Animated Solution for Chemistry - Organic Chemistry: The compounds A and B in the following reaction are, respectively

Imagine you are a detective trying to trace the path of a chemical reaction. You start with a simple, elegant molecule: benzene. It's stable, happy, and minding its own business. But then, you introduce a mixture of formaldehyde ($HCHO$) and hydrogen chloride ($HCl$). This is where the plot thickens.

This specific combination of reagents is famous in organic chemistry. It's known as the Blanc chloromethylation reaction. But how does it work? Let's break it down.

Benzene is an electron-rich ring, which means it loves to attack electrophiles (electron-poor species). However, formaldehyde and $HCl$ on their own aren't quite electrophilic enough to tempt benzene. They need to team up.

When formaldehyde reacts with $HCl$, the oxygen atom gets protonated. This creates a highly reactive intermediate: the hydroxymethyl cation ($C{H}_2=O^{+}H$). This cation is the perfect bait for our benzene ring.

Now, the electron-rich $\pi$-cloud of benzene reaches out and attacks the hydroxymethyl cation. Through a classic electrophilic aromatic substitution, the ring temporarily loses its aromaticity, only to quickly regain it by kicking out a proton ($H^{+}$).

The result? Benzyl alcohol ($C_6{H}_{5}C{H}_{2}OH$).

But wait! If you look at the options, you might be tempted to choose benzyl alcohol as compound A. Here is where mistakes happen. We are in an acidic medium (remember the $HCl$?). Alcohols in the presence of strong acids don't just sit around. The hydroxyl group ($-OH$) gets protonated to form $-O{H}_2^{+}$, which is an excellent leaving group (water).

The chloride ion ($C{l}^{-}$) floating around seizes the opportunity, attacks the benzylic carbon, and kicks out the water molecule. This rapid nucleophilic substitution transforms our intermediate into benzyl chloride ($C_6{H}_{5}C{H}_{2}Cl$). This is our true Compound A.

Now that we have benzyl chloride, we move to the second phase of our reaction. We introduce silver cyanide ($AgCN$).

This is a classic trap in JEE chemistry. You might think, "Cyanide is $C{N}^{-}$, so it will attack and form a cyanide (nitrile)." But you must look closely at the nature of the reagent.

Silver cyanide is predominantly covalent. The bond between silver and carbon is strong, meaning the carbon atom is not free to act as a nucleophile. Instead, the lone pair of electrons on the nitrogen atom steps up to the plate.

Nitrogen attacks the benzylic carbon in an $S_{N}2$ fashion, displacing the chloride ion. Because the nitrogen atom forms the new bond with the carbon, the resulting functional group is an isocyanide ($-NC$), not a cyanide ($-CN$).

Therefore, the reaction of benzyl chloride with $AgCN$ yields benzyl isocyanide ($C_6{H}_{5}C{H}_{2}NC$). This is our Compound B.

If the reaction had used potassium cyanide ($KCN$) instead, the story would have ended differently. $KCN$ is ionic, providing free $C{N}^{-}$ ions where the carbon acts as the nucleophile, leading to benzyl cyanide.

By understanding the subtle differences in reagent properties, we've successfully navigated the traps and solved the mystery. Compound A is benzyl chloride, and Compound B is benzyl isocyanide!