0
$\begingroup$

It is given that due to Steric Inhibition of Resonance (SIR), the acidity of ortho-substituted benzoic acid increases with respect to benzoic acid. However, I'm unable to understand the reason behind this fact.

We know that resonance stabilises a molecule and the absence of which decreases the stability. In case of ortho substituted benzoic acids, resonance is inhibited due to the loss of planarity due to the steric interactions from the ortho substituent.

If this is the case then the conjugate base of the acid will not be stabilized by resonance with the benzene ring. So it seems the ortho-substituted benzoic acid will not readily donate H+ and is less acidic than benzoic acid, contrary to what is being given in various sources.

Then kindly explain why is ortho-substituted benzoic acid is more acidic than benzoic acid.

Improve this question
$\endgroup$
8
  • $\begingroup$ which ortho substituent are you refering to? $\endgroup$
    – Waylander
    Aug 1, 2019 at 12:09
  • $\begingroup$ Think of a localized, in plane, benzoate conjugate base. The carbonyl can busy itself by resonance with the ring or resonate with the localized charge-bearing oxygen. The carbonyl is splitting its effort. But in the out-of-plane conformation (say, from o-toluic acid) the ring resonance is lost and the carbonyl can now focus on stabilizing the charge on oxygen. On this basis, the out-of-plane acid would be predicted to be more acidic.. $\endgroup$
    – user55119
    Aug 1, 2019 at 12:37
  • $\begingroup$ @Waylander Ortho-substituents like alkyl group(methyl,ethyl,etc.), halogens(F,Cl,Br,I), etc., where the SIR effect depends on the size of the substituent. The substituent I referred does not include small ones like hydroxyl or amine groups. $\endgroup$
    – Vishnu
    Aug 1, 2019 at 13:20
  • $\begingroup$ @user55119 Ok I understood your point. But we know resonance is a stabilizing phenomenon and more the number of resonance structures, more must be the stability right. This one confuses me. Kindly guide me. $\endgroup$
    – Vishnu
    Aug 1, 2019 at 13:23
  • 1
    $\begingroup$ In case of ortho substituted benzoic acids, resonance is inhibited due to the loss of planarity due to the steric interactions from the ortho substituent. This is incorrect for most substituents in 2 position. SIR is a very fuzzy subject and it is based on false premises. All steric effects are essentially a consequence of electronic effects; it is therefore a balance between electronic effects to begin with. It is unfortunately more complicated than what some books make it out to be. $\endgroup$
    – Martin - マーチン ♦
    Feb 10, 2020 at 14:08

1 Answer 1

Reset to default
-1
$\begingroup$

Important information before understanding the reason for steric inhibition of resonance and the ortho effect

In general, resonance stabilizes a molecule. But in the case of carboxylic acids, the conjugate base (anion after losing $\ce{H+}$) there are two equivalent resonance structures. Take the example of acetic acid, $\ce{CH3COOH}$, and thus acetate ion, $\ce{CH3COO-}$:

Resonance Structures of Acetate Ion

The hybrid would look something like this:

RH of Acetate Ion

It could be easily seen that equivalent resonance structures mostly stabilise the hybrid greatly. Therefore, equivalent resonance tends to stabilize a molecule/compound more than normal resonance. (in most cases)

Now let's take the case of o-toluic acid.

enter image description here

There are two ways by which resonance can take place:

enter image description here

  • In (I), it takes place through $\ce{-COOH}$. Its ion is expected to be more stable due to the above result. Also, here $\ce{OH}$ is highly polar and it will tend to lose $\ce{H+}$ to become uncharged and thus be more stable (as compared to $\ce{O-}$).
  • In (II), it takes place through the ring. Its ion is less stable due to more formal charge and less aromatic character (due to + on the ring). Also, $\ce{OH}$ is less polar than $\ce{H+}$ due to the above point.

Therefore, it will be better to prevent the taking place of resonance through ring.

That is done by $\ce{-CH3}$ group which provides bulky group repulsion and takes the ring out of the plane of $\ce{-COOH}$ and thus providing condition for equivalent resonance in $\ce{-COOH}$.

Improve this answer
$\endgroup$
7
  • $\begingroup$ If you want more explanation on Ortho-Effect. Visit here:- chemistry.stackexchange.com/questions/7683/… $\endgroup$
    – S K
    Feb 10, 2020 at 6:22
  • $\begingroup$ I've learnt that more the number of resonance structures more stable the compound is. I understood equivalent resonance structures are possible in all circumstances (whether there is SIR or not). But I don't understand why the molecule gets destabilised if we count the resonance between the carboxylic group and the ring. However, I'm glad that you understood my question properly. $\endgroup$
    – Vishnu
    Feb 10, 2020 at 6:40
  • $\begingroup$ Its not destabilised, its just less stabilised. The compound would prefer to be more stable rather than being less stable. $\endgroup$
    – S K
    Feb 10, 2020 at 8:04
  • 2
    $\begingroup$ No problem, unfortunately I have to tell you, that what you write is not correct. The 'equivalent resonance' is a widely propagated myth, it doesn't hold up to any scientific rigor. Resonance in itself is only a result from our inability to grasp the actual bonding description. $\endgroup$
    – Martin - マーチン ♦
    Feb 10, 2020 at 13:45
  • 2
    $\begingroup$ It is very unfortunate that this has not yet been purged from the syllabus and textbooks. It remains a popular concept, even though it is of little to no use. It is a post-hoc rationalisation based on a few cases, which cannot be generalised; it uses reverse argumentation (i.e. it presents the reasons for something as the conclusion of something else). $\endgroup$
    – Martin - マーチン ♦
    Feb 10, 2020 at 13:53

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.