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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?

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    $\begingroup$ If you carry it out in alcohol, the LiAlH4 will decompose, perhaps more violently than you might like. $\endgroup$ Feb 16, 2016 at 17:46
  • $\begingroup$ @ 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 $\endgroup$ Feb 17, 2016 at 3:37
  • $\begingroup$ What? There's no mechanism for it. It's just an acid-base reaction. $\endgroup$ Feb 17, 2016 at 9:16
  • $\begingroup$ @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. $\endgroup$ Apr 3, 2017 at 8:12
  • $\begingroup$ @Martin-マーチン Oops, the %edit should have gone behind the }. $\endgroup$
    – mhchem
    Apr 3, 2017 at 11:57

1 Answer 1

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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.

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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

enter image description here

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    $\begingroup$ The side reaction between alcohol and LAH is certainly not minimal. It can be violent and cause your reaction mixture to catch fire. $\endgroup$
    – bon
    Feb 16, 2016 at 22:10

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