# Solubility of Amines

1. Jul 26, 2007

### Soaring Crane

1. The problem statement, all variables and given/known data

See below.

2. Relevant equations

See below.

3. The attempt at a solution

For 2-aminobutane, four solubility tests were done with the following solvents:

NaOH
NaHCO3
HCl
H2O

It was soluble in all of them. Is this correct? Is it soluble in H2O and NaOH because of its low molecular weight?

Are these the correct equations for aminobutane’s results?

CH3CH2CH(NH2)CH3 + HCl ->
_________H
_________|
______H-N-H
________|
CH3CH2CHCH3 + Cl-

where N is +.

For NaOH,

CH3CH2CH(NH2)CH3 + NaOH -> ??

I’m unsure at exactly how to express the solubility in an equation.

In H2O,

CH3CH2CH(NH2)CH3 + H2O ->
_________H
_________|
______H-N-H
_________| (+ for N)
CH3CH2CHCH3 + OH- ??

In NaHCO3,

CH3CH2CH(NH2)CH3 + NaHCO3 -> ???

Again, I’m confused at expressing the experienced solubility.

Any help is appreciated.

Thanks.

2. Jul 27, 2007

### chemisttree

Rethink this. It is not true that low molecular weight compounds will be soluble in water due to low molecular weight. It is more related to polar/nonpolar.

Here you are asked if the amine will exist either as a free base or as an ammonium compound.

$$CH_3-CH(NH_2)-CH_2-CH_3$$ vs. $$CH_3-CH(N^+H_3)-CH_2-CH_3$$

3. Jul 29, 2007

### Soaring Crane

Yes, but what does 2-aminobutane exist as in NaOH and NaHCO3? I found it to be soluble in both, and I don't think it exists as the ammonium compound. (If 2-aminobutane does exist as an ammonium compound, then how does this come about?) How is it soluble in these bases if it exists as a free base?

Last edited: Jul 29, 2007
4. Jul 30, 2007

### chemisttree

Apparently sec.aminobutane is soluble in water. In water the compound is only sparingly protonated and so its solubility must be due strictly to its polar nature and not the fact that it will be protonated or not.

the equilibrium expression for the protonation would be

$$CH_3CH_2CH(NH_2)CH_3 + H_2O \rightleftharpoons CH_3CH_2CH(N^+H_3)CH_3 + OH^-$$

the Ka would be products/reactants and pKa would be -log[products/reactants]

5. Jul 31, 2007

### Soaring Crane

Yes, but why does 2-aminobutane appear to be soluble in NaOH and NaHCO3? Can the equilibrium expression above be applied in its solubility in NaOH?

6. Jul 31, 2007

### chemisttree

The solubility must be solely due to the polar nature of the free base (as opposed to what I wrote in my first post in this thread). Adding NaOH or NaHCO3 will not change that. Usually, NaOH is added to amines to convert them to their free base and often this is enough to render them insoluble in water. It doesn't seem to be enough to do that in this case. It is unusual IMO.