The reaction of CH_3-CH=CH_2 (propene) with H | vedclass

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Question

(A) The addition of $HI$ to propene follows Markovnikov's rule. In the first step,the electrophile $H^+$ attacks the double bond to form a carbocation

Vedclass · Accepted Answer

The reaction proceeds through a more stable carbocation.

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(A) The addition of $HI$ to propene follows Markovnikov's rule. In the first step,the electrophile $H^+$ attacks the double bond to form a carbocation intermediate. Two possible carbocations can be formed: $1$. $CH_3-CH^+-CH_3$ (secondary carbocation,more stable) $2$. $CH_3-CH_2-CH_2^+$ (primary carbocation,less stable) Since the secondary carbocation is more stable due to the inductive effect and hyperconjugation,the reaction proceeds through this intermediate. Finally,the nucleophile $I^-$ attacks the secondary carbocation to form $CH_3-CH(I)-CH_3$ (isopropyl iodide).

The reaction of $CH_3-CH=CH_2$ (propene) with $HI$ gives isopropyl iodide instead of $n$-propyl iodide because:

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