Question

Consider the compounds butanoic acid, pentanal, n-hexane, and-pentanol. The boiling points of these compounds (in no specific order) are${\mathbf{\text{69}}}^{\mathbf{\text{o}}}{\mathbf{\text{C,103}}}^{\mathbf{\text{o}}}{\mathbf{\text{C,137}}}^{\mathbf{\text{o}}}{\mathbf{\text{Cand 164}}}^{\mathbf{\text{o}}}\mathbf{\text{C}}$. Match the boiling points to the correct compound.

Short answer

From the highest to the lowest boiling point: butanoic acid (164⁰C),1-pentanol (137⁰C), pentanal (103⁰C) and n-hexane (69⁰C).

Step-by-step solution

To match the boiling points

Before we match the boiling point with each compound, we have to determine intermolecular forces for each compound. An increase in intermolecular forces will result in an increase in boiling point.

Explanation

n-hexane has only London dispersion forces because it does not have any functional group.

1-pentanol has a chain that has London dispersion forces and a functional group -OH that can form a hydrogen bond.

Pentanal has a chain where London dispersion forces can occur and functional group -CHO where dipole-dipole forces may occur.

Butanoic acid can form hydrogen bond via a functional group -COOH as well as can form an additional hydrogen bond via the same functional group. Besides that, London disperse forcescan occur.

Hydrogen bonding is the strongest, then dipole-dipole forces and in the end, London disperse forces.

Therefore, the highest boiling point will have butanoic acid (164⁰C), then1-pentanol (137⁰C), then pentanal (103⁰C) and in the end n-hexane (69⁰C).

Hence From the highest to the lowest boiling point: butanoic acid, 1-pentanol, pentanal and n-hexane.