2021 H2 Chem P1 Q7 — Chemical bonding Solution

The answer is B.

Boiling point depends on the strength of the intermolecular forces that must be overcome.

• N, \(\text{CH}_2\text{Cl}\text{CH}_2\text{Cl}\) (\(M_r\) = 99): no H atom bonded to N, O or F, so it cannot form hydrogen bonds. Only id-id attractions (plus weak pd-pd from the polar C-Cl bonds). Although it has the largest electron cloud, these are the weakest intermolecular forces of the four, so N has the lowest boiling point.
• P, \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}\) (\(M_r\) = 74): one OH group, so hydrogen bonding between molecules, on average about one hydrogen bond per molecule.
• Q, \(\text{CH}_3\text{CH}_2\text{CH}_2\text{CO}_2\text{H}\) (\(M_r\) = 88): the \(\text{CO}_2\text{H}\) group has both an O-H and a C=O, so it forms more extensive hydrogen bonding (two hydrogen bonds between a pair of molecules, giving dimers). Q also has a larger electron cloud than P, so stronger id-id. Hence Q boils higher than P.
• M, \(\text{CH}_2(\text{OH})\text{CH}(\text{OH})\text{CH}_2(\text{OH})\) (\(M_r\) = 92): three OH groups per molecule, so the most extensive hydrogen bonding, on average about three hydrogen bonds per molecule. M has the highest boiling point.

Order of increasing boiling point: \(\text{N} < \text{P} < \text{Q} < \text{M}\), which is option B.

• A is wrong: M \(\rightarrow\) Q \(\rightarrow\) P \(\rightarrow\) N is the same order reversed, i.e. decreasing boiling point.
• C is wrong: it puts Q below P, but Q has more extensive hydrogen bonding and a larger electron cloud than P.
• D is wrong: it puts N highest, but N cannot form hydrogen bonds and has the weakest intermolecular forces, and it puts M below N although M has the most extensive hydrogen bonding.