Arrange the following carbocations in decreasing order of stability.

Solution

Chemistry ma’am · 4 steps

Check where the oxygen sits: right next to C⁺ it donates a lone pair, one carbon away it only pulls.

1. Reference cation. A is a plain secondary cyclopentyl cation. Its only support is hyperconjugation and the +I effect of the neighbouring CH₂ groups.
2. Oxygen adjacent. In B the positive carbon is bonded directly to oxygen. Oxygen pushes a lone pair into the empty p orbital (+M effect), giving the resonance form C=O⁺ in which every atom has an octet. This outweighs oxygen's -I pull, so B is the most stable.
3. Oxygen one carbon away. In C a CH₂ sits between C⁺ and oxygen, so the lone pair cannot reach the empty orbital. Only the electron-withdrawing -I effect of oxygen acts, which makes the positive charge worse. C is less stable than A.

   Cation	Position of O	Effect of O on C⁺	Stability
   B	directly attached	+M dominates, octet resonance form	highest
   A	no oxygen	none, only hyperconjugation	middle
   C	one carbon away	only -I, withdraws electrons	lowest

4. So. The order is B > A > C, which is option (a).

The idea to keep. A lone-pair atom directly on C⁺ stabilises it by resonance, even if it is electronegative. Move it one carbon away and only its -I pull remains, which destabilises.

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