1.2 ALKANES AND CYCLOALKANES PREPARED BY SYED ABDUL ILLAH ALYAHYA UNIVERSITI TEKNOLOGI MARA PUNCAK ALAM CAMPUS
1
1.2 ALKANES AND CYCLOALKANES • • • • • • • • • • •
Introduction to Hydrocarbons Alkanes and Cycloalkanes Nomenclature of Alkanes and Cycloalkanes Reactions Reacti ons of Alk Alkanes anes and Cycloalkanes Cycloalkanes Alkenes and Cycloalkenes Nomenclature of Alkenes and Cycloalkanes Reactions Reacti ons of Alk Alkenes enes and Cyc Cycloalk loalkenes enes Alkynes Nomenclatur Nomencla ture e of Alkyn Alkynes es Reactions of Alkynes Aromatic hydrocarbons and Nomenclature 2
Introduction Intr oduction to Hydr Hydrocarbons ocarbons •
•
•
•
Hydrocarbons Hydrocarbons contain only C and H – aliphatic or aromatic Insoluble in water – no polar bonds to attract water molecules Aliphatic hydrocarbons – satura saturated ted or o r unsaturated aliphatics – saturated = alkanes, unsaturated = alkenes or alkynes – may be chains (no rings or acyclic) or rings (cyclic e.g :cycloalkanes is alcyclic). – chains may be straight or branched Aromatic Aromatic hydr h ydrocarbons ocarbons 3
Introduction to Hydrocarbons
4
•
Alkanes are saturated hydrocarbons
•
Single bonds with C-C and C-H
•
Formula :CnH2n+2 n= integer 1,2,3,4….
•
Carbon can form sp 3 hybridized
•
Aliphatic (Greek aleiphas = fat) thus, like fat
5
•
Methane, CH4
H H
C H 1
1
H
•
Ethane, CH3CH3 H
H
H
C
C
H
H
1
2
H
2 1 6
•
Propane, CH3CH2CH3 H H
C
H 1
H •
C H
H 2
C
3
H
2
1
H
3
Butane, CH3CH2CH2CH3
H H H H H
C1 C 2 C 3 C 4 H H H H H
1
2
4 3 7
Straight chain alkanes •
Butane, CH3CH2CH2CH3
•
Pentane, CH3CH2CH2CH2CH3
•
Hexane, CH3CH2CH2CH2CH2CH3
•
Heptane, CH3CH2CH2CH2CH2CH2CH3
•
Octane, CH3CH2CH2CH2CH2CH2CH2CH3
•
Nonane,CH3CH2CH2CH2CH2CH2CH2CH2CH3
•
Decane, CH3CH2CH2CH2CH2CH2CH2CH2CH2CH3
8
Alkanes Branch chain alkanes CH3 CH3CHCHCH3 CH3
CH3 CH3 CH3CH2CHCH2CHCH2CH3
9
•
•
•
Alkyl group – remove one H from an alkane (a part of a structure) General abbreviation “R” (for Radical, an incomplete species or the “rest” of the molecule) Name: replace -ane ending of alkane with - yl ending – CH is “methyl” (from methane) 3 – CH CH is “ethyl” from ethane 2 3
See page 84 - 85
10
Alkyl groups 1
Methyl
Me
-CH3
2
Ethyl
Et
-CH2CH3
3
Propyl
Pr
-CH2CH2CH3
4
But yl
Bu
-CH2CH2CH2CH3
5
Pent yl
-CH2CH2CH2CH2CH3
6
Hexyl
-CH2(CH2)4CH3
7
Hept yl
-CH2(CH2)5CH3
8
Oct yl
-CH2(CH2)6CH3
9
Nonyl
-CH2(CH2)7CH3
10
Decyl
-CH2(CH2)8CH3
See page 84 - 85 11
See page 84 - 85
12
How many carbons?
Prefix
–
Where and what are substituents?
See page 81 - 86
Parent - Suffix
What is the primary functional group?
13
Step 1: Find the parent hydrocarbon chain a) Find the longest continuous chain of carbon atoms as the parent compound. Doesn’t matter whether the carbon skeleton is drawn in an extended straight-chain form or in one with many bends and turns CH2CH3 CH3CH2CH2CH
CH3
Named as a substituted hexane
CH2CH3 H3C CHCH
CH2CH3 Named as a substituted heptane
CH2CH2CH3
14
Step 1: Find the parent hydrocarbon chain b) If two different chains or equal length are present, choose the one with the larger number of branch points as the parent CH3
CH3
CH3CHCHCH2CH2CH3
CH3CHCHCH2CH2CH3
CH2CH3
CH2CH3
Named as a hexane with two substituents
Named as a hexane with one substituent NOT
YES
15
Step 2: Number the atoms in the main chain a) Beginning at the end nearer the first branch point, number each carbon atom in the parent chain. 2 1 CH2CH3 4 H3C CHCH 3
6 7 CH2CH3
CH2CH3
CH2CH2CH3 5 6 7 YES
4 H3C CHCH 5
CH2CH3
CH2CH2CH3 3 2 1 NOT
The first branch occurs at C3 in the proper system of numbering, not at C4
16
Step 2: Number the atoms in the main chain b) If there is branching an equal distance away from both ends of parent chain, begin numbering at the end nearer the second branch point 8 9 CH2CH3
CH3 CH2CH3
H3C CHCH2CH2CH 4 7 6 5
YES
CHCH2CH3 3 2 1
2 1 CH2CH3
CH3 CH2CH3
H3C CHCH2CH2CH 3 4 5 6
CHCH2CH3 7 8 9
NOT 17
Step 3: Identify and number the substituents a) Assign a number, called a locant, to each substituent to locate its point of attachment to the parent chain. 8 9 CH2CH3
CH3 CH2CH3
H3C CHCH2CH2CH 4 7 6 5 Substituents :
CHCH2CH3 3 2 1
Named as a nonane
on C3 , CH2CH3 on C4 , CH3 on C7 , CH3
(3-ethyl) (4-methyl) (7-methyl) 18
Step 3: Identify and number the substituents b) If there are two substituents on the same carbon, give both the same number. There must be as many numbers in the name as there are substituents CH3 CH3 4 CH3CH2CCH2CHCH3 3 2 1 5 6
Named as a hexane
CH2CH3 Substituents :
on C2 , CH3 on C4 , CH3 on C4 , CH2CH3
(3-methyl) (4-methyl) (4-ethyl) 19
Step 4: Write the name as a single word Use hyphens (-) to separate the different prefixes, and use comma to separate numbers If two or more substituents are present, cite them in alphabetical order If two or more identical substituents are present, use prefixes di-, tri-, tetra, and so forth
20
Step 4: Write the name as a single word 8 9 CH2CH3
CH3 3 CH3CHCHCH2CH2CH3 1 2 4 5 6
CH3 CH2CH3
H3C CHCH2CH2CH 7 6 5 4
CHCH2CH3 1 3 2
CH2CH3 3-Ethyl-2-methylhexane
3-Ethyl-4,7-dimethylnonane 2 1 CH2CH3
CH3 CH3
4 H3C CHCH 3
4 CH3CH2CCH2CHCH3 3 2 1 5 6
CH2CH3 4-Ethyl-2,4-dimethylhexane
CH2CH3
CH2CH2CH3 5 6 7
4-Ethyl-3-methylheptane
21
Step 5: Name a complex substituent as though it were itself compound CH3
CH3
6 3 CH3CHCHCH2CH2CH 4 1 2 5
CH3
CH3 CH2CHCH3 1' 2' 3'
CH2CH2CH2CH3 7 8 9 10
2,3-Dimethyl-6-(2-methylpropyl)decane or 2,3-Dimethyl-6-isobutyl decane
H3C
H C 1'
CH3 2'
CH3CH2CH2CHCH2CH2CH3 1 2 3 4 5 6 7
-CH2CHCH3 2-methylpropyl group or Isobutyl
4-(1-Methylethyl)heptane or 4-Isopropyl heptane
22
For historical reasons, some of the simpler branchedchain alkyl groups also have nonsystematic, common names Three-carbon alkyl group
23
Four-carbon alkyl groups
24
When writing an alkane name, the nonhyphenated prefix iso is considered part of the alkyl group name for alphabetizing purpose, but the hyphenated and italicized prefixes, sec and tert are not 2 1 CH2CH3
2' CH3 8 9 CH2CH3 H3C
C 1'
CH3
H3C CHCH2CH2CH2 CHCH2CH3 7 6 5 4 1 3 2 3-(1,1-Dimethylethyl)-7-methylnonane or 3-tert -Butyl-7-methylnonane
4 5 6 7 8 H3C CHCHCH2CH2CH2CH3 3 1'
CHCH2CH3 2' 3' CH3
4-(1-Methylpropyl)-3-methyloctane or 4-sec -Butyl-3-methyloctane25
Exercise 7 :Give IUPAC names for the following compounds (a)
CH2CH(CH3)2
CH3CHCH2CH2CHCH2CH2CH3
CH3
CH3
(CH3)2CHCH2CHCH3
CH3
CH3CHCH2CHCHCH2CHCH2
CH3 (c)
CH2CH3
(b)
(d)
CH2CH3
CH3
C(CH3)3
CH3CH2CHCHCH2CH2CH2CH2CH2CH3 CH2CH3 26
Exercise 8 : Draw structures corresponding to the following IUPAC names (a) 3,4-Dimethylnonane (b) 3-Ethyl-4,4-dimethylheptane (c) 2,2-Dimethyl-4-propyloctane (d) 2,2,4-Trimethylpentane (e) 4-Ethyl-5-isopropyl-3,6-dimethyloctane
27
Saturated cyclic hydrocarbons are called cycloalkanes, or alicyclic compounds (aliphatic cyclic) Because ring is consist of -CH2- units, they have general formula (CH2)n or CnH2n
cyclopropane cyclobutane cyclopentane cyclohexane
28
Step 1: Find the parent hydrocarbon chain Count the number of carbon atoms in the ring and the number in the largest substituent chain If the number of carbon atoms in the ring is equal to or greater than the number in the substituent, the compound is named as an alkyl-substituted cycloalkane. If the number of carbon atoms is the largest substituent is greater than the number in the ring, the compound is named as a cycloalkyl-substituted alkane. 29
Step 1: Find the parent hydrocarbon chain CH2CH2CH2CH3
CH2CH2CH3 5-carbons
3-carbons
3-carbons
4-carbons
1-Cyclopropylbutane
Propylcyclopentane
1-Cyclobutylpentane
Hexylcyclohexane 30
Step 2 : Number the substituents, and write the name For an alkyl- or halo-substituted cycloalkane, choose a point of attachment as carbon 1 and number the substituent on the ring so that the second substituent has as low a number as possible. CH3
CH3
1
1
6
2
2
6
5
3
3
5
CH3
4
4
CH3
1,3-Dimethylcyclohexane
1,5-Dimethylcyclohexane
YES
NOT
Lower
Higher
31
If ambiguity still exists, number so that the third or fourth substituent has a low a number as possible, until a point of difference is found. CH2CH3
CH2CH3 H3C 6
H3C
1
1 2
CH3
1-Ethyl-2,6-dimethylcycloheptane
Higher
3 4
CH3
3-Ethyl-1,4-dimethylcycloheptane
Higher
32
CH2CH3 H3C
2 4
1
CH3
2-Ethyl-1,4-dimethylcycloheptane
Lower
Lower
33
Step 2 : Number the substituents, and write the name (a) When two or more different alkyl groups that could potentially receive the same numbers are present, number them by alphabetical priority. CH3
CH3 3 4
1
2
5 1
CH2CH3
YES 5 1-Ethyl-2-methylcyclopentane
2
4
CH2CH3
3
NOT 2-Ethyl-1-methylcyclopentane 34
Step 2 : Number the substituents, and write the name (b) If halogens are present, treat them just like alkyl groups CH3
CH3 2
1
1
2
Br YES
Br
NOT
1-Bromo-2-methylcyclobutane 2-Bromo-1-methylcyclobutane
35
Step 2 : Number the substituents, and write the name (c) Some additional examples follow: Cl
Br
CH3
H3CH2C
CH3
1-Bromo-3-ethyl-5-methylcyclohexane
CH2CH3 1-Chloro-3-ethyl-2-methylcyclopentane
36
Exercise 9 : Give IUPAC names for the following cycloalkanes Br CH3 CHCH2CH3 H3C
C(CH3)3
Br
CH3
CH2CH3 37
Exercise 10 : Draw structures to the following IUPAC names (a) 1,1-Dimethylcyclooctane (b) 3-Cyclobutylhexane (c) 1,2-Dichlorocyclopentane (d) 1,3-Dibromo-5-methylcyclohexane
38
•
• •
•
•
Alkanes is saturated hydrocarbon therefore they do not react as most chemicals Burning in flame produce carbon dioxide and water React with chlorine Cl2 in a presence of light to replace H’s with Cl’s Boiling point increases as the number of carbon increases Dispersion forces increases as the molecular size increases, resulting melting point and boiling point increases
39
Properties of Alkanes and Cycloalkanes
40
Properties of Alkanes and Cycloalkanes
41
.. .. C l ..
.. .. .. . + . .. Cl Cl .. ..
H C Cl
H .. .. H C . + . Cl ..
.. .. C .. l
H
H
H
H
H
H C Cl
H C
Methyl radical +
H
H
H
H
H C H H
Chlorine radical
Carbocation -.
H C. H
.. . .. + . Cl ..
+
H+
Carbanion
42
•
Free Radical Substitution Reaction – –
replace H with a halogen atom initiated by addition of energy in the form of heat or ultraviolet light •
–
to start break breaking ing bonds
generally get multiple products with multiple substitutions through chain reactions H H C H + H
Cl
Cl
Heat OR UV light
H H C Cl
+
HCl
H 43
• •
Chlorination of Methane Methane reacts reacts with chlorine chlorine in the presence of heat or light to produce a mixture mixture of products products through through a sequence of substitution reactions reactions in which which the H atoms are successively replaced by Cl atoms. atoms. CH4 + Cl2
heat or light
CH3Cl + HCl Cl2
CH2Cl2 + HCl CH3Cl , chloromethane CH2Cl2 , dichloromethane CHCl3 , trichloromethane CCl4 , tetracloromethane
Cl2
CHCl3 + HCl Cl2
CCl4 + HCl 44
Free Radical Substitution •
•
•
The breaking of a covalent bond so that each atom retains one of the shared electrons will produce two . Free radicals are very reactive. They are intermediates in reaction mechanism. The substitution reactions of alkanes follow the free radical mechanism, which involved three major steps: 1. Initiation: Initial production of free radicals 2. Propagation: Free radicals attack molecules to produce another free radical 3. Termination: Two free radicals combine to form a molecule 45
Dissociation of a chlorine molecule into two chlorine atoms
Cl
2 Cl
Cl
The Cl free radical attacks the CH4 molecule to give a methyl free radical Step 1: Hydrogen atom is removed from methane by a chlorine atom H
H H
C
+
H
C
Cl
H
+
H
Cl
H
H
Step 2: Reaction of methyl radical with molecular chlorine H
H Cl
Cl
+
C H
H
Cl
C
H
+
Cl
H 46
Free radicals combine to produce molecules and reaction stops H H
H +
C
C
H
H
H
H
H Cl
+
C H
H
H
C
C
H
H
H
H H
Cl
C
H
H
47
Exercise 11 : Write the free radical substitution mechanism for the monochlorination of the following molecules. (a)Ethane (b)Cyclopropane
48
Also known as olefins •Aliphatic, unsaturated C=C double bonds •Formula for one double bond = subtract 2 H from alkane for each double bond •Trigonal shape around C flat •Polyunsaturated = many double bonds •
49
H
H C
H
H
C H
H C H
H C
C
H
H
propene = propylene used to make polypropylene
ethene = ethylene produced by ripening fruit used to make polyethylene
50
Straight chain alkenes Ethene
CH2=CH2
Propene
CH2=CHCH3
1-Butene
CH2=CHCH2CH3
2-Butene
CH3CH=CHCH3
1-Pentene
CH2=CHCH2CH2CH3
2-Pentene
CH3CH=CHCH2CH3
1-Hexene
CH2=CHCH2CH2CH2CH3
2-Hexene
CH3CH=CHCH3CH2CH3
3-Hexene
CH3CH3CH=CH2CH2CH3 51
Cyclopropene
Cyclopentene
Cyclobutene
Cyclohexene
52
Step 1: Find the parent hydrocarbon chain Find the longest, continuous C chain that contains the double bond and name the compound accordingly, using the suffix -ene H
CH2CH3 C
H
C
H C
YES CH2CH2CH3
Named as pentene
CH2CH3
H
C
NOT CH2CH2CH3
as hexene, since the double bond is not contained in the six-carbon chain
53
Step 2: Number the carbon atoms in the chain Begin at the end nearer the double bond, or if the double bond is equidistant from the two ends, begin at the end nearer the first branch point. CH3 CH3CHCH 2 3 1
CH3
YES CHCH2CH3 4 5 6
CH2CH3
CH3CHCH2CH CHCHCH2CH3 1 2 3 4 5 6 7 8
CH3 CH3CHCH 5 4 6
CH3
NOT CHCH2CH3 3 2 1
CH2CH3
CH3CHCH2CH CHCHCH2CH3 8 7 6 5 4 3 2 1
54
Step 3: Write the full name Indicate the position of the double bond by giving the number of the first alkene carbon and placing that number directly before the parent name. Number the substituents according to their positions in the chain CH3 CH3CHCH 2 3 1
CHCH2CH3 4 5 6
At C3 : double bond, 3-hexene At C2 : substituent, CH3 (2-methyl) 2-Methyl-3-hexene
55
Write the name in the following order If more than one substituent attached to main chain number of first alphabetical substituent: use prefixes to indicate multiple identical substituents repeat for other substituents CH3
CH2CH3
CH3CHCH2CH CHCHCH2CH3 1 2 3 4 5 6 7 8 At C4 : double bond, 8-octene At C2 : substituent, CH3 (2-methyl) At C6 : substituent, CH2CH3 (6-ethyl) 6-Ethyl-2-methyl-4-octene
CH3 CH3CH2CHCH 7 6 5 4
CH3 CHCHCH3 3 2 1
At C3 : double bond, 3-heptene At C2 and C5 substituent, CH3 (2-methyl) 2,5-Dimethyl-3-heptene 56
Newer naming system for alkenes : but this change has not been widely accepted by the chemical community. However we’ll stay with older name. But you may occasionally encounter the newer system. CH3
CH2CH3
CH3CHCH2CH CHCHCH2CH3 1 2 3 4 5 6 7 8
CH3 CH3CH2CHCH 7 6 5 4
CH3 CHCHCH3 3 2 1
Older naming system : 6-Ethyl-2-methyl-4-octene
2,5-Dimethyl-3-heptene
Newer naming system : 6-Ethyl-2-methyloct-4-ene
2,5-Dimethylhept-3-ene
57
Exercise 12 : Give IUPAC names for the following compounds CHCH 3 3 H2C
CH2CH2CH3
CHCHCCH3
CH3CH2C
CH2CH3
CCH2CH2CH3
CH3
CH3
(a)
HC
CCH2CH2CHCH3 CH3
CH3
(b)
(c)
CH3
CH3 CH3
CH3
CH3
CH3 (d)
CH
(e)
(f)
58
Exercise 13 : Draw structures to the following IUPAC names a) 3-Ethyl-2,2-dimethyl-3-heptene b) 3,4-Diisopropyl-2,5-dimethyl-3-hexene c) 3-Ethyl-4,4-dimethylcyclopentene d) 6-Chloro-3,4-dimethylcyclohexene
59
Cycloalkenes are named similarly to open-chain alkenes, but because there is no chain end to begin from, so we number the cycloalkenes so that the double bond between C1 and C2 and the first substituent has as low a number as possible CH3
6 5
CH3
5 4
1
4
1
2 3
3
1-Methylcyclohexene
CH3
2
1,5-Dimethylcyclopentene
60
Geometric Isomerism •
• •
•
Because the rotation around a double bond is highly restricted, you will have different molecules if groups have different spatial orientation about the double bond This is often called When groups on the doubly bonded carbons are cis , they are on the same side of the double bond When groups on the doubly bonded carbons are trans , they are on opposite sides
61
cis/trans Isomers R
R C
H
R
H C
C H
H
C R
62
The cis and trans isomers are different molecules with different properties.
See page 110 - 111
63
Relative Stabilities of Alkenes The stability of alkenes depends upon number of substituents bonded to sp 2 carbon R
H C
H
R
<
C H
H C
R
R
<
C H
R C
H
R
<
C H
H C
H
R
<
C R
H C
R
R
<
C R
R C
R
C R
The more substituents, the more stable
See page 114 - 116
64
Steric repulsion is responsible for energy differences among the disubstituted alkenes
H3C
H C
>
C
H3C
H3C
>
C C
H
H3C
CH3
H
H
1-butene
H C C
H
CH3
trans 2-butene
C C
C C
mono-substituted
See page 114 - 116
disubstituted
65
H3CH2CH2C
CH3
H3CH2CH2C
C C H
H C C
H
Cis-2-hexene
CH3
H
CH3
Trans-2-hexene
CH3 CH2CH3
Trans-1-ethyl-2-methylcyclopent-1-ene
CH2CH3 Cis-1-ethyl-2-methylcyclopent-1-ene
66
How to give a name to the alkenes with more than two substituents
H3CH2CH2C
CH3
H3CH2CH2C
C C H3C
C C H
Cis-3-methyl-2-hexene
This compound is but the groups are
H
H3C
CH3
Trans-3-methyl-2-hexene
This compound is but the groups are 67
HMW
HMW
H3CH2CH2C
HMW
CH3
LMW
H3CH2CH2C
C C LMW
H3C
H C C
H
LMW
Cis-3-methyl-2-hexene
LMW
H3C
CH3
HMW
Trans-3-methyl-2-hexene
HMW is high molecular weight LMW is lower molecular weight This compound named as because the propyl group and methyl group are higher molecular weight and at the same side
This compound named as because the propyl group and methyl group at opposite side 68
Exercise 14 : Tell whether each of the following pairs is isomer or not isomer a H3C
C
=
C
H c
H3C
=
H
C H
CH3
H3CH2C =
CH3
CH3
C
C
H3C
H C C
H
=
C
CH3
CH2CH3 C
CH2CH3
H3CH2C
H
H3C
H C
H3C
C
H
C
H3C CH3
H3C
H C
H
C
d
H
H
b H3 C
CH3 C C
H
H 69
Exercise 15 : Which pair is more stable H3CH2C a
C C H
b
H3CH2C
H
C C
or H
CH3
CH3
CH3 H
CH3 or
70
Exercise 16 : Arrange of isomeric alkenes in order of stability a
b
71
72
Alkenes are relatively more reactive than alkanes. The electron-rich double bond is the ‘active site’ or the of the molecule .
An addition reaction is a reaction in which an unsaturated molecule becomes saturated by the addition of a molecule across a multiple bond.
73
74
1. Catalytic Hydrogenation 2. Halogenation 3. Halohydrin 4. Hydration 5. Hydrohalogenation
C
C
+
X-Y
C
C
X
Y 75
1. Catalytic Hydrogenation of Alkenes Alkenes and hydrogen react in the presence of a catalyst to form alkanes. H
H C C
H
H Ethene
• • • • •
H H
H2
H C C H
Ni
H H
Ethane
Addition of H-H across C=C Reduction in general is addition of H 2 or its equivalent Requires Pt /Pd/Ni/Rh as powders on carbon and H 2 Hydrogen is first adsorbed on catalyst Reaction is heterogeneous (process is not in solution) 76
Mechanism of Catalytic Hydrogenation of Alkenes •
Heterogeneous – reaction between phases
•
Addition of H-H is syn
77
Exercise 17 : The reaction below undergoes catalytic hydrogenation. Draw the product that formed from hydrogenation. H2 Pt
H2 Ni
H2 Pd 78
2. Halogenation of Alkenes Addition of Halogen in inert solvent Halogens dissolved in an inert solvent such as carbon tetrachloride, react readily with alkenes at room temperature and even in the dark. H
H C
H
•
H
Br 2 inert solvent
H
Br C
C
H
H
Br
Bromine and chlorine add rapidly to alkenes to give 1,2-dihalides, a process called halogenation •
•
C
H
F2 is too reactive and I2 does not add
Cl2 reacts as Cl + Cl- same goes to Br2
79
2. Halogenation of Alkenes Addition of Br2 to Cyclopentene • Addition is exclusively trans .. . .B .. r
H
H
.. . . B .. r
H
Br
Br H trans-1,2-Dibromocyclopentane sole product
Br
Br
H H cis-1,2-Dibromocyclopentane NOT formed
80
Exercise 18 : Write the mechanism for the below reaction reaction by showing the formation formation of bromonium bromonium ion
H H3C
C
C CH2 + Br 2
CH3H
+ Br 2
CHCl3 0o
CHCl3 0o
H3C
H
Br Br
C
C CH2
CH3H Br
Br
3. Halohydrin of Alkenes Addition of Halogen in water •
•
This is formally the addition of HO-X to an alkene to give a 1,2-halo alcohol, called a halohydrin The actual reagent is the dihalogen (Br 2 or Cl2 in water in an organic solvent) H
H C
C
H
H Ethene
Br 2 H2 O
H
H
HO C
C
H
H
Br
2-bromoethanol (Bromohydrin)
82
3. Halohydrin of Alkenes Chlorine or bromine dissolved water reacts with alkene to produce halohydrin as the major product. Br2(aq) + H2 O(l)
⇄
HBr(aq) + HOBr(aq)
Note: The red-brown red-brown colour of the bromine solution will disappeared. disappeared.
For asymmetric alkene such as propene, the addition of bromide ion to the carbon less substituent and addition of hydroxide hydroxide ion to the highly substituted substituted carbon
H
CH3 C
H
C H
Br 2 H2O
H Br C H
CH3 C OH H 83
3. Halohydrin of Alkenes •
•
Halogen addition in presence of water is regioselective : Markonikov’s rule The electrophile (Cl or Br) adds to the less substituted end of double bond, and nucleophile (H2O) adds to the more substituted end Carbon with more substituted, OH attached
(H3C)2HC
H C C
H
H
3-methylbutene
OH Br Br 2 H2O
(H3C)2HC C C H H H 1-bromo-3-methyl-2-butanol
Carbon with less substituted, Br attached
84
Exercise 19 : Give the structure of the major product formed when each of the following alkenes react with bromine in water. (a)
Br 2 H2O
(b)
Br 2 H2O
(c)
Br 2 H2O 85
Addition with acidified water (
):
Addition of H and OH groups from a water molecule to a double bond will produce an alcohol. The reaction is facilitated by using an acid as a catalyst.
H
H C C
H
H3O
+
H Ethene
H H H C C OH H H
Ethanol
Note: Hydration of alkenes using concentrated H 2 SO4 followed by hydrolysis, would give the same results. 86
For asymmetric alkene such as propene, the addition of hydrogen atom to the carbon less substituent and addition of hydroxide ion to the highly substituted carbon H3CH2C
H C
H
C H
1-butene
H3O
OH H
+
H3CH2C
C
C
H
H
H
2-butanol
87
Mechanism of Hydration of Alkenes The E+ adds to the sp 2 carbon that H3CH2C H bonded to greater C C numbers of H H hydrogens The Nu- adds to carbocation, forming a protonated alcohol.
H3CH2C H
Protonated alcohol is very strong acid, so it loses a proton, so form final product is alcohol.
H H
H3CH2C
H C+ C
H
H
H O H
H3CH2C O H H
HO
C
H2O
H H
H
H3CH2C C C H H
+
H O H
+
C
H H
H H
H C+ C
H O+ H
H
+
H3O+ 88
Exercise 20 : What is the product formed from each reaction H3O+
H2O H+ H2SO4 Diluted 89
5. Hydrohalogenation Addition of Hydrogen Halides Addition reaction of alkenes with hydrogen halides (HX) form haloalkanes (alkyl halides) Relative reactivity of HX towards alkenes increases in the order of HF < HCl < HBr < HI
H
H C C
H
H Ethene
See page 138 - 141
HBr
H H Br C C H H H
Bromoethane 90
H
H C C
H
H
H
Br
CCl4
H
H
H
H C
C
Br
H H
Br Br
aqueous
H H3CH2C
H C C
See page 138 - 141
H3C
H
Br
H
Br H
H3C C C H
CCl4
C H
H CH3H
91
Mechanism of Hydrohalogenation •
General reaction mechanism:
H3C
H C C
H3C •
•
•
Attack of electrophile (such as HBr) on bond of alkene Produces carbocation and bromide ion is an electrophile, reacting with nucleophilic bromide ion
H Br
H
H C+ C
H3C Carbocation
C+ C
H H
Carbocation
Br H
-
H H
H
H3C
2-methylpropene
H3C
H3C
Br
H3C
C C
H
CH3H 2-bromo-2-methylpropane
92
H
H C C
H
H
H Br CCl4
H
Br
aqueous
See page 138 - 141
.. .. Br -.. ..
H H H C C+ H H
H H H C C
Br
H H
.. .. Br -.. .. +
Br H
H 93
Reactions of Alkenes and Cycloalkenes If HX prepared in aqueous, the same product also formed, CH3
Br H Br H2O
H
CH3 H
H
1-methylcyclohexene
1-bromo-1-methylcyclohexane
H Br
Br
H2O ethylidenecyclopentane
1-bromo-1-ethylcyclopentane
94
Reactions of Alkenes and Cycloalkenes •
The reaction is successful with HCl as well as HI
•
HI is generated from KI and phosphoric acid H3CH2C
H C C
Cl Br H HCl
H Ether
H3C
2-methyl-1-butene
H3CH2CH2C
H3CH2C
C C H CH3H
2-bromo-2-methylbutane
H C
C
H 1-pentene
H
KI H3CH2CH2C H3PO4
I
H
C
C
H
H H
2-iodopentane
95
Orientation of Electrophilic Addition: Markovnikov’s Rule •
•
•
In an unsymmetrical alkene, HX reagents can add in two different ways, but one way may be preferred over the other If one orientation predominates, the reaction is that in the addition of HX to alkene, the H attaches to the carbon with the most H’s and X attaches to the other end (to the one with the most alkyl substituents) – This is Br H
H HBr
CH3CH2CH2CH C H 1-pentene
CH3CH2CH2CH C
H +
H
H
CH3CH2CH2CH C
Br
Ether H 2-bromopentane
H
1-bromopentane 96
Reactions of Alkenes and Cycloalkenes • •
Addition of HBr to 1-pentene Regiospecific – one product forms where two are possible H
H 1-pentene
H
H
Primary carbocation
1-bromopentane
Br
NOT FORMED
HBr
CH3CH2CH2CH C
H
CH3CH2CH2CH C
CH3CH2CH2CH2 C +
H
H
Ether
H CH3CH2CH2C+HCH
Br H CH3CH2CH2CH C
H Secondary carbocation
H
H 2-bromopentane Sole product 97
Markovnikov’s Rule • •
•
Addition HBr to 2-methylpropene More highly substituted carbocation forms as intermediate rather than less highly substituted one Tertiary cations and associated transition states are more stable than primary cations H H3C
C
H
H
CH3 H3C
H C
H3C
Primary carbocation
H
H
C
C
Br
CH3 H 1-bromo-2-methylpropene NOT FORMED
HBr
C
H3C
C+
H
H
Ether H3C
C+
C
Br
H
C
C
H H3C
2-methylpropene
H
CH3 H Tertiary carbocation
CH3 H 2-bromo-2-methylpropane 98
Markovnikov’s Rule • •
Addition HBr to 2-methyl-2-butene If both ends have similar substitution, then not regiospecific – two products formed
This carbon contains two alkyl groups
H
This carbon contains one alkyl group
H3C
C
CH3
H
CH3 H3C
CH3 C
H3C
Secondary carbocation HBr
C H
H3C
C+
H3C
C+
C
CH3
CH3
C
C
Br
CH3 H 2-bromo-3-methylbutane
H
Ether
H
H3C
Br
H
C
C
CH3
2-methyl-2-butene CH3 H Tertiary carbocation
CH3 H 2-bromo-2-methylbutane 99
Carbocation Structure and Stability •
• •
Carbocations are planar and the tricoordinate carbon is surrounded by only 6 electrons in sp 2 orbitals The fourth orbital on carbon is a vacant p -orbital The stability of the carbocation (measured by energy needed to form it from R-X) is increased by the presence of alkyl substituents
100
Inductive Stabilization of Cation Species In tertiary carbocation, all alkyl groups stabilize carbocations by releasing electron density to the positively charged carbon, thereby dispersing the positive charge. However, in secondary carbocation, only two alkyl groups will releases electron but in the presence of one proton will withdraws electron from positively charged carbon.
H3C
+
CH3
C
CH3 Tertiary carbocation CH3 releases electrons, stabilizes carbocation
H3C
C+
CH3
H Secondary carbocation H withdraws electron, destabilizes carbocation 101
Carbocation Structure and Stability
102
Mechanism of Electrophilic Addition: Rearrangements of Carbocations •
Carbocations undergo structural rearrangements following set patterns - hydride shifts occur
•
Goes to give more stable carbocation
•
Can go through less stable ions as intermediates Secondary carbocation
CH3 H3C
CH C H
CH3
H HBr
C H
H3C
Tertiary carbocation
H
CH3 H
H
C+
C
C
C
C+
C
H
H
H
H
H
CH3 Br
H
CH3 H
H
C
C
C
C
C
C
H
H
H
Br
H
H
H3C
H
H
Ether
3-methyl-1-butene
H3C
H3C
H
2-bromo-3-methylbutane
H
2-bromo-2-methylbutane
Predicting Major and Minor Addition Products When an HX molecule is added to an unsymmetrical alkene, two possible products can be formed: If we do not run the experiment, but still we can predict the major and minor from two possible products based on -Markovnikov’s rule -Rearrangement of carbocations (Hydride shifts) H3C HCl C
CHCH3
H3C 2-methyl-2-butene
H3C
H
Cl
C
CHCH3
Ether CH3 2-chloro-3-methylbutane Minor product
H3C
Cl
H
C
CHCH3
CH3 2-chloro-2-methylbutane Major product 104
H C H3C
H
CH3 C
+
CH3
2-methyl-2-butene
H
Cl
H
C
CH3
.. . -. . Cl .. .
C+
CH3
+
CH3
C
H
CH3
2o carbocation
CH3
CH3 CH3
+
Cl 2-chloro-2-methylbutane Major product
See page 138 - 141
C+ H3C
3o carbocation
CH3CH2C
CH3
H
CH3CH C
CH3
Cl H 2-chloro-3-methylbutane Minor product 105
Exercise 21 : What would the major and minor product obtained from the addition of HBr to each the following alkenes? (a)
(c)
(b)
(d)
106
Unsaturation Tests for Alkenes Alkenes or unsaturated organic compounds may be identified using simple lab tests or dilute bromine water are yellow brown in colour. The solution is decolorised when added to an alkene (See : Addition reactions with halogens) Reaction with cold, dilute, alkaline KMnO4 ( ) Alkenes are oxidised readily by the Baeyer’s reagent. The purple colour of the KMnO 4 solution disappears, and a brown precipitate of Mn(IV) oxide appears. The organic product is a diol . H3C
OH
OH
C
CHCH3
KMnO4 /OHC
CHCH3
H3C 2-methyl-2-butene
H3C
CH3 2-methylbutane-2,3-diol
107
Unsaturation Tests for Alkenes Reaction of Alkenes with Vigorous oxidation of alkenes results in the cleavage of the double bonds producing ketones, carboxylic acids or CO2 Potassium permanganate (KMnO 4) works if H’s are present on C=C In this reaction, the KMnO4 solution is decolourised. By identifying the products, the position of the double bond in the alkene can be determined.
108
Unsaturation Tests for Alkenes Example Reaction of Alkenes with hot, acidified KMnO4
H3C
CH2CH3 C
H
C
H3C
CH2CH3 C
H
H
H
KMnO4/H+ H3C
KMnO4/H+
CH2CH3 C
H3C
C
O
O
CH2CH3 O
C OH
C
O
O
C OH
109
Unsaturation Tests for Alkenes - Alkene is oxidised by ozone in tetrachloromethane to form unstable ozonide. - The ozonide is further hydrolysed to produce aldehydes and/or ketones. - Ozonolysis is used to identify the location of the C=C. H3C
CH2CH3 C C
H3C
H
O3
H3C
o coldC CH2Cl2, -78
H3C
O O C
C O
ozonide
CH2CH3 H
Zn/H2O CH2CH3
H3C C O
O C
+
H
H3C 2-propanone
+ byproduct
propanal
110
Exercise 22: What product(s) formed when each of compound treated with the following reagents i)
Cold, dilute alkaline KMnO4
ii) Hot acidified KMnO4 iii) O3/CH2Cl2 (cold) and Zn/H2O
111
Alkynes •
Aliphatic, unsaturated
•
CºC triple bond
•
Formula for one triple bond = Cn H2n −2 –
• •
•
subtract 4 H from alkane for each triple bond
Linear shape Internal alkynes have both triple bond carbons attached to C Terminal alkynes have one carbon attached to H 112
Alkynes
H C C H Ethyne
H3C C C H Propyne
both used in welding torches
113
Straight chain alkynes Ethyne
CHCH
Propyne
CHCCH3
1-Butyne
CHCCH2CH3
2-Butyne
CH3CCCH3
1-Pentyne
CHCCH2CH2CH3
2-Pentyne
CH3CCCH2CH3
1-Hexyne
CHCHCH2CH2CH2CH3
2-Hexyne
CH3CCCH3CH2CH3
3-Hexyne
CH3CH2CCCH2CH3 114
Nomenclature of Alkynes 1. Find the longest, continuous C chain that contains the triple bond and use it to determine the base name
H3CH2C
6
H3CH2CHCH2CHCH2C C C H 8
7
5
4
3
2
1
H3C C CH3 H Longest carbon chain is 8 so parent name given as 1-octyne 2. Number the chain and substituents determine the end closest to the triple bond
if triple bond equidistant from both ends, number from end closest to the substituents 115
Nomenclature of Alkynes 3.
Identify the substituent branches then assign numbers to each substituent based on the number of the main chain C it is attached to.
H3CH2C
6
H3CH2CHCH2CHCH2C C C H 8
7
5
4
3
2
1
H3C C CH3 H there are 2 substituents one is on the C-4 is called ethyl the other substituent on the C-6 is called isopropyl 116
Nomenclature of Alkynes 4.
write the name in the following order a) substituent number of first alphabetical substituent – substituent name of first alphabetical substituent – use prefixes to indicate multiple identical substituents b) repeat for other substituents c) number of first C in triple bond – name of main chain H3CH2C 6
H3CH2CHCH2CHCH2C C C 8
7
H3C C H
5
4
3
2
H
1
CH3
4-ethyl-6-isopropyloctyne
117
Nomenclature of Alkynes 4
CH2CH3
H3CH2CHCHC C C H 3
2
1
H3CHC 5 CH2CH3
6 7 3,4-diethyl-5-methylheptyne
CH3
CH3
H3CH2CC C C CH2CHCH3 8
7
6
5
4
3
2
1
CH3 2,6,6-trimethyl-4-octyne 118
Alkynes undergo many reactions similar to alkenes: 1. Hydrogenation = adding H2 a) alkene or alkyne + H2 → alkane b) generally requires a catalyst 2. Halogenation = adding X2 3. Hydrohalogenation = adding HX a) HX is polar b) when adding a polar reagent to a double or triple bond, the positive part attaches to the carbon with the most H’s 119
1. Hydrogenation of Alkynes •
•
Addition of H2 over a metal catalyst (such as palladium on carbon, Pd/C) converts alkynes to alkanes (complete reduction) The addition of the first equivalent of H2 produces an alkene, which is more reactive than the alkyne so the alkene is not observed H H3CH2C C C CH2CH3 3-hexyne
H2 Pd
H3CH2C C C CH2CH3 H 3-hexene
H
H H
H3CH2C C C CH2CH3 H 3-hexene
H2 Pd
H3CH2C C C CH2CH3 H H hexane 120
2. Halogenation of Alkynes: Addition of X2 •
•
Addition reactions of alkynes are similar to those of alkenes Intermediate alkene reacts further with excess reagent
•
Regiospecificity according to Markovnikov
Initial addition gives trans intermediate
Product with excess reagent is tetrahalide
121
Halogenation of Alkynes : Addition of Bromine and Chlorine H3CH2C C C CH2CH3 3-hexyne
Br 2
Br H3CH2C C C CH2CH3
CH2Cl2
Br 3,4-dibromohex-3-ene
Br H3CH2C
C C CH2CH3 Br
3,4-dibromohex-3-ene
Br 2 CH2Cl2
Br Br H3CH2C
C C CH2CH3 Br Br
3,3,4,4-tetrabromohexane
122
3. Hydrohalogenation of Alkynes : Involves Vinylic Carbocations •
Addition of H-X to alkyne should produce a vinylic carbocation intermediate –
–
•
Secondary vinyl carbocations form less readily than primary alkyl carbocations Primary vinyl carbocations probably do not form at all
Nonethelss, H-Br can add to an alkyne to give a vinyl bromide if the Br is not on a primary carbon
123
Mechanism of Hydrohalogenation of Alkynes Alkynes are electron-rich molecules, thus also takes place.
H B r H C 3
C
C
CH
H
H C 3
3 H C 3
2-butyne
vinyl-carbocation
124
Mechanism of Hydrohalogenation of Alkynes H
H C 3
H +
H C 3
CH
3
B r H C 3
B r
After the addition of the nucleophile we obtain an alkene that still can be attacked by an electrophile .
125
Mechanism of Hydrohalogenation of Alkynes
H
H C 3
CH
3
B r
H B r
H
CH
H C 3
B r
H
3
Formation of a second carbocation
126
Mechanism of Hydrohalogenation of Alkynes
H H
CH3
H H
CH3
H3C
Br
H3C
Br
Br
H
H
H3C
CH3 Br Br
Both nucleophiles end up on the same carbon.
127
Exercise 23 : Proposed the mechanism for the reaction, when 3-methyl-1-pentyne mixed with 2 mol HBr +
2 HBr
128
•
•
Aromatic hydrocarbons contain a ring structure that seems to have C=C, but doesn’t behave that way The most prevalent example is – –
C6H6 Other compounds have the benzene ring with other groups substituted for some of the hydrogens H H
H
H
H H 129
Naming Monosubstituted Benzene Derivatives O HO
Benzene
Methylbenzene
HO
Benzaldehyde
Phenol
O H2N
O
O Benzoic acid -
Aniline
Acetophenone
Nitrobenzene
Propylbenzene
Anisole
O N+
Br O Bromobenzene
Styrene 130
Naming Benzene as a Substituent •
When the benzene ring is not the base name, it is called a group Attach to alkyl group
O OH
2-phenyl-4-octene
6-phenylheptanoic acid
131
Naming Disubstituted Benzene Derivatives
ortho-xylene
meta-xylene
para-xylene
O O HO
m-tert -butylanisole
I
p-Iodophenol
OCH3
4-methoxybenzaldehyde
3-tert -butylanisole 1-tert -butyl-3-methoxybenzene 132
Naming Disubstituted Benzene Derivatives •
Number the ring starting at attachment for first substituent, then move toward second –
order substituents alphabetically
–
use “di” if both substituents are the same CH3
Br
Br Cl Cl
1-chloro-3-methylbenzene 1-bromo-2-chlorobenzene
CHO Br
I
1,2-dibromobenzene 2-iodobenzaldehyde
133
Exercise 24 : Name these compound
CHO
NO2 Br H3C CH3
134
•
To be a aromatic, a compound must follows Huckel’s criteria. 1. cyclic and conjugate 2. all carbon atoms are sp2 hybridized 3. planar or nearly planar 4. the no of pi electrons in the system is (4n + 2) with n = 0,1,2,3…. - (4n + 2) = 2,6,10,14,18 …….. (aromatic) - 4n = 4, 8, 12, 16………(non-aromatic)
135
Cyclobutadiene NOT AROMATIC
AROMATIC
Cyclooctatetraene NOT AROMATIC
NON AROMATIC
Naphthalene AROMATIC
NON AROMATIC
136
