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(D) The ability to act as a hydride donor depends on the stability of the resulting carbonyl compound. $A$ better hydride donor is one that forms a mo

Vedclass · Accepted Answer

Both $(b)$ and $(c)$.

Vedclass · Answer

(D) The ability to act as a hydride donor depends on the stability of the resulting carbonyl compound. $A$ better hydride donor is one that forms a more stable carbonyl compound upon losing a hydride ion. Electron-donating groups (EDGs) destabilize the carbonyl group by increasing electron density,while electron-withdrawing groups (EWGs) stabilize it. However,in the context of the Cannizzaro-like mechanism or hydride transfer,the species that is more electron-rich (due to EDGs) is a better hydride donor because the negative charge is less stabilized by the ring,making the hydride more 'available' to be transferred. The $-OCH_3$ and $-CH_3$ groups are electron-donating groups ($+M$ and $+I$ effects respectively),which increase the electron density on the benzene ring and the carbon atom,making the hydride transfer easier compared to the unsubstituted or $p-NO_2$ substituted species. Thus,both $(b)$ and $(c)$ are better hydride donors than the reference compound.

$x =$ number of compounds that are better hydride donors than $Ph-CH(O^{\Theta})_2$.

Similar Questions

Write the correct order of acidic strength for the following:
$(I) \ CH_3 - NO_2$
$(II) \ NO_2 - CH_2 - NO_2$
$(III) \ CH_3 - CH_2 - NO_2$
$(IV) \ (NO_2)_3CH$

Which of the following does not show hyperconjugation?

What is the correct order of $K_a$ values for the given compounds?

Which of the following represents the hyperconjugation effect?

The nitration will mainly take place at position $.$

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