Reaction of perchloric acid with cyclopropenol

**Question:**



> 
> Identify A and B.
> 
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> [![enter image description here](https://i.stack.imgur.com/GYhjs.png)](https://i.stack.imgur.com/GYhjs.png)
> 
> 
> 


**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:


[![enter image description here](https://i.stack.imgur.com/0JvQj.png)](https://i.stack.imgur.com/0JvQj.png)


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](https://i.stack.imgur.com/AVbVO.png)](https://i.stack.imgur.com/AVbVO.png)


is obtained as the final product **B** (with $\ce{ClO4-}$ being the counteranion). The planar cyclopropenyl cation is a [Hückel aromatic compound](https://en.wikipedia.org/wiki/H%C3%BCckel%27s_rule) with $(4 \cdot 0+2) = 2$ $\pi$ electrons, and delocalization of the positive charge over the two phenyl substituents can provide further stabilization.


 7

[![Reaction Mechanism](https://i.stack.imgur.com/f7Zqk.jpg)](https://i.stack.imgur.com/f7Zqk.jpg)


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.


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