In the Cannizzaro reaction,the intermediate t | vedclass

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(C) In the Cannizzaro reaction,the rate-determining step involves the transfer of a hydride ion $(H^-)$ from the gem-diol dianion intermediate to anot

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(C) In the Cannizzaro reaction,the rate-determining step involves the transfer of a hydride ion $(H^-)$ from the gem-diol dianion intermediate to another molecule of the aldehyde.The ability of the intermediate to act as a hydride donor depends on the electron density at the carbon atom attached to the hydrogen. An electron-donating group (like $-OCH_3$) increases the electron density on the carbon,making the $C-H$ bond more polarized and facilitating the release of the hydride ion.Conversely,an electron-withdrawing group (like $-NO_2$) decreases the electron density,making it a poorer hydride donor.Comparing the options:- Option $A$ is a mono-anion,which is not the primary hydride donor.- Option $B$ is the unsubstituted benzaldehyde gem-diol dianion.- Option $C$ has a methoxy group $(-OCH_3)$,which is an electron-donating group ($+M$ effect),making it the best hydride donor.- Option $D$ has a nitro group $(-NO_2)$,which is an electron-withdrawing group ($-M$ effect),making it the poorest hydride donor.Therefore,the intermediate with the electron-donating group is the best hydride donor.

In the Cannizzaro reaction,the intermediate that will be the best hydride donor is:

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