Question

The compounds n-butane, ch3(ch2)2ch3, and trimethylamine, n(ch3)3, have very similar molecular weights. However, their melting points are appreciably different. Select the compound with the lowest melting point, and the correct reason why.

Answer 1

The lowest boiling point is of n-Butane because it only experiences London Dispersion Forces between molecules.

Step-by-step explanation:

Lets take start with the melting point of both compounds.

n-Butane = - 140 °C

Trimethylamine = - 117 °C

Intermolecular Forces in n-Butane:

As we know n-Butane is made up of Carbon and Hydrogen atoms only bonded via single covalent bonds. The electronegativity difference between C and C atoms is zero while, that between C and H atoms is 0.35 which is less than 0.4. Hence, the bonds in n-Butane are purely non polar in nature. Therefore, only London Dispersion Forces are found in n-Butane which are considered as the weakest intermolecular interactions.

Intermolecular Forces in Trimethylamine:

Trimethylamine (a tertiary amine) is made up of Nitrogen, Carbon and Hydrogen atoms bonded via single covalent bonds. The electronegativity difference between N and C atoms is 0.49 which is greater than 0.4. Hence, the C-N bond is polar in nature. Therefore, Dipole-Dipole interactions will be formed along with London Dispersion Forces which are stronger than Dispersion Forces. Therefore, due to Dipole-Dipole interactions Trimethylamine will have greater melting point than n-Butane.