# Reaction of perchloric acid with cyclopropenol

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 $$\ce{HClO4}$$?

• 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. May 19 '15 at 0:05
• @permeakra it is remarkably stable yes, but getting rid of the strained ring by oxidative cleavage (similar to hot $\ce{KMnO4}$) seems way more plausible. We shouldn't ignore the oxidative capabilities of $\ce{HClO4}$ Jul 25 at 4:31
• The answer provided is wrong . Whenever cyclic conjugated ketones comes in question then the conjugation is the unsaturated part is the point of action. Which means double bond will be attacked. Aug 24 at 4:46
• Can you provide a reference? Aug 24 at 4:57

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.

• Carbocation as the final product, really interesting! Jul 11 '18 at 20:21
• 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 Apr 23 '20 at 16:16
• @Archer I think this answer disregards $\ce{HClO4}$'s oxidative abilities. I think the stability gained by cleaving the alkene is far more than the resonance of the carbocation. Jul 7 at 16:08
• @napstablook $\ce{HClO4}$ is often used in dilute solution where it dissociates, and then the oxidation power of the perchlorate ion is kinetically squashed. Perchloric acid here is acting only as an acid. Aug 24 at 15:59

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.

• Your effort is appreciated, but Grignard reagents with unsaturated ketones undergoes 1,2-addition not 1,4-addition. Refer this: chemistry.stackexchange.com/questions/31618/…
– TRC
Aug 24 at 15:52
• Also, there are a couple more mistakes in your mechanism - after enol is formed it should rapidly tautomerize to ketone. The carbocation you have formed is a vinyl carbocation on a highly strained three membered ring with a double bond, that will never exist. Neither is that carbocation participating in resonance/aromaticity.
– TRC
Aug 24 at 15:54
• Weaker bases do attack a ketone at the double bond, but something as strong as a Grignard reagent still goes for the carbonyl group. Aug 24 at 16:00
• @ TRC and @ Oscar Lanzi,Thanks. Aug 24 at 17:41