Hydroxy Compounds

0. What Examiners Want

Where this topic appears. Recent AS papers have asked for: the mass of product from the oxidation of ethanol with acidified dichromate(VI) with a given percentage yield; the formula of the product when a complex alcohol is refluxed with excess acidified dichromate(VI); which of two isomeric alcohols gives a yellow precipitate with alkaline I₂(aq), using Fehling's solution results to identify them; the conditions for oxidising propan-1-ol to propanoic acid; the effect of an Al₂O₃ catalyst on the dehydration of alcohols using the Boltzmann distribution; and the industrial hydration of ethene (a bond-energy ΔH calculation, and the effect of pressure on the equilibrium). Ester hydrolysis products also appear (Topic 18).

1. Naming, Classes, Hydrogen Bonding and Acidity

Alcohols contain the hydroxyl group, −OH, on an sp³ carbon. The general formula is C_nH_2n+1OH. They are named with the ending -ol and a number for the position of the OH group, counting from the end that gives the smaller number (propan-2-ol, not propan-1-ol for CH₃CH(OH)CH₃). Diols such as ethane-1,2-diol, HOCH₂CH₂OH, have two OH groups.

A diol can have OH groups of different classes: in propane-1,2-diol, HOCH₂CH(OH)CH₃, one is primary and one is secondary.

Hydrogen bonding. The O–H bond is very polar, so alcohol molecules hydrogen-bond to each other (between the δ+ H of one molecule and the lone pair on O of the next). This gives them higher boiling points than alkanes or halogenoalkanes of similar M_r, so even methanol is a liquid at room temperature. Alcohols can also hydrogen-bond to water, so the short-chain alcohols are miscible with water. As the hydrocarbon chain gets longer the non-polar part grows and disrupts the hydrogen bonding in water, so solubility falls (hexan-1-ol is only slightly soluble).

Acidity compared with water. In water, ethanol dissociates only to a very small extent:

C₂H₅OH ⇌ H⁺ + C₂H₅O⁻

Water dissociates much more: H₂O ⇌ H⁺ + OH⁻. So ethanol is a weaker acid than water. The ethyl group donates electron density (positive inductive effect), which concentrates more negative charge on the oxygen of the ethoxide ion. This makes C₂H₅O⁻ less stable than OH⁻, so it is more ready to take back H⁺ and the equilibrium lies further to the left.

2. Making Alcohols

• Hydration of an alkene: CH₂=CH₂ + H₂O(g) ⇌ C₂H₅OH, with steam and a concentrated H₃PO₄ catalyst (electrophilic addition).
• Diol from an alkene: cold, dilute, acidified KMnO₄ gives, for example, ethane-1,2-diol. CH₂=CH₂ + H₂O + [O] → HOCH₂CH₂OH.
• Hydrolysis of a halogenoalkane: heat under reflux with NaOH(aq): CH₃CH₂Br + NaOH → CH₃CH₂OH + NaBr (nucleophilic substitution, Topic 15).
• Reduction of an aldehyde or ketone: with NaBH₄ or LiAlH₄. An aldehyde gives a primary alcohol and a ketone gives a secondary alcohol: CH₃COCH₃ + 2[H] → CH₃CH(OH)CH₃.
• Reduction of a carboxylic acid: with LiAlH₄ in dry ether (not NaBH₄): CH₃COOH + 4[H] → CH₃CH₂OH + H₂O. LiAlH₄ reacts violently with water, so the ether must be dry.
• Hydrolysis of an ester: heating with dilute acid or alkali gives the alcohol and the carboxylic acid (or its salt) (Topic 18).

Ethanol in industry. Ethanol is made by fermenting sugars or by hydrating ethene.

Bond-energy calculation for hydration of ethene (Data Booklet values, kJ mol⁻¹): bonds broken are one C=C (610) and two O–H in water (2 × 460 = 920), total 1530. Bonds made are one C–C (350), one C–H (410), one C–O (360) and one O–H (460), total 1580. ΔH = 1530 − 1580 = −50 kJ mol⁻¹ (exothermic). There are two moles of gas on the left and one on the right, so by Le Chatelier's principle a higher pressure shifts the equilibrium to the right and increases the amount of ethanol.

3. Reactions of Alcohols

Combustion. Alcohols burn completely to carbon dioxide and water: C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O; 2C₃H₇OH + 9O₂ → 6CO₂ + 8H₂O; C₄H₉OH + 6O₂ → 4CO₂ + 5H₂O. Ethanol burns with a clean, blue flame and is used as a biofuel, because the CO₂ it releases was absorbed by the crop while it grew (a nearly carbon-neutral cycle, although fertilisers, harvesting and transport use fossil fuels).

Sodium. The O–H bond breaks: 2C₂H₅OH + 2Na → 2C₂H₅ONa + H₂. Bubbles of hydrogen are seen and the sodium dissolves, leaving sodium ethoxide (a white solid if the excess ethanol evaporates). The reaction is slower than with water, and slower still with longer-chain alcohols.

Substitution to a halogenoalkane. The C–O bond breaks and −OH is replaced by −X:

• With a hydrogen halide: alcohol + HX → halogenoalkane + water, e.g. C₂H₅OH + HBr → C₂H₅Br + H₂O. The HBr is made in the flask from KBr (or NaBr) and concentrated H₃PO₄ or H₂SO₄; the alcohol is first protonated, which makes the C–O bond easier to break.
• With PCl₅ at room temperature: C₂H₅OH + PCl₅ → C₂H₅Cl + HCl + POCl₃. The steamy acidic fumes of HCl are a test for an −OH group.
• With PCl₃ and heat: 3C₂H₅OH + PCl₃ → 3C₂H₅Cl + H₃PO₃.
• With SOCl₂: C₂H₅OH + SOCl₂ → C₂H₅Cl + HCl + SO₂ (both by-products are gases).

Dehydration (elimination). Passing alcohol vapour over hot Al₂O₃ (or heating with concentrated H₂SO₄ or H₃PO₄ at about 170°C, or using hot pumice) removes water and forms an alkene: C₂H₅OH → C₂H₄ + H₂O; CH₃CH₂CH₂OH → CH₃CH=CH₂ + H₂O. Butan-2-ol can lose water in two directions to give but-1-ene and but-2-ene (cis and trans). A catalyst such as Al₂O₃ provides an alternative pathway with a lower activation energy, so more molecules have at least the activation energy (on the Boltzmann distribution the same E_a marker is moved to the left, and the area to the right of it increases).

4. Oxidation of Alcohols

Primary and secondary alcohols are oxidised by warming with acidified potassium dichromate(VI), K₂Cr₂O₇ with dilute H₂SO₄ (or with acidified KMnO₄). The orange Cr₂O₇²⁻ is reduced to green Cr³⁺ (purple MnO₄⁻ to colourless Mn²⁺). In equations the oxidising agent is written as [O].

• Primary alcohol to aldehyde: warm and distil the aldehyde out as it forms. CH₃CH₂CH₂OH + [O] → CH₃CH₂CHO + H₂O. The aldehyde has a lower boiling point than the alcohol (no hydrogen bonding between its molecules), so it distils off before it can be oxidised further.
• Primary alcohol to carboxylic acid: heat under reflux with excess oxidising agent. CH₃CH₂CHO + [O] → CH₃CH₂COOH, so overall CH₃CH₂CH₂OH + 2[O] → CH₃CH₂COOH + H₂O.
• Secondary alcohol to ketone: CH₃CH(OH)CH₃ + [O] → CH₃COCH₃ + H₂O. A ketone is not oxidised further by these reagents.
• Tertiary alcohol: no reaction. There is no hydrogen on the carbon bearing −OH to be removed, so the mixture stays orange.

Worked example (yield). 2.30 g of ethanol (M_r = 46.0) is oxidised and the ethanal (M_r = 44.0) is distilled off with a 60.0% yield. Moles of ethanol = 2.30 / 46.0 = 0.0500 mol; theoretical mass of ethanal = 0.0500 × 44.0 = 2.20 g; actual mass = 2.20 × 0.600 = 1.32 g. If the question says the mixture was refluxed with excess oxidant, the product is ethanoic acid (M_r = 60.0), so read the conditions before choosing the product.

5. Esterification

An alcohol and a carboxylic acid react, heated under reflux with a few drops of concentrated H₂SO₄ catalyst, to form an ester and water. The reaction is a condensation reaction and is reversible, so an equilibrium mixture is formed.

CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O. The ester name has two parts: the first (alkyl) comes from the alcohol and the second (-oate) from the acid. So ethanol with propanoic acid gives ethyl propanoate, CH₃CH₂COOC₂H₅, and not propyl ethanoate. Other examples: butan-1-ol + ethanoic acid gives butyl ethanoate; ethanol + hexanoic acid gives ethyl hexanoate; pentan-1-ol + methanoic acid gives pentyl methanoate. Esters smell sweet and fruity (flavourings, perfumes, solvents).

6. The Tri-iodomethane Test

Warm the compound with alkaline aqueous iodine (I₂ and NaOH). A pale yellow precipitate of tri-iodomethane, CHI₃, with an antiseptic smell, shows that the compound contains the group CH₃CH(OH)– (or CH₃CO–).

The alcohol is first oxidised by the iodine to the carbonyl compound, which then reacts. For ethanol: CH₃CH₂OH + 4I₂ + 6NaOH → CHI₃ + HCOONa + 5NaI + 5H₂O. For propan-2-ol: CH₃CH(OH)CH₃ + 4I₂ + 6NaOH → CHI₃ + CH₃COONa + 5NaI + 5H₂O. Positive: ethanol, propan-2-ol, butan-2-ol, pentan-2-ol. Negative: methanol, propan-1-ol, butan-1-ol, 2-methylpropan-1-ol, 2-methylpropan-2-ol and pentan-3-ol.

7. Distinguishing Alcohols

Worked identification. An alcohol J, C₄H₁₀O, is warmed with acidified K₂Cr₂O₇ and the mixture turns green. The product gives no red precipitate with Fehling's solution. The green colour shows J is primary or secondary, and the negative Fehling's result shows the product is not an aldehyde, so it is a ketone and J is secondary. The only secondary alcohol C₄H₉OH is butan-2-ol, which contains CH₃CH(OH)–, so J would also give a yellow precipitate with alkaline iodine. The primary butanols never do.

8. Quick Sheet and Checklist

Before you leave the question, check:

• Every oxidation says distil or reflux with excess, and names the colour change (orange to green).
• The class of alcohol matches the product (aldehyde, ketone, or no reaction).
• Reduction equations use [H] and oxidation equations use [O], and both balance.
• Esters are named alkyl (from the alcohol) then -oate (from the acid).
• The correct reducing agent is chosen: LiAlH₄ for a carboxylic acid.
• Acidity and boiling-point answers name the inductive effect and hydrogen bonding.