Brilliant STUDY CENTRE LT-2021 (JEE ADV.

) PART II - CHEMISTRY

CHAPTER - 13
ORGANIC CHEMISTRY REACTION MECHANISM

WORK BOOK
1. How many structural isomers are possible for the compound with M.F. C6H14
A) 4 B) 5 C) 6 D) 7
2. Pent-1-yne and 3-methyl-but-1-yne are
A) Chain isomers B) Position isomers
C) Functional isomers D) Metamers
3. Of the isomeric hexanes the isomers that give the minimum and maximum number of monochloro
derivatives are respectively
A) 3-methyl pentane and 2, 3-dimethyl butane
B) 2, 3-dimethyl butane and n-hexane
C) 2, 2-dimethyl butane and 2-methyl petane
D) 2, 3-dimethyl butane and 2-methylpentane
4. 2-butyne and 1,3-butadiene are
A) Position isomers B) chain isomers C) functional isomers D) tautomers
5. Which of the following pairs of compounds are not isomers

OH
OH CH3
A) and B) and
CH3 CH3 OH OH
CH3

OH CH3
CH2

C) and D) and
OCH3 HO CH3
OH

6. Vinyl alcohol and acetaldehyde are

A) Geometrical isomers B) Tautomers
C) Chain isomers D) Position isomers

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 Brilliant STUDY CENTRE LT-2021 (JEE ADV.) PART II - CHEMISTRY

7. In which of the following compounds maximum enolisation takesplace

O O O
O
A) B) C) D)
CH3 C CH3

8. Total number of structural isomers possible for C4H6 is/are

A) 4 B) 5 C) 9 D) 7

9. Maximum number of carbon atoms arranged linearly in the molecule CH 3  C  C  CH  CH 2 is

A) 2 B) 3 C) 4 D) 5
10. The number of 120 angles present in benzene is
A) 6 B) 12 C) 9 D) 18
11. Which of the following acids has the lowest value of acidic strength

A) CH3 CH COOH B) CH2 CH2 COOH

F F

C) CH2 CH2 COOH D) CH3 CH COOH

Br Br

12. Most acidic compound is

A) CH3 COOH B) HO–CH2–COOH C) CN–CH2COOH D) O2N – CH2 – COOH
13. Most basic amine is

A) CF3 NH2 B) CH3 NH2

C) CH3 CH2 NH2 D) CF3 CH2 NH2

14. Which among the following compounds is most stable
A) Ethene B) 1-butene
C) cis-2-butene D) 2, 3-dimethyl but-2-ene
15. Alkyl groups are orthopara directing because of
A) Inductive effect B) Electromeric effect
C) Resonance D) Hyperconjugation
16. The C–C single bond length in propene is less than 1.54 A0. This shrinkage is due to
A) Resonance B) hyper conjugation
C) Inductive effect D) Electromeric effect

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 Brilliant STUDY CENTRE LT-2021 (JEE ADV.) PART II - CHEMISTRY

17. Arrange the following alkenes in the decreasing order of stability

CH 3CH  CH 2 CH 3  CH 2  CH  CH 2  CH3 2 CH  CH  CH 2
(I) (II)  III 

 CH3 3 C  CH  CH 2
 IV 

A) I > II > III > IV B) I > III > II > IV C) I > IV > III > IV D) IV > III > II > I
18. The C – Cl bond length is shortest in

A) CH2 CH Cl B) CH3 – Cl

C) C6H5 – CH2 – Cl D) CH2 = CH – CH2 – Cl

19. Which among the following statements are true with respect to electronic displacements in a covalent
bond
A) Inductive effect operates through  bonds
B) Resonance effect operates through  bonds
C) Iductive effect operates through  bond
D) Resonance effect operates through  bond
5) Resonance and inductive effect operates through  bond
A) 3 & 4 B) 1 & 2 C) 2 & 4 D) 1 & 3
20. The C–C bond length in benzene is
A) 1.54 A0 B) 1.34 A0 C) 1.20 A0 D) 1.39 A0
21. Resonance is not observed in

CH2

A) B) CH2 CH CH2

C) CH2 CH NH3 D)

22. Which of the following is not correctly ordered for resonance stability ?

I II

A) CH 2  CH  N  O  CH 2  CH  N  O ; (II > I)
 
B) H 2 N  C  O  H 2 N  C  O ; (I > II)

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 Brilliant STUDY CENTRE LT-2021 (JEE ADV.) PART II - CHEMISTRY

 
C) H3 C  C  O  H 3C  C  O ; (I > II)
 
D) H 2 N  C  O  H 2 N  C  O ; (II > I)
23. Which one of the following compounds has the greater electron density on its oxygen atom ?

O O

A) B) NH
NH CH3 CH3

C) CH2 CH3 D) All are same

24. Which of the following exhibit electromeric effect

A) Alkanes B) aldehydes C) alkyl halides D) alkyl amines
25. The order of decreasing stability of the following cations is
 
CH3  CH  CH3 CH3  CH  O  CH3 CH3 CH2
I II

A) III > II > I B) I > II > III C) II > I > III D) I > III > II
26. Which among the following is the second most stable carbocation?

CH2 CH CH3
A) B) C) D)

27. Which of the following intermediate have complete octect of electrons around the central carbon
atom?
A) Carbocation B) Carbanion C) Free radical D) Carbene
28. The most stable carbanion is

(-)
CH2CH2 CH2
A) B)

(-) (-)
C) CH3O CH2 D) O2N CH2

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 Brilliant STUDY CENTRE LT-2021 (JEE ADV.) PART II - CHEMISTRY

29. Shape of CH3 and CF3 respectively are

A) Planar and planar B) Planar and pyramidal

C) Pyramidal and planar D) Pyramidal and pyramidal
30. Match the following :

A) C P) Triplet carbene

B) C Q) Singlet carbene

C) Magnetic moment R) sp2 hybridization

D) Ground state of carbene S) sp hybridization

A B C D A B C D
A) B)
(Q, R) (P,S) P R (P,S) (Q, R) P P

A B C D A B C D
C) D)
(P,S) (Q, R) P R (Q, R) (P,S) P P
31. Most stable carbene is :
H Cl NH2
NH2
C C C
A) B) C) C D)
H Cl NH2 H
32. In nitrating mixture nitric acid acts as
A) Electrophile B) Nucleophile C) Acid D) Base

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 Brilliant STUDY CENTRE LT-2021 (JEE ADV.) PART II - CHEMISTRY

CH3

33. + CH3 CH CH2 Cl 

AlCl3
 P . ‘P’ is

CH3 CH3

CH2 CH CH3 CH3 C CH3

A) B)

CH2 CH3
CH3

C) D)

CH3

34. Identify the compound X

O 
anh.AlCl3

X
Cl

A) B) C) D)
O
O
35. Consider the following reaction
OMe
H+ Br2
A B

OH
The product B is

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 Brilliant STUDY CENTRE LT-2021 (JEE ADV.) PART II - CHEMISTRY

OMe OMe

A) B)
Br Br

OMe OMe

C) D)

Br
36. In the following reaction
O

C
N 
con H 2SO 4
 “X”. The structure of the major product “X” is
conc HNO3

O O
NO2
C NO2
N C
A) B) N

H H

O O

C C
C) N D) NO2 N
H NO2 H

37. Which is wrong statement?

A) SN2 mechanism proceeds through T.S
B) SN1 mechanism proceeds through carbonium ion
C) SN1 mechanism is a two step mechanism
D) SN1 mechanism results in complete inversion
38. SN1 reaction at an optically active halide leads to
A) a single stereo isomer
B) racemisation with slight excess of inversion product
C) complete inversion
D) Complete retention
39. SN2 reactivity is maximum in
A) CH3Br B) CH3I C) CH3 CH2 Br D) (CH3)3C Cl
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 Brilliant STUDY CENTRE LT-2021 (JEE ADV.) PART II - CHEMISTRY

H OH
40. T h e p r o d u c t f o r m e d i n t h e r e a c t i o n + SOCl2 
Py
 is

H OCl OCl H H Cl Cl H
A) B) C) D)

41. Chlorination of toluene in presence of sunlight is an example of

1) Free radical substitution 2) Electrophilic substitution
3) Nucleophilic substitution 4) Nucleophilic addition
42. The most reactive of the following towards nucleophilic addition is

CHO CHO CHO

CHO

A) B) C) D)

NO2 Cl CH3 O CH3

43. Dichlorocarbene is generated by the action of potassium t-butoxide on chloroform. This is an example
of

A)  -elimination B)  -elimination C)  -elimination D)  -elimination

44. The main product of the reaction CH3 CH CH2 CH3 + KOH (alc) Product is

Cl
A) CH2 – CH2 – CH = CH2 B) CH3 – CH = CH – CH3

C) CH3 CH CH2 CH3 D) HO CH2 CH CH2 CH3

OH Cl
45. In acid catalysed dehydration of alcohol intermediate is
A) Carbonium ion B) Carban ion C) Free radical D) Carbene

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 Brilliant STUDY CENTRE LT-2021 (JEE ADV.) PART II - CHEMISTRY

More than one correct answer type

46. Which of the following statements are correct for butadiene
1 2 3 4
C H2  C H  C H  C H2
A) The C1 – C2 and C3 – C4 bonds are larger than carbon-carbon double bond
B) The C1 - C2 and C3 – C4 bonds are shorter than carbon-carbon double bond
C) The C2 – C3 bond is shorter than C–C bond
D) The C2 – C3 bond is slightly larger than C=C bond
Matrix Match Type Question
47. Column I Column II

OH
Br
A) P) Elimination

Br OH
B) Q) unimolecular


C2 H5  O
C) CH 3  CH 2  CH 2  Cl   CH 3  CH  CH 2 R) Rearrangment

D) CH3 – Cl  CH3 –OH S) Bimolecular

48. How many position isomers are possible for dichlorobenzene
Passage
Nucleophilic substitution on the saturated carbon usually follows SN1 and SN2 mechanism. Less
sterichindrance, polar aprotic solvent, strong nucleophile is favourable for SN2 reaction. The SN1
mechanism in contrary, is supported by polar protic solvent. This mechanism proceeds through the
formation of a carbocation.
49. The rate of hydrolysis of the following compounds through SN1 is

I) Cl CH2 Cl II) CH2 Cl

OCH3 OCH3

III) IV) CH2Cl

CH2 Cl

A) II > I > III > IV B) III > IV > I > II C) IV > II > I > III D) II > I > III > IV

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 Brilliant STUDY CENTRE LT-2021 (JEE ADV.) PART II - CHEMISTRY

50. Rate the following compounds in order of increasing SN2 reaction rate with NaI/acetone

I) CH3 – CH2 – CH2 – I II) CH3 CH2 CH Cl

CH3

CH3

CH3 C Cl
III)
CH3

A) II > I > III B) I > II > III C) III > I > II D) II > III > I

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 Brilliant STUDY CENTRE LT-2021 (JEE ADV.) PART II - CHEMISTRY

CHAPTER - 13
ORGANIC CHEMISTRY REACTION MECHANISM

WORK BOOK
WORK BOOK
1. B 1) CH3 - CH2 - CH2 - CH2 - CH2 - CH3

CH3

2) CH3 CH CH2 CH2 CH3

CH3

3) CH3 CH2 CH CH2 CH3

CH3

CH3 C CH2 CH3

4)
CH3

CH3 CH3
5)
CH3 CH CH CH3

2. A

1 1 1
CH3 CH3 CH 3
1 1 1
3. D CH3 CH CH CH3 and CH 3 CH CH 2 CH 2 CH 3
2 2 3 4 5
2

4. C Dienes and alkynes are functional isomers

5. D Both molecules are 3-methyl phenol

O H O

CH2 CH 
6. B  CH3 C H
Vinyl alcohol Acetaldehyde

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 Brilliant STUDY CENTRE LT-2021 (JEE ADV.) PART II - CHEMISTRY

OH
O
H
7. D 
H 
Stable due to aromaticity

8. C CH2 = CH - CH = CH2 , CH2 = C = CH - CH3, CH3 – C  C- CH3, CH  C – CH2 - CH3

, , , ,

sp sp
H3C C C CH CH2
9. C
linear
10. D Due to the presence of six sp2 hybridised carbon atoms
11. C
12. D –NO2 is most electron withdrawing group
13. C CH3 – CH2 – group has more +I effect
14. D Due to more number of hyperconjugative structures
15. D
16. B C–C single bond gets partial double bond character by hyperconjugation
17. A Number of  -hydrogens decreases from I to IV
18. A C–Cl bond length decreases due to partial double bond character by resonance
19. A
20. D Partial double bond character by resonance
21. C Octet around N atom is already completed
22. C In structure II all the atoms have complete octet
23. A In compound A the lone pair of electrons on nitrogen atom is always delocalised to the oxygen
aotm by resonance
24. B Due to the presence of C–O multiple bond
25. C In the second case, the lone pair of electrons on ‘O’ atom stabilises the carbocation by resonance.
26. B ‘A’ is most stable due to  -resonacne. B is stable due to its aromaticity..
27. B
28. D –NO2 group at p-position stabilises the carbanion by its -R and –I effect

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 Brilliant STUDY CENTRE LT-2021 (JEE ADV.) PART II - CHEMISTRY


29. B C F3 has pyramidal geometry due to the repulsion between the single electron on carbon and
the lone pairs of electrons on F atom
30. D
31. C The lone pair of electrons on nitrogen atoms stabilises the carbene by resonance.
32. D HNO3 is weak acid than H2SO4.  HNO3 act as base and H2SO4 act as acid

CH3 CH3 CH3

AlCl
Cl   CH3 
1,2H shift
33. B CH3 CH CH2 3
 AlCl4 CH CH2   CH3 C CH3

E+

CH3

H3C C CH3
CH3
+
CH3 C CH3 

34. D Intramolecular Friedel-Craft’s acylation

AlCl3

 AlCl 
 H



4

C C
Cl
O
O O

o,p-directing
OMe OMe

H
35. A 
 H2O

OH

OMe

H

Intramolecular Friedel Craft 's alkynation
 Br2
 

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OMe

Br

O -R
+R

NH C

36. B

Activated ring Deactivated ring

37. D
38. B
39. B SN2 reactivity is CH3–X > 1° > 2° > 3°. If alkyl group is same, the order of S N2 reactivity is
R  I  R  Br  R  Cl  R  F
40. D Reaction between R–OH and SOCl2 in presence of Py follows SN2 mechanism. Stereochemistry
of SN2 reaction is 100% inversion
41. A
42. A –NO2 group present on benzene ring increases the magnitude of partial +ve charge on carbonyl
carbon atom
43. A
44. B Saytzeff rule
45. A
46. A, C, D Due to resonance
47. A  Q, R; B  S; C  P, S; D  S

Cl
Cl Cl

Cl
48. 3 , ,
Cl
Cl

49. C Due to the stability of intermediate carbocation

50. B SN2 reactivity order is 1° > 2° > 3°

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