Describe and explain trends in the physical properties of organic compounds (melting point, boiling point and solubility in water) in terms of molecular structure, functional groups and intermolecular forces

What this dot point is asking

QCAA wants you to predict and explain melting point, boiling point and aqueous solubility for organic compounds by reasoning from their functional groups to the intermolecular forces (IMFs) involved. The dot point underpins many EA Paper 2 short responses ("explain the difference in boiling point between X and Y") and is a routine IA3 secondary-data interpretation step.

The answer

Physical properties of organic compounds are set by the intermolecular forces between molecules. Stronger total IMF means more energy is needed to separate molecules, so higher melting and boiling points and (when matched against water's hydrogen bonds) higher aqueous solubility for polar compounds.

The three intermolecular forces in organic chemistry

These ranges overlap by design: a very large non-polar molecule (e.g. long alkane) can have stronger total dispersion than a small hydrogen-bonded molecule (e.g. methanol). The dot point asks for relative reasoning, not numerical force calculation.

Boiling point trends across homologous series

Within a series, boiling point rises with chain length because dispersion forces scale with surface area and electron count. Each added CH2 raises boiling point by roughly 20 to 30 degrees C in the short-chain region, less above C10.

At matched carbon count, boiling point depends on functional group, in the rough order:

alkane < alkene < haloalkane (Cl) < aldehyde / ketone < amine (1 deg) < alcohol < carboxylic acid < amide

The reasoning runs functional-group by functional-group:

• Alkanes and alkenes are non-polar (C-H, C=C contribute negligible dipole). Only dispersion. Lowest boiling points at matched Mr.
• Haloalkanes have a C-X dipole. Dipole-dipole adds to dispersion; iodo > bromo > chloro > fluoro for dispersion (larger halide, more electrons), but Cl-Cl has stronger dipole than I-I.
• Aldehydes and ketones have a C=O dipole; no O-H donor. Dipole-dipole adds to dispersion.
• Amines (primary) have N-H; weak hydrogen bonding (N is less electronegative than O).
• Alcohols have O-H; strong hydrogen bonding. Substantially higher boiling point than the matched aldehyde or amine.
• Carboxylic acids form hydrogen-bonded cyclic dimers in the liquid (two H-bonds per dimer). Boiling points are noticeably higher than the matched alcohol.
• Amides combine C=O acceptor with N-H donor; primary amides form extensive H-bonded networks. Highest boiling points of the common Unit 4 series.

The classic Unit 4 comparison set (pentane / butanal / butan-1-ol / butanoic acid, all Mr around 72-88) shows the trend cleanly: 36, 75, 118, 164 degrees C respectively.

Branching lowers boiling point

Branched isomers boil lower than straight-chain isomers. Branching reduces molecular surface area and the number of contact points for dispersion. Compare:

• n-pentane: bp 36 degrees C
• 2-methylbutane: bp 28 degrees C
• 2,2-dimethylpropane: bp 9 degrees C

All three are C5H12 (same Mr); only shape differs. This appears in QCAA stimulus comparing isomers.

Aqueous solubility depends on whether the molecule can hydrogen-bond with water

A compound is highly soluble in water if it can hydrogen-bond with water and its non-polar region is small. Two competing factors:

1. Hydrophilic functional groups (-OH, -NH2, -COOH, -CONH2, -COO-, -NH3+) hydrogen-bond to water. Each promotes solubility.
2. Hydrophobic hydrocarbon chain disrupts water-water hydrogen bonding without replacing it. Each added CH2 reduces solubility.

The threshold rule of thumb: alcohols, aldehydes, amines and carboxylic acids up to about C4 are appreciably soluble; C5 to C7 are sparingly soluble; C8 and longer are effectively insoluble. Branching slightly raises solubility (more compact non-polar volume).

For salts of organic acids and bases (sodium ethanoate, ammonium chloride), the ionic interaction with water typically gives full solubility regardless of chain length up to about C12.

Melting point

Melting points follow similar IMF logic, but with an extra crystal-packing factor. Symmetric molecules pack tightly and melt higher than less symmetric isomers of the same Mr. For example, 2,2-dimethylpropane (very symmetric) melts at -17 degrees C, far higher than n-pentane (-130 degrees C), despite n-pentane boiling 27 degrees higher. QCAA EA questions on melting point are less common than on boiling point but follow the same logic plus this symmetry caveat.

Building a model answer

When asked "explain the difference in boiling point between A and B":

1. Identify the dominant IMF in A and in B. Name the force and the structural feature that gives it.
2. State the relative strength order. Hydrogen bonding > dipole-dipole > dispersion (at matched Mr).
3. Acknowledge any matched factor. "Both have similar Mr and so similar dispersion forces, but X also has hydrogen bonding."
4. Conclude. "More energy is required to overcome the stronger total intermolecular forces in X; X has the higher boiling point."

This four-step template earns full marks on QCAA EA and IA3 items.

Why this matters in IA3 and the EA

IA3 research investigations often compare organic compounds in food chemistry, pharmaceutical design, or polymer applications, where solubility and volatility set the practical context. The EA Paper 2 short response set includes one or two property-based reasoning items most years. Memorising the boiling-point order is insufficient; QCAA marks for the IMF-based reasoning behind it.

In one sentence

Physical properties of organic compounds are determined by intermolecular forces (dispersion, dipole-dipole, hydrogen bonding) that depend on functional groups and chain shape; at matched carbon count, boiling point increases in the order alkane < ketone/aldehyde < amine < alcohol < carboxylic acid < amide, and aqueous solubility falls as the non-polar chain lengthens.