Why is biphenyl formed as a by-product in a grignard reaction?

Some biphenyl can form if not all of the phenyl bromide has already reacted with magnesium solid to form the Grignard reagent. In that situation, the Grignard reagent acts as a very strong nucleophile towards the regular phenyl bromide.

First of all, biphenyl looks like this:


Now, when I look at this compound, one reaction immediately comes to mind, and it’s quite an odd one (Organic Chemistry, Bruice, Ch. 16.14).

What you basically have is a nucleophilic aromatic substitution (NAS) where you see a very strong nucleophile (##”NH”_2^(-)##) plucking the proton off of the carbon ##alpha## to an electron-withdrawing, weakly deactivating substituent such as ##”Br”## or ##”Cl”##. It forms an unstable benzyne intermediate.

This can happen because ##”NH”_2^(-)## is such a strong nucleophile. How? The conjugate acid is ammonia, whose pKa is about ##36##. So, the anion has a much higher basicity, and hence is much more nucleophilic (a nucleophile is at its core a lewis base; if it’s a good nucleophile, then it’s just fast).

(Note that a carbon was labeled in this reaction to demonstrate that the benzyne intermediate that forms can get attacked on either carbon that contributes to the triple bond.)


What you should do with the above mechanism is imagine how it would occur if you used phenyl magnesium bromide instead of ##”NH”_2^(-)## as your nucleophile.

The pKa of benzene is ##~43##, whereas that of ammonia is ##~36##. That means the conjugate base of benzene is even more nucleophilic than that of ammonia.

So, if this mechanism can occur (which it does), a very similar mechanism likely will occur to form biphenyl using the analogous Grignard situation.

Essentially, the Grignard reagent takes the role of the ##”NH”_2^(-)##. The final step is electrostatically stabilized by the presence of the magnesium bromide cation.

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