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Question
Benzamide can be converted to benzylamine using

A) $\ce{Br2,\ KOH}$
B) $\ce{PCl5}$
C) $\ce{LiAlH4}$
D) $\ce{NaBH4}$

My answer:

Reducing agent is used so C and D options both must be correct but since its single choice answer, I do not know which is more appropriate. Any help?
Also, what does option A and B reagents do to amides?

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    $\begingroup$ You're on the right track. $\ce{R-CONH2}$ is an electrophile. Is it more or less electrophilic than a ketone? Your reductions are nucleophilic additions to the amide, aren't they? Which of the hydrides is the stronger reagent here? Would the comparison of the electronegativities of boron and aluminium be helpful here? $\endgroup$
    – Klaus-Dieter Warzecha
    Commented Apr 7, 2015 at 11:26
  • $\begingroup$ Less electrophilic! So weaker hydride donor is used and so is the answer $\ce{NaBH4}$ ? @KlausWarzecha $\endgroup$
    – Mixcels
    Commented Apr 7, 2015 at 13:02
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    $\begingroup$ +1 You have correctly assigned the electrophilicity and nucleophilicity but somehow came to the wrong conclusion :) For a ketone or aldehyde, I would indeed use $\ce{NaBH4}$: Offer a reagent that is just strong enough. The less electrophilic carbon in the amide requires the more reactive $\ce{LiAlH4}$ to add a $\ce{H-}$ there: In your face, sluggish amide ;) $\endgroup$
    – Klaus-Dieter Warzecha
    Commented Apr 7, 2015 at 15:35
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    $\begingroup$ @KlausWarzecha Wouldn't the HSAB principle say otherways? $\endgroup$
    – ChemExchange
    Commented Apr 7, 2015 at 15:59
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    $\begingroup$ $\ce{H-}$ is not small at all. Hydrogen atom is as big as oxygen. Its anion is even bigger. Imagine one positive charge vs two negative charge. Thus, $\ce{H-}$ is a very typical soft base: big and easy to be distorted. In this question, hard soft effect is really overwhelmed by the electronic effect as $H$ in $LAH$ has more partial negative charge than $H$ in $\ce{NaBH4}$. $\endgroup$
    – Coconut
    Commented Apr 16, 2015 at 2:10

1 Answer 1

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The more reactive $\ce{LiAlH4}$ is required for the reduction of the amide to the corresponding amine (source). The carbonyl carbon of an amide is less electrophilic compared to a ketone, because electron density is donated by the amide nitrogen via resonance.

Bromine and hydroxide react with an carboxylic acid amide to yield an amine with one fewer carbon atom (Hofmann degradation). $\ce{Br2}$ and $\ce{KOH}$ react to form $\ce{KOBr}$ in situ, which transforms the amide into an intermediate isocyanate ($\ce{R-N=C=O}$). Hydrolysis of the isocyanate yields a carbamic acid $\ce{R-NH-COOH}$, which decomposes to the amine $\ce{R-NH2}$ and $\ce{CO2}$. As described in this paper, the reaction of $\ce{PCl5}$ with carboxylic acid amides produces amidochlorides $\ce{RCCl2NH2}$ as intermediates, which break down to the corresponding nitrile $\ce{RC#N}$ and $\ce{HCl}$.

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  • $\begingroup$ Is that carbamic acid decomposition instantaneous? $\endgroup$
    – Mixcels
    Commented May 26, 2015 at 6:38
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    $\begingroup$ @Mixcels Yes, it is. Carbamic acid and many of its derivatives are generally unstable (en.wikipedia.org/wiki/Carbamic_acid). $\endgroup$
    – Jannis Andreska
    Commented May 26, 2015 at 18:19

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