Identify A and B.

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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 $\ce{HClO4}$?

  • 9
    $\begingroup$ you should read about cyclopropenyl cation in aromaticity section of advanced organic books. HClO4 is a strong protonating agent and should protonate OH-group, that will leave. The remaining cation is known to be remarkably stable in organic chemistry and have stable salts. $\endgroup$ – permeakra May 19 '15 at 0:05

The first step is indeed 1,2-addition of the Grignard reagent to the keto group, and product A is the alcohol:

enter image description here

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:

enter image description here

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.

  • $\begingroup$ Carbocation as the final product, really interesting! $\endgroup$ – Archer Jul 11 '18 at 20:21
  • $\begingroup$ Yes this happens when the intermediate is highly stable with a stable ion like flouro borate, perchlorate etc. (Note that stable counter ion makes the ionic compound to be formed easily according to fajan's rules $\endgroup$ – An enthusiast Apr 23 '20 at 16:16

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