# Reduction of alkyl halides to alkanes

For all the three parts, only methane has been given as the product in my textbook.

In 1) and 2), according to the mechanism I wrote, I get both ethane and methane as products. I think both these products should be formed in good yield. Then, why my products do not match?

In 3) , I got methane as the only product as given in the textbook. Can you please check the mechanism?

• You won't form a free methyl carbanion under those conditions because it is far too unstable. If it did form it's likely reaction would probably be to deprotonate the methyl chloride rather than attack as a nucleophile because it is such as strong base and the methylene chloride anion is somewhat stabilised by the electron withdrawing chlorine.
– bon
Jan 12 '17 at 12:54
• @bon what makes the carbanion unstable under those conditions? Are you suggesting a free radical mechanism in the first two reactions? Jan 12 '17 at 14:18
• Do you mean that the carbanion will deprotonate methyl chloride rather than act as a nucleophile because it is a better base than a nucleophile? If not, how else did you decide that? Jan 12 '17 at 14:20
• It's just a really unstable species in the first place because carbon has relatively high energy orbitals. Yes I think it will be a better base than a nucleophile (if it were ever to exist at all). As Klaus' answer suggests the mechanism involves metal insertion into the C-X bond which I expect does proceed via a radical mechanism.
– bon
Jan 12 '17 at 15:12
• @ApoorvPotnis He deleted it after I posted the comment (I can still see it because I have >10k rep).
– bon
Feb 7 '18 at 12:00

In your notes, it seems that electrons from the metal ($$\ce{Zn}$$ and $$\ce{Mg}$$) somehow "attack" the alkyl halide and generate the methine ion ($$\ce{CH3-}$$). This does not happen for two reasons: Firstly, electrons are not free species in solution (like ions and molecules), they are only transferred from one molecule (reductant) to another (oxidant), so they cannot attack anything by themselves. Secondly, the methine ion is an extremely strong base (the pKa for methane is around 50, so the methine ion, its conjugate base, has to be VERY basic), which could only be produced by using an even stronger base to abstract a proton from it.
What actually happens in both cases is what is called metallic insertion. The metal ($$\ce{Zn}$$ or $$\ce{Mg}$$) is inserted between the carbon-halide bond, forcing the carbon atom to hold a partial negative charge (the author of this picture forgot to include the $$\ce X$$ atom bound to $$\ce M$$ in the organometal compound, but it is not that relevant here).
When it happens, you end up having a nucleophilic carbon, and it could indeed react with a $$\ce{CH3Cl}$$ molecule to yield $$\ce{CH3-CH3}$$ under a unimolecular substitution mechanism ($$\ce{S_N1}$$). However, such substitution reactions are very slow compared to acid-base reactions, and since you have acidic hydrogens available in both cases ($$\ce{HCl}$$ in question #1 and $$\ce{H2O}$$ in #2) the acid-base reaction (abstraction of a H atom from $$\ce{HCl}$$ or water) would happen fast enough to prevent the formation of $$\ce{CH3-CH3}$$ adduct.