JEE Advanced level
A JEE Advanced-level example on General Organic Chemistry (GOC)
Question. Consider the following carbocation intermediates:
(I) (CH₃)₃C⁺ (tert-butyl cation)
(II) CH₂=CH−CH₂⁺ (allyl cation)
(III) C₆H₅−CH₂⁺ (benzyl cation)
(IV) CH₃−CH₂−CH₂⁺ (n-propyl cation)
Arrange these carbocations in order of increasing stability and identify the primary stabilization effect responsible for the stability of carbocation (II).
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- Identify the type of each carbocation: (I) is tertiary, (IV) is primary, (II) is allylic, and (III) is benzylic
- Primary carbocation (IV) is least stable as it has minimal stabilization from weak +I effect and only 2 alpha H atoms for hyperconjugation
- Tertiary carbocation (I) is stabilized by +I effect of three CH₃ groups and hyperconjugation from 9 alpha hydrogen atoms
- Allyl cation (II) is stabilized by resonance: CH₂=CH−CH₂⁺ ↔ ⁺CH₂−CH=CH₂, where positive charge is delocalized over two carbons
- Benzyl cation (III) has the most extensive resonance stabilization as the positive charge delocalizes into the benzene ring at ortho and para positions
- Since resonance effect > hyperconjugation > inductive effect, the stability order is: IV < I < II < III
Answer: Increasing stability order: IV < I < II < III (n-propyl < tert-butyl < allyl < benzyl). The primary stabilization effect for allyl cation (II) is resonance.
This example was solved twice independently and both solutions agree.