Chemoselective Reduction: Which Reagent Touches Which Group
Reduction questions rarely ask what a reagent does to a simple ketone. They hand you a molecule with two or three reducible groups (ketone, ester, nitrile, acid, C=C) and ask which reagent changes only the one you want. Four ideas cover almost every case: the gentle hydride (NaBH₄) versus the thorough one (LiAlH₄), reagents that spare an acid or a C=C, Wolff-Kishner reduction of C=O to CH₂, and the Grignard route that turns a nitrile into a ketone.
LiAlH₄ vs NaBH₄ on a Ketone, Ester and Nitrile is the central idea in this question pattern. Identify the condition in the stem before applying the method below.
NaBH₄ is the gentle hydride, LiAlH₄ is the thorough one
Sodium borohydride in an alcohol solvent at room temperature reduces aldehydes and ketones to alcohols and, in practice, leaves esters, nitriles, amides, carboxylic acids and isolated C=C bonds alone. Lithium aluminium hydride (in dry ether or THF) is a far more powerful hydride donor: it reduces ketones, aldehydes, esters and carboxylic acids to alcohols, and amides and nitriles to amines, while still leaving an isolated C=C untouched. So for a molecule that carries a ketone, a tert-butyl ester and an aryl nitrile, NaBH₄ changes only the ketone (giving the secondary alcohol with the ester and nitrile intact), while LiAlH₄ reduces all three groups (ketone to CH-OH, ester to CH₂OH, nitrile to CH₂NH₂). To answer, list every reducible group first, then ask which of them each reagent can actually reach.
When a carboxylic acid must survive
LiAlH₄ and DIBAL-H are aluminium hydrides strong enough to attack the carboxylic acid as well, and diborane (B₂H₆) is a trap on two counts: it reduces carboxylic acids faster than it reduces ketones, and it also adds across C=C bonds (hydroboration). When a question asks which reagents are NOT suitable, the answer set is usually every reagent that is more powerful, or more specialised, than NaBH₄.
Enone to allylic alcohol: reduce the C=O, keep the C=C
Catalytic hydrogenation (H₂, Pd/C) reduces the C=C bond first, a dissolving metal (Na in liquid NH₃) reduces the conjugated C=C by electron transfer, and Zn-Hg with HCl (Clemmensen) removes the C=O entirely to leave a CH₂ group. A metal hydride such as LiAlH₄ delivers H⁻ to the carbonyl carbon and leaves the C=C in place, so it gives the allylic alcohol (NaBH₄ with CeCl₃, the Luche reduction, is the textbook way to make 1,2-reduction highly selective).
Wolff-Kishner deletes the C=O, and an α-halogen turns it into an alkene
it is the base-mediated counterpart of the acid-mediated Clemmensen reduction. If the ketone carries a leaving group on the α-carbon (such as the chlorine in 2-chloro-1-tetralone), the hydrazone anion expels it instead of simply finishing the reduction, and the product is an alkene (here 1,2-dihydronaphthalene): N₂ is lost, chloride is lost, and a C=C appears between the two carbons. So an α-halo ketone does not give the deoxygenated alkyl chloride, the alcohol or the unchanged ketone.
A nitrile becomes a ketone, then Wolff-Kishner finishes the job
A Grignard reagent adds once to a nitrile C≡N to give an imine salt (a magnesium iminate), which resists a second addition because the nitrogen is negatively charged. Aqueous NH₄Cl or dilute acid then hydrolyses the C=N to C=O, so nitrile plus RMgX followed by aqueous work-up delivers a KETONE, not an aldehyde, acid or alcohol.
Cyclohexanecarbonitrile with isobutylmagnesium bromide, then aqueous work-up, then Wolff-Kishner therefore ends as cyclohexyl-CH₂-CH₂-CH(CH₃)₂: the old nitrile carbon is now the CH₂ attached directly to the ring, followed by the CH₂ and CH(CH₃)₂ of the isobutyl group.
OlympIQ exam lens
- ✓NaBH₄ (MeOH, room temperature) reduces aldehydes and ketones only; esters, nitriles, acids and isolated C=C survive
- ✓LiAlH₄ reduces ketones, esters, acids and amides to alcohol/amine-type products and nitriles to primary amines, but not an isolated C=C
- ✓To keep a COOH while reducing a ketone, use NaBH₄; LiAlH₄, DIBAL-H and B₂H₆ are unsuitable (B₂H₆ also hydroborates C=C)
- ✓Enone to allylic alcohol needs a carbonyl-selective hydride; H2/Pd-C hits the C=C, Na/NH₃ reduces the conjugated C=C, Zn-Hg/HCl removes the C=O
- ✓Wolff-Kishner (N₂H₄, KOH, heat) turns C=O into CH₂; an α-leaving group such as Cl is expelled and an alkene forms instead
- ✓Nitrile + RMgX then aqueous work-up gives a ketone; one more Wolff-Kishner step makes it an alkane
When a question lists several reducible groups, do not pick the reagent you know best. Write down every group that could react and cross each one off only if the reagent truly cannot reach it. LiAlH₄ is the usual wrong answer when an ester, nitrile or acid must survive, and NaBH₄ is the usual wrong answer when the target needs an ester or nitrile reduced.
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