Difficulty: Intermediate | Prerequisites: Part 1 of these notes (alkyl halide classification, intermolecular forces overview)
This is the applied half of Chapter 8. Now that you know what intermolecular forces are (Part 1), this set of notes walks through how to use them to rank real molecules by boiling point, melting point, and solubility. These are the problems that actually appear on exams: "rank these four molecules" or "explain why compound A is more soluble in water than compound B." You need to be comfortable with the three types of IMF and the concept of polarisability before tackling this material. If those feel shaky, review Part 1 first.
Boiling points are predictable because ΔS of vaporisation is roughly constant, so stronger IMFs mean higher boiling points. Melting points are trickier because ΔS of fusion varies with molecular symmetry and rigidity. Solubility in water requires the ability to form hydrogen bonds (both donating and accepting), and "like dissolves like" is the guiding principle.
Boiling point (bp)
The temperature at which a liquid's vapour pressure equals external pressure and it transitions to gas. Higher IMFs mean a higher boiling point. Boiling point is easier to predict than melting point because ΔS of vaporisation is roughly constant.
Melting point (mp)
The temperature at which a solid transitions to liquid. Harder to predict than bp because ΔS of fusion varies: it depends on how much freedom of movement the molecule gains when the crystal lattice breaks apart.
Symmetry (in the context of melting point)
More symmetrical molecules pack more efficiently into a crystal lattice and gain less freedom of movement upon melting, so they tend to have higher melting points than less symmetrical isomers.
Rigidity
How constrained a molecule's shape is in the solid state. A rigid molecule gains more freedom upon melting (large ΔS), which can lower its melting point relative to what you would predict from IMFs alone.
Solubility in water
Requires the solute to both donate an H-bond to water and accept an H-bond from water. Molecules that can do both (alcohols, amines with N–H) are more water-soluble than those that can only accept (ethers, ketones) or do neither (hydrocarbons).
Nonpolar surface area
The portion of a molecule that cannot interact favourably with water. As nonpolar surface area increases relative to polar groups, water solubility drops. Think of it as the "greasy" part of the molecule that water wants to avoid.
Example 1: Amine vs alcohol vs ether
Ranking propylamine (CH₃CH₂CH₂NH₂, bp 49 °C), ethanol-like O–H compound (bp 78 °C), and dimethyl ether-like compound (bp -24 °C):
The O–H compound boils highest. Oxygen is more electronegative than nitrogen, so O–H hydrogen bonds are stronger than N–H hydrogen bonds, even though both can hydrogen-bond.
The amine boils next. It has the same number of H-bond sites as the alcohol but N–H bonds are weaker donors.
The ether boils lowest. It can accept hydrogen bonds (lone pairs on O) but cannot donate them (no O–H or N–H), so its strongest IMF is dipole-dipole, not hydrogen bonding.
Key insight: a small difference in electronegativity between N and O creates a big difference in hydrogen-bond strength.
Example 2: Nonpolar hydrocarbon vs H-bonding molecule
A nonpolar hydrocarbon with higher molecular mass can still have a higher boiling point than a smaller hydrogen-bonding molecule, if the LDF from its large electron cloud outweigh the hydrogen bonds of the smaller molecule. But for molecules of similar size, hydrogen bonding wins.
Example 3: Methyl halides (CH₃F, CH₃Cl, CH₃Br, CH₃I)
Boiling points increase: CH₃F (-78 °C) < CH₃Cl (-24 °C) < CH₃Br (4 °C) < CH₃I (42 °C)
As you go from F to I, molecular mass and polarisability increase, strengthening LDF.
Meanwhile, the dipole moment decreases (C–F is the most polar bond, C–I the least).
LDF dominate here: increasing molecular mass wins over decreasing dipole.
Summary: increasing molecular mass correlates with increasing LDF, and increasing dipole correlates with decreasing polarisability in this series.
Example 4: Alcohol vs ether vs thiol vs thioether
Comparing propanol (bp 78 °C), methyl ethyl ether (bp -24 °C), ethanethiol (bp 35 °C), and dimethyl sulfide (bp 37 °C):
Propanol boils highest: strong O–H hydrogen bonding.
Methyl ethyl ether boils lowest among the oxygen-containing pair: no H-bond donation, only dipole-dipole.
Ethanethiol and dimethyl sulfide boil similarly (~35-37 °C). S–H is a weak hydrogen-bond donor (sulfur is not electronegative enough for strong H-bonds), so thiol hydrogen bonding is weak. The thioether relies on London dispersion. Their boiling points end up close because the weak S–H hydrogen bond and the LDF of the thioether are comparable in strength.
Example 5: Boiling points vs melting points (the disconnect)
Four molecules with similar structures can have boiling points that track their IMF strength in a clean order, but their melting points scramble that order. This happens because ΔS of fusion varies with symmetry and rigidity. Boiling points are easier to predict because ΔS of vaporisation is roughly constant.
Melting points are harder to predict than boiling points. Here is why, and what you can say about them.
Why melting points are tricky
At the melting point, T(mp) = ΔH(fus) / ΔS(fus).
Unlike ΔS(vap), ΔS(fus) is not constant. It depends on how much freedom of movement a molecule gains when it leaves the crystal lattice.
Two factors govern ΔS(fus): symmetry and rigidity.
Symmetry raises the melting point
A more symmetrical molecule packs more tightly into a crystal. When it melts, it gains relatively less freedom (it was already somewhat free to rotate in place), so ΔS(fus) is smaller. A smaller denominator in T = ΔH/ΔS means a higher melting point.
Rigidity matters
A rigid molecule (e.g. one with a ring or multiple double bonds) gains a lot of freedom upon melting, because in the solid it was locked in place. This large ΔS(fus) can lower the melting point relative to what you would predict from enthalpy alone.
General trends
Melting points tend to correlate loosely with boiling points, but not reliably.
More symmetrical molecules tend to have higher melting points than less symmetrical isomers.
Rigidity and how much freedom is gained when the solid melts are very important, and hard to predict without data.
Bottom line for exams: if asked to rank melting points, state that they are harder to predict than boiling points because ΔS of fusion varies. If the question gives you a symmetric vs asymmetric pair, the symmetric one likely melts higher.
The guiding principle: like dissolves like
Substances mix favourably when the solute–solvent IMFs are similar in type and strength to the solute–solute and solvent–solvent IMFs they replace.
Solubility in water specifically
Water is a strong hydrogen-bond donor and acceptor. To dissolve well in water, a molecule must:
Donate an H-bond to water (requires O–H, N–H, or similar)
Accept an H-bond from water (requires lone pairs on O, N, or F)
Molecules that can do both (alcohols, carboxylic acids, amines with N–H) are the most water-soluble. Molecules that can only accept (ethers, ketones) are less soluble. Molecules that can do neither (hydrocarbons, alkyl halides) are essentially insoluble.
Fluorine is a poor partner
C–F bonds make a molecule polar, but fluorine on carbon is not a good hydrogen-bond acceptor and carbon-bonded F does not share electrons well. Fluorinated compounds are often less soluble in water than you would expect from their polarity alone. "F is not polarisable" and "doesn't share electrons well" are the phrases from the lecture notes.
Alcohol vs fluoride solubility
Comparing an alcohol (–OH) with a fluoride (–F) of similar size: the alcohol is more water-soluble. The O–H group both donates and accepts hydrogen bonds with water, while C–F is a weak acceptor and cannot donate.
Oxygen vs sulfur for H-bonding
Comparing an ether (C–O–C) with a thioether (C–S–C): the ether is more water-soluble because oxygen is a better hydrogen-bond acceptor than sulfur. Oxygen's lone pairs are more accessible and its electronegativity is higher.
Nonpolar surface area effect
As the hydrocarbon chain grows, the nonpolar surface area increases relative to the polar functional group. A short-chain alcohol (methanol, ethanol) is miscible with water. A long-chain alcohol (octanol) is not, despite still having an –OH group. The nonpolar tail dominates.
From the lecture notes: a branched alcohol is more water-soluble than a straight-chain alcohol of the same molecular formula, because branching reduces nonpolar surface area and makes the –OH group relatively more prominent.
Melting point:
T_{mp} = \frac{\Delta H_{fus}}{\Delta S_{fus}}Unlike boiling, ΔS(fus) is not constant, which is why melting points are harder to predict.
Solubility (same as Part 1, applied here):
\Delta G_s = \Delta H_{mix} - T \Delta S_{mix}ΔS(mix) > 0 always. Favourable mixing requires that ΔH(mix) is not too large and positive.
The solubility rules here are the reason your laundry detergent has both a polar head and a nonpolar tail: the polar end dissolves in water, the nonpolar end grabs grease, and the whole assembly bridges the two. Pharmacologists use the same logic when designing prodrugs that are water-soluble enough to reach the bloodstream but lipophilic enough to cross cell membranes. Melting-point data is used routinely in labs to confirm the identity and purity of a synthesised compound; an impure sample melts lower and over a wider range.
Students often think melting points should follow the same order as boiling points. They frequently do not, because ΔS of fusion varies with symmetry and rigidity while ΔS of vaporisation is roughly constant.
Students assume that any molecule with a polar bond is soluble in water. A molecule can have polar bonds (like C–F) and still be poorly water-soluble if it cannot donate or accept hydrogen bonds effectively.
Students confuse hydrogen-bond donation with hydrogen-bond acceptance. Ethers can accept H-bonds (lone pairs on O) but cannot donate them (no O–H). You need both for strong water solubility.
Students forget the nonpolar surface area effect. A long-chain alcohol has an –OH group, but the long hydrocarbon tail makes it behave more like a hydrocarbon than like methanol.
⚠️ "Rank these molecules by boiling point" is one of the most common exam formats in this chapter. Practise the three-step hierarchy (hydrogen bonding, dipole-dipole, LDF) until it is automatic.
⚠️ If an exam question asks you to rank melting points, note that this is harder and state why (ΔS of fusion varies). Then use symmetry as your tiebreaker.
⚠️ Solubility questions often ask you to compare an alcohol with an ether, or an O-containing compound with an S-containing one. Know the hydrogen-bond donor/acceptor distinction cold.
⚠️ The CH₃F < CH₃Cl < CH₃Br < CH₃I boiling-point series is a favourite exam example. Be ready to explain why LDF win over dipole in this case.
True or false: Melting points are easier to predict than boiling points.
False. Boiling points are easier because ΔS of vaporisation is roughly constant.
Fill in the blank: To dissolve in water, a molecule must both ____ and ____ hydrogen bonds.
Donate and accept.
True or false: A branched alcohol is less water-soluble than a straight-chain alcohol of the same formula.
False. Branching reduces nonpolar surface area, making it more water-soluble.
Fill in the blank: More symmetrical molecules tend to have ____ melting points.
Higher.
True or false: CH₃I has a higher boiling point than CH₃F because C–I has a larger dipole than C–F.
False. C–F has the larger dipole. CH₃I boils higher because of stronger London dispersion forces (more electrons, higher polarisability).
Q: Rank the following by boiling point (lowest to highest): propan-1-ol, methoxyethane (methyl ethyl ether), propanethiol, dimethyl sulfide.
A: Methoxyethane < dimethyl sulfide ≈ propanethiol < propan-1-ol. The ether has no H-bond donation capability. The sulfur compounds have weak H-bonding (thiol) or rely on LDF (thioether), so they boil similarly. The alcohol boils highest due to strong O–H hydrogen bonding.
Q: Why does neopentane (2,2-dimethylpropane) have a higher melting point than n-pentane, even though n-pentane has the higher boiling point?
A: Neopentane is highly symmetrical (tetrahedral shape) and packs efficiently into a crystal. Its ΔS of fusion is small, raising T(mp). n-Pentane is a flexible chain with more surface area (higher bp due to LDF), but it packs poorly in the solid and gains a lot of freedom on melting (large ΔS(fus)), so its melting point is lower.
Q: Predict whether 1-butanol or diethyl ether is more soluble in water. Explain.
A: 1-Butanol is more soluble. It has an –OH group that both donates and accepts hydrogen bonds with water. Diethyl ether can accept H-bonds (lone pairs on O) but cannot donate, so it interacts less strongly with water.
Q: Explain why perfluorohexane (C₆F₁₄) is essentially insoluble in water despite being highly polar.
A: Although C–F bonds are polar, fluorine on carbon is a poor hydrogen-bond acceptor and cannot donate H-bonds. Perfluorohexane also has low polarisability. It cannot replace the strong water–water hydrogen bonds it would disrupt, so ΔH of mixing is large and positive, making ΔG positive.
Q: Two structural isomers both have the formula C₅H₁₂O. One is 1-pentanol and the other is 2-methyl-2-butanol (tert-pentyl alcohol). Which is more soluble in water?
A: 2-Methyl-2-butanol. Its branched structure gives it less nonpolar surface area relative to its –OH group, so the polar group has more influence on solubility. 1-Pentanol has a longer, more exposed hydrocarbon chain.
The boiling-point ranking skills here are used again in distillation and separation techniques (lab courses). Solubility principles connect to extraction in the lab: you choose solvents based on "like dissolves like." The melting-point discussion ties back to crystal packing and X-ray crystallography concepts in advanced courses. In biochemistry, the balance between polar and nonpolar surface area determines protein folding, membrane permeability, and drug bioavailability.
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