Among the following ethers, the one 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 (SN2): For primary or secondary alkyl groups, the iodide ion ($I^-$) is a strong nucleophile and attacks the less sterically hindered alkyl group. In a methyl ether ($R-O-CH_3$), $I^-$ would typically attack the methyl group to form methyl iodide ($CH_3I$) and the corresponding alcohol ($R-OH$). This does not yield methyl alcohol.
2. Exception (SN1): If one of the alkyl groups is tertiary (e.g., tert-butyl), the reaction proceeds via an SN1 mechanism due to the stability of the tertiary carbocation .

• Protonation: The ether oxygen is protonated.
• Cleavage: The bond between the oxygen and the tertiary carbon breaks, forming a stable tertiary carbocation and methyl alcohol ($CH_3OH$).
• Substitution: The iodide ion then attacks the tertiary carbocation to form tert-butyl iodide.

Therefore, only an ether with a tertiary group, like tert-butyl methyl ether ($(CH_3)_3C-O-CH_3$), will yield methyl alcohol ($CH_3OH$) and the tertiary alkyl iodide.