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I came across this question in three different textbooks.

The question was: What is the major product obtained on dehydration (treatment with concentrated $\ce{H2SO4}$) of cyclohexane-1,2-diol.

I figured it would undergo a pinacole-pinacolone rearrangement. I expected the product to be cyclohexanone but to my surprise the answer was 1-cyclopentylmethanal.

What can be the driving factor for the ring to preferentially contract? Shouldn't the intermediate carbocation be stable in a six membered ring rather than a five membered ring?

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  • $\begingroup$ Your question is essentially why the C-C bond migrates instead of the C-H bond, right? You may want to check out this answer. $\endgroup$ – orthocresol Dec 14 '16 at 5:01
  • $\begingroup$ @orthocresol I actually went through the answers and the links therein. However, in the cases discussed where the ring degraded, there was not an option for hydride shift as the alpha carbon containing hydroxyl group was lacking an H atom. But in my case, when there is a H shift possible, why does it not happen? $\endgroup$ – Amritansh Singhal Dec 14 '16 at 5:42
  • $\begingroup$ Why do you think the bond angles in a 6-member ring are $120^{\circ}$? $\endgroup$ – Zhe Dec 14 '16 at 15:40
  • $\begingroup$ @Zhe A mistake on my part. I forgot that the six membered ring would be puckered and not planar. but anyways, there needs to be a driving force for the ring to degrade and I'm interested in knowing that reason. $\endgroup$ – Amritansh Singhal Dec 14 '16 at 15:47
  • $\begingroup$ It could be equilibrium driven. You immediately form the aldehyde after ring contraction. $\endgroup$ – Zhe Dec 14 '16 at 16:34
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Background

  • You don't specify whether we are starting with cis- or trans-cyclohexane-1,2-diol, but it doesn't matter since both isomers will interconvert under the reaction conditions. Therefore we will need to consider the reaction products from both isomers.
  • These reactions run in strong acid are never clean. I'd bet that some cyclohexanone was also formed, but that it is the minor product. Here is a link (see section 2. Pinacol Rearrangement, example #3) showing that 1,2-dimethylcyclohexan-1,2-diol produces >90% of the corresponding cyclopentyl analogue and around 6% of the cylcohexanone analogue.
  • Predominant formation of the cyclopentyl carbaldehyde indicates that our reaction must be kinetically driven. 5-membered rings have ~6 kcal/m more strain than 6-membered rings. Therefore, if the reaction were thermodynamically controlled the 6-membered ring would predominate.
  • In the case of cyclohexane-1,2-diol, if a protonated hydroxyl group departed, a seconday carbocation would form. Secondary hydrocarbon carbocations are not particularly stable, so it is likely that rather than generate a secondary carbocation there is some migrating group migration at the same time the protonated hydroxyl is departing. Such migrations have a strong preference for occurring by way of an anti-periplanar geometry.

enter image description here

Solution

In the figure below I've drawn the 3 isomers of cyclohexane-1,2-diol. The isomer on the left is the axial, axial trans isomer, the isomer in the middle is the equatorial, equatorial trans isomer, and then on the right is the axial, equatorial cis isomer. I've also labeled the hydroxyls (A and B) for clarity. In an equilibrium mixture of these isomers we would expect the equatorial, equatorial isomer to predominate because both hydroxyl groups are in the sterically less encumbered equatorial position and both hydroxyls can hydrogen bond with each other in this conformation.

enter image description here

Examining the isomer on the left, neither the $\ce{C-H}$ bond nor the $\ce{C-C}$ bond is aligned anti-periplanar to either hydroxyl A or B, consequently this isomer should be relatively unreactive.

Examining the middle isomer only $\ce{C-C}$ bonds are aligned anti-periplanar to the hydroxyl groups. Rearrangement of this isomer can only produce the cyclopentyl derivative.

Examining the isomer on the right we see that a $\ce{C-H}$ bond is aligned anti-periplanar to hydroxyl A, while a $\ce{C-C}$ bond is aligned anti-periplanar to hydroxyl B. Hence, this isomer can produce both the cyclopently carbaldehyde and cyclohexanone.

We see that there are 3 pathways leading to the 5-membered ring product and just one leading to the 6-membered ring product. This suggests that the 5-membered ring product should predominate. If we further consider that, as explained above, the middle isomer should predominate at equilibrium, then our product mixture should be even further enriched in the 5-membered ring product just as you reported.

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    $\begingroup$ I hadn’t seen new answers of your’s in a while which I felt was a shame. Nice to ‘see’ you’re ‘back’ (if you ever ‘left’ which I can’t verify). =) $\endgroup$ – Jan Dec 14 '16 at 23:46
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    $\begingroup$ @Jan Hi Jan...There just don't seem to be as many interesting organic questions these days. Then when one does come along, few people seem really interested in the question or answer, so I've kinda become more of an observer. I enjoy your answers, you seem to have a passion for chemistry. $\endgroup$ – ron Dec 15 '16 at 2:53
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    $\begingroup$ a really good explanation Ron. I never considered the cis-trans isomers. $\endgroup$ – Amritansh Singhal Dec 15 '16 at 4:37
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    $\begingroup$ Absolutely magnificent!;-) $\endgroup$ – tatan May 22 '18 at 15:34
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    $\begingroup$ This is perhaps one of the most beautiful explanations in organic chemistry I have seen recently... $\endgroup$ – tatan May 22 '18 at 15:40

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