The ether shown reacts with one equivalent of $HI$ . By convention, a methyl alkyl ether is cleaved by $S_{N} 2$ attack of iodide at the methyl carbon. Choose all correct statements.

Answer: A, B, D

The ether is methyl isopropyl ether. Iodide attacks the methyl carbon of the protonated ether, giving

CH X_{3} I

and propan-2-ol (A, B), so 2-iodopropane is not the main iodide (C false). With excess

HI

the alcohol is converted to 2-iodopropane, so both groups end up as iodides (D).