Animated Solution for Chemistry - Organic Chemistry: 'A' and 'B' in the following reaction are

From Aniline to Benzaldehyde

Organic chemistry is often like solving a beautiful puzzle where each reagent acts as a specific key to unlock the next molecular structure. In this problem, we are given a two-step reaction sequence starting from Aniline, and our goal is to identify the intermediate product A and the final product B. Let's embark on this molecular journey and decode the transformations step by step.

Diazotisation - The Gateway Reaction

Our journey begins with Aniline ($C_6{H}_{5}N{H}_2$), a primary aromatic amine. The first set of reagents thrown at it is a mixture of sodium nitrite ($NaN{O}_2$) and hydrochloric acid ($HCl$) at a strictly maintained, ice-cold temperature of $0-5^{\circ }C$.

This specific combination is the hallmark of the Diazotisation reaction. At these low temperatures, the nitrous acid generated in situ reacts with the primary amine group, converting it into a highly reactive diazonium group ($-N_2^{+}C{l}^{-}$).

The resulting compound is Benzene diazonium chloride. This diazonium salt is incredibly versatile but highly unstable at room temperature, which is exactly why the ice bath is non-negotiable!

Nucleophilic Substitution - Forming Cyanobenzene

Without isolating the diazonium salt, the reaction mixture is immediately treated with potassium cyanide ($KCN$). The diazonium group ($-N_2^{+}$) is one of the best leaving groups in organic chemistry because it leaves as stable, neutral nitrogen gas ($N_2$).

The cyanide ion ($C{N}^{-}$) acts as a strong nucleophile, attacking the benzene ring and displacing the nitrogen gas. This type of substitution is closely related to the Sandmeyer reaction.

This substitution yields our first unknown, Product A, which is Cyanobenzene (also known as benzonitrile).

Stephen's Reduction - The Art of Partial Reduction

Now that we have Cyanobenzene, we move to the second phase of the sequence. The reagents provided are stannous chloride ($SnC{l}_2$) and hydrochloric acid ($HCl$).

Whenever you see a nitrile ($-C\equiv N$) reacting with $SnC{l}_2/HCl$, your mind should immediately jump to Stephen's Reduction. This is a highly selective reaction designed to partially reduce nitriles into aldehydes.

In the first stage of Stephen's reduction, the stannous chloride provides electrons, and the acid provides protons, reducing the triple bond of the nitrile to a double bond, forming an imine hydrochloride intermediate.

Acidic Hydrolysis - The Final Touch

The final piece of the puzzle is the hydronium ion ($H_3{O}^{+}$), which signifies acidic hydrolysis. The imine intermediate is highly susceptible to hydrolysis.

When water attacks the carbon-nitrogen double bond in an acidic medium, the bond cleaves completely. The nitrogen atom is expelled as an ammonium ion ($N{H}_4^{+}$), and an oxygen atom takes its place, forming a carbonyl group ($C=O$).

This elegant hydrolysis transforms the imine into Benzaldehyde, which is our Product B.

By carefully analyzing the reagents and recalling our classic name reactions, we have successfully mapped the entire pathway.

- Product A is Cyanobenzene. - Product B is Benzaldehyde.

Matching these findings with the given options, we can confidently conclude that Option (d) is the correct answer. This problem beautifully illustrates how sequential functional group transformations can build complex molecules from simple starting materials.