Question

Reduction of alkyl halide in presence of zinc and ethanoic acid

Answer 1

Answer:

CH3-Cl on reaction with zinc in the presence of acid nd h2 gives CH3-H+HCl

Answer 2

Answer:

Explanation:

Alkyl halides are reduced by zinc and dil. HCl into alkanes

CH3Cl+H2−→−−−dil. HCl Zn CH4+HCl

The part that I do not understand is the underlying mechanism behind this reaction, is it nucleophilic or free radical substitution. If it is nucleophilic, then why isn't CH3−CH3 a product? I would appreciate if someone could further explain the entire mechanism behind this reaction.

The action of HCl on Zn results in the formation of hydrogen.

Then, the action of a hydrogen radical on an alkyl halide:

H+CH3Cl⟶HCl+CH3 (Source on radical reactions)

H+CH3⟶CH4

I think that more reactions are possible resulting in minor products, for example:

CH3+CH3⟶C2H6

H+CH4⟶H2+CH3

The final products include HCl, CH4 and other minor products.