enter image description here I was practising some problems on hydride shift and got stuck in this particular question. I wanted to know if hydride shift would take place in this particular case and which carbon would get the positive charge. Will a hydrogen from left or right carbon would shift??

  • $\begingroup$ The molecule is symmetric? $\endgroup$ – Karl Sep 26 '18 at 21:08

I strongly suspect that a hydride shift is not observed in this example.

The reason is that it seems much more likely that you will have an alkyl shift instead, accompanied by ring contraction and formation of a carbonyl. This is essentially a pinacol rearrangement.

EDIT: Adding more details per @OscarLanzi comment...

I would think that this is the expected product after proton transfer.


It looks like the pinacol arrangement of 4,8-decalindiol has been known for quite some time. Hückel, Danneel, Schwartz, Gercke, Ann. 1929, 474, 121. and Meiser. Chem. Ber. 1899, 32, 2055.

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  • $\begingroup$ Could you render the expected product? Your description seems to suggest a spiro-type product and that does not go well with my intuition. Thanks. $\endgroup$ – Oscar Lanzi Sep 26 '18 at 14:36
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    $\begingroup$ @OscarLanzi That's exactly what I'm suggesting. Let me edit with details. $\endgroup$ – Zhe Sep 26 '18 at 15:48
  • $\begingroup$ @zhe but why hydride shift would not take place?? I thought that hydride shift would have been easier than an alkyl shift?? And why ring contraction would happen?? Please explain . Thanks $\endgroup$ – gks Sep 26 '18 at 17:18
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    $\begingroup$ Well, for one thing, a hydride shift still gives a secondary cation, but a alkyl shift provides a carbonyl group and a smaller, not-very-strained ring. I'm not sure if this is a thermodynamic or kinetic effect, but it seems reasonable that this difference can lead to a preference in either. $\endgroup$ – Zhe Sep 26 '18 at 17:30

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