Question

Arrange the following in the decreasing order of their boiling points.
(I) n-Butane
(II) 2-Methylbutane
(III) n-Pentane
(IV) 2,2-Dimethylpropane

Answer 1

Hint: Revise the boiling point concept. As the branching in the molecule increases, vapour pressure of the compound increases.

Complete step by step answer:
-Boiling point is the temperature at the vapour pressure of the compound is equal to the atmospheric pressure that is, 1atm.
-It can also be defined as the temperature at which a compound in its liquid phase gets transformed into a gaseous phase. More precisely, the temperature at which there is equilibrium between liquid phase and gaseous phase of a compound is known as its boiling point.
-At boiling point, the intermolecular forces between molecules are broken and molecules are isolated and loosely packed so they get converted into gaseous phase. The energy required to break the intermolecular forces is provided by increasing the temperature.
-Boiling point increases with increase in molecular weight because more energy is required to break the intermolecular forces of interaction such as Vander Waals forces, dipole-dipole interaction, hydrogen bonding, etc. present in the compound.
-Boiling point is more for compounds having straight chains in case of isomers. As the branching in the molecules increases, the intermolecular forces between the molecules decrease due to steric effect and hence, less energy is required to break those interactions and boiling point decreases.
- In case of branched chains, vapour pressure is high due to less intermolecular forces present in the compound and hence, boiling point is less.

-Now, let’s have a look at compounds given in the question, n-butane, 2-methylbutane, n-pentane and 2,2-dimethylpropane. Let’s start by counting the number of carbon atoms in each molecule.
-In n-butane, there are 4 carbon atoms. In 2-methylbutane, n-pentane and 2,2-dimethylpropane, there are 5 carbon atoms in each compound. So, now we know, a smaller number of carbon atoms means less molecular weight. So, n-butane has the lowest boiling point.

-2-methylbutane, n-pentane and 2,2-methylpropane are isomers of each other. Among them n-pentane has a straight chain, so it will have the highest boiling point.

-In 2-methylbutane, there is one branched methyl group whereas in case of 2,2-dimethylpropane there are two branched methyl groups. This indicates, 2-methylbutane will have a higher boiling point than 2,2-dimethylpropane.
-Therefore, the decreasing order of boiling points is given as,
n-Pentane > 2-methylbutane > 2,2-dimethylpropane > n-butane

Note:Keep in mind the relation between vapour pressure and boiling point. As the vapour pressure of compounds increases, boiling point decreases. Also, as the number of carbon atoms increases, boiling point increases.