LiAlH₄ reduction of an ester template

Hydride adds to methyl benzoate, methoxide leaves to give benzaldehyde, a second hydride adds and acid work-up gives benzyl alcohol and methanol.

Mechanism step by step

Two hydrides: addition–elimination gives an aldehyde, which is reduced at once to the primary alcohol

1. Hydride adds – Al–H electrons attack the C=O carbon
2. Methoxide leaves – C=O re-forms: an aldehyde
3. Second hydride – The aldehyde is more reactive than the ester
4. Acid work-up – The alkoxide is protonated

Key points

• The aldehyde cannot be isolated: it is reduced faster than the ester (DIBAL-H at −78 °C stops there).
• NaBH₄ does not reduce esters or acids; LiAlH₄ also reduces acids to 1° alcohols and amides to amines.
• Dry ether or THF: LiAlH₄ reacts violently with water. Add the work-up acid afterwards.
• Li⁺ binds the carbonyl O and speeds the hydride attack (not drawn).

Drawn with 101 atoms, 13 arrows, 8 shapes – every one editable in MolDraw.

Sources

• Clayden, Greeves & Warren, Organic Chemistry, 2nd ed., OUP 2012, ch. 10 Textbook
• Smith, March's Advanced Organic Chemistry, 8th ed., Wiley 2020, ch. 19 Textbook
• Nystrom & Brown, J. Am. Chem. Soc. 1947, 69, 1197 Original paper