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How would you estimate the $\mathrm pK_\mathrm a$ values of the nitrogen atoms in 1,4-dimethyl-1,2,3,4-tetrahydroquinoxaline?

1,4-dimethyl-1,2,3,4-tetrahydroquinoxaline

I struggle with it because I am unsure whether to classify them as tertiary amines (with a $\mathrm pK_\mathrm a$ of around 9) or whether they can be considered as aromatic amine (similar to aniline), which would result in a lower $\mathrm pK_\mathrm a$ value (around 5ish?) due to the delocalisation of the aromatic ring.

Can someone explain to me which is the correct idea? I really struggle with the concept.

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    $\begingroup$ They don't have a hydrogen attached, so they don't have a p$K_\mathrm{A}$. However, you can discuss the acidity of conjugate acid, the ammonium $\ce{NR3H+}$. $\endgroup$ Jun 16 at 22:00
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    $\begingroup$ @KarstenTheis That's what OP's discussing, but I guess it technically could act as a C-acid (now that would be more interesting!). $\endgroup$
    – Mithoron
    Jun 16 at 22:06
  • $\begingroup$ See chemistry.stackexchange.com/a/162479 $\endgroup$ Jun 16 at 22:16
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    $\begingroup$ Thanks for the link, but its not really what I am looking for. @Mithoron is right! I also meant to be discussing the effect of the benzene ring on it. And whether you would still consider this tertiary amine to be aliphatic or aromatic (in terms of pka) $\endgroup$
    – Laura
    Jun 16 at 22:29
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    $\begingroup$ @tess are you are interested in the ability of your molecule to donate protons from the C atoms (what Mithoron described), or to accept protons onto the N atoms? The link Karsten Theis shared is helpful in understanding the ambiguity of "pKa". $\endgroup$ Jun 17 at 0:12

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The nitrogen atoms are attached to the aromatic ring, and this placement is the dominant factor. The conjugate acid formed by protonating either nitrogen atom would be expected to have a $\mathrm pK_\mathrm a$ similar to that of protonated aniline, not a protonated aliphatic amine.

We might expect this constant to be modified by the second nitrogen atom. Let us explore this.

The second nitrogen atom actually has two conflicting influences:

  • Electron-withdrawing inductive effect, which would decrease the conjugate acid $\mathrm pK_\mathrm a$ (stronger conjugate acid, weaker base). We expect this to be strongest in the ortho position as in the structure given in the original problem.

  • Electron-releasing mesomeric effect, which would increase the conjugate-acid $\mathrm pK_\mathrm a$ (weaker conjugate acid, stronger base). As with similar effects acting through a benzenoid ring, this would be stronger in the ortho and para positions, weaker in the meta position.

Wikipedia gives the following conjugate-acid $\mathrm pK_\mathrm a$ values:

Amine Structure $\mathrm pK_\mathrm a$
Aniline aniline 4.63
o-Phenylenediamine o-phenylenediamine 4.57
m-Phenylenediamine m-phenylenediamine 5.11
p-Phenylenediamine p-phenylenediamine 6.31

Compared with the aniline base case, the ortho-diamine (matching the arrangement of nitrogen atoms in the original problem) shows very nearly the same acid-base balance, as the two effects described above appear to balance out. The meta arrangement gives a slightly stronger base, but it is the para isomer that shows a clearly stronger basic behavior: the mesomeric electron donation is relatively strong with this arrangement while the induction effect is minimal.

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