Question

4. Arrange the following substances in order of increasing boiling point: CH3CH2OH, HOCH2CH2OH, CH3CH2Cl , and ClCH2CH2OH A. CH3CH2OH < HOCH2CH2OH < CH3CH2Cl < ClCH2CH2OH B. CH3CH2Cl < CH3CH2OH < ClCH2CH2OH < HOCH2CH2OH C. CH3CH2OH < CH3CH2Cl < HOCH2CH2OH < ClCH2CH2OH D. CH3CH2Cl < ClCH2CH2OH < CH3CH2OH < HOCH2CH2OH E. CH3CH2OH < ClCH2CH2OH < CH3CH2Cl < HOCH2CH2OH

Answer 1

Answer: Option (B) is the correct answer.

Explanation:

A molecules that will have more number of hydrogen bonding will has highest boiling point because to break the hydrogen bonds high heat needs to be provided.

So, in the molecule $HOCH_{2}CH_{2}OH$ there is presence of two alcoholic groups. Hence, it will have strongest hydrogen bonding as compared to the rest of molecules.

In the molecule $ClCH_{2}CH_{2}OH$ there will be hydrogen bonding and dipole-dipole interactions. Hence, it boiling point will be slightly less than $HOCH_{2}CH_{2}OH$.

In the molecule $CH_{3}CH_{2}OH$, there will be only hydrogen bonding. Hence, its boiling point is less than $ClCH_{2}CH_{2}OH$.

In the molecule $CH_{3}CH_{2}Cl$, there is no hydrogen bonding but there will be only dipole-dipole interactions. Hence, its boiling point will be the least.

Therefore, we can conclude that increasing order of boiling point will be as follows.

        $CH_{3}CH_{2}Cl$ < $CH_{3}CH_{2}OH$ < $ClCH_{2}CH_{2}OH$ < $HOCH_{2}CH_{2}OH$