Animated Solution for Chemistry - Organic Chemistry: Trichloroacetaldehyde was subjected to Cannizzaro's reaction by using NaOH. The mixture of the products contains sodium trichloroacetate ion and another compound. The other compound is

The Cannizzaro reaction is a classic organic chemistry transformation that beautifully demonstrates the concept of disproportionation. Let's dive into the mechanics of this reaction using trichloroacetaldehyde as our subject.

Our starting material is trichloroacetaldehyde ($CC{l}_{3}CHO$). The most critical feature to observe here is the $\alpha$-carbon—the carbon atom directly adjacent to the carbonyl group. In trichloroacetaldehyde, this $\alpha$-carbon is bonded to three chlorine atoms.

Why is this important? Because it means there are absolutely zero $\alpha$-hydrogens. Aldehydes that possess $\alpha$-hydrogens typically undergo the aldol condensation when treated with a base. However, without these acidic protons, the molecule is forced down a different reaction pathway when exposed to a strong base like $NaOH$.

When an aldehyde lacking $\alpha$-hydrogens is treated with concentrated alkali, it undergoes the Cannizzaro reaction. This is a redox disproportionation, meaning that out of two reacting molecules of the aldehyde, one is oxidized and the other is reduced simultaneously.

The reaction begins with the nucleophilic attack of the hydroxide ion ($O{H}^{-}$) on the electrophilic carbonyl carbon of the first aldehyde molecule. This forms a tetrahedral intermediate.

The magic happens in the next step: the intermediate collapses, reforming the carbon-oxygen double bond, and expels a hydride ion ($H^{-}$). This hydride ion is a very strong nucleophile and immediately attacks the carbonyl carbon of a second aldehyde molecule. This hydride transfer is the rate-determining step of the reaction.

The molecule that lost the hydride ion is oxidized to a carboxylic acid. However, because we are in a basic medium ($NaOH$), it immediately deprotonates to form the carboxylate salt. In our case, this yields sodium trichloroacetate ($CC{l}_{3}COONa$).

The second molecule, which accepted the hydride ion, is reduced to an alkoxide ion, which then picks up a proton from the solvent to become a primary alcohol. For trichloroacetaldehyde, this reduction product is 2,2,2-trichloroethanol ($CC{l}_{3}C{H}_{2}OH$).

Therefore, the mixture of products contains sodium trichloroacetate and 2,2,2-trichloroethanol. The correct option is (a).