# How does the position of an electron donating group affect acidity in benzoic acid?

I know that the closer the electron withdrawing group to the $\ce{-COOH}$, the stronger is the acid. I am interested as to how this relationship works out in the case of an electron donating group.

• This is the exact goal of Hammett plots and analysis of linear free energy relationships (LFERs). – Zhe May 12 '17 at 13:30
• @Zhe, judging by the question I guess your attempt here is to further confuse the person asking the question.. :P – Red Floyd May 12 '17 at 13:34
• @AlphaRomeo Yeah, that wasn't super helpful... Let's actually read the question and not just the title and try again: Electron donation groups increase the negative charge density on the conjugate base, which destabilizes the conjugate base. The less stable the conjugate base, the less acidic the corresponding acid. The effect is exactly opposite with electron withdrawing groups. The closer the substituent is on the ring, the stronger the inductive effects. Just be careful if you're working with substituents that can participate heavily via resonance. – Zhe May 12 '17 at 14:15
• Keep in mind, resonance effects do not matter much, since the $\ce{-COOH}$ isn't in conjugation with the ring. – Pritt Balagopal May 12 '17 at 15:26
• chemistry.stackexchange.com/questions/12309/inductive-effect – Mithoron May 12 '17 at 22:20