# Reduction of amides

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?
• 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. – permeakra Feb 14 '15 at 21:25
• @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 ? – Del Pate Feb 15 '15 at 12:57
• @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. – permeakra Feb 15 '15 at 13:48
• @permeakra Why do you say unneeded ? If the mechanism liberates H2 gas , wouldnt it be dangerous or explosive ? – Del Pate Feb 16 '15 at 9:07
• 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. – 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.