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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.

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  • $\begingroup$ Epoxide formation under dilute conditions? $\endgroup$
    – Waylander
    Jul 13 '18 at 15:33
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    $\begingroup$ 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
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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 ?

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  • $\begingroup$ Is it possible for both the OH groups to eliminate? $\endgroup$
    – epsilon-emperor
    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. $\endgroup$
    – epsilon-emperor
    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

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