Reduction of esters to primary alcohols using LiAlH4
Description: Addition of lithium aluminum hydride to esters leads to the formation of primary alcohols (after addition of acid)
Notes: It’s a good assumption that at least 2 equivalents of LiAlH4 are necessary
Notes: Note that the last example shows reduction of a cyclic ester (lactone) that results in a linear product. The byproducts here are AlH3 and a lithium salt.
Mechanism: LiAlH4 is a source of hydride (H-) and the reaction begins with 1,2-addition of hydride to the carbonyl of the ester (Step 1, arrows A and B). The second step is 1,2-elimination of the alkoxide (Step 2, arrows C and D) to give an aldehyde. The aldehyde then undergoes a second 1,2-addition by hydride (Step 3, arrows E and F) to give an alkoxide. Protonation of the oxygen then gives the alcohol (Step 4, arrows G and H).
Notes: The exact identity of the acid in the last step is not crucial, it’s just important to convert the alkoxide to the alcohol.