Structural Isomerism
ISOMERISM
ISOMERISM
Structural Isomerism
(Constitutional isomerism)
Stereoisomerism
Configurational Isomeris
Chain isomerism
Position Isomerism
Ring-Chain Isomerism
Functional Isomerism
M etamerism
Optical Isomerism
Conformation
Isomerism
Geometrical Isomerism
Tautomerism
What are Isomers: Isomers have same molecular formula but have different from each other either in terms of
structure(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 the
principal carbon chain or at branch points changes the carbon skeleton and make them chain isomers.
Chain Isomerism in Alkanes: (CnH2n+2)
Methane, ethane and propane do not have any chain isomers. However from butane onwards, the alkanes
show chain isomerism.
C4H10: 2 chain isomers.
CH3
CH3 CH2 CH2 CH3
butane (n-butane)
CH3 CH
CH3
2-methylpropane(isobutan
N.B: For writing the structures of chain isomers, first keep the principal chain fixed and then change the
position of branches to different non-equivalent positions to get a new name. After finding all the different
isomers belonging to the same principal carbon chain, the chain length is reduced by one carbon atom and the
rest carbon atoms are put as brances in as many different ways as possible so as to get different names. If some
name is repeated when, then its not a new isomer. Delete it from your list.
Dr. S. S. Tripathy
Structural Isomerism
C5H12: 3 isomers
CH3
CH3
CH3
CH3 CH2 CH2 CH2 CH3
pentane (n-pentane)
C6H14:
CH3 CH
CH2 CH3
2-methylbutane(isopentane
C
CH3
CH3
2,2-dimethylpropane
(neopentane)
5 isomers
CH3
CH3 CH2 CH2 CH2 CH2
hexane (n-hexane)
CH3
CH3 CH
CH2 CH2 CH3
2-methylpentane
CH3
CH3
CH3
CH3
CH2
CH CH2 CH3
3-methylpentane
CH3
CH
CH CH3
2,3-dimethylbutane
CH3
CH3
C
CH2
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 which
you have already completed.
N.B: 2-methylpentane and 3-methylpentane are not positional isomers. They are chain isomers. Only when
any functional group including –X(halo) changes its position on the same carbon skeleton, we call them positional
isomers. In benzene ring, of course, it is different. Even if there are two Me– branches, if their positions(o-, mp) are changed w.r.t each other, they become positional isomers, as benzene ring itself is treated as functional
group. Some authors might consider 2-methylpentane and 3-methylpentane as positional isomers, but this
auhtor does not agree on this. IUPAC is also not clear on this point. Hence students are advised not be much
concerned about this.
C7H16 :
9 isomers
CH3
CH3 CH2 CH2 CH2 CH2 CH2
heptane( (n -heptane)
CH2
CH CH2 CH2
3-methylhexane
CH3
CH
CH2 CH2 CH2
2-methylhexane
CH3 CH3
CH3
CH3
CH3
CH3
CH3 CH2
CH CH CH3
2,3-dimethylpentane
Dr. S. S. Tripathy
CH3
Structural Isomerism
CH3
CH3
CH3
CH3
CH3
CH
CH2 CH CH3
2,4-dimethylpentane
CH2
C
CH2
C CH3
CH3
2,2-dimethylpentane
CH3
CH3
CH2
CH3
CH CH3
CH2
CH3
CH3
3,3-dimethylpentane
CH2 CH CH CH3
3-ethylpentane
CH3 CH3
CH3
C
CH
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. Then
we reduced to 5- carbon chain got 5 isomers and finally 4-carbon chain, got 1 isomer, all total 9 isomers. Can
you 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
including stereoisomers
C4H10
2
2(no stereoisomersim)
C5H12
3
3 (no streoisomerism)
C6H14
5
5 (no stereoisomerism)
C7H16
9
11
C8H18
18
24
C9H20
35
55
C10H22
75
136
C11H24
159
345
C12H26
355
900
C13H28
802
2412
C14H30
1858
6563
C15H32
4347
18,127
C16H34
10,359
50,699
C17H36
24,894
1,43,255
C18H38
60,523
4,08,429
C19H40
1,48,284
11, 73,770
C20H42
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 out
all possibilities with the help of a computer, of course. For still higher alkanes from C21 and above you can refer
the website ( https://en.wikipedia.org/wiki/List_of_straight-chain_alkanes. )
Dr. S. S. Tripathy
No. of isomers
Structural Isomerism
C9H20:
(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; 2,2,4Trimethylhexane; 2,2,5-Trimethylhexane; 2,3,3-Trimethylhexane; 2,3,4-Trimethylhexane; 2,3,5Trimethylhexane; 2,4,4-Trimethylhexane; 3,3,4-Trimethylhexane; 3-Ethyl-2-methylhexane; 3-Ethyl-3methylhexane; 3-Ethyl-4-methylhexane; 4-Ethyl-2-methylhexane; 2,2,3,3-Tetramethylpentane; 2,2,3,4Tetramethylpentane; 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
C10H22: (75)
Decane; 2-Methylnonane; 3-Methylnonane; 4-Methylnonane; 5-Methylnonane; 2,2-Dimethyloctane; 2,3Dimethyloctane; 2,4-Dimethyloctane; 2,5-Dimethyloctane; 2,6-Dimethyloctane; 2,7-Dimethyloctane; 3,3Dimethyloctane; 3,4-Dimethyloctane; 3,5-Dimethyloctane; 3,6-Dimethyloctane; 4,4-Dimethyloctane; 4,5Dimethyloctane; 3-Ethyloctane; 4-Ethyloctane; 2,2,3-Trimethylheptane; 2,2,4-Trimethylheptane; 2,2,5Trimethylheptane; 2,2,6-Trimethylheptane; 2,3,3-Trimethylheptane; 2,3,4-Trimethylheptane; 2,3,5Trimethylheptane; 2,3,6-Trimethylheptane; 2,4,4-Trimethylheptane; 2,4,5-Trimethylheptane; 2,4,6Trimethylheptane; 2,5,5-Trimethylheptane; 3,3,4-Trimethylheptane; 3,3,5-Trimethylheptane; 3,4,4Trimethylheptane; 3,4,5-Trimethylheptane; 3-Ethyl-2-methylheptane; 3-Ethyl-3-methylheptane; 3-Ethyl-4methylheptane; 3-Ethyl-5-methylheptane; 4-Ethyl-2-methylheptane; 4-Ethyl-3-methylheptane; 4-Ethyl-4methylheptane; 5-Ethyl-2-methylheptane; 4-Propylheptane; 4-(1-Methylethyl)heptane; 2,2,3,3Tetramethylhexane; 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; 2,3,3,5Tetramethylhexane; 2,3,4,4-Tetramethylhexane; 2,3,4,5-Tetramethylhexane; 3,3,4,4-Tetramethylhexane; 3Ethyl-2,2-dimethylhexane; 3-Ethyl-2,3-dimethylhexane; 3-Ethyl-2,4-dimethylhexane; 3-Ethyl-2,5dimethylhexane; 3-Ethyl-3,4-dimethylhexane; 4-Ethyl-2,2-dimethylhexane; 4-Ethyl-2,3-dimethylhexane; 4Ethyl-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-2,2,3trimethylpentane; 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) is
different, then such structural isomers are called positional isomers.
Examples:
CH2 CH CH2
but-1-ene
CH3
and
OH
CH3 CH CH3
propan-2-ol
CH3 CH CH CH3
but-2-ene
OH
and
CH3 CH2 CH2
propan-1-ol
Dr. S. S. Tripathy
Structural Isomerism
O
O
CH3 C CH2 CH2 CH3 and
pentan-2-one
CH3 CH2 C CH2 CH3
pentan-3-one
In the above cases, the carbon skeletons are the same in each pair, but the position of the functional groups
namely C=C, -OH, -CO- are different.
IMP: Note that if two isomers are chain isomers, they can never be positional isomers even if the
locants of the functional groups are different.
CH3 OH
OH
CH3 CH2 CH CH3
butan-2-ol
and
CH3 CH CH2
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 an
exercise to know the different possible acyclic structures.
C4H8:
(3)
CH2 CH CH2
but-1-ene
CH3
and
CH3 CH CH CH3
but-2-ene
CH3
CH3 C CH2
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 chain
isomers. 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.
C5H10:
(5)
CH2 CH CH2 CH2 CH3
pent-1-ene
pent-1-ene and pent-2-ene are positional isomers.
CH3
CH2 C CH2 CH3
2-methlbut-1-ene
CH3 CH CH CH2 CH3
pent-2-ene
CH3
CH3
C CH CH3
2-methylbut-2-ene
2-methylbut-1-ene and 2-methylbut-2-ene are also positional isomers. But any one from the previous pair and
any one from the above pair are chain isomers.
Dr. S. S. Tripathy
Structural Isomerism
CH3
CH3
CH CH CH2
3-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 its
relationship 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 be
different.
Can we make a chain of propene and keep two Me- branches in the middle carbon ? No. So all together 5
structural isomers(excluding ring-chain and stereoisomers) for C5H10.
C6H12:
(13 acycli structural isomers excluding stereoisomers)
CH2 CH
CH2 CH2 CH2 CH3
hex-1-ene
CH3
CH3
CH
CH CH2 CH2 CH3
hex-2-ene
CH2
CH CH CH2 CH3
hex-3-ene
Hex-1-ene, hex-2-ene and hex-3-ene are positional isomers.
CH3
CH2 C
CH3
CH2
CH2 CH3
CH3
C
CH3
CH3
CH CH CH CH3
4-methylpent-2-ene
CH3
CH CH2 CH CH2
4-methylpent-1-ene
CH3
CH3
CH2
CH2 CH3
2-methylpent-2-ene
2-methylpent-1-ene
CH3
CH
CH3
CH CH CH2 CH3
3-methylpent-1-ene
CH C CH2 CH3
3-methylpent-2-ene
CH2
CH3
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-methylpent1-ene and 2-methylpent-2-ene are positional isomers. So also 3-methylpent-1-ene and 3-methylpent-2-ene
can 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
Structural Isomerism
CH3
CH3 CH3
CH2 CH
CH2 C
CH CH3
2,3-dimethylbut-1-ene
C CH3
CH3
3,3-dimethylbut-1-ene
CH3 CH3
CH3 C
C CH3
2,3-dimethylbut-2-ene
All these three are chain isomers to first 3 and the second 7 isomers. But within them 2,3-dimethylbut-1-ene
and 2,3-dimethylbut-2-ene are positional isomers. The third one(3,3-dimethylbut-1-ene) is a chain isomer to
former two positional isomers.
N.B: Student to note that in many cases, IUPAC has not spelt out a sharp border line between chain and
positional 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 chain
and then put that carbon as branch at a particular position. Keeping this skeleton fixed, the position of double
bond are changed to generate different isomers. Then change the branch point to another fixed position and
with this change the position of the double bond. Never forget that carbon atom cannot be attached to terminal
position lest it will produce a higher principal chain already completed. Also you have to see that every time the
valency of carbon is to be 4.
(3) Ring-Chain Isomerism:
Double Bond Equivalent(DBE):
nH n X n N


1
2
2
2
nC = number of carbon atoms; nH = number of H atoms
nN = number of N atoms.
Examples:
C4H8 :
DBE = 4 –(8/2) + 1 = 1;
C3H4:
DBE = 3 – (4/2) +1 = 2;
C6H6:
DBE = 6 –(6/2) + 1 = 4
C3H6O:DBE = 3 – (6/2) + 1 = 1; and so on.
This formula will be of great use in solving organic problems later.
DBE  nC 
DBE = 1 :
One C=C or 1 cyclic ring
DBE = 2;
one C
C or
2 C=C or
Note that a cyclic ring is equivalent to a C=C. One C
nX = number of halogen atoms
(one C=C + one ring) or ( 2 rings)
C is equivalent to two C=C or one C=C + a ring or
two rings.
C4H8:
CH2=CH–CH2–CH3
CH2
CH2
CH2
CH2
and
have same formula.
Benzene(C6H6) has a DBE=4, three for 3 C=C and one for the ring.
Dr. S. S. Tripathy
Structural Isomerism
Cyclic Isomers:
C3H6:
(1)
cyclopropane
C4H8:
(2)
cyclobutane
C5H10:
methylcyclopropane
(5)
cyclopentane
1,1-dimethylcyclopropane
methylcyclobutane
1,2-dimethylcyclopropane
ethylcyclopropane
C6H12: (10)
cyclohexane
methylcyclopentane
1,2-dimethylcyclobutane
Dr. S. S. Tripathy
Structural Isomerism
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)
C4H8:
acyclic = 3;
CH2 CH CH2
but-1-ene
cyclic =2
CH3
Total = 5
CH3 CH CH CH3
but-2-ene
CH3
CH3 C CH2
2-methylprop-1-ene
(isobutene)
Dr. S. S. Tripathy
cyclobutane
methylcyclopropane
Structural Isomerism
C5H10:
acyclic: 5;
cyclic: 5
Total = 10
(refer the all these structures given before)
cyclic = 10
Total = 23
C6H12: acyclic = 13;
N.B: We have not included stereoisomers to these formulae. Later when we discuss this isomerism, then we
shall 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 and
greater without following any special rule.
Structural Isomers of Alkynes(alkadienes):
C4H6:
acyclic : 4;
cyclic: 5
CH
C CH2 CH3
but-1-yne
CH3
C C CH3
but-2-yne
CH2
C CH CH3
buta-1,2-diene
CH2
CH CH CH2
buta-1,3-diene
cyclobutene
1-methylcyclopropene
3-methylcyclopropene
bicyclo[1.1.0]butane
methylidenecyclopropane
SAQ: Draw all the structural isomers from formula C5H8:
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-diene
Cyclic: 11 : cyclopentene, 1-methylcyclobutene, 3-methylcyclobutene, methylidenecyclobutane, 1,2dimethylcyclopropene, 1,3-dimethylcyclopropene, 1-ethylcyclopropene,
3-ethylcyclopropene,ethylidenecyclpropane, ethenylcyclopropane, bicyclo[2.1.0]pentane, 1methylbicyclo[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
Structural Isomerism
Haloalkanes(Alkyl halides): CnH2n+1X; CnH2nX2 etc.
DBE = 0 (saturated)
C3H7Cl: (2)
Cl
Cl
CH3 CH2 CH2
1-chloropropane
(n-propyl chloride)
CH3 CH CH3
2-chloropropane
(isopropyl chloride)
These are positional isomers.
C3H6Cl2: (4)
Cl
Cl
CH3
CH3 CH2 CH Cl
1,1-dichloropropane
Cl
C CH3
Cl
CH3 CH CH2
1,2-dichloropropane
Cl
2,2-dichloropropane
Cl
Cl
CH2 CH2 CH2
1,3-dichloropropane
C4H9Br: (4)
n-butyl bromide, sec-butyl bromide, isobutyl bromide, tert-butyl bromide
C4H8Cl2: (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: CnH2n–1X, CnH2n–2X2 etc.
DBE = 1; So we can have cyclic isomers also.
C3H5Br: (4)
Br
CH2 CH CH2
3-bromopropene
Br
CH2 C CH3
2-bromopropene
Br
CH CH CH3
1-bromopropene
Br
bromocyclopropane
(N.B: We shall find later that 1-bromopropene exists as two stereoisomers(geometrical isomres) E and Z. We
are not including stereoisomers to any formula now, only finding the structural isomers)
Dr. S. S. Tripathy
Structural Isomerism
C3H4Cl2: (7)
Cl
Cl
Cl
Cl
Cl
CH CH CH2
1,3-dichloropropene
CH2 C CH2
2,3-dichloropropene
Cl
CH C CH3
1,2-dichloropropene
Cl
Cl
Cl C CH CH3
1,1-dichloropropene
CH2 CH CH Cl
3,3-dichloropropene
Cl
Cl
Cl
1,2-dichlorocyclopropane
Cl
1,1-dichlorocyclopropane
(N.B : We have not considered the stereoisomers that exist in some of them)
C4H7Cl: (12)
acyclic : (8): 1-chlorobut-1-ene; 2-chlorobut-1-ene; 3-chlorobut-1-ene; 4-chlorobut-1-ene; 1chlorobut-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)
C3H3Br: ( 3 + 2 =5)
DBE=2
alknyl bromide, dienyl bromide
Br
Br
CH C CH2
3-bromoproyne
Br
C C CH3
1-bromopropyne
CH C CH2
1-bromopropadiene
Br
Br
1-bromocyclopropene
3-bromocyclopropene
Dr. S. S. Tripathy
Structural Isomerism
(C) FUNCTIONAL ISOMERISM:
The isomers have different functional groups and hence differ from each other in their properties, mostly in their
chemical properties.
(I) CnH2n+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.
C3H8O :
Alcohol:
OH
OH
(2 alcohols)
CH3 CH
CH3
propan-2-ol
CH3 CH2 CH2
propan-1-ol
(positional isomers)
Ether:
CH3 O CH2 CH3
methoxyethane
(ethyl methyl ether)
( 1 ether)
Ether and alcohol are functional isomers.
C4H10O :
Alcohols:
(4)
OH
OH
CH3
CH2 CH2 CH2
butan-1-ol
(n-butyl alcohol)
CH3
CH2 CH CH3
butan-2-ol
(sec-butyl alcohol)
butan-1-ol and butan-2-ol are positional isomers.
OH
CH3
CH
CH2
CH3
2-methylpropan-1-ol
(isobutyl alcohol)
OH
CH3
C
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 the
above 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 and
then 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
Structural Isomerism
Ethers:
(3)
CH3
CH3 O CH2 CH2 CH3
1-methoxypropane
(methyl n-propyl ether)
CH3 O CH CH3
2-methoxypropane
(isopropyl methyl ether
CH3 CH2 O CH2
ethoxyethane
(diethyl ether)
CH3
N.B: Note that three carbon atoms can be arranged in two ways i.e n-propyl and isopropyl. But how are they
related with each other ? Are they positional or chain isomers ? 1-methoxypropane and 2-methoxypropane
can be called positional isomers, if you consider alkoxy as a functional group. Ethoxyethane can be treated as
chain isomer of both 1-methoxypropane and 2-methoxypropane. But in these cases, definitely there is certain
degree of ambiguity as there is a bivalent –O– group in the chain containing carbon atoms. That is why the best
isomerism to be coined with these is “Metamerism” to be discussed later.
C5H12O:
Alcohols:
(8)
OH
OH
CH3
CH2
CH2 CH2
pentan-1-ol
CH2
CH3
CH2 CH2 CH
pentan-2-ol
CH3
OH
CH3
CH3
CH2 CH CH2
pentan-3-ol
OH
CH2
OH
CH
CH2
CH3
CH3
C
CH
`
OH
OH
CH
CH3
CH3
2-methylbutan-2-ol
CH3
2-methylbutan-1-ol
CH3
CH2
CH3
CH3
3-methylbutan-2-ol
CH3
CH
CH2
CH3
3-methylbutan-1-ol
CH2
CH3 OH
CH3
C
CH2
CH3
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 3methylbutan-2-ol are positional isomers. In between them, the relationship is chain isomerism. The last one i.e
neopentyl alcohol is a chain isomer to all the rest.
Dr. S. S. Tripathy
Structural Isomerism
Ethers:
(6)
CH3
CH3 O CH2 CH2 CH2 CH3
1-methoxybutane
(n-butyl methyl ether)
CH3 O
CH
CH2
CH3
2-methoxybutane
(sec-butyl methyl ether)
The above two can be called positional isomers.
CH3
CH3
CH3 O CH2 CH CH3
1-methoxy-2-methylpropane
(isobutyl methyl ether)
CH3 O
C
CH3
CH3
2-methoxy-2-methylpropane
(tert-butyl methyl ether)
The above two can also be called positional isomers.
CH3
CH3 CH2 O CH2 CH2 CH3
1-ethoxypropane
(ethyl n-propyl ether)
CH3 CH2 O CH
CH3
2-ethoxypropane
(ethyl isopropyl ether)
The above two can also be called positional isomers. But in between the two pairs , the relatioship is chain
isomerism. But they are best exressed by Metamerism(See later).
C6H14O:
Alchols: 17
hexan-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-3ol, 2-ethylbutan-1-ol; 2,3-dimethylbutan-1-ol; 2,3-dimethylbutan-2-ol; 2,2-dimethylbutan-1-ol; 3,3dimethylbutan-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/l
pair(optically active). Hence there are 23 isomeric alochols inclusive of stereoisomers. Do not poke your nose
now until you study stereoisomerism)
Ethers: (15)
1-methoxypentane; 2-methoxypentane, 3-methoxypentane; 1-methoxy-2-methylbutane, 2-methoxy-2methylbutane, 2-methoxy-3-methylbutane; 1-methoxy-3-methylbutane; 1-methoxy-2,2-dimethylpropane; 1ethoxybutane; 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 18
ethers inclusive of stereoisomers.
Dr. S. S. Tripathy
Structural Isomerism
II: CnH2nO 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 few
examples.
C2H4O:
O
OH
O
CH3 C H
ethanal
(acetaldehyde)
CH2
CH
ethenol
(vinyl alcohol)
oxirane
(ethylene oxide
With this formula there is one aldehyde, one enol and one cyclic ether. There is no ketone or unsaturated ether
with this formula. Later when we shall study ‘Tautomerism’ ,we shall know that vinyl alcohol is unstable and it
tautomerizes readily to acetaldehyde.
C3H6O:
Aldehyde: (1)
Ketone: (1)
O
O
CH3 C CH3
propan-2-one
(acetone)
CH3 CH2 C H
propanal
Enols:
(3)
OH
OH
OH
CH2 C CH3
prop-1-en-2-ol
CH2 CH CH2
prop-2-en-1-ol
(allyl alcohol)
CH CH CH3
prop-1-en-1-ol
Note that the first two enols are unstable, as we shall know soon while learning tautomerism, and change to
propanone and propanal respectively. But the third one is stable(allyl alcohol).
Unsaturated ether:
CH2 CH O CH3
methoxyethene
(methyl vinyl ether)
Cylcic isomers:
(2)
OH
O
oxetane
cyclopropanol
Dr. S. S. Tripathy
Structural Isomerism
C4H8O:
Aldehyde: (2)
CH3 O
O
CH3
CH2 CH2 C H
butanal
(butyraldehyde)
CH3 CH C H
2-methylpropanal
(isobutyraldehyde)
Ketone: (1)
O
CH3
CH2 C CH3
butan-2-one
(butanone)
Enols: (8)
OH
OH
OH
CH
CH CH2
but-1-en-1-ol
CH3
CH2
C CH2
but-1-en-2-ol
CH3
CH2
CH CH CH3
but-3-en-2-ol
OH
CH2
CH CH2
but-3-en-1-ol
CH2
Out of the above enols, the first two are unstable which readily tautomerises to butanal and butanone respectively(
we shall study this soon).
OH
OH
CH CH
CH2
but-2-en-1-ol
CH3
C CH CH3
CH3
but-2-en-2-ol
In this pair, the second one is also unstable and will tautomerize to butanone.
CH3 OH
CH2 C
CH2
2-methylprp-2-en-1-ol
CH3
HO CH C
CH3
2-methylprop-1-en-1-ol
The second one also is unstable and will tautomerize to 2-methylpropanal. In all these, I have forgotten to
remind you whether a pair belongs chain isomers or positional. You can scrutinize that. If the everthing else
remain the same, only the locant of functional group differs then they are positional, otherwise chain isomers.
Unsaturated acyclic ether: (4)
CH2
Dr. S. S. Tripathy
CH O CH2 CH3
ethoxyethene
(ethyl vinyl ether)
CH3 CH CH O CH3
1-methoxyprop-1-ene
Structural Isomerism
CH2
C
CH2 CH CH2 O CH3
3-methoxyprop-1-ene
(allyl methyl ether)
O
CH3
CH3
2-methoxyprop-1-ene
Cyclic isomers: (11)
OH
O
oxolane
(tetrahydrofuran,THF
cyclobutanol
OH
2-methylcyclopropanol
HO
OH
1-methylcyclopropanol
cyclopropylmethanol
(hydroxymethylcyclopropane)
O
O
CH3
O
methoxycyclopropane
O
2,3-dimethyloxirane
2-methyloxetane
O
2,2-dimethyloxirane
3-methyloxetane
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 have
not tallied with any source, hopefully not available.
From the above extensive discussion on structural isomerism, i feel like dropping enols, unsaturated ethers and
cyclic isomes for higher formulae.
Dr. S. S. Tripathy
Structural Isomerism
C5H10O: (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 many
number of aldehydes and ketones possible from the formula.
Aldehydes:
(4)
CH3
CH3 CH2 CH2 CH2
pentanal
CHO
CH3
CH2
CH CHO
2-methylbutanal
CH3
CH3
CH3
CH3 CH CH2 CHO
3-methylbutanal
Ketones:
C
CHO
CH3
2,2-dimethylpropanal
(3)
O
O
CH3 C CH2 CH2 CH3
pentan-2-one
O
CH3 CH2 C CH2 CH3
pentan-3-one
CH3
CH3 C CH CH3
3-methylbutan-2-one
(III) CnH2nO2 : (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 in
which we shall draw all structural isomers possible. But for higher formula, we shall not venture to do so, on the
fear of missing an isomer and also to avoid monotony. For higher formulae, we shall concentrate on only
carboxylic acids and esters.
C2H4O2:
(8)
In this case too, DBE = 1.
OH
CH3 COOH
ethanoic acid
(acetic acid)
OH
CH2 CHO
2-hydroxyethanal
Dr. S. S. Tripathy
HCOOCH3
methyl methanoate
(methyl formate)
O
O
1,2-dioxetane
OH
OH
CH CH
ethene-1,2-diol
O
O
1,3-dioxetane
C
CH2
OH
ethene-1,1-diol
O
oxiran-2-ol
OH
Structural Isomerism
So we are disgusted in writing so many isomers including cyclic ones. Let us concentrate on carboxylic acids
and esters only for higher formulae.
C3H6O2 :
Carboxylic acids: (1)
CH3 CH2 COOH
propanoic acid
(propionic acid)
Esters : (2)
O
CH3 C OCH3
methyl ethanoate
(methyl acetate)
O
H C O CH2 CH3
ethyl methanoate
(ethyl formate)
C4H8O2:
Carboxylic acids: (2)
CH3
CH2 CH2 COOH
butanoic acid
(butyric acid)
CH3
CH COOH
CH3
2-methylpropanoic acid
(isobutyric acid)
Esers: (4)
O
CH3 CH2 COOCH3
methyl propanoate
(methyl propionate)
CH3 COOCH2CH3
ethyl ethanoate
(ethyl acetate)
H
C O CH2 CH2 CH3
propyl methanoate
(n-propyl formate)
O
H
C
O CH CH3
CH3
propan-2-yl methanoate
(isopropyl formate)
SAQ: Write the structures of all C. acids and esters from the formula C5H10O.
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 acid
Esters: methyl butanoate, methyl 2-methylpropanoate; ethyl propanoate; propyl ethanoate; propan-2-yl
ethanoate; butyl methanoate, butan-2-yl methanoate, 2-methylpropyl methanoate; 2-methylpropan-2-yl
methanoate
Dr. S. S. Tripathy
Structural Isomerism
(IV) AMINES [CnH2n+3 N]
DBE =0, so these are alkyl amines(10, 20 and 30).
C3H9N:
10: (2)
CH3
CH3 CH2 CH2 NH2
propan-1-amine
(n-propyl amine)
20( 1)
CH3 CH NH2
propan-2-amine
(isopropyl amine)
30 (1)
CH3
CH3
CH2 NH CH3
N-methylethanamine
(ethyl methyl amine)
CH3 N CH3
N,N dimethylmethanamine
(trimethyl amine)
-
Total = 4
C4H11N:
10: (4)
CH3
CH3 CH2 CH2 CH2
butan-1-amine
(n-butyl amine)
NH2
CH3 CH2 CH NH2
butan-2-amine
(sec-butyl amine)
CH3
CH3
CH3 C NH2
CH3 CH CH2 NH2
2-methylpropan-1-amine
(isobutyl amine)
CH3
2-methylpropan-2-amine
(tert-butyl amine)
20: (3)
CH3 CH2 CH2 NH CH3
N-methylpropan-1-amine
(methyl n-propyl amine)
CH3
Dr. S. S. Tripathy
CH3
CH3 CH NH CH3
N-methylpropan-2-amine
(isopropyl methyl amine)
CH2 NH CH2 CH3
N-ethylethanamine
(diethyl amine)
Structural Isomerism
30: (1)
CH3
CH3 N CH2 CH3
N,N-dimethylethanamine
(ethyl dimethyl amine)
Total isomers: 8
C5H13N :
10 amine: (6) : hexan-1-amine, hexan-2-amine; 2-methylbutan-1-amine; 3-methylbutan-1-amine; 2,2dimethylpropan-1-amine
20 amine: (6) : N-methylbutan-1-amine; N-methylbutan-2-amine; N,2-dimethylpropan-1-amine; N,2dimethylpropan-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-Nmethylethanamine
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)
C8H10: (4)
DBE = 8 – (10/2) + 1 = 4; Already it has been told that for benzene ring DBE requirement is
4. So no more unsaturation is present.
CH3
CH3
CH3
CH3
1,2-dimethylbenzene
(o-xylene)
CH3
1,3-dimethylbenzene
(m-xylene)
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
Structural Isomerism
C7H8O: (5)
DBE = 7 – 4 + 1 = 4 ; That means it has benzene ring only and no more unsaturation.
OH
OH
OH
CH3
CH3
3-methylphenol
(m-cresol)
2-methylphenol
(o-cresol)
CH3
4-methylphenol
(p-cresol)
Cresols are positional isomers of each other.
OCH3
CH2OH
methoxybenzene
(anisole)
benzyl alcohol
Anisole and benzyl alcohol are functional isomers of each other and each one of them is a functional isomer of
each of the xylenes.
N.B: Theoretically we can draw more acyclic isomers having four C=C or their equivalent and increase the
number of isomers to a very large extent. However, once we get DBE is 4 or more, then the most likely isomers
are those which contain benzene ring. Other purely acyclic isomers are ignored for practical point of view.
C8H8O: (11)
DBE = 8 – 4 +1 = 5; In this, there is one more unsaturation (double bond) in addition to the
benzene ring.
O
CHO
CH2 CHO
C CH3
CH3
acetophenone
phenylacetic acid
CHO
CH3
3-methylbenzaldehyde
Dr. S. S. Tripathy
2-methylbenzaldehyde
CHO
CH3
4-methylbenzaldehyde
OH
C CH2
1-phenylethenol
Structural Isomerism
OH
CH
CH
CH
CH
CH2
CH2
OH
2-phenylethenol
CH
OH
3-ethenylphenol
2-ethenylphenol
CH2
O CH CH2
OH
4-ethenylphenol
(ethenyloxy)benzene
Note that 1-phenylethenol and 2-phenylethenol will tautomerise to give acetophenone and phenylacetic acid
respectively. We shall learn it very soon.
SAQ: Draw all the isomers of C8H8O2 and C7H7Cl 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 the
isomers 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 other
names i.e positional or chain isomers. ‘Metamers’ term is used for them preferentially.
CH3 NH CH2 CH2 CH3
N-methylpropan-1-amine
(methyl n-propyl amine)
CH3
CH3
CH3 NH CH CH3
N-methylpropan-2-amine
(isopropyl methyl amine)
CH2 NH CH2 CH3
N-ethylethanamine
(diethyl amine)
The first two can be called positional isomers. But how is 3rd one related to any one of the first two ? Here the
principal chain carries a bivalent group –NH–. Hence for all these isomers, a speical isomeism has been
assigned. These are metamers, in which the one or more alkyl groups attached to the divalent –NH– group is/
are different.
Dr. S. S. Tripathy
Structural Isomerism
O CH3
O
(I)
CH3
CH2 C CH2 CH2
hexan-3-one
(ethyl n-propyl ketone)
CH3
(II)
CH3 CH2 C CH CH3
2-methylpentan-3-one
(ethyl isopropyl ketone)
O
(III)
CH3 C CH2 CH2 CH2 CH3
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 no
ambiguity as the divalent –CO– group is a carbon bearing group. All the the three are metamers.
CH3
CH3 O CH2 CH2 CH3
1-methoxypropane
(I)
(methyl n-propyl ether)
CH3
(III)
(II)
CH3 O CH CH3
2-methoxypropane
(isopropyl methyl ether
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 a
divalent –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 C3H3Cl
(e) functional isomer of (i) propan-1-ol and (ii) butanal
(iii) ethyl propanoate
(f) 2-methylaniline
Solution:
OH
(b) CH3
(a) CH3 CH2 CH2 CH3 (butane)
C
CH3
( tert-butyl alcohol)
CH3
(c) pentan-2-one
Cl
(d) DBE = 3 – 1.5 – 0.5 + 1 = 2
(e)
(i) methoxyethane
(f) 3-methylaniline
Dr. S. S. Tripathy
(ii) butan-2-one
( 3-chlorocyclopropene)
(iii) pentanoic acid
Structural Isomerism
TAUTOMERISM:
This is a case of dynamic functional isomerism, where the ismers(called tautomers) differ from each other only
in the position of one (rarely more) mobile atom and in electron distribution. The special case of tautomerism is
PROTOTROPY, in which the mobile species is a H+ ion. The tautomers are basically functional isomers which
remain always in dynamic equilibrium with each other in the liquid state. Tautomerism is exhiibited in the
liquid state only. Any attempt to separate one tautomer from the other will be countered by the formation of the
same equilibrium mixture(%) in each with passage of time.
Types : (a) Triad System
(b) Dyad sytem
Triad System: In which the mobile species(H+) shifts through 3 atoms.
H
CH2
1
H
3
O
C
2
CH2
CH3
3
O1
C
2
CH3
II
I
In the above example belonging to triad system, one H+ ion leaves the carbon atom in I and shifts through three
atoms and bonds with O atom, thus converting to the isomer II. Similarly one H+ ion leaves O atom in II and
shifts through three atoms and bonds with C atom, thus converting II back to I. This is called keto-enol
tautomerism as the isomer I is in KETO form and isomer II is in ENOL form. Keto-enol tautomerism is most
common and widely discussed tautomerism(prototropy) in organic chemistry. Other cases of prototopy belonging
to 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
1
2
I
C
2
N1
H
II
In this case H+ shifts through two atoms, hence belongs to dyad system. HCN and HNC are functional isomers
as well as tautomers.
R C N
cynide form
C
N R
isocyanide
Similary alkyl cyanide and alkyl cyanide are structural isomers and also can be called tautomers under
dyad category.
In both the cases, independently each tautomer is a stable compound and the other remains almost in negligible
amount in equilibrium. Hence such tautomerism has limited practical significance.
The most useful tautomerism belongs to the TRIAD category.
Dr. S. S. Tripathy
Structural Isomerism
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 electron
withdrawing effect.
(b) at least one α H atom with respect to the multiple bonded functional group such as
–CO–, –CN; –NO2 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 resonance
stabilised. 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 more
polar 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:
OH
O
CH2
CH3 C CH3
Keto form
(99.99975%)
C
CH3
Enol form
(0.00025%)
Enolisation:
The negligible % of enol form performs magic when we undertake addition reaction with Br2. A quick and small
addition reaction is observed due to the presence of C=C in the minute quantity of enol form. But slowly further
addtion of Br2 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 ???? A
ketone which is not supposed to show any addition reaction with Br2, 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 D2O(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 with
D. Only O–H bond is polar enough to undergo deutermium exchange. Since there is always a dynamic equilbrium
between the forms, all the H atoms are unltimately replaced by D atoms.
CH3
C
CH2
CH3
C
C
D2O
CH3
OH
O
CH2D
OD
OH
O
CH3
CHD
C
CH2
C
CH3
O
D2O
CH3
CD3
C
Acetaldehyde:
O
CH3 C H
(keto form)
nearly 100%
Dr. S. S. Tripathy
OH
CH2 CH
(enol form)
negligible %
(similar to acetone)
CD3
Structural Isomerism
Acetaldehyde also can be slowly brominated and deuterated like acetone due to the presnce of the dynamic
enol form, though in negligible percentage in the pure compound, but shows its beautiful appearance during
reaction.
Butanone:
OH
OH
O
CH2
CH3 C CH2 CH3
keto form
(nearly 100%)
C
CH3 C
CH2 CH3
CH
CH3
two enol forms
(negligible %)
Butanone has α H atom on either side of carbonyl group. Hence there are two enol forms, both together make
negligible % in the pure liquid form.
Cyclic ketones however have relatively greater percentage of enol form compared to acyclic carbonyl
compounds. Cyclohexanone has 0.02% of enol form as against 0.00025% in acetone. This is due to the
restricted conformation of a cyclic ketone which stabilises the enol form bit more relative to that in acyclic
compound.
(B) β-diketones and β-keto esters:
These two types of compounds have a higher stablity of enol form and therefore the % of enol form is
appreciably larger. In β-keto ester, the enol % though less ( 7.5%) than keto %, but not negligible like simple
ketones and aldehydes. In β-diketones the % of enol is significantly high(> 80%). First see the structures.
β-diketones:
+
H
active methylene group
O
-
O
O
O
CH3 C
CH2 C
CH3
acetyl acetone(pentane-2,4-dione)
keto form
(20%)
CH3
C
C
CH3
CH
enol form
(80%)
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 this
active methylene position( α to both the carbonyl group), in stead of the α H atom present on the other side of
each – 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%.
active methy lene group
O
CH3
C
OH
O
CH2
C
keto form
CH2 CH3
CH3
C
O
CH
O
CH3 C
C
CH2
CH3
OH
CH
C
CH2 CH3
2 enol forms
Dr. S. S. Tripathy
Structural Isomerism
O
N.B: CH3
O
C
CH2
(acetyl acetophenone) has nearly 99 % of enol form. Can you
C
say why and which enol form is predominantly present ? It is due to further resonance effect shown by
benzene ring to C=C in addition to C=O.
O
CH3
C
O
CH2
OH
O
C
CH3
C
CH
C
major enol form
(other enol form not shown)
(2) β-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.
active methylene group
O
CH3
+
H
O
O
C
CH2 C
OEt
ethyl acetoacetate
keto form
(92.5%)
CH3
-
O
C
C
OEt
CH
enol form
(7.5%)
The enol form is stablised in this case by intramolecular H-bonding. The resonance stabilisation is almost
negligible as there is an in-built resonance between C=O and OEt in the ester function. So the resonance
between C=C of enol form with C=O of ester function is negligible. That is why, we do not have a second enol
form 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 keto
form. Make sure that you cannot keep a single form without the attainment of equilbrium mixture for a long
time.
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 such
case is less stable and the precentage of enol form is diminished. This decrease is more profound in βdiketones than in β-keto esters, may be due to the fact that in the latter the one Et group of ester function is little
away from the alkyl substitution at the active postion. Hence SIR is not as high as in β-diketones.
CH3
O
CH3 O
C
CH
C
CH3
(% of enol =33% as against 80% in unsubstituted compound)
Dr. S. S. Tripathy
Structural Isomerism
CH3
O
iPr
O
C
CH
C
OEt
(% of enol = 5% as against 7.5% in unsubstituted compound)
Cyclic β-diketones:
Due to greater planarity at the conjugated postions for restricted conformation, the enol form consitutes
nearly 100%. Ironically, in cyclic compound, the enol form cannot have intramolecular H-bonding as the
geometry prohibits it. Despite this also, the enol is almost 100%. The effect of cyclisation has been found to be
remarkably surprising.
O
O
O
OH
5,5-dimethylcyclohexane-1,3-dione
(dimedon)
enol form
(nearly 100%)
α-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) ?
O
CH3
OH
O
C
C
CH3
O
biacetyl
(nearly 100%)
O
cyclopentane-1,2-dione
1%
O
enol form
99%
In biacetyl, the two C=O functions remain anti to each other in the most stable conformation( s-trans) to
minimize dipolar strain. The enol form is far less stable that this and hence is totally eliminated. In cyclopentane1,2-dione, however the enol form is much more stable than the keto form due to the rigdity of the 5-membered
ring 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 makes
resonance stabilisation in enol form less dominating than stablity of the keto form. The reader to ensure that the
exact cause of this unexpected observation in cyclization can’t be said with certaintly. Its just a speculation.
Dr. S. S. Tripathy
Structural Isomerism
Keto-Enol contents(%) in neat liquids:
Compound
Acetone
cyclohexanone
CH3COCH2COCH3
CH3COCH(CH3)COCH3
CH3COCH2COOEt
CH3COCH(i-Pr)COOEt
PhCOCH2COCH3
Cyclohexane-1,3-dione
CH3COCOCH3
Cyclopentane-1,2-dione
cyclohexane-1,2-dione
CH2(COOEt)2
% of keto form
99.99975
99.98
20
67
92.5
95
1.0
5.0
99.9944
1.0
60
99.9
% enol form
0.00025
0.02
80
33
7.5
5.0
99.0
95.0
0.0056
99.0
40
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 enol
form and hence the % of keto form increases. Conversely a less polar solvent will make the keto form less
stable and hence the % of enol form increases. In the above data table or whatever analysis we have already
made, there was no solvent effect. We considered the neat liquid. Acetoacetic ester(ethyl acetoacetate) remians
only 7.5 % in enol form as a neat pure liquid, however in polar solvent like H2O, 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 effect
of solvent polarity on the enol content of acetoacetic ester.
Enol % of ethylacetoacetate in different solvents:
Solvent
pure(neat)
H2O
AcOH
% of enol
7.5
0.4
5.7
EtOH Benzene
10.5 16.2
cyclohexane
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:
H
O
CH3
C
CH3
keto
O
+
H
+
(H3O )
CH3
C
H
CH2
OH
H2O
- H3O +
CH3
C
CH2
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 second
step, deprotonation will occur from O atom thus restoring neutrality. Note that the conjugate acid of keto form
and enol form are the same, hence the two forms equilbriate with each other in presence of acid.
Dr. S. S. Tripathy
Structural Isomerism
Base Catalysis:
O
O
O
H
CH3
C
-
CH2
OH
- H2O
CH3
C
CH2
CH3
C
CH2
-
- OH
H OH
OH
CH3 C
CH2
The enol also changes to keto form by base in the opposite direction. In fact the conjugate base of both the
keto 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 enolate
ion is in resonance with the carbanion, but the contribution of enolate ion is greater, as you know from the
theory of resonance.
SAQ: Give the tautomer of cyclohexa-2,4-dien-1-one.
OH
O
enol form
phenol
(nearly 100%)
keto form
(nearly 0%
In this example, the aromatic structure of phonol makse the enol form so stable, that the keto form almost is
extinct.
Another example where the compound exists in 100% keto form and a case of pseudotropism is cyclohexa2,5-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 phenomenon
is called pseudotropism.
O
O
This does not have a enol form as H- cannot shift from a sp2 hybrid carbon.
Dr. S. S. Tripathy
Structural Isomerism
Other Triad Systems:
(ii) Nitro-aci-nitro tautomerism:
H
CH3
OH
O
CH
N
nitro form
O
CH3 CH
N
aci-nitro form O
Though nitro form is more stable and is the predominant contributor to the equilibrium, the addition of a little
base, shifts the equilibrium toward right i.e more aci-nitro form is formed. Note that aci-nitro compound is
more acidic than nitro compound. Also note that the aci-nitro form is both an electrophile for CH=N and a
nucleophile due to lone pair on O atom.
(iii) Nitroso-oxime tatutomerism:
H
R
C
R'
N
O
niroso form
R C
N
OH
R
oxime form
(more stable)
Oxime form is moe stable and equilibrium lies almost entirely to the right.
(iv) Enamine-imine tautomerism:
C
C
CH
C
N R
H
enamine form
N R
imine form
Imine form is more stable than enamine. Enamine from 30 amine cannot form imine, in stead can form iminium
salt. Imines from 10 amines are easily hydrolysed to carbonyl componds and ammonia(or ammonium
ion). Imines are often called schiff bases.
(v) Amide-iminol Form:
O
R
C NH H
amide form
OH
R
C
NH
iminol
Here you find that H atom moves from N atom to O atom. In all the earlier triad systems, H atom moved from
C 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 electronic
arrangements. 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
Structural Isomerism
(3) Tautomers are functional isomers. Since they remain in dynamic equilibrium, they are addressed by a
different unique name.
SAQ: Give the tautomer of the following and indicate which is one is more stable and more abundant/
OH
OH
(a)CH3CH2CHO
(b)
Answer:
CH3 CH2
C
CH2
CH
C
CH
O
OH
O
OH
(d)
(c)
CH3
(i) CH3CH=CH(OH) (unstable)
(e)
(f)
O
(b) Cyclohexanone (stable)
O
(c) CH3CHO(stable) (d) CH3CH2COCH2COCH3 (less stable) (e)
(more stable)
O
OH
(f)
(more stable as it is a conjugated dienol)
Dr. S. S. Tripathy