NSWChemistrySyllabus dot point

Inquiry Question 3: How do alcohols form, react, and how does their structure affect their properties?

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

A focused answer to the HSC Chemistry Module 7 dot point on alcohols. Classifying primary, secondary and tertiary alcohols, the oxidation pathway with acidified dichromate or permanganate, hydration of alkenes to form alcohols, and worked HSC past exam questions.

Generated by Claude OpusReviewed by Better Tuition Academy9 min answerUpdated 2026-05-18

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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:

R-CH_2OH \xrightarrow{[O]} R-CHO \xrightarrow{[O]} R-COOH

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:

R_2CH-OH \xrightarrow{[O]} R_2C=O

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:

CH_2=CH_{2(g)} + H_2O_{(g)} \xrightarrow{H_2SO_4, 300 \text{ degrees C}, 70 \text{ atm}} CH_3CH_2OH_{(g)}

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:

CH_3CH=CH_2 + H_2O \xrightarrow{H_2SO_4} CH_3CH(OH)CH_3

The other industrial route is fermentation of glucose by yeast:

C_6H_{12}O_{6(aq)} \xrightarrow{\text{yeast, 25-37 degrees C, anaerobic}} 2C_2H_5OH_{(aq)} + 2CO_{2(g)}

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:

C_2H_5OH_{(l)} + 3O_{2(g)} \rightarrow 2CO_{2(g)} + 3H_2O_{(l)} \quad \Delta H = -1367 \text{ kJ/mol}

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

Common traps

Oxidising a tertiary alcohol. It does not happen. If you write a product, you have made an error. The dichromate stays orange.

Stopping a primary alcohol oxidation at the aldehyde when refluxing. Under reflux with excess oxidant, you get the carboxylic acid. The aldehyde is only isolable by distilling it out as it forms.

Markovnikov direction in hydration. Propene plus water gives propan-2-ol (secondary), not propan-1-ol.

Forgetting the colour change. Dichromate: orange to green. Permanganate: purple to colourless. State the change when asked about an observation.

Confusing fermentation conditions with hydration conditions. Fermentation is yeast, 25 to 37 degrees C, anaerobic. Hydration is $H_2SO_4$ catalyst, 300 degrees C, high pressure. Do not mix them.

In one sentence

Classify an alcohol by counting carbons on the C-OH carbon (1, 2, 3 give primary, secondary, tertiary), oxidise primaries to aldehydes then acids, secondaries to ketones, tertiaries not at all, and produce alcohols industrially by Markovnikov hydration of an alkene with $H_2SO_4$ or by yeast fermentation of glucose.

Past exam questions, worked

Real questions from past NESA papers on this dot point, with our answer explainer.

2022 HSC4 marksCompare the products of the oxidation of propan-1-ol and propan-2-ol with acidified potassium dichromate. Write equations using [O] to represent the oxidising agent.

Show worked answer →

A 4 mark answer needs both alcohols classified, both oxidation products named, both equations, and the colour change.

Propan-1-ol is a primary alcohol ( $-OH$ on a terminal carbon). It oxidises in two steps: first to an aldehyde, then on further oxidation to a carboxylic acid.

CH_3CH_2CH_2OH + [O] \rightarrow CH_3CH_2CHO + H_2O \text{ (propanal)}

CH_3CH_2CHO + [O] \rightarrow CH_3CH_2COOH \text{ (propanoic acid)}

Under reflux with excess acidified dichromate, the final product is propanoic acid. To isolate propanal, distil it off as it forms.

Propan-2-ol is a secondary alcohol ( $-OH$ on a middle carbon). It oxidises in one step to a ketone and no further (the carbon has no $H$ left to lose):

CH_3CH(OH)CH_3 + [O] \rightarrow CH_3COCH_3 + H_2O \text{ (propan-2-one)}

Observation. Acidified dichromate is orange ( $Cr_2O_7^{2-}$ ) and turns green ( $Cr^{3+}$ ) as it is reduced. Both alcohols give the same colour change; the products differ.

Markers reward (1) correct classification, (2) the two-step pathway for the primary, (3) the single-step ketone for the secondary, (4) the visual colour change.

2019 HSC3 marksExplain why the boiling point of ethanol (78°C) is significantly higher than that of ethane (-89°C), despite both having two carbons.

Show worked answer →

Both molecules have similar molar masses (ethanol 46 g/mol, ethane 30 g/mol), so dispersion forces alone cannot account for the 167 degrees C difference.

Ethanol contains a polar $O-H$ bond. The hydrogen, bonded to highly electronegative oxygen, can hydrogen-bond to the oxygen of a neighbouring ethanol molecule. Hydrogen bonds are roughly ten times stronger than dispersion forces.

Ethane has only C-H and C-C bonds, both nearly non-polar, so the only intermolecular force is weak dispersion. The molecules separate easily, giving a very low boiling point.

To boil ethanol, the energy supplied must overcome hydrogen bonds and dispersion. Hence the dramatically higher boiling point.

Markers reward (1) identifying the polar $O-H$ and hydrogen bonding, (2) stating that ethane only has dispersion forces, (3) linking IMF strength to boiling point.