[shon1802]

Hi, the question goes like that:
How would you separate a mixture of benzoic acid, p-nitroaniline, azobenzene, and ethylbenzene, using acid base extraction.
My attempt: well, since all the compounds are aromatic ones, i would begin by dissolving the mixture in some non-polar organic solvent, like benzene (it should be non-polar so that the salts will go to the aqueous phase). Then, add some NaOH, which will react with the benzoic acid to give sodium benzoate that will go to the aqueous phase. Separate the aqueous phase. Then add some HCl, which will protonate the nitroaniline, and it will be transferred to the aqueous phase. Again, separate the aqueous phase. And here i got stuck. I am left with azobenzene and ethylbenzene. As far as i think, the first is basic because of the non-bonding electrons on the nitrogens. So if it is basic, it would be protonated just like nitroaniline when acid is added, and i have no idea how to separate those two. On the other hand, if it's neither acidic nor basic, how can i separate it from ethylbenzene? Could it have something to do with H₂SO₄?
Azobenzene:
benzoic acid:
p-nitroaniline:
ethylbenzene:

Thanks

[AWK]

Did you check boiling point of ethylbenzene and azobenzene?

[kamiyu]

My suggestions are:

Step 1: remove benzoic acid (as shon1802 said)

Step 2: Add HCl to remove ethylbenzene (because both azobenzene and aniline are basic)

Step 3: buffer the resulting aqueous phase (which contains the salt of azobenzene and aniline) such that neutral aniline is generated while azobenzene is still charged (you can check their pKa to know which one is more basic)

BTW, the nitro group is wrongly drawn

[AWK]

Step 2: Add HCl to remove ethylbenzene (because both azobenzene and aniline are basic)

Do you know a real pKa or pKb of azobenzene (forget wikipedia)?

[kamiyu]

Sorry, actually I do not know.

I think one of them will be more basic than the other anyway, thus controlling the pH of the aqueous solution should work, I guess

[AWK]

pKa of  "trans-azobenzene (-2.6 ± 0.2)" (but not 3.3 as in wikipedia)
G. E. LEWIS. J. Org. Chem. , 1960, 25 (12), pp 2193–2195
If this value is true, what does it mean for your suggestion?

[kamiyu]

Hi AWK, Thanks for your input

I personally think the exact pKa values of aniline and azobenzene are not very important in this case as long as their pKa are different.

Of course, I think distillation also works, but I think we have to suggest within the scope of the question (acid-base extraction)

[AWK]

What do you know about The levelling effect?

[shon1802]

Did you check boiling point of ethylbenzene and azobenzene?

Yes, they are quite different, but it's irrelevant to the question since i have to use acid-base extraction.

What do you know about The levelling effect?

Not much...

Well, I asked my Professor, and the answer is the following: there was a mistake in this question, since both azobenzene and aniline are basic. But, for the purpose of the question, azobenzene is not basic. Therefore, first, we add NaOH and remove benzoic acid. Then, add HCl to remove nitroaniline. We are remained with azobenzene and ethylbenzene, and to separate them add cold and concentrated H₂SO₄. The sulfuric acid reacts with the azobenzene (because of the N's), but not with the ethylbenzene.
I still remain with a question: how to remove the sulfuric acid from the water, since they are miscible?

[AWK]

Levelling effect causes that for pKa less than -1.74 sulfuric acid in water solution will not protonate azobenzene (pka -2.6) . Protonation is possible in anhydrous conditions (anhydrous sulfuric acid and anhydrous mixture azobenzene-ethylbenzene). May be azobenzene hydrogen sulfate is a solid in ethylbenzene. Addition of water to solid decompose protonation (since levelling effect) and azobenzene precipitates.

how to remove the sulfuric acid from the water, since they are miscible?

Why for?
You should separate benzoic acid, p-nitroaniline, azobenzene and ethylbenzene.