Question

Why is the boiling point of ethyl bromide higher than that of ethyl chloride?

Answer 1

Polarisation in a molecule can be explained as the creation of a dipole within a molecule. This dipole is formed due to the difference in electronegativity of the constituting atoms. The more electronegative atom tends to pull the bonding electron cloud towards itself to create a shift in charge within the molecule. This shift in charge is responsible for the formation of the dipole.

Complete Step-by-Step Answer:
Before we move forward with the solution of the given question, let us first understand some important basic concepts.
As the size of the atom increases, the degree of Van der Waals forces also increases. The size of the bromine atom is greater than that of chlorine. Hence, it exerts greater Van der Waals forces than chlorine when bonded with an ethyl molecule. This makes the bonding in the case bromine stronger as compared to chlorine. Hence, more energy would be required to break the bonds in case of ethyl bromide.
On the other hand, the electronegative character of the functional group also determines the strength of the bond it forms. Here too, the electronegative character of bromine is greater than chlorine. Hence, here too, more energy would be required to break the bonds in case of ethyl bromide.
We know that temperature is a direct translation of energy. Hence high energy required resembles higher temperatures to be reached.
Hence, because of the above reasons, the boiling point of ethyl bromide higher than that of ethyl chloride

Note: In the case of alkyl halides, they contain carbon and halogen atoms and which exhibit polar character in their bonds, and a type of dipole – dipole interaction is observed. Van der Waals forces can be explained as the forces that account for the attraction as well as repulsion between the constituent atoms and molecules. These forces are caused by the fluctuations in the polarisations of nearby particles.