Question

what will be product on reduction of methyl acetate in presence of LiAlH4

CH3COOCH3 +LiAlH4-------> ??
dry
ether​

Answer 1

Explanation:

Step 1:

The nucleophilic H from the hydride reagent adds to the electrophilic C in the polar carbonyl group of the ester. Electrons from the C=O move to the electronegative O creating the tetrahedral intermediate a metal alkoxide complex.

reduction of an ester using hydride

Step 2:

The tetrahedral intermediate collapses and displaces the alcohol portion of the ester as a leaving group, in the form of the alkoxide, RO-. This produces an aldehyde as an intermediate.

Step 3:

Now we are reducing an aldehyde

The nucleophilic H from the hydride reagent adds to the electrophilic C in the polar carbonyl group of the aldehyde. Electrons from the C=O move to the electronegative O creating an intermediate metal alkoxide complex.

Step 4:

This is the work-up step, a simple acid/base reaction. Protonation of the alkoxide oxygen creates the primary alcohol product from the intermediate complex