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I have to compare the acidic strength of these compounds:

A: 2‐hydroxybenzoic acid; B: 2,6‐dihydroxybenzoic acid; C: 2‐methylbenzoic acid

Now, in A and B hydrogen bonding will be present which will help stabilize the negative charge of the conjugate base. However, the hydroxyl group is not bulky and won't cause steric inhibition of resonance, hence the +M effect of phenyl will still be present.

In C, methyl will cause steric inhibition of resonance and hence stop/reduce the +M effect of phenyl which will stabilize the conjugate base.

How do I compare the relative stabilization offered by these two effects (H-bonding and SIR)?

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  • $\begingroup$ I think the main effect in play here is straight electron withdrawing potential of the oxygens. In my opinion this is a classical BSc question that tries to avoid the grey zones of pka eye balling. I would invite you to review the concept of steric inhibition of resonance since I'm not entirely sure I understand your argument there. Here's a good starting point if you do not have a text book covering the subject quora.com/What-is-steric-inhibition-of-resonance $\endgroup$
    – user46680
    Commented Jun 22, 2017 at 11:06
  • $\begingroup$ chemistry.stackexchange.com/questions/7683/… $\endgroup$
    – Mithoron
    Commented Jun 22, 2017 at 16:30
  • $\begingroup$ chemistry.stackexchange.com/questions/20190/… $\endgroup$
    – Mithoron
    Commented Jun 22, 2017 at 16:35
  • $\begingroup$ I think the answer is C>B>A. In C due to SIR effect the conjugate base is more stabilised whereas in A and B no SIR takes place as OH- isn't bulky. But in B there are 2 OH- present for hydrogen bonding compared to A so conjugate base of B is more stable than A $\endgroup$
    – user130760
    Commented Jan 25, 2023 at 10:40

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