Reducing carbonyls — Oxidation & Reduction

Reducing a carbonyl is the reverse of the last section, and mechanically it is one idea: a hydride — H with two electrons — attacks the electrophilic carbonyl carbon, the π electrons go up onto oxygen, and the resulting alkoxide is protonated on workup to give an alcohol. The whole subject is which reagent delivers that hydride, and to what.

Two reagents, very different appetites

Substrate	NaBH₄	LiAlH₄
Aldehyde	1° alcohol	1° alcohol
Ketone	2° alcohol	2° alcohol
Ester	very slow / no	1° alcohol
Carboxylic acid	no reaction	1° alcohol
Amide	no reaction	amine
Nitrile	no reaction	1° amine
Alkene, alkyne	no reaction	no reaction

Two rows in that table are worth reading twice. An amide reduced by LiAlH₄ gives an amine, not an alcohol — the nitrogen stays attached and the oxygen leaves entirely, which makes it the standard route from an amide to an amine. And neither reagent touches an isolated alkene or alkyne: hydride reduction needs a polarized π bond to attack, and C=C has no electrophilic end. Reducing an alkene is catalytic hydrogenation's job, in the next section.

Why the difference, and how to use it

Aluminum is less electronegative than boron, so the Al–H bond is more polarized and LiAlH₄ is by far the more reactive hydride source. That single fact produces the practical differences:

NaBH₄ is mild enough to use in methanol or ethanol. It reacts with protic solvents only slowly, so you can simply dissolve it and go. That convenience is real and is part of why it is the default for a simple ketone.
LiAlH₄ reacts violently with water, including atmospheric moisture, and must be used in an aprotic ether solvent — dry THF or diethyl ether — with a separate aqueous workup afterwards to protonate the alkoxide.
Selectivity comes from choosing the weaker reagent. A molecule containing both a ketone and an ester is reduced at the ketone alone by NaBH₄, while LiAlH₄ would reduce both. Deliberately reaching for the less powerful reagent is the commonest way this material appears in a synthesis question.

Hydride reagents are nucleophiles, not acids or bases in the working sense, and they attack the carbonyl carbon. That is the same electrophilic carbon the nucleophilic addition chapter is about — so this is not a new mechanism, it is the general carbonyl addition with H⁻ as the nucleophile.

Esters and acids take two hydrides

Reducing a ketone delivers one hydride and stops, because the alkoxide formed has nothing to expel. An ester is different: the first hydride gives a tetrahedral intermediate that does have a leaving group, the alkoxide of the alcohol half. It collapses, expelling that alkoxide and giving an aldehyde — which is more electrophilic than the ester was, and is immediately reduced again by a second hydride.

So an ester goes to a primary alcohol and you cannot normally stop at the aldehyde, because the intermediate is more reactive than the starting material. Stopping there requires a deliberately weakened reagent (DIBAL-H at low temperature), which is worth knowing exists even if the details belong to a later course.

Worked example — reading a molecule before choosing

A compound contains a ketone, an ester and a terminal alkene. What does each reagent give?

NaBH₄: the ketone becomes a secondary alcohol. The ester and the alkene are untouched.

LiAlH₄: the ketone becomes a secondary alcohol and the ester becomes a primary alcohol. The alkene is still untouched — no hydride reagent reduces an isolated C=C.

H₂ over Pd/C: the alkene is reduced to an alkane and both carbonyls survive. Three reagents, three different products, and the molecule did not change.

Reduction to a methylene

One hydride or two, settled by the question that settles every carbonyl reaction: does the tetrahedral intermediate have anything it can throw out? A ketone’s does not, so it stops. An ester’s has an alkoxide — and what it collapses to is an aldehyde.The reason you cannot stop an ester at that aldehyde is in the bottom row: the aldehyde is a better electrophile than the ester it came from, so it is consumed faster than it accumulates. Stopping there means crippling the reagent rather than rationing it, which is what DIBAL-H at low temperature is for. The same reading runs down the table above — an acid chloride and an ester both give tetrahedral intermediates with an alkoxide or chloride to expel, which is why LiAlH₄ takes them past the aldehyde every time. An amide is the one that does not fit, and it is worth keeping separate: R₂N− is far too strong a base to leave, so the intermediate expels its oxygen instead and the product is an amine.

Hydride reduction takes a ketone down one rung, to an alcohol. Two named reactions take it down two, replacing C=O with CH₂ entirely, and they exist as a pair because they tolerate opposite conditions:

Clemmensen reduction — Zn(Hg), concentrated HCl. Strongly acidic, so use it when the rest of the molecule can survive acid but not base.
Wolff–Kishner reduction — hydrazine then hot KOH. Strongly basic, for molecules that cannot survive acid.

The pairing is the point: whichever one your substrate tolerates, the other is available. Both come up again in aromatic chemistry, where a Friedel–Crafts acylation followed by one of these is the standard way to attach an unrearranged alkyl chain to a ring.

What carries forward

Hydride addition to a carbonyl is the same mechanism as every other nucleophilic addition in the carbonyl chapters, with H⁻ standing in for the nucleophile — so if you can draw a Grignard addition, you can already draw this. The selectivity table is the part to hold: it is the source of most "which reagent" questions in synthesis, and the amide-to-amine row is the one most often missed.