GENERAL ORGANIC CHEMISTRY & ISOMERISM
Inductive and resonance effects on acidity and basicity of acids and bases
respectively. Reactive intermediates produced during homolytic and heterolytic
cleavage. Formation, structure, stability and reactivity of alkyl carbocations and
free radicals .
The Breaking and Forming of Bonds
A covalent bond between two atoms can be broken in essentially the following ways:
R. .
X
R:X R :-
X
R
:X-
In the first case each atom separates with one electron, leading to the formation of highly
reactive entities called radicals, owing their reactivity to their unpaired electron; this is
referred to as homolytic fission of the bond. Alternatively, one atom may hold on to both
electrons, leaving none for the other, the result in the above case being a negative and
positive ion, respectively. Where R and X are not identical, the fission, can, of course, take
place in either of two ways, as shown above, depending on whether R or X retains the
electron pair. Either of these processes is referred to as heterolytic fission, the result being
the formation of an ion pair. Formation of a covalent bond can take place by the reversal of
any of these processes, and also, of course, by the attack of first- formed radicals or ions on
other species:
R
+ Br - Br  R - Br + Br
R
+ H2 O  R - OH + H
Such radicals or ion pairs are formed transiently as reactive intermediates in a very wide
variety of organic reactions, as will be shown below. Reactions involving radicals tend to
occur in the gas phase and in solution in non-polar solvents, and to be catalysed by light
and by the addition of other radicals. Reactions involving ionic intermediates take place
more readily in solution in polar solvents, because of the greater ease of separation of
charges therein and very often because of the stabilisation of the resultant ion pairs through
solvation. Many of these ionic intermediaries can be considered as carrying their charge on
a carbon atom, though the ion is often stabilised by delocalisation of the charge, to a greater
or lesser extent, over other carbon atoms, or atoms of different elements:
CH2=CHCH2
CH2 = CHCH2OH CH2=CHCH2OH2
CH2CH=CH2
O O
|| ||
CH3CCH3 CH3CCH2
+H2O
O–
|
CH3C=CH2
-
OH
-H2O
H
-




When a positive charge is carried on carbon the entity is known as a carbocation, and when
a negative charge, a carbanion. Though such ions may be formed only transiently and be
present only in minute concentration, they are nevertheless often of paramount importance
in controlling the reactions in which they participate.
These three types, radicals, carbocations and carbanions, by no means exhaust the
possibilities of transient intermediates in which carbon is the active centre: others include
the electron-deficient species carbenes,
Electronic Displacement in Covalent Bonds
The following four types of electronic effects operates in covalent bonds
i) Inductive effect
ii) Mesomeric and Resonance effect
iii) Electronic effects
iv) Hyperconjugation
Inductive Effect
In a covalent bond between the two dissimilar atoms, the electron pair forming the bond is
never shared absolutely equally between the two atoms but is attracted a little more towards
the more electronegative atom of the two, eg. The electron pair forming the C–X bond is
somewhat more attracted towards the atom X with the result – it attains a partial negative
charge (–) while the carbon atoms attain a partial positive charge (+)




 X
C
or
X
:
C
On the other hand, in compounds like C–Y, where Y in an electropositive element or group
i.e., C is more electronegative than Y, the electron pair forming the C–Y bond is somewhat
displaced towards the carbon atom and thus C and Y attain partial negative and partial
positively charges respectively.
C: Y or




 Y
C
According to Ingold sign convention, the former is called as (–I) effect and the later is called
as (+I) effect.
The inductive effect causes certain degree of polarity in the bond which in term renders the
bond much more liable to be attacked by other charged atoms or group.
Thus, inductive effect may be defined as the permanent displacement of electron forming a
covalent bond towards the more electronegative element or group.
The inductive effect is represented by the symbol , the arrow pointing towards the more
electronegative element or group of elements eg. N – butyl chloride















Cl
CH
CH
CH
CH 2
2
2
3
The extent of positive charge keeps on decreasing away from Cl atom and at third and
fourth carbon it is almost zero for all practical purposes.
..

Examples of I effect groups
a) (–I) effect group (electron attracting)
H
CH
CH
H
C
OH
OR
COOR
OAr
I
Br
Cl
F
COOH
CN
NO
H
N
Me
N
2
5
6
2
3
3


















b) (+I) effect – group (electron – repelling)
H
CH
R
CH
CHR
C
R
COO
O
H
C 3
2
2
3
5
6 




 

Applications of Inductive effect
i) Effect on Bond lengths: Since the inductive effect leads to ionic character in the
bond, the increase in –I effect usually decreases the bond length.
Increasing –I effect
CH3  F CH3Br CH3 Cl CH3F
2.14Å 1.94 Å 1.78Å 1.38 Å
Decrease in bond length
ii) Dipole moment : Since, inductive effect leads to a dipolar character in the molecule,
it develops some dipole moment in the molecule, which increases with the increase
in the inductive effect.
CH3 – I, CH3 — Br, CH3 — Cl
Increasing dipole moment
iii) Reactivity of alkyl halides: Alkyl halides are more reactive than the corresponding
alkanes due to presence of C––X bond which is polar due to I effect, furthermore
reactivity increases with increase of branching.
H3C—CH2  X CH – X
H3C
H3C
C— X
H3C
H3C
H3C
 
Increasing reactivity due to increasing C — X bond polarity
iv) Strength of Carboxylic Acids: Strength of an acid depends upon the ease with which
an acid ionises to give proton. A molecule of carboxylic acid can be represented as a
resonance hybrid of the following structures.
R—C——O—H  R— C = O — H
O
 
 
 
O




 
+
I II
In the II structure, the oxygen atom of the hydroxyl group has a positive charge due
to which it has a tendency to attract electron pair (inductive effect) of the O—H bond
towards itself, which results in the removal of hydrogen atom as proton and hence
carboxylic acids behave as acids.

Once, the carboxylate anion is formed, it is stabilised more easily by resonance than
undissociated acid.
R—C
O
O–
R—C
O–
O
R—C
O
O
Thus, the acidity of carboxylic acid is due to inductive effect and resonance stabilisation of
the carboxylate anion. Thus any group or atom, which is highly electronegative help in
removing the hydrogen atom as proton and the group or atom which is less electronegative
than C makes the removal of proton difficult.
Hence (–I) effect group increases acidic strength and (+I) effect groups decreases the acidic
strength of carboxylic acid.
Exercise 1: i) a) Why guanidine is basic in nature. Explain the site of protonation
and provide resonating structures.
b) Alkenes undergo electrophilic addition reaction and benzene
undergoes electrophilic substitution whereas both proceeds
through carbocation intermediate. Explain.
ii) (a) Arrange the various resonating structures of formic acid in order
of decreasing stability
b) The order of basicity for the following compound is
N
|
H
(i) (iii) (iv)
N
|
H
(ii)
O
N N
iii) a) 4- nitrophenol is more acidic than 3,5 - dimethyl –4-nitro-phenol.
Explain
b) The order of acidic strength of the following compound is
OH
(i)
C2H5OH
(ii)
OH
NO2
(iii) (iv)
OH
C
CH
||
O
3 

Illustration 1: Arrange the following in the increasing order of their acidic strength.
CH3COOH, FCH2COOH, Cl – CH2COOH BrCH2COOH, I – CH2–COOH
Solution: Acidic strength increase with increase in strength of (–I) group hence, the
correct order is
CH3COOH<ICH2COOH<BrCH2COOH<ClCH2COOH<F–CH2COOH
Illustration 2: Compare the acidic strength of the following.
CH3COOH ClCH2COOH Cl2HCCOOH CCl3COOH

Solution:
CH3COOH Cl C — C O H Cl C — C O H
Cl C — C O H
O
Cl
O
H
O
H
H
Cl
Cl
 

Increasing acidic strength due to increasing number of (–I) effect groups
v) Basic strength of Amines
The basic character of amines is due to presence of unshared electron pair on
nitrogen atom which accepts proton; the readiness with which the lone pair of
electrons available for protonotion determines the relative strength of amines.
Due to +I effect of alkyl group, the nitrogen atom becomes rich in electrons with the
result the lone pair of electron on nitrogen atom in amines is more easily available
than in ammonia and hence generally, amines are stronger bases than ammonia.
On the other (–I) groups or electron groups attached to nitrogen atom makes it
difficult for protonation.
Note: Relative basic strength of amines is not in total accordance with the inductive effect,
other factors like steric effect and stabilisation of cation by hydration also play
important role to determine the basic strength of amines.
Illustration 3: Compare the basic strength of the following
NH3, CH3NH2 (CH3)2NH (CF3)3N
Solution: The increasing order of basic strength is as follows.
F3C N:
CF3
CF3
H — N
H
H
H — N
H
H3C
H
CH3
N
CH3
  
CF3 group being – I effect groups attracts the electron pair of nitrogen
makes protonation difficult in primary and secondary amine due to
presence of +I effect group (CH3) which make protonation easy. Hence
the increasing order of basic strength is as follows.
Exercise 2: a) Write the following Alkenes in increasing order of their stability with
explanation
R2C=CR2, R2C=CHR, R2C=CH2, RCH=CH2, CH2=CH2
b) Can you think acidic nature of
(a) F—CH2COOH (b) Cl—CH2COOH
(c) Br—CH2COOH (d) I —CH2COOH

Resonance or Mesomeric Effect
The phenomenon in which two or more structures, involving identical position of atoms, can
be written for a particular compound, is called resonance. The various structures are called
as resonating structures or contributing structures.
The benzene molecule can be expressed as a resonance hybrid of the two contributing
Kekule structures. Such contributing structures are commonly known as canonical
structures.
or
This canonical structures of a system is a set of various structures which are sufficient to
define all the possible electron distributions.
i) Presence of a conjugate system (alternate double, single bond or single, multiple
bond)
C = C — C = C  C — C =C — C+
–
ii) Conjugated system, attached to electron deficient atom with vacant p-orbital.
C = C — C = C — G
iii) Conjugated system attached to electron rich atom or that atom should have filled
orbital or free lone pair
C = C — C = C— G
 
iv) All resonating structures must have same number of unpaired electrons
H2C
CH
CH2
 

= CH2
CH
CH2
This is not a proper R-structure of allyl radical 








2
2 H
C
CH
CH
v) All atoms that are a part of delocalized system must lie in a plane or be nearly planar
C —— C
CH2
C(CH3)3
CH2
(CH3)3C
vi) Equivalent resonance structures make equal contributions to the hybrid and a
system described by them has a large resonance stabilization

Cl
 



 Cl
 


 
Cl
 
Cl
 
Cl



 
Cl
I II III IV V VI
+ + +
Structure I and V are equivalent and have equal contribution in resonance
hybrid (VI)
vii) The more stable a structure is the greater is its contribution to the hybrid
H3C — C — CH — CH2 
CH3
+ –
H3C—C—CH = CH2  H3C—C = CH — CH2
CH3
+
CH3
+
I II
Structure (I) makes a larger contribution
Relative stability of Resonating structure:
a) The more covalent bonds a structure has, more stable it is
CH2 = CH — CH = CH2  CH2 — CH = CH — CH2  CH2 — CH = CH — CH2
+
–
–
+
I II III
b) Structures in which all of the atoms have a complete valence shell of electrons are
especially stable and make large contribution to the hybrid
CH2 — O — CH3  CH2 = O — CH3
+  
 
+
 
6 electron 8 electron
(I) (II)
Structure (II) is more stable
Mesomerism: It is the resonance effect of certain compounds under the influence of
particular group attached to the conjugated system M – effect is very popular in aromatic
compounds having one or more groups attached with aromatic nucleus It is of two types.
+M: When movement of electron starts from the group. (When group has a lone pair or
an extra electron).
A
 
A = – NH2,–
OH, X, —OR etc.
–M: When the movement of electrons takes place towards the group

B
B = –NO2, –SO3H, –C =O, –C N, CHO etc.
Note: Mesomeric effect is permanent in nature
Rules for writing Resonance structures
a) No Real existence: Resonance structures exist only on paper. Resonance structures
are useful because they allow us to describe molecules, radicals and ions for which a
single lewis structure is inadequate. We write two or more lewis structures, calling
them resonance structures or resonance contributors. We connect these structures
by double headed arrows () and we say that the real molecule, radical or ion is
a hybrid of all of them.
b) In writing resonance structures we are only allowed to move electrons
Position of nuclei of the atoms must remain same in all resonance structures eg.
CH3 — CH — CH = CH2  CH3— CH = CH — CH2
+ +
CH2 — CH2 — CH = CH2
+
R — Structures of allylic cation
c) Lewis Structures: All of the structures must be proper lewis structures
H — C = O — H
 
+
H
H
not a proper R- structure because
carbon is Penta valent
d) Charge separation should be low since, to separate, charge energy is required,
therefore, structure in which opposite charges are separated have greater energy
and hence less stable.
CH2 = CH — Cl  CH2 — CH = Cl
 
 




+
I II
(more stable)
Resonance Energy
The difference in energy between the hybrid and the most stable canonical structure is
called as Resonance energy
36 kcal
Total
energetic
structure
Resonance energy of benzene
Other canonical structure

Effect of Resonance
a) Dipole moment: Dipole moment of certain compounds can be explained by
resonance eg. Vinylchloride
CH2 = CH — Cl  H2C — CH = Cl
+
 = 1.4D
b) Bond length: the phenomenon of resonance explain the abnormal bond length
between C—C, C = C, C = O, etc in compounds exhibiting resonance e.g. in
benzene C—C bond length acquires a value which lies between C—C single bond
length (1.54Å) and C = C (double bond) length (1.33Å)
c) Strength of acids and bases: The concept of resonance explain clearly the acidic
character of acids and basic character of bases eg. resonance explains why the
alcohols are neutral and carboxylic acids are strong acids.
ROH RO–
+ H+
R—C
O
OH
R—C
O
O
–
+ H
+
But after loss of H+
carboxylate ion
R – C
O
O–
undergo resonance and stabilised,
hence it will favour the loss of H+
ions
R—C
O
O
–
R—C
O
–
O
d) Stability of free radicals and carbonium ions
(i) C6H5CH2

(ii) C6H5—C
(iii) C6H5—C
H
C6H5 C6H5
C6H5
(ii) 9–R- Structure
(iii) 13–R structure
Since number of resonating structures increases from I to II, hence stability also increases in
the same order.
Similarly we can explain the stability of carbonium ions.
Illustration 4: Explain, why phenol is acid while aliphatic alcohols are not.
Solution: After loss of H+
ion from OH group of phenol the remaining part
(Phenoxide ion) stabilizes by resonance, hence it will favour the loss of H+
ion and hence acidic in nature.

OH

 OH
+
 
3 more R-structure
–H+
(Charge separation is present. Not much stable)
O
 



 O
 


 
O
 


 
O


 
 
O




+ + +
(No charge separation, more stabilization by Resonance)
Illustration 5: Explain why aniline is less basic than ammonia or aliphatic amines
Solution: The lone pair present at the nitrogen in aniline is delocalised in the ring
(by resonance) and hence, it is not free for protonation, while in ammonia
or aliphatic amines it is present at nitrogen all the time, hence it is readily
available for protonation.
Electromeric Effect
It is a temporary effect in which a shared pair of electron ( - electron pair) is completely
transferred from a double bond or triple bond to one of the atoms joined by the bond at the
requirement of attacking reagent.
Case 1: When multiple bond is present between two similar atom (symmetric alkenes or
alkynes) electronic shift can take place in any direction
2
H
C
C
2
H
HBr
2
CH
C
2
H 


 



Note 1: If two carbon atoms are different (asymmetric alkenes or alkynes) then the
direction of electronic shift is determined by the direction of the inductive effect of
the group present at doubly or triply bonded atom e.g.,.
H3C  CH = CH2  CH3—CH —CH2 (Favoured by +I effect)
+ –
H3C  CH = CH2  CH3—CH —CH2
– +
(Opposite shift is not
possible because it will be
opposed by +I – effect)
Note 2: When inductive and electromeric effects oppose each other, in such cases,
electromeric effect usually overcome inductive effect eg.
H2C = CH — Br  Br
H
C
C
H2 


( E-effect operate in the direction of I-effect)
(I)
H2C = CH — Br H2C — CH = Br ( E-effect operate opposite to I-effect)
II
+

Conjugate effect
H2C = CH — Br H2C — CH — Br
  +
 
Since conjugate effect is stronger, it suppresses the I – effect and hence vinyl bromide
undergoes electromeric effect as (II)
Conjugate effect: When an electron transfer takes place from octet of an atom to that of
another atom without breaking its bond, it is known as conjugate effect
Case – II: When multiple bond is present between two different atoms electromeric shift
takes place towards more electronegative atom.
 C = O 







 

reagent
Attacking
:
O
C


+E and – E effect: When electron displacement takes place away from the atom or group it
is said to be +E and if it is towards the group, it is –E.
X — C = C— +E
 
X — C = C— –E
Hyperconjugation: it is the delocalisation of sigma electron. Also known as sigma-pi
conjugation or no bond resonance
Occurrence: Alkene, alkynes. Free radicals (saturated type) carbonium ions (saturated
type)
Condition: Presence of –H with respect to double bond, triple bond carbon containing
positive charge (in carbonium ion) or unpaired electron (in free radicals)
Example
H—C—CH = CH2
H
H
H—C—CH = CH2
H
H+
H+
C—CH = CH2
H
H
H —C—CH = CH2
H+
H
(I) (II) (III) (IV)
Note: Number of hyperconjugative structures = number of -Hydrogen hence, in above
examples structures ii,iii,iv are hyperconjugative structures (H-structures).
Hyperconjugation is a permanent effect
Effects of hyperconjugation
i) Bond Length: Like resonance, hyperconjugation also affects bond lengths because
during the process the single bond in compound acquires some double bond
character and vice-versa. E.g. C—C bond length in propene is 1.488 Å as compared
to 1.334Å in ethylene .

H—C——CH = CH2
H
H
1.488Å
1.353 Å
H——C =CH—CH2
H
–
H+
1.353 Å
1.488 Å
ii) Dipole moment : Since hyperconjugation causes the development of charges, it also
affects the dipole moment of the molecule.
iii) Stability of carbonium Ions
The order of stability of carbonium ions is as follows
Tertiary  Secondary  Primary
above order of stability can be explained by hyperconjugation. In general greater the
number of hydrogen atoms attached to -carbon atoms, the more hyperconjugative
forms can be written and thus greater will be the stability of carbonium ions.
Tertiary carbonium ion H3C—C+
CH3
CH3
H3C—C
CH3
CH2H+
(9 equivalent forms)
Secondary carbonium Ion H3C—C+
H
CH3
H3C—C
H
CH2H+
(6 equivalent form)
Primary carbonium Ion H—C+
H
CH3
H—C
H
CH2H+
(3 equivalent form)
(iv) Stability of Free radicals:
Stability of Free radicals can also be explained as that of carbonium ion
3
2
3
2
3
3
3
H
C
H
C
CH
H
C
)
CH
(
C
)
CH
(







v) Orientation influence of methyl group: The o,p-directing influence of the methyl group
in methyl benzenes is attributed partly to inductive and party of hyperconjugation
effect.

CH3
CH3 CH3
(orientation influence of the methyl group due to +I effect )
The role of hyperconjugation in o,p,-directing influence of methyl group is evidenced
by the fact that nitration of p-isopropyl toluene and p-tert-butyl toluene from the
product in which —NO2 group is introduced in the ortho position with respect to
methyl group and not to isopropyl or t-butyl group although the latter groups are
more electron donating than
H—C—H
H
H—C—CH3
CH3
H—C—H
H
H3C—C—CH3
CH3
Methyl groups i.e.., The substitution takes place contrary to inductive effect. Actually
this constitutes an example where hyperconjugation overpowers inductive effect.
Reactive Intermediates
Synthetic intermediate are stable products which are prepared, isolated and purified and
subsequently used as starting materials in a synthetic sequence. Reactive intermediate, on
the other hand, are short lived and their importance lies in the assignment of reaction
mechanisms on the pathway from the starting substrate to stable products. These reactive
intermediates are not isolated, but are detected by spectroscopic methods, or trapped
chemically or their presence is confirmed by indirect evidence.
Carbocations
Carbocations are the key intermediates in several reactions and particularly in nucleophilic
substitution reactions.
a) Structure: Generally, in the carbocations the positively charged carbon atom is
bonded to three other atoms and has no nonbonding electrons. It is sp2
hybridized
with a planar structure and bond angles of about 120°. There is a vacant
unhybridized p orbital which in the case of CH3
+
lies perpendicular to the plane of
C—H bonds.
b) Stability:There is an increase in carbocation stability with additional alkyl
substitution. Thus one finds that addition of HX to three typical olefins decreases in

the order (CH3)2C=CH2CH3—CH = CH2  CH2 = CH2. This is due to the relative
stabilities of the carbocations formed in the rate determining step which in turn
follows from the fact that the stability is increased by the electron releasing methyl
group (+I), three such groups being more effective than two, and two more effective
than one.
CH3 C+
 CH3—CH—CH3—CH2—CH3
+ +
CH3
CH3
Stabilized by three
electron – releasing
groups
Stability of carbocations 3° 2°  1°  CH3
+
Electron release : Disperses charge, stabilizes ion
Further, any structural feature which tends to reduce the electron deficiency at the
tricoordinate carbon stabilizes the carbocation. Thus when the positive carbon is in
conjugation with a double bond, the stability is more. This is so, because due to
resonance the positive charge is spread over two atoms instead of being
concentrated on only one. This explains the stability associated with the allylic cation.
The benzylic cations are stable, since one can draw canonical forms as for allylic
cations.
CH2
+
CH2
+
CH2
+
CH2
+
The benzyl cation stability is affected by the presence of substituents on the ring.
Electron donating p-methoxy and p-amino groups stabilize the carbocation by 14
and 26 kcal/mole, respectively. The electron withdrawing groups like p-nitro
destabilize by 20 kcal/mol.
A hetero atom with an unshared pair of electrons when present adjacent to the
cationic centre strongly stabilizes the carbocation. The methoxymethyl cation has
been obtained as a stable solid 

6
2
3
SbF
H
C
O
CH . Cyclopropylmethyl cations are even
more stable than the benzyl cations. This special stability is a result of conjugation
between the bent orbitals of the cyclopropyl ring and the vacant p orbital of the
cationic carbon. That the carbocations are planar is shown by the fact that these are
difficult or impossible to form at bridgeheads, where they cannot be planar.
The stability order of carbocation is explained by hyperconjugation. In vinyl cations
(CH2 = C+
H), resonance stability lacks completely and these therefore are very much
less stable.

Carbanions
a) Structure: A carbanion possesses an unshared pair of electron and thus represents
a base. The best likely description is that the central carbon atom is sp3
hybridized
with the unshared pair occupying one apex of the tetrahedron. Carbanions would
thus have pyramidal structures similar to those of amines. It is believed that
carbanions undergo a rapid interconversion between two pyramidal forms.
There is evidence for the sp3
nature of the central carbon and for its tetrahedral
structure. At bridgehead a carbon does not undergo reactions in which it must be
converted to a carbocation. However, the reactions which involve carbanions at such
centres take place with ease, and stable bridgehead carbanions are known. In case
this structure is correct and if all three R groups on a carbanions are different, the
carbanion should be chiral. All reactions therefore, which involve the formation of
chiral carbanion should give retention of configuration. However, this never happens
and has been explained due to an umbrella effect as in amines. Thus the unshared
pair and the central carbon rapidly oscillate from one side of the plane to the other.
b) Stability and Generation: The Grignard reagent is the best known member of a
broad class of substances, called organometallic compounds where carbon is
bonded to a metal lithium, potassium sodium, zinc, mercury, lead, thallium – almost
any metal known. Whatever the metal it is less electronegative than carbon, and the
carbon metal bond like the one in the Grignard reagent highly polar. Although the
organic group is not a full-fledged carbanion – an anion in which carbon carries
negative charge, it however, has carbanion character. Or organometallic compounds
can serve as a source from which carbon is readily transferred with its electrons. On
treatment with a metal, in RX the direction of the original dipole moment is reversed
(reverse polarization).
CH3CH2—Br+Mg  CH3CH2—Mg—Br
+ – – +
considerable carbanion character
Acetylene is ionized on treatment with amide ion in liquid ammonia to form a sodium
acetylide; this has a little covalent character and may be regarded as a true
carbanion. This property is used in making substituted alkynes. The stability order of
carbanions points to their high electron density. Alkyl groups and other electron –
donating groups in fact destabilize a carbanion. The order of stability is the opposite
of that for carbocations and free radicals, which are electron deficient and are
stabilized by alkyl groups. Based on this stability order it is easy to understand that
carbanions that occur as intermediates in organic reactions are almost always
bonded to stabilizing groups. An imporant method of preparation thus involves a
loss of proton from a haloform to afford a stabilized carbanion. Another factor which
leads to stability is resonance e.g., a carbonyl group stabilizes an adjacent carbanion
via resonance e.g., a carbonyl group stabilizes an adjacent carbanion via overlap of
its pi bond with the nonbonding electrons of the carbanion. Carbanions derived from
carbonyl compounds are often called enolate anions. Among the other functional

groups which exert a strong stabilizing effect on carbanions are nitro and cyano
groups. The second row elements, particularly phosphorus and sulphur stabilize the
adjacent carbanions. A very important nucleophilic carbon species constitute the
phosphorus and sulphur ylide. The preparation of ylide is a two stage process, each
state of which belongs to a familiar reaction type: nucleophilic attack on an alkyl
halide, and abstraction of a proton by a base.
R—CHX + Ph3P  Ph3
+
P—CH—R + X–

 
base
Ph3P = C—R
R R R
Triphenyl-phosphine Phosphonium salt An ylide
Ph3P=C—R  Ph3P—C—R
 
R R
+
The phosphorus ylide have hybrid structure, and it is the negative charge on carbon
i.e, the carbanion character of ylide which is responsible for their characteristic
reactions. The sulphur atoms stabilize carbanions. When a double or triple bond is
located  to the carbanionic carbon the ion is stabilized by resonance as in the case
of benzylic type carbanions .
c) Properties: Carbanions are nucleophilic and basic and in this behaviour these are
similar to amines, since the carbanion has a negative charge on its carbon, to make
it a powerful base and a stronger nucleophile than an amine. Consequently a
carbanion is enough basic to remove a proton from ammonia.
Free Radicals
a) Structure and Geometry: A free radical is a species which has one or more
unpaired electrons. In the species where all electrons are paired the total magnetic
moment is zero. In radicals, however, since there are one or more unpaired
electrons, there is a net magnetic moment and the radicals as a result are
paramagnetic. Free radicals are usually detected by electron spin resonance, which
is also termed electron paramagnetic resonance.
Simple alkyl radicals have a planar (trigonal) structure, i.e., these have sp2
bonding
with the odd electron in a p orbital. The pyramidal structure is another possibility
when the bonding may be sp3
and the odd electron is in an sp3
orbital. The planar
structure is in keeping with loss of activity when a free radical is generated at a chiral
center. Thus, a planar radical will be attacked at either face after its formation with
equal probability to give enantiomers. Unlike carbocations, the free radicals can be
generated at bridgehead shows that pyramidal geometry for radicals is also possible
and that free radicals need to be planar.
C
H
H
H
p-orbital
C
H
H
H
sp
3
hybridized orbital
H
Trigonal structure Pyramidal structure

b) Stability: As in the case of carbocation, the stability of free radicals is tertiary 
secondary  primary and is explained on the basis of hyperconjugation. The
stabilizing effects in allylic radicals and benzyl radicals is due to vinyl and phenyl
groups in terms of resonance structures. Bond dissociation energies shown that 19
kcal / mol less energy is needed to form the benzyl radical from toluene than the
formation of methyl radical from methane. The triphenyl methyl type radicals are no
doubt stabilized by resonance, however, the major cause of their stability is the steric
hindrance to dimerization.
C6H5CH3  C6H5CH2 + H H = + 85 kcal
Toluene Benzyl radical
Ease of formation
of free radicals Benzyl  allyl  3°  2° 1°  CH3
°
 vinyl
Carbenes
Carbenes are neutral intermediates having bivalent carbon, in which a carbon atom is
covalently bonded to two other groups and has two valency electrons distributed between
two non bonding orbitals. When the two electrons are spin paired the carbene is a singlet, if
the spins of the electrons are parallel it is a triplet.
a) Structure: A singlet carbene is thought to possess a bent sp2
hybrid structure in
which the paired electrons occupy the vacant sp2
orbital. A triplet carbene can be
either bent sp2
hybrid with an electron in each unoccupied orbital, or a linear sp
hybrid with an electron in each of the unoccupied p-orbital. It has however, been
shown that several carbenes are in a non-linear triplet ground state. However, the
dihalogenocarbenes and carbenes with oxygen, nitrogen and sulphur atoms
attached to the bivalent carbon, exist probably as singlets. The singlet and triplet
state of a carbene display different chemical behaviour. Thus addition of singlet
carbenes to olefinic double bond to form cyclopropane derivatives is much more
stereoselective than addition of triplet carbenes.
b) Generation: Carbenes are obtained by thermal or photochemical decomposition of
diazoalkanes. These can also be obtained by -elimination of a hydrogen halide
from a haloform with base, or of a halogen from a gem dihalide with a metal.
c) Reactions: These add to carbon double bonds and also to aromatic systems and in
the later case the initial product rearranges to give ring enlargement products (a car-
benoids –oranometallic or complexed intermediates which, while not free carbenes
afford products expected from carbenes are usually called carbenoids).
When a carbene is generated in a three membered ring allenes are formed by
rearrangement. However, a similar formation at a cyclopropylmethyl carbon gives
ring expansion. Carbenes are also involved in Reimer —Tiemann reaction .
Carbene CH2
electrocylic
rearrangement
benzene norcaradiene
intermediate
cycloheptatriene

Nitrenes
a) Structure: The nitrenes R—N represent the nitrogen analogous of carbenes and
may be generated in the singlet 












N
R or triplet state 




 




N
R . A nitrene can be
generated via elimination or by the thermal decomposition of azides (R—N=N+
=N–

RN + N2).
R—N—OSO2Ar 
 
base
R—N + B + ArSO2O
H
b) Reactions: In their chemical behaviour, nitrenes are similar to carbenes. Nitrenes,
(in particular acyl nitrenes) get inserted into some bonds e.g. a C—H bond to give an
amide. Aziridines are formed when nitrenes add to C = C bonds.
Arenium Ions
A considerable amount of experimental evidence indicates that electrophiles attack the 
system of benzene to form a delocalized non-aromatic carbocation known as arenium ion or
sometimes a  complex CMR spectroscopic evidence is available in favour of  complex.
Benzynes
It is a reactive intermediate in some nucleophilic aromatic substitutions. It is a benzene with
two hydrogen atoms removed. It is usually drawn with a highly strained triple bond in the six
membered ring. Benzyne intermediate has been observed spectroscopically and trapped .
Exercise-3: a) Predict the preferred regiochemistry for the addition of HCl to each of the
following compounds on the basis of carbocation stability.
(a) (b) (c) (d)
Ph
b) Why benzyl carbonium ion is more stable than ethyl carbonium ion
c) Addition of HCl on 1,3 butadiene gives two products. Explain
d) Explain acidic nature of vinyl alcohol
ISOMERISM
In the study of organic chemistry we come across many cases when two or more
compounds are made of equal number of like atoms. A molecular formula does not tell the
nature of organic compound; sometimes several organic compounds may have same
molecular formula. These compounds possess the same molecular formula but differ from
each other in physical or chemical properties, are called isomers and the phenomenon is
termed isomerism (Greek, isos = equal; meros = parts). Since isomers have the same
molecular formula, the difference in their properties must be due to different modes of the
combination or arrangement of atoms within the molecule. Broadly speaking, isomerism is of
two types.

i) Structural Isomerism
ii) Stereoisomerism
i) Structural isomerism: When the isomerism is simply due to difference in the
arrangement of atoms within the molecule without any reference to space, the
phenomenon is termed structural isomerism. In other words, while they have same
molecular formulas they possess different structural formulas. This type of isomerism
which arises from difference in the structure of molecules, includes:
a) Chain or Nuclear Isomerism;
b) Positional Isomerism
c) Functional Isomerism
d) Metamerism and
e) Tautomerism
ii) Stereoisomerism: When isomerism is caused by the different arrangements of
atoms or groups in space, the phenomenon is called Stereoisomerism (Greek,
Stereos = occupying space). The stereoisomers have the same structural formulas
but differ in the spatial arrangement of atoms or groups in the molecule. In other
words, stereoisomerism is exhibited by such compounds which have identical
molecular structure but different configurations.
Stereoisomerism is of two types :
a) Geometrical or cis-trans isomerism ; and
b) Optical Isomerism.
Thus various types of isomerism could be summarised as follows.
ISOMERISM
STRUCTURAL
ISOMERISM
STEREOISOMERISM
GEOMETRICAL
ISOMERISM
FUNCTIONAL
ISOMERISM
POSITIONAL
ISOMERISM
CHAIN
ISOMERISM
METAMERISM TAUTOMERISM
OPTICAL
ISOMERISM

Chain or Nuclear Isomerism
This type of isomerism arises from the difference in the structure of carbon chain which
forms the nucleus of the molecule. It is, therefore, named as chain, nuclear isomerism or
Skeletal isomerism. For example, there are known two butanes which have the same
molecular formula (C4H10) but differ in the structure of the carbon chains in their molecules.
CH3–CH2–CH2–CH3 H3C–CH–CH3
n-butane |
CH3
isobutane
While n-butane has a continuous chain of four carbon atoms, isobutane has a branched
chain. These chain isomers have somewhat different physical and chemical properties, n-
butane boiling at -0.5o
and isobutane at -10.2o
. This kind of isomerism is also shown by
other classes of compounds. Thus n-butyl alcohol and isobutyl alcohol having the same
molecular formula C4H9OH are chain isomers.
CH3–CH2–CH2–CH2OH CH3–CH–CH2OH
n-butyl alcohol |
CH3
isobutyl alcohol
It may be understood clearly that the molecules of chain isomers differ only in respect of the
linking of the carbon atoms in the alkanes or in the alkyl radicals present in other
compounds.
Illustration 6: How many chain isomers does butane have?
Solution: CH3—CH2—CH2=CH2
n-Butylene (1-Butene)
CH3—C=CH2
CH3
Isobutylene (2-Methylpropene)
Illustration 7 : How many chain isomers does propyl benzene have?
Solution:
CH2CH2CH3
n-Propylbenzene
CH—CH3
Isopropylbenzene
CH3
Illustration 8: Give the possible chain isomers for ethyl benzene.
Solution: CH2CH3
Ethylbenzene
CH3
Xylene (o,m,p)
CH3

Positional Isomerism
It is the type of isomerism in which the compounds possessing same molecular formula
differ in their properties due to the difference in their properties due to difference in the
position of either the functional group or the multiple bond or the branched chain attached to
the main carbon chain. For example, n-propyl alcohol and isopropyl alcohol are the
positional isomers.
OH
|
CH3–CH2–CH2–OH CH3–CH–CH3
n-propyl alcohol isopropyl alcohol
Butene also has two positional isomers:
CH2=CH–CH2–CH3 CH3–CH=CH–CH3
1-butene 2-butene
1-Chlrobutane and 3-Chlorobutane are also the positional isomers:
e
tan
Chlorobu
1
Cl
CH
CH
CH
CH 2
2
2
3




e
tan
Chlorobu
2
3
2
3 CH
CHCl
CH
CH



Methylpentane too has two positional isomers:
CH3—CH2—CH2—CH—CH3
CH3
2-Methylpentane
CH3—CH2—CH—CH2—CH3
CH3
3-Methylpentane
In the aromatic series, the disubstitution products of benzene also exhibit positional
isomerism due to different relative positions occupied by the two substituents on the
benzene ring. Thus xylene, C6H4(CH3)2, exists in the following three forms which are
positional isomers.
CH3
CH3
o-Xylene
CH3
CH3
m-Xylene
CH3
CH3
p-Xylene
Functional Isomerism
When any two compounds have the same molecular formula but possess different
functional groups, they are called functional isomers and the phenomenon is termed
functional isomerism. In other words substances with the same molecular formula but
belonging to different classes of compounds exhibit functional isomerism. Thus,

1. Diethyl ether and butyl alcohol both have the molecular formula C4H6O, but contain
different functional groups.
C2H5–O–C2H5 C4H9–OH
diethyl ether butyl alcohol
The functional group in diethyl ether is (–O–), while is butyl alcohol it is (–OH).
2. Acetone and propionaldehyde both with the molecular formula C3H6O are functional
isomers.
CH3–CO–CH3 CH3–CH2–CHO
acetone acetaldehyde
In acetone the functional group is (–CO–), while in acetaldehyde it is (–CHO).
3. Cyanides are isomeric with isocyanides:
cyanide
Alkyl
RCN
isocyanide
Alkyl
RNC
4. Carboxylic acids are isomeric with esters.
acid
opanoic
Pr
COOH
CH
CH 2
3
ethanoate
l
Methy
3
3COOCH
CH
5. Nitroalkanes are isomeric with alkyl nitrites:
R—N
Nitroalkane
R—O—N=O
Alkyl nitrite
O
O
6. Sometimes a double bond containing compound may be isomeric with a triple bond
containing compound. This also is called as functional isomerism. Thus, butyne is
isomeric with butadiene (molecular formula C4H6).
Butyne
1
CH
C
CH
CH 2
3



Butadiene
3
,
1
CH
CH
CH
CH 2
2




7. Unsaturated alcohols are isomeric with aldehydes. Thus,
alcohol
Vinyl
OH
CH
CH2 

de
Acetaldehy
CHO
CH3
8. Unsaturated alcohols containing three or more carbon atoms are isomeric to
aldehydes as well as ketones:
alcohol
Allyl
OH
CH
CH
CH 2
2 

yde
opionaldeh
Pr
CHO
CH
CH 2
3
Acetone
COCH
CH 3
3
9. Aromatic alcohols may be isomeric with phenols:
CH2OH
Benzyl alcohol
CH3
o-Cresol
OH

10. Primary,secondary and tertiary amines of same molecular formula are also the
functional isomers.
)
o
1
(
e
min
propyla
n
2
2
2
3 NH
CH
CH
CH
 )
o
(2
e
min
la
Ethylmethy
5
2
3 H
C
NH
CH 

Trimethylamine(3°)
CH3—N—CH3
CH3
11. Alkenes are isomeric with cycloalkanes:
Butene
CH3CH2CH=CH2
Cyclobutane
CH3
Methylcyclopropane
Such isomers in which one is cyclic and other is open chain, are called ring-chain isomers.
Alkynes and alkadienes are isomeric with cycloalkenes.
1-Butyne
CH3CH2CCH
Cyclobutene
CH2=CH—CH=CH2
1,3-Butadiene
Metamerism
This type of isomerism is due to the unequal distribution of carbon atoms on either side of
the functional group in the molecule of compounds belonging to the same class. For
example, methyl propyl ether and diethyl ether both have the same molecular formula.
CH3–CO–C3H7 C2H5–O–C2H5
methyl propyl ether diethyl ether
in methyl propyl ether the chain is 1 and 3, while in diethyl ether it is 2 and 2. This isomerism
known as Metamerism is shown by members of classes such as ethers, and amines where
the central functional group is flanked by two chains. The individual isomers are known as
Metamers.
Examples:
Triethylamine
C2H5—N—C2H5
C2H5
Ethylmethylpropylamine
C3H7—N—CH3
C2H5
Butyldimethylamine
C4H9—N—CH3
CH3
Tautomerism
It is the type of isomerism in which two functional isomers exist together in equilibrium. The
two forms existing in equilibrium are called as tautomers. For example, the compound
acetoacetic ester has two tautomers – one has a keto group and other has an enol group:

Keto-form
CH3—C—CH2—COOC2H5
enol-form
CH3—C=CH—COOC2H5
OH
O
Out of the two tautomeric forms, one is more stable and exists in larger proportion. In above,
normally 93% of the keto form (more stable) and only 7% of the enol form (less stable i.e.
labile) exist.
The equilibrium between the two forms is dynamic, i.e., if one form is somehow removed by
making a reaction, some of the amount of the other form changes into the first form so that
similar equilibrium exists again. Thus, whole of the acetoacetic ester shows the properties of
both ketonic group as well as the enolic group. Thus, it adds on HCN, NaHSO3 etc. due to
the presence of C=O group and it decolourises bromine water and gives dark colouration
with FeCl3 due the presence of C—OH group. Due to the presence of keto and enol form
this type of tautomerism is known as keto-enol tautomerism. It is the most commonly
observed type of tautomerism.
Keto-enol tautomerism is generally observed in those compounds in which either a methyl
(CH3—), methylene (—CH2—), or a methyne ( 
 H
C
|
) group is present adjacent to a
carbonyl (—CO—) group as in acetoacetic ester above. In other words, it can be said that
keto-enol tautomerism is possible in only those carbonyl compounds in which atleast one -
hydrogen atom is present so that it may convert the carbonyl group to enol group.
Another example of keto-enol tautomerism is :
Keto-form
CH3—C—CH2—C—CH3
2,4 - Pentanedione
CH3—C=CH—C—CH3
OH
O O O
It is found that if the -hydrogen atoms are present on both the carbons attached to carbonyl
group, more stable is the enol form and hence more its content. Thus, larger the number of
-hydrogens in a ketone, more is enol content. Also, if number of-hydrogen containing
carbonyl groups is more, again more is the enol content. Thus the order of enol content is:
CH3CHO  CH3COCH3  CH3COCH2CHO  CH3COCH2COCH3
CH3COCH2COH3 has about 75% enol content. Moreover, the enol form shows acidic nature
due to the tendency to liberate proton(H+
) from the enol ( C—OH) group. This H is the -
hydrogen in keto – form. Therefore, -H in carbonyl compound is acidic in nature. More the
enol content, more is the acidic nature of -hydrogens in a carbonyl compound. Thus above
is also the increasing order of acidity of -hydrogens.
An interesting observation about the enol content in acetoacetic ester is the fact that above
mentioned percentage ratio (93:7) is in its aqueous solution. In liquid state this ratio is nearly
25 : 75. It is because in liquid state the enol form is much stabilised by intramolecular H-
bonding.

O
H3C—C O—OC2H5
O
CH2
O
H3C—C O—OC2H5
O
CH
H
keto enol
In aqueous solution the intermolecular H-bonding with water takes place and it dominates
the intramolecular H-bonding, resulting in lower enol content.
Keto-enol tautomerism exists in cyclic carbonyl compounds also if they fulfil the condition of
presence of -H. Thus, we have
O —OH
O
O
O
—OH O
OH
O
O OH
OH
O OH
O
But the compound cannot exist as tautomers since -H is already at unsaturated carbon.
O
T
automerism is also termed as desmotropism (desomo= bond, tropis = turn) because in
tautomers the bonding changes.
Every compound having skeleton 

 2
CH
C
||
O
has its tautomer skeleton 

 CH
C
H
|
O
e.g., in
the preparation of 3
3 CH
C
CH
||
O

 a very small amount of enol isomers 2
3 CH
C
CH
H
|
O

 also
forms which can be isolated. The two exist in equilibrium with each other and can be
separated by suitable methods.
A hydroxy group attached to a carbon which is itself attached to another carbon atom by a
double bond is known as enolic (en for double bond, ol for alcohol). Its nature becomes
acidic as in phenol and unlike OH group in alcohol which is neutral or only very slightly
acidic (C2H5OH + Na  C2H5ONa +
2
1
H2). In the above two examples migration of a proton
from one carbon atom to another takes place with simultaneous shifting of bonds.
Hydrocyanic acid, H – C  N and Isohydrocyanic acid H — N  C are also tautomeric
isomers or tautomers.

Difference between Tautomerism & Resonance
a) In tautomerism, an atom changes place but resonance involves a change of position
of pi-electrons or unshared electrons.
b) Tautomers are different compounds and they can be separated by suitable methods
but resonating structures cannot be separated as they are imaginary structures of
the same compound.
c) Two tautomers have different functional groups but there is same functional group in
all canonical structures of a resonance hybrid.
d) Two tautomers are in dynamic equilibrium but in resonance only one compound
exists.
e) Resonance in a molecule lowers the energy and thus stabilises a compound and
decreases its reactivity. But no such effects occur in tautomerism.
f) In resonance, bond length of single bond decreases and that of double bond
increases e.g. all six C—C bonds in benzene are equal and length is in between the
length of a single and a double bond.
g) Resonance occurs in planar molecule but atoms of tautomers may remain in different
planes as well.
h) Tautomers are indicated by double arrow in between the two isomers but
double headed single arrow  is put between the canonical (resonating)
structures of a resonating molecule.
Stereoisomerism
The isomers which differ only in the orientation of atoms in space are known as
stereoisomerism. It’s of two types.
a) Geometrical isomerism: Isomers which posses the same molecular and structural
formula but differ in arrangement of atoms or groups in space around the double
bonds, are known as geometrical isomers and the phenomenon is known as
geometrical isomerism. Geometrical isomerism are show by the compounds having
the structure.
C = C
X
Y
X
Y
i) Cis – trans isomerism: When similar groups are on the same side it is cis and it
same groups are on the opposite side it is trans isomerism.
C = C
COOH
H
HOOC
H
(Maleic acid)
(cis fom)
C = C
COOH
H
HOOC
H
Fumeric acid
(trans fom)

ii) Syn - anti
C
H
(Syn)
(aldoxime)
N
CH3
OH
C
H
(anti)
N
CH3
OH
C6H5—N
(Syn)
C6H5—N
C6H5—N
(anti)
N— C6H5
Syn anti isomerism is not possible in ketoxime since only one from is possible
two —CH3 groups are at one C.
C
N
CH3
OH
CH3
iii) In cyclic compounds:
Cis
OH
OH
trans
OH
OH
Differentiating properties of cis-trans isomerism
i) Dipolemoment : Usually dipole moment of cis is larger than the trans-isomer.
C
C
Cl
H
Cl
H
 = 1.84D
C
C
Cl H
Cl
H
 = 0
ii) Melting point: The steric repulsion of the group (same) makes the cis isomer less
stable than the trans isomers hence trans form has higher melting point than cis.
iii) Different chemical properties: Syn-addition makes cis forms into meso and trans into
d and l, anti addition makes cis into d – l and trans into meso.
meso
addn)
,
syn
(
KMnO
.
alc 4




 

C
H
C
C
OH
OH
H
H
COOH
COOH
C
COOH
H COOH
addn.
anti
water
Br2 



 
 C
C
Br
H
H
COOH
COOH
Br
d & l

d & l
addn)
,
syn
(
KMnO
.
alc 4




 

C
H
C
C
OH
HO H
H
COOH
COOH
C
COOH
H
HOOC
addn.
anti
water
Br2 



 
 C
C
Br
H
H
COOH
COOH
Br
meso
H—C—COOH
H—C—COOH


Maleic acid
C
C
H
H
CO
CO
O + H2O
Maleic anhydride
H—C—COOH
HOOC—C—H


no reaction
b) Optical isomerism: Any substance which rotates the plane polarised light (PPL) is
said to be optically active. If a substance is optically active, it is non - superimposable
on its mirror image. If a molecule of substance is superimposable on its mirror image,
it cannot rotate PPL and hence optically inactive. The property of non-
superimposability on mirror image is called chirality. The ultimate criterion for optical
activity is chirality , i.e., non-superimposibility on its mirror image.
If a molecule of organic compound contains 'one' chiral carbon, it must be chiral and
hence optically active.
Chiral carbon: If all the four bonds of carbon are satisfied by four different atoms / groups, it
is chiral. Here it should be noted that isotopes are regarded as different atoms / groups )
chiral carbon is designated by an asterisk(*).
Optical isomerism in bromochloroiodomethane: The structural formula of
bromochloroiodomethane is
H—C—Cl
I
Br
*
The molecule has one chiral carbon as designated by star. So molecule is chiral. It is non -
superimposable on its mirror image.
According to Van't Hoff rule,
Total number of optical isomers should be = 2n
; when n is number of chiral centre
The Fischer projections of the two isomers are
(i)
Br
H——— I
Cl
Br
I——— H
Cl
(ii)
Mirror

 Non-super imposable
 Can rotate PPL
 Optically active
Stereoisomers which are mirror - image of each other are called enantiomers or
enantiomorphs. (i) and (ii) are enantiomers. All the physical and chemical properties of
enantiomers are same except two:
i) They rotate PPL to the same extent but in opposite direction. One which rotates PPL in
clockwise direction is called dextro-rotatory (dextro is Latin word meaning thereby right)
and is designated by d or (+). One which rotates PPL in anti-clockwise direction is called
laevo rotatory (means towards left) and designated by  or (–).
ii) they react with optically active compounds with different rates.
Illustration 9: State whether the following compound is optically active or inactive.
H3C—C—CH2—CH3
H
H
2
Solution:
H3C—C—CH2—CH3
H
H
2
or H3C—C—CH2—CH3
D
H
*
The molecule contains one chiral carbon and so must be optically active.
Isotopes are regarded as different atom.
Optical isomerism in compounds having more than one chiral carbons
If an organic molecule contains more than one chiral carbons then the molecule may be
chiral or achiral depending whether it has element of symmetry or not.
Elements of symmetry: If a molecule have either
a) a plane of symmetry, and / or
b) centre of symmetry, and / or
c) n-fold alternating axis of symmetry
If an object is superimposable on its mirror image; it cannot rotate PPL and hence optically
inactive. If an object can be cut exactly into two equal halves so that half of its become
mirror image of other half., it has plane of symmetry.

Centre of symmetry: It is a point inside a molecule from which on travelling equal distance
in opposite directions one takes equal time.
A
B
B
C
A
B
B
C
Thus, if an organic molecule contains more than one chiral carbons but also have any
elements of symmetry, it is superimposable on its mirror - image, cannot rotate PPL and
optically inactive. If the molecule have more than one chiral centres but not have any
element of symmetry, it must be chiral.
Stereoisomerism in 2,3-dibromopentane
The structural formula of 2,3-dibromopentane is
H3C—C—C—CH2—CH3
H
*
Br Br
H
*
The molecule contains two chiral carbons and hence according to Van't Hoff rule the total
number of optical isomers should be 2n
= 22
= 4 and it is. The four optical isomers are.
(i)
H———Br
CH3
(ii)
H———Br
CH2CH3
Br———H
H3C
Br———H
H3CH2C
H———Br
CH3
Br———H
CH2CH3
(iii)
Br———H
H3C
H———Br
H3CH2C
(iv)

 Nonsuperimposable  Non -superimposable
 Rotate PPL  Rotate PPL
 Optically active  Optically active
I,II,III and IV are four stereoisomers of 2,3-dibromopentane.
I and III are enantiomers.
III and IV are also enantiomers
What is relation between I and III; or I and IV; or II and III; or II and IV?
All these pairs are diastereomers, stereoisomers which are not mirror - image of each other
are called diastereomers.
Therefore,
I and III are diastereomers
I and IV are diastereomers
II and III are diastereomers
II and IV are diastereomers
Stereoisomerism in Tartaric Acid
The IUPAC name of tartaric acid is 2,3 - dihydroxy butandioic acid. The structural formula is
COOH
HOH
C
HOH
C
HOOC
*
*



The molecule contains two chiral carbon and the number of optical isomers should be
2n
=22
=4; but number of optical isomers reduces to 3 because one molecule has plane of
symmetry.
Stereoisomers of Tartaric Acid
(i)
H———OH
COOH
HO———H
COOH
(ii)
HO———H
HOOC
H———OH
HOOC
(iii)
H———OH
COOH
H———OH
HOOC
---------------- Plane of symmetry
HO———H
HOOC
HO———H
HOOC
(iv)
 Non - superimposable  Superimposable
 Can rotate PPL III and IV are same
 Optically active Rotation of IV by 180° yield III.
Superimposability means same in identity
I and II are enantiomers
III is meso-form of tartaric acid.

A meso compound is one which is optically inactive although have more than one chiral
carbons.
Number of Optical Isomers
Number of possible optical isomers in compounds containing different no. of asymmetric
atoms.
1. The molecule has no symmetry
The no. of d and l – forms a = 2n
n = no. of asymetric atoms
The no. of meso l- forms m = 0
Total no. of optical isomers = a + m = 2n
2. The molecule has symmetry
The no. of d and l forms a = 2n -1
Meso forms m = 2n–1
Total = a + m
It is add a = 2n–1
, m = 2
1
n
2

e.g. Lactic acid
H—C—OH
COOH
CH3
n = 1 no symmetry
d & l forms = 2n
= 2
meso form = 0
total = 2
H—C—OH
COOH
CH3
OH—C—H
COOH
CH3
d-form l-form
Mirror images are also
called as enantiomers
3. Tartaric acid
H—C—OH
COOH
Plane of syn n =2
H—C—OH
COOH

a = 2n–1
= 22–1
m =
1
2
n
2

= 21–1
= 2° = 1
Total = 2 +1 = 3
H—C—OH
COOH
(meso forms)
HO—C—H
COOH
I
COOH
H—C—OH
COOH
HO—C—H H—C—OH
COOH
H—C—OH
COOH
II III
One of asymmetric centre is rotating plane polarised light towards right other towards
left so total optical rotation is zero.
I & III or II & III which are not mirror image to each other are called diastereomers.
Difference between racemic mixture and meso compound
A racemic mixture contains equimolar amounts of enantiomers. It is optically inactive due to
external compensation. It can be resolved into optically active forms. A meso compound is
optically inactive due to internal compensation.
Optically active compounds having no chiral carbon
The presence of chiral carbon is neither a necessary nor a sufficient condition for optical
activity, since optical activity may be present in molecules with no chiral atom and since
molecules with two or more chiral carbon atoms are superimposable on their mirror images
and hence inactive.
i) Any molecule containing an atom that has four bonds pointing to the corners of a
tetrahedron will be optically active if the four groups are different.
—CH2—S— —CH3
18O
16
O
ii) Atoms with pyramidal bonding might be expected to give rise to optical activity if the
atom is connected to three different groups, since the unshared pair of electron is
analogous to a fourth group.
N
Z
Y
X
Many attempts have been made to resolve such compounds, but until recently all failed
because of umbrella effect, also called pyramidal inversion. The umbrella effect is rapid
oscillation of the unshared pair form one side of XYZ plane to the other.

iii) Biphenyls, containing four large groups in the ortho position so that there is restricted
rotation, are optically active if the rings are asymmetrical.
If either or both rings are symmetrical, the molecule has plane of symmetry and optically
inactive.
NO2
HOOC
NO2
COOH
Cl
'A'
'B'
Mirror
NO2
COOH
Cl
'B'
NO2
Cl
COOH
Ring B is symmetrical (having plane of symmetry) superimposable optically inactive.
NO2
NO2
COOH
Non-superimposable
optically active.
COOH
NO2
Cl
NO2
COOH
COOH
Allenes, with even number of cumulative double bonds are optically active if both sides
are dissymmetric.
H3C
C=C=C
Inactive
H
H
H3C
H3C
C=C=C
Active
H
H CH3
Specific Rotation
The specific rotation  T

 is an inherent physical property of an enantiomer, which varies
with the solvent used, temp (in °C) and wavelength of the light used. It is calculated from the
observed rotation  as follows.
 T

 =
c
l

Where  = length of tube in decimeters (dm)
C = Concentration in gram cm–3
, for a solution density in gcm–3
, for a pure luqid.
Illustration 10: The specific rotation of R— (—) –2– bromooctane is –36°. What is
percentage composition of a mixture of enantiomers of 2-bromooctane
where rotation is +18°.

Solution: Let X = mole fraction of R,
 (1–X) = mole fraction S.
X(–36°) + (1–X) 36° = 18°
or, X =
4
1
The mixture has 25% R and 75%, it is 50% racemic and 50% S.
E and Z nomenclature of Geometrical Isomerism
If all the four groups/ atoms attached to C = C double bond are different, then Cis and trans
nomenclature fails in such cases and a new nomenclature called E and Z system of
nomenclature replace it.
C
C
H I
Br Cl
higher priority
higher priority
C
C
H I
Br
Cl
higher ranked
E Z
I II
The group / atom attached to carbon - carbon double bond is given to higher rank, whose
atomic weight is higher. If the two higher ranked group are across, it is called E form (E
stands for the German word entgeger meaning thereby opposite) and the two higher ranked
groups are on the same side, they are called Z-form (Z stands for German word Zusammen
meaning thereby on the same side).
In general trans - isomer is more stable then cis isomer because in cis from, there will be
more interaction in between groups.
Besides suitably substituted alkene and cycloalkane, suitably substituted oximes and azo
compounds also exhibit geometrical isomerism.
H
C=N
OH
R
syn-oxime (Aldoxime)
H
C=N
R OH
Anti-oxime (Aldoxime)
R
C=N
R
R-syn-oxime or
R-anti-oxime
R
C=N
R
OH
R-Syn-oxime or
R-anti-oxime
OH
N=N
R
Cis-azo
N=N
R
anti-azo
R

Nomenclature of Optical Isomers
1. Absolute and Relative Configuration
While discussing optical isomerism, we must distinguish between relative and
absolute configuration (arrangement of atoms or groups) about the asymmetric
carbon atom. Let us consider a pair of enantiomers, say (+)-
and (–)-
lactic acid.
COOH
H––– C–––OH
CH3
COOH
HO–––C–––H
CH3
We know that they differ from one another in the direction in which they rotate the
plane of polarised light. In other words, we know their relative configuration in the
sense that one is of opposite configuration to the other. But we have no
knowledge of the absolute configuration of the either isomer. That is, we cannot
tell as to which of the two possible configuration corresponds to (+) - acid and
which to the (–) - acid.
D and L system
The sign of rotation of plane-polarized light by an enantiomer cannot be easily
related to either its absolute or relative configuration. Compounds with similar
configuration at the asymmetric carbon atom may have opposite sign of rotations
and compounds with different configuration may have same sign of rotation.
Thus d-lactic acid with a specific rotation + 3.82o
gives l-methyl lactate with a
specific rotation -8.25°, although the configuration (or arrangement) about the
asymmetric carbon atom remains the same during the change.
CO2H CO2CH3
| |
H––C––OH H––C––OH
| |
CH3 CH3
+ 3.82 - 8.25o
Obviously there appears to be no relation between configuration and sign of
rotation. Thus D-L-system has been used to specify the configuration at the
asymmetric carbon atom. In this system, the configuration of an enantiomer is
related to a standard, glyceraldehyde. The two forms of glyceraldehyde were
arbitrarily assigned the absolute configurations as shown below.
CHO CHO
| |
H––C––OH HO––C––H
| |
CH2OH CH2OH
(+)-glyceraldehyde (–)-glyceraldehyde
D configuration L configuration
(–)-lactic acid
(+)-lactic acid

If the configuration at the asymmetric carbon atom of a compound can be related
to D (+)-glyceraldehyde, it belongs to D-series; and if it can be related to L(–)-
glyceraldehyde, the compound belongs to L-series. Thus many of the naturally
occurring -amino acids have been correlated with glyceraldehyde by chemical
transformations. For example, natural alanine (2-aminopropanoic acid) has been
related to L(+)-lactic acid which is related to L(–)-glyceraldehyde. Alanine,
therefore, belongs to the L-series. In general, the absolute configuration of a
substituent (X) at the asymmetric centre is specified by writing the projection
formula with the carbon chain vertical and the lowest number carbon at the top.
The D configuration is then the one that has the substituent 'X' on the bond
extending to the 'right' of the asymmetric carbon, whereas the L configuration has
the substituent 'X' on the 'left'. Thus,
R1 R1
| |
R2 ––C––X X––C––R2
| |
R3 R3
D configuration L configuration
When there are several asymmetric carbon atoms in a
molecule, the configuration at one centre is usually related
directly or indirectly to glyceraldehyde, and the
configurations at the natural (+)-glucose there are four
asymmetric centres (marked by asterisk). By convention for
sugars, the configuration of the highest numbered
asymmetric carbon is referred to glyceraldehyde to
determine the overall configuration of the molecule. For
glucose, this atom is C–5 and, therefore, OH on it is to the
right. Hence the naturally occurring glucose belongs to the
D-series and is named as D-glucose.
1
CHO
|
H – C2*
– OH
|
HO – C3*
– H
|
H – C4*
– OH
|
H – C5*
– OH
|
6
CH2OH
D(+)–glucose
However, the above system of nomenclature based on Fischer projection
formulae, has certain disadvantages. Firstly before a name can be assigned to a
compound, we must specify how its projection formula is oriented.
Secondly, sometimes the two asymmetric carbon atoms having the same kind of
arrangements of substituents are assigned opposite configurational symbols.
Thus for (–)-2, 3-butanediol we have
4
CH3
|
HO – C3
– H
|
H – C2
– OH
|
1
CH3

Answers to Exercise
Exercise 1: i) a)

 


H
C
H2N NH2
NH
C
H2N NH2
NH2
+
C
H2N NH2
NH2
+
C
H2N NH2
NH2
+
Guanidine behaves as strong base because it can provide electron
pair easily resulting in three identical resonating structure. Site of
protonation is sp2
hybridised N atom rather than sp3
hybridise N atom
resulting in three symmetrical structures.
ii) a)
H—C—O—H
O
+
H
O
C
H
||
O


 H—C—O—H
O
+
– H—C=O+
—H
O
  
b) (iv)  (iii)  (ii)  (i)
iii) b) (iv)  (iii)  (i)  (ii)
Exercise2: (A) Let R = CH3
(a) CH2 = CH2 (no. -H. no. Hyperconjugation)
(b) (3-H, 3-Hyperconjugative structures)
C = C
H
H
H
H3C
(c) (6-H, 6H–structures)
C = C
H
H
CH3
H3C
(d) (9–H, 9H–structures)
C = C
H
CH3
CH3
H3C
(e) (12–H, 12 H structures)
C = C
CH3
CH3
H3C
CH3
From (a) to (e) number of hyperconjugative structures increases, hence
stability of alkene also increases, hence the correct order is
a  b  c  d  e
B) As we have stated (–I) effect (electron- withdrawing nature) is in order
F  Cl  Br  I,
hence (a) is strongest acid and (d) is weakest acid out of (a), (b),(c)
and (d).
FCH2COOH  Cl—CH2COOH  Br—CH2COOH
> I —CH2COOH

Exercise 3:
(a) HCl
–
Cl
(b)
HCl Cl–
Cl
(c) HCl Cl–
Cl
(d) HCl Cl–
Ph
Cl
Ph
Ph
(b)
(a)
(b)
(c)
(d)
(c) Mechanism of addition of HCl on 1,3-butadiene is as follows
Step :1
H+
+ CH2 = CH—CH=CH2CH3—+
CH—CH=CH2CH3—CH=CH–CH2
An allylic cation equivalent to
CH3—CH— CH—CH2
+ +
(Resonance hybrid)
+
Step:2
CH3—CH— CH—CH2 + :Cl: —
+ + . .
. .
CH3CH—CH=CH2 1,2 addition
Cl (78%)
(a)
(b)
CH3CH=CHCH2Cl 1,4 addition (22%)
In step 1 proton adds to one of the terminal cabons of 1,3 butadiene to
form, as usual, the more stable carbonium ion, in this case a resonance
stabilized allylic cation. Addition of one of the inner carbon atoms would
have produced a much less stable primary cation that could not be
stabilized by resonance;
CH2=CH—CH=CH2 
+
CH2—CH2—CH=CH2
H
+
In step (2), a chloride ion forms a bond to one of the carbon atoms of the
allylic cation that bears a partial positive charge. Reaction at one carbon

atom results in the 1,2 – addition product, reaction at the other gives that
1,4 addition product.
(d)
CH2 = CH—OH  CH2—CH = OH
  +
 
less stable due to charge separation.
–H
+
CH2=CH—O  CH2—CH = O
 
 
More stable due to absence of charge
separation.
After loss of H+
ion vinyl alcohol stabilizes by resonance, hence it will
favour loss of H+
and hence acidic in nature.

Solved Problems
Subjective
Problem 1: Can you think acidic nature of
(a) F—CH2COOH (b) Cl—CH2COOH
(c) Br—CH2COOH (d) I —CH2COOH
Solution: As we have stated (–I) effect (electron- withdrawing nature) is in order
F  Cl  Br  I,
hence (a) is strongest acid and (d) is weakest acid out of (a), (b),(c) and (d).
FCH2COOHCl—CH2COOHBr—CH2COOH > I —
CH2COOH
Problem 2: Take the following isomeric , ,  chlorobutyric acid
)
i
(
COOH
C
CH
CH
Cl
|
2
3


)
II
(
COOH
HCH
C
CH 2
3
Cl
|


)
III
(
COOH
CH
CH
H
C 2
2
2
Cl
|


Solution: As we have stated, as we go away from the source, electron - withdrawing
tendency decreases, hence acidic nature also decrease. Thus
Problem 3: Compare the basic strength of the following
NH3, CH3NH2 (CH3)2NH (CF3)3N
Solution: The increasing order of basic strength is as follows
F3C N:
CF3
CF3
H — N
H
H
H — N
H
H3C
H
CH3
N
CH3
  
CF3 group being – I effect groups attracts the electron pair of nitrogen makes
protonation difficult in primary and secondary amine due to presence of +I
effect group (CH3) which make protonation easy. Hence the increasing order
of basic strength is as follows.
weakest
strongest
III
II
I 

Problem 4: C4H8 can have so many isomers. Write their structures. What are different tests
to make distinction between them .
Solution: C4H8 can have following structures:
(a) CH3—CH2—Ch=CH2 1-butene
(b) CH3—CH=CH—CH3 2-butene, (cis and trans)
(c) CH3—C=CH2 2-methyl-1-propene
|
CH3

(d) Cyclobutane
(e) methyl cyclopropane
Distinctions:
i) Ozonolysis of (a) forms CH3CH2CHO and HCHO or (b) forms CH3CHO of
(c) forms
CO and HCHO
CH3
CH3
ii)
CH3—C—H
CH3—C—H
+ alkaline KMnO4 
CH3—C—OH
CH3—C—OH
H
H
Cis-2-butene
Meso-isomer
CH3—C—H
H—C—CH3
+ alkaline KMnO4 
CH3—C—OH
HO—C—CH3
H
H
trans-2-butene
d-and l-
Problem 5: C2H4Cl2 can have two position iosomers 1,2- and 1,1-dichloro ethane. Each is
subjected to following sequence of reactions:




 



 
 

O
H
KCN 3
2
4
2 Cl
H
C
What is the end product in each case?
Solution:
CH2Cl
CH2Cl
KCN
CH2CN
CH2CN
H3O+
CH2COOH

CH2CO
CH2CO
O
CH3—CH
Cl
Cl
KCN
CH3CH
CN
CN
H3O+
CH3—CH
COOH
COOH

CH3CH2COOH
succinic anhydride
propanoic acid
(if two –COOH groups are
at adjacent C-atoms
heating would eliminate
H2O)
(if two—COOH groups are
at same C-atom heating
would eliminate CO2)
CH2COOH
Problem 6: C4H4O4 can have isomers A,B,C and D
(a) is cyclic ester
(b) is dicarboxylic acid giving racemic tartaric and with alkaline KMnO4
(c) is dicarboxylic acid giving meso tartaric acid with alkaline KMnO4
(d) is also dicarboxylic acid giving another monobasic acid on heating.
Identify A,B,C and D.

Solution: a) Cyclic ester is obtained from dicarboxylic acid and diol. Given formula
shows A is cyclic ester of oxalic acid and glycol.
O
O
O=C
O=C CH2
CH2
(b), (c) and (d) contain two —COOH groups. Possible structures are
COOH
H—C
H—C
COOH
COOH
H—C
C—H
COOH
CH2=C
COOH
COOH
(maleic acid) (fumaric acid) gem
(ethene dicarboxylic acid)
On heating d forms another monobasic acid hence d is
CH2=C
COOH
COOH
gem
(ethene dicarboxylic acid)


CH2=CH—COOH
when alkaline KMnO4 reacts with maleic acid, there is formation of
meso-tartaric acid (by syn addition)
COOH
H—C
H—C
COOH
(C)
H2O + O
alkaline KMnO4
H—C—OH
COOH
H—C—OH
COOH
meso
Hence this represents isomer C
Alkaline KMnO4 converts fumaric acid into a mixture of d- and l-
isomers hence mixture is optically inactive (racemic)
(e)
CH3—CH2—C————–––
C—CH2CH3
CH2CH3 CH2CH3




 
 Zn
/
O
H
/
O 2
3
2 3
2
2
3 CH
CH
C
CH
CH
||
O



Problem 7: Identify the 1,2- and 1-4 addition products of Free Radical addition of CBrCl3 to
1,3-butadiene.

Solution:
CH2=CH–CH=CH2 

 
 3
CCl
Cl3C–CH2–CH–CH=CH2
(I)
1,2 addition product
Cl3C–CH2–CH–CH=CH2
Br
Cl3C–CH2–CH=CH–CH2Br
Br
Cl3C–CH2–CH=CH–CH2
1,4 addition product
Br
—CH3
N(CH3)2
Br2
AlBr3
CH3
Br
Br
Br
–N(CH3)2 and –CH3 both groups are o,p-directing but products are
formed only under direction of –(CH3)2. Explain.
–N(CH3)2 group as well as –CH3 group are o–,p– directing and if we
want to place both groups, none of the position will be available for
SE.
—CH3
N(CH3)2
CH3
o–w.r.t
m-w.r.t. N(CH3)2 and so on.
But o,p-directive effect of the stronger donor (–N(CH3)2) dominates
over that of the weaker donor (–CH3), hence SE takes place at o-and
p-position w.r.t. –N (CH3)2.
—CH3
N(CH3)2
Br2
O
P
—CH3
N(CH3)2
Br
CH3
Problem 8: Write all possible structural isomers of the compound with molecular formula
C5H12O.

Solution Total number of structural isomers = 7
CH3—CH2CH2CH2—OH CH3—CH—CH2OH
|
(i) CH3 (ii)
CH3
|
CH3—CH—CH2CH3 H3C—C—OH
| |
OH CH3
(iii) (iv)
CH3CH2—O—CH2CH3 CH3—O—CH2—CH2—CH3
(v) (vi)
CH3—O—CH—CH3
|
CH3
(vii)
Problem 9: Write the following Alkenes in increasing order of their stability with
explanation
R2C=CR2, R2C=CHR, R2C=CH2, RCH=CH2, CH2=CH2
Solution: Let R = CH3
(a) CH2 = CH2 (no. -H. no. Hyperconjugation)
(b) (3-H, 3-Hyperconjugative structures)
C = C
H
H
H
H3C
(c) (6-H, 6H–structures)
C = C
H
H
CH3
H3C
(d) (9–H, 9H–structures)
C = C
H
CH3
CH3
H3C
(e) (12–H, 12 H structures)
C = C
CH3
CH3
H3C
CH3
From (a) to (e) number of hyperconjugative structures increases,
hence stability of alkene also increases, hence the correct order is
a  b  c  d  e
Problem 10: Arrange the following species in increasing order of dipole moment.
)
i
(
butene
2
Dichloro
3
,
2
Cis 


 ,
)
ii
(
hene
dichloroet
2
,
1
Cis 
 ,

)
iii
(
dibromo
2
,
1
Cis 
 -1,2-dichloroethene,
)
iv
(
ene
dichloreth
2
,
1
trans 

Solution:
C
C
H3C Cl
H3C Cl
I
Resultant vector
C
C
H Cl
H Cl
II
Resultant vector
C
C
Br Cl
Br Cl
III
Resultant vector
C
C
Cl H
H Cl
I
Nonet resultant
vector
In I, there is addition of vector
In II, there will neither be addition nor subtraction of vector
In III, there is subtraction of vector
In IV, the vectors almost cancel each other. So the increasing order of dipole
moment is IV  III  II  I.
Objective
Problem 1: Nitrophenol, C6H4(OH) (NO2) can have:
(A) No isomer ( only a single compound is possible)
(B) Two isomers
(C) Three isomers
(D) Four isomers
Solution: o-,m- and p-isomers, i.e., position isomers
(C)
Problem 2: Geometrical isomerism is possible in:
(A) Butene-2 (B) Ethene
(C) Propane (D) Propene
Solution:
C = C
H
CH3
H
CH3
If either of the two doubly bonded carbon atoms has same group or
atoms attached on it, it will not show geometrical isomerism.
 (A)
Problem 3: Which of the following compounds will exhibit cis-trans isomerism?
(A) 2-butene (B) 2-butyne
(B 2-butanol (D) Butanol
Solution: Due to the presence of asymmetric carbon atom.
(B)
Problem 4: Dichloro ethylene shows.

(a) Geometrical isomerism (b) Position isomerism
(c) Both (d) None
Solution: CH2 = CCl2 and CHCl = CHCl are position isomers; CHCl = CHCl also
show geometrical isomerism.
(C)
Problem 5: The compound having molecular formula C4H10O can show
(A) Metamerism (B) Functional isomerism
(C) Positional isomerism (D) All
Solution: Alcohols show position isomerism; Ethers show metamerism; Alcohol and
ethers shows functional isomerism.
(D)
Problem 6: A compound contains 2 dissimilar asymmetric carbon atoms. The number
of optical isomers is:
(A) 2 (B) 3
(C) 4 (D) 5
Solution: a = 2n
; where n is no. of dissimilar asymmetric carbon atoms.
(C)
Problem 7: The greater the s-character in an orbital the ------------ is its energy
(A) Greater (B) Lower
(C) Both (D) None
Solution: Bond energy order
sp–sp  sp2
-sp2
 sp3
-sp3
(A)
Problem 8: The type of isomerism observed in urea molecule is
(A) Chain (B) Position
(C) Geometrical (D) Functional
Solution: NH4CNO is functional isomer of urea.
(D)
Problem 9: Number of possible isomers of glucose are
(A) 10 (B) 14
(B) 16 (D) 20
Solution: Glucose has four dissimilar asymmetric carbon atoms; a = 24
.
 (C)
Problem 10: Which of the following statement is correct?
(A) Allyl carbonium ion (CH2=CH– 2
H
C

) is more stable than propyl
carbonium ion
(B) Propyl carbonium ion is more stable than allyl carbonium ion
(C) Both are equally stable
(D) None
Solution: Allyl carbonium undergoes resonance stabilization.
(A)

Assignments
Subjective
LEVEL – I
1. Identify the stable carbocation in each pair
a) CH3 – CH2 – CH2
+
and CH3 – H
C

– CH3
b) CH3 –
3
CH
|
CH – H
C

– CH3 and CH3 – CH2 – H
C

– CH3
c) CH3 –CH2
+
and CH3 – O – CH2
+
2. Identify the stable carbanion in each pair
a) CH3 – CH2 – CH2
–
and CH3 – H
C

– CH3
b) 2
H
C

–
||
O
C – CH3 and 2
H
C

– CH2 –
||
O
C – H
c) Ph – CH2
–
and Ph – CH2 – CH2
–
3. Which of the following intermediates is stable
+
(a)
–
(b)
4. Benzylamine is a stronger base than aniline. Explain.
5. Which of the following carbocations is most stable.
+
(a)
+
(b)
6. Why is always the resonance effect dominating over the inductive effect?
7. Write the isomeric structures of an alkene having m.f. C4H8. Indicate which is more
stable.
8. Which of the following C – Cl bond is weaker.
a) Ph –
|
Cl
CH– CH3 b) Ph – CH2 – CH2 – Cl

9. Which of the following carbonyl compound is more acidic.
a) CH3 –
||
O
C – CH3 b) CH3 –
||
O
C – H
10. Identify the effects operating in each of the molecule.
a) CH3Cl
b) CH3 – CH = CH2
c) CH3 – CH = CH –
||
O
C – H
d) CH2 = CH – CH = CH2

LEVEL – II
1. CH3CO2H is a stronger acid than CH3CH2 – OH. Explain
2. Which of the following will be a stronger acid and why? Cl3CH and F3CH
3. HC  C–
is a weaker base than CH2 = CH–
. Why?
4. Why does 2-butene exhibit cis-trans isomerism but but-2-yne does not?
5. How will you distinguish between maleic acid and fumaric acid?
6. Arrange the following in increasing order of reactivity towards H+
addition.
a) CH = CH2 c)
CH = CH2
Me
b)
CH = CH2
O2N
d)
CH = CH2
MeO
7. Arrange the following in the increasing order of C – Br bond energy.
a) CH3 –
|
Br
CH
|
C
3
 CH2 – CH3
b) CH3 –
|
Br
CH– CH2 – CH2 – CH3
c) CH3 – CH2 – CH2 – CH2 – CH2 – Br
d) CH2 = CH – Br
e) Ph – CH2 – Br
8. Identify the isomeric and diasterreomeric pairs among following
CH3
CH3
HO
H
H
OH
(I)
CH3
CH3
HO
H
H
OH
(II)
CH3
CH3
H OH
(III)
H OH
9. Arrange the following alcohols in increasing order of acidity.
a) CH2 = CH – OH
b) CH3 – CH2 – OH

10. Which of the following pairs show tautomerism.
(a)
CH3 – C – NMe2 and CH3 – C – NMe2
OCH3 OMe
+
b) O
N
|
H
and
OH
N

LEVEL - III
1. Why is Guanidine a stronger base?
NH2 –
||
NH
C  NH2 (Guanidine)
2. Among orthochlorophenol and orthofluorophenol, which will be a stronger acid and
why?
3. Unlike other aromatic amines, why is the following amine strongly basic?
NMe2
NO2
NO2
4. Explain the following observations.
i) CH3 – I 
 


OH
CH3OH + I–
ii) CF3 – I 
 


OH
CF3H + IO–
5. Which of the following C – I bond is weak and why?
i) CH3 – O – CH2 – I
ii) CH3 –
|
H
N – CH2 – I

Objective
LEVEL – I
1. Fischer projection indicates
(A) Horizontal substituents above the plane
(B) Vertical substituents above the plane
(C) Both horizontal and vertical substituents below the plane
(D) Both horizontal and vertical substituents above the plane
2. 2 - Methylpenta - 2, 3 diene is achiral because it has :
(A) a centre of symmetry
(B) a plane of symmetry
(C) a C2 axis of symmetry
(D) both centre and a plane of symmetry.
3. Which of the following structures are super impossible
H
Br
HO
Me
(1)
Et
Me
Br
H
Et
Me
(2)
OH
Me
Br
Me
Et
OH
(3)
Me
H
Me
Br
HO
Me
(4)
Et
H
(A) 1 and 2 (B) 2 and 3
(C) 1 and 4 (D)1 and 3
4. The racemisation of optically active compounds is driven by
(A) entropy (B) enthalpy
(C) entropy and enthalpy (D) element of symmetry
5. The following compound can exhibit
C = C H
H
C
COOH
CH3
CH3
CH3
(A) geometrical isomerism
(B) geometrical and optical isomerisms
(C) optical isomerism
(D) tautomerism
6. The number of isomers for the aromatic compounds having M. F. C7H8O is :
(A) two (B) three
(C) four (D) five
7. The total number of isomeric optically active monochloro isopentanes is :
(A) two (B) three
(C) four (D) one

8. The spatial arrangement of atoms or groups in a stereoisomers is called :
(A) conformation (B) configuration
(C) inversion (D) none of these
9. Which of the following species is most stable ?
 
(A) p-NO2C6H4-CH2 , (B) C6H5CH2 ,
(C) p-


l
C l-C6H4- 2
H
C

, (D) H3CO-C6H4- 2
H
C

.
10. Resonance structures of a molecule should have
(A) identical arrangement of atoms (B) nearly the same energy content
(C) the same number of paired electrons (D) identical bonding

LEVEL – II
1. Which is the strongest carboxylic acid among the following ?
(A) Cl3CCO2H (B) Br3CCO2H
(C) F3CCO2H (D) Cl2CH-CO2H
2. The compounds C2H5OC2H5 and CH3OCH2CH3 are
(A) enantiomers (B) geometrical isomers
(C) metamers (D) conformational isomers
3. Which of the following, compounds displays geometrical isomerism ?
(A) CH2=CHBr (B) CH2=CBr2
(C) ClCH=CHBr (D) Br2C=CCl2
4. Among the following the compounds having the most acidic alpha - hydrogen is :
(A) CH3CHO (B) CH3COCH3
(C) CH3-C-CH2CHO (D) CH3-CO-CH2-CO2CH3
||
O
5. The number of optically active isomers observed in 2,3, - dichlorobutane is :
(A) 0 (B) 2
(C) 3 (D) 4
6. Allyl isocyanide has:
(A) 9 sigma bonds and 4 pi bonds
(B) 9 sigma bonds, 3 pi bonds and 2 non-bonding electrons
(C) 8 sigma bonds and 5 pi bonds
(D) 8 sigma bonds, 3 pi bonds and 4 non-bonding electrons
7. The percentage of `s' character of the hybrid orbital of carbon in ethane, ethene and
ethyne respectively are
(A) 25, 33, 50 (B) 20, 50, 33
(C) 25, 50, 75 (D) 33, 66, 99
8. Isomers which can be interconverted through rotation around a single bond are
(A) conformers (B) diastereoisomers
(C) enantiomers (D) positional isomers
9. Keto - enol tautomerism is observed in
(A) H5C6-CHO (B) H5C6-CO-CH3
(C) H5C6-CO-C6H5 (D) H5C6-CO-CH2-CO-CH3
10. The compound which has one isopropyl group is :
(A) 2,2,3,3 -tetramethyl pentane (B) 2,2 - dimethyl pentane
(C) 2,2,3 - trimethyl pentane (D) 2-methyl pentane

Answers
Subjective
LEVEL – I
1. a) CH3 – H
C

– CH3 (because of + I effect and hyperconjugation)
b) CH3 – CH2 H
C

– CH3 (because of hyperconjugation)
c) CH3 – O – 2
H
C

(because of resonance due to l.p. on oxygen)
2. a) CH3 – CH2 – CH2
–
(because of inductive effect)
b) 2
H
C

–
||
O
C – CH3 (because of mesomeric effect)
c) Ph – CH2
–
(because of resonance)
3. (b) is stable because carbocations can’t form in the bridge head position.
4. In aniline the lone pair of electrons on N atom are delocalised in the benzene ring
and hence the basicity decreases.
5. (a) is most stable as it is a 3° carbocation as well as allylic.
6. Resonance effect operates thro’ the -electrons and inductive effect operates thro’
the -electrons. -electrons can be easily perturbed and hence that effect is
dominating.
7.
C = CH2;
CH3
H3C
(I)
C = C ;
CH3
H
(II)
H
CH3
C = C ;
CH3
H
(III)
CH3
H
The stability order is I  III  II
8. a) In this structure if we break C – Cl bond heterolytically we end up with a C+
ion; a
benzyl carbocation.
9. b) The acidity of -hydrogen depends on +ve charge on the carbon atom to which it
is attached. In an aldehyde the carbonyl carbon has more positive charge and
hence more –I effect.
10. a) Inductive effect
b) Inductive and hyperconjugative effects
c) Inductive, hyperconjugative and mesomeric effects
d) Mesomeric effect and inductive effect

LEVEL – II
1. CH3 – CO2H  CH3 –
||
O
C – O–
+ H+
CH3 – CH2 – OH CH3 – CH2 – O–
+ H+
In EtO–
the negative charge on O atom is increased by +I effect, whereas in
CH3CO2
–
, the negative charge on O atom is decreased by delocalisation. Lesser the
charge density on negatively charged atom, weaker is the base.
2. CHCl3 is a stronger acid. Cl3CH  Cl3C–
+ H+
In this case the negative charge on C is delocalised in the vacant orbitals of Cl,
which is absent in F3C–
.
3. HC  C–
, in this the negative charge is on a sp hybridised carbon atom which has
less tendency to give electrons.
4. Alkynes are all linear molecules and hence there is no chance of exhibiting
geometrical isomerism.
5. Maleic acid forms an anhydride where as fumaric acid does not.
HC
HC
CO2H
CO2H
Δ
H




H
H
O
||
C
C
||
C
no anhydride
CO2H
CO2H
Δ
H




C
||
C
HO2C
H
6. d  c  a  b
The order depends on the carbocation stability.
CH –CH3
MeO
+
is a stable carbocation. It is stabilized both by
+I effect and resonance.
7. e  a  b  c  d
The order depends on carbocation stability
8. I and II are isomeric pair
I, III and II, III are diastereomeric pairs

9. a  b
CH2 = CH – O–
the negative charge on ‘O’ atom is delocalized to a larger extent.
CH3 – CH2 – O–
, the negative charge on ‘O’ atom is increased by +I effect.
10. (b). In (a) there is simple shifting of charge and hence it is resonance and not
tautomerism.

LEVEL – III
1. In Guanidine lone pair on any one of the nitrogen atoms is always available for
donation as a base.
NH
H2N NH2
NH
H2N NH2
+
–
If any of the nitrogen gets protonated, even then also we have a lone pair available
for donation as a base.
NH2
H2N NH2
NH2
H2N NH2
+
+
Hence Guanidine is a stronger base
2. The one having a weaker conjugate base will be a stronger acid. If the conjugate
base has to be weak, the negative charge has to be delocalised to a larger extent.
In o-chlorophenol
O
Cl
O
Cl
O
Cl
–
–
Due to the availability of vacant orbitals in chlorine, the negative charge is
delocalised to a larger extent. The same cannot take place in case of F as F does
not have vacant orbitals. So, o-chlorophenol, having a weaker conjugate base,
becomes a stronger acid.
3. Due to the presence of bulky –NO2 groups on its ortho positions, the –NMe2
group goes outside the plane of resonance to avoid steric repulsion. The C – N bond
rotates and hence the lone pair of N goes perpendicular to the plane of benzene
ring. As a result the resonance is stopped and hence the lone pair is readily available
as a base.
4. In the first case I is more electronegative than C and hence the positive charge is an
the C atom. So OH–
attacks C and displaces I–
.
In the second case due to the presence of three F atoms, the electron deficiency of
C increases. Hence it gives a positive charge on I. Hence OH–
attacks I and
displaces CF3.

CF3 – I OH  CF3
–
+ IOH
–
CF3H + IO–
5. In the first case the carbocation generated is stabilized by resonance due to the lone
pair of O. The same happens in the second case, but now due to the lone pair of N.
N being less electronegative can provide electrons easily. Hence the C – I bond in
second case is weak.
CH3 – NH – CH2 – I  [CH3 –
|
H
N – CH2  CH3 –
H
|
N = CH2]
+
+

Objective
LEVEL – I
1. A 2. C
3. D 4. A
5. C 6. D
7. A 8. B
9. D 10. A, B, C
LEVEL – II
1. C 2. C
3. C 4. C
5. B 6. B
7. A 8. A
9. B, D 10. D
6