Question

Arrange the compounds in each set in the order of increasing boiling point. (a) \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{5} \mathrm{COOH}\) \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{6} \mathrm{CHO}\) \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{6} \mathrm{CH}_{2} \mathrm{OH}\)

Short answer

Question: Arrange the following compounds in order of increasing boiling point: an eight-carbon carboxylic acid, an eight-carbon aldehyde, and an eight-carbon alcohol. Answer: The order of increasing boiling point is: eight-carbon aldehyde < eight-carbon alcohol < eight-carbon carboxylic acid.

Step-by-step solution

1. Identify compound structures

First, write down the structure of each compound: (a) \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{5} \mathrm{COOH}\) is an eight-carbon carboxylic acid. \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{6} \mathrm{CHO}\) is an eight-carbon aldehyde. \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{6} \mathrm{CH}_{2} \mathrm{OH}\) is an eight-carbon alcohol.

2. Identify intermolecular forces

Now, determine the types of intermolecular forces present in each compound: (a) The carboxylic acid has hydrogen bonding due to the \(\mathrm{OH}\) group and the carbonyl group (\(\mathrm{C=O}\)) has a dipole moment. The aldehyde has a dipole moment due to the carbonyl group (\(\mathrm{C=O}\)), but no hydrogen bonding. The alcohol has hydrogen bonding due to the presence of the \(\mathrm{OH}\) group.

3. Compare van der Waals forces

Since all the compounds have the same number of carbon atoms, their van der Waals forces will be quite similar, and we can focus on hydrogen bonding and dipole-dipole interactions to arrange them according to boiling points.

4. Arrange compounds according to intermolecular forces

Since hydrogen bonding is a stronger intermolecular force than dipole-dipole interactions and van der Waals forces, we can arrange the compounds in the following order: The aldehyde (\(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{6} \mathrm{CHO}\)) has the lowest boiling point due to only having dipole-dipole interactions from the polar carbonyl group. The alcohol (\(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{6} \mathrm{CH}_{2} \mathrm{OH}\)) has a higher boiling point than the aldehyde because it has hydrogen bonding in addition to van der Waals forces, but its hydrogen bonding is less extensive than that of the carboxylic acid because it only has one \(\mathrm{OH}\) group. The carboxylic acid (\(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{5} \mathrm{COOH}\)) has the highest boiling point because it has the most extensive hydrogen bonding, both from the \(\mathrm{OH}\) group and the carbonyl group. So, the correct order of increasing boiling point is: \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{6} \mathrm{CHO} \lt \mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{6} \mathrm{CH}_{2} \mathrm{OH} \lt \mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{5} \mathrm{COOH}\)