Major product of the given reaction is CH_2=C | vedclass

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(B) The reaction of allyl iodide with excess $HI$ proceeds as follows: $CH_2=CH-CH_2-I + HI \rightarrow CH_3-CHI-CH_2-I$. Vicinal diiodides are unstab

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$CH_3-CHI-CH_3$

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(B) The reaction of allyl iodide with excess $HI$ proceeds as follows: $CH_2=CH-CH_2-I + HI \rightarrow CH_3-CHI-CH_2-I$. Vicinal diiodides are unstable and undergo deiodination to form propene: $CH_3-CHI-CH_2-I \rightarrow CH_3-CH=CH_2 + I_2$. Propene then reacts with $HI$ according to Markovnikov's rule to give $2$-iodopropane as the major product: $CH_3-CH=CH_2 + HI \rightarrow CH_3-CHI-CH_3$.

Major product of the given reaction is $CH_2=CH-CH_2-I \xrightarrow[CCl_4]{HI(excess)}$

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