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Arrange the following carbocations in decreasing order of stability: I: CH2=CH-CH2+ (allyl) II: CH3-CH2-CH2+ (n-propyl, primary) III: C6H5-CH2+ (benzyl)

1. III > II > I
2. II > III > I
3. I > II > III
4. III > I > II

Correct answer: III > I > II

Solution

Benzyl cation (III) is stabilised by resonance delocalisation into the aromatic ring (several resonance forms), making it the most stable. Allyl cation (I) is resonance-stabilised over one C=C, less than benzyl. n-Propyl cation (II) is an unstabilised primary cation, the least stable. Order: III > I > II.

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