Reduction of amides

The reduction of an amide using LAH yields a primary amine . I looked up the mechanism [over here](http://chemwiki.ucdavis.edu/Organic_Chemistry/Amides/Reactions_of_Amides/Conversion_of_Amides_into_Amines_with_LiAlH4).


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](http://www.chemguide.co.uk/organicprops/amides/other.html)'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?




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. 


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