The following compounds have the same molecular formulas (C4H10). (a) h5bur1220 (b) h5bur122002 Which would you expect to have a higher boiling point?

Option

A) butane

B) 2-methylpropane

The answer is A has a higher boiling point than B.

Explanation:

The options are shown in the figure below.

The stronger the intermolecular force, the higher the boiling point. Usually hydrogen bonding is stronger than dipole-dipole attractions, which are stronger than London dispersion forces.

The boiling point depends in the intermolecular force within the molecule thus the stronger there is the force within the molecules the higher the resultant boiling point.

The order of the type of force defines the variation in force observed. The order of molecular strength shows that hydrogen bonding is relatively stronger to dipole-dipole attractions, and these are stronger than London dispersion forces.

For the organics samples here we see two isomers, butane (the unbranched structure) has higher boiling point than 2-methylpropane (branched structure).

Butane is of the higher boiling point because the intermolecular dispersion forces are greater in it.

Note: That the surface area of the butane molecules are pack together and make contacts. Butane molecules set up bigger temporary dipoles because they are longer and by this can lie closer together than the shorter and ball like fat 2-methylpropane molecules.

As a result, it takes more energy by reason of having a higher boiling point for the butane molecules to change phase from the liquid state for to enter the gaseous state relative to the 2-methylpropane.