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What is the enol form of this compound?

(4S)-4-methylcyclohex-2-en-1-one

I have made an attempt, but it doesn't look quite right.

enter image description here

Which hydrogen should be taken off (left or right hand side of the carbonyl)? And I don't think the ring is aromatic so I don't know how it is stabilized.

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    $\begingroup$ Why should it be stabilised? I doubt this enol form is more stable than the α,β-unsaturated carbonyl. $\endgroup$
    – orthocresol
    Oct 19, 2015 at 7:57
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    $\begingroup$ 1) There are two possible enol forms of this compound. 2) Being an enol is not intrinsicly more stable than being a ketone (in fact, ketones are usually more stable than enols) but with a strong base you can deprotonate the system. The enolate will then be more stable. 3) Which hydrogen you take off depends on your base. $\endgroup$
    – Jan
    Oct 20, 2015 at 15:48
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    $\begingroup$ As @Jan points out, there are two possible enols. One (the one you've drawn above) has a hydroxy group located on the diene double bonds; the other (produced by removing the proton on the methylated carbon) has both a hydroxy and a methyl group attached to the diene double bonds. One enol has one more substituent, the methyl group, located on the diene. Methyl groups stabilize double bonds. Therefore the enol with both the hydroxy and methyl group located on the diene will be the more stable enol and will predominate at equilibrium. $\endgroup$
    – ron
    Oct 21, 2015 at 19:09
  • $\begingroup$ More substitutes alkenes are more stable. $\endgroup$
    – Aditya Dev
    Oct 22, 2015 at 16:44

3 Answers 3

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There are two possible enol forms of this ketone. The first one is:

enter image description here

This is the one you drew, which shows the α hydrogen being abstracted. It is perfectly valid.

The other one is:

enter image description here

Note how the alkene next to the carbonyl group "extends" the range of the carbonyl group. This allows the γ hydrogen (i.e. the one three carbons away) to be removed, and then you can "push" electrons into the carbonyl group through the alkene group. In general, this is a property of α,β-unsaturated ketones (i.e. ketones which have a double bond between the α and β carbons).

Note that in this enol form, the methyl substituent is now in the same plane as the C=C next to it. This means that, if you were to reverse the tautomerisation to get back the keto form, there's an equal chance of the hydrogen coming in from above the plane and from below the plane, and racemisation would occur. In practice, there is a rapid equilibrium between keto and enol forms, and this means that if you were to dissolve this compound in any kind of acid (or base), the racemisation would occur pretty quickly.

Here, I have drawn the mechanisms for acid-catalysed tautomerisation. I assume you are now able to draw the mechanisms for the reaction in base (which in general forms an enolate instead of an enol).

At this stage, you might wonder why we cannot abstract the other α hydrogen. There's a couple of reasons. Firstly, it's attached to a $\mathrm{sp^2}$ carbon, and the C-H bond is therefore somewhat stronger. Secondly, even if you could take off that proton and form the "enol", you would have two double bonds next to each other. That's a huge problem because those double bonds want to be linear, but don't quite have the space to do so since there's a ring:

enter image description here

Regarding the stability of the two legitimate enol forms, I would not say that they are any more stable than the keto form. In general, the keto form is more stable. There are cases in which the enol form is relatively stabilised, most often when there is aromaticity or extra conjugation in the enol form (other factors such as solvent will also influence the position of equilibrium):

enter image description here

However, that is not the case here. The starting keto form is as conjugated as any enol you can form.

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  • $\begingroup$ I was taught that it depends on the solvent whether acetoacetate adopts a diketo or enol form: Diketo in hydrogen-bonding solvents and enol in those solvents that cannot be hydrogen bond donors. $\endgroup$
    – Jan
    Oct 22, 2015 at 16:31
  • $\begingroup$ @Jan hmm, I know it does depend on solvent, but never learnt that little detail. I'll go look it up when I have time. $\endgroup$
    – orthocresol
    Oct 22, 2015 at 16:34
  • $\begingroup$ Enol form of aceto acetic ester is more stable than the diketo compound because of intramolecular hydrogen bonding between OH and carbonyl Oxygen. The enol form is only stable in non polar medium. In polar solvent, the solvent interrupts the intramolecular Hydrogen bonding. I don't think it's because of conjugation $\endgroup$
    – Aditya Dev
    Oct 22, 2015 at 16:49
  • $\begingroup$ Ok, I looked it up: en.wikipedia.org/wiki/Dimedone has a 67:33 ratio of keto:enol in $\ce{CDCl3}$. In this compound, there's no possibility of intramolecular H-bonding, so the extra stability must come from conjugation (normally, the ratio is $10^5:1$). Presumably, the intramolecular H-bonding in ethyl acetoacetate further stabilises the enol form in a solvent that cannot form hydrogen bonds. $\endgroup$
    – orthocresol
    Oct 22, 2015 at 16:55
  • $\begingroup$ In basic medium, one possibility is for hydrogen four in the ring (according to iupac name) to get abstracted by the base to form a carbanion at position 4, which is in resonance with the alkene and the carbonyl group, hence in one resonating structure the carbanion will be at carbon number 2 (which was originally sp2 hybridised, the one you talk about). In this position the carbanion could abstract an hydrogen to form a second keto form (4-methylcyclohex-3-en-1-one); although this form is less stable than the other keto/enol forms, it should surely exist in equilibrium, correct? $\endgroup$
    – Box Box Box Box
    Jul 25, 2021 at 12:21
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Also which hydrogen should be taken off (left or right hand side of the carbonyl)? And i dont think the ring is aromatic so i dont know how it is stabilized...

When creating an enol the hydrogen alpha relative to the carbonyl is generally ionized. The alpha hydrogens are somewhat more acidic than the other hydrogens because of the possibility for resonance withdrawal of electron density by the carbonyl oxygen.

The ring is conjugated, but not fully conjugated, so the ring isn't aromatic. But partial conjugation is stabilizing.

Whether the enol or the keto form of this molecule is more stable requires further consideration; we cannot simply say that one is conjugated and the other is less conjugated and the one that is more conjugated must be more stable.

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    $\begingroup$ It does not have to be $\alpha$ in the presence of an enone system … $\endgroup$
    – Jan
    Oct 21, 2015 at 17:33
  • $\begingroup$ @Jan edited to reflect that fact $\endgroup$
    – Dissenter
    Oct 21, 2015 at 22:19
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    $\begingroup$ But Dissenter, isn't the OP really asking which of the 2 possible enols is more stable ("What is the enol form of this compound")? $\endgroup$
    – ron
    Oct 21, 2015 at 23:25
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yes your answer is right. this is more stable as it is a conjugated diene. but, initially the H+ attacks on the lone pair of =O then the oxygen takes up the pie bond to neutralize the formal + charge on it.

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    $\begingroup$ I wouldn't say that an electrophile (i.e. H+) attacks anything. $\endgroup$
    – Dissenter
    Oct 19, 2015 at 8:02
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    $\begingroup$ @Dissenter If it is acid catalysed, it totally does. If not, then there is no electrophile. $\endgroup$
    – Martin - マーチン ♦
    Oct 19, 2015 at 8:05
  • $\begingroup$ Also it is not necessarily more stable. The carbonyl is also conjugated so just saying that it is a conjugated diene is not enough. $\endgroup$
    – bon
    Oct 19, 2015 at 9:34
  • $\begingroup$ @Dissenter yes, it will. otherwise its illogical to show that the pi- bond to shift directly to an H+. if an electrophile hadnt attacked anything, how would any reaction proceed? $\endgroup$
    – royred_98
    Oct 20, 2015 at 14:58
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    $\begingroup$ Guys please do note again that wrong answer (if this is one) isn't equal to VLQ. $\endgroup$
    – M.A.R.
    Oct 20, 2015 at 19:13

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