In Cannizaro reaction given below: 2PhCHO:OH→PhCH2OH+PhCOO− the slowest step is:
A
the attack of :OH at the carbonyl group
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B
the transfer of hydride to the carbonyl group
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C
the abstraction of proton from the carboxylic group
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D
the deprotonation of PhCH2OH.
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Solution
The correct option is B the transfer of hydride to the carbonyl group Transfer of hydride ion from formed carboxylic acid to carbonyl carbon and it forms a stable product.