In this website it is claimed that "[e]thene reacts explosively with fluorine to give carbon and hydrogen fluoride gas", via this equation:

$$\ce{CH2=CH2 + 2F2 -> 2C + 4HF}$$

Why does the fluorine hog the hydrogen atom instead of attacking the $\pi$-bond?

Additionally, may I ask for the mechanism of this reaction?

  • 2
    $\begingroup$ It's seems wrong because fluorine reacts very vigorously to form not only fluorinated but also several oxidation products. tetrafluoromethane also formed during the reaction. The exact mechanism is unknow $\endgroup$ – Khan Sep 4 '16 at 3:48
  • $\begingroup$ @N.ALodhi Would I be correct in saying that the tetrafluoromethane may have formed from the C in the product above? $\ce{C +2F2 -> CF4}$ $\endgroup$ – DHMO Sep 4 '16 at 3:49
  • $\begingroup$ yes, you are right, here is the link link.springer.com/article/10.1023/B:RJAC.0000024584.69292.6d $\endgroup$ – Khan Sep 4 '16 at 3:53
  • $\begingroup$ @N.ALodhi However I do suspect that there could be a $\ce{CH2F2}$ intermediate, meaning that the carbon atom is not completely separated before the tetrafluoromethane starts forming. $\endgroup$ – DHMO Sep 4 '16 at 4:16
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    $\begingroup$ There is no mechanism, in the sense that there are too many things happening all at once. Name just about any remotely sensible particle composed of C, H, F, and it would be there. $\endgroup$ – Ivan Neretin Sep 4 '16 at 6:54

The equation gives only the reactants and final products and tells us nothing about the mechanism; stoichiometric equations rarely do.
This is an explosive reaction so it will proceed by many steps usually involving radicals. To get feedback, and so explosion, one step must be of the branched chain form $$\ce{R_1^. + M -> R_2^. + R_3^.}$$ where $\ce{R^.}$ is a radical and $\ce{M}$ a reactant species (e.g. $\ce{CH2=CH2}$). As two radicals are produced for one reactant radical a chain reaction ensues, the reaction rate accelerates and so bang!

Any explosive reaction has initiation, propagation, branching and termination steps, some also have an inhibition step.
The initiation step in your reaction probably forms the radical $\ce{FCH2-CH2^.}$ although I don't know the mechanism.

  • $\begingroup$ Is there a typo in your equation? $\endgroup$ – DHMO Sep 4 '16 at 9:43
  • $\begingroup$ I did not downvote. $\endgroup$ – DHMO Sep 4 '16 at 13:19
  • $\begingroup$ ok, one can't tell who does this. I hope that you found the answer helpful. You can always up vote! $\endgroup$ – porphyrin Sep 4 '16 at 13:21
  • $\begingroup$ My answer is, in principle, correct so why down vote? It does not make sense to do this and no, no typo. Two radicals are produced, one can be the same and one different, this is a generic way of illustrating that two radicals are produced from one initial one. If you feel happier the right hand side can be written as R2⋅+R3⋅ which I have changed in my answer. $\endgroup$ – porphyrin Sep 4 '16 at 17:59
  • $\begingroup$ I'd make one other point. To be "explosive" the reaction is certainly producing heat. So the rate of reaction starts off slow and as the system heats up the rate of reaction increasing dramatically. $\endgroup$ – MaxW Sep 5 '16 at 5:57

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