Why do we have an alkyl shift rather than epoxidation in a pinacol reaction? I don't understand why the hydroxyl group doesn't attack the positive charge.



1 Answer 1


Excellent question.

When the first Pinacol rearrangements were being studied, there was a suggestion that the product of the reaction was indeed the epoxide (migrations/shifts weren't well known reactions at this point in time, and methods of characterisaton were limited). The reaction that was being attempted is shown below (source: Wikipedia), however at this point there was even dispute over the structure of acetone, further adding to the complications.

One of the first reports of a Pinacol rearrangement enter image description here

Today, we of course know that Pinacol and semi-Pinacol rearrangements occur when vicinal diols are treated with protic acid (or Lewis acids) to give the carbonyl product as a result of a [1,2]- alkyl shift.

enter image description here Source: Strategic Applications Of Named Reactions. Wiley.

Looking at the mechanism (below), we first have a rate-determining loss of water via protonation-elimination. This generates the all important cationic intermediate.

The simple answer to your question is that rearrangement occurs faster than epoxidation (which is never observed to any significant extent), this is intuitive if we consider that:

  1. Under the acidic conditions, the alcohol required to form the epoxide is fully protonated, and hence not a very good nucleophile (in agreement with R-OH generally being a poor nucleophile except in basic conditions where we get the anionic alkoxide RO(-)

  2. The migration generates a more stabilised carbocation by the influence of the oxygen LP donating into the carbocation to form the resulting ketone.

enter image description here

Source: Strategic Applications Of Named Reactions. Wiley.

As an aside, if you'd kept reading through Clayden by another page or so, you would have discovered that epoxides also undergo Pinacol-type rearrangements, so realistically even if the epoxide did form, it would simply re-open under the acidic/Lewis-acidic conditions before undergoing a rearrangement reaction.

enter image description here

Source: Organic Chemistry, 1 ed. Oxford University Press (Warren & Clayden)

  • 1
    $\begingroup$ great answer, thanks! But why does the OH group stabilize the positve charge , i always thought an OH group on an alkyl is electron withdrawing? $\endgroup$
    – hanna
    Jan 18, 2016 at 13:03
  • 3
    $\begingroup$ It's not so much that the oxygen is 'stabilising' the positive charge, its that the lone pair of the electrons are kicking down to form the pi-bond, forming the ketone/aldehyde. I'll try and dig up a study of the kinetics, but IIRC, once the water has left, the [1,2] shift and subsequent formation of the pi bond are incredibly rapid (i.e. the cation adjacent to the oxygen is very short lived indeed) $\endgroup$
    – NotEvans.
    Jan 18, 2016 at 19:40
  • $\begingroup$ Hanna, it is often said that oxygen is electron-withdrawing by induction and electron-donating by resonance. Here the resonance of the carbocation to the carbonyl is a much more significant factor than the induction. $\endgroup$ Dec 23, 2016 at 22:43

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