Given below are two statements :
Statement I : In
, the carbocation is stabilised by +R effect of –OCH₃ group.
Statement II : In
, the carbanion is stabilised by –R effect of –NO₂ group.
In the light of the above statements, choose the correct answer from the options given below :

Solution

Chemistry ma’am · 4 steps

Both groups sit para to the charged carbon, so each one talks to that carbon through the ring by resonance.

1. Read the structures. Both species are a CH carbon joining two benzene rings. One ring has -NO₂ and the other has -OCH₃, each para to that carbon. Para groups are in full conjugation with the charged carbon, so their resonance effects reach it.
2. Statement I. The O of -OCH₃ has a lone pair. It pushes this pair into the ring, and the electron density moves to the para position until a canonical form has a C=O⁺ bond and a neutral benzylic carbon. That spreads the positive charge onto oxygen, so the +R effect stabilises the carbocation. Statement I is true.
3. Statement II. In the carbanion, the lone pair on the benzylic carbon flows into its ring and on to the para -NO₂. It ends up on an oxygen of the nitro group. Spreading the negative charge onto electronegative oxygen is the −R effect stabilising the carbanion. Statement II is true.

   Intermediate	Group named	Effect needed	Does it work?
   Carbocation	-OCH₃ (para)	electron donation, +R	yes, + charge reaches O
   Carbanion	-NO₂ (para)	electron withdrawal, −R	yes, − charge reaches O of NO₂

4. So. Each statement pairs the right group with the right charge, and both groups are in resonance with the charged carbon. Both statements are true, which is option (a).

The idea to keep. Donors (+R, like -OCH₃) stabilise carbocations. Acceptors (−R, like -NO₂) stabilise carbanions. At ortho or para, the resonance effect wins over the inductive effect.

Answer from the NTA final answer key. Solution by PrepWiser.