The major products formed in the following reaction are: [Reaction Structure Missing]

The reaction of ethers with hydroiodic acid (HI) involves the cleavage of the C–O bond. The products depend on the nature of the alkyl groups:

1. Standard $S_N2$ Mechanism: If the alkyl groups are primary or secondary, the iodide ion ($I^-$) attacks the less sterically hindered alkyl group. The products are the corresponding alkyl iodide and an alcohol.

• Example: $CH_3-O-CH_2CH_3 + HI \rightarrow CH_3I + CH_3CH_2OH$

2. Tertiary/Benzylic Groups ($S_N1$ Mechanism): If one of the alkyl groups is tertiary, the reaction proceeds via a stable carbocation intermediate. The iodide attacks the tertiary carbon. The products are a tertiary alkyl iodide and an alcohol.

• Example: $(CH_3)_3C-O-CH_3 + HI \rightarrow (CH_3)_3C-I + CH_3OH$

3. Aryl Ethers: The $O-C_{aryl}$ bond has partial double bond character and is strong. Cleavage occurs at the $O-C_{alkyl}$ bond. The products are Phenol and an Alkyl Iodide.

• Example: $C_6H_5-O-CH_3 + HI \rightarrow C_6H_5OH + CH_3I$