1

JEE Advanced 2020 Paper 1 Offline

MCQ (More than One Correct Answer)
With respect to the compounds I-V, choose the correct statement(s).

A
The acidity of compound I is due to delocalisation in the conjugate base.
B
The conjugate base of compound IV is aromatic.
C
Compound II becomes more acidic, when it has a $$-$$NO2 substituent.
D
The acidity of compounds follows the order

I > IV > V > II > III.

Explanation

Triphenylmethane (I) is acidic because its conjugate base is stabilised by resonance.

$${(Ph)_3}CH\buildrel {} \over \longrightarrow \mathop {{{(Ph)}_3}C}\limits_{Triphenylmethyl\,carbanion} + {H^ + }$$

Cyclopentadiene (IV) is acidic because its conjugate base is aromatic.



Nitrobenzene is more acidic than benzene because nitro group is electron withdrawing. It will stabilise the conjugate base of benzene by $$-$$R and $$-$$I effect.

The acidic strength order on the basis of pKa data is

IV > V > I > II > III.

Hence, the correct options are (a), (b) and (c) only.
2

JEE Advanced 2020 Paper 2 Offline

MCQ (More than One Correct Answer)
Consider the following four compounds, I, II, III, and IV.



Choose the correct statement(s)
A
The order of basicity is II > I > III > IV.
B
The magnitude of pKb difference between I and II is more than that between III and IV.
C
Resonance effect is more in III than in IV.
D
Steric effect makes compound IV more basic than III.

Explanation

The correct basic strength order is

(IV) > (II) > (I) > (III);

(IV) is strongest base due to steric inhibition to resonance effect.

(III) is weakest base due to $$-$$M group of three nitro groups present at ortho and para positions.

(II) is stronger than (I) since (III) is tertiary and (I) primary aromatic amine.

So, option (a) is incorrect.

(b) pKb different between I and II is 0.53 and that of III and IV is 4.6. So, option (b) is incorrect.

(c) and (d) In 2, 4, 6-trinitro aniline (III) due to strong $$-$$R effect of $$-$$NO2 groups, the lone pair of $$-$$NH2 is more involved with benzene ring hence it has least basic strength. Whereas (IV) N, N-dimethyl 2, 4, 6-trinitro aniline, due to steric inhibition to resonance (SIR) effect; the lone pair of nitrogen is not in the plane of benzene, hence makes it lone pair more free to protonate.
3

JEE Advanced 2017 Paper 2 Offline

MCQ (More than One Correct Answer)
For the following compounds, the correct statement(s) with respect to nucleophilic substitution reaction is (are)

A
$${\rm I}$$ and $${\rm I}$$$${\rm I}$$$${\rm I}$$ follow $${S_N}1$$ mechanism
B
$${\rm I}$$ and $${\rm II}$$ follow $${S_N}2$$ mechanism
C
Compound $${\rm IV}$$ undergoes inversion of configuration
D
The order of reactivity for $${\rm I},$$ $${\rm I}{\rm I}{\rm I}$$ and $${\rm IV}$$ is : $${\rm I}V > {\rm I} > {\rm I}{\rm I}{\rm I}$$

Explanation

(a) I is a benzylic halide, thus, it undergoes SN1 reaction easily as benzylic carbocation is resonance stabilized. III also follows SN1 mechanism as it is 3$$^\circ$$ alkyl halide.

(b) Compounds I and II are 1$$^\circ$$ alkyl halides, then they undergo SN2 mechanism.

(c) Correct. In SN1 reaction, the nucleophile approaches the substrate carbon from the back side with respect to the leaving group. Nucleophilic displacement of the leaving group in an SN2 reaction causes inversion of configuration at the substrate carbon.

(d) Stability of carbocations follows the order:

$$\mathop {2^\circ \,benzylic}\limits_{(IV)} > \mathop {3^\circ \,alkyl}\limits_{(III)} > \mathop {1^\circ \,benzylic}\limits_{(I)} $$

4

JEE Advanced 2017 Paper 1 Offline

MCQ (More than One Correct Answer)
The IUPAC name(s) of the following compound is (are)

A
$$1$$-chloro-$$4$$-methylbenzene
B
$$4$$-chlorotoluene
C
$$4$$-methylchlorobenzene
D
$$1$$-methyl-$$4$$-chlorobenzene

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