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Rank the following compounds in decreasing order of acidity:

  • 2-Fluorobenzoic acid
  • 2-Chlorobenzoic acid
  • 2-Bromobenzoic acid
  • 2-Iodobenzoic acid

My approach

Since the ortho effect is proportional to the size of the substituent, it will be the greatest in the case of iodine and smallest in the case of fluorine. So, the molecule will be more non-planar in 2-iodobenzoic acid.

So resonance will be inhibited in all the above compounds. But what will happen to the acidity?

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Acidity data:

The true order of acidity is $\ce{Br = I > Cl >> F}$. Relative to unsubstituted benzoic acid, all of the ortho-halogen substituted benzoic acid derivatives have a greater acidity, clearly having some stabilising effect on the resulting benzoate anion.

enter image description here

Data retrieved from Journal of Research of the National Bureau of Standards, 1958, 60, p569 - research paper 271. $\mathrm{p}K_\mathrm{a}$ values measured in water.

Rationalisation:

  • Polar effects All of the halogens exert a negative inductive (-I) effect, removing electron density from the ring. This makes the benzoic acid derivatives more acidic in all 4 cases since the resulting conjugate base is stabilised to a certain extent by the presence of the electron withdrawing halogen.

  • Steric effects As the halogen gets bigger, strain may force the carboxylate out of planarity with the aromatic ring. Whilst fluorine is roughly the same size as the hydrogen atom, bromine and iodine are both sterically more demanding.

  • Internal H-bonding This is suggested by another answer below, whilst this has been studied and shown to operate for protic groups such as alcohols and amines, in halogens it has little impact on stabilising the negative charge.

Overall, steric and polar effects operate in tandem to give an overall acidity for the benzoic acid. It's perhaps not intuitive which one will dominate (e.g. fluorine is far more electronegative, but also far smaller than iodine, so which is more acidic).

The table below, taken from the cited paper, attempts to separate out steric and polar effects through calculations on both the acid and its conjugate base. Sadly, they only look into $\ce{F}$ and $\ce{Cl}$ (no $\ce{Br}$ or $\ce{I}$), but it can be clearly seen that:

  1. Hydrogen bonding is making no significant contribution to the acidity of the substituted benzoic acid
  2. Polar and steric effects are both in operation, with sterics being far more important in the ground state (the carboxylic acid) and polar effects dominating for the conjugate base (the carboxylate).

    enter image description here

    Table copied from New. J. Chem., 2004, 28, p67

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At least 3 (or 4) facts are required to be considered for the acidic strengths of halogen acids:

  1. The resonance stabilization of carboxylate anion is poor in benzoic acid due to cross conjugation, wherein the pi system of phenyl ring also 'competes for' the same $\ce{C=O}$, where negative charge of one of the $\ce{O-}$ would have been shifted enter image description here

  2. Steric effects of any group, irrespective of its +M/-M effect, will allow $\ce{-COOH}$ to go perpendicular to the plane, destroying the cross conjugation; which give a more stable carboxylate, which means more acidic substituted $\ce{Ph-COOH}$ (This is what they say is, ortho effect)

enter image description here

  1. Halogens have dominant negative inductive effect over their positive mesomeric effect.

  2. H-bonding would have played some role in stabilizing carboxylate ion in halogen acid.

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  • $\begingroup$ Could you please cite the source from where you got these images, or are these self-made? Thanks! $\endgroup$ – Gaurang Tandon Apr 20 '18 at 16:29
  • $\begingroup$ Gaurang, this is my work only $\endgroup$ – Che Mistry Jan 20 at 5:56
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The intramolecular hydrogen bonding between the halogen substituent and the adjacent carboxyl group (as shown below) would decrease the acidity of the halobenzoic acids:

$\hspace{60 mm}$H bonding

Since the strength of the hydrogen bonding decreases in the order $\ce{F}>\ce{Cl}>\ce{Br}>\ce{I}$, we can arrive at the order of acidity $4 > 3 > 2 > 1$.

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