Answer
The boiling point of butanoic acid will be $164^{\circ}C$
The boiling point of pentanal will be $ 103^{\circ}C$
The boiling point of n-hexane will be $69^{\circ}C$
The boiling point of 1-pentanol will be $137^{\circ}C$
Work Step by Step
First find which of those molecules participate in intermolecular H bonding. Compounds participate in H bonding will have higher boiling point than other. Here, butanoic acid and 1-pentanol are compounds that participate in intermolecular H bonding due to presence of OH group in those molecules. As H is bonded with O atom, so polarity of H is increased and as a result intermolecular H bonding occurs. Now, between these two compounds butanoic acid will have more boiling point than that of 1-pentanol because of presence of -CO- group in butanoic acid. So, it can participate in dipole dipole interaction and as a result boiling point increases. So, butanoic acid have the boiling point $164^{\circ}C$ and 1-pentanol have the boiling point $137^{\circ}C$.
Now, the remaining two compounds do not participate in H bonding. But in pentanal, there is CO group present and dipole dipole interaction between molecules occur here. So, the molecules of pentanal are joined together with that interaction and as a result boiling point increases ($ 103^{\circ}C$) and in hexane, only london dispersion force between molecules are present and it will have the lowest boiling point among other compounds. ($69^{\circ}C$)
