Boiling Points, Chain Length and Branching

FREE CHEMISTRY REVISION · TOPIC 55

Boiling Points, Chain Length and Branching

Pearson BTEC Level 3 AAQ Applied Science
Unit 2: Principles and Applications of Chemistry · E1.6

What you will learn

Use molecular size and shape to explain boiling-point trends within the alkanes. Compare unbranched members of a homologous series, then compare branched structural isomers with the same molecular formula. Always compare at the same pressure and identify which variable changes.

1. What happens when an alkane boils?

Schematic neighbouring butane molecules. London attractions act between all molecular regions; dashed marks identify example intermolecular contacts and are not extra chemical bonds.

Alkanes are essentially non-polar molecules. Their intermolecular attractions are London dispersion forces: fluctuations in electron distribution create instantaneous dipoles, which induce dipoles in neighbouring molecules. These attractions occur even without permanent molecular dipoles.

At the boiling point, the liquid’s vapour pressure equals the external pressure. Molecules gain enough energy to separate into the gas phase. The C–C and C–H covalent bonds inside each molecule normally remain intact. A higher boiling point therefore indicates stronger overall intermolecular attractions for these comparisons, not necessarily stronger covalent bonds.

2. Increasing chain length

Across unbranched alkanes, increasing chain length increases molecular size and electron count. Larger electron clouds are more polarisable. Longer molecules also offer more surface over which neighbouring molecules can interact. The combined London attractions become stronger, so more energy is needed to separate molecules and boiling point generally rises.

Original plot of the table values at 101.325 kPa. Lines guide the eye between three discrete compounds; they are not a continuous equation or a universal straight-line law.

Compound	Formula	Normal b.p. / K	Approx. normal b.p. / °C
Butane	C₄H₁₀	272.65	−0.5
Pentane	C₅H₁₂	309.2	36.1
Hexane	C₆H₁₄	341.9	68.8
2-Methylbutane	C₅H₁₂	301.1	28.0
2,2-Dimethylpropane	C₅H₁₂	282.6	9.5

Data source: NIST Chemistry WebBook SRD 69. Values refer to normal boiling at 101.325 kPa (1 atm), not arbitrary room pressure. Butane uses the 272.65 K value reported by Desty and Whyman (1957); other entries use NIST’s displayed compiled averages. Celsius values are calculated using θ/°C = T/K − 273.15 and rounded to one decimal place; the final digit is not a claim of greater measurement precision.

Source records: Butane, Pentane, Hexane, 2-Methylbutane, 2,2-Dimethylpropane.

3. Branching: compare like with like

All three contain five carbon and twelve hydrogen atoms. The central C in the last structure has four C–C bonds and no attached H. Diagrams show connectivity, not flat molecular shapes. Boiling points are approximate normal values.

Structural isomers have the same molecular formula and electron count, so molecular size alone cannot explain their different boiling points. Greater branching makes an alkane molecule more compact. Compared with its less-branched isomers, it generally has less effective surface contact with neighbouring molecules, reducing the overall London attractions. Its boiling point is consequently lower.

These are average interactions among moving three-dimensional molecules in a liquid. Do not explain boiling trends solely by how neatly molecules pack into a crystal: that concerns melting and can behave differently. Do not apply the alkane branching trend uncritically across compounds with different functional groups, hydrogen bonding or different pressures.

4. Worked comparisons

Worked example 1 — hexane versus pentane

Hexane, C₆H₁₄, boils at about 68.8 °C; pentane, C₅H₁₂, at about 36.1 °C. Hexane has more electrons and a larger, more polarisable electron cloud, with more possible intermolecular contact. Stronger London attractions require more energy to overcome. Difference from the rounded table: 68.8−36.1=32.7 °C. Do not write “hexane has stronger C–C bonds”.

Worked example 2 — rank the C₅H₁₂ isomers

Descending boiling point: pentane (36.1 °C) > 2-methylbutane (28.0 °C) > 2,2-dimethylpropane (9.5 °C). All have the same formula and electron count. Increasing branching reduces effective intermolecular contact; dispersion attractions are weaker overall. This isolates shape more effectively than comparing compounds with different carbon numbers.

Worked example 3 — calculate a branching effect

Compare pentane and 2,2-dimethylpropane at the same pressure. Using the rounded Celsius values, Δθ=36.1−9.5=26.6 °C. A temperature difference of 26.6 °C is also 26.6 K. This is a difference between boiling temperatures, not an energy or a percentage reduction in bond strength. Using unrounded data gives 309.2−282.6=26.6 K, consistent with the rounded result.

Worked example 4 — predict before consulting data

Predict pentane versus 2-methylbutane: both C₅H₁₂; pentane is unbranched, so expect greater contact, stronger overall London attractions and the higher boiling point. The data support this: 36.1 °C versus 28.0 °C. The 8.1 °C difference uses rounded values; uncertainty and rounding limit its precision.

Worked example 5 — identify an unfair comparison

A student compares hexane with 2-methylbutane and attributes the entire difference to branching. Both carbon number and shape changed, so the data cannot isolate branching alone. Compare pentane and 2-methylbutane to hold molecular formula constant. To isolate chain length approximately, compare unbranched butane, pentane and hexane at the same pressure.

Worked example 6 — interpret physical state

At 20 °C and 1 atm, 2,2-dimethylpropane is above its normal boiling point (about 9.5 °C), whereas pentane is below its normal boiling point (36.1 °C) and above its melting point. Thus neopentane is a gas and pentane a liquid under these conditions. Their equal molecular formula does not force identical physical properties. This conclusion assumes equilibrium and the stated pressure.

Practice and examination applications

Task	Answer
Explain the upward trend on the graph	Larger, more polarisable molecules have stronger overall London attractions and require more energy to separate.
Does branching remove electrons from C₅H₁₂?	No. The isomers have equal electron counts; their shapes and effective intermolecular contacts differ.
Which bonds break when pentane boils?	No covalent bonds normally break; molecules separate against intermolecular attractions.
Why state the pressure?	Boiling occurs when vapour pressure equals external pressure, so boiling temperature depends on pressure.

Common misconceptions

Alkanes have no intermolecular forces because they are non-polar: London forces are present.
Longer chains boil higher because extra C–C bonds must break: covalent bonds stay intact during ordinary boiling.
Branched isomers have fewer electrons: their formulae and electron counts are the same.
A more compact molecule always boils higher: compact branching generally lowers boiling point among alkane isomers.
Every alkane trend predicts every organic compound: functional groups, pressure and other interactions matter.
Boiling-point and melting-point trends must match: liquid attractions and crystal packing are different considerations.

Examination tips

State whether you are comparing chain length or branching.
Use the phrase stronger overall London dispersion forces, then explain the molecular reason.
Connect stronger attractions to more energy needed for separation and a higher boiling temperature.
Use matched isomers to isolate branching and a constant pressure for all numerical comparisons.
Quote units and avoid false precision when subtracting rounded data.

Check your understanding

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