enter image description here

What happens when the above pinacol is treated with - (a) conc. sulphuric acid (b) dil. sulphuric acid

I'm confused between pinacol rearrangement and dehydration (E1 or E2?), as both possibilities exist. I think the concentration of the reagent might have a role to play, but I'm not sure. Please help.

P.S. It'd be great if you could quote data and sources with your answer.

  • $\begingroup$ Epoxide formation under dilute conditions? $\endgroup$ – Waylander Jul 13 '18 at 15:33
  • 1
    $\begingroup$ I wonder if treatment with conc. sulphuric acid would form some octamethyl-1,4-dioxan (in analogy to glycol). $\endgroup$ – aventurin Jul 13 '18 at 23:02

When you treat pinacol with sulfuric acid, an E1 elimination will take place ( having a carbocation as intermediate ) and it forms pinacolone ( 3,3-dimethyl-2-butanone ). I am not aware of how concentrated the acid should be. Do you require a photo depicting the mechanism ?

  • $\begingroup$ Is it possible for both the OH groups to eliminate? $\endgroup$ – strawberry-sunshine Jul 13 '18 at 15:30
  • $\begingroup$ No. If that happened, you would technically end up with 2 positive charges next to each other and that is not favourable. When a proton “attacks” one of the hydroxyl group, it leaves as water and a positive charge is left on the carbon and that rearranges. In all the books, this is the mechanism that appears. $\endgroup$ – AndrewB Jul 14 '18 at 14:33
  • $\begingroup$ That wouldn't happen, elimination could happen one after the other. $\endgroup$ – strawberry-sunshine Jul 14 '18 at 14:54
  • $\begingroup$ How could they happen one after the other when, after the first one, a ketone is formed ? How can a ketone eliminate water like that ? $\endgroup$ – AndrewB Jul 14 '18 at 18:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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