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 pK_a data is

IV > V > I > II > III.

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

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) pK_b 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.

(a) I is a benzylic halide, thus, it undergoes S_N1 reaction easily as benzylic carbocation is resonance stabilized. III also follows S_N1 mechanism as it is 3$$^\circ$$ alkyl halide.

(b) Compounds I and II are 1$$^\circ$$ alkyl halides, then they undergo S_N2 mechanism.

(c) Correct. In S_N1 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 S_N2 reaction causes inversion of configuration at the substrate carbon.

(d) Stability of carbocations follows the order:

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