# Para effects on substituents in aromatic compounds

I wanted to know how to compare the following compounds on basis of acidic strength. I have to find the acidic strength of following compounds with $$\mathrm{X}=\ce{CH3, \ OCH3,}$$ and $$\ce{NO2}$$.

For ease of explaining, denote the first compunds as (A), the second as (B), and the third as (C)

Due to steric inhibition of Resonance (SIR) effect, compound (A) will be the most acidic, as its anion will be the least stable.

Now for ordering the other two, I know that there are two things affecting acidic character:

1. Bond strength; the weaker the bond, the higher its acidic strength
2. Charge on hydrogen; the more positive charge their is on the hydrogen, the more its acidic strength.

So now if we take $$\mathrm{X}=\ce{CH3}$$, we should have that $$\mathrm{(A)>(C)>(B)}$$; as $$\ce{CH3}$$ is an electron donating group and it would make the hydrogen less positive, and inductive effect is distance dependent. This answer is incorrect as per my textbook, which says $$\mathrm{(A)>(B)>(C)}$$.

If we take $$\mathrm{X}=\ce{OCH3}$$ which is an electron withdrawing group, we should have $$\mathrm{(A)>(B)>(C)}$$. This answer is correct.

If we take $$\mathrm{X}=\ce{NO2}$$, which is also an electron withdrawing, we should have $$\mathrm{(A)>(B)>(C)}$$. But my textbook says the answer is $$\mathrm{(A)>(C)>(B)}$$.

I wanted to know where am I going wrong, and if there is some other effect I am not taking into account.

Note: my earlier question was closed due to the similar question: Ortho-effect in substituted aromatic acids and bases. But that question only deals with the ortho effects of substituents, not the para.

• Actually, where are your structures (A), (B), and (C)? May 1, 2020 at 4:29

I have "Solomons, Fryhle, Snyder Organic Chemistry" open in front of me, and it seems there are a few difference between the book's order and the order given by you.

Acidic strength order for

• $$\mathrm{X}=\ce{CH3}$$: m-dimethylbenzene > p-dimethylbenzene > o-dimethylbenzene
• $$\mathrm{X}=\ce{OCH3}$$: m-methoxyphenol > o-m-methoxyphenol > p-m-methoxyphenol
• $$\mathrm{X}=\ce{NO2}$$: o-nitrotoluene > p-nitrotoluene > m-nitrotoluene

$$\ce{CH3}$$ shows +I and hyperconjugation, $$\ce{OCH3}$$ shows +M and -I and $$ce{NO2}$$ shows -R and -I effect.

The dominating orders of effects is: mesomeric>hyperconjugation>inductive.

This and the answer by Bitthal Maheshwari should explain the order.

For the halotoluenes but it came out m-halotoluene > o-halotoluene > p-halotoluene for me but I do not have the data so I cannot confirm it. Perhaps someone else could do it.

At meta position the M effect is not considered. The reason behind is that during resonance of a six membered cyclic compound charges never reaches at meta position that why it is so try to proof it by drawing resonance structure and if you didn't get us then tell me in the comments below.

• I still don't get why there would be no resonance, could you please elaborate? May 1, 2020 at 2:40
• There would be resonance but when you draw resonance hybrid(expecting you know how to draw it) of benzene you will observe that the slight(+ or -) charges will be at ortho and para position not at meta position so that is the reason that resonance effect is not observed at meta position. May 1, 2020 at 3:36