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)
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Notes: It’s a good assumption that at least 2 equivalents of LiAlH4 are necessary
Examples:
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.
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{ 2 comments… read them below or add one }
What about with the cyclic ester? I am wondering about the alkoxide intermediate. Is it two separate LAH molecules, one acting at each end?
It’s not 2 separate LAH, but rather 2 separate H+ in the acid workup. If you look at the mechanism, when bond D, the one on the “leaving” oxygen, breaks, the electrons go toward the oxygen. It’s not shown, but this just makes another oxygen anion. In the case of the cyclic ester, the mechanism is the same, but if you draw it out, you’ll see that the “leaving” oxygen is just stuck at the other end of the chain. In the acid workup, the H+ protonates both ends.