1. Recall the mechanism of Cannizzaro reaction

For benzaldehyde ($\text{PhCHO}$), which has no $\alpha$-hydrogen, Cannizzaro reaction occurs in concentrated base.

Overall reaction: $2\text{PhCHO} \xrightarrow{\,\text{OH}^-\,} \text{PhCH}_2\text{OH} + \text{PhCOO}^-$

The mechanism is:

Step 1: Nucleophilic addition of hydroxide to benzaldehyde $\text{PhCHO} + \text{OH}^- \rightarrow \text{PhCH(OH)O}^-$

This gives a tetrahedral alkoxide intermediate.

Step 2: Hydride transfer from this intermediate to a second molecule of benzaldehyde $\text{PhCH(OH)O}^- + \text{PhCHO} \rightarrow \text{PhCOO}^- + \text{PhCH}_2\text{O}^-$

This is the characteristic disproportionation step.

Step 3: Proton transfer $\text{PhCH}_2\text{O}^- + \text{H}_2\text{O} \rightarrow \text{PhCH}_2\text{OH} + \text{OH}^-$

2. Identify the rate-determining (slowest) step

In the Cannizzaro reaction, the difficult step is the transfer of hydride from the tetrahedral intermediate to another carbonyl carbon. This intermolecular hydride migration has the highest activation energy and is known to be the rate-determining step.

So the slowest step is: $\boxed{\text{hydride transfer to the carbonyl group}}$

3. Check the options

• A: the transfer of hydride to the carbonyl group
  ✅ Correct. This is the slowest step.

• B: the abstraction of proton from the carboxylic group
  ❌ Not the slow step; proton transfer steps are generally fast.

• C: the attack of $\text{OH}^-$ at the carboxyl group
  ❌ Not relevant here; hydroxide attacks the aldehyde carbonyl, not a carboxyl group.

• D: the deprotonation of $\text{PhCH}_2\text{OH}$
  ❌ Not the rate-determining step; also the key proton-transfer processes are fast.

4. Final answer

The slowest step is: $\boxed{\text{A}}$

5. Comparison with stored correct answer

Stored correct answer = A
Derived answer = A
So they agree.