Compound among the following that will produce methyl alcohol on treatment with hot concentrated HI is:

The reaction of ethers with hot concentrated HI involves the cleavage of the C-O bond. The mechanism depends on the nature of the alkyl groups attached to the oxygen.

1. General Rule ($S_N2$): If the alkyl groups are primary or secondary, the reaction proceeds via an $S_N2$ mechanism. The iodide ion ($I^-$) attacks the less sterically hindered alkyl group. For methyl ethers ($R-O-CH_3$), the $I^-$ attacks the methyl group, forming methyl iodide ($CH_3I$) and the corresponding alcohol ($ROH$). This does not yield methyl alcohol.
2. Exception ($S_N1$): If one of the alkyl groups is tertiary (e.g., tert-butyl), the reaction proceeds via an $S_N1$ mechanism. The reaction is governed by the stability of the carbocation intermediate .

• Protonation: The ether oxygen is protonated.
• Cleavage: The weak C-O bond breaks to form the stable tertiary carbocation ($(CH_3)_3C^+$) and methyl alcohol ($CH_3OH$).
• Substitution: The iodide ion then attacks the carbocation to form tert-butyl iodide.

Therefore, only the ether with the tertiary group, tert-butyl methyl ether ($(CH_3)_3C-O-CH_3$), will produce methyl alcohol.