Alcohols, Phenols and Ethers · Lesson 2 of 12
Bonding, Hydrogen Bonds and Physical Properties
“Connect oxygen geometry and intermolecular interactions with boiling points, aqueous solubility and steam separation.”
• Explain geometry and bond-length differences around oxygen. • Compare boiling points using hydrogen bonding, size and branching. • Predict water solubility from polar and nonpolar molecular regions. • Distinguish intramolecular and intermolecular hydrogen bonding.
A bent oxygen centre with polar bonds
Oxygen has six valence electrons. In a neutral alcohol or ether, it normally makes two single bonds and retains two lone pairs. Four electron-pair regions give an approximately tetrahedral electron-pair arrangement, but the two visible bonds form a bent shape. The oxygen is not a straight connector. Since oxygen attracts bonding electrons more strongly than carbon or hydrogen, C–O and O–H bonds are polar: oxygen carries partial negative character, while an OH hydrogen carries partial positive character.
In a localised bonding model, an alcohol has an sp³ carbon bonded to oxygen through a sigma bond. Methanol has an approximate C–O length of 142 pm and a C–O–H angle of 108.9°. Dimethyl ether has C–O near 141 pm and C–O–C near 111.7°. Lone pairs, substituent repulsions and the actual electron distribution all affect angles; “tetrahedral” does not mean that every measured angle equals 109.5°. These representative values should not be applied as exact constants to every member of the class.
Phenol has oxygen attached to an sp² ring carbon. A lone pair can overlap with the aromatic π system, giving the C–O bond partial double-bond character; the sp² carbon also contributes to a shorter bond. The approximate phenolic C–O length is 136 pm, shorter than methanol’s. Resonance describes one delocalised molecule, not rapid switching between independently existing bond pictures. This distinction will explain both acidity and ring activation.
Why ethanol boils far above its ether isomer
Boiling separates molecules into a vapour; it does not normally break their internal C–O bonds. The energy requirement depends on intermolecular attractions and the entropy change of vaporisation. All molecules have dispersion attractions. Polar molecules also have dipole interactions. Alcohols and phenols can additionally form extended hydrogen-bond networks because each molecule has an O–H donor and oxygen lone-pair acceptors.
A hydrogen bond links an OH hydrogen on one molecule with a lone pair on another oxygen. In a diagram, a dashed O···H contact represents this intermolecular attraction; the solid O–H line is a covalent bond. Methoxymethane contains oxygen lone pairs but has no O–H group, so pure methoxymethane cannot build the corresponding donor–acceptor network with itself. It remains polar; lack of self hydrogen bonding does not mean lack of all intermolecular attraction.
| Compound | Approximate normal boiling point | Main comparison |
|---|---|---|
| Propane | 231 K | Dispersion; no oxygen or OH |
| Methoxymethane | 248 K | Polar ether; no OH donor |
| Ethanol | 351 K | Alcohol hydrogen-bond network |
| n-Pentane | 309.1 K | Nonpolar reference of comparable mass to ethoxyethane |
| Ethoxyethane | 307.6 K | Ether close to the hydrocarbon value |
| Butan-1-ol | 390 K | OH hydrogen bonding raises boiling temperature |
Compare like with like. Along a straight-chain alcohol series, increasing carbon number increases the size and polarizability of the electron cloud, usually increasing boiling point. Among alcohols with the same formula, branching generally lowers boiling point because compact molecules make less extensive contact. This is a trend, not a universal numerical law: functional group position and the details of association also matter. A very large ether need not boil below a very small alcohol merely because it lacks OH.
Problem
Arrange methanol, ethanol, propan-1-ol, butan-2-ol, butan-1-ol and pentan-1-ol in increasing boiling point.
- 1.All can self hydrogen-bond, so first compare the hydrocarbon portions. A longer unbranched chain normally gives stronger dispersion and a higher boiling point.
- 2.For the two butanols, the more internally placed OH and molecular shape lower the boiling point of butan-2-ol relative to butan-1-ol.
- 3.The order is methanol < ethanol < propan-1-ol < butan-2-ol < butan-1-ol < pentan-1-ol. The ranking combines related molecular comparisons rather than assuming molecular mass is the only variable.
Problem
Order n-butane, ethoxyethane, pentanal and pentan-1-ol by increasing boiling point.
- 1.n-Butane has only dispersion and is small enough to boil lowest. Ethoxyethane is heavier and polar but cannot self hydrogen-bond.
- 2.Pentanal has a larger carbon skeleton and a strongly polar carbonyl group. Pentan-1-ol has a similar-sized skeleton plus self hydrogen bonding.
- 3.The order is n-butane < ethoxyethane < pentanal < pentan-1-ol. The explanation compares size, dipole interactions and association together.
Dissolving requires replacing old interactions
When an alcohol dissolves in water, some water–water and alcohol–alcohol contacts are replaced by water–alcohol contacts. Alcohol oxygen can accept a hydrogen bond from water, and its OH can donate one to water. Small alcohols such as methanol and ethanol mix with water in all proportions. As the nonpolar alkyl region grows, its unfavourable exposure to water increasingly offsets the benefit of the OH group, so water solubility generally decreases.
An ether can accept hydrogen bonds from water even though it cannot donate them. Water supplies the O–H donor that the ether lacks. Thus an ether can have a low boiling point yet appreciable water solubility: the pure-liquid and mixed-liquid interactions are different questions. At comparable molecular mass, ethoxyethane and butan-1-ol have limited but broadly similar water solubilities, approximately 7.5 and 9 g per 100 mL water in the tabulated comparison. These are not unlimited miscibilities. Solubility depends on temperature, and these approximate values do not establish equality for every ether/alcohol pair.
Phenol also hydrogen-bonds to water, but its six-carbon aromatic region limits neutral-phenol solubility. Converting phenol to an ionic phenoxide in alkaline solution can greatly increase its apparent solubility; that involves a chemical acid–base reaction in addition to neutral dissolution. More hydroxyl groups can improve water compatibility, but their number must be weighed against the entire nonpolar skeleton and any intramolecular association.
One molecule can satisfy its own hydrogen bond
In 2-nitrophenol, OH lies beside NO₂, allowing the OH hydrogen to interact with a nitro oxygen within the same molecule. This intramolecular hydrogen bond reduces its ability to form the same extended intermolecular network. In 4-nitrophenol the groups are too far apart for that local contact, and intermolecular association is stronger. Consequently 2-nitrophenol is more steam volatile and can be carried over preferentially in a steam separation of the two isomers.
Steam distillation is not the same as heating each pure substance to its normal boiling point. The vapour pressures of the components contribute to the total pressure; a sufficiently steam-volatile component can distil with water at a lower temperature than its own normal boiling point. Hydrogen bonding helps explain the contrasting volatility, but it is not a claim that the para isomer has absolutely zero vapour pressure.
Quick Quiz
Quiz
Why does ethanol boil above methoxymethane?
Explanation: The isomers have equal molecular mass, but only ethanol supplies an OH donor.
Why can ethoxyethane hydrogen-bond with water?
Explanation: An acceptor need not contain its own O–H bond.
Which is preferentially steam volatile in a nitrophenol mixture?
Explanation: Intramolecular hydrogen bonding limits extended self-association in the ortho isomer.
What does increasing straight-chain length usually do within a monohydric alcohol series?
Explanation: Larger chains strengthen dispersion while increasing the hydrophobic region.
Why is the phenolic C–O bond relatively short?
Explanation: Both conjugation and the bonding environment matter.
Practice Problems
- Explain why propan-1-ol boils higher than butane despite their similar masses.
- Compare water solubility of ethanol and pentan-1-ol.
- A student says ethers cannot hydrogen-bond because they have no OH. Correct the statement.
- Predict the boiling-point order of butan-1-ol, 2-methylpropan-1-ol and ethoxyethane.
- Explain the different hydrogen-bond patterns of 2- and 4-nitrophenol and their separation.
- Why can converting phenol to sodium phenoxide increase its solubility?
Propan-1-ol has an OH donor and oxygen acceptor, enabling self hydrogen bonding in addition to dispersion. Butane has no corresponding polar group. Boiling disrupts these intermolecular contacts, so the alcohol requires a higher temperature; no covalent-bond cleavage is needed.
Key Takeaways
• Oxygen in alcohols and ethers has two bonds and two lone pairs; the visible geometry is bent. • Phenolic conjugation shortens C–O relative to ordinary saturated alcohols. • Boiling normally disrupts intermolecular contacts rather than covalent bonds. • Alcohols and phenols donate and accept hydrogen bonds; ethers accept from suitable partners. • Chain size and branching modify dispersion and boiling trends. • A larger nonpolar region generally reduces water solubility for a fixed OH count. • Intramolecular hydrogen bonding makes 2-nitrophenol more steam volatile than the para isomer.