The reduction of an amide using LAH yields a primary amine . I looked up the mechanism over here.

The mechanism suggests that the final main product (amine) is produced before giving the products a water bath (hydrolyze) .

  • So why do we need to add water at the end of the reaction?

This page's last paragraph suggests that 4[H] are required for the reaction.

  • Two [H] are used up to make the amine (according to the mechanism), so what happens to the other two nascent hydrogens?
  • $\begingroup$ 4 [H] in the second reference are general one-electron reducing agent equivalents. They are not 'literal' hydrogens. If you need water for the reaction is somewhat debatable, but since primary amines are very weak acids and hydrides are extremely strong bases, the primary amine product likely will form a complex with aluminium, that can be destroyed by water. $\endgroup$
    – permeakra
    Feb 14 '15 at 21:25
  • $\begingroup$ @permeakra This(curvedarrowpress.com/partd/aminefamide.html) provides the reason for the workup but with a completely different mechanism . Also check out my comment on J.LS's answer . Which is the correct mechanism ? $\endgroup$
    – Del Pate
    Feb 15 '15 at 12:57
  • $\begingroup$ @DelPate Likely, you won't find proven correct mechanism for the reaction. It certainly includes oxygen coordination by aluminium and hydrogen attack on carbon, but proving anything else is hard and unneeded. $\endgroup$
    – permeakra
    Feb 15 '15 at 13:48
  • $\begingroup$ @permeakra Why do you say unneeded ? If the mechanism liberates H2 gas , wouldnt it be dangerous or explosive ? $\endgroup$
    – Del Pate
    Feb 16 '15 at 9:07
  • 1
    $\begingroup$ Technically speaking, LAH liberates hydrogen on contact with acids, so yeah, unsubstituted amides may produce some hydrogen (this is unrelated to the reduction itself, though). However, LAH is so active it is always used under fume hood and in (weak) argon stream, so accumulation of any significant amount of hydrogen is not an issue. $\endgroup$
    – permeakra
    Feb 16 '15 at 10:07

You don't need to add water to produce the amine, but it's the simplest way of destroying any remaining hydride reducing agent (to give aluminium hydroxide, lithium hydroxide and hydrogen). Presuming you're doing the reaction in ether, the $\ce{LiOH}$ will all go to the aqueous layer, whereas $\ce{Al(OH)3}$ is pretty much insoluble in any solvent you're likely to use in organic synthesis, so it's easy to extract the amine from the aqeuous solvent.

The "chemguide" page is a bit confusing. The reaction they write out isn't a great description of what's happening to the oxygen in the amide (it's probably going to be incorporated into $\ce{Al(OH)3}$ but the other two "$\ce{[H]}$"s just go to making the carbonyl oxygen into water.

  • $\begingroup$ Thank you . I think the mechanism of that link(that I mentioned) is wrong . In the 1st step itself , a hydride is given out for attack and AlH3 must remain but they seem to have converted the H to CH3 . This (m.youtube.com/watch?v=CJU4NIFac9k) provides an altogether different mechanism which involves H2 release . And this mechanism (curvedarrowpress.com/partd/aminefamide.html) actually provides the use of the workup . Which is the correct mechanism ? $\endgroup$
    – Del Pate
    Feb 15 '15 at 12:46
  • $\begingroup$ You'd really have to look at the research literature to check what people think at the moment. I suspecct that as permeakra said, there's no "established" route because while AlH3 is pretty reactive, it is deactivated by oxygen coordination so it's hard to tell exactly what processes are going to dominate. $\endgroup$
    – J. LS
    Feb 15 '15 at 14:09

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