Alcohols — OCR A-Level Chemistry

Alcohols contain the hydroxyl group, written as −OH, covalently bonded to a carbon chain. The −OH group is polar, so hydrogen bonding occurs between alcohol molecules, raising boiling points above those of comparable alkanes. Hydrogen bonding to water molecules also makes short-chain alcohols such as methanol and ethanol fully miscible with water, while longer chains become less soluble as the hydrocarbon part grows. Alcohols are neutral, unlike carboxylic acids, but they react with sodium to release hydrogen and can be oxidised by acidified potassium dichromate(VI). Primary alcohols oxidise to aldehydes and then carboxylic acids, secondary alcohols to ketones, and tertiary alcohols resist oxidation under these conditions.

Alcohols such as ethanol, CH₃CH₂OH, contain an O–H bond in which oxygen is far more electronegative than hydrogen, so the bond is polar and the oxygen carries a partial negative charge (δ⁻) while the hydrogen carries a partial positive charge (δ⁺). This makes the molecule polar overall. The δ⁺ hydrogen of one alcohol molecule is attracted to a lone pair on the oxygen of a neighbouring molecule, forming hydrogen bonds. These intermolecular hydrogen bonds must be overcome to separate molecules, so alcohols have higher boiling points and are less volatile than alkanes of similar Mr, whose only intermolecular forces are weaker London dispersion forces. Hydrogen bonding to water molecules also lets small alcohols dissolve readily in water, because the alcohol can both donate and accept hydrogen bonds.

Alcohols are classified by the number of carbon atoms bonded directly to the carbon that carries the –OH group. In a primary alcohol the carbon bearing the –OH is attached to one other carbon, as in propan-1-ol, CH₃CH₂CH₂OH. In a secondary alcohol that carbon is attached to two other carbons, as in propan-2-ol, CH₃CH(OH)CH₃. In a tertiary alcohol it is attached to three other carbons, as in 2-methylpropan-2-ol, (CH₃)₃COH. The classification depends only on the carbon bonded to the hydroxyl group, not on the total number of carbons in the chain, so a long-chain alcohol can still be primary. Drawing the structure and counting the carbon neighbours of the –OH-bearing carbon gives a reliable classification.

This row is a topic heading, so treat it as a reading guide rather than an assessed answer target. As you read the alcohols section, organise the reactions into families: combustion, oxidation, dehydration to alkenes, and substitution to form haloalkanes. For oxidation, note that a primary alcohol is oxidised by acidified potassium dichromate(VI) first to an aldehyde and then to a carboxylic acid, a secondary alcohol to a ketone, and a tertiary alcohol is not readily oxidised. Record the colour change of the dichromate(VI) oxidising agent as it is reduced. For dehydration, note the need for a heated catalyst such as concentrated sulfuric acid or aluminium oxide. Build a summary table linking each alcohol class to its oxidation product, and check each equation balances.

Combustion of alcohols means burning them in excess oxygen to give carbon dioxide and water, releasing energy. For a primary alcohol such as ethanol the equation is C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O; check it by counting carbon, hydrogen and oxygen on each side. Learn to balance the equation for any given alcohol formula by first balancing carbon as CO₂, then hydrogen as H₂O, then oxygen with O₂, adjusting the O₂ coefficient last. Recognise that incomplete combustion in limited oxygen gives carbon monoxide or soot, and that alcohols are used as fuels because they burn cleanly. In an MCQ, expect to select the correct balanced equation, the correct products, or the correct volume or mole relationship from a given amount of alcohol.

Acidified dichromate(VI) is the standard oxidising agent for alcohols in this topic. The ion is written Cr₂O₇²⁻ and the reagent is potassium dichromate(VI), K₂Cr₂O₇, dissolved in dilute sulfuric acid, H₂SO₄, which supplies the H⁺ ions. Chromium in dichromate(VI) is in oxidation state +6, and it is reduced to Cr³⁺, giving the colour change from orange to green that signals oxidation. A primary alcohol is first oxidised to an aldehyde, which distils off, and prolonged excess oxidant gives the carboxylic acid. A secondary alcohol gives a ketone, which resists further oxidation. A tertiary alcohol has no hydrogen on the carbon bearing the –OH group, so it is not oxidised and the orange colour persists. Write the half-equation Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O and balance the organic half-equation before combining them.

A primary alcohol such as ethanol can be oxidised by acidified dichromate(VI) to either an aldehyde or a carboxylic acid, and the product is controlled by the reaction conditions. Gentle oxidation with distillation removes the aldehyde as it forms, because the aldehyde has a lower boiling point than the alcohol, so it escapes before further oxidation. Heating under reflux with excess oxidising agent keeps the aldehyde in contact with the reagent and oxidises it on to the carboxylic acid. Write the equations: ethanol to ethanal, CH₃CH₂OH + [O] → CH₃CHO + H₂O, and ethanal to ethanoic acid, CH₃CHO + [O] → CH₃COOH. The aldehyde functional group is –CHO and the carboxylic acid group is –COOH. Recognise that distillation gives the aldehyde and reflux with excess oxidising agent gives the carboxylic acid.

A secondary alcohol has the –OH group on a carbon that is bonded to two other carbon atoms and to one hydrogen atom. On oxidation with acidified dichromate(VI), the alcohol loses one hydrogen atom from the carbon bearing the –OH group and one hydrogen atom from the –OH group itself, so a C=O double bond forms and a ketone results. For example, propan-2-ol, CH₃CH(OH)CH₃, is oxidised to propanone, CH₃COCH₃. The ketone has no hydrogen on its carbonyl carbon, so it resists further oxidation and the reaction stops there. The orange dichromate(VI) turns green as Cr(VI) is reduced to Cr(III). Distinguish this from a primary alcohol, which can be oxidised further to a carboxylic acid.

A tertiary alcohol has the –OH group on a carbon bonded to three other carbon atoms, for example 2-methylpropan-2-ol, (CH₃)₃COH. Oxidation of an alcohol requires the carbon bearing the –OH group to lose a hydrogen atom as well as the hydrogen of the –OH group, so that a C=O double bond can form. In a tertiary alcohol that carbon carries no hydrogen atom, so this loss cannot happen. Consequently tertiary alcohols resist oxidation by acidified potassium dichromate(VI): the orange solution stays orange on heating, showing no reduction to green Cr³⁺ ions. This contrasts with primary and secondary alcohols, which are oxidised. The resistance is a structural feature, not a matter of reaction conditions.

This is acid-catalysed dehydration of an alcohol. Heat the alcohol with a concentrated acid catalyst such as H₃PO₄ or H₂SO₄; the acid protonates the –OH group, a water molecule leaves and a carbocation forms, then a neighbouring C–H bond breaks to give a C=C and regenerate the catalyst. Ethanol, for example, gives ethene plus water. The organic product is an alkene, so the key structural change is loss of H and OH from adjacent carbon atoms to create the double bond. Recognise that the acid is a catalyst, that heat is needed, and that the product contains a C=C bond rather than a new functional group such as a halogen.

This is conversion of an alcohol into a haloalkane by substitution. A mixture such as NaBr with H₂SO₄ generates hydrogen bromide in situ; the acid protonates the alcohol's –OH group so it becomes a good leaving group, a halide ion then attacks the carbon and displaces water, giving a haloalkane. The net change is replacement of –OH by –X (for example –Br), so the carbon skeleton is retained and one functional group is swapped for another. Distinguish this from elimination: substitution keeps the single bonds and adds a halogen, whereas elimination forms a C=C. Recognise that acid is present to generate the hydrogen halide and to protonate the alcohol.