isomerism - the uraniumtheuranium.org/content-images/11-structural-isomerism.pdfstructural isomerism...

Structural Isomerism

Dr. S. S. Tripathy

ISOMERISM

ISOMERISM

Structural Isomerism Stereoisomerism(Constitutional isomerism)

Chain isomerism

Position Isomerism

Ring-Chain Isomerism

Functional Isomerism

Metamerism

Tautomerism

Configurational Isomeris Conformation

Optical Isomerism Geometrical Isomerism

Isomerism

What are Isomers: Isomers have same molecular formula but have different from each other either in terms ofstructure(hence name) or spatial relationship.Structural(Constututional) Isomerism:The isomers differ from each other in structure and hence they have different names and properties.

CHAIN ISOMERISM(Skeletal Isomerism):

The difference arises in the nature of carbon skeleton. Any change in the arrangement of carbon atoms in theprincipal carbon chain or at branch points changes the carbon skeleton and make them chain isomers.Chain Isomerism in Alkanes: (C

nH

2n+2)

Methane, ethane and propane do not have any chain isomers. However from butane onwards, the alkanesshow chain isomerism.C

4H

10: 2 chain isomers.

CH3 CH2 CH2 CH3

butane (n-butane) CH3 CH

CH3

CH32-methylpropane(isobutan

N.B: For writing the structures of chain isomers, first keep the principal chain fixed and then change theposition of branches to different non-equivalent positions to get a new name. After finding all the differentisomers belonging to the same principal carbon chain, the chain length is reduced by one carbon atom and therest carbon atoms are put as brances in as many different ways as possible so as to get different names. If somename is repeated when, then its not a new isomer. Delete it from your list.


Dr. S. S. Tripathy

C5H

12: 3 isomers

CH3 CH2 CH2 CH2 CH3pentane (n-pentane) CH3 CH

CH3

CH2 CH32-methylbutane(isopentane

CH3 C

CH3

CH3

CH3

2,2-dimethylpropane (neopentane)

C6H

14: 5 isomers

CH3 CH2 CH2 CH2 CH2 CH3

hexane (n-hexane)CH3 CH

CH3

CH2 CH2 CH3

2-methylpentane

CH3 CH2 CH CH2 CH3

CH3

3-methylpentaneCH3 CH CH CH3

CH3CH3

2,3-dimethylbutane

CH3 C CH2 CH3

CH3

CH3

2,2-dimethylbutane

(You can try to get a new structure it will be a mere repetation)N.B: Do not put any carbon on the terminal carbon atom of a new chain. It will extend the chain length whichyou have already completed.N.B: 2-methylpentane and 3-methylpentane are not positional isomers. They are chain isomers. Only whenany functional group including –X(halo) changes its position on the same carbon skeleton, we call them positionalisomers. In benzene ring, of course, it is different. Even if there are two Me– branches, if their positions(o-, m-p) are changed w.r.t each other, they become positional isomers, as benzene ring itself is treated as functionalgroup. Some authors might consider 2-methylpentane and 3-methylpentane as positional isomers, but thisauhtor does not agree on this. IUPAC is also not clear on this point. Hence students are advised not be muchconcerned about this.

C7H

16 : 9 isomers

CH3 CH2 CH2 CH2 CH2 CH2 CH3

heptane( (n -heptane) CH3 CH

CH3

CH2 CH2 CH2 CH32-methylhexane

CH3 CH2 CH CH2 CH2 CH3

CH3

3-methylhexaneCH3 CH2 CH CH CH3

CH3 CH3

2,3-dimethylpentane


Dr. S. S. Tripathy

CH3 CH CH2 CH CH3

CH3CH3

2,4-dimethylpentane

CH3 CH2 CH2 C CH3

CH3

CH32,2-dimethylpentane

CH3 CH2 C CH CH3

CH3

CH3

3,3-dimethylpentane

CH3 CH2 CH CH CH3

CH2

CH3

3-ethylpentane

CH3 C CH CH3

CH3

CH3

CH3

2,2,3-trimethylbutane

Note that we have started from 7-carbon chain and got 1 isomer, then 6- carbon chain, got 2 isomers. Thenwe reduced to 5- carbon chain got 5 isomers and finally 4-carbon chain, got 1 isomer, all total 9 isomers. Canyou get one more isomer by changing the carbon skeleton ? The answer is NO. You can try.

List of number of isomers of alkanes upto C20

.

Alkane No. of isomers No. of isomersincluding stereoisomers

C4H

102 2(no stereoisomersim)

C5H

123 3 (no streoisomerism)

C6H

145 5 (no stereoisomerism)

C7H

169 11

C8H

1818 24

C9H

2035 55

C10

H22

75 136C

11H

24159 345

C12

H26

355 900C

13H

28802 2412

C14

H30

1858 6563C

15H

324347 18,127

C16

H34

10,359 50,699C

17H

3624,894 1,43,255

C18

H38

60,523 4,08,429C

19H

401,48,284 11, 73,770

C20

H42

3,66,319 33, 96,844 (Incredible !!!!!!) Ha Ha......N.B: There is no guiding rule to find out the numbr of structural isomers for a given alkane. One has to find outall possibilities with the help of a computer, of course. For still higher alkanes from C

21 and above you can refer

the website ( https://en.wikipedia.org/wiki/List_of_straight-chain_alkanes. )


Dr. S. S. Tripathy

C9H

20: (35)

Nonane; 2-Methyloctane; 3-Methyloctane; 4-Methyloctane; 2,2-Dimethylheptane; 2,3-Dimethylheptane;2,4-Dimethylheptane; 2,5-Dimethylheptane; 2,6-Dimethylheptane; 3,3-Dimethylheptane; 3,4-Dimethylheptane;3,5-Dimethylheptane; 4,4-Dimethylheptane; 3-Ethylheptane; 4-Ethylheptane; 2,2,3-Trimethylhexane; 2,2,4-Trimethylhexane; 2,2,5-Trimethylhexane; 2,3,3-Trimethylhexane; 2,3,4-Trimethylhexane; 2,3,5-Trimethylhexane; 2,4,4-Trimethylhexane; 3,3,4-Trimethylhexane; 3-Ethyl-2-methylhexane; 3-Ethyl-3-methylhexane; 3-Ethyl-4-methylhexane; 4-Ethyl-2-methylhexane; 2,2,3,3-Tetramethylpentane; 2,2,3,4-Tetramethylpentane; 2,2,4,4-Tetramethylpentane; 2,3,3,4-Tetramethylpentane; 3-Ethyl-2,2-dimethylpentane;3-Ethyl-2,3-dimethylpentane; 3-Ethyl-2,4-dimethylpentane; 3,3-Diethylpentane

C10

H22

: (75)Decane; 2-Methylnonane; 3-Methylnonane; 4-Methylnonane; 5-Methylnonane; 2,2-Dimethyloctane; 2,3-Dimethyloctane; 2,4-Dimethyloctane; 2,5-Dimethyloctane; 2,6-Dimethyloctane; 2,7-Dimethyloctane; 3,3-Dimethyloctane; 3,4-Dimethyloctane; 3,5-Dimethyloctane; 3,6-Dimethyloctane; 4,4-Dimethyloctane; 4,5-Dimethyloctane; 3-Ethyloctane; 4-Ethyloctane; 2,2,3-Trimethylheptane; 2,2,4-Trimethylheptane; 2,2,5-Trimethylheptane; 2,2,6-Trimethylheptane; 2,3,3-Trimethylheptane; 2,3,4-Trimethylheptane; 2,3,5-Trimethylheptane; 2,3,6-Trimethylheptane; 2,4,4-Trimethylheptane; 2,4,5-Trimethylheptane; 2,4,6-Trimethylheptane; 2,5,5-Trimethylheptane; 3,3,4-Trimethylheptane; 3,3,5-Trimethylheptane; 3,4,4-Trimethylheptane; 3,4,5-Trimethylheptane; 3-Ethyl-2-methylheptane; 3-Ethyl-3-methylheptane; 3-Ethyl-4-methylheptane; 3-Ethyl-5-methylheptane; 4-Ethyl-2-methylheptane; 4-Ethyl-3-methylheptane; 4-Ethyl-4-methylheptane; 5-Ethyl-2-methylheptane; 4-Propylheptane; 4-(1-Methylethyl)heptane; 2,2,3,3-Tetramethylhexane; 2,2,3,4-Tetramethylhexane; 2,2,3,5-Tetramethylhexane; 2,2,4,4-Tetramethylhexane;2,2,4,5-Tetramethylhexane; 2,2,5,5-Tetramethylhexane; 2,3,3,4-Tetramethylhexane; 2,3,3,5-Tetramethylhexane; 2,3,4,4-Tetramethylhexane; 2,3,4,5-Tetramethylhexane; 3,3,4,4-Tetramethylhexane; 3-Ethyl-2,2-dimethylhexane; 3-Ethyl-2,3-dimethylhexane; 3-Ethyl-2,4-dimethylhexane; 3-Ethyl-2,5-dimethylhexane; 3-Ethyl-3,4-dimethylhexane; 4-Ethyl-2,2-dimethylhexane; 4-Ethyl-2,3-dimethylhexane; 4-Ethyl-2,4-dimethylhexane; 4-Ethyl-3,3-dimethylhexane; 3,3-Diethylhexane; 3,4-Diethylhexane; 2-Methyl-3-(1-methylethyl)hexane; 2,2,3,3,4-Pentamethylpentane; 2,2,3,4,4-Pentamethylpentane; 3-Ethyl-2,2,3-trimethylpentane; 3-Ethyl-2,2,4-trimethylpentane; 3-Ethyl-2,3,4-trimethylpentane; 3,3-Diethyl-2-methylpentane;2,4-Dimethyl-3-(1-methylethyl)pentane

(N.B: For still higher alkane ie. C11

, C12

, C13

, C14

you can visit the website http://www.kentchemistry.com/links/organic/isomersofalkanes.htm )

(2) POSITIONAL ISOMERISM(REGIOIOSOERISM):Carbon skeleton remaining same, if the position of any functional group or important branch(not alkyl) isdifferent, then such structural isomers are called positional isomers.

Examples:

CH2 CH CH2 CH3but-1-ene

and CH3 CH CH CH3but-2-ene

CH3 CH

OH

CH3propan-2-ol

and CH3 CH2 CH2

OH

propan-1-ol


Dr. S. S. Tripathy

CH3 C

O

CH2 CH2 CH3pentan-2-one

and CH3 CH2 C CH2 CH3

O

pentan-3-one

In the above cases, the carbon skeletons are the same in each pair, but the position of the functional groupsnamely C=C, -OH, -CO- are different.IMP: Note that if two isomers are chain isomers, they can never be positional isomers even if thelocants of the functional groups are different.

CH3 CH2 CH

OH

CH3butan-2-ol

and CH3 CH

CH3

CH2

OH

2-methylpropan-1-ol

are never positional isomers. They are chain isomers as the nature of their carbon skeleton are different.Note that chain and positional isomers are often considered together with a particular functional group.

Alkenes: (chain and positional positional) (Stereoisomers and ring-chain isomers are not included)In fact, we shall finalise the total nmber of isomers of all types later from such formulae. For now, just anexercise to know the different possible acyclic structures.

C4H

8: (3)


and CH3 CH CH CH3but-2-ene

CH3 C CH2

CH3

2-methylprop-1-ene(isobutene)

but-1-ene and 2-methylprop-1-ene are chain isomers, so also but-2-ene and 2-methylprop-1-ene are chainisomers. Once you find a pair of isomers differ in their carbon skeletons, then declare them as chain isomers.Forget about the position of their functional groups.Note that chain and positional isomers differ in their properties marginally, mostly in physical properties.C

5H

10: (5)

CH2 CH CH2 CH2 CH3pent-1-ene

CH3 CH CH CH2 CH3pent-2-ene

pent-1-ene and pent-2-ene are positional isomers.

CH2 C

CH3

CH2 CH32-methlbut-1-ene

C

CH3

CH CH3CH32-methylbut-2-ene

2-methylbut-1-ene and 2-methylbut-2-ene are also positional isomers. But any one from the previous pair andany one from the above pair are chain isomers.


Dr. S. S. Tripathy

CH

CH3

CH3 CH2CH3-methylbut-1-ene

3-methylbut-1-ene is another structural isomer of the above five. No doubt, it is a chain isomer of the first pair.But how it is related to the second pair ; is very difficult to answer. No rule can unambiguously tell about itsrelationship with the previous pair. 3-methylbut-1-ene and 2-methylbut-1-ene differ in their carbon skeleton,hence can be called chain isomers. 3-methylbut-1-ene and 2-methylbut-2-ene can also be called chain isomers.For positional isomers, the locants all branches should be same, only position of functional group should bedifferent.Can we make a chain of propene and keep two Me- branches in the middle carbon ? No. So all together 5structural isomers(excluding ring-chain and stereoisomers) for C

5H

10.

C6H

12: (13 acycli structural isomers excluding stereoisomers)

CH2 CH CH2 CH2 CH2 CH3hex-1-ene

CH CH CH2 CH2 CH3CH3hex-2-ene

CH2 CH CH CH2 CH3CH3hex-3-ene

Hex-1-ene, hex-2-ene and hex-3-ene are positional isomers.

CH2 C CH3CH2CH2

CH3

2-methylpent-1-ene

C CH3CH2CH

CH3

CH3

2-methylpent-2-ene

CH CH3CHCHCH3

CH3

4-methylpent-2-ene

CH CH2CHCH2CH3

CH3

4-methylpent-1-ene

CH2 CH CH3CH2CH

CH3

3-methylpent-1-eneCH CH3CH2C

CH3

CH3

3-methylpent-2-ene

CH2CH3

CH2

CH CH2 CH3

3-methylidenepentane

All the above 7 isomers are chain isomers to the first three. But between them, the first two namely 2-methylpent-1-ene and 2-methylpent-2-ene are positional isomers. So also 3-methylpent-1-ene and 3-methylpent-2-enecan be called positional isomers. So also 4-methylpent-1-ene and 4-methylpent-2-ene are positional isomers.Others are chain isomers to each other.


Dr. S. S. Tripathy

CH2 C

CH3

CH

CH3

CH3

2,3-dimethylbut-1-ene

CH2 CH C

CH3

CH3

CH3


CH3 C

CH3

C

CH3

CH3


All these three are chain isomers to first 3 and the second 7 isomers. But within them 2,3-dimethylbut-1-eneand 2,3-dimethylbut-2-ene are positional isomers. The third one(3,3-dimethylbut-1-ene) is a chain isomer toformer two positional isomers.N.B: Student to note that in many cases, IUPAC has not spelt out a sharp border line between chain andpositional isomerism. That is why in cases of ambiguity, it is better to address them as structural isomers.* To generate the structures, in each step, we had to reduce one carbon atom to make the principal chainand then put that carbon as branch at a particular position. Keeping this skeleton fixed, the position of doublebond are changed to generate different isomers. Then change the branch point to another fixed position andwith this change the position of the double bond. Never forget that carbon atom cannot be attached to terminalposition lest it will produce a higher principal chain already completed. Also you have to see that every time thevalency of carbon is to be 4.

(3) Ring-Chain Isomerism:Double Bond Equivalent(DBE):

1222 NXH

C

nnnnDBE

nC = number of carbon atoms; n

H = number of H atoms n

X = number of halogen atoms

nN = number of N atoms.

Examples: C4H

8 : DBE = 4 –(8/2) + 1 = 1;

C3H

4: DBE = 3 – (4/2) +1 = 2;

C6H

6: DBE = 6 –(6/2) + 1 = 4

C3H

6O:DBE = 3 – (6/2) + 1 = 1; and so on.

This formula will be of great use in solving organic problems later.

DBE = 1 : One C=C or 1 cyclic ring

DBE = 2; one C C or 2 C=C or (one C=C + one ring) or ( 2 rings)

Note that a cyclic ring is equivalent to a C=C. One C C is equivalent to two C=C or one C=C + a ring or

two rings.

C4H

8: CH

2=CH–CH

2–CH

3and

CH2 CH2

CH2CH2

have same formula.

Benzene(C6H

6) has a DBE=4, three for 3 C=C and one for the ring.


Dr. S. S. Tripathy

Cyclic Isomers:

C3H

6:

cyclopropane

(1)

C4H

8:

cyclobutanemethylcyclopropane

(2)

C5H

10: (5)

cyclopentane methylcyclobutane 1,2-dimethylcyclopropane

1,1-dimethylcyclopropane ethylcyclopropane

C6H

12: (10)

cyclohexane methylcyclopentane 1,2-dimethylcyclobutane


Dr. S. S. Tripathy

1,3-dimethylcyclobutane ethylcyclobutane 1,2,3-trimethylcyclopropane

1,1,2-trimethylcyclopropane 1-ethyl-2-methylcyclopropane propylcyclopropane

propan-2-ylcyclopropane

Acyclic + Cyclic Isomers: (excluding stereoisomers)

C4H

8: acyclic = 3; cyclic =2 Total = 5


CH3 CH CH CH3but-2-ene

CH3 C CH2

CH3

2-methylprop-1-ene(isobutene) cyclobutane

methylcyclopropane


Dr. S. S. Tripathy

C5H

10:

acyclic: 5; cyclic: 5 Total = 10(refer the all these structures given before)C

6H

12: acyclic = 13; cyclic = 10 Total = 23

N.B: We have not included stereoisomers to these formulae. Later when we discuss this isomerism, then weshall be really knowing the total number of isomers which can be possible from a formula.Also note that as the number of carbon atoms increases, the total number of isomers become greater andgreater without following any special rule.

Structural Isomers of Alkynes(alkadienes):

C4H

6:

acyclic : 4; cyclic: 5

CH C CH2 CH3but-1-yne

CH3 C C CH3but-2-yne

CH2 C CH CH3buta-1,2-diene

CH2 CH CH CH2

buta-1,3-diene

cyclobutene 1-methylcyclopropene3-methylcyclopropene

methylidenecyclopropane

bicyclo[1.1.0]butane

SAQ: Draw all the structural isomers from formula C5H

8:

Acyclic: 8 : pent-1-yne, pent-2-yne, 3-methylbut-1-yne, penta-1,2-diene, penta-1,3-diene, penta-1,4-diene,penta-2,3-diene, 2-methylbuta-1,3-dieneCyclic: 11 : cyclopentene, 1-methylcyclobutene, 3-methylcyclobutene, methylidenecyclobutane, 1,2-dimethylcyclopropene, 1,3-dimethylcyclopropene, 1-ethylcyclopropene,3-ethylcyclopropene,ethylidenecyclpropane, ethenylcyclopropane, bicyclo[2.1.0]pentane, 1-methylbicyclo[1.1.0]butane, 2-methylbicyclo[1..1.0]butane(Reader to check if i have missed any more isomer and if so, please inform me for inclusion:([email protected])


Dr. S. S. Tripathy

Haloalkanes(Alkyl halides): CnH

2n+1X; C

nH

2nX

2 etc.

DBE = 0 (saturated)C

3H

7Cl: (2)

CH3 CH2 CH2

Cl

1-chloropropane(n-propyl chloride)

CH3 CH CH3

Cl

2-chloropropane(isopropyl chloride)

These are positional isomers.C

3H

6Cl

2: (4)

CH3 CH2 CH

Cl

Cl1,1-dichloropropane

CH3 C CH3

Cl

Cl2,2-dichloropropane

CH3 CH CH2

Cl Cl

1,2-dichloropropane

CH2 CH2 CH2

ClCl

1,3-dichloropropane

C4H

9Br: (4)

n-butyl bromide, sec-butyl bromide, isobutyl bromide, tert-butyl bromideC

4H

8Cl

2: (7)

1,1-dichlorobutane, 2,2-dichlorobutane; 1,2-dichlorobutane, 1,3-dichlorobutane; 2,3-dichlorobutane,1,2-dichloro-2-methylpropane; 1,1-dichloro-2-methylpropane

Haloalkenes: CnH

2n–1X, C

nH

2n–2X

2 etc.

DBE = 1; So we can have cyclic isomers also.C

3H

5Br: (4)

CH2 CH CH2

Br

3-bromopropeneCH2 C CH3

Br

2-bromopropeneCH CH CH3

Br

1-bromopropene

Br

bromocyclopropane

(N.B: We shall find later that 1-bromopropene exists as two stereoisomers(geometrical isomres) E and Z. Weare not including stereoisomers to any formula now, only finding the structural isomers)


Dr. S. S. Tripathy

C3H

4Cl

2: (7)

CH2 C CH2

ClCl

2,3-dichloropropeneCH CH CH2

ClCl

1,3-dichloropropeneCH C CH3

ClCl

1,2-dichloropropene

CH2 CH CH

Cl

Cl3,3-dichloropropene

C CH CH3

Cl

Cl1,1-dichloropropene

Cl

Cl1,2-dichlorocyclopropane

Cl Cl

1,1-dichlorocyclopropane

(N.B : We have not considered the stereoisomers that exist in some of them)C

4H

7Cl: (12)

acyclic : (8): 1-chlorobut-1-ene; 2-chlorobut-1-ene; 3-chlorobut-1-ene; 4-chlorobut-1-ene; 1-chlorobut-2-ene; 2-chlorobut-2-ene; 3-chloro-2-methylpropene; 1-chloro-2-methylpropene

cyclic: (4) : chlorocyclobutane; 1-chloro-2-methylcyclopropane; 1-chloro-1-methylcyclopropane;cyclobutylchloromethane(chloromethylcyclobutane)C

3H

3Br: ( 3 + 2 =5)

DBE=2alknyl bromide, dienyl bromide

CH C CH2

Br

3-bromoproyneC C CH3

Br

1-bromopropyneCH

Br

C CH21-bromopropadiene

Br1-bromocyclopropene

Br

3-bromocyclopropene


Dr. S. S. Tripathy

(C) FUNCTIONAL ISOMERISM:

The isomers have different functional groups and hence differ from each other in their properties, mostly in theirchemical properties.

(I) CnH

2n+2O category:

This formula corresponds a saturated compound(DBE=0). Alchol and ether are two two functional groups,the compounds bearing this formula can have.C

3H

8O :

Alcohol:

CH3 CH2 CH2

OH

propan-1-olCH3 CH CH3

OH

propan-2-ol(2 alcohols)

(positional isomers)

Ether:

CH3 O CH2 CH3methoxyethane(ethyl methyl ether)

( 1 ether)

Ether and alcohol are functional isomers.

C4H

10O :

Alcohols: (4)

CH3 CH2 CH2 CH2

OH

butan-1-ol(n-butyl alcohol)

CH3 CH2 CH CH3

OH

butan-2-ol(sec-butyl alcohol)

butan-1-ol and butan-2-ol are positional isomers.

CH3 CH CH2

OH

CH3

2-methylpropan-1-ol(isobutyl alcohol)

CH3 C

OH

CH3

CH3

2-methylpropan-2-ol(tert-butyl alcohol)

2-methylpropan-1-ol and 2-methylpropan-2-ol are also positional isomers. But one from each of theabove pairs constitutes chain isomers.N.B: Note that the first two were obtained by keeping the parent chain 4 and then changing the position of–OH. The last two are obtained by taking the parent chain of 3 with one Me-branch at the 2nd carbon andthen changing the position of –OH group. Note that four carbon atoms can be arranged in four ways, n-, sec-, iso and tert- ways.


Dr. S. S. Tripathy

Ethers: (3)

CH3 O CH2 CH2 CH3

1-methoxypropane(methyl n-propyl ether)

CH3 O CH CH3

CH3

2-methoxypropane(isopropyl methyl ether

CH3 CH2 O CH2 CH3

ethoxyethane(diethyl ether)

N.B: Note that three carbon atoms can be arranged in two ways i.e n-propyl and isopropyl. But how are theyrelated with each other ? Are they positional or chain isomers ? 1-methoxypropane and 2-methoxypropanecan be called positional isomers, if you consider alkoxy as a functional group. Ethoxyethane can be treated aschain isomer of both 1-methoxypropane and 2-methoxypropane. But in these cases, definitely there is certaindegree of ambiguity as there is a bivalent –O– group in the chain containing carbon atoms. That is why the bestisomerism to be coined with these is “Metamerism” to be discussed later.

C5H

12O:

Alcohols: (8)

CH2 CH2 CH2 CH2

OH

CH3pentan-1-ol

CH2 CH2 CH CH3

OH

CH3

pentan-2-ol

CH3 CH2 CH CH2

OH

CH3

pentan-3-ol

CH2 CH CH2

CH3

CH3

OH

2-methylbutan-1-ol

CH3 C CH2

CH3

CH3

OH

2-methylbutan-2-ol

`

CH3 CH CH

CH3

CH3

OH

3-methylbutan-2-ol

CH3 CH CH2

CH3

CH2

OH

3-methylbutan-1-ol

CH3 C

CH3

CH3

CH2

OH

2,2-dimethylpropan-1-ol(neopentyl alcohol)

The first three namely pentan-1-ol, pentan-2-ol and pentan-3-ol are positional isomers. Out of the next four,2-methylbutan-1-ol and 2-methylbutan-2-ol are also positional isomers, so also 3-methylbutan-1-ol and 3-methylbutan-2-ol are positional isomers. In between them, the relationship is chain isomerism. The last one i.eneopentyl alcohol is a chain isomer to all the rest.


Dr. S. S. Tripathy

Ethers: (6)

CH3 O CH2 CH2 CH2 CH31-methoxybutane(n-butyl methyl ether)

CH3 O CH CH2

CH3

CH3

2-methoxybutane(sec-butyl methyl ether)

The above two can be called positional isomers.

CH3 O CH2 CH CH3

CH3

1-methoxy-2-methylpropane(isobutyl methyl ether)

CH3 O C

CH3

CH3

CH32-methoxy-2-methylpropane(tert-butyl methyl ether)

The above two can also be called positional isomers.

CH3 CH2 O CH2 CH2 CH3

1-ethoxypropane(ethyl n-propyl ether)

CH3 CH2 O CH CH3

CH3

2-ethoxypropane(ethyl isopropyl ether)

The above two can also be called positional isomers. But in between the two pairs , the relatioship is chainisomerism. But they are best exressed by Metamerism(See later).

C6H

14O:

Alchols: 17hexan-1-ol, hexan-2-ol, hexan-3-ol, 2-methylpentan-1-ol, 2-methylpentan-2-ol; 2-methylpentan-3-ol,4-methylpentan-2-ol, 4-methylpentan-1-ol, 3-methylpentan-1-ol, 3-methylpentan-2-ol, 3-methylpentan-3-ol, 2-ethylbutan-1-ol; 2,3-dimethylbutan-1-ol; 2,3-dimethylbutan-2-ol; 2,2-dimethylbutan-1-ol; 3,3-dimethylbutan-2-ol, 3,3-dimethylbutan-1-ol; (17)(N.B: Subsequently we shall find that out of 17 alcohols, 11 are achiral(optically inactive) and six exist as d/lpair(optically active). Hence there are 23 isomeric alochols inclusive of stereoisomers. Do not poke your nosenow until you study stereoisomerism)

Ethers: (15)1-methoxypentane; 2-methoxypentane, 3-methoxypentane; 1-methoxy-2-methylbutane, 2-methoxy-2-methylbutane, 2-methoxy-3-methylbutane; 1-methoxy-3-methylbutane; 1-methoxy-2,2-dimethylpropane; 1-ethoxybutane; 2-ethoxybutane; 1-ethoxy-2-methylpropane; 2-ethoxy-2-methylpropane; 1-propoxypropane,2-propoxypropane; 2-(propan-2-yloxy)propane (15)(N.B: Subsequently we shall fnd that out of 15 ethers, 12 are achiral and 3 exist as d/l pair. Hence there are 18ethers inclusive of stereoisomers.


Dr. S. S. Tripathy

II: CnH

2nO Type : (Aldehyde, Ketone, enol, ene-ether, cycloalkanol, cyclic ether)

DBE = 1, Hence in addition to acyclic isomers, there can be cyclic isomers. Lets see this with a fewexamples.

C2H

4O:

CH3 C

O

Hethanal(acetaldehyde)

CH2 CH

OH

ethenol(vinyl alcohol)

O

oxirane(ethylene oxide

With this formula there is one aldehyde, one enol and one cyclic ether. There is no ketone or unsaturated etherwith this formula. Later when we shall study ‘Tautomerism’ ,we shall know that vinyl alcohol is unstable and ittautomerizes readily to acetaldehyde.C

3H

6O:

Aldehyde: (1) Ketone: (1)

CH2 C

O

HCH3

propanal

CH3 C

O

CH3propan-2-one(acetone)

Enols: (3)

CH2 C

OH

CH3prop-1-en-2-ol

CH CH CH3

OH

prop-1-en-1-ol

CH2 CH CH2

OH

prop-2-en-1-ol(allyl alcohol)

Note that the first two enols are unstable, as we shall know soon while learning tautomerism, and change topropanone and propanal respectively. But the third one is stable(allyl alcohol).Unsaturated ether:

CH2 CH O CH3

methoxyethene(methyl vinyl ether)

Cylcic isomers: (2)

O

oxetane

OH

cyclopropanol


Dr. S. S. Tripathy

C4H

8O:

Aldehyde: (2)

CH2 C

O

HCH2CH3

butanal(butyraldehyde)

CH C

O

HCH3

CH3

2-methylpropanal(isobutyraldehyde)

Ketone: (1)

CH2 C

O

CH3CH3butan-2-one(butanone)

Enols: (8)

CH CH CH2 CH3

OH

but-1-en-1-olCH2 C CH2 CH3

OH

but-1-en-2-olCH2 CH CH CH3

OH

but-3-en-2-ol

CH2 CH CH2 CH2

OH

but-3-en-1-ol

Out of the above enols, the first two are unstable which readily tautomerises to butanal and butanone respectively(we shall study this soon).

but-2-en-1-olCH CH CH3

OH

CH2but-2-en-2-ol

C CH CH3CH3

OH

In this pair, the second one is also unstable and will tautomerize to butanone.

CH2 C CH2

CH3 OH

2-methylprp-2-en-1-olCH C CH3

CH3

HO2-methylprop-1-en-1-ol

The second one also is unstable and will tautomerize to 2-methylpropanal. In all these, I have forgotten toremind you whether a pair belongs chain isomers or positional. You can scrutinize that. If the everthing elseremain the same, only the locant of functional group differs then they are positional, otherwise chain isomers.Unsaturated acyclic ether: (4)

CH2 CH O CH2 CH3

ethoxyethene(ethyl vinyl ether)

CH CH O CH3CH31-methoxyprop-1-ene


Dr. S. S. Tripathy

CH2 CH CH2 O CH3

3-methoxyprop-1-ene(allyl methyl ether)

OCCH2

CH3

CH3

2-methoxyprop-1-ene

Cyclic isomers: (11)

O

oxolane(tetrahydrofuran,THF

OH

cyclobutanol

OH

2-methylcyclopropanol

OH

1-methylcyclopropanol

HO

cyclopropylmethanol(hydroxymethylcyclopropane)

OCH3

methoxycyclopropane

O

2-methyloxetane

O

3-methyloxetane

O

2,3-dimethyloxirane

O

2,2-dimethyloxirane

O

2-ethyloxirane

Total structural isomers(excluding stereoisomers):1 + 1 + 8 + 4 + 11 = 25

N.B: Please see, if i have missed any structural isomer. If you find any missing isomer, please inform me. I havenot tallied with any source, hopefully not available.From the above extensive discussion on structural isomerism, i feel like dropping enols, unsaturated ethers andcyclic isomes for higher formulae.


Dr. S. S. Tripathy

C5H

10O: (only find aldehydes and ketones)

N.B: Let us not find enols, ene-ethers and cyclic isomers as their numbers will be very large. Only find as manynumber of aldehydes and ketones possible from the formula.Aldehydes: (4)

CH3 CH2 CH2 CH2 CHOpentanal

CH3 CH2 CH

CH3

CHO2-methylbutanal

CH3 CH CH2 CHO

CH3

3-methylbutanal

CH3 C

CH3

CH3

CHO

2,2-dimethylpropanal

Ketones: (3)

CH3 C

O

CH2 CH2 CH3

pentan-2-oneCH3 CH2 C CH2 CH3

O

pentan-3-one

CH3 C

O

CH

CH3

CH3

3-methylbutan-2-one

(III) CnH

2nO

2 : (Carboxylic acid-Ester Type)

In addition to carboxylic acid and ester, it will include unsaturated diols, hydroxyketones and hydroxyaldehydes,alkoxyalkenols, unsaturated diethers and several types of cyclic compounds. We will take only one example inwhich we shall draw all structural isomers possible. But for higher formula, we shall not venture to do so, on thefear of missing an isomer and also to avoid monotony. For higher formulae, we shall concentrate on onlycarboxylic acids and esters.C

2H

4O

2: (8)

In this case too, DBE = 1.

CH3 COOHethanoic acid(acetic acid)

HCOOCH3

methyl methanoate(methyl formate)

CH

OH

CH

OH

ethene-1,2-diol

C

OH

CH2

OHethene-1,1-diol

CH2

OH

CHO2-hydroxyethanal

OO

1,2-dioxetaneO

O

1,3-dioxetane

O

OHoxiran-2-ol


Dr. S. S. Tripathy

So we are disgusted in writing so many isomers including cyclic ones. Let us concentrate on carboxylic acidsand esters only for higher formulae.C

3H

6O

2 :

Carboxylic acids: (1)

CH2 COOHCH3propanoic acid(propionic acid)

Esters : (2)

CH3 C

O

OCH3methyl ethanoate(methyl acetate)

H C

O

O CH2 CH3

ethyl methanoate(ethyl formate)

C4H

8O

2:

Carboxylic acids: (2)

CH2 COOHCH2CH3

butanoic acid(butyric acid)

CH COOHCH3

CH32-methylpropanoic acid(isobutyric acid)

Esers: (4)

CH2 COOCH3CH3

methyl propanoate(methyl propionate)

CH3 COOCH2CH3

ethyl ethanoate(ethyl acetate)

H C

O

O CH2 CH2 CH3propyl methanoate(n-propyl formate)

H C

O

O CH CH3

CH3

propan-2-yl methanoate(isopropyl formate)

SAQ: Write the structures of all C. acids and esters from the formula C5H

10O.

Answer: 4 carboxylic acids and 9 esters. Draw the structures in a systematic manner.Acids: pentanoic acid, 2-methylbutanoic acid, 3-methylbutanoic acid, 2,2-dimethylpropanoic acidEsters: methyl butanoate, methyl 2-methylpropanoate; ethyl propanoate; propyl ethanoate; propan-2-ylethanoate; butyl methanoate, butan-2-yl methanoate, 2-methylpropyl methanoate; 2-methylpropan-2-ylmethanoate


Dr. S. S. Tripathy

(IV) AMINES [CnH

2n+3 N]

DBE =0, so these are alkyl amines(10, 20 and 30).C

3H

9N:

10: (2)

CH3 CH2 CH2 NH2

propan-1-amine(n-propyl amine)

CH3 CH NH2

CH3

propan-2-amine(isopropyl amine)

20( 1) 30 (1)

CH3 CH2 NH CH3N-methylethanamine(ethyl methyl amine)

NCH3 CH3

CH3

N,N-dimethylmethanamine

(trimethyl amine)

Total = 4

C4H

11N:

10: (4)

CH2 CH2 CH2 NH2CH3butan-1-amine(n-butyl amine)

CH3 CH2 CH NH2

CH3

butan-2-amine(sec-butyl amine)

CH3 CH CH2 NH2

CH3

2-methylpropan-1-amine(isobutyl amine)

CH3 C NH2

CH3

CH32-methylpropan-2-amine(tert-butyl amine)

20: (3)

NH CH3CH2CH2CH3

N-methylpropan-1-amine(methyl n-propyl amine)

NH CH3CHCH3

CH3

N-methylpropan-2-amine(isopropyl methyl amine)

CH3 CH2 NH CH2 CH3N-ethylethanamine(diethyl amine)


Dr. S. S. Tripathy

30: (1)

NCH3 CH2

CH3

CH3

N,N-dimethylethanamine(ethyl dimethyl amine)

Total isomers: 8

C5H

13N :

10 amine: (6) : hexan-1-amine, hexan-2-amine; 2-methylbutan-1-amine; 3-methylbutan-1-amine; 2,2-dimethylpropan-1-amine

20 amine: (6) : N-methylbutan-1-amine; N-methylbutan-2-amine; N,2-dimethylpropan-1-amine; N,2-dimethylpropan-2-amine; N-ethylpropan-1-amine; N-ethylpropan-2-amine

30 amine: (3) : N,N-dimethylpropan-1-amine; N,N-dimethylpropan-2-amine; N-ethyl-N-methylethanamine

Total isomers: 15.(V) Alkynes and Dienes:Butynes and butadienes are functional isomers. We have already discussed about them before.

(V) Aromatic compounds (show both positional and functional isomerism)

C8H

10: (4)

DBE = 8 – (10/2) + 1 = 4; Already it has been told that for benzene ring DBE requirement is4. So no more unsaturation is present.

CH3

CH3

1,2-dimethylbenzene(o-xylene)

CH3

CH31,3-dimethylbenzene(m-xylene)

CH3

CH3

1,4-dimethylbenzene(p-xylene)

The above three xylenes are positional isomers.

CH2 CH3

ethylbenzene

Ethyl benzene is a chain isomer of the xylenes.


Dr. S. S. Tripathy

C7H

8O: (5)

DBE = 7 – 4 + 1 = 4 ; That means it has benzene ring only and no more unsaturation.

OH

CH3

2-methylphenol(o-cresol)

OH

CH3

3-methylphenol(m-cresol)

OH

CH3

4-methylphenol(p-cresol)

Cresols are positional isomers of each other.

OCH3

methoxybenzene(anisole)

CH2OH

benzyl alcohol

Anisole and benzyl alcohol are functional isomers of each other and each one of them is a functional isomer ofeach of the xylenes.N.B: Theoretically we can draw more acyclic isomers having four C=C or their equivalent and increase thenumber of isomers to a very large extent. However, once we get DBE is 4 or more, then the most likely isomersare those which contain benzene ring. Other purely acyclic isomers are ignored for practical point of view.

C8H

8O: (11)

DBE = 8 – 4 +1 = 5; In this, there is one more unsaturation (double bond) in addition to thebenzene ring.

C

O

CH3

acetophenone

CH2 CHO

phenylacetic acid

CHO

CH3

2-methylbenzaldehyde

CHO

CH33-methylbenzaldehyde

CHO

CH34-methylbenzaldehyde

C CH2

OH

1-phenylethenol


Dr. S. S. Tripathy

CH CH

OH

2-phenylethenol

CH CH2

OH

2-ethenylphenol

CH CH2

OH3-ethenylphenol

CH CH2

OH4-ethenylphenol

O CH CH2

(ethenyloxy)benzene

Note that 1-phenylethenol and 2-phenylethenol will tautomerise to give acetophenone and phenylacetic acidrespectively. We shall learn it very soon.SAQ: Draw all the isomers of C

8H

8O

2 and C

7H

7Cl containing benzene ring.

N.B: You draw independently. I shall not give its answer and hope you can do it.

METAMERISM:This is applicable for ethers, ketones, 20 and 30 amines where it becomes ambiguous to ascertain whether theisomers belong to positional or chain category.Definition: The isomers differ in the nature of alkyl group bonded to a polyvalent atom or group like –O–,–CO–, –NH– or 30 amine.Sometimes metamerism is a special case of either positional or chain isomerism.If two or structural isomers belong to metamerism i.e they are metamers, they are not addressed by othernames i.e positional or chain isomers. ‘Metamers’ term is used for them preferentially.

CH3 NH CH2 CH2 CH3

(methyl n-propyl amine)N-methylpropan-1-amine CH3 NH CH CH3

CH3

(isopropyl methyl amine)N-methylpropan-2-amine

CH3 CH2 NH CH2 CH3

(diethyl amine)N-ethylethanamine

The first two can be called positional isomers. But how is 3rd one related to any one of the first two ? Here theprincipal chain carries a bivalent group –NH–. Hence for all these isomers, a speical isomeism has beenassigned. These are metamers, in which the one or more alkyl groups attached to the divalent –NH– group is/are different.


Dr. S. S. Tripathy

(I) CH3 CH2 C

O

CH2 CH2 CH3hexan-3-one(ethyl n-propyl ketone)

(II) CH3 CH2 C CH CH3

O CH3

2-methylpentan-3-one(ethyl isopropyl ketone)

(III) CH3 C CH2CH2 CH2 CH3

O

hexan-2-one(n-butyl methl ketone)

I and II are chain isomers, so also II and III, while I and III are position isomers. In this case there is noambiguity as the divalent –CO– group is a carbon bearing group. All the the three are metamers.

(I)CH3 O CH2 CH2 CH3

1-methoxypropane(methyl n-propyl ether)

(II)CH3 O CH

CH3

CH32-methoxypropane(isopropyl methyl ether

(III)

CH3 CH2 O CH2 CH3

ethoxyethane(diethyl ether)

Like sec-amines, I and II can be called positional isomers, but how do you related III with others as there is adivalent –O– group inside the carbon chain. Hence ‘metamers’ is the right term for them.SAQ: Write structure of one isomer as indicated.

(a) Chain isomer of 2-methylpropane(b) position isomer of isobutyl alcohol(c) metamer of 3-methylbutan-2-one(d) cyclic isomer of C

3H

3Cl

(e) functional isomer of (i) propan-1-ol and (ii) butanal (iii) ethyl propanoate(f) 2-methylaniline

Solution:

(a) CH3 CH2 CH2 CH3 (butane) (b) CH3 C

OH

CH3

CH3 ( tert-butyl alcohol)

(c) pentan-2-one

(d) DBE = 3 – 1.5 – 0.5 + 1 = 2

Cl

( 3-chlorocyclopropene)

(e) (i) methoxyethane (ii) butan-2-one (iii) pentanoic acid(f) 3-methylaniline


Dr. S. S. Tripathy

TAUTOMERISM:

This is a case of dynamic functional isomerism, where the ismers(called tautomers) differ from each other onlyin the position of one (rarely more) mobile atom and in electron distribution. The special case of tautomerism isPROTOTROPY, in which the mobile species is a H+ ion. The tautomers are basically functional isomers whichremain always in dynamic equilibrium with each other in the liquid state. Tautomerism is exhiibited in theliquid state only. Any attempt to separate one tautomer from the other will be countered by the formation of thesame equilibrium mixture(%) in each with passage of time.Types : (a) Triad System (b) Dyad sytemTriad System: In which the mobile species(H+) shifts through 3 atoms.

CH2

H

C

O

CH3 CH2 C

OH

CH31 2

3 1

23I II

In the above example belonging to triad system, one H+ ion leaves the carbon atom in I and shifts through threeatoms and bonds with O atom, thus converting to the isomer II. Similarly one H+ ion leaves O atom in II andshifts through three atoms and bonds with C atom, thus converting II back to I. This is called keto-enoltautomerism as the isomer I is in KETO form and isomer II is in ENOL form. Keto-enol tautomerism is mostcommon and widely discussed tautomerism(prototropy) in organic chemistry. Other cases of prototopy belongingto triad systems are very much similar to this.Dyad system: When the H+ ion shifts through two atoms, it belongs to dyad system.

H C N C N H1 2

12

I II

In this case H+ shifts through two atoms, hence belongs to dyad system. HCN and HNC are functional isomersas well as tautomers.

R C N C N Rcynide form isocyanide

Similary alkyl cyanide and alkyl cyanide are structural isomers and also can be called tautomers underdyad category.In both the cases, independently each tautomer is a stable compound and the other remains almost in negligibleamount in equilibrium. Hence such tautomerism has limited practical significance.

The most useful tautomerism belongs to the TRIAD category.


Dr. S. S. Tripathy

Structural Requirement to show Tautomerism(Triad System):The compond must have

(a) multiple bond(double/triple) with a hetero atom like O/N/S for producing electronwithdrawing effect.

(b) at least one α H atom with respect to the multiple bonded functional group such as–CO–, –CN; –NO

2 etc. You will study in General Organic Chemistry(GOC) that α H atom with respect to a

–R/–M (negative Resonance) group is acidic in nature, because its conjugate base at the α position is resonancestabilised. If you cannot understand now about this, just forget.

Keto-Enol Tautomerism:(1) Simple Ketones and Aldehydes:The keto form for simple ketones and aldehydes is much more highly stable(stronger C=O BE) and morepolar than the enol form and hence keto form constitutes nearly 100% (99.99975%)of the eqilibrium mixture.Only 0.00025% exist as enol form. But this 0.00025% of enol form plays magical role in reactions. Just wait,we shall see this.Acetone:

CH3 C

O

CH3 CH2 C

OH

CH3

Keto form(99.99975%)

Enol form(0.00025%)

Enolisation:The negligible % of enol form performs magic when we undertake addition reaction with Br

2. A quick and small

addition reaction is observed due to the presence of C=C in the minute quantity of enol form. But slowly furtheraddtion of Br

2 takes place as the equilibrium is continuously driven to the right( Le Chatelier’s principle) and all

the keto form ultimately is converted to enol form and undergoes bromination. Is it not interesting ???? Aketone which is not supposed to show any addition reaction with Br

2, does undergo reaction though slowly till

the complete exhaustion of the ketone. This is possible due to this dynamic form of isomerism(tautomerism).Similarly when actone is treated with D

2O(heavy water), slowly all the six H atoms are deuterated. This

happens only due the presence of enol form. C–H bond is almost nonpolar and cannot undergo exhange withD. Only O–H bond is polar enough to undergo deutermium exchange. Since there is always a dynamic equilbriumbetween the forms, all the H atoms are unltimately replaced by D atoms.

CH3 C

O

CH3 CH2 C

OH

CH3

D2OCH2 C

OD

CH3

CH2D C

O

CH3 CHD C

OH

CH3 CD3 C

O

CD3

D2O

Acetaldehyde:

CH3 C

O

H CH2 CH

OH

nearly 100% negligible %(enol form)(keto form) (similar to acetone)


Dr. S. S. Tripathy

Acetaldehyde also can be slowly brominated and deuterated like acetone due to the presnce of the dynamicenol form, though in negligible percentage in the pure compound, but shows its beautiful appearance duringreaction.Butanone:

CH3 C

O

CH2 CH3 CH2 C

OH

CH2 CH3 CH3 C

OH

CH CH3

two enol forms(negligible %)

keto form(nearly 100%)

Butanone has α H atom on either side of carbonyl group. Hence there are two enol forms, both together makenegligible % in the pure liquid form.

Cyclic ketones however have relatively greater percentage of enol form compared to acyclic carbonylcompounds. Cyclohexanone has 0.02% of enol form as against 0.00025% in acetone. This is due to therestricted conformation of a cyclic ketone which stabilises the enol form bit more relative to that in acycliccompound.(B) βββββ-diketones and βββββ-keto esters:

These two types of compounds have a higher stablity of enol form and therefore the % of enol form isappreciably larger. In β-keto ester, the enol % though less ( 7.5%) than keto %, but not negligible like simpleketones and aldehydes. In β-diketones the % of enol is significantly high(> 80%). First see the structures.β-diketones:

CH3 C

O

CH2 C

O

CH3 CH3 C

OH

CH

C

O

CH3

active methylene group +-

enol form(80%)

keto form(20%)

acetyl acetone(pentane-2,4-dione)

The –CH2– group flanked between two –M group like –CO– is called an active methylene group. The carbanion

at this position is more stabilised by the R-effect on both the sides. Hence proton transfer takes place from thisactive methylene position( α to both the carbonyl group), in stead of the α H atom present on the other side ofeach – CO- group.The enol form is stabilised by two factors.

(i) R-effect : C=C is in conjugation with C=O in the enol form.(ii) Intramolecular H-bonding in the enol form.

Out of the two stabilising effects, the resonance stabilisation contributes more.Unsymmetrical β-diketones have two enol forms which together consitute more than 80%.

CH3 C

O

CH2 C

O

CH2 CH3

active methylene group

CH3 C

OH

CH C

O

CH2 CH3

CH3 C

O

CH C

OH

CH2 CH3

2 enol forms

keto form


Dr. S. S. Tripathy

N.B: CH3 C

O

CH2 C

O

(acetyl acetophenone) has nearly 99 % of enol form. Can you

say why and which enol form is predominantly present ? It is due to further resonance effect shown bybenzene ring to C=C in addition to C=O.

CH3 C

O

CH2 C

O

CH3 C

O

CH C

OH

major enol form

(other enol form not shown)

(2) βββββ-keto esters:Aceto acetic ester(ethyl acetoacetate) is a classic example of this type of compounds. In such compounds

enol form is appreciable but still keto form is more abundant.

CH3 C

OH

CH

C

O

OEt

active methylene group +-

keto form(92.5%)

CH3 C

O

CH2 C

O

OEtethyl acetoacetate

enol form(7.5%)

The enol form is stablised in this case by intramolecular H-bonding. The resonance stabilisation is almostnegligible as there is an in-built resonance between C=O and OEt in the ester function. So the resonancebetween C=C of enol form with C=O of ester function is negligible. That is why, we do not have a second enolform where C=C can remain adjacent to the OEt group and form a hemiacetal which is unstable.Change in the structure of β-ketoesters also change the enol-keto ratio.

IMP: Enol form has lower polarity than keto form, primarily due to the presence of intramolecular H-bonding.That is why pure enol form when separated from the mixture has a greater volatility(lower bp) than the ketoform. Make sure that you cannot keep a single form without the attainment of equilbrium mixture for a longtime.Effect of alkyl substitution on active methylene group:Due to steric inhibition of resonance(SIR), effected by destruction of complete planarity, the enol form in suchcase is less stable and the precentage of enol form is diminished. This decrease is more profound in β-diketones than in β-keto esters, may be due to the fact that in the latter the one Et group of ester function is littleaway from the alkyl substitution at the active postion. Hence SIR is not as high as in β-diketones.

CH3 C

O

CH C

O

CH3

CH3

(% of enol =33% as against 80% in unsubstituted compound)


Dr. S. S. Tripathy

CH3 C

O

CH C

O

OEt

iPr

(% of enol = 5% as against 7.5% in unsubstituted compound)Cyclic βββββ-diketones:

Due to greater planarity at the conjugated postions for restricted conformation, the enol form consitutesnearly 100%. Ironically, in cyclic compound, the enol form cannot have intramolecular H-bonding as thegeometry prohibits it. Despite this also, the enol is almost 100%. The effect of cyclisation has been found to beremarkably surprising.

O

O

5,5-dimethylcyclohexane-1,3-dione(dimedon)

O

OH

enol form(nearly 100%)

ααααα-diketones: (Look to the paradox)acyclic α- diketone remains nearly 100% in keto form while cyclic α-diketone remains nearly 100%

in enol form. Is it not paradoxical (contray to our expectation) ?

CH3

C

O

C

O

CH3

(nearly 100%)biacetyl

OH

OO

O

99%1%cyclopentane-1,2-dione enol form

In biacetyl, the two C=O functions remain anti to each other in the most stable conformation( s-trans) tominimize dipolar strain. The enol form is far less stable that this and hence is totally eliminated. In cyclopentane-1,2-dione, however the enol form is much more stable than the keto form due to the rigdity of the 5-memberedring which promotes resonance stabilisation to a large extent.However in cyclohexane-1,2-dione, the % of enol is 40% due to greater flexibility of the ring which makesresonance stabilisation in enol form less dominating than stablity of the keto form. The reader to ensure that theexact cause of this unexpected observation in cyclization can’t be said with certaintly. Its just a speculation.


Dr. S. S. Tripathy

Keto-Enol contents(%) in neat liquids:

Compound % of keto form % enol formAcetone 99.99975 0.00025cyclohexanone 99.98 0.02CH

3COCH

2COCH

320 80

CH3COCH(CH

3)COCH

367 33

CH3COCH

2COOEt 92.5 7.5

CH3COCH(i-Pr)COOEt 95 5.0

PhCOCH2COCH

31.0 99.0

Cyclohexane-1,3-dione 5.0 95.0CH

3COCOCH

399.9944 0.0056

Cyclopentane-1,2-dione 1.0 99.0cyclohexane-1,2-dione 60 40CH

2(COOEt)

299.9 0.1

Effect of Solvent:Since the keto form is more polar than enol form, a more polar solvent stabilises the keto form more than enolform and hence the % of keto form increases. Conversely a less polar solvent will make the keto form lessstable and hence the % of enol form increases. In the above data table or whatever analysis we have alreadymade, there was no solvent effect. We considered the neat liquid. Acetoacetic ester(ethyl acetoacetate) remiansonly 7.5 % in enol form as a neat pure liquid, however in polar solvent like H

2O, the enol % is reduced to

0.4%, while in nonpolar solvent like n-hexane the enol % increases to 46.4%. The table below gives the effectof solvent polarity on the enol content of acetoacetic ester.

Enol % of ethylacetoacetate in different solvents:Solvent pure(neat) H

2O AcOH EtOH Benzene cyclohexane

% of enol 7.5 0.4 5.7 10.5 16.2 46.4

Similar analogy can be extended to any keto-enol mixture in different solvents.

Base/Acid Catalysed Enolisation:The keto to enol conversion is catalysed by either acid or base.Acid catalysis:

CH3 C

O

CH3

H+

CH3 C

O

CH2

H

H H2OCH3 C

OH

CH2(H3O

+) H3O

+-keto enol

If you go in the opposite direction to ensure dynamic equilibrium between two, H+ addition will make a 30

carbocation. Carbocation will be stabilised by the movement of lone on O atom(+R effect). Then in the secondstep, deprotonation will occur from O atom thus restoring neutrality. Note that the conjugate acid of keto formand enol form are the same, hence the two forms equilbriate with each other in presence of acid.


Dr. S. S. Tripathy

Base Catalysis:

CH3 C

O

CH2

H

OH- CH3 C

O

CH2 CH3 C CH2

O

H2O-

H OH

CH3 C

OH

CH2

OH-

-

The enol also changes to keto form by base in the opposite direction. In fact the conjugate base of both theketo and enol forms are the same. Resonance makes all the job.N.B: For β-diketones, β-keto esters and other compounds where enol form remains in appreciable quantity,the enol form remains as stable enoloate ion, in stead of neutral enol in presence of base. Of course, the enolateion is in resonance with the carbanion, but the contribution of enolate ion is greater, as you know from thetheory of resonance.SAQ: Give the tautomer of cyclohexa-2,4-dien-1-one.

O OH

keto form(nearly 0%

enol formphenol(nearly 100%)

In this example, the aromatic structure of phonol makse the enol form so stable, that the keto form almost isextinct.Another example where the compound exists in 100% keto form and a case of pseudotropism is cyclohexa-2,5-diene-1,4-dione, which does not have any enol form

Pseudotropism: When only one tautomer exists in 100% abundace as in the previous case, the phenomenonis called pseudotropism.

O

O

This does not have a enol form as H- cannot shift from a sp2 hybrid carbon.


Dr. S. S. Tripathy

Other Triad Systems:(ii) Nitro-aci-nitro tautomerism:

CH3 CH

H

N

O

O

CH3 CH N

OH

Onitro form aci-nitro form

Though nitro form is more stable and is the predominant contributor to the equilibrium, the addition of a littlebase, shifts the equilibrium toward right i.e more aci-nitro form is formed. Note that aci-nitro compound ismore acidic than nitro compound. Also note that the aci-nitro form is both an electrophile for CH=N and anucleophile due to lone pair on O atom.(iii) Nitroso-oxime tatutomerism:

C

H

N OR

R'niroso form oxime form

(more stable)

C N OHR

R

Oxime form is moe stable and equilibrium lies almost entirely to the right.(iv) Enamine-imine tautomerism:

C C

N

H

R

CH

RN

C

enamine formimine form

Imine form is more stable than enamine. Enamine from 30 amine cannot form imine, in stead can form iminiumsalt. Imines from 10 amines are easily hydrolysed to carbonyl componds and ammonia(or ammoniumion). Imines are often called schiff bases.(v) Amide-iminol Form:

C

O

NH HR R C

OH

NHamide form iminol

Here you find that H atom moves from N atom to O atom. In all the earlier triad systems, H atom moved fromC atom another hetero atom. The amide from is highly stable and constitute 100%. However in some reactions,first iminol is formed which tautomerises to amide form.

Conclusion:(i) Always α− H atom with respect to a –M group takes part in 1-3 shift to form the other tautomer.(2) Tautomers are not resonating structures, as their atomic arrangement get changed alongiwth electronicarrangements. But one the tautomers which has conjugating units can have more than one resonating structures.For example the enol form of β-diketone has two resonating structures.


Dr. S. S. Tripathy

(3) Tautomers are functional isomers. Since they remain in dynamic equilibrium, they are addressed by adifferent unique name.

SAQ: Give the tautomer of the following and indicate which is one is more stable and more abundant/

(a)CH3CH

2CHO (b)

OH

(c) CH2 CH

OH

(d) CH3 CH2 C

OH

CH C

O

CH3(e)

OH

O(f)

O

Answer: (i) CH3CH=CH(OH) (unstable) (b) Cyclohexanone (stable)

(c) CH3CHO(stable) (d) CH3CH2COCH2COCH3 (less stable) (e)

O

O(more stable)

(f)

OH

(more stable as it is a conjugated dienol)

isomerism - the uraniumtheuranium.org/content-images/11-structural-isomerism.pdfstructural isomerism...

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