Animated Solution for Chemistry - Alcohols, Phenols and Ethers: The major product of the following reaction is

We are given an ether molecule, specifically $p$-cyanophenyl 2-phenylethyl ether, and we are treating it with an excess of hydrogen iodide ($HI$) under heating conditions. The core question here is to determine the regioselectivity of the ether cleavage. When an asymmetric ether is subjected to strong acidic conditions, the molecule must decide which of the two carbon-oxygen bonds will break.

The reaction begins with the protonation of the ether oxygen by the strong acid $HI$, forming an oxonium ion intermediate. This protonation transforms the oxygen into a much better leaving group.

Now, the iodide ion ($I^{-}$) must act as a nucleophile and attack one of the adjacent carbon atoms. We have two choices: the aromatic carbon of the benzene ring or the aliphatic $s{p}^3$ hybridized carbon of the alkyl chain.

Here is the critical catch: the bond between the oxygen atom and the aromatic ring ($Ar-O$) possesses partial double bond character. This is due to the delocalization of the oxygen's lone pair electrons into the $\pi$-system of the benzene ring. Because of this resonance stabilization, the $Ar-O$ bond is exceptionally strong and is almost never broken during standard ether cleavage reactions.

Since the $Ar-O$ bond is off-limits, the iodide ion has no choice but to attack the aliphatic carbon. It approaches the less sterically hindered $s{p}^3$ carbon via an $S_{N}2$ mechanism. As the $I-C$ bond forms, the $O-C$ bond breaks, releasing the aromatic portion as a stable phenol derivative.

Therefore, the cleavage exclusively yields $p$-cyanophenol ($p-NC-C_6{H}_4-OH$) and 1-iodo-2-phenylethane ($Ph-C{H}_2-C{H}_2-I$). Looking at our given choices, this perfectly aligns with option (d).