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 $\endgroup$ – Mockingbird 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
  • $\begingroup$ There are two answers there ^ so it's not OK to ask again just because you're not satisfied. BTW you can use simple rationalisations for aliphatic acids but for aromatic they just don't work properly. $\endgroup$ – Mithoron Feb 6 '18 at 17:58

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.

  • $\begingroup$ Why is there relatively higher -I for chlorine? Isn't F more electronegative? $\endgroup$ – Tan Yong Boon 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

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. $\endgroup$ – Vaibhav Dixit Jun 23 '17 at 8:28

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. $\endgroup$ – Tan Yong Boon Feb 22 '18 at 23:01
  • $\begingroup$ "due to their +I effect. " You definitely meant the -I effect instead... $\endgroup$ – Gaurang Tandon Feb 23 '18 at 3:00

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