Question:
Identify A and B.
My attempt
The first step will be the attack of $\ce{PhMgX}$ on carbonyl carbon, that is, 1,2 position followed by hydrolysis, yielding an alcohol.
But how does the reaction proceed with $\ce{HClO4}$?
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Identify A and B.
My attempt
The first step will be the attack of $\ce{PhMgX}$ on carbonyl carbon, that is, 1,2 position followed by hydrolysis, yielding an alcohol.
But how does the reaction proceed with $\ce{HClO4}$?
The first step is indeed 1,2-addition of the Grignard reagent to the keto group, and product A is the alcohol:
Treatment of the alcohol with the strong acid $\ce{HClO4}$ cannot yield the usual dehydration product (alkene) in this case, as the corresponding alkene would have two double bonds in an already highly strained three-membered ring, which is highly unfavorable. Instead, after protonation of the $\ce{OH}$ group and subsequent loss of $\ce{H2O}$, the carbocation:
is obtained as the final product B (with $\ce{ClO4-}$ being the counteranion). The planar cyclopropenyl cation is a Hückel aromatic compound with $(4 \cdot 0+2) = 2$ $\pi$ electrons, and delocalization of the positive charge over the two phenyl substituents can provide further stabilization.
As it is an unsaturated cyclic ketone,it will undergo 1,2-addition.
Then $\ce{HClO4}$ will dehydrate the alcohol but will not yield the usual product due to the high angle strain of unsaturated three-membered ring. Instead, the carbocation will be the final product. with $\ce{ClO4−}$ as the counteranion. The carbocation is aromatic, and because of the two phenyl groups it has extended resonance.