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Choose the Best Answer · Q20

Q.Among the following ethers, which one will produce methyl alcohol on treatment with hot HI?

a) (CH3)3C-O-CH3
b) (CH3)2CH-CH2-O-CH3
c) CH3-(CH2)3-O-CH3
d) CH3-CH2-CH(CH3)-O-CH3
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Step 1. Ether cleavage by hot HI proceeds with iodide attacking the LESS sterically hindered of the two carbons attached to the ether oxygen (for a primary-vs-more-hindered pairing), or via SN1 at a tertiary centre if one side is tertiary.

Step 2. In (CH3)3C-O-CH3 (tert-butyl methyl ether), one side is a TERTIARY carbon and the other is a plain METHYL group; the tertiary C-O bond ionises readily (SN1, stable tertiary carbocation) while the methyl-oxygen bond simply releases as methanol -- i.e. the ether splits to give tert-butyl iodide + METHANOL directly. …

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