Question

4-nitroaniline + Acid?

Answer 1

What would result if an acid, say, HCl were added to 4-nitroaniline?

Answer 2

is the molecule

Answer 3

Umm, I'm not quite sure what either of those words mean. We're trying to extract it from other components, including 2-naphthol, benzoic acid, and naphthalene

Answer 4

I'm going to use 2 bases of different strengths to separate the 2-naphthol and benzoic acid, and be left with the naphthalene at the end

Answer 5

But I'm not entirely sure about when I should do the extraction of this thing, nor what happens when Hcl is added (its in the required supplied, and I'm not using it elsewhere in the experiment, so I'm assuming its used here)

Answer 6

What? Lol.
Addition of HCl to 4-nitrotoluene??? (by the way, ortho, meta, para, is just the positions of di-substituted benzene rings)
Is it just HCl in something else?

Answer 7

Sorry if I'm describing this poorly. I don't entirely get the experiment, so am likely misrepresenting some things.

We're using a separatory funnel, and adding various compounds to a mixture to separate the various parts of it.

Answer 8

For example, the mixture contains both benzoic acid and 2napthol. So, I'm going to first add sodium bicarbonate to protonate the benzoic acid but not the 2naphtol, and then separate it. After that, NaOH will be used to remove hte 2naphthol from the (neutral?) naphthalene.

Answer 9

Now, what would I do if 4-nitroaniline were also in that mixture of naphthalene, 2-naphthol, and benzoic acid? HCl is certainly involved.

Answer 10

As for position of the di-substituted benzene rings, its just like it is in that structure I drew. that was provided.

Answer 11

I understand what you're doing, but you're kind of jumbling it all together... First of all, the sodium bicarbonate would not be the one protinating the benzoic acid. the benzoic acid would already BE protonated.

Answer 12

Ah. I think that was a typo, cause the sodium bicarbonate deprotonates the benzoic acid right?

Answer 13

I see

Answer 14

You would form water. The reason for this is that you want to make it more soluble so that the benzoic acid could easily be separated.

Answer 15

Idk if you already did the experiment or if you're doing it right now, but you should see two layers form.

Answer 16

Yeah. There was a demo and that's what's supposed to happen. How do I make that happen with the 4-nitroaniline?

Answer 17

(by the way, thanks a ton for the help/ being so patient with me)

Answer 18

wait wait a minute, give me a moment to read everything you wrote.

Answer 19

The question is "Consider how you would extract 4-nitroaniline if it were included in the mixture to be extracted in this experiment"

Answer 20

THEN, you would use the NaOH to reprotonate the banzoate ion.

Answer 21

Alright

Answer 22

sorry, not NaOH, the HCL!! Sorry sorry.

Answer 23

Oh. The banzoate ion is the 4-nitroaniline?

Answer 24

What solvent are you using for this separation?

Answer 25

Umm let me check

Answer 26

Diethyl Ether I believe

Answer 27

benzoate ion:

Answer 28

Alright

Answer 29

Yeah, I'm pretty sure its Diethyl Ether being used.

Answer 30

So the 4-nitroaniline extraction would be after the 2 acids?

Answer 31

If the 4-nitroaniline was in the mixture containing benzoic acid and 2-napthanol???

Answer 32

I believe so? That's what the questions asking.

Answer 33

I'm guessing its not strong enough to deprotonate the others, so extraction of it would be possible.

Answer 34

We're not actually using it in the experiment, this is a "what if" type thing to make sure we have the concepts down (Which as you can probably tell, I don't.)

Answer 35

Ohhh. I see. Well, what you're basically doing here is taking two neutral compounds, and dissolving them in your solvent, which you said was diethyl ethere. wHAT they are trying 2 get u to understand here is principle that that many neutral organic compounds are not soluble in water but are soluble in organic solvents!

Answer 36

I see. So adding the Hcl will make it soluble in water?

Answer 37

by making the benzoic acid into an \(\sf \color{red}{ion}\) which is the structure i drew above...

Answer 38

Oh okay, I see. I think I understand now.

Answer 39

u make it highly polar.

Answer 40

So we're doing the opposite with this then, by protonating it were making this polar by giving it a positively charged end?

Answer 41

NOW that it is polar, it can react with other polar substances, like water...

Answer 42

this is how i was saying above, you should see the two layers. they seperate.

Answer 43

Yeah. I understand. That actually makes a lot of sense.

Answer 44

So does it really matter when the Hcl is added if its only going to be protonating the 4-nitroaniline?

Answer 45

yes. you add the HCl after you separated them both.

Answer 46

Alright. That makes a lot of sense. Thanks a ton for your help.

Answer 47

I know it took a while, I really appreciate the effort.

Answer 48

because when u separated them, you have it's conjugate bsse, which is the benzoate ion (molecule i drew above)