JEE Main 2021 · 17 March · evening · Single correct · +4 / −1
Choose the correct statement regarding the formation of carbocations A and B given.
CH₃-CH₂-CH=CH₂+HBr {→ CH₃-CH₂-CH₂-C(+)H₂+Br⁻; → CH₃-CH₂-C(+)H-CH₃+Br⁻
- (a) Carbocation A is more stable and formed relatively at faster rate
- (b) Carbocation B is more stable and formed relatively at slow rate
- (c) Carbocation B is more stable and formed relatively at faster rate
- (d) Carbocation A is more stable and formed relatively at slow rate
Answer and solution
Answer (c) Carbocation B is more stable and formed relatively at faster rate
Solution
Chemistry ma’am · 4 steps
Count the alkyl groups on the positive carbon: more neighbours, more stability, faster formation.
- Classify each cation. In A, the positive carbon is the end CH₂, bonded to only one other carbon, so A is primary (1°). In B, the positive carbon carries a methyl group and an ethyl group, so B is secondary (2°).
- Compare stability. Alkyl groups push electron density towards the positive carbon by the +I effect and by hyperconjugation. B has 5 α-hydrogens available for hyperconjugation and A has only 2, so B spreads its charge better and is more stable.
- Link stability to rate. Making a carbocation is the slow, uphill step of HBr addition, and its transition state looks a lot like the carbocation (Hammond postulate). A more stable cation therefore sits behind a lower activation energy, so B forms faster.
- So. B is more stable and is formed at the faster rate, which is option (c). This is also why the major product is 2-bromobutane, as Markovnikov's rule predicts.
The idea to keep. Carbocation stability runs 3° > 2° > 1°. In electrophilic addition, the more stable cation also has the lower energy barrier, so it is the one that forms faster.
Answer from the NTA final answer key. Solution by PrepWiser.