The Prins reaction – stepwise addition of an alkene to a carbonyl compound – leads to the same product as a concerted ene reaction. This is illustrated here with tin tetrachloride as a Lewis acid, which (as far as I know) lowers the activation energy of both pathways:

Mechanistic pathways of Prins and ene reactions

How can these two mechanistic pathways be differentiated experimentally?

One thing that I can think of right now would be the stereochemistry of the alkene formed - presumably in a concerted pathway, the cyclic transition state (envelope-like, if I remember correctly?) the substituent R' would be positioned equatorial or pseudo-equatorial. However, I'm not confident enough to undertake an analysis of the stepwise reaction.

I'm open to hearing about other possibilities, e.g. kinetic studies or spectroscopic studies. I suppose the free carbocation in the Prins reaction could lead to a more negative coefficient $\rho$ in a Hammett analysis, compared with a stepwise mechanism?

I'd also be glad if somebody could provide an example in the literature, if it exists. (I confess to not having looked this up - been very busy.)

  • 1
    $\begingroup$ Determine the kinetic isotope effect for the $\ce{C-H}$ bond that is broken. $\endgroup$
    – ron
    Dec 6, 2017 at 0:06

1 Answer 1


The traditional Prins reaction was run in aqueous acid (H2SO4, HCl) with formaldehyde. The carbocation formed by addition of the alkene was trapped by a nucleophile, or if E1 elimination occurred, the most stable alkene formed. The Lewis acid-catalyzed ene reaction is intramolecular without nucleophile capture and the less stable alkene is formed, at least when an isopropenyl group is present in the reactant. Two examples follow:

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Malonaldehyde 1, which has two enantiotopic aldehyde groups, reacts via transition state 2 to produce alcohol 3 as the major product. The cyclopentadiene ring is in effect a t-butyl group that prefers being equatorial in the transition state. Note the exocyclic double bond as opposed to a more stable endocyclic position. The hydroxyl group is axial. Aldehyde 4 follows the same type of pattern forming axial alcohol 5. If an open transition state were operable, the equatorial alcohol would be anticipated. [PS: You can Google much of this information.]

1) J. Am. Chem. Soc., 1990, 112, 2149-2158

2) J. Am. Chem. Soc. 1984, 106, 718-721

  • $\begingroup$ I probably think it does answer the question. The main point is that the transition state for a cyclic mechanism leads to different stereocontrol. $\endgroup$ Dec 7, 2017 at 14:06

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