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In the Cannizzaro reaction given : $2C_6H_5-CHO\quad \underrightarrow {OH^-}\quad C_6H_5-CH_2OH +C_6H_5COO^-$ the slowest step is

$\begin{array}{1 1}(a)\;\text{The abstraction of proton from the caboxylic acid}\\(b)\;\text{the deprotonation of }C_6H_5-CH_2OH\\(c)\;\text{the transfer of hydride to the carbonyl group}\\(d)\;\text{the attack of }OH^-\text{at the carbonyl groups}\end{array}$

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Slowest step is the transfer of Hydride to the carbonyl group because the hydride ion is directly transferred from one molecule of the aldehyde to the other and does not become free in solution.
Hence (c) is the correct answer.
answered Feb 18, 2014 by sreemathi.v
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