5
$\begingroup$

I was wondering, what would happen if I treated the following compound i.e. trispiro[2.1.25.1.29.13]dodecane-4,8,12-triol with sulfuric acid[$\ce{H_2SO_4}$] to eliminate OH group to form double bond:

molecule

Will the double bond form? if yes then where and why? If no then what other compound will form and why?

Improve this question
$\endgroup$
8
  • 3
    $\begingroup$ Well, there's gonna be various rearrangements $\endgroup$
    – Mithoron
    Commented Oct 8, 2016 at 12:42
  • 3
    $\begingroup$ Agree, probably a lot of 1,2-alkyl shifts. $\endgroup$
    – orthocresol
    Commented Oct 8, 2016 at 13:04
  • 1
    $\begingroup$ You might get a carbonyl from a hydride shift... $\endgroup$
    – Zhe
    Commented Oct 8, 2016 at 15:59
  • $\begingroup$ Have you tried to draw the structure of a compound resulting from elimination? $\endgroup$
    – jerepierre
    Commented Oct 19, 2016 at 15:14
  • $\begingroup$ @jerepierre I am really confused about what would happen. I think that there will be a ring expansion for each of those cyclopropane rings, forming a four member bridged ring. But then where would the double bonds form? Bredt's rule disallows formation of double bonds at bridgehead. $\endgroup$
    – S R Maiti
    Commented Oct 20, 2016 at 8:47

1 Answer 1

Reset to default
4
$\begingroup$

This question has been unanswered for quite long, and meanwhile, I have found an answer, so I wanted to share it.

On treating the compound with sulfuric acid, the $\ce{OH}$ groups will be protonated to $\ce{OH_2^+}$.

protonation of the OH groups

The $\ce{OH_2^+}$ groups are good leaving groups, so they will leave as $\ce{H_2O}$, forming carbocation.(one by one) formation of carbocation

As cyclopropane is under significant ring strain, it will undergo ring expansion to form cyclobutane ring, which is somewhat less strained. At the same time, $\ce{H^+}$ ions will be expelled, forming double bonds. (It should also take place step by step.) formation of double bonds

In this case I don't think that the double bond will not cause any significant barrier to the formation of the molecules, as the double bond is here the bridging bond, not at the bridgehead[formation of double bonds at bridgehead is not possible for smaller rings(<8 carbon rings), as it causes very high strain: Bredt's rule]

I have considered only elimination in this reaction scheme. Other compounds may also be formed. Feel free to post answers if you think other compounds will be formed or if you think the above mechanism is wrong in any way.

Improve this answer
$\endgroup$
11
  • 1
    $\begingroup$ I don’t think the steps would happen to each carbon simultaneously but other than that this sounds reasonable. $\endgroup$
    – Jan
    Commented Dec 3, 2016 at 18:46
  • $\begingroup$ @Jan Why not simultaneously? $\endgroup$
    – S R Maiti
    Commented Dec 4, 2016 at 4:32
  • 4
    $\begingroup$ Carbotrications seem like a far-fetched idea to me... $\endgroup$
    – Curt F.
    Commented Dec 4, 2016 at 7:10
  • 1
    $\begingroup$ sequential not simultaneous reaction $\endgroup$
    – Curt F.
    Commented Dec 5, 2016 at 15:24
  • 1
    $\begingroup$ Double bonds won't form in this scenario. It still would be a triol probably. $\endgroup$
    – permeakra
    Commented Dec 5, 2016 at 15:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.