I wanted to ask something more in reference to this question.

Structures of 2- and 3-methoxyanilines

Suppose instead of toluidine, 2-methoxyaniline and 3-methoxyaniline are present. Now, why is 2-methoxyaniline more basic than 3-methoxyaniline (even though 2-methoxyaniline has reduction of basicity due to ortho-effect but 3-methoxyaniline does not)?

Also what would be the correct order of basicity if nitro group was present in place of methoxy group? Would ortho-effect still take place?

  • $\begingroup$ Why would ortho effect reduce basicity? Shouldn't it increase? $\endgroup$
    – Papul
    Jun 15 '15 at 19:44

Take home message: Just because you have an ortho substituent does not mean you will have an ortho effect.

To have an ortho effect two conditions must be met:

  • you need an ortho substituent
  • and, importantly, it must be of sufficient steric bulk to disturb the adjacent substituent.

One measure of steric bulk is the cyclohexane $A$-Value, a table of which can be found here. From the table we see that methoxy $(A=0.6)$ and nitro $(A=1.1)$ groups have a smaller steric size than that of a methyl $(A=1.7)$ group. Methoxy and nitro groups generally do not produce an ortho effect.

Therefore, the basicities of the various methoxy anilines (the anisidines) are controlled by a combination of resonance and inductive effects.

\begin{array}{l|r} \hline \text{Substituent} & \mathrm{p}K_\mathrm{b} \\ \hline \ce{-H} & 9.13 \\ \ce{o-methoxy} & 9.5 \\ \ce{m-methoxy} & 9.8 \\ \ce{p-methoxy} & 8.7 \\ \hline \end{array}

(data source)

The basicity of p-methoxy aniline is controlled primarily by resonance effects (too far away for inductive effects to be significant). Drawing resonance structures we see that the p-methoxy group is pushing electron density into the ring, this tends to keep the nitrogen lone pair more localized on nitrogen (if we also delocalize the nitrogen lone pair into the ring like charges repel leading to destabilization). This results in p-methoxy aniline being more basic than aniline.

The basicity of m-methoxy aniline is controlled purely by inductive effects (resonance is not possible from this position). The inductively electron withdrawing methoxy group removes electron density from nitrogen, making it less basic than aniline.

The basicity of o-methoxy aniline is controlled by a mix of resonance and inductive effects operating in opposite directions. Apparently the inductive effect is stronger and wins out (this shouldn't be surprising, note that the inductive effect from the meta position had a 0.7 unit effect on the basicity, and now in the ortho position it will be even stronger, while the resonance effect in the para isomer only changed the basicity by 0.4 units) making o-methoxy aniline less basic than aniline.