Dicumyl synthesis

Question

2,3-Dimethyl-2,3-diphenylbutane (Dicumyl) (63). A solution of 2-bromo-2-phenylpropane (10.0 g, 50 mmol) in anhydrous diethyl ether (25 mL) was stirred with magnesium turnings (0.60 g, 0.025 g atom) overnight. The reaction mixture was poured into aqueous $\ce{NH4Cl}$ solution (100 mL, 5%) and extracted with dichloromethane (100 mL). The solvent was evaporated, and the solid residue was recrystallized from 95% ethanol to give 2,3-dimethyl-2,3-diphenylbutane (2.4 g, 47%).

Does anyone have any intuition as to how this reaction is working mechanistically? Is it a Grignard formation and then homocoupling event? Can anyone see a reason for why an ortho methyl substituent on the phenyl ring might interfere with this reaction?

Answer 1

Well, that's interesting as nothing is added here. Today Palladium-catalyzed cross-couplings are quite common, but on a historical side, they took their idea from the very first reactions of sodium and organo-chlorides (Wurtz-coupling) and the magnesium mediated formation of biaryls. Back then transition metal salts were added which lead to the idea to use $\ce{Pd}$ which is way more carbophilic and can easily enter and leave the catalytic cycle. I have heard that the idea for the cross-coupling derived from the Schlenk-equilibrium, so your Grignard-reagent dimerizes and then equilibrates to from a $\ce{MgBr2}$ and a $\ce{MgR2}$. I'd love to say now that this simply reductively eliminates to form $\ce{Mg}$ and your product, but I have my doubts (although I have heard people talk about it before).

So in your case, we probably have to go the Wurtz path and say that the negatively polarized carbon on your Grignard-reagent nucleophilically attacks the $\ce{C-Br}$ bond in an $\ce{S_{N}2}$ reaction. If you used sodium the bond would be much more ionic and you'd probably also end up in simply deprotonating the second equivalent of starting material forming a 1-propene derivate.

Answer 2

The ammount of magnesium was 25 mmol, so only half of the 2-bromo-2-phenylpropane was turned into magnesium bromide.

After a molecule of magnesium bromide formed, I would guess it undergo a elimination of $\ce{MgBr^+}$ moity since it is on a tertiary (and also benzyl) carbon and form a relatively stable carboanion. The carboanion then attacks one of the 2-bromo-2-phenylpropane molecules forming the product. But I also wouldn't rule out $\ce{S_{N}2}$ type substitution.