SolutionsPhysical and Bond Properties of Amines
1. Boiling points
Order from lowest to highest boiling point: propane, trimethylamine, propylamine and 1-propanol. Explain the order.
Propane (−42 °C) < trimethylamine (3 °C) < propylamine (48 °C) < 1-propanol (97 °C).
Trimethylamine is polar but has no N–H bonds, so it does not form hydrogen bonds between its molecules. Propylamine does form them, although they are weaker than those of the alcohol because nitrogen is less electronegative than oxygen.
2. Nitrogen inversion
N-Methylethylamine has a nitrogen with four different substituents (counting the lone pair). Why can its enantiomers not be separated?
Because the nitrogen inverts quickly through a planar transition state (sp2 hybridization), with a barrier of only 5-7 kcal/mol. At room temperature the two enantiomers interconvert millions of times per second and the sample is always racemic.
3. Chiral or not?
Which of these compounds can be isolated as a pure enantiomer?
- N-Ethyl-N-methylpropylamine
- Ethylmethylpropylbenzylammonium iodide
- A quinuclidine with a substituent on an asymmetric carbon
- No: the nitrogen inverts quickly.
- Yes: in the quaternary ammonium salt the nitrogen has no lone pair and cannot invert; with four different groups it is a stable stereocentre.
- Yes, because of the asymmetric carbon; moreover, the quinuclidine nitrogen cannot invert because the bicyclic framework prevents it.
4. Solubility
Why are low-molecular-weight amines soluble in water, and why does their smell disappear when hydrochloric acid is added?
They form hydrogen bonds with water (the nitrogen lone pair accepts the hydrogen of water). With HCl they form ammonium salts (RNH3+ Cl−), which are ionic, non-volatile and very soluble, so the amine no longer escapes into the air.