Hello.
I seem to recall in my undergraduate organic course that the reduction of an ester to an alcohol via LiAlH
4 could be halted at the aldehyde step.
1. LiAlH4
(CH
3CH
2)
2O
The reaction was: Ester
Aldehyde
2. H
2O
If the second step were left out, then, if my memory serves correctly, the reaction would proceed to the alcohol. Is this true?
I searched namely for this reaction to double check, and I find primarily the DibAl (Di-isobutyl Aluminum Hydride) reagent is used to achieve isolation of the Aldehyde.
Yet I am still not certain if it is impossible for LiAlH
4, because if I look at the mechanism for LiAlH
4, I see that after the first hydride reduction, the (R-O-AlH
3)
-Li
+ salt forms. If water were added, would the salt partition into the aqueous phase, leaving the aldehyde behind without reducing it further? This is assuming the aldehyde had a reasonable alkyl chain attached to hinder aqueous solubility.
Or maybe adding the water would quench the aluminum hydride. I also read that protic solvents resulted release of H
2 gas.
Thank you!