Organic Chemistry
Question Bank Target IIT JEE-2023
Rank Boosting Course (RBC)
GOC
1.
Determine the inductive effect of phenyl ring on attached group in the (a) and (b) molecules respectively
O
(a)
(a) a : + I, b : + I
2.
(b)
(b) a : – I , b : + I
(c) a : + I , b : – I
(d) a : – I , b : – I
Which is the least stable resonating structure of the given molecule?
O
O

O
O
(a)
O
(c) 
(b)
(d)


3.
Identify the resonance hybrid of the given ion :
O
O
(a)
4.
–1/4
O
–1/4
S
–1/4
(b)
–1/2
O
S
O
O
–1/2
O
O
–1/2
(c)
–3/4
O
–1/2
–3/4
S
O–3/4
O
–3/4
O
–1
(d) –1 O S O
–1
–1
O
Which of the following is the most stable?
(b)
(c)
(d)
Which of the following has minimum heat of hydrogenation?
(a)
6.
O
O–1/4
(a)
5.
O
O
S
O
(b)
(c)
(d)
The decreasing order of bond dissociation energies amongst the C–H bonds labelled x, y, z (in homolysis)
is:
H
H
z
x
y
H
(a) x > y > z
(b) x > z > y
(c) z > x > y
(d) y > z > x
7.
Determine the stability order of given intermediates.
Me
OMe

Me
OMe
H2C 
OMe

H2C — OMe

Me
(I)
(a) I > II > III > IV
8.
(b) I > II > IV > III
2H2
2H2
(b)
(a) a>c>b
(b) c>a>b
(c) a>b>c
(d) b>c>a
Which of the following bonds is the longest ?
O
(b)
NH2
(a)
NH2
(a) a
(c)
CH2 = NH
(b) b
(c) c
(d)
NH2
(d) d
What is the correct order of stability among the given carbocations?
CH2
CH2
(I)
(a) II > I > IV > III
11.
(d) III > I > II > IV
2H2
(c)
10.
(c) IV > I > II > III
Rank the following reactions in DECREASING order of heat of hydrogenation.
(a)
9
(IV)
(III)
(II)
(II)
(b) II > I > III > IV
(III)
(IV)
(c) I > II > IV > III
(d) I > II > III > IV
Which of the following are not resonating structures of each other?
(a) CH3 – N = C = S & CH3 – S – C  N


(b) CH 3  C  O & CH 3  C  O
O
O
(c) CH3–C–OH & CH 3–C=O–H
(d) CH2 = CH – C  N & CH2–CH=C=N
12.
Find correct order of resonance energy for
(I)
OCH3
(II)
(III)
OCH 3
OCH 3
(a) III > I > II
13.
(b) II > I > III
(c) I > II > III (d) III > II > I
Which of the following substituents will decrease the acidic strength of phenol?
(a) – NO2
(b) – CN
(c) – CH3
(d) – CHO
14.
Correct order of stability of carbocations:
(i) CH 3  CH 2  CH 2
(a) (iii) > (ii) > (i)
15.
(ii)
CH 3
|
CH 3  CH  CH 2
(b) (ii) > (iii) > (i)
CH 3
|
CH

C
 CH 2
(iii) 3
|
CH 3
(c) (i) > (iii) > (ii)
(d) (i) > (ii) > (iii)
(c)
(d)
Which of the following is an aromatic specie?
O
(a)
16.
17.
(b)
Which is not a valid resonating structure of H2C = CH – CH = CH – O – CH3 ?
(a) H2C–CH=CH–CH=O–CH3
(b) H2C=CH–O=CH–CH–CH3
(c) H2C=CH–CH–CH=O–CH3
(d) H3C–O–CH–CH=CH–CH2
What is C – O bond order for H 3C–CH 2–C–O ?
O
(a) 2
18.
(b) 1.5
(c) 0.5
Solubility order in H2O for the following compounds is :
(I) H2C – CH – CH – CH2
(II) H3C – CH2 – CH2 – CH2
OH OH
OH
OH
OH
(III) H3C – CH2 – CH2 – CH3
(a) I > III > II
(b) I > II > III
19.
(c) II > III > I
(b)
(c)
(a) I > II

(II) CD3  C  CD3
|
CD3
(b) II > I
(c) I = II
(d) Stability can't be predicted
Which of the isomer of C4H8 is most stable?
(a)
22.
(d)
Which is more stable?

(I) CH 3  C  CH 3
|
CH 3
21.
(d) III > II > I
Which of the following alkene has minimum heat of hydrogenation.
(a)
20.
(d) 1
(b)
(c)
(d)
Which one of the following will have the longest C=C bond length?
(a)
(b)
(c)
(d)
23.
In which of the following, the 2nd structure has more resonance energy.
(a)
vs
(b)
O
CH3
(c) CH2=CH–CH=CH– CH2 vs
24.
vs
N
H
(d) All
Which compound has identical C–C bond length?
(a)
(b)
(c)
(d)
O
25.
Which compound is most basic among the following?
N
(I)
(II)
N
(a) I
26.
(III)
(IV)
N
N
(b) II
(c) III
(d) IV
(c) b = d > a = c
(d) d < a = b = c
What is bond length order ?
a
b
c
d
(a) d > b > a = c
27.
Increasing order of stability of following resonating structures is :
N
C
O


N


N
C
C
(I)
(a) I < II < III
28.
(b) d > a = b = c
O
(II)
(b) III < II < I
Which group has greater +M effect than —N
(a) –NH2
(b) –NH–C–Me
O
O
(c) II < III < I
(III)
(d) II < I < III
?
(c)  O
(d) –OH
29.
Which of the following will have lowest dissociation constant (Ka) ?
COOH
Cl
COOH
(d)
Select the one which is most basic among the following:
N–H
(a)
31.
(b)
COOH
(c)
30.
NO2
COOH
(a)
Me

(b)
N
Me
(c) O
N–H
O
(d)
Which of the following pair having same functional group?
(a) CH3–CH2 – NH2 , CH3 – CH – NH2
CH3
OH
(b)
OH
,
(c) CH3 – C – O – CH3 , CH3 – C – CH2 – CH2– O – CH3
O
O
(d) CH3 – CH2 – C – H , CH3 – C – CH3
O
O
32.
33.
34.
Which of the following is not + I group?
(a) –SiMe3
(b) –CMe3
(c) –COO¯
(d) None of these
The correct order of decreasing basicity of the compounds is:
(I) Et  N  Et
|
Et
(II)
(a) I > II > III > IV
(b) III > I > II > IV
(III)
N
(IV)
N
N
H
(c) I > III > II > IV
(d) I > III > IV > II
Which compound has lesser C–O bond length in comparison to C–O bond length in
O
?
OH
O
(a)
(b)
O
O
(c)
(d)
N
H
35.
COOH
OH
OH
Ka value order for
will be
(I)
(II)
(IV)
NO2
(III)
(a) III > II > IV > I
(c) III > IV > II > I
36.
COOH
(b) IV > III > II > I
(d) III < IV < II < I
Which of the following is not aromatic compound /ion?
OH
(a)
37.
NH2
(b)
(c)
(d)
O
Which of the following reaction is most favourable?

(a) CH3–C CH + NaOH  CH3 – C  C Na + H2O

(b) CH4 + NaOH  CH3Na + H2O
OH
(c)

ONa
+ NaHCO3 
+ H2O + CO2
(d) EtONa + H2O  EtOH + NaOH
38.
Arrange the following compounds in order of increasing basic strength.
CH2–NH2
NH2
N
H
(I)
(a) III > II > I > IV
(c) IV > III > II > I
39.
N
(IV)
(III)
(II)
(b) III > IV > II > I
(d) IV > II > III > I
Correct increasing order of (C–C) bond length is
CH3
NH2
(y)
(x)
CHO
CHO
(a) x < z < y
40.
(b) y < z < x
(z)
CHO
(c) z < y < x
Which among following is correct order of –M strength?
(a) – CHO <
O
||
 C  OR
O
||
 C  NH 2
<
(b) –CN > –CH = NH > –CHO
(d) x < y < z
(c) –Ph <
NO2
<
NH2
O
O
||
||
(d)  C  NH  C 2 H 5 <  C  OCH 3
<
O
||
CH
41.
1 mol of which compound can release maximum heat when it reacts with air?
(a) E-cyclo decene
(b) Z–cyclo decene
(c) E–cyclo pentadecene
(d) Z– cyclo pentadecene
42.
Which of the following is a +I group?
O
O
(b) –C–Me
(c) –C–NH2
O
(a) –C–H
43.
O
(d) –C–O
Which of the following statement is correct?
(a) The relative order of + I groups is
–O > NH > –CH2
(b) The relative order of – I groups is



– NF3 > – NH 3 >  N Me3 > –NO2
(c) The relative order of basic strength in aqueous solution is
NH3 < MeNH2 < Me2NH < Me3N
(d) None of these
44.
Which of the following compound is least basic?
O
(a)
N
(b)
(c)
N
H
O
O
(d)
N
H
O
N
H
H
45.
Which of the following is least stable carbanion?
OCH2
(a) CH 3  OCH 3
(b)
(c)
NO2
46.
OCH2
OCH2
(d)
OCH3
CN
Arrange the following in decreasing order of acidic strength.
COOH
COOH
COOH
COOH
(I)
(II)
(III)
NO2
(a) I > II > III > IV
(c) III > II > I > IV
(IV)
SO3H
(b) II > III > IV > I
(d) III > I > II > IV
CH3
47.
Find the number of sp2 hybridised atoms in the following compound.
C N
CH=CH–CH=CH2
(a) 12
48.
(b) 14
(c) 15
Which of the following resonating structure is most stable?








(c) CH 3  O  CH  CH  CH 2
(d) CH 3  O  CH  CH  CH 2
Which of the following can show resonance with the alkene?
O
(a)
50.

(b) CH 3  O  CH  CH  CH 2
(a) CH  O  CH  CH  CH
3 
2
49.
(d) 11
CN
(b)
COOH
(c)
(d)
Which of the following is non-aromatic?
(a)
(b)
(c)
(d)
N
51.
In the following compounds:
OH
(I)
OH
OH
CH3
(II)
(III)
The order of acidity is:
(a) III > IV > I > II
(b) I > IV > III > II
52.
NO2
NO2
(IV)
(c) II > I > III > IV
(d) IV > III > I > II
Which of the following compound is most basic?
O
(a) CH3– C – NH2
53.
OH
NH
(b) H2N– C – NH2
NH
(c) CH3– C – NH2
O
(d) Ph – C – NH2
Select the incorrect statement for
(a) 4 electrons are participating in resonance
(c) It is a Homocyclic compound
(b) Compound is anti aromatic
(d) Compound is non aromatic
54.
Which of the following is most basic?
NH2
NH2
NH
NH2
OCH
(a)
(b)
(c)
(d)
NO2
OCH
55.
What is correct order for bond length for mentioned bonds?
a
b CH
H—C—O—
3
c
(a) a = b = c
56.
(b) c > a > b
(c) b > c > a
(d) b > a > c
Which specie is aromatic?
O
(b)
: :
(a)
(c)
(d)
O
O
57.
H
Most stable carbocation amongst the following is :
CH2
(a)
(b)
OCH3
CH
2
CH
2


CH2
58.
O
N
(c)
(d)
NO2
OCH3
CH3
Least stable carbanion amongst the following is :
CH2
CH2
(a)
CH2
CH2
(b)
(c)
(d)
CH3
CN
NO2
59.
Consider following compounds:
a
b
c
Which bond is strongest?
(a) a
(b) b
60.
(d) Can not be predicted
Which compound has highest electron density in benzene ring?
(a) I
61.
(c) c
CH3
NO2
(I)
(II)
(b) III
OH
(III)
NH2
(IV)
(c) IV
(d) II
Which of the following is most basic compound ?
(a) H3C – C  N
(b) H3C – CH = NH (c) H3C – C = N
(d) H2C = C = N
62.
Which of the following compound has least value of pKa?
HO
OH
(a)
(b)
(c)
O
O
(d)
O
OH
O
OH
63.
In which pairs, specie-I is more stable than specie-II?
O
O
(a) H–C–CH2–O , CH3–C–O
O
O
O
O
(b) CH3–C–CH–CH2–C–CH3, CH3–C–CH–C–CH3
O
O
(c) CH3–CH–CH2–C–CH3, CH3–CH2–CH–C–CH3
O
O
N
(d)
,
N
O
64.
Which of the following is least acidic compound ?
OH
COOH
(a)
SH
OH
(b)
(c)
(d)
NO2
CH2
65.
Which resonating structure is most stable for
?

N
O
O
CH2
(a)
(b)
(c)
N
O
66.
N
N
O
O

(d)
O
O

N
O
O

Which of the following alkene is least stable ?
(a)
67.

CH2
CH2
CH2
(b)
(c)
(d)
(c)
(d)
Heat of combustion is be maximum for
(a)
(b)
O
68.
Which of the following has the highest heat of hydrogenation ?
(a)
69.
(b)
(c)
Which of the following carbanion is most stable ?
(a)
(b)
(c)
(d)
O
O
O
O
70.
(d)
mole
C 6 H12 1

H2
(A)
Which isomer of A has maximum heat of combustion?
(a)
71.
(b)
(c)
(d)
Which of the following pairs are resonance structures of each other?
O
OH
C
H3C
(I)
H3C
CH3

(III)
(a) I, II, III, IV
72.

(IV)
(b) I, IV
(c) I, II, III
(d) I, III, IV
‘y’ = Number of compound having –I group directly attached to benzene.
SO3H
CN
CH3
CF3
Find the value of ‘y’?
(a) 5
73.
CH2
(II)
COOH
NO2
OH
(b) 6
(c) 7
(d) 8
Which of the following is not an aromatic compound?
O
(a)
S
(b)
(c)
B
H
(d)
74.
In which of the following pair Ist compound has more resonance energy than IInd compound?
(a)
SH
(b)
(I)
(II)
S
(II)
(I)
O
(c)
75.
O
(I)
(d)
(I)
(II)
(II)
Correct order of acidic strength among the following compounds is
OH
OH
OH
OH
NO2
NO2
(II)
(I)
(a) I > II > III > IV
76.
(III)
(b) II > III > IV > I
(IV)
(c) II > III > I > IV
Correct order of Basic Strength (Kb order) is :
NH2
NH2
NH
1
(a) 4 > 3 > 2 > 1
77.
(d) III > II > IV > I
NH
CH3
2
NH2
H2N
3
(b) 1 > 2 > 3 > 4
NH
4
(c) 4 > 3 > 1 > 2
(d) 1 > 4 > 3 > 2
In which of the following molecules, all atoms are coplanar?
CH3
(a)
(b)
H3C
CH3
(c)
H3C
CH2
CN
C=C
(d)
CH3
CN
H2C
CH2
78.
CH3
(I)
CH3
(II)
CH2
(III)
The enthalpy of hydrogenation of these compounds will be in the order as
(a) I > II > III
(b) III > II > I
(c) II > III > I
(d) II > I > III
79.
Which among the given resonance structures is the most stable ?
N
N
N
C
C
C
(a)
(b)
O
N
C
(c)
O
(d)
O
O
80.
Determine the correct order of acidity of the following compounds.
O
O
O
O
(i)
O
HO
O
(ii)
(a) i > ii > iii
81.
HO
(iii)
(b) ii > iii > i
(c) iii > i > ii
Which of the following is most stable resonance structure?


(a) CH 2  CH  C OH
|
CH
||
CH  NH 2
(b) CH 2  CH  C  OH
|
CH
||
CH  NH 2

(d) CH 2  CH  C  OH
||
CH
|
CH
||

NH 2
(c) C H 2  CH  C  OH
|
CH
||
CH  NH 2
82.
Compare rotational energy barrier about indicated C = C between given compounds.
n-C3H7
n-C3H7
(I)
(a) I > II > III
CHO
(b) III > II > I
n-C3H7
CHO
(II)
n-C3H7
83.
(d) iii > ii > i
(III)
n-C3H7
n-C3H7
CHO
(c) III > I > II
(d) I = II = III
Which of the following is correct order of – I ?
(a) –F > – NO2 > – CN > Br
(b) –NH3 > NO2 > –CN > –C–H
O
(c) –NH3 > –NH2Me > – NHMe2 > – NMe3 (d) –NH2 > – OH > –Cl > –Br
84.
Compare bond length of the indicated bond.
(i)
NH–Me
(ii)
(a) (i) > (ii) > (iii) > (iv)
(c) (iv) > (ii) > (iii) > (i)
NH–Me
(iv)
CN
(b) (iv) > (iii) > (ii) > (i)
(d) (i) > (iii) > (ii) > (iv)
(iii)
85.
Which amongst the following has the most extensive hydrogen bonding in aqueous medium?
(a) Ethanol
(b) Acetaldehyde
(c) Acetone
(d) Acetic acid
86.
Compare acidic strength among the given compounds.
H
H
Cl
Cl
(I)
(II)
COOH
COOH
(a) I > II
87.
H
H
Cl
Cl
(b) II > I
(c) I = II
(d) Can not be compared
Which of the order is incorrect?
(a) Acidic strength :
O – C – C – OH
O – C – C – OH <
O O
O O
F
(b) Acidic strength :
F
C –OH
F
>
C –OH
F O
F
O


<
(c) Stability order :
<
N
N
N
CH3
CH3
CH3

(d) Basic Strength order : Et2NH > Et3N > Et–NH2 > NH3 (In H2O)
88.
The correct order of basic strength for the following bases is / are :
(a) CH3 > NH2 > OH > F
NH2
NH2
NH2
CH3
(b)
CH3
CH3
(c) Both are correct
(d) Both are incorrect
89.
Which of the following has highest heat of hydrogenation?
(a)
(b)
(c)
(d)
90.
Which of the following is aromatic ?
H
H
(a)
91.
(b)
(c)
B
(d)
B
Which of the following statement is incorrect?
(a) p-nitro phenol is more acidic than p-flouro phenol.


(b) – N F3 is stronger – I group than – N H3
(c) Acidity order : o-nitrobenzoic acid > p-nitrobenzoic acid > m-nitrobenzoic acid
O
||
(d) Acidity order : H3C – COOH > H  C  OH
92.
Among the following, the least stable canonical structure is :
O
(a)
O
(b)
N
N
O
O
O
(c)
(d)
N
N
O
O
93.
O
Find the correct acidic strength order :
NH3 (c)
NH3
(a)
N
H (b)
(a) a > b > c
94.
(b) c > b > a
(c) b > c > a
(d) c > a > b
The correct order of basic strength of the following compounds is
H2N
NH
H2N
NH
O
O
(1)
(2)
(3)
NH2
NH
(4)
O
OCH3
(a) 1 > 2 > 3 > 4
CN
(b) 2 > 1 >3 > 4
(c) 3 > 4 > 1 > 2
(d) 1 > 3 > 2 > 4
OH
95.
2 moles
NaNH
SO3H  
2  X
HOOC
NO2
OH
The product X will be
OH
OH
(a) HOOC
SO3
(b) OOC
SO3
NO2
NO2
O
OH
OH
O
(c) OOC
SO3H
(d) OOC
SO3
NO2
NO2
O
96.
O
Identify the compound having highest basic strength.
OCH3
CHO
CH3
NH2
(a)
(b)
(c)
(d)
H2N
NH
NH
NH2
H2N
NH
NH
H2N
NH
97.
Which of the following compounds will have highest thermodynamic acidic strength?
SO3H
98.
(a)
(b) p-nitro phenol
(c) p-Sulpho Benzoic acid
(d) 2,4-Dinitro phenol
Arrange the compounds in decreasing acidic strength order:
O
O
||
||
CH 3  C  CH 2  C  CH 3
(I)
(III) CHCH
(IV) CH3–CHO
(a) I > IV> III > II
99.
(II)
O
||
CH 3  C  CH 3
(b) I > IV > II >III (c) III > I > IV > II
(d) II > IV > I > III
The correct basic strength order of the following bases is:
O
NH
(I)
C
CH3
C
(II)
CH3
NH
O
O
O
C
(III)
CH2 NH2
(a) I > II > IV > III
(c) III > II > IV > I
100.
(IV)
H2N
C
CH3
(b) IV > III > II > I
(d) III > IV > II > I
Which of the following hydrocarbon is most acidic ?
(a)
(b)
(c)
(d)
ANSWER AND SOLUTION
1.
(b)
O –I
+I
Sol.
–I
+I
(a)
2.
(b)
(c)
O
Sol.
has negative charge on carbon atom which is less stable than negative charge on oxygen
atom in option A, B & D.
3.
Sol.
4.
Sol.
(b)
The molecule has four resonating structures and each oxygen atom has negative charge in 2 out of 4
1
resonating structures. Thus each oxygen atom has average of
charge.
2
(b)
(b) and (d) have resonance, thus more stable than (a) and (c). (b) is more stable than (d) because of
more hyperconjugation (more –H).
5.
Sol.
(d)
c has highest HOH because of most -bonds. (d) has least HOH because of more hyperconjugation
(more –H) thus more stable -bond than a and b.
6.
Sol.
(c)
Free radical formed at y is most stable because of resonance thus it has least bond dissociation energy.
Free radical formed at z is vinyl free radical, thus least stable.
7.
Sol.
(b)
I, II and IV carbocations are resonance stabilised and can be compared on the basis of degree of
carbocations. In III only the – I effect of O operates thus it is least stable.
8.
Sol.
(d)
(a) and (c) Dienes are resonance stabilised while (b) is not. (a) is more stable than (c) because of
greater hyperconjugation. Heat of hydrogenation is inversely proportional to stability for isomers.
9
Sol.
(d)
(d) has the lowest % double bond character.
10.
(c)
2

vacant sp

Sol.
Carbocation stability :

>
(-bond resonance)
(Dancing resonance)
>

>>
(2RS)
(5RS)
(unstable)
(Aryl carbocation)
11.
Sol.
(a)
First pair is not resonating structure because the position of atoms is not same.
12.
Sol.
(c)
The order of resonance energy is I > II > III greater the resonance, greater will be resonance energy
and thus III has the lowest resonance energy then I(Extended resonance) > II(cross resonance).
13.
Sol.
(c)
Acidic strength of phenol will decrease when electron releasing group will be substituted.
14.
Sol.
(d)
Greater the number of hyperconjugation greater will be the stability of carbocation.
15.
(a)
Sol.
is an aromatic compound since it is cylic planar, conjugated system and follow Huckel’s rule. It
has (4n + 2)  - electron where n = 1
16.
Sol.
(b)
Structure can not be changed in resonance.
17.
(b)
Sol.
H3C–CH2–C–O
O
Bond order =
18.
Sol.
CH3–CH2–C=O
O
Number of bonds between atoms 3
 = 1.5
Resonating structures
2
(b)
More the hydroxy groups, more is H-bonding with solvent water and thus more is the solubility.
19.
(c)
Sol.
HOHhyd. 
1
stability
(1) 3  H
(2) 6  H
HOH (1 > 4 > 2 > 3)
20.
Sol.
(a)
(I) 9  H
Bond strength order : C – D > C–H
+H order : C – D < C–H
21.
(d)
Sol.
H
(II) 9  D
H
H
6 H
6H
+
+
H
H
H
(4) 4  H
H
H
H
(3) 10  H
H
H
H
CH2
Hyperconjugation structures of isobutene and trans-2- butene respectively have 1° anion and 2° anion,
1° anion is more stable than 2° anion.
22.
(c)
Sol.
more -hydrogen, more will be the hyperconjugation, more will be the [C=C] bond length.i.e.
it will have more single bond character
23.
Sol.
(d)
In (a) N is better donor than O due to lesser electronegativity. In (b) and (c), the second compound is
aromatic.
24.
(d)
Sol.
Compound
has identical C – C bond length as the negative charge participates in resonance
and molecule is aromatic.
25.
Sol.
(b)
(b) It is most base as the lone pair of N is not delocalised and also the N atom is sp3 hybridised.
26.
Sol.
(a)
Partial double bond character appears in bond b due to resonance while there will be pure
predomindently double bond character single bond character in d bond.
27.
(d)
Sol.


N
N
C
O
Octet incomplete
O
N
C
C
(I)


(II)
Increased polarity
decreases stability
O
Octet complete
28.
(c)
Sol.
 O has greater (+M) effect due to greater charge density..
(c)
29.
Sol.
COOH
(III)
is weakest acid among given acids & weakest acid will have lowest Ka value.
30.
Sol.
(a)
(a) is the most basic due to highest electron density availability for donation.
31.
Sol.
(a)
In structure (a) 1°-amine is present. So both figures are same.
32.
Sol.
(d)
All the groups show +I effect.
33.
Sol.
(b)
In (III), lone pair present in sp3 hybridisation 'N' atom so not delocalised.
In (I), bulky group present on sp3 hybridisation 'N' atom.
In (II), lone pair present on sp2 hybridisation 'N' atom.
In (IV), lone pair present in aromatic compound so least basic.
34.
Sol.
(a)
(a) has shorter C–O bond length due to pure double bond.
35.
Sol.
(c)
III and IV are carboxylic acids, thus more acidic than I and II.
I and II
III > IV due to –M of –NO2 in III
II > I due to resonance stabilised phenoxide ion.
36.
(c)
Sol.
37.
Sol.
O
 Incomplete cyclic conjugation.
(d)
Strong acid + Strong base l weak conjugate acid + weak conjugate base. For forward direction
reaction conjugate acid-base should be weaker be than acid- base on left hand side.
(a)

CH3–C CH + NaOH
Acid
Base
Acidic strength
CH3 – C  C Na
Conjugate base
+
H2O
Conjugate acid
CH3 – C  CH < H2O

(b)
(c)
(d)
38.
(d)
Sol.
(I)
Basic strength :
NaOH < CH3 C C Na
Spontaneous in backward direction
Acidic strength CH4 < H2O in (Reaction backward direction)
Acidic strength Ph–OH < H2CO3 (Reaction in backward direction)
Acidic strength H2O > EtOH (Reaction in forward direction)
Lone pair is in resonance as part of aromaticity
N
H
(II)
N
(More Electronegativity
& localized lone pair)
NH2
(III)
 lone pair is delocalised
CH2NH2
(IV)
39.
Sol.
Lone pair is localised
(a)
As –CHO group is –M, the +M group will decrease C–C bond length, by increasing the double bond
character.
40.
Sol.
(d)
–M strength order
O
(a) – C – H >
O
||
 C  OR
>
O
||
 C  NH 2
(b) –CN > > –CHO > –CH = NH
NO2
(c)
> –Ph >
NH 2
O
O
O
||
||
||
(d)  C  H >  C  OCH 3 >  C  NH  C 2 H 5
41.
Sol.
(d)
HOC  number of CH2 units
1

stability
Also for ring size greater than 11, cis form is less stable than trans.
42.
Sol.
(d)
(a), (b), (c) have –I effect. In 'd' due to strong +I of –O–, it has + I effect.
43.
(d)
Sol.
(a) The relative order of +I groups is CH2 > –NH > –O
(b) The relative order of – I groups is



NF3 >  N Me3 > – NH 3 > –NO2
(c) The relative order of basic strength in aqueous solution is
Me2NH > MeNH2 > Me3N > NH3
44.
Sol.
(d)
It is least basic due to maximum –I effect.
45.
Sol.
(a)
In first compound the negative charge is not in resonance.
46.
(c)
COOH
Sol.
COOH
is more acidic than
SO3H
because of highly acidic –SO3H group.
NO2
47.
(a)
C N
Sol.
CH=CH–CH=CH2
48.
Sol.
(d)
(d) Complete octet RS
49.
Sol.
(c)
Only in (c) the alkene is in conjugation.
50.
Sol.
(c)
(c) Cyclooctatetraene is non-aromatic due to non-planar tub shape.
51.
Sol.
(d)
II : (+I) and (+H) of CH3, III : (–I) of NO2, IV : (–M) and (–I) of NO2.
52.
(b)
Sol.
(1)
O
O
||

 CH3 – C = NH2
CH3 – C – NH2 
NH
(2)
NH
NH


 NH2 – C = NH2 
 NH2 = C – NH2
NH2– C – NH2 
More basic due to 3 equivalent RS in conjugat acid.
(3)
NH
NH
||

 CH3 – C = NH2
CH 3  C  NH 2 
O
(4)
O
..
 Ph – C = NH2
Ph – C – NH2 
53.
Sol.
(b)
as the compound is not completely conjugated, therefore it is non aromatic.
54.
Sol.
(c)
(c) has (+M) equal (+M) effect of OCH3 as (b) but user (–I) effect due to greater distance. Mesomeric
effect is distance indendent while inductive effect is distance dependent.
Thus (c) has maximum electron density and is most basic.
55.
(d)
Sol.
b
H —C — O—
CH3 H — C = O — CH3

a
c
O
O
Because of resonance single bond converts to double bond and vice versa. Hence the order of bond
length is b > a > c
56.
(c)
Sol.

N

aromatic
N
H
H
57.
Sol.
(c)
Greater the + M greater the stability
58.
(c)
Sol.
Stability of carbanion 
59.
Sol.
(c)
The bond which has highest double bond character will be strongest and double bond character
decreases with increase in hyperconjugation.
60.
Sol.
(c)
Greater the +M effect greater the electron density in the ring.
Order of + M effect : – NH2 > –OH > –CH3 > NO2
61.
Sol.
(c)
Greater the availability of electrons greater will be the basicity.
62.
(c)
1
 EWG
EDG
–H
Sol.
O
O
OH
O
O
O
O
O
[– ve charge is delocalised on both oxygen and thus charge is more stable]
63.
Sol.
(d)
Greater the resonance greater will be the stability of negative charge.
64.
Sol.
(b)
The order of acidic strength of the following:
COOH
OH
>
SH
>
NO2
OH
>
65.
Sol.
(c)
(c) is most stable due to –ve charge on more electronegative O atom.
66.
(c)
Sol.
is least stable because double bond at bridge head is unstable. (Bredt's rule)
67.
Sol.
(c)
Greater the number of carbon atoms in a molecule, greater will be the heat of combustion.
68.
(c)
Sol.
HOH 
69.
Sol.
(a)
Greater the conjugation, greater will be the stability.
70.
(c)
Sol.
Heat of combustion 
71.
(I)
1
Stability of alkene (where no. of -bonds are same)
1
Stability
R.S.
O
OH
C
(II)
H3C
H3C
C–H2–H
CH2
H-atom migrating not possible in resonance
Number of unpaired e–s are not same.
(III)
2 unpaired
electrons
(IV)

72.

(c)
SO3H–I
CN –I
Sol.
CF3 –I
OH –I
–I
COOH –I
NO2 –I
73.
(c)
3
sp
O
Sol.
(a)
(b)
S
(c)
B
6e
Aromatic
74.
(d)
Non-aromatic
H
–
6e
Aromatic
(c)
Sol.
75.
(b)
OH
O
O
O
Sol.
(Resonance
in one ring)
NO2
N
O
O
(–M)
(highly extended
resonance)
76.
(Extended
resonance
with (–I) of
–NO2)
(Extended
resonance only)
NH2
NH2
(c)
Sol.
NH2
NH
'N' is sp3
hybridised
'N' is sp2
hybridised
CH3
NH
H2N
NH
(+M) Guanidine
77.
Sol.
(a)
All atoms are sp2 hybridisation, so coplanar.
78.
(b)
79.
Sol.
(c)
(c) is most stable due to negative charge on more electronegative O atom.
6e–
in one loop
Aromatic
80.
(c)
OH
O
O
HO
Sol.
>
HO
>
O
OH
Aromatic
O
Antiaromatic
81.
Sol.
(d)
The octet of every element is complete & the positive charge is on nitrogen is more stable than oxygen.
82.
Sol.
(a)
The electron with drawing CHO group would further increase the polarisation of H bond by stabilising
the –ve change, then increasing the single bond character and decreasing the rotational barrier.
CHO
83.
Sol.
(b)
Correct orders are as follws :
(a) – NO2 > –CN > –F > –Br




(c) – N Me3 > – NH3 > – NH 2 Me > – NH Me2
(d) – Cl > –Br > –OH > –NH2
84.
(c)
Sol.
NH–Me
(i)
(Less Hyperconjugation)
(iii)
(More Hyperconjugation)
85.
(d)
86.
(b)
(ii)
(Resonance)
(iv)
C N
NH–Me
(More Resonance)
87.
(c)
Sol.
(a)
O – C – C – OH
O O
<
H – C – OH
O
+ I effect
(b) I more acidic due to –I of F.
(c) III (resonance stabilised) > I (Lesser –I) > II (More I).
(d) Order is a combined effect of +I effect of ethylgroup and hydration of conjugate acid.
28.
Sol.
(c)
In a period, lesser the electronegativity, greater the basic strength.
NH2
CH3
is least basic because of SIP. The para isomer is more basic than meta due to +H effect
of CH3 in para isomer.
89.
Sol.
(c)
Lesser the stability of alkene, greater will be, the heat of hydrogenation.
(c) is least stable as only two  bonds are in conjugation.
90.
(d)
H
B
Sol.
91.
Sol.
is aromatic since the empty P orbital of B completes the cyclic conjugation.
(d)
(a) Stronger (–M) of NO2 than (–I) of F makes p-nitro phenol more acidic.

(b) – N F3 is stronger –I group due to (–I) of F..
COOH
COOH
COOH
(c)
NO2
SIR, –M
92.
Sol.
NO2
NO2
–I
–M
(a)
O
N
O
Highly unstable
Like charges are close and undergo repulsion.
(b)
Acidic strength order is opposite of basic strength order
:
93.
Sol.
NH3
:
NH2 (c)
NH3
NH2
(a)
N
:
N (b)
Basic strength
a>b>c
H
94.
Sol.
(a)
(1) and (2) are highly basic due to (+M) effect of NH2 (1) > (2) due to (+M) of OCH3 as opposed
to (–M) of CN.
(3) > (4) since in (4) the lone pair of N undergoes resonance with two C = O groups and is thus less
available for donation.
95.
Sol.
(b)
The two most acidic H atoms will be abstracted.
96.
Sol.
(c)
(c) is Guanidine, one of the strongest organic base.It is highly basic due to 3 equivalent RS in its
conjugate acid.
(b) is less basic than c due to lesser +M of the top NH group.
97.
Sol.
(c)
Sulphonic acid is more acidic than carboxylic acid (c) is more acidic than (a) due to electron withdrawing
effect of the –COOH group.
98.
(b)
O
Sol.
O
O
(I) CH3– C – CH2– C – CH3
C – C – CH – C – CH3
Most stable
Most acidic
O
O
(II) CH – C – CH
3
(III) CH  CH
Least acidic
(IV) CH3– CHO
O
3
CH3– C – CH2
CH  C
Least stable
CH2– C – H
O
99.
Sol.
(d)
(III) is most basic as the lone pair of N is localised IV > II > I as per extent of involvement of lone pair
of N in resonance. More resonance decreases basic strength due to decreased availability of lone pair
for line is donation.
100.
(b)
Sol.

H



Aromatic