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Could someone please explain the order of acidity of ortho, para and meta substituted benzoic acids with the substituent as methyl group? Methyl group operates +I and +R effect, I believe.

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  • $\begingroup$ Related: Why is o-toluic acid (2-methylbenzoic acid) more acidic than benzoic acid? $\endgroup$ – Loong Apr 16 '16 at 10:54
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The ortho substituted benzoic acid is most acidic, followed by the meta substituted and finally para substituted benzoic acid.

The methyl group, when at ortho position, causes steric inhibition of resonance, which essentially means that the $\ce{-COOH}$ group goes out of the plane of the benzene ring (due to steric hindrance from the $\ce{-CH3}$ group). Due to this, the benzene's $\pi$ electrons can no longer be transferred to the central carbon of the $\ce{-COOH}$ group by resonance, hence increasing the positive charge on it. This makes the hydrogen of the $\ce{-COOH}$ group more acidic.

When the methyl group is on meta position, it operates by +I, thus reducing its acidity. When on para position, it operates by both +I and +R (actually +H, which means hyperconjugation effect), hence decreasing the acidic strength even more.

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