Hydrogen Bonding
Hydrogen bonding is a special type of intermolecular (and sometimes intramolecular) dipole–dipole attraction that occurs when a hydrogen atom is covalently bonded to a highly electronegative atom — specifically nitrogen (N), oxygen (O), or fluorine (F) — and is simultaneously attracted to a lone pair on another electronegative atom nearby. Because N, O, and F are so electronegative, the shared electrons in the X–H bond are pulled strongly toward the electronegative atom, leaving the hydrogen nucleus almost bare. This creates a large partial positive charge (δ+) on H and a large partial negative charge (δ−) on the electronegative atom, producing a bond dipole far stronger than an ordinary dipole–dipole interaction. The resulting electrostatic attraction between the δ+ hydrogen of one molecule and the lone pair of an N, O, or F atom on a neighbouring molecule is called a hydrogen bond, often written X–H···Y, where X is the atom bonded to H, and Y is the atom donating the lone pair. Hydrogen bonds are typically 10–40 kJ mol⁻¹ in strength — about 5–10 times weaker than a typical covalent bond (~350 kJ mol⁻¹) but 5–10 times stronger than ordinary London dispersion forces between small molecules. This intermediate strength makes hydrogen bonds enormously important in biology (DNA base pairing, protein α-helices), materials science, and everyday chemistry. The physical consequences of hydrogen bonding are dramatic and measurable. Water (H₂O) has a boiling point of 100 °C, far above what its molar mass would predict compared to H₂S (−60 °C), H₂Se (−41 °C), and H₂Te (+2 °C), which lack O–H groups and cannot form strong hydrogen bonds. Similarly, hydrogen bonding between solute and solvent molecules dramatically increases solubility: ethanol (CH₃CH₂OH) is infinitely miscible with water because both molecules form O–H···O hydrogen bonds with each other, while diethyl ether (CH₃CH₂OCH₂CH₃) is far less soluble because only weak C–H interactions remain after the ether oxygen's lone pairs have been saturated.
- The Three Requirements for Hydrogen Bond Formation
- Intermolecular vs Intramolecular Hydrogen Bonds
- Effect on Boiling Point
- Effect on Solubility
The Three Requirements for Hydrogen Bond Formation
A hydrogen bond forms ONLY when three conditions are simultaneously met: (1) A hydrogen atom must be covalently bonded directly to N, O, or F — no other elements qualify because their electronegativity is insufficient to polarise the X–H bond strongly enough. Carbon, for example, has electronegativity 2.5, far below N (3.0), O (3.5), and F (4.0). (2) The H atom must carry a significant δ+ charge resulting from the bond polarisation. (3) A neighbouring molecule (or a distant part of the same molecule) must provide a lone pair on an N, O, or F atom to act as the hydrogen bond acceptor. The geometry is nearly linear: X–H···Y bond angles are typically 160–180°, maximising orbital overlap between the H and the lone pair.
Intermolecular vs Intramolecular Hydrogen Bonds
Intermolecular hydrogen bonds form between two separate molecules and are responsible for raising boiling points and increasing viscosity. Classic examples: water (each molecule can donate two O–H bonds and accept two lone pairs, forming a tetrahedral network), HF (chains: F–H···F–H···F), and ammonia (N–H···N). Intramolecular hydrogen bonds form within a single molecule when a suitable H donor and lone-pair acceptor are geometrically close. A classic example is 2-nitrophenol: the –OH group and the adjacent –NO₂ group form an internal O–H···O bond, creating a six-membered ring. This internal bond satisfies the H's attraction without needing a neighbouring molecule, so 2-nitrophenol has a LOWER boiling point (216 °C) than 4-nitrophenol (279 °C), where the –OH and –NO₂ groups are far apart and must form intermolecular H-bonds instead. Intramolecular H-bonds also reduce aqueous solubility because the molecule is internally satisfied and cannot form as many H-bonds with water.
Effect on Boiling Point
Boiling a liquid requires supplying enough energy to overcome intermolecular forces. Hydrogen bonds add to the energy required, raising the boiling point significantly above predictions based on molar mass alone. Comparing Group 16 hydrides: H₂O (BP 100 °C, MM 18 g mol⁻¹) vs H₂S (BP −60 °C, MM 34 g mol⁻¹). Despite being lighter, water boils 160 °C higher because of its extensive O–H···O network. Each mole of water forms roughly 2 hydrogen bonds per molecule simultaneously (network solid-like structure even in liquid phase). The same trend appears in Group 15: NH₃ (BP −33 °C) vs PH₃ (BP −87 °C), and Group 17: HF (BP 19 °C) vs HCl (BP −85 °C). In all cases, only the Period 2 hydride (the one with N, O, or F) deviates anomalously upward.
Effect on Solubility
The rule 'like dissolves like' operates largely through intermolecular forces. A solute dissolves in a solvent when solute–solvent intermolecular attractions are comparable to or stronger than solute–solute and solvent–solvent forces. Hydrogen bonding between solute and water greatly increases aqueous solubility. Methanol (CH₃OH), ethanol (C₂H₅OH), propanol (C₃H₇OH), and glucose (C₆H₁₂O₆, with five –OH groups) are all highly or infinitely miscible with water because they form O–H···O bonds with water molecules. As the non-polar hydrocarbon portion of an alcohol grows (butanol, pentanol…), solubility decreases because the non-polar tail disrupts the water H-bond network without replacing it with equivalent attractions — an entropic and enthalpic penalty. Proteins and DNA are soluble in water partly because their amide (N–H) and hydroxyl (O–H) groups form extensive H-bonds with surrounding water molecules.
The Key Equation
- \delta^+ — Partial positive charge (dimensionless): Charge on the hydrogen atom resulting from electron density being pulled toward the electronegative O, N, or F atom.
- \delta^- — Partial negative charge (dimensionless): Charge on the electronegative donor/acceptor atom (N, O, or F) due to its high electronegativity withdrawing electron density.
- E_{HB} — Hydrogen bond energy (kJ mol⁻¹): Energy required to break one mole of hydrogen bonds; typically 10–40 kJ mol⁻¹ for O–H···O and N–H···N interactions.
- \theta — Bond angle X–H···Y (degrees (°)): The angle at the hydrogen atom between the covalent bond and the hydrogen bond; strongest when close to 180° (linear geometry).
Worked Example
Problem. Explain why propan-1-ol (CH₃CH₂CH₂OH, MM = 60 g mol⁻¹) has a boiling point of 97 °C, while its structural isomer methoxyethane (CH₃OCH₂CH₃, MM = 60 g mol⁻¹) boils at only 11 °C. Both molecules have identical molar masses, so what accounts for the 86 °C difference?
Answer. Propan-1-ol has O–H groups enabling intermolecular hydrogen bonds (~20 kJ mol⁻¹ each) between molecules. Methoxyethane has no O–H groups and cannot donate hydrogen bonds, so only weaker London and dipole–dipole forces act between its molecules. The extra energy needed to break H-bonds raises propan-1-ol's BP by ~86 °C.
Origins · 1 minLinus Pauling and how this was discovered
Pauling pioneered the modern understanding of chemical bonding, including hydrogen bonds, in his landmark 1939 textbook 'The Nature of the Chemical Bond.' He formally defined the criteria for hydrogen bond formation, showed that the bond is primarily electrostatic in origin, and demonstrated its critical role in protein secondary structure (the α-helix and β-sheet). He also used hydrogen bonding theory to explain the anomalous properties of water and nucleic acid structure. He received the Nobel Prize in Chemistry in 1954 for his work on the nature of the chemical bond. Pauling is one of only four people to have won two Nobel Prizes — Chemistry (1954) and Peace (1962) — and the only person to have won two unshared Nobel Prizes.
Related Concepts
- London Dispersion Forces — London dispersion forces are the weakest intermolecular force and act alongside hydrogen bonds in all molecules; understanding their relative weakness helps explain why hydrogen bonding causes anomalously high boiling points.
- Dipole–Dipole Interactions — Hydrogen bonds are a special, stronger case of dipole–dipole interaction; understanding ordinary dipole–dipole forces provides the conceptual foundation for recognising why X–H···Y bonds are uniquely strong.
- Electronegativity and Bond Polarity — The requirement that H must be bonded to N, O, or F is directly explained by the high electronegativity of these atoms, which creates the large δ+ on H necessary for hydrogen bonding.
- Molecular Geometry (VSEPR Theory) — The geometry of lone pairs on N, O, and F — determined by VSEPR theory — controls the directionality and strength of hydrogen bonds, including why intramolecular H-bonds require specific ring geometries.
- Solubility and the Like-Dissolves-Like Rule — Hydrogen bonding between solute and solvent is the primary driver of aqueous solubility for polar, hydroxyl-containing compounds such as alcohols, sugars, and amino acids.