Rearrangement of cycloheptatrienyl carbene yields heptafulvene via a hydrogen shift.

Heptafulvene from Rearrangement of cycloheptatrienyl carbene

I see that the reaction is favorable because there is a conjugative effect with the $\pi$ system of the ring in the product.

Does this rearrangement happen by a radical transition state (TS) or is the mechanism more of a sigmatropic rearrangement?

My arguments for

  1. radical TS: The electron of the carbene (let's assume a singlet carbene here) lies close to the hydrogen, it could be a planar cyclic transition state.

  2. [1,2]-Hydrogen shift: From my chemical intuition I think the $\ce{C-H}$ bond is more a polar bond and therefore I would describe hydrogen more as a proton (positive partial charge). I would say it's a [1,2]-sigmatropic rearrangement, 2 electrons are involved, and therefore suprafacial and describing the carbene $\mathrm{sp^2}$ as HOMO and the hydrogen as LUMO.

What's the correct mechanism for this rearrangement?


1 Answer 1


This reaction can be viewed as a 1,2-sigmatropic hydrogen shift or as a 2+1 (if we are counting atoms, 2+0 if we are counting electrons) cycloaddition (adding the carbene's empty $\mathrm{p}$-orbital across the $\ce{C-H}$ bond - analogous to its addition to an olefinic pi bond to produce a cyclopropane). However the reaction is most commonly viewed as and referred to as an "insertion" reaction because the carbenic carbon is inserting itself in between the carbon and hydrogen atoms that comprised the $\ce{C-H}$ bond.

Carbene addition-1

(image source)

Here's what is known.

  • Thermal decomposition of most diazo compounds produce singlet carbenes.

Singlet carbene

  • Singlet carbenes add stereospecifically to olefins. That is, addition of a singlet methylene carbene to trans-2-butene produces only the corresponding trans-dimethylcyclopropane

Carbene addition-2

(image source)

Further, insertion of singlet carbenes into stereogenic centers results in retention of stereochemistry at the chiral center. These facts suggest that both the cycloaddition and insertion reactions of simple singlet carbenes either proceed in a concerted fashion, or if any intermediate is formed it is extremely shorted-lived in order that stereochemical information be preserved.

  • Simple carbenes are typically mildly electrophilic. The HOMO (the filled $\mathrm{sp^2}$ orbitals) and LUMO (the empty $\mathrm{p}$-orbital) for a singlet carbene are shown on the left side of the following figure.

Carbene energy diagram

(image source)

Therefore, we would expect the interaction of the carbene LUMO (the empty $\mathrm{p}$-orbital) and the $\ce{C-H}$ bond HOMO to be controlling, as the energy separation here is much less than the other possible HOMO-LUMO interaction between the carbene HOMO (the filled $\mathrm{sp^2}$ orbital) and the much higher in energy $\ce{C-H}$ antibonding orbital, the LUMO.

This controlling HOMO-LUMO interaction involving the empty carbenic $\mathrm{p}$-orbital in the rate determining step is consistent with the observed electrophilic nature of the carbene.

All of this information taken together suggests that interaction of the empty carbenic $\mathrm{p}$-orbital with the electrons in the $\ce{C-H}$ bond is the critical step. A geometry like that shown below is likely preferred. As the insertion proceeds, the trigonal carbenic carbon rotates. The rotation starts to bring the carbenic $\mathrm{sp^2}$ orbital into alignment with the p-orbitals in the 7-membered ring. Also during the rotation the $\mathrm{sp^2}$ orbital starts its rehybridization into a $\mathrm{p}$-orbital. When completed we have created the final product, heptafulvene. The reaction is likely concerted and certainly does not involve any long-lived intermediates. Note too that the heptafulvene is formed with increased electron density on the former carbenic carbon in accord with the dipolar nature (dipole moment ~$\pu{0.5 D}$ due to resonance stabilization) of the molecule.

Carbene addition-3

  • $\begingroup$ Thank you for your comprehensive answer! I understand your answer in the way that the carbene makes an insertion into the CH bond and the empty p orbital helps to stabilize this conformation with the hyperconjugative effect. I thought that the empty p-orbital is already perfectly aligned with the π system in the planar ring in your drawing, therefore I'm wondering why we still need a rotation. I would prefer a bisected conformation of the p & carben sp2 with the $\sigma_{CH}$ bond. Is the roation in clockwise or anticlockwise direction if we look from the carbenic carbon onto the CHR2 carbon? $\endgroup$
    – laminin
    Commented Aug 23, 2015 at 12:53
  • 1
    $\begingroup$ In the early stage of the reaction, the carbene p-orbital is separated from the ring's pi system by an insulating sp3 carbon so there is minimal interaction between the carbenic p-orbital and the ring's pi system. The conformation I drew up above with the p-orbital eclipsing the C-H bond seems best aligned for the insertion. The p-orbital becomes a C-H bond and then upon rotation the filled sp2 orbital rehybridizes and becomes the terminal p-orbital in the exo double bond. The direction of rotation does not matter; because of the p-orbital symmetry, $\endgroup$
    – ron
    Commented Aug 23, 2015 at 14:43
  • 1
    $\begingroup$ rotation in either direction produces the same result. $\endgroup$
    – ron
    Commented Aug 23, 2015 at 14:44

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