Chemistry
Session
GENERAL ORGANIC CHEMISTRY - 4
Session Objectives
Session Objectives
1. Resonance or Mesomeric effect
2. Inductive effect
3. Electromeric effect
4. Hydrogen bond
5. Hyper conjugation
6. Steric effect
Resonance or Mesomerism
1. All the properties of a compound
cannot be explained by single
structure.
2. Canonical structures or resonance
contributing structures–differ in position
of electrons.
3. Delocalisation of electrons leads to decrease in
potential energy of molecule.
canonical structures of
benezene
Resonance hybrid
structure of benezene
Resonance or Mesomerism
4. Resonance hybrid is more stable than
canonical structures.
5. Resonance structures are imaginary.
6. Resonance energy = Actual energy of hybrid–energy of
most stable contributing structure.
7. Resonance is measure of stability.
Rules for Drawing Resonance Structure
1. The molecule should be planar.
2. It contains an alternating system
of single and double bonds
(a conjugated system).
3. The relative positions of nuclei should remain unchanged
(e.g. tautomerism).
4. The negative charge must preferably lie on the most
electronegative atom.
5. The charge needs to be preserved in all the resonating
structures.
6. The electrons always move away from a negative charge.
7. Arrows should be drawn to indicate the direction of the
movement of electrons.
Types of Resonance
+R or +M effect
C
C
–
—C—C
OH
+
OH
(+M or +R effect of –OH group)
–R or –M effect
+
C
C
CH
O
(–R effect of –CHO group)
—C—C
CH — O
–
Types of Resonance
For substituted benzene
+
NH2
NH2
+
NH2
–
etc.
+R effect of –NH2 group.
–
O
+
N
O
–
O
+
N
–
O
–
O
+
N
+
–
–
O
etc.
+
–R effect of –NO2 group.
Inductive Effect
1. Permanent effect in saturated
carbon chain compounds.
2. Group attached to carbon chain should
have tendency to release or withdraw
electrons.
Types of inductive effect
+ I effect effect –electron donating groups
e.g., CH3 , C2H5
– I effect effect –electron withdrawing groups
e.g., - NO2 , –CN
Features of Inductive Effect
1. Chloroacetic acid is a stronger acid
than acetic acid because
O
Cl - CH2 C
O
Cl-CH2 - C
OH
+
+
-I
H
O
Ka = 1.4 × 10-3
O
CH3-C
O
+
CH3-C
OH
Ka = 1.75 × 10-5
+I
O
+
H
Features of Inductive Effect
2. The larger is the electron-withdrawing
effect of a group, the greater is the
–I (inductive) effect.
F CH2 COOH
Ka
2.5×10-3
Br CH2 COOH
1.3×10-3
3. Inductive effect is additive
Cl3CCOOH
Ka
2.3×10-1
Cl2CHCOOH
5.4×10-2
Features of Inductive Effect
4. Since this effect is transmitted
through a chain it becomes less
effective with distance
ClCH2CH2COOH
Ka
8.32×10-4
ClCH2CH2CH2COOH
3.02×10-5
Electromeric Effect
Temporary effect which is observed
in presence of reagents involving
transfer of electrons in an unsaturated
system.
in presence
of reagent
X
Y
in absence
of reagent
+
X
–
Y
Electromeric Effect
Addition of HBr to an alkene
R
1
C
H
R — CH — CH3
2
H
C
R
HBr
H
Br
–
H
R
H
Br
H
C—C
+
–
+
C — CH3
H
H
+
Hyperconjugation or no bond
resonance
(a) Involves  and  bond orbitals
(b) More the number of hyperconjugative
structures, more will be the stability of
ion or molecule
H
H
H – C – C+
H
H
+
H
H
H–C=C–H
H
H
Structure of ethyl carbonium ion
H
H
H–C=C
+
H H
+
H C=C
H
H
H
Hyperconjugation or no bond
resonance
(c)
The number of hyperconjugative
structures in an alkene is obtained by the
number of C — H bonds attached to the
carbon bonded directly to the double
bonded carbon atoms.
H+
H
H
C
H
CH
CH2
H
C
H
–
CH
H
CH2
+
C
CH
–
CH2
H
H
H
H
C
CH
+
H
–
CH2
Significance of Hyperconjugation
H3C — CH — CH
CH2
1–butene
H3C — CH
CH — CH3
H3C — CH
+
H
CH — CH2
(2 hyperconjugative structures)
–
H3C — CH — CH
CH2
H
2–butene
+
(6 hyperconjugative structures)
More stable
Relative strength of organic acids
O
R-C
O
R-C
OH
+
+
H
O
R-C
O
O
Resonance structures
Class Exercise
Class exercise 1
The hybridization of carbon
atoms C — C single bond in
vinylacetylene is
(a) sp3 - sp3 (b) sp - sp2
(c) sp2 - sp (d) sp3 - sp
Solution :
1
H2 C
2
3
CH — C
Vinylacetylene
4
CH
Hence answer is (c).
Class exercise 2
Allyl isocyanide has
(a) 9  bonds and 4  bonds
(b) 8  bonds and 5  bonds
(c) 8  bonds, 5  bonds and 4 non-bonding electrons
(d) 9  bonds, 2  bonds and 2 non-bonding electrons
Solution:
+
H2C
CH — CH2 — N
–
C
Allyl isocyanide
The compound has 3  bonds and
one lone pair, i.e. two non-bonding
electrons. It also contains 9 bonds.
Hence answer is (d).
Class exercise 3
Among the following which has the
most acidic  -hydrogen?
O
O
O
(a) CH3CCH2CHO
(b) CH3CCH2CCH3
O
(c) CH CCH COOCH
3
2
3
(d) CH3CHO
Solution
O
O

H3 C — C — CH2 — CH
O
O

H3 C — C — CH2 C — CH3
O
one keto and one
aldehydic carbonyl group.
two e-withdrawing keto
groups.
O

H3 C — C — CH2 C — OCH3
keto and ester group.
O

H3 C CH
one aldehyde group.
Solution
Since e-withdrawing nature of C
O
gas varies as aldehyde > keto > ester
Then most acidic a-H atom is present in
O
O
H3C C — CH2 — CH
Hence answer is (a).
Class exercise 4
The decreasing order of acidity among
phenol, p-methylphenol, m-nitrophenol
and p-nitrophenol is
(a) m-nitrophenol, p-nitrophenol, phenol,
p-methylphenol
(b) p-nitrophenol, m-nitrophenol, phenol,
p-methylphenol
(c) p-methylphenol, phenol, m-nitrophenol,
p-nitrophenol
(d) phenol, p-methyl phenol, p-nitrophenol,
m-nitrophenol
Solution
OH
OH
OH
OH
NO2
CH3
–I, –R
–I
+I
NO2
Electron withdrawing groups increase acidic strength
while electron donating group decreases the same.
So the proper decreasing order of acidic strength is
OH
>
NO2
OH
OH
OH
>
NO2
>
Hence answer is (b).
CH3
Class exercise 5
In the following compounds, the order
of basicity is
O
N
N
H
H
(I)
N
(II)
(III)
N
H
(IV)
(a) I > IV > II > I
(b) II > I > IV > III
(c) III > I > IV > II
(d) IV > I > III > II
Solution
O
N
H
I
N
sp3
N
sp2
II
H
IV
N
sp3
H
III
sp2
Between I and IV, IV is less basic because of the –I
effect of oxygen atom.
II is more basic than III as the lone pair on N-atom in
III is not available for protonation as it is involved in
resonance.
Therefore, the correct order is I > IV > II > III
Hence answer is (a).
Class exercise 6
Account for the order acidity in the
following compounds.
O
(i) CH3CCH2COOH
(ii) HC
CCH2COOH
CH3CH2COOH
H2C = CHCH2COOH
Solution
O


(i) H3C C CH2 COOH > CH3 CH2 COOH
(a)
(b)
In compound (a), electron-withdrawing keto group increases
the acidic strength by decreasing the O — H bond strength,
while no such effect is there in compound (b).
(ii) HC
C CH2 COOH > CH2
CH CH2 COOH
Carbon atoms attached to triple bond is sp hybridised and
more electron-withdrawing than sp2 hybridised carbon atom.
Hence, such order in acidic strength is observed.
Class exercise 7
Which of the following two amines is
more basic and why?
CCl3CH2CH2CH2NH2
or CCl3CH2CH2NH2
Solution
Electron-withdrawing groups decrease
the charge density on N-atom of
organic amines and hence decrease the
basic strength. In Cl3C CH2 CH2 CH2
NH2, the electron withdrawing — CCl3 is
far apart from — NH2 group as
compared to Cl3C CH2 CH2 NH2.
Hence, the former is more basic in
nature.
Class exercise 8
Do you think CH3CH
CH2
OH
can show tautomerism?
Solution:
Yes. The tautomeric form is CH3COCH3.
Class exercise 9
Which one will be more acidic ?
OH
CH3
NO2
OH
CH3
CH3
CH3
C
N
I
II
Solution
Because of steric inhibition of
resonance conjugate base of I will not
be stabilised by resonance. But for II
there is no such steric inhibition of
resonance.
Class exercise 10
Which hydrogen is maximum
acidic in the following compound?
H—C
C
NO2
O—H
COOH
Solution:
Carboxylic hydrogen is maximum acidic.
Thank you