# Why can't LiAlH4 react with alcohols?

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 $\ce{O-AlH2-}$?

• Alcohols need to be turned into much better leaving groups. – Mithoron Jul 22 '15 at 13:32
• 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

• 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