Reduction of carboxylic acids by LiAlH4 I've read that when a carboxylic acid reacts with $\ce{LiAlH4}$ the corresponding alcohol is formed:



But when I try to think of the mechanism, I get stuck here:

$\ce{LiAlH4}$ produces $\ce{H-}$. Since $\ce{H-}$ is a strong base it should immediately abstract a proton from the carboxylic acid to give the corresponding carboxylate ion (just like in the reaction of carboxylic acids with Grignard reagents), instead of undergoing nucleophilic addition to give the alcohol.

How do I resolve this?
The $\ce{Li^+}$ ion substitutes the $\ce{OH}$ in $\ce{COOH}$ first. The $\ce{C=O}$ is then reduced by $\ce{AlH_3}$ to give an aldehyde. See below:



Another $\ce{LiAlH_4}$ then reduces the aldehyde to alcohol:


I found the pictures here.