# Reaction of coumarin with phenylmagnesium bromide

I came across this problem..

I thought that the Grignard reagent, PhMgBr, would cleave the lactone and then subsequently add to the ketone formed, giving an allylic tertiary alcohol after aqueous workup. Treatment with acid would then lead to intramolecular etherification, giving 2,2-diphenyl-2​H-1-benzopyran:

But then our chemistry professor gave out a different mechanism for this reaction as follows.

What is the correct product?

• I immediately call BS on this. You can argue for the conjugate addition step, but you shouldn't argue for a $\mathrm{S_N2}$ at a $\mathrm{sp^2}$ carbon except under very special circumstances (which this reaction does not fit)... – Zhe May 10 '17 at 3:23
• Unfortunately there aren't any examples of coumarins being treated with aryl Grignards, so this is again a made up question. I much prefer your mechanism, though. – orthocresol Aug 14 '18 at 9:38

In terms of regioselectivity, at least for the first addition of the Grignard reagent, I suggest to have a look on the cumarin, that may be seen as an α,β-unsaturated carbonyl compound. Hard nucleophiles (in terms of HSAB principle) like $\ce{MeLi}$ would almost exclusively react with the carbonyl carbon, as you drew in your reaction equation. Soft nucleophiles, such as cuprates, would prefer the 1,4-addition.
Going from $\ce{MeMgBr}$, to $\ce{EtMgBr}$, $\ce{i-PrMgBr}$, and $\ce{t-BuMgBr}$ gradually lowers the polarity of the bond between the metal (magnesium) and the directly attached carbon atom. Hence, along with this line (in other words, as a trend), the preference to react in 1,4-position (more like Michael reaction) increases on expense of the reaction of this nucleophile on the carbonyl carbon. I speculate the bond in the Grignard reagent, between $\ce{Mg}$ and the phenyl group in $\ce{PhMgBr}$, is even less polarised, than in the instance of $\ce{t-BuMgBr}$, and hence conclude at least a significant portion of the product of the first addition formed will be the one of the 1,4-addition (second drawing).
As @Zhe, I however still have a difficulty to understand the subsequently suggested second addition of $\ce{PhMgBr}$ in the lower drawing, because the activation of the double bond once in conjugation with the carbonyl group is gone.