7
$\begingroup$

What is the mechanism of the conversion of cyclopropane-1,2,3-triyltrimethanol to benzene in in acidic medium?

conversion of cyclopropane-1,2,3-triyltrimethanol to benzene in in acidic medium

After performing E1 thrice I was expecting 1,2,3-trimethylidenecyclopropane to form instead:

1,2,3-trimethylidenecyclopropane

I know it's highly strained, but I couldn't adopt any other pathway to reach any other product. Is the triene product not formed at all? If cyclopropane-1,2,3-trione can exist, I think even the triene product can.

This reaction is from my exercise book, couldn't find it on OrgSyn.

$\endgroup$
2
  • 3
    $\begingroup$ This reaction does not show up in Scifinder. The triol is known though: Org Lett, 2007, 9, 2617 DOI:10.1021/ol070707r $\endgroup$
    – Dennis Cao
    Jul 10, 2018 at 17:01
  • 4
    $\begingroup$ No hits on Reaxys either. I strongly doubt it is real. $\endgroup$ Aug 1, 2018 at 6:21

2 Answers 2

13
$\begingroup$

I have not written the steps for carbocation formation, simple dehydration.enter image description here

$\endgroup$
6
  • 9
    $\begingroup$ The mechanism is plausible, but I think it is still important to emphasise that this reaction is not known. There are zero reports of anybody getting benzene by treating the starting material with acid. Probably somebody made it up. $\endgroup$ Aug 1, 2018 at 6:21
  • 13
    $\begingroup$ @ortho With a little foresight into this whole JEE business, you'll know that it's almost definitely made up ;-) $\endgroup$ Aug 1, 2018 at 7:20
  • 7
    $\begingroup$ It looks nice on the paper. $\endgroup$
    – mykhal
    Aug 1, 2018 at 12:07
  • $\begingroup$ Hey primary cations are too unstable. $\endgroup$
    – Tech kidda
    Aug 12, 2022 at 10:14
  • $\begingroup$ @Tech kidda: They may be unstable but that doesn't mean they don't form. I also checked Chem. Absts. NADA! $\endgroup$
    – user55119
    Aug 12, 2022 at 23:21
0
$\begingroup$

As per my knowledge, in the second step there should be migration of Hydrogen to form a 3° carbocation and then the double bond will form between the two methanol branches.This will entirely change the product but as per rule the 1° cation has to be further stabilized by shifting of hydrogen.

$\endgroup$

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.