Reduction of oximes with lithium aluminium hydride

I know that $\ce{LiAlH4}$ is good a reducing agent and it donates hydride ions but then I am not able to understand the mechanism of this reaction.

I attempted this way

I am not understanding what to do with the extra charge on nitrogen as there is no acidic medium so that I can donate $\ce{H+}$ to it or can remove ($-\ \ce{OH}$) by any protonation.

• You are leaving out one positively charged species that remains after it looses it's hydride ion. Jan 10, 2017 at 4:46
• You might be saying about Li+ and it will get donated by N-. But how will it help in getting there the group (-NH2)
– Avi
Jan 10, 2017 at 12:14
• How about LiAlH3+ ? Jan 10, 2017 at 23:28
• If LiAlH3+ is formed it will get donated by N- but what about further reaction? How will it help to get -NH2 there?
– Avi
Jan 11, 2017 at 12:21

I would guess that after the addition of hydride to the $\ce{C=N}$ bond (incidentally, you should not draw it as free hydride) the next step is a substitution at nitrogen. $\ce{X-Y}$ $\sigma^*$ orbitals ($\ce{X},\ce{Y}$ being heteroatoms) tend to be low in energy and rather electrophilic, so this is probably no different.
If it helps you to feel better about the negative charges, you should probably draw the nitrogen as being coordinated to a lithium and the oxygen as being coordinated to an $\ce{AlH3}$ fragment, or some permutation thereof. The close association of the heteroatoms with metal fragments also encourages "hydride" to attack. Again it's not really hydride attacking but rather the $\ce{Al-H}$ $\sigma$ bond.