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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.
Comprehensive Organic Synthesis, 2nd ed. writes (emphasis mine)1
The reduction of oximes or oxime ethers and esters to primary amines is a useful synthetic transformation that is used as a substitute for direct reductive amination to convert ketones or aldehydes to primary amines. The reaction involves reduction of C=N bond and reductive cleavage of N–O bond. Thus, the transformation requires the use of strong hydride reagents or reagent combinations.
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.
1 Abdel-Magid, A. F. Reduction of C=N to CH–NH by Metal Hydrides. In Comprehensive Organic Synthesis, 2nd ed.; Knochel, P., Molander, G. A., Eds.; DOI: 10.1016/B978-0-08-097742-3.00802-8.
The intermediate hydroxylamine is further deoxidatively reduced to the amine. LAH is a blunt instrument; aluminum(III) is a potent oxophile. Cf: ketone or aldehyde plus amine plus cyanoborohydride in mildly acidic buffer. It only takes a tickle to reduce a Schiff base.
