LiAlH4 Reduction: Which Carbonyls It Reduces

Quick answer Lithium aluminum hydride (LiAlH4) is a strong hydride (H-) donor that reduces carboxylic acids, esters and aldehydes to 1° alcohols, ketones to 2° alcohols, and amides and nitriles to amines. It is far more powerful than NaBH4 — which reaches only aldehydes and ketones — and it reacts violently with water, so it is used in dry ether and then quenched with aqueous acid.

⚙ Mechanism · Hydride Reduction (LiAlH₄)2 steps

Step 1 — hydride is delivered to the carbonyl carbon.

The Al–H bond is polarised towards hydrogen, so what leaves is hydride (H⁻) — a hydrogen with both electrons. That is a nucleophile, which is why the arrow starts at the Al–H bond.

Step 2 — workup protonates the alkoxide.

LiAlH₄ is strong enough to reduce esters and carboxylic acids as well; NaBH₄ is milder and stops at aldehydes and ketones.

1. LiAlH4 delivers hydride (H-) to an electrophilic carbonyl carbon, then aqueous acid protonates the product.

The two-step notation "1) LiAlH4 2) H3O+" means the reduction happens in dry ether, and water is added only at the end.

Acetaldehyde → ethanol, a primary alcohol.

2. Carboxylic acids and esters are both reduced all the way to primary alcohols.

Acetic acid → ethanol; methyl acetate reduces to ethanol + methanol the same way.

Methyl acetate (ester)

Ethanol (1° alcohol)

Methanol (from OR group)

3. A ketone gives a secondary alcohol, because its carbonyl carbon already carries two carbon groups.

Acetone → 2-propanol, a secondary alcohol.

4. Amides and nitriles are reduced to amines, not alcohols, because nitrogen stays bonded to carbon.

Acetamide → ethylamine; nitriles behave the same, gaining a carbon.

Propanenitrile

Propylamine (1° amine)

5. LiAlH4 leaves isolated C=C alkenes alone, and it is stronger than the milder reagent NaBH4.

NaBH4 reduces only aldehydes and ketones, so use LiAlH4 when you must reduce an acid, ester, amide or nitrile.

Benzoic acid → benzyl alcohol; the aromatic ring is untouched.

6. Summary

LiAlH4 = strong H- donor · acids, esters, aldehydes → 1° alcohols · ketones → 2° alcohols · amides, nitriles → amines · ignores isolated C=C · stronger than NaBH4 (aldehydes/ketones only) · dry ether, then quench with H3O+.

Worked example

Problem. What does LiAlH₄ (then acid) give with butan-2-one?

LiAlH₄ is a strong hydride (H⁻) donor.
Hydride adds to the carbonyl carbon; the π electrons move to oxygen → an alkoxide.
Acidic workup protonates it. A ketone gives a 2° alcohol (LiAlH₄ also reduces aldehydes, esters, acids).

Answer. Butan-2-ol.

How each reagent works — the arrow pushing

Electron flow only. Follow the arrows; the structures do the talking.

LiAlH₄ / NaBH₄ — hydride reduction2 steps

Why it works · Both deliver hydride, H:⁻ — a hydrogen carrying the bonding pair, so it is δ– and nucleophilic. It attacks the δ+ carbonyl carbon; the π electrons go to oxygen (alkoxide); workup gives the alcohol. Carbon gains an H and electron density — this is a reduction. LiAlH₄ is strong (reduces esters, acids, amides too); NaBH₄ is mild and selective (aldehydes/ketones only).

Quiz yourself

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A carboxylic acid is reduced all the way to a primary (1°) alcohol (acetic acid → ethanol), while a ketone gives a secondary (2°) alcohol (acetone → 2-propanol).

LiAlH4 reacts violently with water, so the reduction is run in dry ether; the aqueous acid (H3O+) is added only afterward to protonate the alkoxide and give the neutral alcohol or amine.

Yes. LiAlH4 reduces the ester to a 1° alcohol but does not touch an isolated C=C alkene, so the double bond survives.

LiAlH4. It reduces amides and nitriles to amines (acetamide → ethylamine), whereas NaBH4 is too mild and reduces only aldehydes and ketones.

Draw this on the whiteboard

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LiAlH4 reduction of carbonyl groups