Bondline / Reactions

Reduction of an ester with LiAlH₄

Hydride adds to the ester, the tetrahedral intermediate expels alkoxide to give an aldehyde, and a second hydride reduces that straight away. Two primary alcohols form after workup.

Acyl substitution · 4 steps

Conditions

LiAlH₄ in dry ether or THF, then H₃O⁺ workup. NaBH₄ is too weak to reduce esters.

Mechanism in 4 steps

Red curly arrows show where each pair of electrons moves.

HAl−OOOO−+
Step 1An Al–H bond delivers hydride to the ester carbonyl carbon.
OO−HAl−O+
Step 2The intermediate collapses, expelling methoxide and giving an aldehyde.
HAl−OO−HOH2+++
Step 3The aldehyde is more reactive than the ester, so a second hydride adds at once.
O−HOH2+OHH2OH₃O⁺++
Step 4Acidic workup protonates the alkoxide.

About acyl substitution reactions

Carboxylic acid derivatives react by nucleophilic addition followed by elimination: the nucleophile adds to C=O, then the leaving group is expelled and the C=O reforms.

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