Question

JEE ADVANCED CHEMISTRY QUESTION SEPTEMBER 2020

Answer 1

Answer:

C,D

Explanation:

Steric effect in 4 results in rotation of C--N bond thus breaking the resonance due to break of planarity.

Thus lone pair can be easily given.

In 3 due to resonance , basicity decreases.

in 1,2 --> both have resonance , hence their basicity will not differ much , but in 3 resonance is there and 4 no resonance --> high difference in basicity

Answer 2

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.

solution : The basic strength depends on tendency to accept protic ion ( H⁺), higher the neutrophilic tendency of functional group, higher the order of basicity will be ...

so the most basic is compound II and least basic is compound (III)

here compound III ; 2, 4, 6 trinitro aniline due to strong Resonance effect of NO2 group , the lone pair of NH2 is more involved with benzene ring hence it has least basic strength.

so option (A) is incorrect.

option (C) is correct, resonance effect is more in III than in IV.

pKb different between I and II is just 0.53 and that of III to IV is 4.6 so option (B) is incorrect.

compound IV , N, N - dimethyl 2, 4, 6 trinitro aniline , due to steric inhibition to resonance effect; the line pair of nitrogen is not in the plane of benzene, hence make it lone pair more free to protonate as shown in figure.

so steric effect makes compound IV more basic than III, option (D) is correct choice.

Therefore option (C) and (D) are correct choices.