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When mixed ethers with two distinct alkyl groups undergo cleavage, the resulting alcohol and alkyl iodide are determined by the type of alkyl groups. For primary or secondary alkyl groups, the smaller alkyl group typically forms the alkyl iodide. Which of the following reactions illustrates this process?

1. CH3—O—CH(CH3)2 reacts with HI to produce CH3I and (CH3)2CHOH.
2. Methyl iodide is formed.
3. Isopropyl alcohol is produced.
4. The alkyl halide is always derived from the smaller alkyl group.

Correct answer: CH3—O—CH(CH3)2 reacts with HI to produce CH3I and (CH3)2CHOH.

Solution

In the cleavage of mixed ethers with HI, the smaller alkyl group (here CH3) forms the alkyl iodide (CH3I), while the larger group forms the alcohol ((CH3)2CHOH).

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