In the given question, the solution (according to the book) says that the electron withdrawing group will facilitate hydride transfer (not mentioning how so) and therefore, said the answer to be (D). But should not (C) be the answer, as the electron donating $\ce {-OCH3}$ group creates a "pushing" effect?

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I would like to say that you must be knowing that the best hydride donor is almost always the entity with most partial positive charge, which is most in $\bf D$ due to strong electron withdrawing group $\bf \ce{-NO_2}$ both via inductive and resonance effects,. Remember $\bf\ce{-OCH3}$ in $\bf C$ is actually an electron donating group via resonance and electron withdrawing group via inductive effect of which former dominates largely. I think the rest two are pretty simple.

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  • $\begingroup$ That, in fact, was my doubt- why is the compound with greatest partial positive charge the best hydride donor? $\endgroup$ – Abhishek Nov 30 '14 at 15:00
  • $\begingroup$ By the way -O- is an electron donating group, not electron withdrawing. $\endgroup$ – Abhishek Nov 30 '14 at 15:14
  • $\begingroup$ @Abhishek the reason of first point is mechanism and second point is false, consider -I $\endgroup$ – RE60K Dec 1 '14 at 5:37
  • $\begingroup$ @Abhishek the O- thing was correct by you $\endgroup$ – RE60K Dec 12 '14 at 8:21

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