Investigate the structural formulae, properties, classification (primary, secondary, tertiary), oxidation reactions and production by hydration of alkenes for alcohols up to C8

What this dot point is asking

NESA wants you to identify and draw primary, secondary and tertiary alcohols, predict their oxidation products with acidified $KMnO_4$ or $K_2Cr_2O_7$, explain why alcohols have high boiling points and full water solubility for short chains, and write the equation for hydration of an alkene to produce an alcohol.

The answer

Structure and the -OH group

An alcohol has the general formula $R-OH$ where $R$ is an alkyl group. The functional group is the hydroxyl $-OH$. Alcohols are named using the suffix $-ol$ with a locant for the carbon bearing the OH.

The $O-H$ bond is polar (oxygen electronegativity 3.44, hydrogen 2.20), giving alcohols hydrogen-bonding capability. The lone pairs on oxygen can also accept hydrogen bonds. This dual donor-acceptor capacity makes alcohols very soluble in water (short chains) and high-boiling compared with alkanes of the same molar mass.

Classification: primary, secondary, tertiary

Classify by counting how many other carbon atoms are bonded to the $C-OH$ carbon.

• Primary (1 degrees C): the OH-bearing carbon is attached to one other carbon. Example: ethanol $CH_3CH_2OH$, propan-1-ol $CH_3CH_2CH_2OH$.
• Secondary (2 degrees C): attached to two other carbons. Example: propan-2-ol $CH_3CH(OH)CH_3$, butan-2-ol.
• Tertiary (3 degrees C): attached to three other carbons. Example: 2-methylpropan-2-ol $(CH_3)_3COH$.

Methanol $CH_3OH$ is technically a special case (zero other carbons) but is usually grouped with primaries for oxidation purposes.

Physical properties

Boiling point

Hydrogen bonds dominate, so alcohols boil far above alkanes of the same molar mass. Within the alcohol series, boiling point rises with chain length (more dispersion) and falls slightly with branching (less surface contact).

Solubility in water

Methanol, ethanol, propan-1-ol and propan-2-ol are fully miscible with water. From butanol onwards, solubility drops sharply because the non-polar alkyl tail dominates. By octan-1-ol, the alcohol is essentially insoluble.

Viscosity

Increases with chain length and with the number of OH groups (compare ethanol with ethane-1,2-diol or with glycerol).

Oxidation: the central reaction

The oxidising agents are acidified potassium dichromate $K_2Cr_2O_7 / H_2SO_4$ (orange to green) or acidified potassium permanganate $KMnO_4 / H_2SO_4$ (purple to colourless). HSC equations use $[O]$ to represent the oxidising agent.

Primary alcohols oxidise in two steps:

The first step removes two hydrogens to give an aldehyde. The second step adds an oxygen across $C-H$ to give a carboxylic acid.

If you want the aldehyde, distil it off as it forms (aldehydes boil lower than alcohols). If you want the acid, reflux with excess oxidant.

Secondary alcohols oxidise once to a ketone:

The ketone cannot oxidise further under HSC conditions because the carbonyl carbon has no $H$ left to lose without breaking a $C-C$ bond.

Tertiary alcohols do not oxidise. The OH-bearing carbon has no hydrogen to remove. The orange dichromate stays orange.

This is the basis of a classification test: if dichromate goes green on warming, the alcohol is primary or secondary; if it stays orange, it is tertiary. To distinguish primary from secondary, use Tollens' reagent on the oxidation product (aldehyde gives silver mirror, ketone does not).

Production by hydration of alkenes

Industrial ethanol is made by adding water across the double bond of ethene, catalysed by dilute sulfuric acid at high temperature and pressure:

For an asymmetric alkene, Markovnikov's rule dictates that the H of water adds to the carbon with more Hs, and OH adds to the more substituted carbon. So propene gives propan-2-ol as the major product (a secondary alcohol), not propan-1-ol:

The other industrial route is fermentation of glucose by yeast:

Fermentation gives the same product as hydration but is run from a biological feedstock and stops around 15% ethanol because the yeast die at higher concentrations.

Combustion of alcohols

Like hydrocarbons, alcohols burn in excess oxygen to $CO_2$ and water:

Ethanol combustion is the basis of bioethanol fuels and breathalyser-style calculations.

Examples in context

Example 1. Manildra ethanol fermentation in the Riverina. Manildra Group's Bomaderry plant ferments wheat-derived glucose using $Saccharomyces$ yeast: $C_6H_{12}O_6 \rightarrow 2C_2H_5OH + 2CO_2$. The ethanol is distilled to 95 percent purity, then dried to anhydrous fuel grade for blending into E10 petrol sold across NSW. Ethanol is a primary alcohol, so it oxidises in vehicle catalytic converters first to ethanal and then to ethanoic acid before reaching $CO_2$. The fermentation route gives carbon-neutral motor fuel because the $CO_2$ released matches the $CO_2$ fixed by the wheat crop. The HSC classification (primary alcohol, single carbon on the C-OH) is the basis for every emissions calculation in the supply chain.

Example 2. Hydration of ethene at the Botany Industrial Park. Qenos at Botany operates an ethene-hydration unit that produces synthetic ethanol via $C_2H_4 + H_2O \rightarrow C_2H_5OH$ at 300 degrees C, 70 atm, over a phosphoric acid catalyst. The Markovnikov rule applies trivially to ethene (symmetric), but engineers use the same rule for propene, where hydration would give propan-2-ol rather than propan-1-ol because the OH attaches to the more substituted carbon. The synthetic route competes with fermentation: it offers higher throughput but tracks oil price. HSC candidates apply the same Markovnikov rule when predicting the alcohol formed from any unsymmetrical alkene plus water.

Try this

Q1. Classify each of the following as primary, secondary or tertiary: butan-2-ol, 2-methylpropan-2-ol, propan-1-ol. [3 marks]

• Cue. Count substituent carbons on the C-OH carbon: butan-2-ol secondary (2), 2-methylpropan-2-ol tertiary (3), propan-1-ol primary (1).

Q2. Calculate the mass of ethanol produced by complete fermentation of 1.00 kg of glucose. [3 marks]

• Cue. $n(\text{glucose}) = 1000 / 180.16 = 5.55$ mol; $n(\text{ethanol}) = 2 \times 5.55 = 11.1$ mol; mass = $11.1 \times 46.07 = 511$ g.

Q3. Propene undergoes hydration with sulfuric acid catalyst. (a) State the Markovnikov product. (b) Write a balanced equation. (c) Classify the alcohol formed and predict its oxidation product. [1+1+2 marks]

• Cue. (a) Propan-2-ol. (b) $CH_3CH=CH_2 + H_2O \rightarrow CH_3CH(OH)CH_3$. (c) Secondary; oxidises to propan-2-one (acetone).