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In reduction of carboxylic acid with $\ce{LiAlH_4}$ the first step is deprotonation.

But in the mechanism of reduction of amide with $\ce{LiAlH_4}$ (here), the first step is addition of hydride on carbonyl group. Why doesn't the hydride ion abstract a proton from $\ce{-NH2}$? Since acid base reactions are faster than nucleophilic addition.

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  • $\begingroup$ Probably a side reaction which doesn't amount to much . $\endgroup$ – Rajesh Mar 23 '15 at 10:55
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As compared to the carboxylic acid, deprotonation of the amide is negligible.

I don't recall any data for THF or 1,4-dioxane as the solvent, but F.G. Bordwell has determined a lot of $pK_a$ values in DMSO.

In DMSO, the $pK_a$ of acetic acid is 12.6 [1], while the $pK_a$ of acetamide is 25.5 [2].

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  • $\begingroup$ So, what is the maximum pKa upto which LiAlH4 will deprotonate and not add it self? I mean is there any criteria by which we can decide? In both cases deprotonation will cause H2 to release which obviously is much less acidic than either of them. So the equilibrium should well be in forward. $\endgroup$ – Ayush Pateria Mar 23 '15 at 12:57
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In the mechanism on the page you referred to, a positive species ($\ce{HRC=NH2+}$) is formed, which is very unlikely in strongly basic conditions. I do believe the amide would deprotonate at some point, most likely in the first step, similarly to the carboxylic acid. Here is a link to a page which shows a more likely mechanism.

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