ether cleavage

Question

Ether cleavage by HI proceeds to give:

RankGuru · Accepted Answer

1. **Step 1**: The Ether oxygen is protonated by the strong acid $HI$ building a dialkyl oxonium ion.
2. **Step 2 (SN2/SN1)**: The Iodide ion ($I^-$), being a good nucleophile, attacks the less sterically hindered carbon (if $SN2$) or the one forming a stable carbocation (if $SN1$).
3. **Bond Breaking**: This results in the cleavage of the $C-O$ bond on one side.
4. **Fragments**: The products are an **Alkyl Iodide** and an **Alcohol**.
5. **Excess Conditions**: If $HI$ is in excess and heated, the resulting alcohol can further react to form a second molecule of alkyl iodide.

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ether cleavage

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