# Origin of alkenes from Wurtz Reaction

According to Wikipedia:

Also, since the reaction involves free radical species, a side reaction occurs to produce an alkene. This side-reaction becomes more significant when the alkyl halides are bulky at the halogen-attached carbon.

How would a radical reaction end up producing an alkene in the context of the Wurtz reaction?

What I do see happening is an E2 elimination creating an alkene.

Also, this raises another question - can't Grignards also create alkenes "by accident"?

• There is a small, inconsequential typo in the reaction depicted here. The LHS shows the alkyl group as $\ce{R'}$, but the RHS depicts it as $\ce{R}$ only. Mar 12 '18 at 4:51

The best explanation I was able to find is given by Avery A. Morton , John B. Davidson , Barton L. Hakan, J. Am. Chem. Soc., 1942, 64 (10), 2242–2247. If I understand it correctly, this boils down exactly to that you already propose.

Disproportionation has been so universally accepted as the criterion of a free radical that there has been no attempt to show a derivation from other molecular species. Nevertheless, a mechanism for this phenomenon by way of an organometallic intermediate appears exceptionally reasonable. If an alkylsodium meets an alkyl halide in such manner that the two alkyl chains are adjacent to each other while sodium halide is being formed, the two alkyl residues will have unlike charges and a proton will be drawn to the alkyl radical having the two unsaturated electrons. The reaction is merely the conventional prototropic change taking place between adjacent portions of two molecules whose inorganic components are effecting, or have just completed, a union.

If this view be accepted, the olefin produced must come from the alkyl halide rather than at random from both participant alkyl radicals. Though conclusions are to some extent obscured by a metal-halogen interchange (see later), all available facts unquestionably point to the correctness of this deduction.

However, there might be another reason, which I cannot back up with literature. This is probably due to the persistence of the radical species; higher concentration of radicals increases the possibility that they react with each other. In other words, one radical abstracts a hydrogen from another radical, hence creating an alkane and an alkene, in a purely radical reaction. $$\ce{2 R-CH2-CH2. -> R-CH2-CH3 + R-CH=CH2}.$$
But then again radical mechanisms are never simple and it is quite possible that this does not happen at all, yet it would appear to be at least an explanation for the Wikipedia statement.

• Imho, the author of the article you linked to has used quite a convoluted language to describe the simple E2 elimination. Is it possible that E2 elimination wasn't known in the 1940s? Or did proper terminology not exist at that time? What exactly had prevented the author from just writing "E2 elimination occurs"? Do you have any clue? Mar 12 '18 at 4:48
• @GaurangTandon The author gave us a precise description of the mechanism. There is nothing convoluted about it. E2 is a very simple, and crude approximation, or an ideal, ideal case. If you need to break it down for a student that's fine, but for a scientifically accurate description it did not nearly contain enough information. Mar 12 '18 at 4:55

Yes it's possible this way:

Also, this raises another question - can't Grignards also create alkenes "by accident"?

It's very common. Grignard reagents will take away any acidic proton available. That's why we use Grignard reagents in anhydrous conditions, else it would take up water's proton.

• What evidence is there for this mechanism? Apr 22 '15 at 14:22
• @jerepierre I remeber it from my teacher's notes, and if my teacher said it, it is a universal fact. He is a very respected person in my view and he really should be. He doesn't teach us anything without concerting various books and thus we only read the relevant and don't need to read the same thing from 50 books. Apr 22 '15 at 15:06