We were given the following carbocations and asked to compare their resonance energies.

$\ce{L-CH2+}$, where $\ce{L}$ is:

  1. $\ce{NH2}$
  2. $\ce{OCH3}$
  3. $\ce{Ph}$
  4. $\ce{F}$
  5. $\ce{Cl}$
  6. $\ce{NO2}$

and the order was given as:

$1 > 2 > 3 > 5 > 4 > 6$

I am clear with the other orders but I am unsure about $\ce{F}$ and $\ce{Cl}$. Why does $\ce{Cl-CH2+}$ have a higher Resonance Energy than its $\ce{F-CH2+}$ counterpart even though in resonance, $\ce{Cl}$ forms $\ce{{3p}-{2p}}\ \pi$ bonds compared to $\ce{{2p}-{2p}}$ ones in $\ce{F-CH2+}$, which are less stable due to less efficient overlapping? $\ce{F-CH2+}$ does have a higher electronegativity, but is it enough to overcome the higher stability of the $\ce{{2p}-{2p}}$ bond over the $\ce{{3p}-{2p}}$ one?


1 Answer 1


It is not like π bond between fluorine and carbon is very stable and that is where your line of thinking is wrong. While it is true that $\ce{{2p}-{2p}}$ π bonds are more stable in general than their $\ce{{2p}-{3p}}$ counterparts, one should not take that alone as a decisive factor.

The question is comparing the electron-donating capabilities of fluorine and chlorine, and that italicised word there alone should ring a bell. Fluorine is much more electronegative than chlorine which means that a carbon bound to fluorine has a much smaller electron density which destabilises cations much more than could be offset by the overall weak π donating capacity of fluorine.

Remember that for electrophilic aromatic substitutions, halide substituents are classified as deactivating — mainly due to their $-I$ effect — and only weakly ortho/para-directing — due to the smaller $+M$ effect. This should serve to show which effect predominates and it only gets more pronounced if instead of a large aromatic system you only consider the atom directly adjacent.

  • $\begingroup$ So, is the bond stability factor overlooked when dealing with such examples $\endgroup$
    – Ayushmaan
    Sep 26, 2017 at 13:38
  • 1
    $\begingroup$ @Ayushmaan It’s not overlooked, it’s simply minor. $\endgroup$
    – Jan
    Sep 27, 2017 at 8:48

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