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In the Cannizzaro reaction shown below, 2 PhCHO —(OH−)→ PhCH2OH + PhCO2(−) which step occurs most slowly?

1. Nucleophilic attack of OH− on the carboxyl group
2. Hydride transfer to the carbonyl carbon
3. Removal of a proton from the carboxylic acid group
4. Deprotonation of PhCH2OH

Correct answer: Hydride transfer to the carbonyl carbon

Solution

The hydride transfer to the carbonyl carbon is the rate-determining step because it involves the formation of a new bond and is typically slower than the other steps, which are more straightforward proton transfers or nucleophilic attacks.

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