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Inquiry Question 2: How is information about the reactivity and structure of organic compounds obtained?

Conduct qualitative investigations to test for the presence in organic molecules of carbon-carbon double bonds, hydroxyl groups and carboxylic acids

A focused answer to the HSC Chemistry Module 8 dot point on qualitative tests for organic functional groups. The bromine water and acidified permanganate tests for C=C, the sodium and acidified dichromate tests for hydroxyl, the sodium carbonate and reactive metal tests for carboxylic acids, a flowchart for an unknown, and worked HSC past exam questions.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-22

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
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In plain English

Imagine you are handed four unlabelled bottles and told each holds a different type of organic liquid: one has a "kink" (a carbon-carbon double bond), one has a slippery alcohol group, one has a sour acid group, and one is a plain, unreactive hydrocarbon. You cannot see the difference, so you run a short series of chemical "litmus tests", each one designed to react with only ONE of these features. Bromine water is like a paint that only sticks to the kink, so if the orange colour vanishes, you have found the double bond. Sodium carbonate is like a fizz-detector that only fizzes for the sour, acidic group. Sodium metal is like a gentle spark-test that only lights up (releases hydrogen) for an acidic O-H, whether that is the alcohol or the acid. The trick to getting full marks is running these tests in the right order, most selective first, so an earlier positive result does not get mistaken for a different group later on. Whatever is left over once every test comes back negative must be the plain, unreactive hydrocarbon.

What this dot point is asking

NESA wants you to know the standard chemical tests that confirm a carbon-carbon double bond (unsaturation), a hydroxyl ( $OH$ ) group on an alcohol and a carboxyl ( $COOH$ ) group on a carboxylic acid, with the reagents, observations, equations and the limits of selectivity for each test.

The answer

Tests for C=C (unsaturation)

Bromine water. Shake the unknown with a few drops of orange bromine water. An alkene rapidly decolourises the orange to colourless by adding $Br_2$ across the double bond:

RCH=CHR' + Br_2 \rightarrow RCHBr-CHBrR'

The reaction is electrophilic addition. Saturated hydrocarbons (alkanes) do not react with bromine water in the dark; they require UV light to undergo radical substitution.

Acidified potassium permanganate (cold dilute, room temperature). Purple $MnO_4^-$ is decolourised by an alkene as the double bond is oxidised to a 1,2-diol. The observation is purple to colourless (or to a brown $MnO_2$ precipitate in neutral conditions):

3RCH=CHR' + 2KMnO_4 + 4H_2O \rightarrow 3RCH(OH)-CH(OH)R' + 2MnO_2 + 2KOH

Under hot acidified conditions the diol oxidises further and the C-C bond is cleaved into two carboxylic acids (or carbon dioxide if a terminal $=CH_2$ is present).

Selectivity caveat. Permanganate also oxidises aldehydes, primary and secondary alcohols, and some aromatic side chains. So a positive permanganate test alone does not prove an alkene. Bromine water at room temperature is more specific.

Tests for hydroxyl $-OH$ (alcohol)

Sodium metal. Drop a small piece of clean sodium into the unknown (dried). An alcohol releases hydrogen gas and forms a sodium alkoxide:

2R-OH + 2Na \rightarrow 2R-O^-Na^+ + H_2

Test the gas with a lit splint (pop). The reaction is calmer than the reaction of sodium with water because alcohols are weaker acids. Tertiary alcohols react more slowly than primary, but all react.

Selectivity caveat. Carboxylic acids also react with sodium (more vigorously). The test confirms an acidic O-H, not specifically an alcohol; combine with the carbonate test (next section) to discriminate.

Acidified potassium dichromate (orange $Cr_2O_7^{2-}/H_2SO_4$ ). Primary and secondary alcohols reduce dichromate from orange to green:

3R-CH_2OH + 2Cr_2O_7^{2-} + 16H^+ \rightarrow 3R-COOH + 4Cr^{3+} + 11H_2O

The orange-to-green colour change is the classical alcohol test. Tertiary alcohols do not react (no $H$ on the OH-bearing carbon to lose).

Ester formation. Heat the unknown with a carboxylic acid (ethanoic acid) and a few drops of concentrated $H_2SO_4$ . A sweet, fruity smell indicates ester formation, confirming an alcohol. Reverse the reagents to confirm a carboxylic acid.

Tests for carboxylic acid $-COOH$

Sodium carbonate (or sodium bicarbonate). Add solid $Na_2CO_3$ or $NaHCO_3$ to the unknown dissolved in water (or to the neat liquid). Brisk effervescence of $CO_2$ indicates a carboxylic acid:

2R-COOH + Na_2CO_3 \rightarrow 2R-COO^-Na^+ + H_2O + CO_2

Confirm the gas by passing it through limewater (turns milky). Alcohols are too weak an acid to react with carbonate; phenols can react with strong base but not with carbonate. So a positive carbonate test is highly diagnostic of a carboxylic acid.

Litmus or universal indicator. A carboxylic acid solution has pH around 3 to 4 (depending on concentration). Litmus turns red. Alcohols are essentially neutral.

Reactive metal. Magnesium ribbon dissolves in a carboxylic acid solution with hydrogen evolution:

2R-COOH + Mg \rightarrow (R-COO)_2Mg + H_2

The bubbling is brisker than with an alcohol (the acid is fully ionised in solution; the alcohol's O-H is barely acidic in water).

Esterification. Warm with an alcohol and concentrated $H_2SO_4$ ; the fruity ester smell confirms the carboxyl group.

Worked identification examples

Worked example 1: identifying but-1-ene, butan-1-ol, butanoic acid and butane.

Add cold bromine water to each. Only but-1-ene decolourises the orange bromine rapidly, by addition across the C=C:

CH_3CH_2CH=CH_2 + Br_2 \rightarrow CH_3CH_2CHBrCH_2Br

To the three remaining liquids, add solid $Na_2CO_3$ . Only butanoic acid effervesces:

2CH_3CH_2CH_2COOH + Na_2CO_3 \rightarrow 2CH_3CH_2CH_2COONa + H_2O + CO_2

To the two remaining liquids, add a small piece of dry sodium metal. Only butan-1-ol releases hydrogen:

2CH_3CH_2CH_2CH_2OH + 2Na \rightarrow 2CH_3CH_2CH_2CH_2ONa + H_2

Whatever gives no reaction to any of the three tests is butane, identified by exclusion.

Marker's note: full marks require the tests IN this order (alkene, then carbonate, then sodium) with the reasoning for the order, at least one balanced equation, and an explicit statement that the alkane is identified by exclusion rather than by a positive test of its own.

Worked example 2: distinguishing propan-2-ol from propanoic acid using only sodium carbonate.

Add solid $Na_2CO_3$ to each liquid separately.
Propanoic acid effervesces briskly because the carboxylic acid proton is acidic enough to be deprotonated by the weak base carbonate:

2CH_3CH(OH)CH_3 \; (\text{no reaction}) \qquad 2CH_3CH_2COOH + Na_2CO_3 \rightarrow 2CH_3CH_2COONa + H_2O + CO_2

Propan-2-ol shows no effervescence at all, because an alcohol's $O-H$ is far too weakly acidic to react with carbonate ion.
Confirm the $CO_2$ gas by bubbling it through limewater, which turns milky.

Marker's note: this question rewards recognising that a SINGLE test (carbonate) is sufficient here because only one of the two candidates is acidic enough to react; do not overcomplicate the answer with the sodium or bromine water tests, which are unnecessary once the acid/alcohol distinction has already been made unambiguously.

A flowchart for an unknown organic liquid

Suppose you suspect an alkane, an alkene, an alcohol or a carboxylic acid:

Bromine water. Decolourised rapidly at room temperature? $\Rightarrow$ alkene.
If no decolourisation, add sodium carbonate. Effervescence? $\Rightarrow$ carboxylic acid.
If no effervescence, add sodium metal. Hydrogen evolved? $\Rightarrow$ alcohol.
If no reaction in any of the above, the unknown is the saturated hydrocarbon (alkane), identified by exclusion.

Sequence the tests in this order to avoid false positives: the alkene test goes first because acidified permanganate would also react with the alcohol; carbonate test before sodium because both alcohols and acids react with sodium.

Summary table

Test	Alkane	Alkene	Alcohol	Carboxylic acid
Bromine water (cold)	No change	Orange to colourless	No change	No change
Acidified $KMnO_4$ (cold)	No change	Purple to colourless	Slow (cold), fast (hot, reflux)	No change
Acidified $K_2Cr_2O_7$ (warm)	No change	(Yes, but used for alcohol)	Orange to green (1, 2 only)	No change
Sodium metal	No change	No change	$H_2$ evolves	$H_2$ evolves (faster)
$Na_2CO_3$ or $NaHCO_3$	No change	No change	No change	$CO_2$ effervescence
Litmus / pH	Neutral	Neutral	Neutral	Red, pH 3 to 4

An owned schematic below shows how the three tests are applied, in the diagnostic order, to a single unknown organic liquid:

Common traps

Treating bromine water and permanganate as equivalent unsaturation tests: Bromine water is more specific; permanganate also reacts with alcohols and aldehydes.
Forgetting to do the carbonate test before the sodium test: A sample of acid plus alcohol would give hydrogen with sodium and confuse the identification. The carbonate test discriminates first.
Using wet sodium: Sodium reacts with water more vigorously than with most alcohols, and any water in the sample will mask the alcohol test. Dry the sample first.
Claiming tertiary alcohols give an orange-to-green dichromate result: They do not. Tertiary alcohols are inert to dichromate.
Saying "an organic acid effervesces with sodium hydroxide": Hydroxide gives a neutralisation, not effervescence. Carbonate gives effervescence because $CO_2$ is released.

Examples in context

Example 1. Margarine quality control by bromine number. Goodman Fielder's margarine plant near Liverpool measures the iodine and bromine number of incoming canola oil to monitor unsaturation, which controls texture and melting point. The bromine value test adds bromine in carbon tetrachloride to a known mass of oil; the volume of bromine consumed gives a quantitative iodine value via stoichiometry. A drop in bromine value flags successful partial hydrogenation. The HSC bromine water test for C=C is the qualitative version of the same chemistry. Students at NSW schools running the test on canola oil watch the bromine colour disappear, exactly the same colour change the QC technician at Goodman Fielder uses.

Example 2. Identifying organic acids in NSW Murray Valley wine. Wineries along the Murray Valley measure tartaric, malic and lactic acid in grape juice by titration, but new juice is first screened qualitatively with sodium hydrogen carbonate to confirm acid presence. A drop of juice on dry $NaHCO_3$ effervesces vigorously if free acid is present, an instant pass-fail screen for spoiled tank deliveries. The acidified dichromate test on the same juice goes green if ethanol contamination from a wild fermentation is present. Both qualitative HSC tests are bench tools winery chemists use before sending samples for HPLC quantification. NESA-style depth study tasks at Riverina and Hunter Valley schools often replicate this winery screening workflow.

Try this

Q1. State the observation for each of: (a) hex-1-ene plus bromine water, (b) ethanol plus acidified dichromate, (c) ethanoic acid plus sodium hydrogen carbonate. [3 marks]

Cue. (a) Orange to colourless. (b) Orange to green. (c) Effervescence (colourless $CO_2$ ).

Q2. A 0.225 g sample of an unknown monoprotic carboxylic acid neutralises 30.40 mL of 0.100 mol L $^{-1}$ NaOH. Calculate the molar mass and identify the most likely candidate (methanoic 46, ethanoic 60, propanoic 74, butanoic 88). [3 marks]

Cue. $n(\text{NaOH}) = 0.100 \times 0.03040 = 3.04 \times 10^{-3}$ mol; 1:1 stoichiometry; $M = 0.225 / 3.04 \times 10^{-3} = 74.0$ g mol $^{-1}$ ; propanoic acid.

Q3. A student is given three unlabelled liquids X, Y and Z. X decolourises bromine water. Y effervesces with sodium hydrogen carbonate. Z gives an orange-to-green colour change with acidified dichromate but does not effervesce. (a) Identify the functional group present in each. (b) Write a balanced equation for the dichromate test on Z. (c) Predict the IR signature of Y. [3+2+1 marks]

Cue. (a) X alkene, Y carboxylic acid, Z alcohol. (b) $3RCH_2OH + Cr_2O_7^{2-} + 8H^+ \rightarrow 3RCHO + 2Cr^{3+} + 7H_2O$ . (c) Broad O-H 2500 to 3300 plus sharp C=O 1700 to 1720 cm $^{-1}$ .

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2022 HSC5 marksAn unknown liquid is one of: pent-1-ene, pentan-1-ol, pentanoic acid or pentane. Describe a sequence of qualitative tests that would identify which compound is present. Write equations where appropriate.

Show worked answer →

A 5 mark answer needs three discriminating tests, the observation for each candidate, and at least one balanced equation.

Test 1: Add bromine water and shake. Pent-1-ene decolourises orange bromine water rapidly to give 1,2-dibromopentane (addition across the C=C). The other three give no change.

CH_3CH_2CH_2CH=CH_2 + Br_2 \rightarrow CH_3CH_2CH_2CHBrCH_2Br

If the bromine water decolourises immediately, the unknown is pent-1-ene.

Test 2: Add sodium carbonate. Pentanoic acid effervesces because $CO_3^{2-}$ is a strong enough base to deprotonate the carboxylic acid:

2CH_3(CH_2)_3COOH + Na_2CO_3 \rightarrow 2CH_3(CH_2)_3COONa + H_2O + CO_2

Pentan-1-ol does not effervesce (it is not acidic enough to react with carbonate). Pentane does not react. If you see effervescence, the unknown is pentanoic acid.

Test 3: Add a small piece of sodium metal. Pentan-1-ol reacts to give hydrogen gas and the sodium alkoxide:

2CH_3(CH_2)_4OH + 2Na \rightarrow 2CH_3(CH_2)_4ONa + H_2

Pentane gives no reaction. If hydrogen evolves but the compound did not react with carbonate, the unknown is pentan-1-ol. If nothing happens at any stage, the unknown is pentane.

Markers reward (1) the bromine water test with equation, (2) carbonate effervescence for the acid with equation, (3) sodium test for the alcohol with equation, (4) a clear logic chain that distinguishes all four, (5) noting pentane is identified by exclusion.

2018 HSC3 marksAcidified potassium permanganate decolourises in the presence of both an alkene and a primary alcohol. Explain how the observations differ and how this allows an alkene to be distinguished from a primary alcohol.

Show worked answer →

Both an alkene and a primary alcohol reduce acidified permanganate from purple to colourless. The chemistry, however, is very different in speed and conditions.

With an alkene (cold dilute $KMnO_4$ in slightly acidic or neutral conditions): decolourisation is essentially instant on shaking at room temperature. The C=C is oxidised to a diol (or further to cleavage products under hot acidic conditions).

With a primary alcohol: cold dilute permanganate decolourises slowly or not at all. The reaction needs warming under reflux with hot acidified permanganate, where the alcohol is oxidised to a carboxylic acid (via the aldehyde).

Discrimination. Use cold dilute permanganate at room temperature. Alkene decolourises in seconds; alcohol does not. To confirm the alcohol, warm a fresh sample with acidified permanganate; the purple fades on standing. The contrast in temperature and rate distinguishes the two.

Markers reward (1) both react, (2) alkene fast and cold, alcohol slow and hot, (3) using temperature/time to discriminate.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation3 marksState the observation when each liquid is treated with cold bromine water: (a) hex-1-ene, (b) hexane, (c) hexan-1-ol.

Show worked solution →

(a) Hex-1-ene: Orange bromine water is decolourised rapidly at room temperature (electrophilic addition of $Br_2$ across the C=C).
(b) Hexane: No change in the dark (saturated alkanes need UV light for radical substitution, which is not the standard cold test).
(c) Hexan-1-ol: No change (an alcohol has no C=C to react with bromine).

Marking criteria: 1 mark per correct observation.

foundation4 marksComplete a results table for four reagents (bromine water, acidified

KMnO_4

cold, sodium metal,

Na_2CO_3

) applied to hex-1-ene, hexan-1-ol and hexanoic acid, stating only the positive observations.

Show worked solution →

Reagent	Hex-1-ene	Hexan-1-ol	Hexanoic acid
Bromine water	Orange to colourless	No change	No change
Acidified $KMnO_4$ (cold)	Purple to colourless	No change	No change
Sodium metal	No change	$H_2$ evolves	$H_2$ evolves (faster)
$Na_2CO_3$	No change	No change	$CO_2$ effervescence

Marking criteria: 1 mark per correctly completed row (bromine/permanganate row counted together as one mark for the alkene column, sodium row, carbonate row), to a maximum of 4.

core4 marksA 0.176 g sample of a pure monoprotic carboxylic acid is dissolved in water and titrated to the equivalence point with 24.00 mL of 0.0800 mol L

^{-1}

NaOH. Calculate the molar mass of the acid to 3 significant figures, and identify it from the candidates: methanoic acid (46.0 g mol

^{-1}

), ethanoic acid (60.0 g mol

^{-1}

), propanoic acid (74.0 g mol

^{-1}

), butanoic acid (88.0 g mol

^{-1}

Show worked solution →

Step 1: moles of NaOH used at the equivalence point.

n(\text{NaOH}) = c \times V = 0.0800\ \text{mol L}^{-1} \times 0.02400\ \text{L} = 1.920 \times 10^{-3}\ \text{mol}

Step 2: moles of acid, using the 1:1 stoichiometry of a monoprotic acid with NaOH.

R\text{-COOH} + NaOH \rightarrow R\text{-COONa} + H_2O

n(\text{acid}) = n(\text{NaOH}) = 1.920 \times 10^{-3}\ \text{mol}

Step 3: molar mass of the acid.

M = \frac{m}{n} = \frac{0.176\ \text{g}}{1.920 \times 10^{-3}\ \text{mol}} = 91.67\ \text{g mol}^{-1}

Step 4: round to 3 significant figures (matching the 3 s.f. of the given mass and volume data).

M(\text{acid}) = 91.7\ \text{g mol}^{-1}

Step 5: identify the closest candidate. 91.7 g mol $^{-1}$ is closest to butanoic acid (88.0 g mol $^{-1}$ ) among the given candidates; the small residual difference is consistent with experimental uncertainty in the mass and volume measurements, so the acid is identified as butanoic acid.

Marking criteria: 1 mark for correct moles of NaOH, 1 mark for the correct 1:1 mole ratio applied to the acid, 1 mark for the molar mass calculation to 3 significant figures with correct units, 1 mark for identifying butanoic acid as the closest candidate with a comment on experimental uncertainty.

core5 marksThe IR spectrum below is an owned illustrative spectrum of organic unknown Y. Two peaks are labelled: a broad peak from about 3200 to 3550

cm^{-1}

, and NO peak in the 1650 to 1750

cm^{-1}

carbonyl region. State the functional group class present, justify your answer using both labelled features, and explain why Y cannot be a carboxylic acid, aldehyde or ketone.

Show worked solution →

Reading the spectrum. A broad peak sits between roughly 3200 and 3550 $cm^{-1}$ . No peak appears in the 1650 to 1750 $cm^{-1}$ carbonyl window.

Interpretation.

A broad peak in the 3200 to 3550 $cm^{-1}$ range is an O-H stretch. Because it is centred noticeably higher than the 2500 to 3300 $cm^{-1}$ range typical of a hydrogen-bonded carboxylic acid O-H, and because there is no carbonyl peak present at all, this O-H is from an alcohol, not an acid.
The absence of any 1650 to 1750 $cm^{-1}$ peak rules out a carbonyl group entirely, which excludes aldehyde, ketone AND carboxylic acid (all three require a $C=O$ stretch in that window).

Conclusion. Unknown Y contains a hydroxyl ( $-OH$ ) group on an alcohol, and is not a carboxylic acid, aldehyde or ketone because none of those carbonyl-containing peaks are present.

Marking criteria: 1 mark for identifying the O-H peak and its region, 1 mark for correctly distinguishing the higher alcohol O-H range from the lower, broader carboxylic acid range, 1 mark for identifying the absence of a carbonyl peak, 1 mark for excluding aldehyde/ketone/acid on that basis, 1 mark for the correct overall conclusion (alcohol).

core6 marksA student is given four unlabelled liquids, W, X, Y and Z, known to be (in some order) hexane, hex-1-ene, hexan-1-ol and hexanoic acid. Design a sequence of tests, stating the order, reagents, expected positive observation and one balanced equation, that unambiguously identifies all four.

Show worked solution →

Test order and reasoning. Run bromine water first (only the alkene reacts, and it does so at room temperature with no interference), then sodium carbonate (only the acid reacts, ruling it out before the sodium test), then sodium metal (only the alcohol reacts once the acid has already been excluded). Whatever remains unreacted at every stage is the alkane, identified by exclusion.

Test 1: cold bromine water, added dropwise with shaking. The liquid that rapidly decolourises the orange bromine is the alkene:

CH_3(CH_2)_3CH=CH_2 + Br_2 \rightarrow CH_3(CH_2)_3CHBrCH_2Br

Test 2: solid $Na_2CO_3$ , added to each remaining liquid. The liquid that effervesces (confirmed with limewater turning milky) is the acid:

2CH_3(CH_2)_4COOH + Na_2CO_3 \rightarrow 2CH_3(CH_2)_4COONa + H_2O + CO_2

Test 3: a small piece of dry sodium metal, added to each remaining liquid. The liquid that evolves hydrogen gas (pop with a lit splint) is the alcohol:

2CH_3(CH_2)_5OH + 2Na \rightarrow 2CH_3(CH_2)_5ONa + H_2

Result by elimination. The one liquid giving no reaction in any of the three tests is hexane, the alkane.

Marking criteria: 1 mark for stating the correct test order with a reason, 1 mark each for the three tests (reagent + positive observation), 1 mark for a correctly balanced equation for at least two of the tests, 1 mark for correctly identifying the alkane by exclusion.

exam7 marksThe graph below shows the volume of hydrogen gas collected over time when excess sodium metal is added separately to equal molar amounts of pentan-1-ol and pentanoic acid under identical conditions, with the acid curve rising far more steeply and reaching a higher plateau sooner. (a) Explain, in terms of acidity, why the two curves differ as described. (b) Evaluate whether the sodium test alone is sufficient to distinguish an alcohol from a carboxylic acid, and justify what additional test should always accompany it.

Show worked solution →

(a) Explanation of the curve shapes. Both hydroxyl-containing compounds react with sodium metal to release hydrogen gas, but a carboxylic acid's $O-H$ is far more acidic (more polarised, and its conjugate base is stabilised by resonance across the carboxylate) than an alcohol's $O-H$ . The more acidic proton reacts with sodium metal faster and more vigorously, so the acid's hydrogen-volume curve rises more steeply and reaches its plateau (all sodium or all substrate consumed) sooner. The alcohol's weaker, less reactive $O-H$ releases hydrogen more slowly, giving a shallower initial slope, though at equal moles and with excess sodium both curves ultimately plateau at a comparable total volume of $H_2$ if the sodium is in excess throughout.

(b) Evaluation. The sodium test alone is NOT sufficient to reliably distinguish an alcohol from a carboxylic acid in a school laboratory: both compounds give a positive result (hydrogen evolution), and while the acid's reaction is visibly brisker, rate differences are qualitative and can be ambiguous, especially for a dilute or impure sample, or when comparing compounds of different chain length where the acid's greater vigour is less obvious. The sodium test only confirms an acidic $O-H$ is present; it does not identify WHICH functional group is responsible.

A sodium carbonate (or sodium bicarbonate) test should always accompany it: effervescence of $CO_2$ occurs with a carboxylic acid but never with an alcohol, because an alcohol's $O-H$ is not acidic enough to react with the weak base carbonate. Running the carbonate test first (before sodium) unambiguously identifies the acid by a clear-cut positive/negative result, after which the sodium test simply confirms the presence of an acidic $O-H$ in whichever compound remains.

Marker's note: full marks in part (a) need the reasoning to reference relative acidity/conjugate-base stability, not just "the acid reacts faster"; part (b) needs an explicit judgement (not sufficient) plus a named confirmatory test (carbonate) with a reason it discriminates unambiguously where the sodium test does not.

What this dot point is asking

The answer

Tests for C=C (unsaturation)

Tests for hydroxyl −OH-OH−OH (alcohol)

Tests for carboxylic acid −COOH-COOH−COOH

A flowchart for an unknown organic liquid

Summary table

Examples in context

Try this

Exam-style practice questions

Practice questions

Tests for hydroxyl $-OH$ (alcohol)

Tests for carboxylic acid $-COOH$