See image — Hydrocarbons Chemistry Question

💡 Solution & Explanation

Diimide (N2H2, sometimes written loosely as N2H4 in certain texts but the active species is diimide N2H2) is a highly selective reducing agent. Concept: Diimide reduces non-polar or symmetrical double bonds via a concerted syn-addition mechanism. It delivers two hydrogen atoms simultaneously in a cyclic transition state to pi bonds that are relatively non-polar. Reasoning: (a) C=O is a polar double bond (carbonyl); diimide does not effectively reduce polar bonds because the cyclic transition state is not favorable for electronegative heteroatom-containing bonds. (b) C≡N (nitrile/imine) is also a polar bond involving nitrogen, and diimide is not used to reduce C≡N triple bonds. (c) -NO2 (nitro group) is a polar functional group; diimide does not reduce nitro groups. (d) -CH=CH- (alkene, carbon-carbon double bond) is a non-polar, symmetrical double bond. Diimide selectively reduces C=C double bonds (alkenes) through a concerted, syn, pericyclic six-membered transition state in which both hydrogens are delivered to the same face of the double bond simultaneously. This selectivity for non-polar pi bonds over polar functional groups is the hallmark of diimide reduction. Why other options fail: Carbonyl, nitrile, and nitro groups all contain electronegative atoms creating polarity; diimide's cyclic transition state is geometrically and electronically suited only for relatively symmetrical pi systems like C=C. Therefore, the correct answer is D.