Addition of HI on double bond of propene yields isopropyl iodide and not n-propyl iodide, because addition proceeds through :

a more stable carbanion

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a more stable carbonium ion

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a more stable free radical

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none of these

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Solution

The correct option is B a more stable carbonium ion
The reaction proceeds through a more stable carbonium ion as shown below:

$C H_{3} - C H = C H_{2} H^{+} - - \to C H_{3} - \oplus C H - C H_{3} I^{-} - \to C H_{3} - C H (I) C H_{3}$

Option B is correct.

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Similar questions

Assertion (A): In addition if HBr to propene, isopropyl bromide is major

product but n - propyl bromide is minor product

Reason(R): Isopropyl carbonium ion is more stable than n - propyl carbonium ion

Electrophilic addition on a carbon-carbon double bond involves the intermediate formation of a more stable carbocation. This statement is called

2^{0}

1^{0}

B r_{2}

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