Coaching Centre
Intermolecular forces and properties
How dispersion forces, dipole–dipole forces and hydrogen bonds explain the boiling points, melting points and solubility of alkanes, alkenes, alcohols, aldehydes, ketones, carboxylic acids, amines and amides, within a series and between series of similar molar mass, with a practical investigation.
Butane, propanol and ethanamide have almost the same molar mass, about 58 to 60 g/mol. Yet butane is the gas in a camping stove, propanol is a liquid that boils at 97 °C, and ethanamide is a crystalline solid that doesn't boil until 221 °C. The molecules weigh the same. What differs is how strongly they cling to each other, and you can predict it from the functional group alone.
Two kinds of bonding
Intramolecular bonds are the covalent bonds inside a molecule. They are strong (typically 300 to 800 kJ/mol) and decide its chemical reactions. Intermolecular forces act between molecules. They are much weaker, and they decide physical properties: melting and boiling points, volatility, viscosity and solubility. When butane boils, no C–H or C–C bonds break; the molecules simply pull apart from each other.
- Dispersion forces act between all molecules. Electrons are always moving, so at any instant a molecule has a slightly uneven charge (a temporary dipole) that induces dipoles in its neighbours. More electrons and more surface contact mean stronger dispersion forces. They are the only intermolecular force in hydrocarbons.
- Dipole–dipole forces act between polar molecules, which have a permanent uneven charge because of a bond between atoms of different electronegativity, such as C=O or C–Cl. The δ+ end of one molecule attracts the δ− end of the next.
- Hydrogen bonds are a strong kind of dipole–dipole attraction. They form when H is bonded to N, O or F (making the H strongly δ+) and is attracted to a lone pair on an N, O or F atom of another molecule. Alcohols, carboxylic acids, primary and secondary amines and primary and secondary amides all hydrogen bond to each other.
Trends within a homologous series
Each extra CH₂ adds eight electrons and more surface for contact, so dispersion forces get stronger and the boiling point climbs steadily, by roughly 20 to 30 °C per carbon in the early members. That is why C₁ to C₄ alkanes are gases at room temperature, C₅ to about C₁₆ are liquids and longer alkanes are waxy solids. Volatility and the ease of lighting a fuel go the other way.
Branching lowers the boiling point. A branched molecule is more compact, so it touches its neighbours over a smaller area and dispersion forces are weaker: pentane 36 °C, 2-methylbutane 28 °C, 2,2-dimethylpropane 10 °C. Melting points are less regular, because they also depend on how neatly molecules pack in a crystal. Alkanes with an even number of carbons pack slightly better, so their melting points zigzag upward.
- Show only the alkanes. By how much does the boiling point rise from C₄ to C₅? From C₇ to C₈? Why does the gap shrink?
- Add the alkan-1-ols. How much higher than the alkane with the same number of carbons is each alcohol? Does the gap grow or shrink with chain length? Explain.
- Switch the x-axis to molar mass. Tap the points near 60 g/mol and rank them. Which functional group wins, and why?
- Compare aldehydes with ketones of the same carbon number. Why are they so close?
- Find the amides. What is unusual about how their boiling points change with chain length?
Comparing different series
To compare the effect of functional groups fairly, compare molecules of similar molar mass, so their dispersion forces are similar. Then the differences come from the polar groups.
| Compound | Series | M (g/mol) | Boiling point | Strongest intermolecular force |
|---|---|---|---|---|
| butane | alkane | 58 | −1 °C | dispersion only |
| propanal | aldehyde | 58 | 48 °C | dipole–dipole (C=O) |
| propanone | ketone | 58 | 56 °C | dipole–dipole (C=O) |
| trimethylamine | tertiary amine | 59 | 3 °C | weak dipole–dipole; no N–H, so no H-bonds between molecules |
| propan-1-amine | primary amine | 59 | 48 °C | hydrogen bonds (N–H···N) |
| propan-1-ol | alcohol | 60 | 97 °C | hydrogen bonds (O–H···O) |
| ethanoic acid | carboxylic acid | 60 | 118 °C | hydrogen bonds, two per pair (dimers) |
| ethanamide | primary amide | 59 | 221 °C | extensive network of N–H···O=C hydrogen bonds |
- Amines vs alcohols: N is less electronegative than O, so N–H bonds are less polar and the hydrogen bonds are weaker.
- Carboxylic acids pair up into dimers held by two hydrogen bonds, which roughly doubles the size of the unit that has to escape the liquid.
- Amides have a very polar C=O (made more polar by the nitrogen next to it) and N–H bonds, so each molecule hydrogen bonds to several neighbours in a network. Most amides are solids at room temperature; methanamide is the only common liquid one.
- Tertiary amines and tertiary amides have no H on the nitrogen, so they can accept hydrogen bonds but cannot donate them to each other.
Solubility
"Like dissolves like": a substance dissolves when the forces it can form with the solvent are about as strong as the ones it breaks. Hydrocarbons are non-polar, so they don't dissolve in water (they would break water's hydrogen bonds and offer only dispersion forces back). They dissolve well in non-polar solvents such as hexane, and they are less dense than water, so they float.
Small alcohols, carboxylic acids, amines, amides, aldehydes and ketones dissolve in water because their polar group hydrogen bonds to water molecules (aldehydes and ketones can only accept hydrogen bonds from water, through the O). As the non-polar chain gets longer, it disrupts more of water's hydrogen bonding and solubility falls. Methanol, ethanol and propan-1-ol mix with water in all proportions; butan-1-ol dissolves only to about 7 g per 100 mL, pentan-1-ol about 2 g and hexan-1-ol about 0.6 g. Longer-chain alcohols dissolve better in hexane than in water.
Test yourself
Which of these compounds, all with M ≈ 72–74 g/mol, has the highest boiling point?
Butan-1-ol is the only one that forms hydrogen bonds between its molecules.
2-Methylbutane boils at a lower temperature than pentane because
Same formula, same electrons; branching reduces contact area.
Propan-1-amine (48 °C) boils at a lower temperature than propan-1-ol (97 °C) because
Both hydrogen bond; the N–H···N bond is weaker.
Which compound is least soluble in water?
Its long non-polar chain outweighs the single OH group.
Label each force.
Order from lowest to highest boiling point (all about 58–60 g/mol).
Forces between molecules are called forces. The only forces between alkane molecules are forces. Hydrogen bonds form when H is bonded to N, O or . Two carboxylic acid molecules can join through two hydrogen bonds to form a . As the carbon chain of an alcohol gets longer, its solubility in water .
The boiling points of butane (−1 °C), propanone (56 °C), propan-1-ol (97 °C) and ethanoic acid (118 °C) are very different, even though all four have a molar mass of 58–60 g/mol. Explain these differences with reference to intermolecular forces.
Because all four compounds have a molar mass of 58–60 g/mol, they have similar numbers of electrons and similar dispersion forces, so the differences in boiling point must come from their functional groups. Butane is non-polar, so dispersion forces are the only intermolecular force; little energy is needed to separate the molecules and it boils below room temperature. Propanone has a polar C=O bond, so dipole–dipole forces act between molecules in addition to dispersion forces. It has no H bonded to O, so it cannot hydrogen bond to itself, and its boiling point (56 °C) is higher than butane's but well below the alcohol's. Propan-1-ol has an O–H group, so its molecules form hydrogen bonds with each other. These are stronger than dipole–dipole forces, so more energy is needed and it boils at 97 °C. Ethanoic acid's carboxyl group has both a C=O and an O–H, so two molecules form two hydrogen bonds with each other to make a dimer. The stronger total attraction, and the larger unit that must escape the liquid, give the highest boiling point, 118 °C.