# What would happen if the reaction of reduction of carboxylic acids by LiAlH4 is carried out in alcohol?

Which is the intermediate compound in the reduction of carboxylic acids by $\ce{LiAlH4}$ and what would happen if the reduction of carboxylic acids is carried out in alcohol?

From my knowledge I know that a geminal diol is formed and then elimination takes place to form $\ce{R-CH=O}$. If this is reacted with alcohol instead of $\ce{LiAlH4}$ again what are the expected products?

• If you carry it out in alcohol, the LiAlH4 will decompose, perhaps more violently than you might like. – orthocresol Feb 16 '16 at 17:46
• @ orthocresol would you please give some kind of the mechanism for the decomposition of LAH in alcohol bcoz the answer below is not that satisfying – Nitro phenol Feb 17 '16 at 3:37
• What? There's no mechanism for it. It's just an acid-base reaction. – orthocresol Feb 17 '16 at 9:16
• @mhchem Your comment %edit uncommented the } and broke the construct. You can always edit out salutations, buzzwords, and unnecessary "statements of weakness" instead, see the edit history for what I mean. – Martin - マーチン Apr 3 '17 at 8:12
• @Martin-マーチン Oops, the %edit should have gone behind the }. – mhchem Apr 3 '17 at 11:57

Alcohols are not an ideal solvent for LiAlH4, but the main issue is that there is no carboxylic center to where the hydride transfer from LiAlH4 can take place.

So in this case, the side-reaction between alcohol and LiAlH4 can be regarded to be minimal and the major product will be the alcohol produced from the carboxylic acid.

P.S. : A diagrammatic representation of the reactions for LiAlH4

• The side reaction between alcohol and LAH is certainly not minimal. It can be violent and cause your reaction mixture to catch fire. – bon Feb 16 '16 at 22:10