Physical Properties of Alkanes: Boiling Points and Van der Waals Forces

Fill a beaker with hexane and a beaker with water side by side. Hexane floats on top, refuses to mix, and evaporates within minutes at room temperature. Water sits underneath, stubborn, boiling only at 100 °C. Both are small molecules of comparable molecular weight. Yet their behavior could not be more different. The explanation is not exotic. Alkane molecules carry no permanent dipoles, no hydrogen-bond donors, nothing polar at all. That absence is exactly the point. What little intermolecular attraction alkanes do experience comes from a purely quantum-mechanical fluctuation called the London dispersion force. The entire empirical trend of alkane boiling points, melting points, densities, and solubilities can be built up mechanistically from that one microscopic phenomenon. This matters because alkanes are the scaffold of essentially all organic and biological molecules. Separating them by boiling point is the single most economically important process in the chemical industry. It is literally how crude oil becomes gasoline, diesel, and jet fuel. Understanding why boiling point rises smoothly through the Constitutional Isomers and the Homologous Series of straight-chain alkanes matters. So does understanding why it drops sharply upon branching. This is a rehearsal for understanding fractional distillation, solubility rules, and separation science. Eventually it applies throughout petroleum refining. The big picture is a competition between two things happening at once as a molecule gets larger. There is more surface area, meaning more places for transient dipoles to attract a neighbor. But there is also more compact shape, meaning less surface area and weaker attraction per collision. Straight-chain alkanes maximize contact area. Branched isomers minimize it by becoming spherical. That single geometric idea — surface area, not molecular weight per se — is the mechanistic key. It explains every trend in this article. What follows treats the dispersion force as a genuine electronic phenomenon. It is derived from an instantaneous-dipole/induced-dipole argument analogous to a mechanism. It is not simply a fact to memorize. From there we build quantitative and semi-quantitative rules. These cover boiling point increments per CH₂, the odd–even alternation in melting points, the density ceiling near 0.78 g/mL, and the thermodynamics of alkane–water immiscibility.

Key takeaways

Alkanes are essentially nonpolar; their only intermolecular attraction is the London dispersion force, arising from instantaneous dipole/induced-dipole coupling of fluctuating electron density.
Boiling point rises smoothly with chain length because dispersion energy scales with the contact surface area between molecules, which grows with each added CH₂.
Branching lowers boiling point because it makes molecules more spherical, shrinking surface area and the number of van der Waals contacts per collision, even though molecular weight is unchanged.
Melting point depends on crystal packing efficiency, not just size, producing a distinctive odd–even zig-zag superimposed on the overall upward trend.
Alkane density rises toward but never exceeds about 0.78–0.79 g/mL, because dispersion forces can never pack chains as tightly as a hydrogen-bonded or ionic lattice.
Alkanes are "hydrophobic" not because water repels them electrically, but because water sacrifices favorable hydrogen bonds to solvate them — an entropy/enthalpy penalty called the hydrophobic effect.
Key idea: every physical trend in alkanes reduces to one variable — the strength and number of transient, surface-area-dependent London dispersion contacts between molecules.

Where this lives today

Every fractional distillation column in a petroleum refinery is, at bottom, an industrial-scale exploitation of the boiling-point trend derived here: straight-run crude oil is separated into gasoline, kerosene, diesel, and heavier fractions purely because their dispersion-force-driven boiling points differ systematically with chain length and branching, a theme picked up directly in the study of Natural Sources of Alkanes: Petroleum and Refining. Octane rating in gasoline is a direct commercial payoff of the branching effect: highly branched alkanes (like isooctane, 2,2,4-trimethylpentane) resist auto-ignition (knocking) better than straight chains, so refiners deliberately isomerize linear alkanes into branched ones — trading away some fraction of dispersion-force cohesion in the liquid for better combustion behavior in the engine. In drug design, the same surface-area logic governs lipophilicity calculations (logP), since a drug's ability to cross a nonpolar cell membrane correlates with how much nonpolar surface area it presents, directly borrowing the alkane dispersion picture. In materials science, polyethylene and other polyolefins derive their mechanical properties — crystallinity, melting point, density — from exactly the packing-efficiency arguments used above for the odd–even melting-point alternation, scaled up to polymer chains thousands of carbons long. In biology, the hydrophobic effect described here is the single dominant force folding proteins into their native three-dimensional structures and assembling lipid bilayer membranes, since nonpolar amino acid side chains and fatty acid tails behave, in aqueous solution, exactly like the alkane molecules analyzed in this article.

Ask the tutor: Why does the boiling-point increment per CH₂ shrink at very long chain lengths instead of staying perfectly constant forever?

The physical picture

Start with a single alkane molecule and ask what its electron cloud is doing. In quantum mechanics, the electron density around any atom is not a static, frozen shell — it is a probability distribution that fluctuates in time. At any given instant, there is a nonzero probability that more electron density sits on one side of a carbon atom than the other, creating a fleeting, asymmetric charge distribution: an instantaneous dipole. This is true even for a perfectly symmetric atom like neon, and it is certainly true for the C–H and C–C framework of an alkane. That instantaneous dipole does not exist in isolation. Its electric field reaches a neighboring molecule and polarizes that molecule's electron cloud — pushing electron density away from the near side and piling it toward the far side, generating an induced dipole whose negative end faces the first molecule's transient positive end. The result is a fleeting, correlated, attractive electrostatic interaction between the two temporary dipoles. A femtosecond later the first molecule's electron distribution fluctuates again, the induced dipole in the neighbor tracks it, and the attraction persists as a statistical average even though any single snapshot shows a random, non-permanent charge pattern. This synchronized fluctuation is the London dispersion force, first treated quantitatively by Fritz London using second-order perturbation theory in 1930. Two structural facts follow immediately, and they are the entire mechanistic content of this article. First, dispersion attraction requires close contact — it falls off as $1/r^6$ , among the steepest of all intermolecular force laws, so only atoms that can actually touch contribute meaningfully. Second, the total attraction between two molecules is roughly proportional to the number of atom-pairs in van der Waals contact, which is set by the molecules' contact surface area, not by their mass or their formula alone. A long, floppy, extended alkane chain lying alongside another chain maximizes the length of that contact; a compact, spherical, highly branched isomer minimizes it. This is why $n$ -pentane (bp 36 °C) boils 27 degrees higher than its isomer neopentane (bp 9.5 °C) despite both being C₅H�12 with identical molecular weight (72.15 g/mol).

Figure

Instantaneous-dipole/induced-dipole mechanism between two alkane chains

Now zoom out to a whole homologous series. Each additional CH₂ unit adds roughly the same increment of surface area to a linear chain, and — crucially — that new CH₂ can lie alongside a CH₂ on a neighboring molecule in the liquid, contributing its own small dispersion contact. Because these contacts are additive along the chain, the total intermolecular attraction per mole grows roughly linearly with carbon number for the unbranched series, which is exactly the empirical pattern seen in the boiling-point table of methane through decane. Branching breaks this additivity: folding the chain into a compact, near-spherical shape reduces the effective surface area available for contact even though the atoms and bonds are all still there, so the dispersion energy — and hence the boiling point — drops.

Figure

Surface-area contact: linear pentane versus spherical neopentane

Finally, consider what happens on cooling to the solid. Melting point depends on an entirely different geometric criterion: how well the molecules can pack into a periodic crystal lattice, maximizing the number and regularity of van der Waals contacts throughout the solid, not merely their sum in a disordered liquid. Straight, symmetric-length chains pack like matchsticks in a box; chains with an awkward shape (e.g., certain odd-numbered chains, or asymmetric branching) pack more loosely. This is the mechanistic root of the well-known odd–even zig-zag in alkane melting points, discussed quantitatively below.

The core ideas

The dispersion (London) interaction energy between two induced dipoles separated by distance $r$ has the classic form:

E_{\text{disp}}(r) = -\frac{3}{4}\,\frac{\alpha_1 \alpha_2}{r^{6}}\left(\frac{I_1 I_2}{I_1+I_2}\right)

where:

$\alpha_1, \alpha_2$ are the electronic polarizabilities of the two interacting groups (how easily their electron clouds distort),
$I_1, I_2$ are their first ionization energies,
$r$ is the instantaneous separation between the interacting centers,
the negative sign shows the interaction is attractive (stabilizing).

Reading this equation: dispersion energy grows with polarizability (more electrons, more diffuse orbitals, larger $\alpha$ ) and falls off extremely steeply with distance, which is why only actual surface contact matters — a chain lying 5 Å away from its neighbor contributes almost nothing compared to one in van der Waals contact at ~3.5–4 Å. Summed over all atom pairs in contact between two molecules, the total intermolecular cohesive energy per mole scales approximately with the accessible surface area $A$ :

E_{\text{cohesive}} \approx -k \cdot A

where:

$A$ is the solvent-accessible (or van der Waals) surface area of the molecule,
$k$ is an empirical proportionality constant reflecting the polarizability density of a CH₂/CH₃ surface (roughly constant across simple alkanes),
the approximation holds because dispersion forces are short-ranged and pairwise-additive, so summing many $1/r^6$ contacts over a contact patch gives an energy roughly proportional to the patch's area.

Boiling point is, to a first approximation, the temperature at which thermal energy $RT_b$ becomes comparable to the cohesive energy holding the liquid together:

T_b \approx \frac{|E_{\text{cohesive}}|}{R} \cdot c

where:

$R$ is the gas constant,
$c$ is a dimensionless entropy-of-vaporization factor (Trouton's constant reasoning: $\Delta S_{vap} \approx 85$ –88 J mol⁻¹K⁻¹ for most nonpolar liquids),
this says directly: more surface area → more cohesive energy → higher boiling point, with molecular weight entering only indirectly (larger molecules usually have more surface area).

Derivation AWhy boiling point increases roughly linearly per CH₂ in a straight chain

Goal. Show mechanistically why each additional methylene unit in an unbranched alkane raises the boiling point by a roughly constant increment (empirically ≈ 20–30 °C for the first few members, converging toward ~20 °C per CH₂ at longer chain length), rather than some erratic or logarithmic pattern.

Step 1 — Model the chain as a sequence of independent contact sites. Treat an unbranched alkane in the liquid as a flexible rod whose CH₂/CH₃ units each present roughly the same van der Waals surface (~ 6–8 Å² per CH₂ group in a fully extended chain) to a neighboring molecule lying alongside it. Because dispersion forces are pairwise-additive and short-ranged, the total attractive energy between two parallel chains is approximately the sum of the pairwise CH₂···CH₂ contact energies:

E_{\text{cohesive}}(n) \approx -\,\varepsilon \cdot n

where $n$ is the number of carbons and $\varepsilon$ is the dispersion energy contributed by one CH₂-to-CH₂ contact (a few tenths of a kJ/mol per contact, summed over many transient neighbor geometries in the liquid).

Step 2 — Translate cohesive energy into a boiling point using Trouton's rule. At the boiling point, $\Delta G_{vap} = 0$ , so $\Delta H_{vap} = T_b \Delta S_{vap}$ . For a family of structurally similar nonpolar liquids, $\Delta S_{vap}$ is nearly constant (Trouton's constant, ≈ 85 J mol⁻¹ K⁻¹), because the disordering achieved on vaporization — going from a liquid with restricted conformational freedom to a gas with free translation/rotation — is chemically similar molecule to molecule. Since $\Delta H_{vap} \approx |E_{\text{cohesive}}|$ (the energy needed to pull a molecule out of its neighbors' dispersion field):

T_b(n) \approx \frac{|E_{\text{cohesive}}(n)|}{\Delta S_{vap}} = \frac{\varepsilon\, n}{\Delta S_{vap}}

Step 3 — Predict a linear (additive) increment per carbon. Differentiating with respect to $n$ :

\frac{dT_b}{dn} \approx \frac{\varepsilon}{\Delta S_{vap}} = \text{constant}

This constant-increment prediction matches experiment closely in the middle of the homologous series (e.g., hexane 69 °C → heptane 98 °C → octane 126 °C, increments of 29 °C and 28 °C), while the increments shrink at very long chain length because the marginal surface area added by one more CH₂ becomes a smaller fractional increase in an already-large molecule (consistent with $A(n)$ growing linearly but $T_b(n)$ empirically approaching $A^{1/2}$ -like saturation for very long chains — the simple additive model is a first approximation that works best for $n < 20$ or so).

Step 4 — Check the limiting case. As $n \to 1$ (methane), the model should reduce to a small residual boiling point set by a single CH₄···CH₄ contact rather than a chain contact; indeed methane boils at −161.5 °C, the lowest in the series, consistent with $\varepsilon \cdot 1$ being a minimal cohesive energy. As $n \to \infty$ , $T_b$ would formally diverge, which is physically sensible: very long-chain alkanes (paraffin waxes, polyethylene) do not boil below their decomposition temperature — they must be heated so strongly that C–C bonds break first, which is exactly what is observed for waxes and polyethylene.

Go deeper · Derivation B · 2 minWhy branching lowers boiling point at constant molecular formula

Derivation BWhy branching lowers boiling point at constant molecular formula

Goal. Derive, from surface area and packing arguments, why boiling point decreases monotonically as a fixed-formula alkane becomes more branched: $n$ -pentane (36 °C) > isopentane (2-methylbutane, 28 °C) > neopentane (2,2-dimethylpropane, 9.5 °C).

Step 1 — Fix the formula, vary only the connectivity. All three C₅H₁₂ isomers have identical atoms, identical numbers of C–H and C–C bonds, and therefore nearly identical total electron count and polarizability volume. What differs is the three-dimensional shape: $n$ -pentane is an extended zig-zag; isopentane has one branch point (a tertiary CH); neopentane has a central quaternary carbon with four methyls radiating outward, making it nearly spherical.

Step 2 — Compute how molecular shape changes the accessible contact area. For a fixed volume, the shape that minimizes external surface area is a sphere (this is the same isoperimetric principle that makes water droplets spherical). Branching pulls mass toward a compact center of the molecule, converting what would be exposed side-chain surface in a linear conformer into "buried," self-contacting surface unavailable to a neighboring molecule. Neopentane's four methyl groups radiate from a single center, giving it the smallest external surface area of the three isomers per unit volume; $n$ -pentane, able to lie extended, exposes the most.

Step 3 — Translate reduced surface area into reduced cohesive energy. Using the surface-area proportionality from the core equation, $E_{\text{cohesive}} \approx -kA$ , a smaller $A$ directly gives a smaller (less negative) cohesive energy, hence smaller $\Delta H_{vap}$ , hence — by the same Trouton argument as above — a lower $T_b$ :

T_b(\text{neopentane}) < T_b(\text{isopentane}) < T_b(n\text{-pentane})

which is exactly the observed order: 9.5 °C < 28 °C < 36 °C.

Step 4 — Quantify with the actual surface-area/boiling-point correlation. Approximate van der Waals surface areas (from standard group contribution tables) are roughly $A(n\text{-pentane}) \approx 260$ Å², $A(\text{isopentane}) \approx 250$ Å², $A(\text{neopentane}) \approx 235$ Å² per molecule. The fractional area drop from $n$ -pentane to neopentane is

\frac{\Delta A}{A} = \frac{260-235}{260} \approx 0.096, \quad \text{i.e., about a 10\% reduction in contact area,}

which, propagated through $T_b \propto A$ (in absolute temperature, using $T_b(n\text{-pentane}) = 309.2$ K), predicts

T_b(\text{neopentane}) \approx 309.2\,\text{K} \times (1-0.096) \approx 279.5\,\text{K} = 6.4\,^{\circ}\text{C},

in good qualitative agreement with the observed 9.5 °C given the crudeness of a single proportionality constant carried across very different shapes — branching effects on packing entropy add a secondary correction that this simple area model does not fully capture.

Step 5 — Check the limiting case. As branching is pushed to its geometric extreme (a perfectly spherical, maximally symmetric isomer, achieved only by neopentane among the C₅H₁₂ set), $T_b$ reaches its minimum for that formula — consistent with the isoperimetric argument that a sphere is the unique minimal-surface-area shape for fixed volume. There is no isomer of C₅H₁₂ with a boiling point lower than neopentane's, exactly as the surface-area model predicts.

Worked examples

Example 1 — Foundational: ranking three isomers of hexane, C₆H�14, by boiling point. The three isomers are $n$ -hexane, 3-methylpentane, and 2,2-dimethylbutane. Applying the surface-area principle: $n$ -hexane is fully linear (maximum surface area, maximum contact), 3-methylpentane has one branch roughly in the middle of the chain (intermediate compactness), and 2,2-dimethylbutane has a quaternary carbon carrying three methyls (most compact, closest to spherical). Predicted order:

T_b(n\text{-hexane}) > T_b(\text{3-methylpentane}) > T_b(\text{2,2-dimethylbutane})

Experimental values confirm this exactly: $n$ -hexane 68.7 °C, 3-methylpentane 63.3 °C, 2,2-dimethylbutane 49.7 °C. Notice the gap between the most-branched and least-branched isomer here (19.0 °C) is smaller in absolute size than the analogous C₅ gap discussed above relative to the fraction of the molecule that is branched, illustrating that the effect strengthens as more branch points are added, not merely as chain length grows — consistent with the surface-area mechanism, since each additional branch point removes more contact area.

Example 2 — Applied: estimating the boiling point of 3-ethylpentane using a group-additivity check. 3-Ethylpentane (C₇H16) is a symmetric, moderately branched isomer of heptane. Using the linear-chain increment derived above, unbranched heptane boils at 98.4 °C, roughly 29–30 °C above hexane (68.7 °C), consistent with $dT_b/dn \approx 29$ –30 °C per carbon in this region of the series. Because 3-ethylpentane carries the same seven carbons as heptane but redistributed into a branched, more compact shape (an ethyl branch at C3 of a pentane chain, giving a "Y"-shaped rather than extended molecule), the surface-area argument predicts a boiling point measurably below heptane's 98.4 °C but above the most compact C₇ isomer (2,2,3-trimethylbutane, bp 80.9 °C, essentially spherical). The observed value, 93.5 °C, sits neatly between these two bounds — closer to heptane because 3-ethylpentane, while branched, still has an extended pentane backbone contributing substantial surface area, whereas 2,2,3-trimethylbutane crowds three methyls onto adjacent quaternary/tertiary centers and is far more spherical. The lesson: branching location and degree of compaction, not merely the presence of a branch, control the magnitude of the boiling-point depression.

Example 3 — Exam/competition-level: rationalizing melting point vs. boiling point trends for $n$ -pentane through $n$ -octane, including the odd–even effect. The data:

Alkane	bp (°C)	mp (°C)
$n$ -pentane	36.1	−129.7
$n$ -hexane	68.7	−95.3
$n$ -heptane	98.4	−90.6
$n$ -octane	125.6	−56.8

Boiling point rises smoothly and almost linearly, exactly as the liquid-state surface-area argument predicts, since the liquid is a disordered phase where each added CH₂ simply adds more contact area with neighbors in any conformation. Melting point, however, jumps erratically: pentane to hexane rises by 34.4 °C, hexane to heptane rises by only 4.7 °C, then heptane to octane rises sharply again by 33.8 °C. This is the classic odd–even alternation: even-numbered chains (hexane, octane) pack into the crystal lattice more efficiently than odd-numbered chains (pentane, heptane), because in the zig-zag all-anti conformation, an even chain has its two terminal methyl groups pointing in opposite directions relative to the chain's zig-zag plane, allowing neighboring chains in the crystal to nest together in a symmetric, densely-packed herringbone arrangement. An odd chain has both terminal methyls on the same side, which frustrates this symmetric nesting and forces a slightly looser, less efficient lattice. Because melting point depends on how well the ordered solid can maximize dispersion contacts throughout a periodic lattice — not merely the total contact summed over a disordered liquid — a poorly-packing odd chain melts at an anomalously low temperature relative to the general upward trend, even though its liquid-state boiling point (which does not care about lattice symmetry) falls right on the smooth curve. This is a clean illustration that boiling point tracks total surface area (a liquid-state, packing-insensitive property) while melting point tracks packing efficiency (a solid-state, lattice-symmetry-sensitive property) — two different geometric criteria built from the same underlying dispersion force.

Limiting cases, scope & common traps

Situation	What actually controls it	Common error
Comparing boiling points of isomers	Surface area / shape (branching)	Assuming molecular weight alone predicts boiling point
Very long-chain alkanes (waxes, PE)	Chains decompose before an ordinary boiling transition is reached	Extrapolating the linear $T_b$ vs. $n$ rule indefinitely
Odd vs. even carbon count, melting point	Crystal-lattice packing symmetry of terminal methyls	Expecting melting point to track boiling point smoothly
Density of liquid alkanes	Packing efficiency in the liquid; caps near 0.78–0.79 g/mL even for very long chains	Assuming density keeps rising indefinitely with chain length
Alkane–water immiscibility	Entropic cost to water of caging a nonpolar solute (hydrophobic effect), not "repulsion"	Saying alkanes and water "repel" each other electrically
Cycloalkanes vs. open-chain isomers	Ring rigidity changes both surface area and packing	Treating a cycloalkane's boiling point as if it were an open-chain homolog

The chemist behind it

The person most responsible for putting the dispersion force on a rigorous theoretical footing is Fritz London, a German-American theoretical physicist. Working in the early 1930s, shortly after quantum mechanics had matured enough to handle multi-electron perturbation problems, London showed — using second-order perturbation theory applied to two interacting, mutually polarizable systems — that even two atoms with zero permanent dipole moment (like two noble gas atoms, or two alkane molecules) must attract one another, because the zero-point quantum fluctuations of their electron clouds are inherently correlated at short range. Before London's 1930 papers, chemists had observed that even "inert" substances like argon and methane liquefy and solidify at low enough temperature, but there was no accepted quantum-mechanical explanation for why — van der Waals's original 1873 equation of state had introduced attractive forces phenomenologically, as a fitting parameter, without a microscopic mechanism. London supplied that mechanism, and the force is accordingly named after him even though it is one member of the broader family of van der Waals forces.

Origins · 1 minFritz Wolfgang London and how this was discovered

Fritz Wolfgang London

1900–1954 · Germany, then Oxford, then Duke University; theoretical physicist working at the intersection of quantum chemistry and condensed matter physics

History

- 1873 — Johannes Diderik van der Waals proposes an equation of state with an empirical attractive-force correction term, without a microscopic mechanism.

- 1916–1920s — G. N. Lewis and Irving Langmuir develop the covalent electron-pair bond model, setting the stage for quantum treatments of molecular interactions.

- 1927 — Fritz London and Walter Heitler apply quantum mechanics to the H₂ bond, founding modern valence-bond theory and establishing London's expertise in electron-pair quantum mechanics.

- 1930 — Fritz London publishes the perturbation-theory derivation of the

1/r^6

dispersion attraction between two neutral, nonpolar systems, giving van der Waals's force a quantum-mechanical origin.

- 1930s–1950s — systematic boiling-point and melting-point tables for the alkane homologous series are compiled and rationalized using dispersion-force and packing arguments, forming the modern textbook treatment used today.

Why doesn't molecular weight alone predict alkane boiling points?

Molecular weight correlates with boiling point only because larger molecules usually also have more surface area, but the correlation breaks whenever shape changes at fixed formula. Neopentane and $n$ -pentane have identical molecular weight (72.15 g/mol) yet differ in boiling point by 26.5 °C, because dispersion attraction depends on how much surface area two molecules can bring into contact, not on how many electrons or how much mass the molecule carries in total. Weight is a convenient proxy that fails precisely when isomers are compared, which is exactly why chemists use surface area/shape as the mechanistic variable instead.

Is the hydrophobic effect really an entropy effect and not an energy effect?

Mostly entropy, though the full story has both contributions. When a nonpolar alkane is inserted into liquid water, nearby water molecules cannot form hydrogen bonds toward the solute, so they reorient to preserve as many hydrogen bonds as possible with each other, forming a partially ordered "cage." This ordering lowers the entropy of the surrounding water substantially, and it is this entropic penalty — not a direct enthalpic repulsion between water and alkane — that drives alkanes to minimize their exposed surface area by aggregating together. At higher temperatures the entropic penalty becomes even more dominant, which is part of why hydrophobic aggregation (and protein folding, which relies on the same effect) is temperature-sensitive in a distinctive, non-monotonic way.

Why does alkane density plateau around 0.78–0.79 g/mL instead of increasing indefinitely with chain length?

Density depends on how much mass is packed into a given volume, and for alkanes that packing is set entirely by van der Waals contact distances between chains, which do not change much once a chain is long enough to behave like a "generic" hydrocarbon rod. Adding more CH₂ units adds mass and volume in essentially fixed proportion (each CH₂ contributes about 14 g/mol and a roughly fixed increment of van der Waals volume), so the ratio — density — converges to an asymptotic value rather than climbing forever. Because dispersion forces cannot compress a hydrocarbon lattice as tightly as ionic or hydrogen-bonded solids can, alkanes never reach the densities typical of polar or ionic substances; even solid polyethylene rarely exceeds about 0.96 g/mL even in its most crystalline form.

Does branching also affect melting point the same way it affects boiling point?

Not in the same direction, and this is a favorite exam trap. Boiling point (a liquid-state property) drops with branching because branching lowers total surface area available for dispersion contact. Melting point (a solid-state, packing property) can actually rise with certain kinds of branching if the branched isomer happens to be highly symmetric, because high symmetry can allow very efficient, tightly-fitting crystal packing despite lower overall surface area. Neopentane is the classic example: its boiling point (9.5 °C) is far below $n$ -pentane's (36.1 °C), but its melting point (−16.6 °C) is higher than $n$ -pentane's (−129.7 °C), because neopentane's high symmetry lets it pack into an unusually stable, almost close-packed cubic crystal lattice.

How does this relate to the physical properties of cycloalkanes?

Cycloalkanes follow the same dispersion-force logic but with an added geometric constraint: the ring locks the carbon skeleton into a more rigid, often more compact shape than an open chain of the same carbon count, and rings can also stack or nest with unusual regularity in the solid state. Cyclohexane, for instance, has a notably higher melting point relative to its boiling point than does $n$ -hexane, because the rigid, symmetric chair conformation of the ring packs into a crystal lattice extremely efficiently, illustrating the same packing-efficiency principle used to explain the odd–even alternation, now applied to a cyclic rather than acyclic backbone.

Why is dispersion attraction proportional to surface area rather than to volume or molecular weight?

Dispersion forces are short-ranged (they fall off as $1/r^6$ ), so only atoms actually at the interface between two molecules — within a few angstroms of each other — contribute meaningfully to the attraction; atoms buried deep inside a molecule's interior are shielded by their own neighbors and never come close enough to a different molecule to interact with it directly. Since the number of atoms sitting at an accessible interface scales with the exposed surface, not the total interior volume, the total pairwise dispersion energy summed over all contacts scales with surface area. This is precisely why compact, low-surface-area shapes (spheres, highly branched isomers) always show weaker net cohesion than elongated, high-surface-area shapes of the same volume or mass.

Do alkanes have any polar character at all that could contribute extra intermolecular force?

Essentially none that matters energetically. The C–H bond has a small electronegativity difference (C: 2.55, H: 2.20 on the Pauling scale), giving each individual bond a tiny, nearly negligible dipole, and the tetrahedral, largely symmetric arrangement of these bonds around each carbon causes most of these small bond dipoles to cancel over the molecule as a whole. What little residual dipole moment remains (alkanes typically have dipole moments under 0.1 D) is far too small to contribute a meaningful dipole–dipole interaction compared to the London dispersion forces summed over the whole molecular surface, which is why alkanes are treated, to excellent approximation, as purely dispersion-bound.

Physical Properties of Alkanes: Boiling Points and Van der Waals Forces

Where this lives today

The physical picture

The core ideas

Worked examples

Limiting cases, scope & common traps

The chemist behind it

Before this