In Cannizzaro reaction given below <br> $2 Ph \; CHO \xrightarrow{\overset{(-)}{: OH}} Ph \; CH_2OH + PhC\overset{..}{O}_2^{(-)}$ the slowest step is :

Question

( A ) the deprotonation of $Ph \; CH_2OH$
( B ) the abstraction of proton from the carboxylic group
( C ) the transfer of hydride to the carbonyl group
( D ) the attack of : $\overset{(-)}{OH}$ at the carboxyl group