Chapter 5: Problem 88
In Cannizzaro reaction given below \(2 \mathrm{Ph} \mathrm{CHO} \stackrel{\stackrel{\ominus}{\mathrm{OH}}}{\longrightarrow} \mathrm{PhCH}_{2} \mathrm{OH}+\mathrm{PhCO}_{2}^{\Theta}\) the slowest step is : (1) the transfer of hydride to the carbonyl group (2) the abstraction of proton from the carboxylic group (3) the deprotonation of \(\mathrm{PhCH}_{2} \mathrm{OH}\) (4) the attack of : \(\stackrel{\Theta}{\mathrm{OH}}\) at the carboxyl group
Short Answer
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