Animated Solution for Chemistry - Organic Chemistry: Identify A in the following chemical reaction.

Crossed Cannizzaro

Our journey begins with $p$-methoxybenzaldehyde. The first reagents thrown into the flask are formaldehyde ($HCHO$) and sodium hydroxide ($NaOH$).

Notice a crucial detail: neither of these aldehydes possesses an $\alpha$-hydrogen. This is the classic signature of a Cannizzaro reaction. Because we have two different aldehydes, it's a crossed Cannizzaro.

Formaldehyde is sterically unhindered and highly susceptible to nucleophilic attack by the hydroxide ion. Therefore, it acts as the reducing agent, getting oxidized to formate ($HCO{O}^{-}$). Meanwhile, our aromatic aldehyde is reduced, transforming its carbonyl group into an alcohol. We now have $p$-methoxybenzyl alcohol.

Williamson Ether Synthesis

In the second act, we introduce sodium hydride ($NaH$) and ethyl bromide ($C{H}_{3}C{H}_{2}Br$).

Sodium hydride is a formidable base. It aggressively plucks the proton off our newly formed alcohol, generating a highly nucleophilic alkoxide ion.

This alkoxide wastes no time. It executes a textbook $S_{N}2$ attack on the electrophilic carbon of ethyl bromide, kicking off the bromide leaving group. This elegant maneuver is the Williamson ether synthesis, and it leaves us with a brand new benzylic ethyl ether linkage.

Selective Ether Cleavage by HI

For the grand finale, we subject our molecule to hydrogen iodide ($HI$) and heat. Hydrogen iodide is notorious for cleaving ethers, but our molecule presents a fascinating dilemma: it has two ether groups!

First, let's examine the methoxy group attached directly to the benzene ring. After protonation, the iodide ion attacks the less hindered methyl group. The bond between the oxygen and the benzene ring is incredibly strong due to partial double bond character from resonance. It refuses to break. Thus, the methoxy group is cleaved to yield a phenol and methyl iodide.

Now, turn your attention to the benzylic ether. Once protonated, the benzylic carbon becomes a prime target. The transition state for nucleophilic attack here is highly stabilized by the adjacent benzene ring. The iodide ion attacks, effortlessly cleaving the $C-O$ bond to form a benzylic iodide and ethanol.

The dust settles, and we can finally observe our masterpiece.

The bottom methoxy group has been unmasked as a phenol ($-OH$). The top benzylic ether has been transformed into an iodomethyl group ($-C{H}_{2}I$).

Our final product is 4-(iodomethyl)phenol. This perfectly matches option (c). A truly beautiful sequence of reduction, etherification, and selective cleavage!