You are viewing a free preview of this lesson.
Subscribe to unlock all 10 lessons in this course and every other course on LearningBro.
Hydrogen bonding is the most important intermolecular force in chemistry and biology. While the previous lesson introduced the basic concept, this lesson explores the profound consequences of hydrogen bonding — from the anomalous properties of water to the structure of DNA. Understanding these applications is essential for A-Level Chemistry.
Water is often described as the most anomalous compound known. Its properties deviate dramatically from what we would predict based on its molecular mass alone, and almost all of these anomalies are explained by hydrogen bonding.
If we extrapolate the trend in boiling points of Group 6 hydrides (H₂Te → H₂Se → H₂S), we would predict that H₂O should boil at approximately −80°C. Instead, it boils at 100°C — about 180°C higher than expected.
This enormous difference is due to the extensive hydrogen bonding network in liquid water. Each water molecule can form up to four hydrogen bonds: two through its two δ+ hydrogen atoms (donating to lone pairs on neighbouring oxygens) and two through the two lone pairs on its oxygen atom (accepting from δ+ hydrogens on other molecules). This creates a dynamic, three-dimensional network of hydrogen bonds that requires substantial energy to disrupt.
This is one of the most unusual properties of any substance. In almost all materials, the solid is denser than the liquid. Water is different.
When water freezes, the hydrogen bonds lock the molecules into a regular, open hexagonal lattice structure. Each oxygen atom is tetrahedrally surrounded by four other molecules, held at fixed distances by hydrogen bonds. This open structure contains large voids, making ice about 9% less dense than liquid water.
When ice melts, some of these hydrogen bonds break, allowing molecules to move closer together and fill some of the voids. The water becomes denser. Maximum density occurs at 4°C, after which thermal expansion outweighs the effect of collapsing structure.
This property is critically important for aquatic life: ice floats, forming an insulating layer on top of lakes and oceans that prevents them from freezing solid.
Water has an exceptionally high specific heat capacity (4.18 J g⁻¹ K⁻¹). This means it requires a large amount of energy to raise its temperature. The reason is that energy supplied to water goes partly into breaking and rearranging hydrogen bonds, and only the remainder goes into increasing the kinetic energy (temperature) of the molecules.
This property makes water an excellent coolant and temperature buffer. Oceans moderate coastal climates because they absorb and release large amounts of heat with relatively small temperature changes.
Water has an unusually high surface tension compared to other liquids. Molecules at the surface form hydrogen bonds with molecules below and beside them, but not above (there is air above). This creates a net inward pull that minimises the surface area, producing surface tension. The strong hydrogen bonds make this inward force particularly large.
| Group | First hydride | bp / °C | Second hydride | bp / °C | Anomaly / °C |
|---|---|---|---|---|---|
| 5 | NH₃ | −33 | PH₃ | −88 | +55 |
| 6 | H₂O | 100 | H₂S | −60 | +160 |
| 7 | HF | 19.5 | HCl | −85 | +105 |
Water shows the largest anomaly because each molecule can form up to 4 hydrogen bonds (2 donated, 2 accepted). NH₃ can form only 2 (1 donated via N–H, 1 accepted via lone pair — though it has 3 N–H bonds, it has only 1 lone pair to accept). HF can form 2 (1 donated, 1 accepted — though it has 3 lone pairs, it has only 1 H–F to donate).
Comparing alcohols (R–OH) with ethers (R–O–R) of similar molecular mass illustrates hydrogen bonding beautifully.
Ethanol (C₂H₅OH, bp 78°C) has an O–H group that can both donate and accept hydrogen bonds. The δ+ hydrogen on one molecule attracts the lone pair on the oxygen of another, creating a chain or network of hydrogen bonds.
Methoxymethane (CH₃OCH₃, bp −24°C) has the same molecular formula (C₂H₆O) but no O–H bond. While the oxygen has lone pairs that could accept a hydrogen bond, there is no sufficiently δ+ hydrogen to donate one. Methoxymethane can form hydrogen bonds with water (using its lone pairs) but cannot form hydrogen bonds with itself. It relies on weaker dipole–dipole and London forces for intermolecular attraction.
The 102°C difference in boiling point is directly attributable to the presence or absence of hydrogen bonding.
| Compound | Formula | Mr | Functional group | Intermolecular forces | bp / °C |
|---|---|---|---|---|---|
| Ethanol | C₂H₅OH | 46 | Alcohol (–OH) | London + dipole + H-bonding | 78 |
| Methoxymethane | CH₃OCH₃ | 46 | Ether (–O–) | London + dipole | −24 |
| Ethanal | CH₃CHO | 44 | Aldehyde (C=O) | London + dipole | 20 |
| Propane | C₃H₈ | 44 | Alkane | London only | −42 |
This table shows clearly how adding stronger intermolecular forces progressively raises the boiling point, even when Mr is similar.
The ability to form hydrogen bonds with water is a key factor in determining whether a substance dissolves in aqueous solution.
Methanol, ethanol, and propanol are all completely miscible with water. They can form hydrogen bonds with water molecules, replacing the water–water hydrogen bonds that are broken when the solute dissolves. The energy released in forming new solute–water hydrogen bonds compensates for the energy required to break water–water hydrogen bonds.
As the carbon chain in alcohols gets longer, solubility in water decreases. Butanol is partially soluble, and hexanol is almost insoluble. This is because the non-polar hydrocarbon chain disrupts the hydrogen bonding network of water without forming hydrogen bonds itself. The larger the non-polar portion, the greater the disruption and the lower the solubility.
Non-polar molecules like hexane and cyclohexane are insoluble in water. Dissolving them would break water–water hydrogen bonds without forming new interactions of comparable strength. The energy cost is too high.
The structure of DNA relies fundamentally on hydrogen bonding between complementary base pairs:
Adenine (A) pairs with Thymine (T) through two hydrogen bonds. Adenine has an N–H group that donates to a C=O lone pair on thymine, and thymine has an N–H group that donates to a nitrogen lone pair on adenine.
Guanine (G) pairs with Cytosine (C) through three hydrogen bonds. The extra hydrogen bond makes G–C pairs slightly more stable than A–T pairs.
These hydrogen bonds are crucial because they are:
Proteins fold into specific three-dimensional shapes determined partly by hydrogen bonding:
Alpha helices are stabilised by hydrogen bonds between the C=O group of one amino acid and the N–H group of an amino acid four residues further along the chain.
Beta sheets are stabilised by hydrogen bonds between C=O and N–H groups on adjacent strands of the polypeptide.
Tertiary structure involves hydrogen bonds between R-groups of amino acids that may be far apart in the primary sequence but close together in the folded structure.
The specificity and directionality of hydrogen bonds are critical for protein function — enzymes work because they have precisely shaped active sites maintained by these interactions.
Not all hydrogen bonds are equal. The strength depends on:
| Interaction | Energy / kJ mol⁻¹ | Context |
|---|---|---|
| O–H···O | ~20 | Water, alcohols |
| O–H···N | ~25 | Alcohol with amine |
| N–H···O | ~15 | Amides, proteins |
| N–H···N | ~12 | DNA base pairs |
| F–H···F | ~29 | Hydrogen fluoride |
For comparison, a typical covalent O–H bond is ~464 kJ mol⁻¹, and a typical London force between small molecules is ~1–5 kJ mol⁻¹. Hydrogen bonds sit between these extremes.
Common exam mistake: Students sometimes write that "hydrogen bonding is the strongest intermolecular force, so molecules with hydrogen bonding always have the highest boiling points." This is not always true. A large non-polar molecule like I₂ (mp 114°C) can have a higher melting point than a small molecule with hydrogen bonding like NH₃ (mp −78°C) because the cumulative London forces in I₂ are very strong. Always consider molecular size alongside the type of intermolecular force.
Edexcel 9CH0 specification, Paper 1 — Advanced Inorganic and Physical Chemistry, Topic 2: Bonding and Structure, sub-topic 2.6 covers intermolecular forces, with hydrogen bonding treated as the strongest of the permanent dipole–dipole class. Candidates must recognise the requirement for a hydrogen atom covalently bonded to a highly electronegative atom (N, O or F) and a lone pair on a neighbouring N, O or F. Cross-paper appearances: Paper 1 Topic 4 (Inorganic chemistry — periodicity) demands explanation of the anomalous boiling points of period 2 hydrides H₂O, HF and NH₃ relative to their respective groups; Paper 2 (Advanced Organic Chemistry) revisits hydrogen bonding when discussing solubility of alcohols, the dimerisation of carboxylic acids in non-polar solvents, and the influence of N–H bonding on amide and amine behaviour; Paper 3 (General and Practical Principles) uses hydrogen bonding synoptically — typical questions ask candidates to predict whether an unfamiliar species can hydrogen-bond, and to interpret IR and physical-property data accordingly (refer to the official Pearson Edexcel specification document for exact wording).
The topic is high-yield: hydrogen bonding appears almost every series, often as a 4–7 mark explanation question rather than a recall task. Examiners expect candidates to handle the concept at three levels — definition, structural drawing with partial charges and lone pairs, and application to physical-property data.
Question (7 marks):
(a) The boiling point of water (H₂O) is 100°C, while the boiling point of hydrogen sulfide (H₂S) is −60°C. Both molecules are bent triatomics with similar permanent dipoles. Explain why H₂O has a much higher boiling point. (3)
(b) Explain why ice floats on liquid water. (2)
(c) The boiling point of hydrogen fluoride (HF) is 19.5°C, lower than that of water (100°C), even though fluorine is more electronegative than oxygen. Explain this observation. (2)
Solution with mark scheme:
(a) M1 — Both molecules have permanent dipole–dipole and London dispersion forces, but H₂O molecules also form hydrogen bonds because hydrogen is bonded to oxygen, a highly electronegative atom (with a lone pair on O of a neighbouring molecule).
M2 — Sulfur is not sufficiently electronegative to support hydrogen bonding, so H₂S has only permanent dipole–dipole and London forces. (Acceptable equivalent: "S–H is not polar enough; sulfur lacks the small-radius, high-electronegativity character of N, O and F.")
A1 — Hydrogen bonds are significantly stronger than the dipole–dipole and London forces in H₂S, so much more energy is required to overcome them, raising the boiling point.
Common error: candidates write "H₂O has hydrogen bonds; H₂S doesn't" without identifying why (the electronegativity / lone-pair requirement). That earns M1 only.
(b) B1 — In ice, each water molecule forms (on average) four hydrogen bonds, two as donor and two as acceptor, arranged in a tetrahedral geometry around oxygen.
B1 — This rigid open lattice contains more empty space per molecule than the disordered, partially collapsed liquid, so ice is less dense and floats.
(c) M1 — A water molecule has two O–H bonds and two lone pairs, allowing it to participate in approximately four hydrogen bonds per molecule (2 donor + 2 acceptor).
A1 — HF has only one H atom (one donor) although three lone pairs on F, so it forms on average only about two hydrogen bonds per molecule. Although each individual H···F bond is slightly stronger than each individual H···O bond, the total hydrogen-bond network in liquid water is more extensive, giving water the higher boiling point.
Total: 7 marks (M3 A2 B2 as shown).
Question (6 marks): Methanol (CH₃OH), ethanol (C₂H₅OH) and propan-1-ol (C₃H₇OH) are completely miscible with water, but butan-1-ol (C₄H₉OH) is only partly miscible, and longer-chain alcohols such as octan-1-ol are essentially insoluble. Explain this trend in terms of intermolecular forces.
Mark scheme decomposition by AO:
Total: 6 marks split AO1 = 2, AO2 = 2, AO3 = 2. This is a classic "trend with explanation" question — examiners reward candidates who explicitly link the structural change (chain length) to a competing-forces argument rather than simply restating the data.
Hydrogen bonding is one of the most synoptically connected topics in 9CH0:
Topic 6/7 (alcohols and halogenoalkanes): the solubility of alcohols, esters and amides in water depends almost entirely on whether the polar –OH, C=O or N–H group can sustain enough hydrogen bonding to compensate for the disruption of water's own network. This explains why methanol mixes freely with water but methanethiol (CH₃SH) does not, and why ethyl ethanoate has limited but non-zero water solubility (its C=O accepts H-bonds from water).
Topic 17 (carboxylic acids): carboxylic acids form cyclic dimers in non-polar solvents (and partially in the gas phase), bound by two O–H···O=C hydrogen bonds in a planar eight-membered ring. This dimerisation roughly doubles the apparent relative molecular mass, as observed historically in freezing-point-depression experiments. It also explains why carboxylic acids have markedly higher boiling points than alcohols of similar Mr.
Topic 16 (carbonyl compounds): aldehydes and ketones cannot hydrogen-bond to themselves (they have no O–H or N–H to donate) but their C=O lone pairs can accept hydrogen bonds from water and alcohols. This is why propanone is fully miscible with water despite having no donor capability — examiners reward candidates who distinguish donor and acceptor roles.
Biological applications (often appearing in synoptic Paper 3 contexts): DNA base pairing relies on hydrogen bonds — adenine and thymine pair via two hydrogen bonds, while guanine and cytosine pair via three, contributing to the higher thermal stability of GC-rich regions. In proteins, the α-helix and β-sheet secondary structures are stabilised by hydrogen bonds along the polypeptide backbone (N–H to C=O). These applications are not assessed for biochemical detail, but candidates may be asked to identify donor and acceptor atoms in a printed structure.
Subscribe to continue reading
Get full access to this lesson and all 10 lessons in this course.