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Predict the major product of this reaction:

enter image description here

The two compounds on the right are the possibilities that I thought of (weren't given in the question). According to me:

  1. attack on carbonyl carbon by $\ce{MeMgBr}$ and then hydrolysis with $\ce{aq. HCl}$
  2. first tautomerise to form phenol. $\ce{MeMgBr}$ will remove acidic proton of phenol. The $\ce{aq. HCl}$ workup will then add the hydrogen back to phenol.

I an confused as to which product would be the major product. I thought phenol would be the major product. But to my surprise, the answer given is 1-methylcyclohex-2-enol. I am not able to understand how it is formed, and need help here.


Source: Joint Entrance Exam (JEE) 1997 India

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  • $\begingroup$ Have you looked at grignard reagents? $\endgroup$ – David Wyn Williams Apr 19 '18 at 15:30
  • $\begingroup$ Yes sir.. I know how grignards react. And drew the products based on that knowledge $\endgroup$ – Kavita Juneja Apr 19 '18 at 15:36
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    $\begingroup$ That's a pretty blatant mistake. Tautomers of phenol enolise so quickly they can't be isolated. $\endgroup$ – Mithoron Apr 19 '18 at 16:01
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    $\begingroup$ Upon further review, maybe they misdrew the substrate structure? The given answer makes sense if we have only one carbon-carbon double bond conjugate to the carbonyl group. $\endgroup$ – Oscar Lanzi Apr 20 '18 at 1:29
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    $\begingroup$ Hmm, I wonder how easily that alcohol decomposes to toluene on acidic workup... $\endgroup$ – Zhe Apr 20 '18 at 13:56
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The given answer makes no sense with the substrate actually presented (but see "alternative possibility" below). You will get phenol.

If you start with phenol then, of course, the Grignard reagent just yanks off the hydroxyl hydrogen, only to have that hydrogen restored by the acid workup. Now suppose you were to have the keto form cyclohexa-2,4-dien-1-one. This tautomerizes to phenol precisely because the methylene hydrogen is more strongly acidic than it would be as a hydroxyl hydrogen in phenol. So the Grignard reagent would if anything, go even faster for the methylene hydrogen in cyclohexa-2,4-dien-1-one than it would go the hydroxyl hydrogen in phenol - unless the ketone enolizes first.

That latter issue is academic. Either way you get the same phenolate ion product, which upon acid workup gives phenol.

An alternative possibility

If the substrate structure in the question was misdrawn, then the given answer could be correct. Cyclohexa-2-en-1-one, with only one double bond in the ring conjugated to the carbonyl group, does not have any (relatively) strongly acidic hydrogens and so the Grignard reagent will add to the carbonyl group. After acid workup the ketone is thereby converted to the tertiary alcohol in the given answer.

Grignard reagents do not react with carbon-carbon double bonds, so after adding to the carbonyl group or extracting an acidic proton that's all there is.

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  • $\begingroup$ Thank you sir, yes maybe they misdrew the answer. I was thinking MeMgBr was in excess.. will the answer be still phenol/phenoxide?? $\endgroup$ – Kavita Juneja Apr 20 '18 at 4:23
  • $\begingroup$ Methyl Grignard will never add to phenoxide $\endgroup$ – Waylander Apr 20 '18 at 11:28
  • $\begingroup$ And it does not in my answer. The "alternative scenario" does not proceed via phenoxide. $\endgroup$ – Oscar Lanzi Apr 20 '18 at 13:07

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