JEE Main 2023 · 10 April · evening · Single correct · +4 / −1
The decreasing order of hydride affinity for following carbocations is:
Choose the correct answer from the options given below:
- (a) A, C, D, B
- (b) C, A, B, D
- (c) A, C, B, D
- (d) C, A, D, B
Answer and solution
Answer (b) C, A, B, D
Solution
Chemistry ma’am · 4 steps
Hydride affinity is stability read backwards: the most stable cation wants an H⁻ the least.
- What it measures. Hydride affinity is how strongly a carbocation grabs an H⁻ to fill its empty p orbital. A cation that is already well stabilised gains little from this, so higher stability means lower hydride affinity.
- Rank the stability. Cyclopropyl rings donate very strongly. Their strained, bent C–C bonds overlap with the empty p orbital, and each ring does better than a phenyl. Three phenyls still beat one vinyl plus two methyls, and an allylic tertiary cation beats a plain tertiary one.
Cation Main stabilisation Stability D, tricyclopropylmethyl 3 cyclopropyl rings (bent-bond overlap) highest B, triphenylmethyl resonance over 3 benzene rings second A, dimethylallyl (tertiary) allylic resonance + 2 methyls third C, tert-butyl hyperconjugation and +I of 3 methyls only lowest - Reverse it. Stability: D > B > A > C. Reading this backwards gives the hydride affinity order C > A > B > D.
- So. The decreasing order of hydride affinity is C, A, B, D, which is option (b).
The idea to keep. Any question on hydride affinity, or on how readily a cation is attacked by a nucleophile, is a stability question in reverse. Remember that cyclopropyl stabilises a cation even better than phenyl.
Answer from the NTA final answer key. Solution by PrepWiser.