I understand that the functional group $\ce{-F}$ is an electron-withdrawing group. It is expected that the difficulties of an electrophile to attack the carbon atoms are $$\ce{PhH}<\ce{2,4of PhF}<\ce{3 of PhF}\tag{1}$$Therefore, I want to demonstrate it quantitatively by computing the charge densities of carbon atoms.
Using command # opt freq hf/3-21g pop=nbo geom=connectivity and the molecule models in Gaussian 09, enter image description hereI obtained the charge densities of $\ce{PhH}$ and $\ce{PhF}$. To my surprise, the result is:
$$\begin{array}c\text{molecule}&\text{ortho}&\text{meta}&\text{para}\\\ce{PhH}&-0.239&-0.239&-0.239\\\ce{PhF}&-0.306&-0.222&-0.261\end{array}$$ By observing this result, the fact that the electrophilic substitution of $\ce{PhF}$ is likely to occur in o- and p- is verified, or the right half of $(1)$ is verified.

But why would the charge density of position 2,4 of $\ce{PhF}$ is more negative than that of benzene, given the fact that $\ce{-F}$ decreases rate of the electrophilic substitution reactions?

My speculation
i) $(1)$ may be false.
ii) The NBO charge densities between two different molecules cannot be compared.
iii) The charge density cannot indicate the difficulty of the attacking of electrophile.

Which one, or something else, can explain this strange observation? Can another method to order electron-withdrawing ability of some functional groups computationally be suggested?


1 Answer 1


enter image description here

This is the way of directing the electrophiles to the ortho and para places of the benzene cycle. As you see the ortho and para places are more negatively charged.That's true. It is due to the ability to donate its electron pair to the benzene cycle.

But an electronegative element can attract the electron cloud to it. Fluorine is the most electonegative element, so it has a greater ability to attract the electron cloud of the benzene cycle. So that makes it relatively resistant to an incoming electrophile.

This does not say that it is not a ortho-para director. It's a ortho-para director but it deactivates the ortho and para places of the benzene cycle.

  • $\begingroup$ I agree with your last sentence. But I still not understand the fact that fluorine atom deactivate the ortho and para places despite that they are more negatively charged. "it has a more ability to attract the electron cloud of the benzene cycle. So that makes relatively resistant for coming for a electrophile." But it does not agree with the charge density calculation result and it contradicts with your first paragraph. $\endgroup$ Sep 22, 2019 at 1:17
  • $\begingroup$ As there is two main factors to decide the charge density of the places,- the ability of donating electrons, and the ability of attract the electron cloud. Diactivating of PhF does not means that the ability of reaching electrophiles has decreased than the PhH. The ability of reaching electrophiles is still more than the PhH but it is slightly low than the other ortho para directors such as fenoxide anion, toluene, etc. For more understand see the highlighted paragraph of pubs.acs.org/doi/10.1021/ed080p679 $\endgroup$ Sep 22, 2019 at 12:42
  • $\begingroup$ Abstractly, This is usually common to helobenzenes, and basing on your question I have to say that the affect of donating electrons in to the ring more affects the attracting of the electron cloud by the atom. So, there is still more net minus charge in the ortho and para places than the benzene. But it is slightly low than the other ortho para directors. (these two factors are known as mesomeric effect and the inducing effect.) $\endgroup$ Sep 22, 2019 at 12:49

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