The most stable carbocation from the following is :

Solution

Chemistry ma’am · 3 steps

All four are benzyl cations, so ask one thing: which group pushes the most electrons into that CH₂⁺?

1. Common base. Each option is a benzyl cation, already stabilised by resonance with the ring. The positive charge spreads to the ortho and para carbons, so a substituent at the para position can interact with the charge directly. A group at the meta position cannot.

   Option	Substituent	Effect on the positive charge	Result
   (a)	OCH₃ at meta	No resonance link to the charge, so only the -I pull of O acts	less stable than benzyl
   (b)	CH₃ at para	+I and hyperconjugation, a mild push	more stable than benzyl
   (c)	OCH₃ at para	+M: the lone pair on O delocalises right onto the benzylic carbon	most stable
   (d)	none	ring resonance only	reference

2. Why para-OMe wins. In (c) one resonance form is CH₃-O+=C₆H₄=CH₂, where every atom has a full octet. That extra contributor makes the +M effect of OCH₃ far stronger than the hyperconjugation from CH₃.
3. So. The stability order is (c) > (b) > (d) > (a), and the most stable carbocation is (c), the p-methoxybenzyl cation.

The idea to keep. For substituted benzyl cations, look at position before group: ortho or para lets +M act on the charge, meta leaves only the inductive effect. A strong +M donor at para beats any alkyl group.

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