Since $\mathrm{p}K_\mathrm{a}$ of primary amides are close to alcohols, will there be an acid-base reaction between amides and $\ce{LiAlH4}$ besides reduction?
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Not according to the commonly accepted mechanism here
The first step is
The nucleophilic H from the hydride reagent adds to the electrophilic C in the polar carbonyl group of the ester.
Note: the pKa of primary amides is in the 22-25 range so not close to alcohols.
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$\begingroup$ For one of the question in Stack exchange "Whether Grignard Reagent react with amides?", It was given that RMgX would deprotonate the amides. Now my doubt is that when alkyl carbanion from RMgX deprotonates the amide at Nitrogen atom, why not the hydride ion(a stronger base than carbanion, i suppose) deprotonates the same. Kinetically an acid base reaction is more feasible than a nucleophilic addition of hydride ion, if iam not confused. $\endgroup$– VMPCommented Nov 10, 2019 at 13:03
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1$\begingroup$ Hydride from LiAlH4 is a very different species than R- from Grignard. What you need to consider is the role of both Al and Li in complexing to the carbonyl group and whether the hydride delivered is truly free H-. I do not think it is. $\endgroup$ Commented Nov 10, 2019 at 13:16
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$\begingroup$ @Waylander: Does it really matter if the anion of the amide is formed? Aluminum complexes with negative oxygen and hydride is delivered to C=N, etc. $\endgroup$ Commented Nov 10, 2019 at 15:51
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$\begingroup$ I don't think it does, but the mechanisms I could find do not involve anion formation $\endgroup$ Commented Nov 10, 2019 at 16:11