What will be the final product when the above compound ((1S,2S,4R,7S)-7-methylbicyclo[2.2.1]heptane-2,7-diol) is heated with concentrated sulfuric acid?

I know that the hydroxyl group in the bridgehead position cannot be protonated and removed to form a carbocation. The other hydroxyl group may be protonated and removed to form a carbocation I suppose. But after that what rearrangements will take place ? What will be the final product(s)?

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    $\begingroup$ There isn't any bridgehead hydroxyl in your compound. $\endgroup$ Dec 29, 2016 at 7:45
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    $\begingroup$ Please have a look at 2-norbornyl cation and/or search for non-classical cation on this site. $\endgroup$ Dec 29, 2016 at 7:48
  • $\begingroup$ @orthocresol I meant the OH on the top of the compound (in digram). That cannot be removed by protonation as a positive charge will be left behind on a non-planar center. $\endgroup$ Dec 29, 2016 at 7:49
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    $\begingroup$ Why can't that carbon be planar? It's not a bridgehead, it's a bridge. $\endgroup$ Dec 29, 2016 at 7:53
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    $\begingroup$ I fail to see why elimination from the tertiary alcohol should not be possible. The resulting cation could either add a nucleophile, which would result in the epimerization of that centre, or undergo elimination. The latter would yield an exomethylene group. $\endgroup$ Dec 29, 2016 at 7:53

1 Answer 1


Protonation of both $\ce{OH}$ groups and subsequent formation of the corresponding cations is possible.

  • The tertiary cation will yield an exomethylene group
  • The "secondary" non-classical cation leads to a $\ce{C=C}$ double bond in the ring. With other words, a norbornene is formed.

The combined eliminations will give 7-methylene bicyclo[2.2.1]hept-2-ene as the product.

In J. Am. Chem. Soc., 1973, 95, 7801-7813, the same product was obtained by elimination from the tosylate of 7-hydroxy-7-methylnorborn-2-ene.


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