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

enter image description here

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?

  • 3
    $\begingroup$ I am curious to know where you read that reduction of the carboxylate to form two alcohols. Do you have a link? The other oxygen becomes attached to the aluminum species as shown in Jerry's answer. Now, a carboxylic acid anhydride is a different beast altogether. Anhydrides will reduce to a pair of alcohols. $\endgroup$ – Ben Norris May 10 '13 at 1:41

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:

enter image description here

Another $\ce{LiAlH_4}$ then reduces the aldehyde to alcohol: enter image description here

I found the pictures here.

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  • 5
    $\begingroup$ $\ce{LiAlH4}$ is such a powerful reducing agent that it will attack carboxylate anions (which are resistant to most other nucleophiles). $\endgroup$ – Ben Norris May 10 '13 at 1:38

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