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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1$\begingroup$ Alcohols need to be turned into much better leaving groups. $\endgroup$– MithoronCommented Jul 22, 2015 at 13:32
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1$\begingroup$ 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. $\endgroup$– CharlesCommented Jul 22, 2015 at 14:26
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1 Answer
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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.
answered Jul 22, 2015 at 13:47
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3$\begingroup$ 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. $\endgroup$– R.M.Commented Jul 22, 2015 at 15:57
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$\begingroup$ 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? $\endgroup$– CharlesCommented Jul 24, 2015 at 14:38