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Arrange the following carbocations in decreasing order of stability: (a) CH3–CH2(+) (b) (CH3)2CH(+) (c) (CH3)3C(+)

1. c > b > a
2. a > b > c
3. b > c > a
4. a > c > b

Correct answer: c > b > a

Solution

Carbocation stability increases with the number of alkyl substituents on the positive carbon because of inductive electron donation (+I) and, more importantly, hyperconjugation. (CH3)3C(+) is tertiary (9 alpha C–H for hyperconjugation), (CH3)2CH(+) is secondary, and CH3–CH2(+) is primary. So stability order is tertiary > secondary > primary, i.e. c > b > a.

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