# How to rationalise the basicity order of phenylenediamine?

Why is the basicity of para- > meta- > aniline > ortho- for phenylenediamine?
($\mathrm pK_\mathrm a$: ortho: 9.53, meta: 9.12, para: 7.96) I tried deducing it using resonance and inductive effects but got stuck.

• – Mithoron Mar 19 '15 at 20:58
• I am not being able to incorporate the ortho effect concept!Let me know please. – user14857 Mar 20 '15 at 5:35
• Isn't the basicity trend listed backwards? The para-analog has the lowest pKa for the conjugate acid (strongest acid), which means that the base is the weakest? – jerepierre Mar 20 '15 at 23:21
• in phenylene-diamine shouldn't the basicity order be p>aniline>o>m since -NH2 is not much of a bulky group and thus doesn't exhibit ortho effect. SO with ortho effect out of the picture -NH2 is very strong +M group and effect of +M>-I (for -NH2). In ortho due to hydrogen bonding its basicity is reduced. Meta position only -I. so by that reasoning para is most basic followed by aniline followed by ortho and then meta....So p>aniline>o>m....Am i going wrong anywhere? – J_B892 Nov 8 '18 at 5:18

## 2 Answers

As Mithoron pointed out in a comment, we had a similar question convering the cases of 2-methylaniline (o-toluidine) and 2-methylbenzoic acid (o-toluic acid) here.

In the case of 1,2-diaminobenzene (o-phenylenediamine), it is conceivable that hydrogen bonding between the two amino groups reduces the electron density on one of the nitrogen atoms, thereby leading to a slight increase on the other. As a result, o-phenylenediamine is significantly less basic than the m- and p-isomer, but a tad more basic than aniline.

Klaus's answer is nice, but is omits one important detail:

The hydrogen bonding would necessarily require that the lone pair of electrons go 'out of conjugation' with the ring system, and hence there can be no donation of electron density via pi-effects as in the para-phenylenediamine case.