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When unsymmetrical ethers undergo cleavage, the position where the bond breaks is influenced by the type of alkyl groups present, e.g., CH3—O—CH(CH3)2 + HI → CH3I + (CH3)2CHOH.

1. Methyl iodide is produced.
2. Isopropyl alcohol is formed.
3. The smaller alkyl group is the source of the alkyl halide.
4. Ethanol's boiling point is significantly higher than diethyl ether due to hydrogen bonding.

Correct answer: The smaller alkyl group is the source of the alkyl halide.

Solution

In the cleavage of unsymmetrical ethers by HI, the alkyl halide is preferentially formed from the smaller alkyl group due to steric hindrance and stability of the carbocation formed during the reaction.

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