JEE Chemistry 2025 Question with Solution

Given: Four carbocations A, B, C, and D are to be arranged in decreasing order of stability.

Find: The correct order of stability.

The solution text identifies:

• A as triphenyl carbocation
• B as diphenyl carbocation
• C as tropylium carbocation
• D as a secondary carbocation

The stability arguments stated in the solution are based on resonance, aromaticity, and inductive / hyperconjugative stabilization.

• C (tropylium carbocation) is stated to be highly stable because it is aromatic with $6\pi$ electrons and has extensive charge delocalization.
• A (triphenyl carbocation) is more stable than B (diphenyl carbocation) because the positive charge can be delocalized over three phenyl rings instead of two.
• D (secondary carbocation) is the least stable because it lacks significant resonance stabilization.

Therefore, the order obtained from the solution working is:

$$C > A > B > D$$

This matches option D.

Note: The answer key marks option B, but the extracted solution explicitly concludes $C > A > B > D$. As per the solution, the correct option is D.

Given: Stability depends on how well the positive charge is dispersed.

Find: Compare the four carbocations.

1. C is the tropylium carbocation. Its positive charge is delocalized over the ring, and the ring has $6\pi$ electrons, so it is aromatic. Aromatic stabilization makes it exceptionally stable.

2. A is the triphenyl carbocation. The positively charged carbon is attached to three phenyl groups, so resonance delocalization is extensive.

3. B is the diphenyl carbocation. It is also resonance-stabilized, but only by two phenyl groups, so it is less stable than A.

4. D is a secondary carbocation. It is stabilized mainly by alkyl effects and hyperconjugation, but it does not have the strong resonance stabilization present in A, B, or C.

Hence,

$$C > A > B > D$$

So, the correct option is D.