Which forms di-iodide on reaction with HI (excess)?

I and II both

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II only

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I only

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none

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Solution

The correct option is A: I only

$I$ will form di-iodide on reaction with excess $H I$ .

In the first step, it will form $O H - C H_{2} - C H_{2} - C H_{2} - C H_{2} - C H_{2} I$

but then again on reaction with $H I$ it will form $I C H_{2} - C H_{2} - C H_{2} - C H_{2} - C H_{2} I$
NOTE:
In the other compound-II, the Lone pair electron is involved in the resonance hence it does not gives an iodized product.

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