2Cr2O7

O H+

CH3 (CH2)4 C OH

36

b) Reduction  Aldehydes are reduced to primary alcohols and ketones to secondary alcohols.

 Reducing agents are : a) H2, Ni @ Pt b) NaBH4 (sodium borohydride) c) LiAlH4 (litium aluminium hydride)

OH

O R

C H

aldehyde

H2, Ni

R

C H H

alcohol 10

37

OH

O R

C R1

H2, Pt

ketone

R

C R1 H

alcohol 20

 Example: 1. O CH3 C H

LiAlH4 H+

ethanal

2. O CH3 C CH3 propanone

NaBH4/H+ or H2/pt

38

Some tests to determine carbonyl compounds Test 1

Purpose

Brady’s H2N

NH O2N

NO2

Observation

To determine carbonyl functional group

+ : yellow / orange precipitate

To determine aliphatic aldehyde

+ : brick-red precipitate

2,4dinitrophenylhydrazine

2

Fehling’s /Benedict Cu2+, OH -

39

3

Tollens’ [Ag(NH3)2] +,OH

4

Schiff’s

5

Iodoform I2, OH-

-

To determine aldehyde

+ : silver mirror

To determine aldehyde & propanone

+ : pinkish purple solution

To determine + : yellow methyl precipitate carbonyl & methyl alcohol structure, CH3 | -C=O

40

CH3 | -C-OH |

41

CARBOXYLIC ACID AND ITS DERIVATIVES

1

Structure O R

C

OH

RCOOH or RCO2H (R ≡ alkyl, aryl or H)

2

NOMENCLATURE

3

IUPAC Nomenclature & Common Name HCOOH Methanoic acid Formic acid

CH3CH2CH2COOH Butanoic acid Butyric acid

CH3COOH Ethanoic acid Acetic acid

CH3CH2COOH Propanoic acid Propionic acid

CH3CH2CH2CH2COOH Pentanoic acid Valeric acid

4

IUPAC Nomenclature •The longest chain must contain the carboxyl group. •The carboxyl group is at the terminal, therefore the carbon of the carboxyl group is not numbered. One COOH – carboxyl group is at one end Two COOH – carboxyl groups are at both ends •Name the compound as alkane, drop ‘e’ in alkane and add ‘oic acid’ (eg: methanoic acid)

5

CH3

Br

CH3

CH

CH

O CH2 C

OH

4-bromo-3-methylpentanoic acid CH3 HO

O

CH CH2 CH2 CH2 C

OH

5-hydroxyhexanoic acid O HO

C CH2 CH

CH3 CH

CH

CH3

5-methyl-3-hexenoic acid 6

•

Two COOH groups, the compound will be named as alkanedioic acid’ (Example: ethanedioic acid, propanedioic acid and etc)

HO

O

O

C

CH2 CH2 CH2 C

OH

pentanedioic acid

7

HO

O

CH3

O

C CH2

CH CH2 CH2 C

OH

3-methylhexanedioic acid

H

CH2 COOH C

HOOC CH2

C H

trans 3-hexenedioic acid 8

• When R is an aryl group, the parent name is benzoic acid

Cl

COOH

4-chlorobenzoic acid

9

•

An aromatic dicarboxylic acid is named as 1,x-benzenedicarboxylic acid

HOOC COOH

1,3-benzenedicarboxylic acid

HOOC CH3 CH CH3 COOH

2-isopropyl-1,4-benzenedicarboxylic acid 10

• A cyclic carboxylic acid is named as cycloalkanecarboxylic acid • The C atom which is attached to —COOH is numbered as C1

1

COOH

cyclopentanecarboxylic acid

11

COOH

1

cyclohexanecarboxylic acid CH3 Br

1

COOH

4-bromo-2-methylcyclohexanecarboxylic acid 12

A cyclic dicarboxylic acid is named as 1,x-cycloalkanedicarboxylic acid COOH

COOH 1,2-cyclohexanedicarboxylic acid COOH

COOH Cl 4-chloro-1,2-cyclohexanedicarboxylic acid 13

• When a compound contains a carboxyl group and other functional group(s), the priority is given to the carboxylic acid as the parent name. COOH

CH3 3-methyl-2-cyclohexenecarboxylic acid

14

PHYSICAL PROPERTIES OF CARBOXYLIC ACIDS

15

Boiling Point The boiling point of carboxylic acid is higher than an alcohol, a ketone or an aldehyde (with Mr that almost the same) because: i. it exists as stable dimers that form hydrogen bond. ii. molecules in dimers are arranged closely packed, therefore the hydrogen bonds are relatively strong. iii. high energy is needed to overcome the intermolecular forces ,  boiling point  16

Hydrogen bond O R

H

O C

C O

H

R

O Hydrogen bond

17

Solubility a) Solubility in water • Carboxylic acids are soluble in water due to the formation of hydrogen bond between the water molecules and carboxylic acid molecules. Hydrogen Bonds

H

O R

H O

H

O

H

C O

H

O

C H

R

O 18

• The solubility of carboxylic acid in water is almost the same as alcohol. • Aliphatic carboxylic acids with C > 5 are insoluble in water. Size of R ↑, hydrophobic area ↑.

O hydrophilic hydrophobic

R

C OH

19

• Aromatic carboxylic acids are slightly soluble in water due to the huge aromatic ring. • Dicarboxylic acids are relatively more soluble since more hydrogen bonds are formed.

20

Example : Descending order of solubility

COOH CH3 CH

COOH

CH2 CH2

COOH

> CH3 CH2 CH2

> COOH

COOH

> CH3 CH2 CH2 CH2

21

b.

Solubility in non polar solvent

•

Carboxylic acids are soluble in non polar solvent such as benzene due to the Van der Waals forces between the benzene and alkyl group of carboxylic acids . O R

H

O C

C O

H

R

O

Van der Waals forces

Van der Waals forces Hydrogen bonds 22

Acidity of Carboxylic Acid • The acidity of carboxylic acid is influenced by: i. Resonance effect ii.Inductive effect

23

Resonance Effect O Carboxylate ion

:

R

C

O-

OPhenoxide ion

:

24

• Carboxylic acids are more acidic due to the resonance stabilisation of the carboxylate ion. • The electrons in carboxylate ion are delocalised between two oxygen atoms, whereas in phenoxide ion, the electrons are delocalized in the benzene ring. • The C=O group in carboxylic acid is a electronwithdrawing group which reduce the electron density of –OH, therefore the –OH bond becomes weaker. • Thus H+ is easily donated and carboxylic acid is more acidic than phenol. 25

• Carboxylic acid is relatively a weak acid, however it is stronger than phenol & alcohol

OH O R

C

OH

>

>

H2O

>

R

OH

26

R

C

O

O

O

+ H2O

⇌

R

C

+ H3O+

-

O O carboxylate ion (resonance structure)

OH carboxylic acid

OH

+ H2O

⇌

R—O—H + H2O

O

-

+ H3O+

phenoxide ion (resonance structure)

phenol

alcohol

R

C

-

⇌

R—O– alkoxide ion

+

H3O+ 27

Inductive Effect O EWG

O

C

EDG O

C

-

An electron withdrawing group (deactivating group) that attached to a carboxylate ion will delocalise the negative charge, thereby stabilises the carboxylate ion and increases acidity.

O

-

An electron donating group, (activating group) will destabilise the carboxylate ion and decreases acidity.

28

i.

The inductive effect electron-withdrawing group in the compound

• electron-withdrawing groups (e.g –NO2 ,-F,-Cl,-Br, -I ) reduce the electron density of –O H. • Thus the O-H bond becomes weaker and H+ can be easily released. • The compound is said to be more acidic  Electron- withdrawing group increases the acidity. 29

• Example: CH3CHCl-COOH and CH3CH2COOH • Cl is an electron-withdrawing groups, therefore reduce the electron density of –OH. • Thus the O-H bond becomes weaker and H+ can be easily released. • Acidity : – CH3CHCl-COOH > CH3CH2COOH • Electron-withdrawing groups increase the acidity. 30

ii) The electronegativity of electronwithdrawing group in the compound • Example: CH3CHF-COOH and CH3CHCl-COOH • Both F and Cl are electron-attracting group. • The electronegativity of F > Cl • The electron density of –OH in CH3CHF-COOH is less, thus the –OH bond is weaker than in CH3CHCl-COOH. Therefore, H+ is easily donated. • Acidity : CH3CHF-COOH > CH3CHCl-COOH 31

iii) Number of electron-attracting group in the compound. • Example: CH3C(Cl)2-COOH and CH3CHCl-COOH • CH3C(Cl)2-COOH contains 2 Cl atoms that make the bond of –OH weaker than CH3CHCl-COOH (with only one Cl atom). Thus, H+ is easily donated. • Acidity : CH3C(Cl)2-COOH > CH3CHCl-COOH

32

iv) The position of electron-attracting group • Example: CH3CH2CH(Cl)COOH and CH2(Cl)CH2CH2COOH • The distance between Cl atom and carboxyl group in CH3CH2CHCl-COOH is nearer compare to in CH2ClCH2CH2-COOH. • The –OH bond in CH3CH2CH(Cl)COOH is weaker than in CH2ClCH2CH2-COOH, so H+ is easily donated. • Acidity : CH3CH2CH(Cl)COOH > CH2(Cl)CH2CH2COOH 33

(v ) The inductive effect of electron- releasing (electron-donating) group in the compound •

Example: CH3COOH and CH3CH2COOH

•

-R is an electron –releasing group.

•

The size of –R group in CH3CH2COOH is larger than in CH3COOH, so CH3CH2- is a stronger releasing group than CH3-.

•

The electron density of –OH in CH3CH2COOH increases and H+ is difficult to be donated.

•

 Electron-releasing groups reduce the acidity of a carboxylic acid. 34

SYNTHESIS OF CARBOXYLIC ACIDS

35

1. Oxidation of primary alcohol and aldehyde H R

C H 1o alcohol

R OH

oxidizing agent

R C

O

oxidizing agent

H aldehyde

C

O

HO carboxylic acid

Common oxidizing agents are : •KMnO4 / H2SO4 potassium permanganate •K2Cr2O7 @ Na2Cr2O7 /H2SO4 potassium /sodium dichromat (VI) 36

2. Oxidation of Alkyl Benzene

R

oxidizing agent

Cl

COOH

Cl CH3 CH

CH3

KMnO4 , H+ Δ

COOH

+ CO2 + H2O 37

3. Formation and Hydrolysis of nitrile R CH2 X

NaCN

CH2 Br

R CH2 CN

H2O,H+

CH2 CN

NaCN

R CH2 COOH

CH2 COOH

H2O,H+

38

4. Carbonation of Grignard Reagents

R—MgX

O

C O H2O, H+

CH2 MgBr

R—COOH + Mg(OH)X

CH2 COOH

CO2 H2O, H+

+ Mg(OH)Br

39

CHEMICAL PROPERTIES OF CARBOXYLIC ACIDS

40

•

Main reactions of carboxylic acid, a. The reaction that involves the donation of H+ from –OH group. b. The reaction that involves the substitution of OH group c. The reaction that involves the reduction with LiAlH4 to primary alcohol

41

a. The reaction that involves the donation of H+ from –OH group 1. Neutralisation • Carboxylic acids are acidic, it can react with base such as NaOH (aq) to give metal carboxylate salts, O R

C

O + NaOH

OH

R

C

+ H2O -

O Na+

42

COO– Na+

COOH

+ NaOH

+ H2O Sodium benzoate

43

2. Reaction with electropositive metals such as Na, K, Ca, Mg and Fe. O

O R

C OH

+

R

M

C O M

+

H2

Exercise: Cl

O C OH

+

K

44

b.

The reaction that involves the substitution of –OH group (to form its derivatives)

1. Acid chloride formation Acid chloride can be prepared from the reaction of carboxylic acids with thionyl chloride, SOCl2 ; phosphorous pentachloride, PCl5 ; phosphorous trichloride, PCl3 O SOCl2

R

O R

C OH

C Cl + SO2 + HCl O

PCl5

R

C Cl + POCl3 + HCl O

PCl3

R

C Cl + H3PO3

45

Exercise :

SOCl2 O CH3 CH C OH

PCl5

CH3 PCl3

46

2. Esterification Carboxylic acids react with alcohols in the presence of mineral acid catalyst to produce esters. O R

C

O OH + H—OR’

carboxylic acid

H+

⇌

R

alcohol

C

O R' + H2O

ester

O

O CH3 CH2 C OH + HOCH2CH3 propanoic acid

ethanol

H+

⇌

CH3 CH2 C OCH2 CH3 ethyl propanoate

+ H2O 47

3. Acid anhydride formation Acid anhydrides can be prepared from carboxylic acids by the loss of water through heating. O R

O

O

C OH + HO C

R

heat

O

R C

O

C R + H2O

acid anhydride

O

O

CH3 C

OH + CH3 C

O OH

heat

CH3 C

O O

C CH3

ethanoic anhydride

+ H2O 48

4. Amides formation Reaction of carboxylic acids with an ammonia or amine give amide. O NH3 R C NH2 + H2O 1o amide

O

O R

C Cl

RNH2

R

(1o amine)

C

NHR + H2O

2o amide

O R2NH (2o

amine)

R

C

NR2 + H2O

(3o amide) 49

Exercise : NH3

CH3

O

CH C

Cl

CH3 NH2

CH3 CH3 NH CH3 50

c. The reaction that involves the reduction with LiAlH4 to primary alcohol Carboxylic acid are reduced to primary alcohols by reaction with lithium aluminium hydride, LiAlH4 . O R

C

O

R'

LiAlH4 ether

R CH2 OH + R’OH 1o alcohol

O CH3 CH C O CH2 CH3 CH3

LiAlH4 ether

CH3 CH CH2 OH CH3 + HO—CH2CH3

51

Methanoic acid, HCOOH as a reducing agent O

• Methanoic acid molecule, H C OH has both O and C OH carboxylic

O H C carbonyl

• It shows the properties of both carboxylic acid and aldehyde. • It also shows reducing properties in reactions with acidified KMnO4 or K2Cr2O7 and Tollens’ reagent. 52

O

KMnO4 / H+

CO2 + H2O + MnO2

(Brown)

H C OH Ag(NH3)2+

Ag + CO2 + H2O

53

DERIVATIVES OF CARBOXYLIC ACIDS

54

Reactions of carboxylic acid derivatives i. Hydrolysis of acid chlorides

O

O

R C Cl

H2O

acid chloride

R C OH + HCl carboxylic acid

ii. Hydrolysis of acid anhydrides

O

O

R C O C R acid anhydride

O H2O

2 R C OH carboxylic acid 55

Reactions of carboxylic acid derivatives iii. Hydrolysis of esters

O

H2O H+

+ ROH carboxylic acid

R C OR ester

alcohol

O H2O + NaOH R C O Na

+ ROH alcohol

56

Example : O C Cl

H2O

benzoyl chloride

O

O

CH3 C O C CH3

H2O

ethanoic anhydride

57

Example :

O CH3 C O CH3

H2O H+

methyl ethanoate

58

Relative Reactivity Of Carboxylic Acid Derivatives • The reactivity of a carboxylic acid derivative depends on the basicity of the substituent (leaving group) that attached to the acyl group • The less basic the substituent, the more reactive the carboxylic acid derivative.

59

Relative basicities of the leaving group (substituent) Cl– < RCOO– < RO– < HO–

O

O

O

O

R C Cl , R C O C R , R C OR , R acid chloride

acid anhydride

ester

< NH2–

O C

O OH , R

carboxylic acid

C

NH2

amide

reactivity increases

60

ACYL CHLORIDE

• Acyl chloride is the most reactive because of its electropositive carbonyl group is attach to the electronegative Cl atom (which is a releasing group).

61

ANHYDRIDE ACID • Anhydride acid is more reactive than ester and amide because the carboxyl group of anhydride is attached to the carbonyl carbon. O R

C

O O

C

R'

• This makes the carbonyl carbon becomes more electropositive and can be easily attack by nucleophile.

62

ESTER • Ester is less reactive towards nucleophile because the delocalization of electron makes the positive charge of carbon can be shifted to oxygen. • That makes the carbonyl carbon less electropositive. H3C

H3C C R

O

+

C

O R

O

-

H3C

O

C R

-

O

+

O

63

AMIDE • Amide is the least reactive because, NH2 group is an electron-donating group that makes the carbonyl less electropositive. • The resonance structure of amide shows that the carbonyl carbon is not electropositive.

R

H

R

+

N

N -

O

H

H

O

H 64

The Uses of Carboxylic Acid Carboxylic acid / derivatives

Uses

Polyamide (Nylon)

carpet, apparel

Ester

Artificial flavors

Acetic acid

Vinegar

Ethanoic anhydride

Drug aspirin

Salicylic acid

analgesic

65

Chapter 19 : Amines 19.1 : Introduction 19.2 : Nomenclature 19.3 : Physical Properties 19.4 : Preparation 19.5 : Chemical properties

19.1

Introduction Amines are organic derivatives of ammonia, NH3 Classification of Amines Primary amines

H

N

H

R

Secondary amines

H

N

R'

R

Tertiary amines

R''

N R

R'


19.1

Classify the following amines: i. CH3CH2CH2NH2

1°

ii. CH3CH2

2°

N

H

Aliphatic amines

CH3

iii. CH3CH2

N

CH3

3°

CH3 iv.

N

CH3

2°

H v.

NH2

vi.

N

1°

CH3

3° CH2CH3

Aromatic amines


19.2

IUPAC Nomenclature In primary amines, the suffix ‘amine’ replaces the ‘e’ in the name of the parent alkane i. CH3CH2CH2NH2 3 2 1

propanee amine

Secondary and tertiary amines are named as N-substituted derivatives of primary amines ii. CH3CH2 2 1

2° N

H

CH3

N-methylethaneamine


19.2

Examples: Structure

CH3NH2 CH 3 CH(NH 2 )CH 3 NH2

CH 3CH 2CHCH2CH3

IUPAC Name Methanamine 2-propanamine Cyclopentanamine N-methylpentanamine

NHCH3 CH3 CH3CH2

N

CH2CH3

N-ethyl-N-methyl ethanamine


19.2

IUPAC Nomenclature When multiple functional groups are present and the –NH2 group does not take priority, it is named as an “amino” substituent Examples:

NH2CH2COOH 2

H 2N

1

4

1 3

2

2-aminoethanoic acid

OH

NH2

2,4-diaminophenol


19.2

IUPAC Nomenclature Aromatic amines are named as derivatives of aniline

N-ethyl-N-methylaniline


19.2

Common Name Most primary amines are named as alkylamines Structure

CH3NH2

IUPAC Name methylamine

CH 3CH 2NH 2

ethylamine

CH3CH2CH2NH2

propylamine

CH 3 NHCH 3

CH3NHCH2CH3

dimethylamine methylethylamine


19.3 Boiling point Solubility Basicity

(i) Boiling Point Both primary and secondary amines can form intermolecular hydrogen bonds. H CH3 N

HCH3 CH3 H CH3 NN CH 3

H hydrogen bonds

CH3

CH3

CH3N NCH H

CH3 N

3

CH3

H

no hydrogen bond

hydrogen bonds

1° amines 3° amines 2° amines Tertiary amines cannot form hydrogen bonds to each other.

CH3

Examples: The table below compares the boiling points of isomeric amines. Amines

Class

Relative molecular mass

Boliling points/oC

CH3CH2CH2NH2

1°

59

49

CH3CH2-NH-CH3 CH3 | CH3-N-CH3

2° 3°

59 59

37 4

Boiling Point : CH3CH2CH2NH2 > CH3CH2NH(CH3) > (CH3)3N



(i) Boiling Point Amines have higher boiling points than alkanes or haloalkanes of similar relative molecular mass due to intermolecular hydrogen bonding Example Amines have lower boiling points than alcohols or carboxylic acids of comparable molecular weight because H- bond in RNH2 is weaker than the H- bond in ROH and RCOOH (Nitrogen is less electronegative than the Oxygen)

Examples: The table below compares the boiling points of organic compounds of comparable relative molecular mass. Amines

Type


Boliling points/oC

CH3CH2CH2CH3 butane CH3CH2Cl chloroethane CH3CH2CH2NH2 1-propanamine

Alkane

58

-0.5

Haloalkane

64.5

12.5

Amine

59

48.6



(i) Boiling Point Amines have higher boiling points than alkanes or haloalkanes of similar relative molecular mass due to intermolecular hydrogen bonding Example Amines have lower boiling points than alcohols or carboxylic acids of comparable molecular weight because H- bond in RNH2 is weaker than the H- bond in ROH and RCOOH (Nitrogen is less electronegative than the Oxygen) Example

Examples: The table below compares the boiling points of organic compounds of comparable relative molecular mass. Amines

Type


Boiling points/oC

CH3CH2CH2CH3 butane CH3CH2Cl chloroethane CH33CH CH22CH22NH CH NH22 1-propanamine CH3CH2CH2OH 1-propanol

Alkane

58

-0.5

Haloalkane

64.5

12.5

Amine

59

48.6

Alcohol

60

97.2

Carboxylic acid

60

118

CH3COOH ethanoic acid

Boiling Point : Carboxylic acid > Alcohol > Amine > Haloalkane > Alkane



(ii) Solubility All amines including tertiary amines, are capable of forming hydrogen bonds with water molecules. Thus simple amines (amines of fewer than 5 carbons) are generally water soluble and dissolve to form basic aqueous solutions

Comparison the solubility between 1°,2° and 3°amine

R N R : N3°H< 2 ° < 1R ° N R TheHsolubility H R R

Increasing2osolubility

o

1 O

H

O

H

3o

H

O

H

H

H R

N

H

H O

O

H

H

R

N R

H

O

H

H

3 hydrogen bonds per a 1° amine molecule

N

R

R

H

H

R

: hydrogen bonds 2 hydrogen bonds per a 2° amine molecule

1 hydrogen bond per a 3° amine molecule


19.3 Boiling point Solubility

Basicity  An amine is a nucleophile (a Lewis base) base because its lone pair of non-bonding electrons on nitrogen

H

H

Basicity

H

N

+ H OH

H Ammonia

H

+

N

H + OH

H Ammonium ion

-



Basicity – (i) Inductive effect  An alkyl group is electron donating group, and it stabilises the alkylammonium ion by dispersing its positive charge H H R

N

+ H OH

H

1° Amine Stabilised by the alkyl group

R

+

N

H + OH

-

H

alkylammonium ion Stronger base


Boiling point

H

H

19.3 R

N

+ H OH

1° Amine

Basicity

R R

N H

N

H + OH

-

H alkylammonium ion

H

Solubility

R

+

R

+ H OH

2° Amine

R

+

N

H

+

OH

H alkylammonium ion

Strength as a base : Methyl, 1°, 2°

Increasing



Basicity – (ii) Resonance effect  Aromatic amines (e.g., aniline) are weaker bases than the corresponding aliphatic and cyclic amines Example:

NH2

Cyclohexylamine pKb = 3.36

NH2

Aniline pKb = 9.42



Basicity – (ii) Resonance effect Aromatic amines are less basic The lone pair electrons of nitrogen atom are delocalised and overlapped with the aromatic ring π electrons system make it less available for bonding to H+ NH2

NH2

NH2

arylamines are stabilised due to the 4 resonance structures

NH2


19.4 Reduction of nitro compound

Reduction of nitro compounds  Aromatic amines can be prepared by reduction of nitro compounds using Zn/H+ or SnCl2/H+

Reduction of nitriles Reduction of amides Hoffmann’s degradation of amides

Zn/H+ or NO2 nitrobenzene

SnCl2/H+

NH2 Aniline anline


19.4 Reduction of nitro compound Reduction of nitriles Reduction of amides Hoffmann’s degradation of amides

Reduction of nitriles  Nitriles are reduced to primary amines by H2 / catalyst, LiAlH4 / H+ or NaBH4 / H+. H2

R C N

catalyst (Pt, Ni, Pd)

RCH2NH2

Example : OH CN

1. LiAlH 4 2. H

+

OH CH 2 NH 2



Reduction of amides  Reduction of an amide using LiAlH4 / H+ or NaBH4 / H+ can yield a primary, secondary, or tertiary amine depending on the type of amide used O

1. LiAlH 4

R C NH 2 2. H +

H R C NH 2 H

+

H 2O


19.4 Reduction of nitro compound

Example : O

i. CH3

Reduction of nitriles Reduction of amides Hoffmann’s degradation of amides

C NH 2

H 1. LiAlH 4 2. H

+

CH3

H

Ethanamide (2o amide)

ii.

Ethanamine (1o amine)

O CH3 1. NaBH4 CH3

C NH

N-methylethanamide (2o amide)

C NH 2 + H2O

2. H

+

CH3

H CH 3 C NH + H2O H

N-methylethanamine (2o amine)



Hoffmann’s degradation of amides  On warming a primary amide with bromine in solution of NaOH, a primary amine is formed.  This reaction is used to synthesise primary alkyl and aryl amines. O R C NH2 amide

Br2 , NaOH H2O

RNH2

+

2CO3

amine

 The elimination of carbonyl group is shortening the length of carbon chain by one carbon atom. atom



Example : CONH2

NH2 Br2, OH-

+ H2O

CO32-


19.5 Reaction with : Acyl chloride Acid anhydride Benzenesulphonyl chloride

Nitrous acid Bromine water Formation of dye

Reaction with acyl chlorides  1° and 2° amines are acylated to form N-substituted amides O

H 2 R N H

+

Cl C CH 3

O

2 R N H + Cl C CH3 2o amine

R N C CH3

+

+

RNH3 Cl -

2o amide

1o amine

R

H O

R O R N C CH3

+

R2NH+2 Cl

-

3o amide

 3° amines are not acylated because they do not have a H atom attached to N atom.



Nitrous acid

Reaction with acid anhydride  1° and 2° amines are acylated to form N-substituted amides

1o amine

H O

O

R N H + CH3 C O C CH 3

R N

O

R 2o amine

+

CH3COO H R O

O

CH 3 C O C CH 3

R N C CH3

acid anhydride

3o amide +

Formation of dye

C CH 3

2o amide

acid anhydride +

R N H

Bromine water

O

H

CH3COO H

 3° amines are not acylated because they do not have a H atom attached to N atom.




Reaction with benzenesulphonyl chloride Hinsberg’s test  This reaction is used to differentiate between 1°, 2° and 3° amines.


19.5 Reaction with : Acyl chloride

1° amines  Benzenesulphonyl chloride reacts with a 1° amine to form a white precipitate (N-substituted sulphonamides)Acidic hydrogen

Acid anhydride

H

Benzenesulphonyl chloride

R-N-H


1o amine

O +

Cl-S-

OH

H O R-N -S-

(-HCl)

O

Na O R-N -S-

O N-substituted benzenesulphonamides (precipitate)

 N-substituted sulphonamides have an O acidic hydrogen, N-H.NaOH wateritsoluble salt in aqueous NaOH.  Therefore, dissolve (clear solution)


19.5 Reaction with : Acyl chloride

2° amines  Benzenesulphonyl chloride reacts with a 2° amines to form a white precipitate (N,N-disubstituted sulphonamide)

Acid anhydride

R'


R-N-H


o

2 amine

O +

Cl-S-

OH

(-HCl) O

R' O R-N -SO N,N-disubstituted benzenesulphonamides (precipitate)

 N,N-disubstituted sulphonamides do not No reaction NaOH have an acidic hydrogen, N-H.  Therefore, it dissolve in aqueous NaOH.


19.5 Reaction with :

3° amines  3° amine do not gives visible reaction with benzenesulphonyl chloride

Acyl chloride Acid anhydride Benzenesulphonyl chloride


Summary of Hinsberg’s test: NaOH

1° amines  white precipitate  solution

clear

NaOH

2° amines  white precipitate  dissolved

3° amines  do not give any visible change

do not




Reaction with nitrous acid (NaNO2 + HCl)  Nitrous acid (HNO2) is a weak and unstable acid.  It is always prepared in situ, by treating cold sodium nitrite (NaNO2) with an aqueous solution of a cold dilute hydrochloride acid (-5°C).  Nitrous acid reacts with all classes of amines Nitrous acid test can be used to distinguish: • 1° aliphatic and 1° aromatic amines • 1° and 2° aliphatic amines


19.5

1° amines (Aliphatic)  Primary aliphatic amines react with nitrous acid to yield highly unstable aliphatic diazonium salts

Reaction with : Acyl chloride

H

Acid anhydride Benzenesulphonyl chloride

R-N-H

NaNO2, HX o

-5 to 0 C

o

1 amine

Nitrous acid

+

[ R-N

NX ]

-N2

aliphatic diazonium salt (unstable)

Bromine water

Observation :

Formation of dye

Formation of gas bubbles (N2)

C=C

+

R

X

+

carbocation

+

ROH

+

RX


19.5 Reaction with :

1° amines (Aromatic)  Primary arylamines react with nitrous acid to give arenediazonium salts

Acyl chloride Acid anhydride Benzenesulphonyl chloride


N2+Cl

NH2 NaNO2, HCl

cold

+ NaCl + H2O

arenediazonium salt




1° aliphatic amines and 2° aliphatic amines Primary aliphatic amines Secondary aliphatic amines

Form a mixture of alkenes, alcohols, alkyl halides and nitrogen gas. Form secondary Nnitrosamines as yellow oil, which is stable under the reaction conditions.



Nitrous acid

Identification test Aniline reacts with aqueous bromine to yield white precipitates NH2 group is an activating and orthopara directors group

Reaction with bromine water

NH2

NH2 3Br2(aq)

Br

Br

+

room temperature

Bromine water Br

Formation of dye

(2,4,6-tribromoaniline) Observation: White precipitate formed

3HBr




Formation of dye Primary arylamines react with nitrous acid to give arenediazonium salts which are stable at 0 C Arenediazonium salts also undergo coupling reaction with aromatic compounds with strong electron donating group, such as –OH and –NR2 at the para position to yield azo compounds Azo compounds are usually intensely coloured and relatively inexpensive compounds, they are used as dyes


19.5 Reaction with : OH

Acyl chloride

N=N

Acid anhydride

OH

(orange) +


N N: X

N


CH3

CH3 CH3

N=N (yellow)

N

CH3

Chapter 20 : Amino Acids and Proteins 20.1 : Introduction

20.2 : Chemical Properties

Chapter 20 : Amino Acids and Proteins 20.1 : Introduction 20.1 -amino acid 20 amino acids Naming

Zwitterion Isoelectric point Structures in: Acidic Basic pI


Chapter 20 : Amino Acids and Proteins 20.1 : Introduction 20.1 -amino acid


Amino acid: acid a compound that contains both an amino group, -NH2 and a carboxyl group, -COOH H

20 amino acids Naming

CH 3 CH 2

C

Zwitterion Isoelectric point Structures in:

NH 2

-amino acid: an amino acid in which the amino group is on the carbon adjacent to the carboxyl group H

Acidic Basic pI

COOH

NH 2

C



COOH

CH 2 CH 3



Name Glycine Gly Alanine Ala Valine Val


Structure H

CH

COOH

NH2 H3C

CH

COOH

NH2 CH 3 HC

CH

CH 3 NH 2

COOH



Name Leucine Leu

Structure C H

C H C H C H


Isoleucine Ile Phenylalanine Phe

3 2

C H N H

3

CH 3CH 2CH

C O O H 2

CH

CH 3

COOH

NH2

CH2

CH COOH NH2


Zwitterion Isoelectric point


Name Threonine Thr Cysteine Cys

Structures in: Acidic Basic pI

Methionine Met

Structure OH CH

3

CH 2

C

CH

H

NH

COOH 2

CH 2

COOH C

CH 2

HO

NH

H

CH2 CH NH2

COOH



Name

Basic pI

Structure O

Tryptophan Trp Proline Pro

Structures in: Acidic


Serine Ser

OH NH2

NH

CH 2

CH 2

COOH C

CH 2

HO

NH

H

CH2 CH NH2

COOH




Name Structure OH Threonine CH3 C Thr H

Cysteine Cys

HS

CH2

CH NH

COOH 2

CH

COOH

NH2

Methionine CH S CH CH CH COOH 3 2 2

Met

NH2



Name

Structure

Tyrosine Tyr

HO

Asparagine Asn

Structures in: Acidic Basic pI


Glutamine Gln

CH2

CH

COOH

NH2

NH2─C─CH2─CH ─COOH ║ │ O NH2 NH2─C─CH2CH2─CH─COOH ║ │ O NH2




Name Structure Aspartic HOOC─CH2─CH─COOH │ acid NH2 Asp Glutamic HOOC─CH2CH2─CH─COOH │ acid NH2 Glu Lysine H2N─CH2CH2CH2CH2─CH ─COOH │ Lys NH2




Name Structure H2N─C─NH─(CH2)3─CH─COOH Arginine │ │ NH NH2 Arg Histidine His

HC─C=CH2─CH ─ COOH │ │ │ N NH NH2 CH



IUPAC Name

H

i) H

20 amino acids

2

3

H

pI

C

2

COOH 2-amino propanoic acid

1

NH2

Structures in:

Basic

1

CH3

ii)

Isoelectric point

Acidic

COOH 2-amino ethanoic acid

NH2

Naming

Zwitterion

C

CH2OH

iii)

3

H

C

2

NH2

COOH

1

2-amino-3-hydroxy propanoic acid



IUPAC Name CH2CH2CH2NH2

iv)

20 amino acids

3

H

4

C

COOH

2

Naming

5

1

NH2

2,5-diamino pentanoic acid


v)

CH2CH2CH2COOH

3

Structures in: H

Acidic Basic pI

C

2

NH2

4

5

6

COOH

1

2-amino hexanedioic acid

Chapter 20 : Amino Acids and Proteins 20.1 : Introduction 20.1

Zwitterion Amino acids can undergo an internal acid-base reaction, in which a proton is transferred from the carboxyl (-COOH) to the amino group (-NH2) to form dipolar ion called zwitterion.

-amino acid 20 amino acids Naming



CH2CH2CH3 H

C

COOH

NH2

(neutral)

CH2CH2CH3 H

C

-

COO

+

NH3

(dipolar ion @ zwitterion)



20.1

Isoelectric Point (pI)

-amino acid

In general, zwitterion is electrically neutral and exists at specific pH

20 amino acids Naming


This particular pH is called isoelectric points Each amino acid has it’s specific pI H2N─(CH2)4─CH ─COOH 9.7 Lysine Example: │

(Lys)

Amino Acid

NH2 Structure

Aspartic Glycine acid (Gly) (Asp)

HOOC─CH ─CH─COOH H CH 2 COOH │ N H 2 NH2

pI 2.9 6.1



In Acidic The cation, H3N+— CH2 — COOH predominates (Amino acids as base) c In Basi The anion, H2N— CH2 — COOpredominates (Amino acids as acid) I n pI The zwitterion, H3N+—CH2—COOpredominates (Amino acids as amphoteric)

20.1 -amino acid 20 amino acids Naming


NH2CH2COOH

Gly (pI = 6.1) NH3CH2COOH ↔NH3CH2COO ↔ NH2CH2COO +

+

at pH < 6.1

at pH = 6.1

at pH > 6.1

Chapter 20 : Amino Acids and Proteins 20.1 : Introduction 20.2 Reactions of –NH2 group: HCl HNO2 Reactions of -COOH group: NaOH ROH Peptide bond Importance




a) Reaction with HCl R' R C NH2

+

HCl

R' _ + R C NH3Cl COOH

COOH Example:

CH2CH3 H C NH2 COOH

CH2CH3 +

HCl

+

H C NH3Cl COOH

_

Chapter 20 : Amino Acids and Proteins 20.1 : Introduction 20.2 Reactions of –NH2 group: HCl


b) With Nitrous Acid (HNO2) R'

R'

R C NH2 COOH

NaNO2 / HCl o

-5 to 0 C

HNO2 Reactions of -COOH group:

R C

N

N

COOH diazonium ion elimination of N2(g)

Observation: Bubbling of gas (N2)

R' R C+ COOH carbocation

NaOH ROH

+

H 2O

Peptide bond

2-hydroxy

Importance

carboxylic acid

+ Cl

-

halocarboxylic acid

-H

+ Alkenoic acid

Chapter 20 : Amino Acids and Proteins Example:

CH3

H C NH2

NaNO2 / HCl

COOH

CH3 H C

+

N

N

COOH diazonium ion elimination of N2(g) CH3 H C+ COOH carbocation H 2O

CH3 H C OH COOH

+ Cl

-H

-

CH3 H C

Cl

COOH

+

CH2 H C COOH



c) Reaction with NaOH R' R C NH2

+

NaOH

R' R C NH2 _

COO Na+

COOH Example:

CH2CH3

CH2CH3 H C NH2 COOH

+

NaOH

H C NH2 _

COO Na+



d) Reaction with ROH + HCl (catalyst) R' R C NH2

+

R''OH

COOH

R' R C NH2

+ H 2O

COOR''

Example: CH2CH3 H C NH2 COOH

+

HCl

CH3OH

R' R C NH2 COOR''

+ H 2O



Peptide Bond 2 amino acids react together, H2O is eliminated. This is condensation reaction. Peptides are amino acid polymers in which the individual amino acid units, are linked together by amide bonds, or peptide bonds •2 amino acids form dipeptide •3 amino acids form tripeptide •15 – 30 amino acids form oligopeptide •> 30 amino acids form polypeptide



Formation of dipeptide bond: O

H H2N CH2 C OH + H N C COOH H CH3 O

Notes:

H

H2N CH2 C

N C COOH + H O N-terminal (with free –NH2) 2is H CH3

always written on the left and C-terminal (with amide bond / free –COOH) at thepeptide right side bond A dipeptide

Chapter 20 : Amino Acids and Proteins Formation of tripeptide bond: O H2N CH2 C OH

H H N C COOH H

Gly

CH2

OH

H N C COOH

CH3

H

H Tyr

Ala

Write the structural formula of tripeptide with the following sequence Gly-Ala-Tyr : O

H O

H2N CH2 C N C C H CH3

CH2 N C COOH H

H

A tripeptide

OH



Importance of amino acids :  Human beings can synthesize about half of the amino acids needed to make proteins.  Other amino acids, called the essential amino acids, acids must be provided in the diet.  The ten essential amino acids are : arginine (Arg) valine (val) methionine (Met) threonine (Thr) leucine (Leu) phenylalanine (Phe) histidine (His) isoleucine (Ile) lysine (Lys) trypthophan (Trp)

Chapter 20 : Amino Acids and Proteins 20.1 : Introduction 20.2 Reactions of –NH2 group: HCl HNO2 Reactions of -COOH group: NaOH


Importance of proteins :  Proteins are the most abundant organic molecules in animals, playing important roles in all aspects of cell structure and function.  Examples of protein functions : Class of protein structural protein

Examples collagen, keratin

Importance

tendons, skin hair, nails

ROH Peptide bond

Functions

enzymes

DNA polymerase

repair DNA

transport protein

hemoglobin

transport O2

CHAPTER 21 POLYMERS FJ/Chemistry Unit, KMPk/Mac 2006

1

21.1 : Introduction -

Polymer is a macro molecule that is made up of many small repeating units called monomers linked together by covalent bond.

nA

-A-A-A-A-A-A-A-A-n

monomer -

polymer

polymer can be represented by their repeating unit in the long chain.

nA

-A-n polymer (repeating unit)

-

Monomer is a basic molecular units that can joined to many others to form a polymer.

2

Homopolymer -

a polymer formed from only one types monomer units.

nA

-

-A-n

Example :

nCH2 CH2 ethene (ethylene)

-CH2CH2-n polyethene (polyethylene)

3

Copolymer -

a polymer formed from two or more different monomers.

nA + nB

-

-A-B-n

Example :

nCH2 CCl2 + nCH2 CHCl 1,1-dichloroethene 1-chloroethene

-CH2-CCl2-CH2-CHCl-n Saran®

4

21.2 : Structures of Polymers 21.2.1 : Linear Polymers Linear polymer is a polymer consist of monomers that are linked in straight and long continuous chain. Linear or straight-chain polymer can be folded back upon themselves in a random fashion. Linear polymer is recycleable because it is soft and can be reformed when heated.

-A-A-A-A-A-A-A-A-n monomers link together in a straight chain folded linear polymer in a random fashion

5

21.2.2 : Cross-linked Polymers Cross-linked polymer contains branches that connect linear polymer chain, as shown in the figures below.

-A-A-A-A-A-A-A-A-A-A-A-AX

X

-A-A-A-A-A-A-A-A-A-A-A-A• Long polymer chain cross-linked by branched Cross-linked polymer is harder (rigid) and more elastic polymer compared to linear polymer. It also cannot be remelted or remolded again. The equation below shows the chemical reaction for the rubber vulcanization process.

6

In a vulcanization process, long chain of polyisoprene are cross-linked by sulphur atoms.

CH3

CH3

-CH2C CH-CH2-

+

-CH2C CH-CH2-

sulphur CH3

CH3

-CHC CH-CH2-CHC CH-CH2S

S

-CHC CH-CH2-CHC CH-CH2CH3

CH3

After vulcanization rubber becomes more stable over wide ranges of temperature and far more durable than natural rubber.

7

21.3 : Types of Polymers 21.3.1 : Natural Polymers Natural polymers are polymers that synthesized in nature or naturally occurring polymers. i.

Proteins Protein are the most abundant organic molecules in animals such as enzymes, hormones, hemoglobin and many other things. It is a natural polymer built from amino acids linked by amide bonds.

H O HN-C-C R

n

8

ii. Carbohydrates Carbohydrates such as starch is also a polymer. i.

Starch (amylose)

H O

CH2OH O HH H OH H O

CH2OH O H H OH H O

H

H

OH

OH

α-(1,4)-glycosidic linkages

n

ii. Starch (amylopectin) -

contains linkages.

α-(1,4)-glycosidic

and

α-(1,6)-glycosidic

9

iii. Natural Rubber Natural rubber is a polymer of 2-methyl-1,3-butadiena, also known as isoprene.

n CH2=C-CH=CH2

-CH2-C=CHCH2-

CH3

CH3

isoprene

cis-polyisoprene

n

Another name for natural rubber is cis-1,4-polyisoprene. All the double configuration.

bond

in

natural

rubber

are

cis-

Natural rubber is soft, not strong or elastic and sticky, that makes it less useful. In order to make it more useful rubber has to undergo a vulcanization process.

10

21.3.2 : Synthetic Polymers Synthetic polymers are polymers that are prepared in industries from monomers that have gone through polymerization process. Polymerization is a process that combines monomers to form polymers. Synthetic polymers can be classified base on their method of preparation ( polymerization process ). 21.3.2.1 : Addition Polymerization Addition polymerization is the addition reaction in which unsaturated monomers are joined together by covalent bonds to form a polymer without elimination of a small molecule. Polymers obtained by addition polymerization are called addition polymers.

11

Therefore, addition polymer always involves the polymerization of monomers which have double bond within the monomers. Peroxide is used as initiator in addition polymerization. i.

Formation of polyethene

n CH2 CH2

ROOR

ethene

-CH ( 2-CH2)n polyethene

ii. Formation of polyvinyl chloride

n HC CH2 Cl

chloroethene

ROOR

( -HC-CH )n 2Cl

polychloroethene (polyvinyl chloride or PVC)

12

iii. Formation of polystyrene

n HC CH2

ROOR

phenylethene (styrene)

( -HC-CH )n 2-

Polyphenylethene (polystyrene)

iv. Formation of polyisopropene

n CH 2 =C-CH=CH 2

-CH 2 -C=CHCH 2-

CH 3

CH 3

isoprene

polyisoprene

n

13

21.3.2.2 : Condensation Polymerization Condensation polymerization is a process that combines the monomers with elimination of a small molecule such as water, methanol, hydrogen chloride or ammonia to form a polymer. The polymers obtained from condensation polymerization are called condensation polymers. The monomers involves in this polymerization must have at least two identical or different functional group in the molecule. When a carboxylic acid with two –COOH group reacts with an amine with two –NH2 groups, a polyamide is formed. When a carboxylic acid with two –COOH group reacts with an alcohol with two –OH groups, a polyester is formed.

14

(A). Polyamides i.

Formation of Nylon 6,6

O

O

n H2N-(CH2)6-NH2 + n HO-C-(CH2)4-C-OH hexane-1,6-diamine

hexane-1,6-dioic acid

O H H O ( -N-(CH )n + n H2O 2)6-N-C-(CH2)4-CNylon 6,6 ii. Formation of Nylon 6

O n H2N-(CH2)5-C-OH 6-aminohexanoic acid

O ( -NH-(CH )n + n H2O 2)5-CNylon 6

15

iii. Formation of Kevlar

H n H-N-

-N-H + n Cl-C-

1,4-diaminobenzene

H N-

O

H

O -C-Cl

terephthalic acid

H O

O

-N-C-

-C n

+ n HCl

Kevlar (B). Polyester The repeating functional groups in this polymer chain are ester. The most familiar polyester is polyethylene terephthalate known as Dacron and Terylene.

16

The polymer is formed by the reaction of ethylene glycol with methyl ester or tetephtalic acid. In this process, a molecule of methanol is split out for each new ester group formed. i.

Formation of Dacron

O n CH3O-C-

O -C-OCH3 + n HO-CH2CH2-OH 1,2-ethanediol (ethylene glycol)

dimethyl terephthalate

O ( -O-CH 2CH2-O-C-

O -C)n + n CH3OH

Dacron or poly(ethylene terephthalate)

metanol

17

ii. Formation of Terylene

O n HO-CH 2CH 2-OH

+ n HO-C-

ethane-1,2-diol

O -C-OH

benzene-1,4-dicarboxylic acid

O ( -O-CH 2CH 2-O-CTerylene

O -C)n + n H 2O water

18

21.4 : Uses of Synthetic Polymers No.

Polymer

Uses

1

Polyethene

plastics, drinking bottles, toys

2

Polyvinyl chloride

3

Polystyrene

4

Nylon 6

Textile

5

Nylon 6,6

Sweater

6

Kevlar

Bullet proof vest

7

Dacron

Fabric

8

Terylene

piping, floor tiles, clothing, toys, wire covering containers, thermal insulation-ice buckets

Fiber-optic material

19

Chemistry Malaysian Matriculation Full Notes & Slides for Semester 1 and 2

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