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I want to compare the relative acidity of p-chlorobenzoic acid and p-flurobenzoic acid. Both of the facts below point to p-chlorobenzoic acid being more acidic:

  1. The carboxylic acid group is too far away from the halogen for inductive effect to have any significance. The +M effect of fluorine is greater than chlorine (due to more effective orbital overlap). +M effect is not distance dependant and hence p-fluoroobenzoic acid should be less acidic since the negative charge on its conjugate base is destabilized.

  2. There is resonance of the negative charge of the carboxylate ion with the empty d-orbitals of chlorine which would stabilize it.

However, in reality, p-chlorobenzoic acid is less acidic than p-fluorobenzoic acid . Why?


  • $\mathrm{p}K_\mathrm{a}$ of para chlorobenzoic acid $\ce{-> 4.03}$.Source

  • $\mathrm{p}K_\mathrm{a}$ of para fluorobenzoic acid $\ce{-> 4.14}$.Source

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    $\begingroup$ What are the pKa values? $\endgroup$
    – orthocresol
    Jun 22 '17 at 8:27
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    $\begingroup$ @orthocrestol I don't know. This was given to me as a problem, and in the answer key it's given that the latter is more acidic. $\endgroup$
    – xasthor
    Jun 22 '17 at 8:37
  • $\begingroup$ @Xashthor You're forgetting that F is more electronegative than Cl $\endgroup$ Jun 22 '17 at 10:48
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    $\begingroup$ +mockingbird the carboxylic acid group is too far away from the halogen for inductive effect to have any significance. $\endgroup$
    – xasthor
    Jun 22 '17 at 11:50
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    $\begingroup$ @orthocresol $\mathrm{p}K_\pu{a}$ values for reference — p-chlorobenzoic acid: 4.03 p-fluorobenzoic acid: 4.14. If the values are correct, then p-chlorobenzoic acid is more acidic than p-fluorobenzoic acid. $\endgroup$ Feb 24 '18 at 13:06
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For acidity of para substituted benzoic acids, -I effect is helpful and +R effect decreases the acidity.

In 4-fluorobenzoic acid, the +R effect is prominent in $\ce{F}$ due to stronger $\ce{2p}\pi - \ce{2p}\pi$ overlap with nearest carbon. Thus, due to stronger +R effect of $\ce{F}$ than -I effect, it creates a negative charge on your starred carbon. This charge delocalises in the carbonyl carbon and the electrophilicity is thus decreased of that carbon. Releasing hydrogen also doesn't help much as the negative charge created over $\ce{O}$ of $\ce{-OH}$ group can only delocalise a little.

Whereas in the $\ce{Cl}$, there is much weaker $\ce{3p}\pi-\ce{2p}\pi$ overlap. So, +R effect of $\ce{Cl}$ is much weaker and due to a relatively higher -I than its +R, there is more positive character on the starred carbon due to $\ce{Cl}$.

And, thus, in reality, acidity of 4-chlorobenzoic acid is a little higher than that of 4-fluorobenzoic acid.

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  • $\begingroup$ Why is there relatively higher -I for chlorine? Isn't F more electronegative? $\endgroup$ Feb 22 '18 at 23:17
  • $\begingroup$ I am saying that in $\ce {Cl} $, $-I $ effect is relatively stronger than $+R $ effect. I am not comparing $\ce{F } $ with $\ce{Cl } $ in -I effect. $\endgroup$
    – Soumik Das
    Feb 23 '18 at 1:40
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In the case of halogens(on ortho/para), the -I effect dominates over the +R effect and hence due to this the negative charge of carboxylate anion is better stabilized by the flourine substituted compound then the chlorine substituted compound. Therefore, the p- flouro benzoic acid is more acidic than the p- chloro benzoic acid.

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    $\begingroup$ Read the entire question. I've addressed this. $\endgroup$
    – xasthor
    Jun 22 '17 at 12:47
  • $\begingroup$ You rather ignored the inductive effect. $\endgroup$ Jun 23 '17 at 8:28
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In most cases, when halogens are attached to aromatic systems, the Inductive effect dominates over the Resonance effect. Also, due to resonance, the carbon at para-position acquires a partial positive charge which the halogens destabilise due to their -I effect. So, the resonance of the C=O bond with benzene weakens and the negative charge delocalisation becomes better. Hence, the correct order should be

p-flourobenzoic acid > p-chlorobenzoic acid > benzoic acid.

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    $\begingroup$ the carboxylic acid group is too far away from the halogen for inductive effect to have any significance $\endgroup$
    – xasthor
    Jul 23 '17 at 16:24
  • $\begingroup$ @xasthor the carboxylic group does not have to be near to the halogen. the +ve charge comes to the para position by resonance. $\endgroup$
    – Ayushmaan
    Jul 23 '17 at 16:25
  • $\begingroup$ Okay, that makes sense. I'll accept this answer once a few more people confirm this(via upvotes) $\endgroup$
    – xasthor
    Jul 23 '17 at 16:30
  • $\begingroup$ I think you meant "-I effect", that means electron-withdrawing by inductive effect. You wrote "+I effect" which refers to electron donation. $\endgroup$ Feb 22 '18 at 23:01
  • $\begingroup$ "due to their +I effect. " You definitely meant the -I effect instead... $\endgroup$ Feb 23 '18 at 3:00

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