Animated Solution for Chemistry - Organic Chemistry: In Cannizzaro reaction given below $2\text{PhCHO}\stackrel{{\text{OH}}^{\ominus }}{\to }{\text{PhCH}}_2\text{OH}+{\text{PhCO}}_2^{\ominus }$ the slowest step is

The Hydride Transfer

The Cannizzaro reaction is a fascinating chemical dance, a classic example of a disproportionation reaction. It occurs when aldehydes that lack an $\alpha$-hydrogen (like benzaldehyde or formaldehyde) are treated with a concentrated strong base. Because they don't have that acidic $\alpha$-hydrogen, they cannot undergo the typical aldol condensation. Instead, they do something remarkable: one molecule sacrifices itself to be oxidized into a carboxylic acid salt, while it reduces another molecule into an alcohol.

To truly understand this reaction, we must look at its mechanism, which unfolds in three distinct acts:

1. The Nucleophilic Attack: The reaction kicks off with the hydroxide ion (${\text{OH}}^{-}$) acting as a nucleophile. It attacks the electrophilic carbonyl carbon of the first aldehyde molecule. This forms a tetrahedral intermediate, an alkoxide anion.

2. The Hydride Transfer: This is where the magic happens. The negatively charged oxygen on the intermediate pushes its electrons back down to reform the carbon-oxygen double bond. But carbon can only have four bonds! To make room, it must kick out a leaving group. It forcefully ejects a hydride ion (${\text{H}}^{-}$), which immediately attacks the carbonyl carbon of a second aldehyde molecule.

3. The Proton Exchange: The hydride transfer creates a carboxylic acid and an alkoxide ion. Because the carboxylic acid is highly acidic and the alkoxide is highly basic, a rapid proton exchange occurs, yielding the final products: a stable carboxylate salt and an alcohol.

In chemical kinetics, the overall speed of a multi-step reaction is bottlenecked by its slowest step, known as the Rate-Determining Step (RDS).

In the Cannizzaro reaction, the hydride transfer is the undisputed RDS. Why? Because a hydride ion (${\text{H}}^{-}$) is an exceptionally poor leaving group. Breaking the strong $\text{C-H}$ bond to expel it requires a massive amount of activation energy. The initial nucleophilic attack is relatively fast, and the final acid-base proton exchange is nearly instantaneous. But forcing that hydride to jump ship takes time and energy, making it the slowest and most critical step of the entire process.

Therefore, the correct answer is the transfer of hydride to the carbonyl group.