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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 answer

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
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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 (OHOH) group on an alcohol and a carboxyl (COOHCOOH) 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 Br2Br_2 across the double bond:

RCH=CHRβ€²+Br2β†’RCHBrβˆ’CHBrRβ€²RCH=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 MnO4βˆ’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 MnO2MnO_2 precipitate in neutral conditions):

3RCH=CHRβ€²+2KMnO4+4H2Oβ†’3RCH(OH)βˆ’CH(OH)Rβ€²+2MnO2+2KOH3RCH=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 =CH2=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 βˆ’OH-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:

2Rβˆ’OH+2Naβ†’2Rβˆ’Oβˆ’Na++H22R-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 Cr2O72βˆ’/H2SO4Cr_2O_7^{2-}/H_2SO_4). Primary and secondary alcohols reduce dichromate from orange to green:

3Rβˆ’CH2OH+2Cr2O72βˆ’+16H+β†’3Rβˆ’COOH+4Cr3++11H2O3R-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 HH on the OH-bearing carbon to lose).

Ester formation. Heat the unknown with a carboxylic acid (ethanoic acid) and a few drops of concentrated H2SO4H_2SO_4. A sweet, fruity smell indicates ester formation, confirming an alcohol. Reverse the reagents to confirm a carboxylic acid.

Tests for carboxylic acid βˆ’COOH-COOH

Sodium carbonate (or sodium bicarbonate). Add solid Na2CO3Na_2CO_3 or NaHCO3NaHCO_3 to the unknown dissolved in water (or to the neat liquid). Brisk effervescence of CO2CO_2 indicates a carboxylic acid:

2Rβˆ’COOH+Na2CO3β†’2Rβˆ’COOβˆ’Na++H2O+CO22R-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:

2Rβˆ’COOH+Mgβ†’(Rβˆ’COO)2Mg+H22R-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 H2SO4H_2SO_4; the fruity ester smell confirms the carboxyl group.

A flowchart for an unknown organic liquid

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

  1. Bromine water. Decolourised rapidly at room temperature? β‡’\Rightarrow alkene.
  2. If no decolourisation, add sodium carbonate. Effervescence? β‡’\Rightarrow carboxylic acid.
  3. If no effervescence, add sodium metal. Hydrogen evolved? β‡’\Rightarrow alcohol.
  4. 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 KMnO4KMnO_4 (cold) No change Purple to colourless Slow (cold), fast (hot, reflux) No change
Acidified K2Cr2O7K_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 H2H_2 evolves H2H_2 evolves (faster)
Na2CO3Na_2CO_3 or NaHCO3NaHCO_3 No change No change No change CO2CO_2 effervescence
Litmus / pH Neutral Neutral Neutral Red, pH 3 to 4

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 NaHCO3NaHCO_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 CO2CO_2).

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

  • Cue. n(NaOH)=0.100Γ—0.03040=3.04Γ—10βˆ’3n(\text{NaOH}) = 0.100 \times 0.03040 = 3.04 \times 10^{-3} mol; 1:1 stoichiometry; M=0.225/3.04Γ—10βˆ’3=74.0M = 0.225 / 3.04 \times 10^{-3} = 74.0 g molβˆ’1^{-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) 3RCH2OH+Cr2O72βˆ’+8H+β†’3RCHO+2Cr3++7H2O3RCH_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 cmβˆ’1^{-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 β†’

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.

CH3CH2CH2CH=CH2+Br2β†’CH3CH2CH2CHBrCH2BrCH_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 CO32βˆ’CO_3^{2-} is a strong enough base to deprotonate the carboxylic acid:

2CH3(CH2)3COOH+Na2CO3β†’2CH3(CH2)3COONa+H2O+CO22CH_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:

2CH3(CH2)4OH+2Na→2CH3(CH2)4ONa+H22CH_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 β†’

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 KMnO4KMnO_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.

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