Why Branching Often Lowers Alkane Isomers’ Boiling Points

n-Hexane and 2,2-dimethylbutane have exactly the same molecular formula, \(\ce{C6H14}\), and the same molar mass. Yet their normal boiling points are very different: n-hexane boils at 68.7 °C, while 2,2-dimethylbutane boils at just 49.7 °C.

Why should rearranging the same six carbon atoms change the boiling point by 19.0 °C?

The short answer is that branching usually makes an alkane more compact. This reduces the effective area over which neighbouring molecules can interact, weakening their overall London dispersion forces. Less energy is then needed for the molecules to separate into a gas, so the boiling point decreases.

Before going further, consider this prediction: if two molecules have the same number of carbon and hydrogen atoms, should they have the same boiling point?

No. Their molecular formula and molar mass are identical, but their three-dimensional shapes are different. That difference changes how strongly their molecules attract one another.

01First, what actually happens when an alkane boils?

To understand the effect of branching, we need to distinguish two types of attraction:

• Intramolecular bonds: These hold atoms together within a molecule. In alkanes, these are covalent C-C and C-H bonds.
• Intermolecular forces: These are attractions between separate molecules.

When liquid hexane boils, its molecules gain enough energy to escape the liquid and move into the gas phase.

The molecules separate from one another, but their individual covalent structures remain intact. The C-C and C-H bonds are not broken.

This distinction matters because a lower boiling point does not mean that a branched alkane has weaker covalent bonds. It means that its molecules are easier to separate from one another.

For a pure liquid, boiling occurs when its vapour pressure equals the external pressure. The normal boiling point is measured at an external pressure of 101.325 kPa.

At the same external pressure, a substance with weaker overall intermolecular attractions will generally reach its boiling point at a lower temperature.

If you need to revise the difference between these two types of bonding, see Bonding in Alkanes: Properties Explained for HSC Chemistry.

02Why branching changes London dispersion forces

Alkanes are non-polar molecules. Their main intermolecular attractions are London dispersion forces.

These forces arise because electrons are constantly moving around a molecule. At any instant, the electron distribution can become slightly uneven, producing a temporary dipole.

This temporary dipole can influence the electron distribution of a neighbouring molecule, creating an induced dipole. The resulting attraction is a London dispersion force.

London dispersion forces exist between all molecules, including non-polar alkanes.

Why molecular shape matters

Imagine placing two long strips of paper alongside one another. They can remain close together along a relatively large area.

Now imagine crumpling both strips into compact balls. Their effective areas of close contact are smaller.

The same general idea helps explain the difference between straight-chain and branched alkanes.

An extended alkane chain can have relatively extensive contact with neighbouring molecules. A compact, highly branched molecule generally has less effective surface contact.

This means the total London dispersion attractions between neighbouring molecules are usually weaker in the more compact structure.

The paper analogy has a limitation: molecules are not sticky surfaces, and their shapes are constantly changing. London dispersion forces arise from interacting electron clouds, not physical adhesion. The analogy simply helps visualise why molecular geometry affects the total attraction.

Connect the shape to the boiling point

The reasoning follows a clear sequence:

1. Branching changes the three-dimensional shape of an alkane.
2. A more compact shape generally reduces effective intermolecular contact.
3. This reduces the overall London dispersion attractions between neighbouring molecules.
4. Less energy is needed for molecules to escape the liquid.
5. The liquid reaches the condition for boiling at a lower temperature.

Notice that branching does not remove carbon atoms or electrons. Two structural isomers have the same molecular formula and the same number of electrons.

The important change is how the molecules interact with one another, not how strongly their atoms are covalently bonded.

03Compare actual boiling points of alkane isomers

Let’s test this explanation using five structural isomers of hexane.

Every molecule in the table has the molecular formula \(\ce{C6H14}\) and a molar mass of approximately 86.18 g mol\(^{-1}\).

All values are normal boiling points at 101.325 kPa, rounded to one decimal place.

Source: NIST Chemistry WebBook, SRD 69: n-hexane, 2-methylpentane, 3-methylpentane, 2,2-dimethylbutane, and 2,3-dimethylbutane.

There is a clear overall pattern.

• n-Hexane, with its unbranched chain, has the highest boiling point.
• The branched isomers generally have lower boiling points.
• The highly compact 2,2-dimethylbutane has the lowest boiling point.

This supports the explanation based on molecular shape and London dispersion forces.

However, the table also reveals something important: simply counting the number of branches does not always predict the exact boiling point.

Why the word “often” matters

Compare 2-methylpentane and 3-methylpentane.

Both molecules have one methyl branch, and both have the formula \(\ce{C6H14}\). However:

• 2-methylpentane boils at 60.3 °C.
• 3-methylpentane boils at 63.3 °C.

The difference is 3.0 °C.

Moving the branch changes the overall molecular geometry. The molecules have different shapes and arrangements, which influence their average intermolecular contact.

The exact boiling point therefore depends on the entire molecular structure, not just the number of branches.

Molecules also rotate and adopt different conformations in a liquid. A simple two-dimensional structural formula cannot show every possible arrangement or calculate the effective contact area directly.

This is why branching gives a useful general trend rather than an infallible rule.

For a broader application of molecular geometry to boiling-point differences, see How Molecular Shape Explains Boiling-Point Differences.

04How could we experimentally test the explanation?

So far, we’ve used measured boiling points to identify a pattern. But how could a chemist obtain a fair comparison between different alkane isomers?

The important question is whether the difference comes from molecular structure or from something else, such as pressure, sample purity, or inaccurate temperature measurements.

The apparatus and measurement

A microscale boiling-point determination can use a small boiling-point tube containing a pure liquid sample.

A narrow capillary tube, sealed at one end, is inverted into the liquid with its open end submerged. The sample tube is positioned beside a calibrated thermometer, with the thermometer bulb level with the sample.

Both are immersed in a stirred, temperature-controlled heating bath.

As the bath heats the sample, vapour bubbles emerge from the capillary and rise through the liquid. When the sample is slowly cooled, the bubbling eventually stops and liquid begins to enter the capillary.

The temperature at this point provides an estimate of the boiling point at the experimental pressure.

The prediction is straightforward.

If the molecular-shape explanation is correct, n-hexane should have a higher boiling point than the more compact 2,2-dimethylbutane when both are measured at the same pressure.

The reference measurements in the table support that prediction.

Why the experimental design matters

Imagine an initial investigation in which four open tubes are heated on separate hotplates. A thermometer is held against the outside of each tube, and the temperature is recorded when bubbles appear.

There are several problems:

• The thermometer may not accurately represent the sample temperature.
• Different heating rates can produce inconsistent readings.
• Volatile samples may evaporate before a reliable measurement is obtained.
• The external pressure may not be identical between measurements.
• Impurities can change the observed boiling point.

A better design uses the same calibrated measurement method for each pure sample, controls the heating and cooling rate, and records the atmospheric pressure.

The independent variable is the molecular structure of the alkane. The dependent variable is its measured boiling point.

Sample purity, pressure, measurement equipment, and experimental conditions must be controlled.

These alkanes are volatile and flammable, so any real investigation requires appropriate laboratory controls, a fume cupboard, and no naked flames.

The apparatus described here is a model investigation, not a claim about a particular historical experiment. The numerical values above are published reference data rather than invented classroom measurements.

05Worked examples

06Questions and solutions

Question 1

n-Hexane and 2,2-dimethylbutane are structural isomers with the molecular formula \(\ce{C6H14}\).

n-Hexane has a normal boiling point of 68.7 °C, while 2,2-dimethylbutane has a normal boiling point of 49.7 °C.

Identify which molecule has the lower boiling point and explain why, referring to molecular shape and intermolecular forces.

Solution 1

2,2-Dimethylbutane has the lower boiling point because its branched structure is more compact than the extended structure of n-hexane.

Both molecules have the same molecular formula, molar mass, and number of electrons. Their main intermolecular attractions are London dispersion forces.

However, the extended n-hexane molecules can have more extensive close contact with neighbouring molecules.

The compact 2,2-dimethylbutane molecules have less effective contact area, resulting in weaker overall dispersion attractions.

Consequently, less energy is needed to separate the branched molecules into the gas phase.

Its boiling point is 19.0 °C lower, demonstrating the general relationship between increased branching and decreased boiling point for comparable alkane isomers.

Question 2

Two structural isomers have the following properties:

Calculate the difference between their boiling points.

A student claims that because both molecules have one methyl branch, they should have identical boiling points.

Evaluate this claim using intermolecular forces and molecular shape.

Solution 2

Isomer B, 3-methylpentane, has a boiling point 3.0 °C higher than isomer A, 2-methylpentane.

The difference is:

\[ \begin{aligned} \Delta T_b &= 63.3-60.3\\ &=3.0^\circ\mathrm{C} \end{aligned} \]

The student’s claim is incorrect.

Although the two molecules have the same molecular formula and number of branches, the branches occupy different positions.

This changes their overall molecular geometry and affects how neighbouring molecules interact in the liquid.

The measured higher boiling point of 3-methylpentane is consistent with stronger average intermolecular attractions in its liquid.

This comparison demonstrates why counting branches is insufficient to predict the exact boiling point. The entire molecular structure must be considered.

Question 3

A student writes the following explanation:

“When 2,2-dimethylbutane boils, its weaker C-C covalent bonds break, allowing the molecule to separate into smaller hydrocarbon molecules. This explains why it has a lower boiling point than n-hexane.”

Identify the errors in this explanation and rewrite it using the correct chemical terminology.

Solution 3

The student’s explanation is incorrect because boiling does not involve breaking the C-C or C-H covalent bonds within an alkane molecule.

Instead, boiling involves overcoming the intermolecular attractions between separate molecules.

A correct explanation is:

2,2-Dimethylbutane has a more compact molecular structure than n-hexane. This reduces the effective surface contact between neighbouring molecules, weakening the overall London dispersion forces.

Less energy is therefore required for the molecules to overcome these intermolecular attractions and enter the gas phase.

The individual \(\ce{C6H14}\) molecules remain intact during boiling.

The key distinction is that intermolecular attractions are overcome during a physical change, while intramolecular covalent bonds remain intact.

If the covalent bonds were broken and smaller hydrocarbons formed, a chemical reaction or decomposition process would be occurring rather than ordinary boiling.

Question 4

A chemistry student wants to investigate whether branching affects the boiling points of alkane isomers.

They place four different alkane samples in separate open tubes, heat each tube using a different hotplate, and record the temperature using a thermometer held against the outside of each tube.

The student concludes that the most branched alkane has the lowest boiling point.

(a) Identify two limitations of this experimental design.

(b) Describe how the experiment could be improved to obtain more reliable boiling-point measurements.

(c) Identify the independent variable, dependent variable, and two controlled variables.

Solution 4

The experiment does not provide sufficient control to establish that molecular branching is responsible for the measured differences.

(a) Limitations

Two major limitations are:

1. The thermometer is held against the outside of the tube, so its reading may not accurately represent the actual sample temperature.
2. Different hotplates may heat the samples at different rates, producing inconsistent measurements.

Other limitations include evaporation of volatile samples, differences in sample purity, and failure to control atmospheric pressure.

(b) Improvements

Use a consistent microscale boiling-point apparatus for every sample.

A small amount of each pure alkane should be placed in a boiling-point tube containing an inverted capillary. Position the sample tube beside a calibrated thermometer and immerse both in a stirred, temperature-controlled bath.

Heat and cool the samples at controlled rates, recording the temperature when bubbling ceases during cooling.

Repeat the measurements to improve reliability.

The atmospheric pressure should also be recorded and kept consistent. This is particularly important because boiling point changes with external pressure.

(c) Variables

• Independent variable: The molecular structure or degree and position of branching of the alkane.
• Dependent variable: The measured boiling point in °C.
• Controlled variables: External pressure and sample purity.

Sample volume, heating rate, thermometer calibration, and measurement apparatus should also be kept consistent.

These improvements make it more reasonable to attribute differences in boiling point to molecular structure rather than experimental errors.

Question 5

Two alkane isomers have the following properties:

Both molecules have the same molecular formula, molar mass, and number of branches.

A student makes two predictions:

• Prediction 1: Isomer A should have the higher boiling point because its more symmetrical structure should allow molecules to pack together more efficiently.
• Prediction 2: Isomer B should have the lower boiling point because its branches are located on adjacent carbon atoms, making it more branched.

(a) Calculate the difference between the normal boiling points.

(b) Evaluate both predictions. Explain why neither symmetry nor branch count alone is sufficient to predict the boiling-point order.

(c) Explain which molecular structure experiences stronger overall intermolecular attractions in the liquid, and why.

(d) A second experiment measures the boiling point of isomer A at 90 kPa but compares it with the original boiling point of isomer B at 101.325 kPa. Explain why this is not a fair comparison.

Solution 5

Isomer B, 2,3-dimethylbutane, has the higher normal boiling point, by 8.3 °C.

This result demonstrates that neither the number of branches nor molecular symmetry alone determines the exact boiling point.

(a) Calculate the difference

Subtract the lower boiling point from the higher boiling point.

\[ \begin{aligned} \Delta T_b &= 58.0-49.7\\ &=8.3^\circ\mathrm{C} \end{aligned} \]

Therefore, 2,3-dimethylbutane boils 8.3 °C higher than 2,2-dimethylbutane at the same external pressure.

(b) Evaluate the predictions

Prediction 1 is incorrect.

Although 2,2-dimethylbutane has a compact and highly symmetrical structure, symmetry does not automatically produce stronger intermolecular attractions in a liquid.

The argument about efficient packing is particularly tempting because molecular symmetry can influence how molecules arrange themselves in a solid crystal. However, boiling point concerns the balance between a liquid and its vapour, not the packing efficiency of a rigid crystal.

In a liquid, molecules move, rotate, and continually change their relative positions.

Prediction 2 is also incorrect.

Having branches on adjacent carbon atoms does not automatically make a molecule more compact or guarantee a lower boiling point.

The overall three-dimensional structure matters more than simply counting branches or identifying their positions in isolation.

(c) Compare intermolecular attractions

The higher boiling point of 2,3-dimethylbutane indicates stronger average cohesive intermolecular attractions in its liquid under the same pressure conditions.

Its overall molecular shape is less compact than that of 2,2-dimethylbutane, allowing more effective close contact between neighbouring molecules.

This produces stronger overall London dispersion attractions.

More energy is therefore required for its molecules to escape into the gas phase, resulting in its higher boiling point.

Importantly, this does not mean that its C-C or C-H covalent bonds are stronger. The difference concerns intermolecular attractions, not the strength of the bonds within individual molecules.

The measured boiling points support this model, although they do not directly measure the strength of an individual molecular contact or establish one fixed arrangement for molecules in the liquid.

(d) Consider external pressure

The second experiment does not provide a fair comparison.

Boiling point depends on external pressure. At a lower external pressure, a liquid reaches the condition where its vapour pressure equals the external pressure at a lower temperature.

Therefore, 2,2-dimethylbutane will have a lower measured boiling point at 90 kPa than its normal boiling point at 101.325 kPa.

Comparing that result directly with the 58.0 °C normal boiling point of 2,3-dimethylbutane mixes two different experimental conditions.

To investigate the effect of molecular structure fairly, both isomers must be measured at the same external pressure.

This question brings together three important ideas: molecular geometry, intermolecular attractions, and experimental control. A reliable explanation must account for all three rather than relying on a single rule about branching.

07Where this fits in HSC Chemistry

For students studying in 2026 and 2027, the NSW Chemistry Stage 6 Syllabus (2017) remains the syllabus being taught. The Chemistry 11-12 Syllabus (2025) begins implementation in 2028.

The explanation in this article applies the existing course ideas about intermolecular and intramolecular forces, molecular structure, and physical properties. The particular boiling-point values of these five isomers are useful reference data, not a claim that NESA prescribes them as a memorisation list.

You can check the current Chemistry Stage 6 syllabus and the implementation information for the 2025 syllabus.

When comparing the boiling points of molecular substances, use a consistent sequence:

1. Check that the substances are being compared at the same external pressure.
2. Compare their molecular formulae and molar masses.
3. Identify their dominant intermolecular forces.
4. Compare their complete molecular structures and three-dimensional shapes.
5. Explain how these features affect the overall attractions between neighbouring molecules.

For alkane isomers, the key relationship is that increased branching generally produces a more compact molecular shape, reducing effective intermolecular contact and weakening overall London dispersion attractions.

The next useful step is to apply the same structure-property reasoning to molecules with different intermolecular forces, where molecular polarity and hydrogen bonding can change the boiling-point trend.