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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10$\begingroup$ 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. $\endgroup$– permeakraMay 19, 2015 at 0:05
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1$\begingroup$ @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 $\ce{HClO4}$ $\endgroup$– napstablookJul 25, 2021 at 4:31
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1$\begingroup$ 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. $\endgroup$– rajendra dubeyAug 24, 2021 at 4:46
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$\begingroup$ Can you provide a reference? $\endgroup$– Nisarg BhavsarAug 24, 2021 at 4:57
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2 Answers
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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.
answered Nov 10, 2015 at 14:15
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$\begingroup$ Carbocation as the final product, really interesting! $\endgroup$– ArcherJul 11, 2018 at 20:21
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1$\begingroup$ 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 $\endgroup$ Apr 23, 2020 at 16:16
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2$\begingroup$ @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. $\endgroup$ Jul 7, 2021 at 16:08
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$\begingroup$ @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. $\endgroup$ Aug 24, 2021 at 15:59
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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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$\begingroup$ 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/… $\endgroup$– TRCAug 24, 2021 at 15:52
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$\begingroup$ 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. $\endgroup$– TRCAug 24, 2021 at 15:54
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$\begingroup$ Weaker bases do attack a ketone at the double bond, but something as strong as a Grignard reagent still goes for the carbonyl group. $\endgroup$ Aug 24, 2021 at 16:00
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2$\begingroup$ How does this answer add anything to the already posted answer? $\endgroup$ Aug 24, 2021 at 19:48