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In the reaction of $\ce{LiAlH4}$ with carboxylic acids, deprotonation is followed by a step in which $\ce{O-AlH2-}$ acts as a leaving group. The mechanism is given here in this answer. $\ce{LiAlH4}$ does not, however, reduce alcohols. Why doesn't a similar reaction occur with alcohols - deprotonation, followed by substitution by a hydride ion and removal of $\ce{O-AlH2-}$?

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Charles
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    Alcohols need to be turned into much better leaving groups. – Mithoron Jul 22 '15 at 13:32
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    I thought so too, but then as I said above, -OAlH2 does leave. And esters are able to react too, where the leaving group is -OR, which is even poorer. – Charles Jul 22 '15 at 14:26

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The main issue is that the Al needs to remove its hydride. With a carboxylic acid and/or an aldehyde, it can stick its hydride onto the carbonyl carbon without issue. But the carbon bonded to the alcohol cannot take on a hydride.

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    This also explains Charles's confusion about the $\ce{-OR}$ group leaving in ester reduction (in the above comments). The hydride from $\ce{LiAlH4}$ isn't displacing the $\ce{-OR}$ group directly in a substitution-like mechanism, instead it's "only" reducing the carbonyl to a (hemi)acetal, and it's the collapse of the hemiacetal which kicks the $\ce{-OR}$ off, rather than the hydride transfer directly. – R.M. Jul 22 '15 at 15:57
  • So in general, it's easier to substitute an ester and remove the -OR than an ether - since the leaving group isn't removed directly? Nucleophilic substitution is easier at carbonyl carbons of acid derivatives than at alcohols, ethers, amines, etc.? In nucleophilic substitution at any acid derivative - which is the RDS - the nucleophile's attack step or the step in which the group leaves? – Charles Jul 24 '15 at 14:38