# Reaction mechanism for reduction of alkyl halide to alkane

Alkyl halides are reduced by zinc and dil. $$\ce{HCl}$$ into alkanes $$\ce{ CH3Cl + H2 ->[Zn][dil\text{.} HCl]CH4 + HCl}$$

What I don't understand is the underlying mechanism behind this reaction, is it free radical or nucleophilic substitution. If it is the latter, then why isn't $$\ce{CH3-CH3}$$ a product? I would appreciate if someone could explain the entire mechanism behind this reaction.

• I was taught the reactants are alkyl halide and nascent hydrogen, along with zinc in a copper couple and acidic medium. Just an observation though. Feb 7 '18 at 7:50
• Feb 7 '18 at 8:16
• It's neither. If I had to guess, I would propose oxidative addition of zinc to form $\ce{R-Zn-Cl}$ as an intermediate, followed by hydrogenolysis.
– Zhe
Feb 7 '18 at 13:37

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

Next, there is some surface chemistry where the diffusion of atomic hydrogen (or hydrogen radical) occurs with select metals. See, for example, this 2008 thesis "Alkaline dissolution of aluminum: surface chemistry and subsurface interfacial phenomena", by Saikat Adhikari, who cites the "ingress of atomic hydrogen into the metal".

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

$$\ce{.H + CH3Cl -> HCl + .CH3}$$ (Source on radical reactions)

$$\ce{.H + .CH3 -> CH4}$$

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

$$\ce{.CH3 + .CH3 -> C2H6}$$

$$\ce{.H + CH4 -> H2 + .CH3}$$

......

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