Reduction with lithium aluminium hydride: practice and mechanism
LiAlH₄ is strong enough to reduce acids and esters all the way to the primary alcohol, and amides and nitriles to the amine.
Example
LiAlH₄ is a very strong reducing agent. Besides aldehydes and ketones it reduces esters and carboxylic acids all the way to the primary alcohol (esters also release the alcohol part). Amides and nitriles become amines.
Mechanism step by step
- 1
A hydride attacks the carbonyl carbon; with an ester a tetrahedral intermediate forms.
- 2
With an ester the alkoxide leaves: an aldehyde forms, which is more reactive than the ester.
- 3
A second hydride immediately reduces the aldehyde to the alkoxide; acidic work-up (H₃O⁺) gives the alcohols.
- 4
With amides the oxygen leaves (as an Al–O species) instead; via an iminium ion the amine forms.
Practice tasks
Which product forms, which reagent is needed? After each answer you see the explanation.
Which main product forms?
Typical mistakes
- With LiAlH₄ the reduction does not stop at the aldehyde; the aldehyde is more reactive than the ester and is reduced on to the alcohol at once. (Only DIBAL-H at −78 °C stops at the aldehyde.)
- An amide forms from the nitrile by hydrolysis, not reduction. LiAlH₄ delivers hydride and turns R–C≡N into the amine R–CH₂–NH₂.
- LiAlH₄ does not stop at the aldehyde: the aldehyde is more reactive than the acid and is reduced on to the primary alcohol at once.
- Too far: LiAlH₄ reduces the carboxyl group to the alcohol, but the C–O bond stays. No CH₃ forms.
- A hydrate (two OH on one carbon) is not a reduction product. Hydride replaces the C=O bonds; you end up with the primary alcohol.
- In the amide it is not the N that leaves: the oxygen goes (as aluminate) and the nitrogen stays. The amine R–CH₂–NH₂ forms.
- An aldehyde only forms from amides with mild hydrides under special conditions. LiAlH₄ reduces the amide all the way to the amine.
- A nitrile would come from dehydration (e.g. P₄O₁₀). LiAlH₄ is a reducing agent and turns the amide into an amine.
- Reducing a nitrile stops at the aldehyde only with DIBAL-H. LiAlH₄ transfers two hydrides and the primary amine forms.
- The imine is only the intermediate after the first hydride. A second hydride follows; after work-up you have the amine.
- A carboxylic acid forms by hydrolysing the nitrile (acid, water, heat), not with the reducing agent LiAlH₄.
Often confused with: Reduction with sodium borohydride
Frequently asked questions
Why must you work water-free?
LiAlH₄ reacts violently with water to give H₂ gas; you work in dry ether/THF and add water only during work-up.
How do you stop at the aldehyde?
With DIBAL-H at −78 °C: the tetrahedral intermediate stays stable until the aqueous work-up.