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In this reaction, why is the $\ce{C-F}$ bond broken instead of the $\ce{C-Cl}$ bond, even though $\ce{C-F}$ bond is stronger than the $\ce{C-Cl}$ bond?

  • 6
    $\begingroup$ The boron probably associates much better with the fluoride, which activates the attached carbon towards substitution. After all, tetrafluoroborate is a very stable anion. $\endgroup$
    – Zhe
    Feb 28, 2018 at 16:18
  • $\begingroup$ @Zhe Could this be because the $\ce{2p}$ orbitals of fluorine have a stronger overlap with the $\ce{2p}$ orbitals of boron, as compared to those with the $\ce{3p}$ orbitals of chlorine? That said however, I am not entirely sure whether the product ratio of 3-chloropropylbenzene to 3-fluropropylbenzene is 40:60 (a minor difference) OR 15:85 (a major difference) $\endgroup$ Mar 1, 2018 at 6:45
  • 1
    $\begingroup$ Could you provide a reference for this reaction? $\endgroup$
    – permeakra
    Mar 1, 2018 at 7:23

1 Answer 1


In Friedel-Crafts alkylation, the halogen-carbon bond breaking depends on the strength of the bond formation between halogen atom and the Lewis acid used. The more stronger is the forming bond, the easier it is to generate a carbocation for the reaction (in all cases, the carbocation explicitly needn't have to generate, a strong $\delta ^+$ charge on the electrophilic carbon is sufficient enough to have a $\ce{S_E2}$ or $\ce{S_E1}$ type reaction).
Here, the Lewis acid used is $\ce{BCl3}$ where there is a vacant $\mathrm{2p_z}$ orbital on the $\ce{B}$ atom (assuming the molecule is oriented in a triangle in the $\mathrm{xy}$-plane), and the Lewis base is the given reagent 1-chloro-3-fluoropropane, which has two basic centres. But donation through $\ce{F}$ will correspond to a stronger $\mathrm{2p_\pi-2p_\pi}$ overlap with Lewis acid, rather than weaker $\mathrm{3p_\pi-2p_\pi}$ overlap in the case of donation through $\ce{Cl}$ atom. Thus, bonding with $\ce{F}$ will correspond to a very huge amount of $\mathrm{-I}$ effect of the adjacent carbon atom, which will weaken the bond between them by pulling the electron pair towards $\ce{F}$ and easily favour the attack by the $\pi$-electron cloud of benzene ring. Also, the repulsion between $\pi$-electron cloud and the large sized $\ce{Cl}$ atom is very less during the formation of the above compound. But, if it would have bonded through $\ce{Cl}$, there won't be a great electrophilic nature of the adjacent carbon due to less effective bonding in the case of $\ce{Cl}$ (corresponding to even lesser $\mathrm{-I}$ effect), so attacking will correspond to a highly energetic transition state, and also, there will be a slight electrostatic repulsion with the $\ce{F}$ atom of the $\pi$-electron cloud.
Due to such aforementioned phenomena and energetically favoured conditions, bonding through cleavage of $\ce{C-F}$ bond is more preferred over cleavage of $\ce{C-Cl}$ bond.


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