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In Cannizzaro's reaction of $C_6H_5CHO$ with concentrated $NaOH$, the products are
A$C_6H_5CH_2OH$ + $C_6H_5COO^-Na^+$
B$C_6H_5OH$ + $HCOO^-Na^+$
C$C_6H_5CH_3$ + $C_6H_5COOH$
D$C_6H_6$ + $CO_2$
Answer & Solution
Correct answer: A. $C_6H_5CH_2OH$ + $C_6H_5COO^-Na^+$
Disproportionation: one molecule of $C_6H_5CHO$ is reduced to $C_6H_5CH_2OH$ (benzyl alcohol), another is oxidised to $C_6H_5COO^-$ (benzoate). Mediated by hydride transfer.
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