1
$\begingroup$

In both chlorobenzene and fluorobenzene, both the halogens show a −I effect where fluorine has more electron withdrawing effect than chlorine.

Now, if we consider the mesomeric/resonance effect, both show +M/+R effect. On comparison of resonance effect and inductive effect, resonance effect dominates over inductive effect. Then:

  1. Why does −I effect dominates over +M/+R effect of chloro and fluoro group in the case of chlorobenzene and fluorobenzene? Give reason.

I know that the −I effect dominates over the +M/+R effect of the chlorine and fluorine substituents in case of chlorobenzene and fluorobenzene. Now, on the basis of this chlorobenzene should be more reactive for electrophilic aromatic substitution reaction than fluorobenzene (as −I effect: $\ce{-F~ >~ -Cl}$). But, it is not true though vice-versa is true.

  1. Why is fluorobenzene more reactive for electrophilic aromatic substitution reaction than chlorobenzene? Give reason.
$\endgroup$
9
$\begingroup$

We need some data to answer your question. As presented in this earlier answer (which is also relevant to this question), here are the relative rates for electrophilic aromatic nitration of the halobenzenes compared to benzene itself.

\begin{array}{|c|c|c|c|} \hline \ce{Ar-X} & \text{Relative Rate} \\ \hline \ce{Ar-H} & 1.0 \\ \hline \ce{Ar-F} & 0.11 \\ \hline \ce{Ar-Cl} & 0.02 \\ \hline \ce{Ar-Br} & 0.06 \\ \hline \ce{Ar-I} & 0.13 \\ \hline \end{array}

You are correct that inductive and resonance effects operate in opposite directions here. All of the halogens are more electronegative than hydrogen so they inductively remove electron density from the aromatic ring and tend to slow the reaction down. Whereas resonance effects would donate electron density into the aromatic ring and speed the reaction up.

However it is not correct to say

On comparison of resonance effect and inductive effect, resonance effect dominates over inductive effect

Here is a series where inductive effects play a larger role than resonance effects. This is why all of the halobenzenes react more slowly than benzene itself.

In fluorobenzene the overlap between carbon 2p and fluorine 2p orbitals is very effective, but the carbon 2p orbital cannot overlap as effectively with higher p orbitals (e.g. chlorine 3p, bromine 4p, etc.). It is the stronger resonance contribution from this effective electron donating resonance interaction with fluorine that makes fluorobenzene more reactive towards electrophilic aromatic substitution than chlorobenzene.

The electrophilic reactivity of the other halobenzenes is also controlled by a blend of resonance and inductive effects. As we move down the series from chlorine to iodine the electronegativity decreases and the strength of the electron withdrawing inductive effect also decreases - this should speed up the rate of reaction. However the strength of the resonance effect also decreases due to less effective orbital overlap between halogen and carbon - this should slow the reaction down. Looking at the data we can see that the reaction rate slowly increases as we move from chlorine to iodine, suggesting that the inductive effect is still the primary effect with resonance, as expected due to poorer overlap, still playing a secondary role.

$\endgroup$
  • $\begingroup$ "It is the stronger resonance contribution from this effective electron donating resonance interaction with fluorine that makes fluorobenzene more reactive" But isn't the electronegativity of fluorine more than that of chlorine? So, shouldn't the "effective electron resonance interaction" be nullified? We already know that all halobenzenes react in EAS at a much lower rate than benzene, so the rings are indeed deactivated. But how can we predict theoretically whether mesomeric effect of F dominates over Cl or whether more EN of F dominates over Cl? $\endgroup$ – Gaurang Tandon Mar 2 '18 at 7:31
  • $\begingroup$ In this answer also, you have said that "we need some data." Apparently there's no way to do this comparison theoretically... $\endgroup$ – Gaurang Tandon Mar 2 '18 at 7:42
0
$\begingroup$

Since in case of halogens, because of their exceptionally high electronegativity, Inductive effect dominates over resonance . In case of flurobenzene though - I effect is the strongest but because of great 2p-2p orbital overlap between F and carbon + r effect decrease the -I effect to a great extent, making it the least deactivating group. In case of idobenzene -I effect though greater than resonance but is not that prominent because of comparirively poor electronegativity and also because of poor 5p-2p overlap with the carbon atom , the group become more deactivating than Flourine. Similary comparing the values of -I and + r effect for chloro and idobenzene the order comes out to be as follows ORDER OF DEACTIVATING GROUPS- Fidobenzene>chlorobenzene>bromobenzene

$\endgroup$
  • 2
    $\begingroup$ Could you revise fidobenzene? $\endgroup$ – A.K. Aug 22 '18 at 15:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.