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.
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.
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+.
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
Open the OChem Board whiteboard — benzene rings, wedge/dash bonds, and a clickable periodic table built in. No account needed.