GOC-1+Lecture+Notes+VIPUL

LECTURE NOTES Session - 2009-10

ORGANIC CHEMISTRY TOPIC : GOC - I CONTENTS :

1

IUPAC Nomenclature

Refer sheet GOC- I JEE Syllabus [2009] Concepts: Hybridisation of carbon; Sigma and pi-bonds; Shapes of molecules; Structural and geometrical isomerism; Optical isomerism of compounds containing up to two asymmetric centers, (R,S and E,Z nomenclature excluded); IUPAC nomenclature of simple organic compounds (only hydrocarbons, mono-functional and bi-functional compounds); Conformations of ethane and butane (Newman projections);

Page No.1

IUPAC NOMENCLATURE 1.1

INTRODUCTION TO ORGANIC COMPOUNDS:Organic compounds are compounds of carbon and hydrogen and the following elements may also be present: (Halogens, N, S, P, O). There are large no. of organic compounds available and large no of organic compounds are synthesized every year.The most important reason for large no of organic compounds is the property of catenation (self–linkage) in carbon. Element

Bond Energy

C

C–C

(strongest bond)

Si

Si–Si

Ge

Ge–Ge

Sn

Sn–Sn

Pb

Pb–Pb

 Decreasing order

Bond energy depends on (i) Size of atom (Inversely proportional)

Size : C–C>Si–Si

(ii) E.N difference along period (Directly)

E.N Diff : C – H < N – H < O – H < H – F

(iii) Bond order (no. of covalet bonds b/w two atoms) (Directly)

Bond order : C – C < C = C < C  C

Catenation in carbon: The element carbon has strongest tendency to show catenation or self–linkage due to –

(a) its tetravalency so that it can form  bonds with many elements as well as carbon itself. (b) It can form multiple bonds (C = C, C  C). Due to its small size, there is efficient colateral overlapping b/w two P–orbitals.

Does not exist (c) High bond energy of C – C bond so that it can form strong bonds in long chains and in cylic compounds.

1.2

CLASSIFICATION OF ORGANIC COMPOUNDS :

Page No.2

Ex.

Compounds

Classification Unsaturated

Heterocyclic , saturated Unsaturated

Saturated

Saturated, Alicylic

1.3

IMPORTANT TERMS: Saturated compounds: When all the valencies of an element are satisfied by  covalent bonds. Unsaturated compound: When a compound contains one or more  bonds (C = C, C = N, N = N, C = O or C  C, C  N, N  N) Molecular Formula (M.F.) : The molecular formula of a compound indicates the actual number of atoms of each element present in one molecule. Structural Formula (S.F.): It indicates the linkage due to covalent bond between different atoms in a molecule. (i) Expanded Structural Formula (E.S.F.) (ii) Condensed Structural Formula (C.S.F.) (iii) Bondline Structural Formula (B.S.F.) Ex.

M.F.

C3H8

Ex. H

Ex.

H

M.F.

C4H10

H

H

H

H

H

E.S.F. H – C – C – C – H H H H C.S.F. CH3 – CH2 – CH3

E.S.F. H – C – C – C – C – H H H H H C.S.F. CH3 – CH2 – CH2 – CH3

B.S.F.

B.S.F.

or

B.S.F.

C.S.F. M.F.

Ex. CH 3 – CH – CH 2 – CH 3 CH3

B.S.F.

M.F.

C6H12

C5H12

Homologous series : Homologous series may be defined as a series of similarly constituted compounds in which the members possess the same functional group, have similar chemical characteristics and have a regular gradation in their physical properties. The two consecutive members differ in their molecular formula by CH2.

Calculation of Degree of Unsaturation (DU):(a) It is the hydrogen deficiency index (HDI) or Double Bond Equivalence (DBE) Page No.3

–2H (b) H3 C – H2C – CH3   (DU  O)

–2H   CH3 – C  CH or CH2 = C = CH2 or

That means Deficieny of 2H is equivalent to 1 DU (c)

(i) 1DU = Presence of 1 Double Bond or Presence of 1 Ring closure (ii) 2DU = Presence of 2 Double bond or 1 Triple bond or two ring closure or 1 double bond + 1 ring.

(d)

G.F.

D.U.

(i) CxHy

y (x + 1) –   2

(ii) CxHyOz

yo  (x + 1) –   2 

(iii) CxHyXs

ys  (x + 1) –   2 

(iv) CxHyNw

y–w  (x + 1) –   2 

(v) CxHyOzXsNw

ys– w  (x + 1) –  2  

Ex

Calculate DU of following compounds (a) C6H6O

DU = 4

(b) C6H5Cl

DU = 4

(c) C6Br6

DU = 4

(d) C5H11OCl

DU = 0

(e) C9H12N2

DU = 5

(f) C6N6

DU = 10

(g) C10H8SO5 N4Cl2 e.g. M.F.

DU = 8 DU

S.F.

H 1.

C4H6

2

H

C–C=C=C

H H

H H C–C–CC C–CC–C C=C–C=C

C

CH2 C

Page No.4

2.

C2H2Cl2

=1

Total isomers = 3

3. C7H6O (Aromatic)

=5

4. C7H8 (Aromatic)

=4

Note : In case of the aromatic Compounds minimum DU = ‘4’. That means at least 1 Benzene ring is present.

Degree of carbon : It is defined as the number of carbon atoms attached to a carbon atom. H |

Primary (1°) carbon

:

| H

1°

Secondary (2°) carbon :

C | Tertiary (3°) carbon

:

| 3° H C | Quaternary (4°) carbon :

| C

Superprimary :

4°

(1°)

Ex.

10 – 5 carbon CH3 CH3 1 2 | 3| 4 5 CH3 – C – CH – CH2 – CH3 | CH3

0

2 – 1 carbon 0

3 – 1 carbon 40 – 1 carbon Page No.5

Ex.

*Degree of hydrogen is same as the degree of carbon to which it is attached. Type of C – C bonds & type of replaceable H-atoms in saturated hydrocarbon. Note : Total no. of monosubstituted product does not depend upon type of C-atoms but depends upon type of replaceable H-atoms.

1.4

IUPAC NOMENCLATURE OF ORGANIC COMPOUNDS: General Scheme of Naming:Secondary Prefix + Primary Prefix _ _Word Root_ _Primary Suffix_ _Secondary suffix The organic compound is always named according to the general scheme as given by IUPAC. In every compound, two parts, viz, word root and primary suffix, always exist.

Word Root: It indicates the no. of carbon atoms present in the main chain. It is represented as Alk. Prefix : It is the first part of the name. (i) Primary Prefix : ‘Cyclo’ (ii) Secondary Prefix : Normal substitutents and junior functional groups are treated as a substituent than their name is treated as secondary prefix. The following substituent groups are always cited in the prefix. (i) R – Alkyl (ii) R – O – Alkoxy (iii) X halo (fluoro, chloro, bromo, iodo) (iv) –NO2 Nitro (v) –N = O Nitroso (vi) Junior functional group  The prefixes are always written in alphabetical order  The position of substituent group in the main carbon chain is mentioned by writing the number just before the name of substituent by writing a small dash (–).

Suffix : (i) Primary Suffix:- It indicates saturation or unsaturation existing in the main chain. ane  single bond (saturated) ene  one = bond diene  If two double bonds Polyene  if plenty of double bond are present. yne  one  bond diyne  two  bond (ii) Secondary Suffix:- It is used for the principal functional group.

1.5

NAMING OF SATURATED HYDROCARBONS Rules :(Branched and substituted Alkanes) (1) Selection of parent chain  (a) Chain with maximum number of ‘C’ atoms (longest chain).

Page No.6

(b) If number of carbon atoms are same in more than one longest chain then more substituted longest chain will be the parent chain.

(c) If number of side chain are also same then that will be parent chain having its substituent at lower number.

C

C

C–C–C–C–C–C–C–C C–C–C C

(2) Numbering (a) Numbering is done from that side of the parent chain having it substituent at lower number (lowest set of locant)

(b) If position of substituent are same from both the end of the parent chain, then numbering is done from alphbetical order.

* If alphabets are also same then numbering is done from that side of the parent chain having its substituent of substituent at lower number.

Methylpropane

CH3 CH3 CH3 – CH – CH – CH3

2, 3-Dimethyl butane

–

CH3 CH3 – C – CH3 CH3 – –

Ex.

–

–

CH3 CH3 – CH – CH3

Dimethyl propane

2-Methylbutane

Page No.7

–

C C–C–C–C–C –

3-Ethyl-2-methyl hexane

C–C–C

4, 4-Diethyl heptane

C C–C–C C–C–C–C–C C–C–C C

4, 4-Diethyl-2, 5-Dimethyl heptane

–

– – –

– –

C–C–C C–C–C–C–C C–C–C

3-Ethyl hexane

CH3 – CH2 – CH2 – CH2Cl

1-Chlorobutane

C–C–C–C–C Cl Br

3-Bromo-2-chloropentane

C–C–C–C–C Br Cl

2-Bromo-4-chloro pentane

–

–

–

–

–

C–C–C–C–C C–C–C

4-Bromo-2, 2-dichloro pentane

C–C–C–C–C–C

5-Bromo-2, 3-dichloro hexane

–

–

–

–

– –

Cl C–C–C–C–C Br Cl

Cl Cl

Br

(3) Rules for writing Alkyl Radicals:Radicals:- (a)

–H CnH2n + 2  CnH2n + 1

Alkyl Radical

–H CnH2n  CnH2n – 1

Alkenyl Radical

–H CnH2n – 2  CnH2n – 3

Alkynyl Radical

R (structural formula)

Name

CH3 –

Methyl

CH3 – CH2 –

Ethyl

Page No.8

CH3 | (b) CH3 – C – CH3 | H

CH3 3 2| 1 –H CH3 – C – CH2 – Isobutyl | H

–H

CH3 | CH3 – C – CH3 Tert butyl (Tertiary butyl) |

(4) Systematic Names of Radicals : Complex Radical : Ex

5 – ( 1,2 – Dimethylpropyl) nonane

5 – (1,1 – Dimethylpropyl) nonane

5 – (1 – Ethylpropyl) nonane

5 – (Dimethylethyl) nonane

Alkene / Alkyne radicals Alkene

Alkenyl

Alkyne CH2 = CH2 CH3 – CH = CH2

Alkynyl CH2 = CH – ethenyl Prop – 1 – enyl Methylethenyl

Prop – 2 – enyl Page No.9

CH3 – C  CH

CH3 – C  C – Prop-1-ynyl

CH3 – C  CH

– CH2 – C  C – H Prop-2-ynyl

(5) Benzene Radical 1.

Phenyl

2.

Benzyl

3.

Benzal

4.

Benzo

5.

Tolyl

CH3 (o, m, p)



Methylene

7.

Alkylidene

8.

- naphthyl

9.

-naphthyl

(6) Numeral Prefixes (1) Following prefixes are considered for alphabetization : (a) Iso

(b) neo

(c) Di, Tri, Tetra of complex radical are considered for alphabetization (2) Following are not considered for alphabetization (a) Di, Tri, Tetra for simple radicals. (b) Sec, Tert are not considered for alphabetization (c) Bis , Tris Page No.10

(7) Retained Names of Alkanes: (1) Normal (n) : Radical or hydrocarbon which has straight chain and if it has free valency it must be present at either of ends. C – C – C – C (n butane) (2) Iso : Two methyl group at the end of linear chain (unbranched chain) Isobutane or methyl propane

(ii)

C C– C –C–C

Isopentane or methylbutane

C C– C –C–C–C

Isohexane or 2-methylpentane

–

–

–

(i)

C C– C –C

(iii) Conditions :-

 There should be 4 to 6 carbon atoms only  There should not be any other alkyl group present in the chain.

Ans.

C C– C–C– C–C C C (3) Neo :

–

Iso-octane (commercial name) in petroleum industry

– –

Q.

IUPAC name is 2, 2, 4-trimethylpentane

 There should be 5 to 6 carbon atoms  2, 2-dimethyl  No other substituent.

– –

C C– C–C C

Neopentane

C C – C – C – C Neohexane C – –

Ex.

Page No.11

(4) Secondary : It is applicable only for radical. Ex.1

(5) Tertiary :

First member

NAMING OF UNSATURATED COMPOUND (ALKENES AND ALKYNES) : (A) General formula:- CnH2n and CnH2n – 2 respectively (B) Rules for selection of main chain: Longest carbon chain with a multiple bond.  Longest carbon chain with maximum number of multiple bonds.  If first and second factor are common then chain with lowest locant (multiple bond) is selected as main chain.  Lowest locant Rule is followed till first point of difference. (Multiple bond prior to substituent).  Alphabetization. (C) Rules for Numbering: Lowest locant Rule till first point of difference (irrespective of double bond or triple bond).  Then double bond is prefer over triple bond in numbering, naming and longest chain selection, If all the other factors are common CH3 – CH2 – CH = CH2

But-1-ene

CH3 – CH = CH – CH3

But-2-ene

CH3 CH3 – CH – CH2 – CH = CH2

4-Methylpent-1-ene

–

Ex.

CH3 | CH3 – CH – CH = CH – CH3

4-Methylpent-2-ene

Cl 5 4 3 2 1 CH3 – C – CH2 – CH2 – CH = CH2 Cl

5, 5-Dichlorohex-1-ene

(a) C = C – C = C

Buta-1,3-diene

(b) C  C – C  C

Butadiyne

(c) C = C – C  C

Butenyne

– –

1.6

6

CC–C–C=C–C 1

2

3

5

4

2 1

4

6

Hex-4-en-1-yne

6

Hexa-1, 3, 5-triene

3 5

Page No.12

Ex.1

3-(2-Methylpropyl) hept-1-ene

Ex.2

4 -(1,1-Dimethylpropyl)-4-ethenylhepta-1,5-diene

Ex.3

3-Ethynylhexa-1, 5-diene

Ex.4

4-Ethenylhept-2-en-5-yne

Ex.5

4-(1, 2-Dimethylbutyl) hept-2-en-5-yne

Ex.6

2, 4-Dimethylpenta-1, 3-diene

Ex.6

2,3-Dimethylhex-1-en-4-yne

1.7

NAMING OF CYCLIC HYDROCARBON (ALICYCLIC COMPOUNDS) : (A) Main chain selection: (a) Multiple Bond > No. of carbon atoms > Maximum no. of substituents > Nearest locant > Alphabetization. (b) If all factors are similar in cyclic and acyclic part, then Cyclic > Acyclic

(B) Numbering: (a) Lowest Locant

(b) Alphabetization

(C) Naming:  Prefix ‘cyclo’ is used just before the word root if it constitutes the main chain.  If cyclic part is the main chain then the prefix ‘cyclo’ is not considered for alphabetical order.. If cyclic part constitutes the side-chain (substituent) then prefix cyclo is considered for alphabetization:-

Ex:-

Cyclopropane

Cyclobutane Page No.13

Cyclohexane

Cycloheptane

Cyclooctane

Methylcyclopropane

Ethylcyclopropane

Propylcyclopropane

1-Cyclopropylbutane

–

Cyclopentane

–

C – C – Cl

2-Chloroethylcyclopropane –

Cl – C – C

Chloroethylcyclopropane

–

Cl C–C–C

–

3

2

1

–

Cl 3 C–C–C

1-Chloro-3-cyclopropylpropane

1

2

–

– –

Cl C–C–C 1

2

3



–

–

C–C–C 1-Chloro-2-propylcyclopropane

Cl –

1

2 3

F

–

4

Br –

–

Cl

2-Bromo-1-chloro-3-fluoro-4-iodocyclohexane

I Methylethylcyclopropane or isopropylcyclopropane

–

Cl

–

Br

4

5

I

–

2

3

1 6

1-Bromo-2-chloro-4-iodocyclohexane

6 5

1 2

3

4

3-Chlorocyclohex-1-ene

Cl

Br

6 1

5

2 3

4

5-Bromo-3-chlorocyclohex-1-ene

Cl Page No.14

1.8

FUNCTIONAL GROUP TABLE (Seniority order):

Class

Name

Suffix

Prefix

1.

R – COOH

Alkanoic acid

– oic acid (carboxylic acid)

Carboxy

2.

R – SO3H

Alkane sulphonic acid

– sulphonic acid

sulpho

3.

R–C–O–C–R || || O O

Alkanoic anhydride

– anhydride

------------

4.

R – COOR

Alkyl alkanoate

– alkanoate (carboxylate)

Alkoxy carbonyl

5.

R–C–X || O

Alkanoyl halide

– oyl halide (carbonyl halide)

halo carbonyl

6.

R – C – NH2 || O

Alkanamide

– amide (carboxamide)

Carbamoyl

7.

R–CN

Alkanenitrile

– nitrile (carbonitrile)

cyano

8.

R–C–H || O

Alkanal

– al (carbaldehyde)

formyl / Oxo

Alkanone

– one

Oxo / Keto

– ol

hydroxy

9.

R–C–R || O

10. R – OH

Alkanol

11. R – SH

Alkanethiol

– thiol

mercapto

12. R – NH2

Alkanamine

– amine

amino

1.9

NAMING OF FUNCTIONAL GROUP CONTAINING COMPOUNDS : (A) Selection of Main Chain: Senior F.G. > Max. no. of F.G. (Similar group) > Multiple bond > Max. no. of ‘C’ atoms > Max. no. of locants > lowest locant > alphabetization

(B) Numbering (See F.G. + Sub.): The carbon atom bearing functional group (or C – atom of terminal functional group) is given lowest possible number. (i) Lowest locant (F.G. > M.B. > substituent > Alphab)

(C) Naming:- General scheme : The senior most functional group constitutes secondary suffix. Other junior F.G.’s are written as prefix in alphabetical order.

Page No.15

1.9.1

Carboxylic acid F.G.

– COOH

Prefix

Suffix

IUPAC name

Oic acid - 'C' of COOH considered in the parent chain

Alkanoic acid

Carboxylic acid - 'C' of COOH is not considerd in parent chain

Alkane carboxylic acid

Carboxy

Rule : If first alphabet of sec. suffix is begin from a, i, o, u, y then ‘e’ of primary suffix will be dropped. Ex.

(1) HCOOH

Methanoic Acid

(2) CH3 – COOH

Ethanoic Acid

(3) C – C – COOH

Propanoic Acid

(4) C – C – C – COOH

Butanoic Acid

2

3

4

5

6

(5) C – C – C – C – C – C – C – C 1 COOH

2-propylhexanoic acid

–

COOH

(6)

Cyclopropanecarboxylic acid

2

1

(7)

–

CH2 – COOH Cyclopropylethanoic acid

–

COOH 3' C–C–C 1'

2'

4-(2-carboxypropyl) cyclohexane-1-carboxylic acid

–

(8)

COOH Ethanedioic acid

(10) HOOC – CH2 – COOH

Propanedioic acid

(11) HOOC – CH2 – CH2 – CH2 – CH2 – COOH

Hexanedioic acid

(12)

Cycloprop-1-ene-1, 2, 3-tricarboxylic acid

Sulphonic acid F.G.

Prefix

– SO3H

Sulpho

Ex.

Suffix Sulphonic acid

IUPAC name Alkanes sulphonic acid

(1) CH3 – SO3H

Methanesulphonic acid

(2) C – C – C – C – C SO3H

Pentane-2-sulphonic acid

–

1.9.2

(9) HOOC – COOH

Page No.16

SO H – 3 (3)

Cyclobutanesulphonic acid 1

2

3

4

5

(4) HOOC – C – C – C – C – SO 3H

1.9.3

5-Sulphopentanoic acid

Anhydride

(a) R – C – O – H + H – O – C – R

Heat –H2O

  

O

O

R– C –O– C –R O

O

Alkanoic anhydride (b) Mixed Anhydride (according to alphabet) CH3 – COOH + Ethanoic acid

CH3 – CH2 – COOH Pr opanoic Acid



CH3 – C – O – C – C2H5 O

O

Ethanoic propanoic anhydride Ex.

(1)

H– C –O– C –H O

(2)

O

C C C–C–C–O–C–C–C O

Methanoic Anhydride (unstable )

2-Methylpropanoic anhydride

O

(3)

Butanedioic anhydride

(4) C – C – C – C – O – C – C – C – C || || O O

Butanoic anhydride

O || C O

(5)

Cyclohexane-1, 2-dicarboxylic anhydride

C || O (6)

–C–O–C– O

(7)

Cyclopropane carboxylic anhydride

O

Benzene dicarboxylic anhydride (Phthalic anhydride)

Page No.17

1.9.4

Ester : F.group

Prefix

Suffix

Name

– C – OR

Alkoxy Carbonyl

Oate

Alkyl....alkanoate

or Alkanoyloxy

or carboxylate

O

(Ester)

or Alkyl.....Alkanecarboxylate

–H2 O R – C – O – R'  R – C – O – H + H – O – R'   O O Example:(1) H – C – O – CH3

Methyl methanoate

O (2) C – C – C – O – C – C – C

Propyl propanoate

O (3) C – C – O – C – C – C – C

Ethyl butanoate

O

–

C (4) C – C – C – O – C – H

Isobutyl methanoate

O 3

2

1

(5) C – O – C – C – C – COOH

3-(Methoxycarbonyl)propanoic acid

O (6) C – C – O – C – C – COOH

3-(Ethanoyloxy)propanoic acid

O

(7)

2-(Methoxycarbonyl)-4-(Propanoyloxy) cyclohexan-1-carboxylic acid O –

–

C

O–

(8)

Cyclopropyl benzene carboxylate

O (9)

– O – C – CH2 – CH3

Methyl-3-propanoyloxy cyclohexane carboxylate

–

–

O=C OCH3

1.9.5 Acid halide:F.Group

Prefix

–C–X

Halo Carbonyl

O

Suffix

Name

Alkanoyl halide Oyl-halide or or Carbonylhalide Alkane carbonyl halide

–

O = C – Cl Example:-

(1)

Benzene carbonyl chloride Page No.18

(2) CH3 – C – Cl

Ethanoyl chloride

O

(3)

Prop-2-enoyl bromide

(4)

6-Methyl hept-4-en-2-ynoyl chloride

O

–

Cl C (5) CH2 – CH2 – CH2 – CH – C – Cl COOH O

–

–

5, 5-Dichlorocarbonylpentanoic acid

–

COOH

COOH

–

–

(6)

3-Chlorocarbonyl cyclopropane-1, 2-dicarboxlic acid C – Cl

O O

(7)

5-Methylhept-3-ene-1, 7-dioyl Chloride

–

–

C – CH2 – CH – CH = CH – CH 2 – COCl

Cl

CH 3

(8)

4-Bromocarbonyl-2-chlorocarbonyl-5-propanoyloxycycloexane-1-carboxylic acid

1.9.6 Amide:F.Group

Prefix

Suffix

Name

– C – NH2

Carbamoyl

amide or Carboxamide

Alkanamide or Alkane Carboxamide

O

Example:-

(1) H – C – NH2

Methanamide (Formamide)

O

(2) CH3 – C – NH2

Ethanamide (Acetamide)

O

(3) CH3 – C – NH – CH3

N-methylethanamide

O

–

(4) CH3 – C – N – C – C – C

O

N-butyl-N-propylethanamide

C C C C

Page No.19

–

O = C – NH2 (5)

Benzene carboxamide

–

O = C – NH2 Benzene-1, 3-dicarboxamide

–

(6)

C – NH2 O –

COOH (7)

3-Carbamoyl cyclohexane-1- carboxylic acid

–

C – NH2

O

O C – NH – CH3 (8) CH2

N, N’-Dimethyl propane-1, 3-diamide

C – NH – CH3 O

Nitrile:-

(1) H – CN

Methanenitrile

(2) CH3 – CN

Ethanenitrile

(3) CH3 – CH – CH2 – CN CH3 –

Example:-

3-Methyl butanenitrile

–

CN

(4)

Cyclopropane-1, 2, 3-tricarbonitrile –

– CN

CN

CN –

1.9.7

(5)

Benzene carbonitrile

(7)

CH2 – CH – CH2 CN CN CN

Propane-1,2,3-tricarbonitrile

–

–

3-Cyano propanoic acid

–

CH2 – CH2 – COOH CN

–

(6)

Page No.20

1.9.8

Isocyanide :-

F.Group

Prefix

Suffix

Name

–NC

Isocyano

Isocyanide

Alkyl Isocyanide

Example:-

(1) CH3 – N  C

Methyl isocyanide

(2) CH3 – CH2 – N  C

Ethyl isocyanide

–

NC

(3)

Aldehyde:Prefix

F.Group

Name

Formyl or oxo al or carbaldehyde Alkanal or Alkane Carbaldehyde

– CH = O

(1) HCHO

Methanal

(2) CH3 – CH2 – CH = O

Propanal

(3) CCl3 – CH2 – CH2 – CH2 – CH = O

5, 5, 5-Trichloro pentanal

–

CH = O (4)

Cyclohexane carbaldehyde

–

CH = O

(5)

Benzaldehyde/Benzene carbaldehyde

Ph (6)

–

CH – CH Ph –

CH = O –

2-Chloromethyl-3, 3-diphenyl propanal

CH2Cl

–

CH = O (7)

Benzene-1, 3-dicarbaldehyde

–

CH = O (8) HOOC – CH2 – CH2 – CH2 – CH = O –

(9) HOOC – CH2 – CH – CH2 – COOH CH2 – CH = O

5-Oxopentanoic acid 3-(2-Oxoethyl) pentane-1, 5-dioic acid

–

COOC2H5 (10)

–

Example:-

Suffix

–

1.9.9

Phenyl isocyanide

`

Ethyl 3-(3-oxopropyl) cyclohexane-1- carboxylate

CH2– CH2 – CH = O Page No.21

1.9.10 Ketone:Prefix

F.Group

Name

one

Alkanone

–

Keto or oxo

C=O –

Example:-

Suffix

(1) CH3 – C – CH3

Propanone

O

O (2)

Cyclohexanone

O (3)

Cyclohexane-1, 3, 5-trione

O

O

O

–

CH2 – C – CH3

(4)

2-(2-Oxopropyl) cyclohexanone

O

–

COOH 3-(3-Oxobutyl) cyclohexane -1-carboxylic acid

–

(5)

CH2 – CH2 – C – CH3 O (6) OHC – CH2 – C – CH3

3-Oxobutanal

O

1.9.11 Alcohol:F.Group

Prefix

Suffix

Name

–OH

Hydroxy

ol

Alkanol 2-Methyl Propanol

(2) CH 3 – CH – CH 2 – CH 2 – CH – CH 3 CH 2 – OH CH 2 – OH

2, 5-Dimethylhexane-1, 6-diol

–

–

–

(1) CH3 – CH – CH2 – OH CH3

–

HO

–

–

OH

–

(3)

–

OH HO

–

Cyclohexane-1, 2, 3, 4, 5, 6-hexaol

OH

OH (Glyceraldehyde) 2, 3-Dihydroxy propanal

– –

(4) CHO CH – OH CH2 – OH

–

OH Cl – CH2 – CH – CH2

–

CH – CH2 – CH2 – CH2 – CH2 – OH

–

(5)

–

CH3 – CH2 – CH2 OH Cl –

Example:-

7-Chloro-5-(3-Chloro-2-hydroxy propyl) octane-1, 6-diol Page No.22

O OH

–

(6)

–

HO

2, 4, 6-trihydroxy cyclohexane-1, 3, 5-trione

–

O

OH

O

(Glycerol) Propane-1, 2, 3-triol

–

–

–

(7) CH2 – CH – CH2 OH OH OH

1.9.12 Amines:F.Group

Prefix

Suffix

Name

–NH2

amino

amine

Alkan amine

(a) R – NH2

1º = amine

Alkanamine

(b) R – N – R' 2º = amine H

N  Alkylalkan a min e (R' ) (R )

(c) R – N – R' 3º = amine

N, N-Dialkylalkanamine

R'

(d) R – N – CH3

N-Ethyl-N-methylalkanamine

CH2 – CH3

(1) CH3 – NH2

Methanamine

(2) C – C – C – C – C NH2

Pentan-3-amine

(3) C – C – C – NH – C – C

N-Ethylpropan-1-amine

(4) C – C – C – N – C

N-Ethyl-N-methylpropan-1-amine

–

–

Ex:-

C C N-Ethylpropan-2-amine

–

(5) C – C – C

NH – C – C

–

OH (6) CH2 – CH = CH – CH – CH3 NH – CH2 – CH3

–

–

(7) OHC – CH2 – CH – CH2 – CH3

4-(N-Ethyl amino) pent-2-en-1-ol

3-(N-Phenyl amino) pentanal

NH – Ph If hetero atom’s are count as a ‘C’ atom in the parent chain then they are written as (1) –NH–  Aza (2) –O–  Oxa (3) –S–  Thia (4) –Se  Selena

(1) HOOC – CH2 – NH – CH2 – CH3

3-Aza pentanoic acid

H N

–

Example:-

(2)

Aza Cyclo pentane Page No.23

–

H N

(3)

3-Methyl aza cyclo hexane –

CH3

1.10 ETHERS (R – O – R’):F.Group

Prefix

Suffix

Name

R – O – R'

alkoxy

–

Alkoxy alkane

Alkoxy + Alkane   less no. more no. of ' C' of ' C'

Methoxymethane

(2) C – O – C – C

Methoxyethane

(3) C – O – C – C – C

1-Methoxypropane

(4) C – C – C O C

2-Methoxypropane

–

(i) Acyclic Ethers:(1) C – O – C

–

–

C (5) C – C – C – O – C – C

1-Isopropoxypropane 1-(Methylethoxy) propane

–

O–C–C

(6)

Ethoxycyclohexane

–

O–

(7)

Cyclopropoxycyclohexane

–

Br O–C–C–C

–

3

(8)

4

2

5

3-Bromo-6-chloro-2-propoxycyclohexan-1-ol

1

–

–

6

OH

Cl  In cyclic system, numbering always starts from senior most functional group.

–

OH –

Br

1

2

6

(9)

–

3

5

4

6-Bromo-5-methoxycyclohex-2-en-1-ol

O – CH3

–

OH 1

Br

–

2

6

(10) 3

–

5

4

6-Bromo-5-ethoxycyclohex-3-en-1-ol

O–C–C

Page No.24

(ii) Cyclic Ether: (3-membered Ring) Hetero cyclic compounds (1) C – C O

Oxirane or Epoxyethane

(2) C – C – C O

1, 3-Epoxypropane

(3) C – C – C O

1, 2-Epoxypropane

–

Cl (4) C – C – C O

1-Chloro-2, 3-Epoxypropane

 Polyethers:(1) C – C – O – C – C – O – C – C 1

2

1, 2-Diethoxyethane 1

2

3

4

5

6

7

8

9

10

11

1

2

3

4

5

6

7

8

9

10

11

(2) C – C – O – C – C – O – C – C – O – C – C 3, 6, 9-Trioxaundecane 12

13

14

15

(3) C – O – C – C – C – O – C – C – O – C – C – C – O – C – C 2, 6, 9, 13-Tetraoxapentadecane 2

3

1 12

4

O

O

O

O7

5

(4) 11

10

6

8

9

1, 4, 7, 10-Tetraoxacyclododecane

1.11 AROMATIC COMPOUNDS: Classification:



H Cation

2 e–

Cyclopropenium ion

Cation

2 e–

Cyclobutenium dication

–

H

–

Non-Benzenoid:

–

+ H –



–

+ +

–

H

H

–

H

H

Page No.25

–

Anion –

– .. H H

–



H

–

H

–



+

6 e–

Cycloheptatrienyl cation (Tropylium cation)

H

–

–

–

H

Cyclopentadienyl anion

H

H – –H

–

H H

6 e–

H

Benzenoid Aromatic Compounds: All organic compounds which contain atleast one benzene ring are known as benzenoid aromatic compounds.

Naming of Aromatic Hydrocarbons (Arenes): –

CH3 (1)

Methylbenzene (Toluene)

–

CH3 –

CH3

(2)

(Ortho) 1, 2-Dimethylbenzene (Ortho-xylene)

–

CH3 (3)

(Meta) 1, 3-Dimethylbenzene (Meta-xylene)

–

CH3

–

CH3 (4)

(Para) 1, 4-Dimethylbenzene (Para-xylene)

– CH3 Write structures of

–

C–C (i) Metadiethylbenzene

–

C–C

–

C–C (ii) 1-ethyl-4-isopropylbenzene

– C–C–C (iii) Orthodiethenylbenzene

–

–

C=C C=C

Trialkyl Substituted Benzene: If all the three substituents are similar, then only 3 trisubstituted benzene derivative are possible. –

–

C 1, 2, 3-Trimethyl benzene (Vicinal Trimethyl benzene)

C –

–

–

C –

C 1, 2, 4-Trimethyl benzene (Unsymmetrical Trimethyl benzene)

C

–

C

C

–

Q.

– C C 1, 3, 5-Trimethyl benzene (Symmetrical Trimethyl benzene)

Page No.26

Examples : Isopropylbenzene (Cumene)

– –

1.

C–C C –

C

1-Chloro-4-methyl-2-nitrobenzene

–

–

2.

NO2

Cl C=C

3.

Phenylethene (Double bond > Phenyl) (Common name : Styrene)

–

C–C=C

4.

3-Phenylprop-1-ene (Allylbenzene)

–

CH = CH – CH3

5.

1-Phenylprop-1-ene

–

COOH

6.

Benzene carboxylic acid (Benzoic acid)

O

O –

–

C–O–C

7.

Benzene carboxylic anhydride (Benzoic anhydride)

–

SO3H

8.

Benzene sulphonic acid

O –

C – O – C2H5

9.

Ethyl benzene carboxylate (Ethyl Benzoate)

–

O – C – CH3

10.

O

Phenyl ethanoate

O –

C – Cl

11.

Benzene carbonyl chloride (Benzoyl chloride)

O –

NH – C – CH3

12.

N-Phenylethanamide

O –

C – NH – CH3

13.

N-Methylbenzene carboxamide

Page No.27

–

CN

14.

Benzene carbonitrile (Benzonitrile – popular)

–

CH = O

15.

Benzene carbaldehyde (Benzaldehyde – popular) O

– –

C

16.

Diphenyl ketone (Benzophenone)

O

– –

C

17.

CH3

Methyl phenyl ketone (Acetophenone)

–

OH

18.

Phenol

–

OH

19.

––– CH3

Methylphenol [o-Cresol, m-cresol, p-cresol]

–

OH OH –

20.

o-Hydroxyphenol (Catechol)

–

OH

m-Hydroxyphenol (Resorcinol) –

21. OH

–

OH

22.

p-Hydroxyphenol (Quinol)(Hydroquinone)

– OH –

NH2

23.

Benzenamine (Aniline)

–

SH

24.

Benzenethiol

O O – C – CH3 O 25.

–C–O–H

2-Ethanoyloxy (Acetoxy) benzene -1-carboxylic acid (Aspirin)

Page No.28



1

Isomerism * Structure Isomerism * Stereo Isomerism Geomertical Optical Conformational

Refer sheet GOC- I JEE Syllabus [2009] Concepts: Hybridisation of carbon; Sigma and pi-bonds; Shapes of molecules; Structural and geometrical isomerism; Optical isomerism of compounds containing up to two asymmetric centers, (R,S and E,Z nomenclature excluded); IUPAC nomenclature of simple organic compounds (only hydrocarbons, mono-functional and bi-functional compounds); Conformations of ethane and butane (Newman projections);

Page-29

Isomerism Isomers: Compounds with same general formula or molecular formula but different physical and chemical property. Ex:-

CH3 – CH2 – OH and CH3 – O – CH3

Ex:-

CH3–COOH and HCOOCH3

Homologs: Compounds with same general formula differing by same structural unit – CH2 – or molecular weight by 14 unit. Ex:- CH3 – OH and CH3 – CH2 – OH Difference between isomers and homologs:-

Page-30

Structural isomers: When two or more number of organic compounds have same molecular formula but different structural formula these are called structural isomers.

Stereo isomers: When two or more compounds have same Molecular Formual (M.F.) and same Structural Formula (S.F.) but have different stereochemical formula (S.C.F.), these are called stereoisomers.

Stereo Chemical Formula (S.C.F) : It indicates different arrangements of atoms or groups in space around a stereo centre or it indicates different spatial orientations of atoms or groups around a stereo centre. C4H8 CH3 (iii) CH 3 – C = CH 2 –

M.F.

S.F. (i) CH3 – CH2 – CH = CH2 These are structural isomers. For (ii), S.C.F. are H3C H

CH3

H

H3C and

C=C

(ii) CH3 – CH = CH – CH3

C=C

H CH3 Stereocentre These are stereoisomers. H

Structural isomers are also known as:(i) Skeletal isomers (ii) Linkage isomers

(iii) Constitutional isomers

Chain isomers : They have different size of main carbon chain and / or side alkyl chain (i) The two chain isomers should have same nature of F.G./multiple bonds/substituents (except –R group) (ii) The position of F.G./M.B./substituent (locants) is not. considered here. Ex:-

Alkanes:(a) C1 – C3 (b) C4H10

chain isomers not possible H3C – CH2 – CH2 – CH3 and CH3 –

(c) C5H12

CH3 – CH2 – CH2 – CH2 – CH3 , CH3 – CH2 –

(d) C6H14

(i) CH3 – CH2 – CH2 – CH2 – CH2 – CH3, (ii) CH3 – CH – CH2 – (iii) CH3 – CH2 –

(v) CH3 – CH2 –

– CH3

– CH2 – CH3, (iv) CH3 –

– CH3 ,

–

CH3 –

– CH3 – CH3,

– CH3,

– CH3

(i) and (ii) – Chain isomers (i) and (iii) – Chain isomers (i), (iv) – Chain isomers (i), (v) – Chain isomers (ii), (v) – Chain isomers (ii), (iii) – Position isomers (iv), (v) – Position isomers

Ex.

Ans.

and

shows which types of isomerism ?

Chain Isomers Page-31

Q.

Write chain isomers of N-alkanamine (1º) containing 4 carbon atoms ?

Ans.

C – C – C – C – NH2,

–

C C – C – C – NH2

Positional isomers: They have different position of locants Functional group (F.G.) or Multiple Bond (M.B.) or substituents in the same skeleton of C-atoms. Nature of F.G. or M.B. should not change. The skeleton of C-atom should not change. (1) C – C – C – C – C ,

C–C– C –C–C

C

C

–

–

(2) C = C – C – C ,

C–C=C–C

(3) C  C – C – C – C ,

C–CC–C–C

(4) C – C – C – C , C – C – C – C – OH OH Write all Positional isomers of Dichlorobenzene ? –

Ex.

Cl

Cl

–

–

Cl

–

Q.

–

Cl

Q.

,

, Cl

–

Cl– Cl –

Ans.

–

, Cl

Write all Positional isomers of Dichlorocyclobutane ? Cl Cl– ,

–

Cl–

Ans.

, –

Cl– Cl –

Cl

Write all Positional isomers of Dichlorocyclopentane ?

Cl

Cl Cl

–

–

–

Cl

–

Q.

Cl

Write all Positional isomers of Dichlorocyclopropane ? Cl

Q.

–

–

Ans.

–

Cl

Ans.

,

,

–

Cl C

–

C

(iii) C – C – C – C – C – C –

C–C

–

(ii) C – C – C – C – C – C, –

(i) C – C – C – C – C – C –

Ex.

C

C

(i), (ii) – Chain (ii), (iii) – Position (i), (iii) – Chain

–

C6H12 (Cycloalkanes):(2)

,

(3)

– ,

(4)

, –

,

–

(1)

–

Ex.

Page-32

–

–

–

(10)

– –

–

Ans.

,

–

– –

–

(6)

–

,

C–C

(9)

C–C–C

–

–

(5)

C C–C

,

(7)

,

(11)

,

(8)

,

C–C –

(3) and (4) – Positional isomer (3) and (6) – Positional isomer (4) and (6) – Positional isomer (5) and (10) – Positional isomer (All are chain isomers of 1) Remaining are chain isomers

Functional isomers: They have different nature of functional group (F.G.). The chain and positional isomerism is ignored (not considered). Compound Functional isomer C–C–C–C

Nil

C–C–C=C Compound C–C–CC

Isomer Functional (a) C = C – C = C Alkadiene

Remarks (a), (b) are not functional isomers

Alkyne

(b)

Cycloalkene

among themselve

(c)

Bicyclo

(2)

C – C – C – OH Alcohol

C–C–O–C Ethers

(3)

C – C – CH = O

C– C –C

Aldehydes

O Ketones

C – C – COOH

C– C –O–C

Carboxylic acids

O Esters

C – C – C – NH2

.. (a) C – C – NH – C

1º amine

2º amine

C–C–CN Cyanide (Propanenitrile)

C – C – NC Isocyanide (Ethane isocyanide)

(4)

(5)

(6)

.. (b) C – N – C C 3º amine –

Ex. (1)

(Ring-chain isomers are Functional isomers)

1º, 2º, 3º amines are functional isomers

–

Important point :- Following compounds don’t exist at room temperature therefore not consider as a structure isomer (ii) – C  C – OH

(iii) – C – OH –

–

–

(i) – C = C – OH

–

(iv) – C – OH OR

–

(v) – C – O – C = C –

–

OH

(vi) Any peroxy compound

OH

Page-33

Write acyclic isomers of C5H12O

Ans.

(i) C – C – C – C – C – OH,

C C – C – C – C – OH,

–

Q.

C – C – C – C – OH,

(7 alcohols) C C – C – O – C, C – –

–

–

C – C – C – O – C, –

C – C – C – O – C, –

C – C – C – C – O – C,

C – C – C – C – C, OH C C – C – C – OH C – –

– –

–

C

C – C – C – C – C, OH C C – C – C – OH, C

C

C – C – C – O – C – C,

C C– C –O–C–C

(6 ethers).

G.F. CnH2n + 3N

n (1) 3

M.F. C3H9N

2º Amine (R – NH – R)

3º Amine R – N – R R

G.F. CnH2n + 1N 2

n (1) 2

M.F. C2H5N

1º Amine (1) C = C – NH2 Ethenamine

2º Amine 3º Amine C–C=N C–N=C Ethanimine N-Methylene methanamine (Unstable at room temperature)

Q.

1º Amine (R – NH2) Q

–

–

C

–

H2C– – CH2 Azine N H –

Q.

G.F. CnH2n – 1N

n 2

M.F. C2H3N

Ans.

Cyanides and isocyanides H3 C – C  N

CH2 = C = NH

HC  C – NH2

H3 C – N



C

Metamers: When two isomers have same functional group (containing a hetero atom –O, N, S) but have different nature of alkyl or aryl (aromatic radical) group attached to hetero atom, then these are called metamers.

Conditions: (a) Same functional group (b) Chain or position isomerism is not considered. (c) >C = O (keto) group does not show metamerism Following functional groups show metamerism Ethers (All isomeric ethers are metamers)

(b) R’ – NH – R

2º Amines

(c) R’ – N – R R'' (d) R – S – R’

3º Amines

–

(a) R – O – R’

Thioethers Page-34

(e) R – C – O – R’

Esters

O

(f) R – C – O – C – R’ Anhydrides O

O

O (g) R – S – O – R’

Sulphonate esters

O Q.

How many esters are possible for C3H6O2 ?

Ans.

C  C  O  C  H and C  C  O  C are metamers || || O O

Q.

3º Amines of M.F. C5H13N (All metamers) C– C– N–C

O–C

–

–

O–C–C

Q.

C–C–C– N–C

C

–

C–C– N–C–C

C

–

Ans.

–

C

–

C

and

Show which type of isomerism ? –

C

Q.

Acyclic compound (A) contains 18 1º H atoms, two types of ‘C’ atoms. All H are identical. Identify (A).

Ans.

C C C – C – C – C  C8H18 C C

Q.

Cyclic compound (P) contains 18 1º H atoms, two types of ‘C’ atoms. All H are identical. Identify (P).

– –

Metamerism

– –

Ans.

–

–

or

–

CH3 H3C – O – C – CH3 – –

Ans.

C –

C C Compound X is an ether. It has 12 1º H atoms. It has 2 types of H-atoms and 3 types of C-atoms. Write it Structural formula ? C

Q.

–

Ans.

–

C C

CH3

Page-35

Stereoisomers (Classification)

Stereoisomers : The stereoisomers has different orientation of groups along a stereo centre. A stereocentre can be C = C (any double bond), a ring structure, asymmetric carbon atom (*Cabcd). These isomers has same general formula, structural formula and molecular formula but different stereochemical formula. E.g.

CH3

C=C

H CH3

CH3 H H

(I)

C=C

(II)

H CH3 CH3–CH2–CH=CH2

(III)

I, III II, III I, II

  

Positional Positional Stereoisomer

Configurational isomers : Configurational isomerism arises due to different orientations along a stereocentre and these isomers can be seperated and these isomers do not convert into one-another at room temperature. Therefore, they are true isomers. They can separated by physical and chemical method. E.g. Cis–2–Butene Trans-2-Butene

Conformational isomers : When different orientations arise due to the free-rotation along a sigma covalent bond. Such isomers are called conformational isomers. E.g. eclipsed ethane and staggered ethane These isomers change into each other at room temperature and can never be isolated. So these are not considered as true isomers.

Geometrical Isomerism 1.

Cause of Geometrical isomerism : Geometrical isomerism arises due to the presence of a double bond or a ring structure C = C, C = N, N = N, or Ring structure (Stereo centres) Due to the rigidity of double bond or the ring structure to rotate at the room temperature the molecule exist in two or more orientations. This rigidity to rotation is described as restricted rotation/hindered rotation/no rotation. E.g.

a

a C=C b (I) b The root form of geometrical isomers lie in restricted rotation.

a b

a (I)

b

Page-36

Condition (i) Restricted rotation (ii) The two groups at each end of restricted bond must be different. 1 3 C=C 2 4 Caa Caa X Caa Cbd X Cae Cbb X Cab Cab  Cab . Cbd Cab Cde

 

(iii) In two geometrical isomer the distance between two particular groups at the ends of the restricted bond must be changed.

Q.

Which of the following compounds show G.I. (1) Ethene

(2) Propene

(3) 2-Methylbut-2-ene CH3  C  CH  CH3 | CH3

(4) But-2-ene CH3 – CH = CH – CH3

(5) Penta-1, 3-diene C = C – C = C – C

(6) 1, 2-Dideuteroethene

H2C = CH (7) Phenylethene Ans.

(8) Buta-1, 3-diene C = C – C = C

4, 5, 6

1. Geometrical isomerism across

– Nil

and

By E / Z

2. Geometrical isomerism across (a) Imine ( ) Imine compounds are produced from carbonyl compounds on reaction with ammonia.

(Syn and anti) Imines prepared from unsymmetrical addehydes and ketones, always show geometrical isomerism. Page-37

Q.

Which of the following compounds show geometrical isomerism after reaction with NH3.

O || (a) CH3  C  CH3 O || (d) CH3  C  H

(g)

Ans.

O || (b) H  C  H

O || (c) H  C  D

(e) Ph  C  CH3 || O

(f) Ph  C  Ph || O

(h)

(i)

c, d, e, g, i (b) Oximes

C = N – OH :

These are prepared by reacting carbonyl compuond with hydroxyl amine (NH2 – OH)

R C=O H

H2O + H2 N–OH   

R

R C = N–OH H (Aldoxime) OH

R

C=N H

(I) (syn)

and

OH

C=N H

(II) (anti)

Syn and anti in aldehyde only not for ketones. * Except formaldehyde (CH2O) All other aldehyde form two oximes. * Unsymmetrical ketoes form two oximes. Eg.

Which of the following ketones will form two oximes.

(1) Propanone

O || CH3  C  CH3

(2) Butanone

O || CH3  CH2 – C  CH3

(3) 3-Pentanone

C–C–C–C–C

(4) Acetophenone

C 6H 5–C–CH 3

O

O

O (5) Benzophenone

C 6H 5–C–C 6H 5

(6) Cyclohexanone

O

O Me (7)

Methyl Cyclohexanone Ans.

2, 4, 7

Q.

The lowest molecular weight of acyclic ketone and its next homologue are mixed with excess of NH2 – OH to react. How many oximes are formed after the reaction ? 3

Ans.

Page-38

(c) Hydrazones

C = N – NH2 :

R

R

H2O C = O + H2N.NH2    H hydrazine

C = N–NH2 H R

NH2

R +

C=N H

(I)

NH2

C=N H

(II)

(Geo. diastereomers) Q.

Ans.

Two chain isomer of a cycloalkanone which are next higher homologue of lowest molecular weight, cycloalkanone reacted with hydrazine. Identify the structure and number of isomer of hydrazones prepared ? .. .. N NH2 NH 2 N N .. O + + H2N – NH2 + + NH2 CH 3 CH3 O CH 3

Number of isomer = 3 (3) Geometrical isomerism across azo compounds (– N = N – ) (i) H – N = N – H (H2N2)

(ii) Ph2 N2 (Azobenzene)

..

.. N=N Ph

Ph syn

..

Ph ..

N=N anti

Ph

(4) Geometrical isomerism across ring structure

(i)

Restricted rotation

(ii)

(iii)

Page-39

(5) Geometrical isomerism in cycloalkenes across double bonds : In cycloalkenes, G.I. exists across double bonds with ring size equal to or greater then 8 carbon atoms (due to ring strain)

Stereocentre:An atom or bond across which stereoisomerism exists (either G.I. or optical isomerism)

Those stereoisomers which are not mirror images of each other are called diastereomers. Those compounds which are non-superimposable mirror-images of each other are called enantiomers

are enantiomers

E/Z Nomenclature : Z (Zussamen = together) a > b and e > d

E (Entegegen = opposite) a > b and e > d Rules :(i) The group with the first atom having higher atomic number is senior. Thus – F > – OH > – NH2 > – CH3 (ii) If the first atom is identical, then second atom is observed for deciding the seniority of the group.

(a)

(b)

<

(c)

<

(d)

> Page-40

Ex.

Z

Ex.

E (iii) If the first atom has same atomic number but different atomic mass, that is isotopes. Then heavier isotope has higher seniority. (iv) If the group has unsaturation, then a hypothetical hypothetical equivalent in drawn for it and it is compared with other group for seniority.

(1) (2) – CH = CH2 < – C  CH

(Hypothetical)

(v) Bond pair is always senior to lone pair. CH 2 = CH

Ex.

(1)

HC

C

C(CH3)3 C == C

E CH2 – CH = CH2

(2)

Z

N (3)

CH2 – C(CH3)3

C

H2C = C = CH

C == C

E

C

CH

(4)

Z

(5)

E

(6)

E

Page-41

Number of geometrical isomers Case I : Compounds having dissimilar ends No. of G.I. = 2n where n = number of stereocentre Ex. Ans.

Stereocentre = 2 Geometrical isomerism = 4

(a)

(b)

(c)

(d)

Case II : Compounds with similar ends & even no. of stereocentre. n

Ex.1 Ans.

No. of Geometrical isomerism = 2n–1 + 2 2 CH3 – CH = CH – CH = CH – CH3 Stereocentre = 2 Geometrical isomerism = 21 + 20 = 3

1

Case III : Compounds with similar ends but odd number of stereocentre. No. of Geometrical isomerism = 2n–1 + 2 Ex.1 Ans.

n 1 2

CH3 – CH = CH – CH = CH – CH = CH – CH3 Stereocentre = 3 Geometrical isomerism = 6

Physical properties of Geometrical isomers : The physical properties of organic compounds can be compared through the knowledge of their molecular formula, structural formula and stereochemical formula (S.C.F.) More Polar geometrical isomer is more soluble in water.

(i)

>

(ii)

Cis > Trans

<

COOH

COOH C=C

(iii) H

Trans > Cis

H

COOH

H C=C

> COOH

Cis > Trans H

Page-42

Page-43

Optical Isomerism Some organic compounds can rotate the plane of plane-polarised light. Such compounds are called optically active compounds. The optically active compound can show optical isomerism. Measurement of optical activity It is measured by an instrument called polarimeter.

= (1)

(5)

(3)

(6)

Recorder (8)

(1)  Source of light (2)  Polychromatic Non-polarised light (3)  Slit (Monochromator) (4)  Monochromatic Non-polarised light (5)  Polariser (Prism-setting) (6)  Monochromatic plane-polarised light (7)  Sample tube (  = path length) (8)  Recorder for measurement of optical rotation Observations:- When plane polarised light is passed through sample tube, then following changes can be observed in the recorder:Angle of rotation () (1)  = 0º (2)  = +xº (clockwise rotation) (3)  = –xº (anticlockwise rotation)

Inference Compound is optically inactive. Optically active, dextrorotatory or d or (+). Optically active, laevorotatory or l or (–).

Specific Rotation:- The specific rotation of a compound indicates the optical rotation of unit concentration (1 g/mL) present in a sample tube of 1 dm of path-length at given temperature and given wavelength of light. It is represented as follows:-

t  25 º C

[ ]  580 nm  where

 c

 = observed angle of rotation c = concentration in g/mL (or density)  = Path length of Sample tube in dm

Q.

Compound ‘X’ has  = +70º for 2 g/mL solution in sample tube of  = 1 dm. Calculate  ?

Ans.

=

(1) (2)

70 = +35º 2 Its concentration is made twice, the observed rotation() will be 140º but  = 35º If  = +70º, it will be a (i) d compound or (ii)  compound of  = –(360º – 70º) = –290º It can be decided by changing the concentration or by changing the length of the tube (c or  ). If concentration is reduced to half, d will have +35º and  will have –145º (not distinguish) still halfed, it will give +17.5º and –72.5º to distinguish.

(I) (II) With symmetrical With asymmetric molecule molecule [] = 0 [] 0 Optically inactive compounds:For symmetrical molecule optical rotation observed after interaction of light is zero. Page-44

But in case of optically active compounds. The molecules are asymmetric in nature and show non-zero optical rotation.

Symmetry of elements :(1) Centre of symmetry (C.S.) (2) Plane of symmetry (P.S.) (3) Axis of symmetry (4) Alternating axis of symmetry Plane of symmetry : - The imaginary plane which divides a molecule into two equal halves which are related as mirror image is known as plane of symmetry

Centre of symmetry : - A centre of symmetry in a molecule is said to exist if a line is drawn from any atom or group to this points and then extended to an equal distance beyond this point, meets the identical atom or group.

Cl H

R

H

C=C H R Centre of symmetry

H

Cl Centre of symmetry

Axis of symmetry : It is defined as Cn axis of symmetry that means of the molecule is rotated by

360 º n

angle, then its original or identical or superimposable. The molecule is said to have n-fold axis of symmetry

Ex.

C

2   

(180 ºrot )

Ex.

(C4 axis)

Ex.

(axis)

There is no relation whatsoever with chirality and axis of symmetry. Meso molecule does not have axis of symmetry

 no axis of symmetry

.

Page-45

Alternating Axis of Symmetry : When a molecule is rotated by

360º angle and its mirror image is taken in the perpendicular plane of n

rotation, then its original molecule is obtained. It is represented by Sn

Ex.

H H

Me

Me

H

Me

180º rotation

(I)

H

Me Me

H

H

Reflextion

(II)

Me

(III)

I and III are equivatent. Dissymmetry:- The molecules or objects in which minimum two elements of symmetry (Centre of symmetry and Plane of symmetry) are absent are called dissymetric molecules/objects. Asymmetry:- When all the elements of symmetry (23 including Centre of symmetry and Plane of symmetry are absent) the molecule is said to be asymmetry. So, all asymmetric molecules are always dissymmetric, but the reverse is not true. Dissymmetry and chirality:- All dissymmetric molecules are chiral and are optically active. Dissymmetry, Chirality and Optical Activity:- Dissymmetry/Chirality is the minimum and sufficient condition for a molecule to be optically active. So, if in a molecule, a C.S. and P.S. are absent, the molecule will be optically active. All asymmetric molecules are also optically active. Chirality (Hand-like property or Handedness):- The human hand does not have Centre of symmetry and Plane of symmetry , so it is dissymmetry. It does not have any element of symmetry, so it can be called asymmetry. Any molecule which has this property is called chiral. Note : All dissymmetric compounds are chiral.

Optical Isomers:Enantiomers:- The mirror-image stereoisomers are called enantiomers. Two types:(a) Dextrorotatory (b) Laevorotatory

       

Enantiomers are always mirror image isomers. They have same molecular formula, structural formula but have different orientation in space. They have dissymmetry/chirality. Every enantiomer is optically active. The two enantiomers can rotate the plane-polarised light with equal magnitude and opposite signs. They have similar physical properties except the sign of optical rotation. These are the isomers which have maximum resemblance with each other. These can be distinguished only by polarimeter.

Chirality (Dissymmetry) and Optical Isomers:(a) (b) (c) (d) (e)

The dissymmetric molecules have two orientations in space. These two orientations are called stereoisomers, optical isomers, enantiomers. These are mirror images (enantiomers). Non-superimposability of enantiomers:- The dissymmetric molecules are always non-superimposable on their mirror-image orientations. The non-superimposable orientations are non-identical orientations, so are called isomers. Because of dissymmetry, these two isomers are capable of rotating plane-polarised light. So, these are called optical isomers. Page-46

Optically Active Carbon Compounds:If a carbon-atom is attached with four different group, then if does not have any element of symmetry. It is known as asymmetric carbon atom, which is represented as *Cabde. a C* e d If a molecule contains only one asymmetric carbon atom, then the molecule as a whole becomes chiral and optically active and show optical isomers. b

Centre of symmetry

Plane of symmetry

Optical active

H C

(1)

H

H

H

Absent

Yes

No

Absent

Yes

No

Absent

Yes

No

Absent

No

Yes

H C

(2)

H

H

Cl

H C

(3)

Br

H

Cl

H C

(4)

Br

Cl

F Ex.1:- Mark the chiral objects:(i) Cup (ii) Plate (vii) Shoe (viii) Glove Ans.

(iii) Letter A

(iv) Letter G

(v) Fan

(vi) Door

IV, VII VIII

Projection Formula of Chiral Molecules:(i) Wedge-Dash Projection formulae down up

Ex:-

(i) Butan-2-ol

CH3 C H

C2H5 OH

(ii) Fisher Projection formula (i) Butan-2-ol

(CH3 – CH2 – CH – CH3) –

Ex:-

CH3 H

OH CH2CH3

///////////////////////

OH Rules of writing Fisher Projection formula :(i) It is represented by a cross (+). (ii) Groups at Vertical line are away from observer. (iii) Groups at Horizontal line are towards the observer. (iv) Central ‘C’ atom of the cross is chiral. (v) High priority group lies at the top of vertical line (IUPAC Numbering starts from top).

Page-47

Draw Fisher Projection formula of following molecules:-

(1) 2-chlorobutane

(2) Pentan-2-ol

–

* (3) CH2 – CH – CH OH OH O –

Glyceraldehyde

1

2

3

4

5

–

* * (4) CH3 – CH – CH – CH2 – CH3 OH OH –

Ex:-

,

,

,

(I, II), (III, IV)  Enantiomers. All other diastereomers. They are true isomers (not superimposable)

CHO CHOH CHOH CHOH CHOH CH2OH

– – – – –

(5) Glucose

Mark:- (i) No. of C*  4 (ii) Draw one Fisher Projection Formula CH = O H

OH

H

OH

H

OH

H

OH

Page-48

Configuration nomenclature in Optical Isomers :Relative configuration : The experimentally determined relationship between the configurations of two molecules, even though we may not know the absolute configuration of either. Relative configuration is expressed by D-L system. Absolute configuration : The detailed stereochemical picture of a molecule, including how the atoms are arranged in space. Alternatively the (R) or (S) configuration at each chirality centre. (I) D - L System (Relative configuration) : Application on correct Fisher Projection Formula This method is used to relate the configuration of sugars and amino acids to the enantiomers of glyceraldehyde. The configuration of (+)-glyceraldenyde has been assigned as D and the compounds with the same relative configuration are also assigned as D, & those with (-) glyceraldehyde are assigned as L.

Examples :

Sugars have several asymmetric carbons. A sugar whose highest numbered chiral centre (the penultimate carbon) has the same configuration as D-(+)-glyceraldehyde (– OH group on right side) is designated as a Dsugar, one whose highest numbered chiral centre has the same configuration as L-glyceraldehyde is designated as an L-sugar.

e.g.

(I) R/S Configuration : R  Rectus

S  Sinister

Examples :

a b

c

d

(1) Seniority order a > b > c > d (2) Put junior most group at dotted line b



(3)

Q.

c a d Now go from a to b, b to c. (i) If it follows clockwise route, it is (R). (ii) If it follows anticlockwise route, its configuration is (S). Br Br

Cl

I

F (R)

Cl F (S)

I

Page-49

R/S Configuration in Fisher Projection Formula :(c)

Me(c)

Ex:-

H(d)

Et(b)  (d) Pr (a)

(b) 

abc     Clock wise

R

(a)

 

If the junior most group is at horizontal line clockwise  S & anticlockwise  R If juniormost at vertical line, then clockwise  R & anticlockwise  S

Q.

(1)

Ans. R

(2)

Ans. R

(3)

Ans. S

Properties of enantiomers :One chiral carbon : (i) Number of optical isomer = 2 (d or  ) (ii) Number of Racemic Mixture  Equimolar mixture of d and  . [] = 0 due to external compensation of optical rotation.

Properties:(i) Dipole moment (ii) Boiling Point (iii) Melting Point (iv) Solubility (v) Specific rotation []

same same same same different

Molecules with more than one chiral carbons:(I) Calculation of No. of optical isomer :Case I:- When all Chiral carbon atoms are differently substituted (all are dissimilar C*) optical isomer = (2)n Case II:- When all the Chiral carbon atoms are similar at ends. n

(i) n = even

 x = 2n – 1 + 2 2

–1

(ii) n = odd

 x = 2n – 1 Page-50

CH3

CH3

–

–

Molecules with two Asymmetric Carbon Atoms of Dissimilar Nature:(i) Structural formula CH3 – CH – CH – CH2 – CH3 Cl Cl n 2 (ii) Optical isomer = 2 = 2 = 4 (iii) Stereochemical Formula:-

CH3

CH3

H

Cl

Cl

H H

Cl

Cl

H

H

Cl

Cl

H; Cl

H

H

Cl

C2H5

C2H5

I [] = +xº

C2H5

II –xº

C2H5

III +yº

Analysis:(a) I, II – Enantiomers I, III – Diasteromers II, III – Diasteromers

IV –yº

III, IV – Enantiomers I, IV – Diasteromers II, IV – Diasteromers

(b) No. of Racemic Mixture – Two (I + II, III + IV).

n = 3, x = 23 = 8 (4 d-  pairs)

–

Ans.

–

CH3 – CH – CH – CH – CH2 – CH3 Cl Cl Cl Calculate total number of optical isomer. –

Q.

(i)

(ii)

(ii)

(iii)

(iv)

and their mirror images. 4 Racemic Mixtures (4 different fractions on fractional distillation)

(i) n = 2

–

–

Compounds with 2 Asymmetric Carbons of Similar Nature:Ex:CH3 – CH – CH – CH3 OH OH n

(ii) x = 2n – 1 + 2 2

–1

= 3 (Optical stereoisomers).

(iii) Stereo chemical formula :-

Me H

OH

H

OH Me (I) (II)  Plane of symmetry present. In (I and II)  Superimposable on its mirror image.  Thus, I and II are identical.  [] = 0, optically inactive.  Meso isomer In (III)  Specific rotation [] = +xº In (IV)  Specific rotation [] = –xº

(III)

(IV)

Page-51

No. of d –  pairs = 1 (III + IV) = Racemic mixtures In (I) and (II) : [] = 0 due to internal compensation and Non-Resolvable In (III) and (IV) : Resolvable [can be separated into two isomers (enantiomers)]

Meso isomers: The optical stereoisomers which have more than one asymmetric carbon atoms but have a plane of symmetry are called meso compound.  They are achiral (optical rotation = 0).  They have [] = 0 due to internal compensation of optical rotation.  They are diastereomer of d –  pair. So, it has different physical properties than d –  -pair..  Presence of more than one asymmetric ‘C’ atoms.  They are non resolvable.

Q.

Mark meso compound among following

H

H

COOH OH

(1)

(2)

H Ans.

OH COOH

Cl Me

Cl

(3)

(4)

Me

OH H

(5)

* OH

(1) (2) (4) and (5)

–

–

–

Molecules with three similar chiral carbon: Ex:CH3 – CH – CH – CH – CH3  M.F.. Cl Cl Cl n=3 x = 2n – 1 = 4 stereoisomers. Stereo Chemical Formula :-

\\\\\\\\\\\\\\\\\\\\

(I) 2S 3S 4R

(II) 2S 3R 4R

(III) 2S 3  achiral 4S

(IV) 2R 3  achiral 4R Optically active

(i) & (ii) Achiral due to Plane of symmetry. (Optically inactive).

Properties of Optical Diastereomers:The optical isomer which are neither mirror image nor superimpossible to each other are called optical diastereomers.

Conclusion:- Optical Diastereomers have different physical properties, so these can be separated by normal physical methods of separation. (i) By fractional distillation (different B.P.) (ii) By fractional crystallization (different solubility) (iii) Chromatography (different solubility) (iv) Differential Melting (M.P. diff.) Page-52

 Assertion:- All diastereomers have different polarities (Both Geometrical and Optical) – True  Assertion:- Geometrical isomer are always diastereomers – True

Chemical method of separation (resolution) by using optically active reagent:Separation of Racemic Mixture:- The enantiomers in a d –  pair have identical physical properties, so these cannot be separated by normal physical methods of separation. The special method is used for separation of d –  pair known as optical resolution.  General scheme of resolution:-

d+ + d' Racemic Optically mixture active Reagent

Hydrolysis

Hydrolysis

d + d'   (I) Resolution (by using Esterification) of RCOOH and R'OH

Example:-

 Esterifica tion   General Reaction:- R – C – O – H + H – O – R '      R* – C – O – R' (H2 SO 4 ) O O – H2 O

Me H

COOH (d) Et Et

H

Me

+ OH COOH ( )

H D (d')

H

H + H (dd')

C–O O

(d')

O D ( d ')

(dd')

Me

H (d' )

C–O Me

D

Et

Me

Et

Me

Me

Pair of Diastereomers Ester upon still hydrolysis will give back the carboxylic acid and alcohol. (II) Separation of (d ) pair of alcohol by using optically active acid:-

D

Ph +

H

H (dd')

O–C O Me

Me

Ph

Me

+

H

H

C–O O D

(d' )

Me

Page-53

(III) By salt formation (RCOOH + R’NH2)

+

+

Diastereomers (Separable)

Optical Purity or Enantiomeric Excess (O.P. or E.E.):Ex:Compound X has [] =  70º We have mixture of enantiomers in different percentage. %E.E. 

(obs)  100% []specific

Page-54

Optically active compounds without a chiral carbon:Rapid R R Inversion –   Asymmetric N:R' – N N – R' R'' R''  A nitrogen atom attached with 3 different groups (sp3) is chiral, asymmetric.  In case of acyclic compounds, the Nitrogen, having trigonal pyramidal shape undergoes rapid inversion of shape at room temperature. So at room temperature, everytime a racemic mixture of ‘d’ and ‘  ’ forms exists. This racemic mixture can never be resolved at room temperature.  Nitrogen converts into its enantiomer..

–

–



–

So, acyclic molecules with chiral nitrogen are chiral but optically inactive ([] = 0) and are non-resolvable. If asymmetric Nitrogen atom is present in ring structure, then inversion does not take place and such molecules are optically active [ ]  0  . (otherwise ring will break)

Ex:-

 Me N

[ ]  0

Optical activity without asymmetric carbon :

CCC–C–C (Cumulated double bond)  One after another

CC–CC–C (Conjugated double bond)  Alternate

C C–C–C C (Isolated double bond)  Separated

(I) Case of allene : (a) Allenes with even  bonds :

e.g.

the orbital diagram of this structure will be Since the groups at the end of allene are in perpendicular plane, it will not show geometrical isomerism. The molecule lacks centre of symmetry as well as plane of symmetry. Overall the structure has molecular dissymmetry which is the sufficient condition for optical activity. The molecule will exist in two enantiomeric forms.

(b) Allenes with odd  bonds :

e.g.

the orbital diagram of this structure will be

-

The groups at the end of allene structure lie in same plane (ZX plane). Therefore it will have a plane of symmetry (ZX plane). The molecules lacks molecular dissymmetry & it will not show optical activity hence optical isomerism. But the compound will exist in two geometrical diastereomeric forms.

Page-55

(II) Case of spiranes : A similar case like allenes is observed in spiranes. The spiranes with even rings and different groups at terminal carbons show optical activity & optical isomerism, while the spiranes with odd rings shows geometrical isomerism. (a) spiranes with even rings : shows optical isomerism.

(b) spiranes with odd rings : shows geometrical isomerism. (III) Case of cycloalkylidene :

(IV) Case of ortho-ortho-tetrasubstituted biphenyls : becomes non-planar at room temperature in order to have minimum electronic repulsion among the substituent. In this orientation (phenyl planes perpendicular to each other) the free rotation of C – C single bond is restricted and molecule shows optical activity due to molecular disymmetry. e.g.

Page-56

CONFORMATIONAL ISOMERISM : 1.

Free Rotation : A sigma covalent bond undergoes free rotation at room temperature. (C–C, C –O, C–N, N–N, O–O)

2.

Conformers / Rotamers or conformations : The infinite number of spatial oreintation of molecule arises due to free rotation along a sigma covalent bond

3.

Conformational Isomers : Those conformations which are most stable and have minimum P.E. are defined as conformational isomers. These are not true isomers and can never be isolated.

4.

Conformational Energy : The rotational energy barrier is known as conformational energy. It is the P.E. difference between conformational at potential energy minima and maxima.

5.

Angle of rotation/Dihedral Angle (D.H.A.)/Torsion Angle : The interfacial angle between the groups attach at two -bonded atoms is defined as dihedral angle. Projection formula (a) Saw horse Projection formula Ex.

CH3 – CH3

H(a) H(c) rotation    H (b)

Eclipsed

H (c)

H(b)

H(a)

Staggered

(b) Newmann Projection formula

      1 12.5 kJ / mol

Energy - Level Diagram of ethane

Ex.

CH3–CH2–CH3

Page-57

Saw horse Projection formula

Eclipsed

Staggered

Newman Projection Formula CH3 H

CH3

  60 º

H

H

H

H

120 º  

H II Staggered

Eclipsed

HH

H III Eclipsed

CH3 H H H



H

H

H H IV Staggered

Propane :

Butane : Ex.

Ethyl-hydrogen repulsion is less than methyl-methyl repulsion. So draw the newman between C2 - C3 CH3

CH3 CH3

60 º

CH3 CH3

H

120 º  

 

H H

H (I)

H

H

H (II)

180 º  

H

H

H 240 º  

H

CH3

(IV)

300º

 

CH3 H

H

H CH3 H H (V)

360º

 

Page-58

I/VII = Fully eclipsed II/IV = Gauche form III/V = Partially eclipsed IV = Anti form Stability Order : IV > II > III > I

Anti > Gauche > Partially eclipsed > Fully eclipsed

P.E. Order : IV < II < III < I Potential Energy Diagram of butane :

Conformational Isomers : Due to free rotation from 0º to 360º those conformations which are most stable and have minimum P.E. are defined as conformational isomers.If these conformers have same energy then there isomers are called degenerate isomers or equienergic isomers. The conformational isomers with different energies are called non-degenerate isomers. The conformational isomes lie at the P.E. minima in the P.E. diagram with respect to rotational angle.

Strains : (i) Torsional Stran (eclipsing strain) : It is defined as the electronic repulsion between the bond-pairs electrons of two adjacent eclipsed bonds. It is active at torsional angles 0º, 120º, 240º in which the molecule has eclipsed conformation. It is considered almost zero in the staggered conformation. (D.H.A. = 60º, 180º, 300º) (ii) Vander Waal strain : It is the repulsion between the group attached at adjacent bonds. The vander waal strain is maximum in eclipsed conformation, and minimum in anti conformation, while intermediate in gauche conformations. It is almost zero for H-atom since the sum of vander waal radii is less than internuclear distance between two H atom. (iii) Angle Strain : It arises due to distortion in normal bond-angles. In acyclic compounds, there is no distortion of bond-angle due to free rotation. So, all conformations have zero angle strain in acyclic compound

Page-59

Ques. Draw conformation isomers of following compound (a) CH3 – CH – CH3

(b) CH3 – (CH2)3 – CH3

CH3 Ques. Draw conformation isomers of following compound CH3 – CH –CH – CH3 with respect to C2 and C3 carbon CH3 CH3 atoms. H H

H 60 º

H3C

180 º



CH3

CH3 CH3

CH3

CH3

120 º

 

CH CH3 CH3 3 Eclipsed

H3C

H

CH3

H

H



H3CCH

3

H

CH3

H3C

CH3

H Staggered

Page-60

Case of intramolecular hydrogen bonding : In case of G – CH2 – CH2 – OH, where G = – OH, – NH2 , – F, – NR2, – NO2, – COOH, – CHO the Gauche form is more stable than the anti form due to intramolecular hydrogen bonding i.e.

stability : Gauche form > anti form. Example 1.

CH2–OH (Glycol). CH2OH OH H

OH



 H



H

H (II) Gauche

Hydrogen Bonding : (II)

Gauche form most stable due to hydrogen bonding. Stability Order : II > IV > III > I Example 2.

O2N – CH2 – CH2 – OH

Interconversion of projection formulae : Example

Tartaric Acid : COOH – CHOH – CHOH – COOH

no. of Sterioisomers : 3 COOH

1. Meso

H

OH

H

OH



COOH

Page-61

2. d/



COOH

COOH H

Meso



COOH H

OH

H

OH COOH



OH

OH



H H

COOH OH

COOH

COOH COOH

H H

COOH OH

H

OH

COOH

 

OH

H

OH

H

OH

d/ COOH

COOH COOH OH H

OH

COOH

 H OH

 H

H

OH

COOH

COOH OH

HH

COOH  H OH

COOH

COOH

  H OH

OH

OH COOH

H

H

COOH

OH

H

OH H COOH OH

Conformation isomers of cyclohexane : Chair form and Boat form

Page-62



1

Structure Identification * Monochlorination * Catalytic Hydrogenation * Ozonolysis * Elements detection * Identification of Functional Group by Lab. Test

Refer sheet GOC- I JEE Syllabus [2009] Practical organic chemistry: Detection of elements (N, S, halogens); Detection and identification of the following functional groups: hydroxyl (alcoholic and phenolic), carbonyl (aldehyde and ketone), carboxyl, amino and nitro; Chemical methods of separation of mono-functional organic compounds from binary mixtures.

63

STRUCTURE INDENTIFICATION Monochlorination:(a)

Cl / h  CH3Cl + HCl (i) CH4 2  Cl / Sunlight (ii) CH3 – CH3  2    CH3 – CH2Cl + HCl –

Cl

Cl / h 2  

(iii)

+ HCl – –

– –

C C Cl2 / h  C – C – C – Cl + HCl (iv) C – C – C    C C Cl –

Cl / h 2  

(v)

+ HCl

Remarks:- When an alkane or a cycloalkane is treated with halogen (Cl2, Br2, F2, I2), a photochemical reaction takes place and a C – H bond cleaves and a C – Cl bond is formed. If one H-atom is substituted by one halogen atom. This is known as monohalogenation reaction. Application:- If a molecule has more than one type of H-atom, then on monochlorination, it forms a mixture of monochloroisomers. All these isomers are position isomers. Conclusion:- Hence, it can be concluded that the total no. of position isomers (structural) of monochloro compounds is equal to the number of different types of H-atoms present in the reactant. The different type of H-atoms are also known as non-identical Hydrogens or non-equivalent Hydrogens or chemically different Hydrogens.

Ex.

Monochlori nation        2 Products (structure isomers)

(b) C – C – C – C


(c) C – C – C – C – C


(d) C – C – C – C


–

(a) C – C – C

–

C CH3 Monochlori nation        5 Products (structure isomers)

(e)

Q.

Cl / h  only one monochloro isomer.. X(C5H12) 2 

Ans.

X = Neopentane

Q.

Cl / h  Two monochloro P(C6H14) 2 

How many isomers of P will give two monochloro compounds ? Ans.

C

–

–

C

C – C – C – C only one isomers

Remark : In aromatic hydrocarbons, the hydrogen atoms of the side-chain are chlorinated, but H-atoms of Benzene ring are stable.

CH3 Ex.

–

–

CH2Cl Cl / h 2  

64

Q.

Cl / h  Two mohochloro X(C8H10) (Aromatic) 2  Cl / h  One monochloro Y(C8H10) (Aromatic) 2 

Ans.

(X)

(Y)

Catalytic Hydrogenation of C = C; C  C General reaction:Ni  R – CH2 – CH2 – R (a) R – CH = CH – R + H2 

Ni / Pt / Pd  R – CH2 – CH2 – R (b) R – C  C – R + 2H2     H2

H2 R – CH  CH – R  R – CH2 – CH2 – R (Not isolated) 2H2 / Ni   CH3 – CH2 – CH2 – CH3 (c) CH2 = CH – CH = CH2  

(d)

(e)

3H2 / Ni    

CH = CH2 –

H2 / Ni       room temperature CH – CH3 – 2

H 2/Ni (100 – 150ºC)

(f)

H / Ni,  2  

CH – CH3 – 2

[Reaction cannot be stopped at any intermediate stage]

Remarks:(a) Alkenes, Alkynes, polyenes or polyynes can be hydrogenated by using catalysts Ni/Pt/Pd at room temperature. (b) All C – C  bonds(C = C, C  C) are hydrogenated. The reaction can’t be stopped at any intermediate stage.

Exceptions:Aromatic  bonds which are stable at room temperature but can be hydrogenated at high temperature. 

It can be concluded that the hydrogenation product of an alkene or alkyne or any unsaturated compound is always a saturated compound.



The no. of moles of H2 consumed by 1 mole of compounds is equal to the no. of  bonds presents.



All positional isomers of alkenes or alkynes (due to multiple bond) always give same product on hydrogenation. 65

During catalytic hydrogenation and monochlorination, carbon skeleton remain unchanged.

Ex:- C = C – C = C + Cl2 X

Ex.

C– C–C C – Cl –



2H2 / Ni    

(1)

Cl

Y =

(X)

Cl / h 2   Y + Z

Z =

Cl CH3

CH3 H / Ni 2

(2)

CH3

CH3 H / Ni 2

(3)

Q.

Cl / h 2   5 Monochloro product

5 Monochloro product

H / Ni Cl / h  Z(only one monochloro product) X(C4H6) 2 Y 2 

Identify X, Y, Z Ans.

DU = 2 X

Y

,

Z

Identify the lowest molecular weight alkane which gives four structural isomeric monochloro products ? C– C –C–C –

Q. Ans.

,

Cl –

C

C5H12

= 72g

Identify the structure of hexane which gives 3 monochloro products ?

Ans.

C C – C – C – C, C

Q.

Find the no. of monochloro products of a fully saturated isomer of C4H6.

Ans.

DU = 2

– –

Q.

C–C–C–C–C–C

 2 monochloro product Find the structural isomers of product?

fully saturated cycloalkane of M.F. C6H12 which gives two monochloro

–

Q.

– Q.

–

Ans.

Cl / h 2   A C 8H17 Cl (C8H18 ) (Only one type )

Identify A ? 66

– –

Q.

– –

Ans.

C C C– C– C–C C C

Write all isomeric alkynes which produce an isomer of heptane which on further monochlorination gives (a) three monochloro products.

Ans.

C–C (i) C  C – C – C – C

(b) Ans.

two monochloro Nil

Q.

Find the structure of lowest molecular weight hydrocarbon and maximum unsaturation which on hydrogenation produce such an alkane which gives two monochloro products ?

Ans.

C = C = C or C – C  C

Q.

Determine the M.W. of maximum unsaturated hydrocarbon which on hydrogenation gives C6H12 which on further chlorination gives two monochloro.

– –

–

C (ii) C  C – C – C – C C

CH2

Ans. CH2

CH2

C6H6 = 78g

Ozonolysis: It tells about position of unsaturation. Remarks:(1)

Alkene and polyalkene on ozonolysis undergo oxidative cleavage.

(2)

(a) The reagent of reductive ozonolysis is (i) O3 (ozone) (ii) Zn and H2O or Zn and CH3COOH or (CH3)2 S (b) The reagent of oxidative ozonolysis is O3 and H2O2.

(3)

The products are carbonyl compounds (aldehydes or ketones). This type of ozonolysis is known as reductive ozonolysis.

(4)

Ozonolysis does not interfere with other F.G.s.

General Reaction:- R – CH = CH – R

Ex:-

(1) CH2 = CH2

(1) O 3 R – CH = O + O = CH – R + ZnO + H2O (2) Zn/H2O

(1) O 3 CH2 = O + CH2 = O (2) Zn/H2O

(2) CH3 – CH2 – CH = CH2

(1) O 3 CH3 – CH2 – CH = O + O = CH2 (2) Zn/H2O

(3) CH2 = CH – CH2 – CH = CH – CH3

O3 / Zn

CH2 = O + O = CH – CH2 – CH = O + O = CH – CH3 67

O / Zn 3  + OHC – CH2 – CHO (Propandial)

(4)

Applications:  The process is used to determine the position of C = C in a molecule.  If the products are rejoined, the position of C = C can be determined in the reactant molecule. All C = C (except aromatic ones) undergo oxidative cleavage under normal conditions.  At higher temperature, the aromatic double bonds can also undergo ozonolysis.

C

O3   O = C – C – C – C – C = O + O = CH2

(1)

Zn

O

C

O () 3

(2)

(4)

Q.

CH = CH – CH3 –

low temperatur e

CH = O

–

(3)

Zn

       O3 / Zn

+ O = CH – CH3

O3   C6H5 – CH = O

– CH = CH –

Zn

H / Ni Cl / h 2 CnH2n  2 2   CnH2n1Cl (P) CnH2n – 2 (Q ) (m products) Single Compound m3 (no isomer )

–

–

CH3 HCOOH + CH3 – C – COOH CH3

Identify P ?

Ans.

P=

Q.

An unsaturated hydrocarbon on ozonolysis produces 1 mole of

, 1 mol CO2, 1 mol

Find the structure of the hydrocarbon and the no. of monochloro products formed followed by hydrogenation. Ans.

,

5 monochloro product

68

– X (Unsat. hydrocarbon)

H / Ni 2

–

Q.

Cl2   m products h (m  3 )

O 3 (Zn/H2O)

O identify structure of X ?

O

Ans.

Q.

X =

H / Ni 2 X (Unsat. H.C.)

Cl / h 2   m products (m  7 )

O 3 (Zn/H2O)

HCHO + 4(1-oxoethyl) Cyclohexan-1-one. Identify X ?

C X

–

Ans.

C=C–C

Q.

Identify structure of X ?

Ans.

X is

Q.

H2 C–C–C–C–C–C X 

O3 (Zn)

CH3CHO + CHO – CHO Identify structure of X ? Ans.

X is C – C = C – C = C – C

69

Q.

O3   Zn, H2O

–

CH3

CH – 3

O3   Zn, H2O

–

CH3

2

CH3

+

CH3 C – C– +2

–

CH – 3

Sol.

O O

Q.

A

Methyl glyoxal + Formaldehyde

Ex.1

Identify A,B & C with the help of following reactions. Cl2 / h

(A) (C 9H18 )

    Single monochloro produc

( Saturated Hydrocarbon )

t Cl2 / h (B) (C8H18)    Single monochloro product

Cl / h

(C) (C 7H14 )  2  s  Two monochloro products ( Saturated Hydrocarbon)

Sol.

A=

or

Ex.2

C C | | B = CCCC | | C C

C =

Identify -Terpinene and P-Menthane.

Sol.

-Terpinene

Ex.3

Identify A & B

Sol.

A = CH3  C  CH  CH  CH2 | CH3

B = CH3  CH  CH2  CH2 – CH3 | CH3 70

Ex.4

Identify A & m

A = Ph  C  C  Ph | | CH3 CH3

Sol.

m=3

Ex.5

H 2 / Ni A  

Cl / h n-products 2  

O3 Zn/H 2O

+

H– C –H +

H – C – C – CH2 – CH2 – C – C – H

Sol.

A=

n=7

71

1.1

Identification of Elements in Organic Compounds Element

1. Nitrogen

Test / Reaction

Remark

Lassaigne’s test

The appearance of green or

Na + C + N  NaCN

prussian blue colour confirms

FeSO4 + 6NaCN  Na4 [Fe(CN)6] + Na2SO4

the persence of nitrogen.

3Na4[Fe(CN)6] + 4FeCl3  Fe4[Fe(CN)6]3 + 12NaCl

2. Sulphur

Formation of a white ppt.

(a) Oxidation test

indicates presence of sulphur

3KNO3  3KNO2 + 3[O] Na2CO3 + S + 3[O]  Na2SO4 + CO2 BaCl2(aq) + Na2SO4(aq)  BaSO4  + 2NaCl(aq)

Appearance

(b) Lassaigne’s test

purple

colouration confirms the

2Na + S  Na2S Na2S + Na2[Fe(CN)5NO]  Na4[Fe(CN)5NO.S]

3. Halogens

of

presence of sulphur

Lassaigne’s test

A white ppt. soluble in NH4OH

X + Na  NaX

solution indicates chlorine.

NaX + AgNO3  NaNO3 + Ag X 

A dull yellow ppt. partly soluble in NH4OH solution indicates bromine. A yellow ppt. completely in-

,

soluble in NH4OH solution indicates iodine

A white ppt. of magnesium

4. Phosphorus

pyrophosphate indicates phosphorus H3PO4 + Magnesia mixture  MgP2O7 + H2O 2MgNH4PO4  Mg2P2O7 + 2NH3 + H2O

5. Nitrogen and Sulphur

Blood red colouration confirms

Lassaigne’s test FeCl

Na + C + N + S  NaSCN   3 Fe(SCN)3

presence of both nitrogen & sulphur

72

73

Bubbles of H2 come out (3)° Cloudiness appears immidiately (2°) Cloudiness appears within 5 min. (1°) Cloudiness appear after 30 min.

Lucas Reagent [Conc. HCl + anhyd. ZnCl2]

ROH 3° 2° 1°

White ppt.

R – OH + HCl

white ppt

+ H2 O cloudiness

R – C  C Ag  (white)

R – C  C Cu  (red)

RCOOH + RCOOH

2HCHO

2ROH + Na  2RONa + H2

R – C  CH + Ag+

R – C  CH + CuCl

Red ppt.

Na

(b) AgNO3 + NH4OH

(a) Cuprous chloride + NH4OH

R – C  C – R

Acid formed.

= O Compounds

O3(ozone)

O3

Red colour decolourises

H2C = CH2 + O3

--------------

NR NR NR NR Pink colour Disappears

Reaction

Observation

Br2 / H2O

[Bayer’s reagent] alk. dil. cold KMNO4

conc. H2SO4 conc. NaOH KMnO4 LiAlH4

Reagent

(R – OH)

R – C  CH

CC

CC

C=C/ CC

C–C

Functional Groups

1.2 Identification of Functional Groups by Laboratory Tests

Lucas Test I. ter.alcohol II. sec. alcohol III. pri.alcohol

Presence of active ‘H’

Ozonolysis

Ozonolysis

Bromination

Hydroxylation

Inert paraffins

Remarks

74

Amides

Ester

CH3CHO

or ArCOCH3 or

R – COCH3

R – CHO

Ar – OH Enols

Functional Groups

 (yellow orange ppt.)

Sodium bicarbonate test

Effervescence evolve.

Pink colour  disappear on heating.

Smell of NH3

Conc. NaHCO3 solution

NaOH, phenophthalein.

Conc. NaOH, 

RCOONa + NH3 

Schiff’s reagent : p-Rosiniline hydrochloride saturated with SO2 so it is colourless. The pink colour is resumed by RCHO.

RCONH2 + NaOH

R COOH + R’ OH (Colourless solution)

(pink)

R COOR’ + NaOH + Phenophthalein

Litmus test.

Iodoform reaction

Tollen’s test

Fehling’s test

DNP-test

Test of enols / phenols

Remarks

Litmus change to red.

H2O + CO2

RCHO + Ag+  RCOOH + 2Ag (Silver mirror)

RCHO + Cu+2  RCOOH + Cu2O  + 2H2O Fehling soln. Red

H

+ H2

Reaction

Blue litmus

Yellow ppt of CHI3 (iodoform)

Pink colour resume

Schift’s Reagent *

I2 / NaOH

Black ppt. or silver mirror

Tollen’s reagent

Red ppt.

Yellow orange ppt.

2, 4-Dinitrophenyl hydrazine (2, 4-DNP) solution

Fehling solution A&B

Coloured ppt. (violet, blue, green buff)

Observation

FeCl3 (Neutral)

Reagent

75

Violet colour Blue colour

Ninhydrin reagent (0.2 % sol.n)

Amino acids

Reddish violet colour.

Molisch’s reagent (10% -naphthol in alcohol).

(i) NaNO2 + H2SO4 (ii) Phenol

R2NH Sec. Amines

red colouration Liebermann test

Orange red dye is formed

Effervescence of N2

Nauseating odour (Carbylamine)

black ppt

Observation

Carbohydrate

HNO2 (NaNO2 + HCl) + -Naphthol

HNO2 (NaNO2 + HCl)

CHCl3, KOH

Mulliken’s test

Reagent

Ar. amines. ArNH2

Amines (pri.) RNH2

Nitro Compounds (RCH2NO2) or ArNO2

Functional Groups

ArNHOH

C

C=N – C (Blue colour)

CO

CO

C OH

CO

+ H2N.CHR.COOH (Amino acid) OH

OH

(Ninhydrin)

CO

CO

Benzenediazonium chloride  – Naphthol

N=N-Cl +

OH N=N

OH

+ CO2+ RCHO + H2O

orange-red dye

+ 2H2O

N2Cl

NH2.HCl

ROH + N2 + H2O NaCl + HNO2

+ HNO2

Ag Ag

RNC + 3KCl + 3H2O

NaNO2 + HCl

R NH2 + HONO

R NH2 + CHCl3 + 3KOH

Reaction

Ninhydrin test

Dye test

Carbylamine Reaction

Remarks

Examples 1.

: Structure Determination

Identification of Organic Compounds on the bacis of Physical Properties (a) Physical state (b) Odour (c) Water solubility (d) m.p. / b.p. (e) relative

Ex. 1

Ex. 2

Ex. 3

Ex. 4.

2.

Identification of Organic Compounds on the bacis of Chemical Properties

Ex.5

Which of the following will not give (+ve) L.S. test for Nitrogen. (A) CH3–CH2–NH2 (B) (C)

(D) NH4NO3 ( Here is no carbon for the formation of CN–)

76

Instant turbidity with Lucas reagent 

H2C = CH – CH2 – OH

CH2 = CH – CH+  CH2+ – CH = CH2 Allyl C+ , 1°

 Benzyl C+ , 1°

 Angle strain / less stable / unstable C+

Although 3° , but has angle strain , slow r×n with HCl/ZnCl2

Other Examples

Ex.1

Ans.

Ex.2

Ans.

Ex.3

Ans.

Ex.4

-

CH3 – CH2 – C  C – CH2 – CH3

Ans.

77

Ex.5

Ans.

CH3 – CH2 – CHO

Ex.6

C4H8

Ans.

CH3 – CH = CH – CH3

Ex.7

(A)

Ans.

A=

Ans.

CH3 – CH2 – COOH

Ans.

O || CH3  C  O  CH3

Ans.

A=

Ex.8

Ex.9

Na metal –ve Ex.10

A (C7H8O)

FeCl3 (neutral)

–ve

Lucas –ve Reagent Ex.11

R gives following tests.

Ans.

78

1.

How will you distinguish the following pair of compounds. Compounds Isomers

Reagents

(I) (a) CH3 – CH2 – COOH (b) CH3 – C – O – CH3

1

2

3

NaHCO3 (+) NaHCO3 (–)

Acidic odour Fruity/Sweet odour

Na metal (+) Na metal (–)

O (II) (a) Ph – CH2 – C – OH Same

O (b) Ph – C – O – CH3

Same

Same

O (III) (a) Ph – CH2 – CH = CH2

Br2/H2 O (+)

Dil KMnO4 (+)

(b) Ph –

Br2/H2O (–)

Dil KMnO4 (–)

(IV) (a) CH3 – CH2 – CH2 – OH

Na metal (+)

(b) CH3 – CH2 – O – CH3

Na metal (–)

Ceric Ammonium Nitrate (+) ,, (–)

(V)

– –

CH3 (a) CH3 – C – OH CH3

Lucas Reagent instant turbidity

Na metal (+)

Ceric Ammonium (+)

In 5 min.

Na metal (+)

(+)

In 30 min.

Na metal (+)

(+)

(VI) (a) CH3 – CH2 – CH2 – CH2 – NH2

NaNO2/HCl (N2 gas )

Na metal (+)

(b) CH3 – CH2 – CH2 – NH – CH3

N2 gas not liberated

Na metal (+)

–

(b) CH3 – CH – CH2 – CH3 OH (c) CH3 – CH2 – CH2 – CH2 – OH

2.

Tick mark the reagents which will give positive response with the following compound. (A) Na metal O O H2N

O

Ans. A C D F G H I J

(B) NaHCO3 (C) 2, 4-DNP (D) AgNO3 + NH4OH

OH

(E) Fehling solution (F) NaNO2/HCl (cold) (G) ZnCl2 (Anhyahous)/HCl (H) Cu2Cl2 + NH4OH (I) Br2/H2O

3.

(J) Dil KMnO4 (cold) T (C7H6O2) is an aromatic white solid which liberates a colourless odourless gas on heating with NaHCO3. Write S.F. of X and its all possible functional isomers (all aromatic)

Sol.

T =

;

,

79

POC - II Separation of binary mixture of organic compounds Theory of separation : Organic compounds have different solubilities in different solvents. So they can be separated by use of appropriate solvents.

Purification of organic compounds : 

The organic compounds derived from natural sources or prepared in the laboratory are seldom pure. They are usually contaminated with other substances.



Purification means the removal of undesirable impurities associated with a particular organic compound, i.e to obtain the organic compound in pure state.



Various methods have been developed to purify organic compounds

1.

Physical methods : (i) Crystallisation (ii) Sublimation (iii) Distillation (iv) Solvent extraction (v) Chromatography

2. 

Chemical methods:Chemical methods of separation depend upon the nature of the functional group present in the component. Hence these can be applied to solid as well as liquid compounds.



A chemical method can be applied only when one of the components of the mixture is soluble in a particular solvent while the other is insoluble in the same solvent .



Separation is the first step during the actual analysis of organic mixture. It is the most important step in the sense that if separation is incomplete the result will not be correct because the impure compound will give tests of different functional group and its melting point will also be very much different from that of the pure compound obtained from complete separation.

(i) (ii) (iii) (iv)

Separation of Binary mixtures of organic compounds. The usual systematic scheme for separating a solid binary mixture is discussed below. Separation with water Separation with sodium bicarbonate Separation with sodium hydroxide Separation with hydrochloric acid Solubility of two components. Separation Scheme Solvent H O

2   

(1)

Some important point :

The mixture of organic compounds can be separated by using appropriate solvent.



Most of the aromatic compounds are water insoluble due to large hydrophobic group of six carbon atom



Aromatic acids are insoluble in water but soluble in aqueous NaHCO3 solution or NaOH solution, due to salt formation.



Aromatic hydroxy compounds are water insoluble but are soluble in aqueous NaOH solution due to salt formation.



Aromatic amine (Aniline 1º, 2º, 3º) are organic base and water insoluble but are soluble in aqueous HCl solution due to salt formation.



Aliphatic compoud with atleast two functional group ( which can form H-bonding) are water soluble. Ex. Diacids, diols. diamines, hydroxy acids (– OH, –COOH), Amino acids (– NH2 , –COOH) . oxalic acid , malonic acid, maleic acid, fumaric acid, glycol, glycerol, sucrose, glucose. 80

Ex.1

Binary mixtures - (Two components)

Compound A

Compound B

Appropriate Solvent

(1)

+

H2O

(2)

+

H2O

81

(3)

Fructose

+

H2O

(4)

+

aq. NaHCO3

(5)

+

aq. NaOH

(6)

+

aq. HCl

Ex.2

Sol.

+

Identify I & II. I  A, II  B

Ex.3

+

Identify I & II Sol.

Q,

  P

Ex.4

+

+

P+Q+R

Identify P, Q, R Sol.

PI

Q  II

R  III

82

Ex. 5. Mixture separation

(1)

(2)

(3)

Succinic Acid + Phthalic Acid + ortho–Cresol

+

+

III

83

GOC-1+Lecture+Notes+VIPUL

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