Question

Arrange the following in increasing order of boiling points.
i) $(C{H}_3{)}_{2}CHC{H}_{2}I$
ii) $C{H}_{3}C{H}_{2}C{H}_{2}Cl$
iii) $C{H}_{3}C{H}_{2}C{H}_{2}C{H}_{2}I$

Answer 1

Hint: The given options come under the term isomeric alkyl halides. In alkyl halides, boiling point decreases with increase in the branching on parent carbon chain as in case of tertiary and secondary alkyl halides which have low boiling points. The reason behind this is boiling point depends on the surface area i.e. higher the surface area, higher will be the boiling point.

Complete answer:
- As we know, isomeric alkyl halides are organic compounds that have the same halogen functional groups and same molecular formula but they have different structures and orientation.
- Thus, among isomeric alkyl halides, the boiling point inversely depends on the alkyl group present, which means that the boiling point decreases with the increase in the number of alkyl groups.
- The decrease in boiling point is seen due to increased branching because of which the molecule attains a spherical shape with less surface area available. Due to less surface area available, intermolecular forces or van der Waal forces become weaker in the alkyl halides leading to a decrease in boiling point.
- Therefore, the correct order of increasing boiling points of the given alkyl halides will be:
 $C{H}_{3}C{H}_{2}C{H}_{2}C{H}_{2}I>(C{H}_3{)}_{2}CHC{H}_{2}I>C{H}_{3}C{H}_{2}C{H}_{2}Cl$

Hence, the correct answer is iii>i>ii i.e. $C{H}_{3}C{H}_{2}C{H}_{2}C{H}_{2}I>(C{H}_3{)}_{2}CHC{H}_{2}I>C{H}_{3}C{H}_{2}C{H}_{2}Cl$.

Note: We can easily remember this trend keeping in mind that the trend is followed as,
Due to branching the given molecule attains a spherical shape with less surface area $\to$ intermolecular forces become weaker $\to$ and the boiling point decreases.