Question

In the cannizzaro reaction given below $2 Ph - CHO \ce{->[{ OH ^{-}}]} Ph - CH - OH + PhCO _{2}^{-}$the slowest step is

• A. the attack of $OH$ at the carbonyl group
• B. the transfer of hydride ion to the carbonyl group
• C. the abstraction of a proton from the carboxylic acid
• D. the deprotonation of $Ph - CH _{2} OH$

Answer 1

Answer: (B)

Transfer of hydride ion to the carbonyl group is the slowest or the rate determining step

$ \underset{\overset{+}{Ph - C H_{2}OH}}{Ph-\overset{\underset{||}{O}}{C}-OH} \ce{<=>} \left[ Ph - \overset{\underset{||}{O}}{C} OH + ph - \underset{\overset{|}{H}}{\overset{\underset{|}{O^{-}}}{C}}- H\right]$