[johnj7]

Hello, I'm struggling with an extraction concept:

There is a mixture of aniline and naphthalene in ether.
Aniline = 4g/100mL of water; miscible with ether
Naphthalene = insoluble in water; 30 g/100 mL in ether
Extract with aq. NaHCO3

Why would you extract with aq. NaHCO3?
What exactly would happen?

This is a part of my lecture guide, and in my notes I wrote that aniline became protonated.  Is this proton from NaHCO3 or from an acid?

Would adding an acid do anything?

Heres my guess as to whats going on:
so we add an acid such as HCL, which will protonate aniline and create a salt, which is soluble in NaHCO3.  Then aniline can be extracted in that salt water, leaving napthalene in the ether.  Then a base, like NaOH (would NaOH be too strong of a base?)  can be added to obtain aniline. 

Is this correct?  Any help would be much appreciated.

[camp]

You pretty much got it.  Protonate the amide and it becomes a salt.  Nothing happens to the naphalene.  They separate into a polar and organic layer which then can be separated with a sep. funnel.  If you had a carboxilic acid you could put a base in and pull a hydrogen off it and get the same result.  Neutralize and evaporate

[johnj7]

hmm gotcha, thanks

so the only purpose of the NaHCO3 is to create a polar aqueous layer in which the aniline salt can be extracted into?  Why wouldn't you just use water, for instance?

 

[azmanam]

***SAFETY WARNING***

Mix sodium bicarbonate (NaHCO₃) and acid (especially HCl) SLOWLY!  Otherwise you'll get a science-fair worthy volcano - potentially in your face.  Wear safety goggles.  (What do you suppose the products are of mixing sodium bicarbonate with a strong acid?)

***SAFETY WARNING***

--------

This statement is not technically correct:

we add an acid such as HCL, which will protonate aniline and create a salt, which is soluble in NaHCO3.

Let's look more closely at the various acid/base equilibria in this problem.  We have several potential acids:

Compound		pKa
HCl		~ -7
aniline-H+		~ 4
(conj. acid of aniline)
NaHCO3		~ 10.5
aniline		~ 30
naphthalene		~ 40

Remebering that acid/base equilibra favor the weaker acid (the compound with the highest pK_a, do you predict the following equilibra favor the right side of the arrow, or the left side of the arrow?

HCl  +  aniline  --><--  aniline-H⁺  +  Cl^-
NaHCO₃  +  aniline  --><--  aniline-H⁺  +  NaCO₃^-
aniline-H⁺  +  NaHCO₃  --><--  aniline  +  H₂CO₃

With those answers in mind:


--What happens when you add aqueous HCl to a mixture of naphthalene/aniline in ether? 
--Do you think your HCl/ether naphthalene/aniline mixture will be one phase or biphasic? 
--What happens if you add NaHCO₃  to that resulting mixture of HCl/ether naphthalene/aniline, as your "will be soluble in NaHCO₃" statement implies? 
--What happens after you separate your aniline from the naphthalene  (is the aniline in the water layer or the ether layer?  Is the aniline protonated or not?) 
--And what happens when you add NaHCO₃ to that separated aniline?  What if you added HCl to that separated aniline?



Hopefully these questions will help you understand exactly what is going on in your acid/base extraction.  Enjoy the smell of mothballs

[johnj7]

ahh okay, i think i understand now

HCl  +  aniline  --><--  aniline-H+  +  Cl-
NaHCO3  +  aniline  --><--  aniline-H+  +  NaCO3-
aniline-H+  +  NaHCO3  --><--  aniline  +  H2CO3

the second equilibrium stays to the left, as NaHCO3 is a weak acid.
We are concerned with only the first and last equilibrium equations.
So dilute aq HCl is added, protonating aniline. 
- The HCl/ether mixture will be biphasic.

I'm still confused as to what the NaHCO3 would do,
is this correct?
- extract aq protonated aniline layer,
- add NaHCO3 to depronate and get aniline back
- add fresh ether to get aniline in organic layer
- evaporate ether

[azmanam]

Well done.

Just make sure you add that bicarb slowly.  How do you suppose you'll know when you've added enough bicarb?