Year 12 Chemistry Module 8 ⏱ ~35 min 5 MC · 3 Short Answer Lesson 11 of 16

Chemical Tests for Functional Groups

You already know how alkenes, alcohols and carboxylic acids react, you met those reactions in Module 7. This lesson turns those same reactions into a detective's toolkit: a colour change or a fizz becomes evidence that identifies the functional group hiding in an unknown bottle.

Today's hook: A laboratory hands you an unlabelled colourless liquid and asks one question: which functional group does it contain? You cannot see bonds, but you can watch bromine water lose its colour, smell a fruity ester, or see brisk bubbling over a carbonate. Each observation rules options in or out. How would you sequence three simple tests so that, whatever the liquid is, you can name its functional group with confidence?
0/5TASKS
Before you read

An unknown colourless liquid rapidly decolourises bromine water in the dark, but produces no fizzing when sodium carbonate is added.

  • What does the bromine water result tell you is present?
  • What does the absence of fizzing rule out, and what could the liquid still be?
Learning Intentions

Know

  • The standard qualitative tests for C=C double bonds, hydroxyl (–OH) groups and carboxylic acid (–COOH) groups
  • The positive and negative observation for each test

Understand

  • Why each test works, addition across C=C decolourises bromine water; –COOH is acidic enough to release CO₂ from a carbonate; –OH is not
  • Why bromine water alone cannot separate an alkene from an unsaturated alcohol

Can Do

  • Select and sequence tests to distinguish an alkene, an alcohol and a carboxylic acid
  • Write balanced equations and state the safety controls for each test
Key Terms
Qualitative testA test that detects the presence or absence of a substance or functional group (an observation), not its concentration.
UnsaturationThe presence of a C=C double bond; unsaturated compounds undergo addition reactions.
Bromine waterBr₂ dissolved in water (orange/red-brown); decolourised by addition across a C=C double bond.
Hydroxyl group (–OH)The functional group that defines alcohols.
Carboxylic acid group (–COOH)A weak-acid functional group; reacts with carbonates and hydrogencarbonates to release CO₂.
EffervescenceVisible bubbling as a gas (here CO₂) is released from solution.
You met these reactions in Module 7, here we use them as tests: bromine water adding across C=C (M7 L05, M7 L06), structure and oxidation of alcohols (M7 L09, M7 L12), aldehyde/ketone tests (M7 L13), carboxylic acids (M7 L14) and esterification (M7 L15). This lesson focuses on the analytical logic how to tell the groups apart, not on re-deriving the reactions.

Core Content

1

Testing for C=C Double Bonds: Bromine Water

+5 XP

A colour that disappears when there is a double bond to add across

Add orange bromine water dropwise to the unknown and shake, in the absence of UV light (so a saturated compound cannot undergo a slow substitution reaction and give a false positive).

  • Positive (unsaturated): the bromine water is decolourised (orange → colourless) rapidly at room temperature, because an addition reaction adds bromine across the double bond:
    CH₂=CH₂ + Br₂ → CH₂BrCH₂Br (1,2-dibromoethane)
  • Negative (saturated, e.g. an alkane): no rapid colour change in the dark.

An alternative C=C test is acidified potassium permanganate (purple KMnO₄): an alkene decolourises the purple colour and a brown MnO₂ precipitate may form. Bromine water is the syllabus-default test; permanganate is an accepted alternative.

C=C test: add orange bromine water in the dark. Positive = rapid decolourisation (addition: CH₂=CH₂ + Br₂ → CH₂BrCH₂Br). Negative = no change. Acidified KMnO₄ (purple → colourless, brown MnO₂) is an alternative.

Pause, copy the highlighted C=C test summary into your book.

Why in the dark? Without UV light, an alkane cannot undergo the radical substitution that would slowly decolourise bromine, so a rapid decolourisation in the dark is specific to an addition across C=C.
Rapid decolourisation of bromine water in the dark indicates the presence of:
2

Testing for Hydroxyl Groups: Esterification & Oxidation

+5 XP

A fruity smell, or an orange-to-green colour change

We just saw that bromine water flags a C=C double bond. That raises a question: how do you detect an –OH group, which has no double bond to add across? This card answers it → use reactions of alcohols you met in Module 7 as identification tests.

Esterification: warm the unknown with a carboxylic acid and a few drops of concentrated H₂SO₄ (catalyst) under reflux, then neutralise excess acid with sodium hydrogencarbonate. A sweet, fruity smell indicates an ester formed, so an –OH group was present:

CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O  (ethyl ethanoate, fruity)

Oxidation (also distinguishes 1°/2°/3°): acidified dichromate (K₂Cr₂O₇/H⁺, orange) is reduced by primary and secondary alcohols, turning orange → green (Cr³⁺). Tertiary alcohols are not readily oxidised (no colour change). The oxidation behaviour of 1°/2°/3° alcohols is developed in Module 7 L12.

–OH test: esterification with a carboxylic acid + conc. H₂SO₄ under reflux gives a fruity ester smell. Acidified dichromate (orange → green) confirms a primary or secondary alcohol; tertiary alcohols give no change.

Pause, copy the highlighted –OH test summary into your book.

Reframe, don't re-derive: esterification and alcohol oxidation are Module 7 reactions. Here you use the observation (fruity smell / colour change) as evidence of an –OH group.
Warming an unknown with acidified dichromate turns it from orange to green. This indicates:
3

Testing for Carboxylic Acids: Carbonate Effervescence

+5 XP

The one group acidic enough to fizz over a carbonate

We just saw tests for C=C and –OH. That raises a question: how do you pick out a carboxylic acid, which also contains an –OH within its –COOH group? This card answers it → use its acidity, which alcohols and alkenes do not share.

Add solid or aqueous sodium carbonate (Na₂CO₃) or sodium hydrogencarbonate (NaHCO₃) to the unknown:

  • Positive (carboxylic acid): brisk effervescence CO₂ is released:
    2CH₃COOH + Na₂CO₃ → 2CH₃COONa + H₂O + CO₂↑
    CH₃COOH + NaHCO₃ → CH₃COONa + H₂O + CO₂↑
  • Confirm the gas is CO₂: bubble it through limewater (Ca(OH)₂), it turns milky/cloudy as CaCO₃ precipitates:
    CO₂ + Ca(OH)₂ → CaCO₃ + H₂O
  • Negative (alcohol or alkene): no effervescence, an –OH group is far too weakly acidic to react with a carbonate.

–COOH test: add Na₂CO₃ or NaHCO₃ → brisk effervescence of CO₂ (2CH₃COOH + Na₂CO₃ → 2CH₃COONa + H₂O + CO₂). Confirm CO₂ with limewater (turns milky). Alcohols/alkenes give no fizz.

Pause, copy the highlighted –COOH test summary into your book.

Common error: assuming any –OH-containing molecule will fizz with carbonate. Only the carboxylic acid –COOH is acidic enough; a plain alcohol –OH does not react.
True or false: an alcohol such as ethanol produces brisk effervescence when added to sodium carbonate.
4

The Trap, and a Decision Tree

+5 XP

No single test is enough, sequence them

We just saw three individual tests. That raises a question: can one test alone identify an unknown? This card answers it → no, an unsaturated alcohol decolourises bromine water too, so you must combine and sequence the tests.

The trap: if an unknown alcohol also contains a C=C, it will also decolourise bromine water. So bromine water alone cannot prove "alkene versus alcohol", you must combine tests.

A reliable sequence tests for the most distinctive group first:

  1. Add Na₂CO₃/NaHCO₃, fizz? Yes → carboxylic acid. No → step 2.
  2. Add bromine water in the dark, decolourises? Yes → contains C=C (an alkene, or an unsaturated alcohol). No → step 3.
  3. Esterification / acidified dichromate, fruity smell or orange → green → alcohol (–OH) (dichromate also classifies 1°/2°/3°).
1. Add carbonate fizz? 2. Bromine water decolourises? 3. Ester / dichromate smell / colour? carboxylic acid C=C present alcohol (–OH) 1°/2°/3° from dichromate no no yes yes

Test for the most distinctive group (the acid) first, then resolve the alkene-versus-alcohol ambiguity with bromine water and an –OH test.

No single test identifies an unknown: an unsaturated alcohol also decolourises bromine water. Sequence: carbonate (acid?) → bromine water (C=C?) → esterification/dichromate (–OH?). Test the most distinctive group first.

Pause, copy the highlighted decision tree into your book.

Why can bromine water decolourisation alone fail to distinguish an alkene from an alcohol?
5

Safety & Writing Up the Analysis

+5 XP

Controls that mark up, and what a test can and cannot prove

We just saw how to sequence the tests. That raises a question: what safety controls and what reasoning does a full-mark answer need? This card answers it → name the specific hazards and state both the positive and the negative result.

Commonly examined safety controls:

  • Bromine water is toxic and corrosive, use a fume cupboard, gloves and eye protection; keep away from UV/strong light.
  • Concentrated H₂SO₄ is corrosive and the esterification mixture is heated, use reflux, anti-bumping granules and eye protection.
  • Acidified dichromate is toxic, oxidising and a suspected carcinogen, gloves and minimal quantities.

A strong analytical answer always states both the positive and the negative observation, because an absence of change is also evidence (e.g. no fizz with carbonate rules out a carboxylic acid).

Safety: bromine water and dichromate are toxic/corrosive (fume cupboard, gloves, eye protection); conc. H₂SO₄ heated under reflux. Always report both the positive and the negative result, a "no change" is also evidence.

Pause, copy the highlighted safety and reasoning note into your book.

Analysis mindset: these tests confirm which functional group is present, they do not, on their own, give the full structure. Module 8 IQ2 pairs them with mass spectrometry, IR and NMR (the next four lessons) to pin down a complete structure.
Which control is essential when using bromine water for the C=C test?
🔬Predict, Then Reveal+8 XP
A colourless liquid gives brisk effervescence with sodium carbonate and also decolourises bromine water in the dark. Predict which functional group(s) it contains and what further test (if any) you would do.
Your predictionExpert answerCompare

Complete the Learn phase to unlock Practice.

Activities

ACTIVITY 1, Equations & Observations

For each test, write a balanced equation and state the positive observation.

1. The reaction of ethanoic acid with sodium carbonate.

2. The addition of bromine across ethene.

3. The reaction used to confirm the gas released in the carboxylic-acid test.

A2

Activity 2, Identify the Unknown

Apply the decision tree to reason about unknown samples.

1. Sample X: no fizz with Na₂CO₃; does not decolourise bromine water; turns acidified dichromate orange → green. Identify the functional group and the alcohol class.

2. Sample Y: brisk fizz with NaHCO₃. Which group is present, and which groups are ruled out?

3. Explain why you would test with carbonate before bromine water when screening an unknown.

Check Your Understanding

MC

Multiple Choice

1. An unknown liquid rapidly decolourises bromine water in the dark but produces no effervescence with sodium carbonate. The unknown most likely contains:

2. Brisk effervescence when sodium hydrogencarbonate is added indicates the presence of:

3. Why is the bromine water test performed in the absence of UV light?

4. A student finds an unknown gives no fizz with Na₂CO₃, does not decolourise bromine water, but turns acidified dichromate from orange to green on warming. The unknown is most likely:

5. Which observation confirms the gas released in the carboxylic-acid test is carbon dioxide?

SA

Short Answer

1. A colourless liquid X decolourises bromine water in the dark and produces no effervescence with sodium carbonate. Outline, with reasoning, two tests you would perform next to identify the functional group(s) present, and the expected results. (4 marks)

2. Write balanced equations for (i) the reaction used to test for a carboxylic acid with sodium carbonate, and (ii) the reaction used to confirm the gas produced. State one observation for each. (3 marks)

3. Evaluate the usefulness of these qualitative tests for fully identifying an unknown organic compound. In your answer, refer to what the tests can establish and what they cannot. (5 marks)

Show All Answers

Activity 1

1. 2CH₃COOH + Na₂CO₃ → 2CH₃COONa + H₂O + CO₂, observation: brisk effervescence.

2. CH₂=CH₂ + Br₂ → CH₂BrCH₂Br, observation: orange bromine water is decolourised.

3. CO₂ + Ca(OH)₂ → CaCO₃ + H₂O, observation: limewater turns milky/cloudy.

Activity 2

1. X is a primary or secondary alcohol: no acid (no fizz), no C=C (no decolourisation), but the orange → green dichromate change shows an oxidisable –OH.

2. Y contains a carboxylic acid (–COOH); the fizz with hydrogencarbonate rules out a simple alcohol and an alkene as the acidic group.

3. The carbonate test for the acid is the most distinctive (only –COOH fizzes), so testing it first cleanly separates acids before resolving the alkene/alcohol ambiguity.

Multiple Choice

1. C decolourising bromine water (with no acid fizz) indicates a C=C double bond.

2. C only a carboxylic acid is acidic enough to release CO₂ from hydrogencarbonate.

3. B in UV light an alkane can undergo substitution and slowly decolourise bromine, a false positive.

4. B orange → green with dichromate, with no acid and no C=C, indicates a primary or secondary alcohol.

5. B CO₂ turns limewater milky (CaCO₃ precipitate).

Short Answer Model Answers

Q1 (4 marks): Decolourising bromine water in the dark shows X contains a C=C double bond (an addition reaction removes the coloured Br₂). The lack of effervescence with Na₂CO₃ rules out a carboxylic acid. To check whether X is also an alcohol (an unsaturated alcohol would still decolourise bromine water), warm X with ethanoic acid and a few drops of concentrated H₂SO₄ under reflux: a sweet, fruity ester smell after neutralising excess acid with NaHCO₃ indicates an –OH group. Separately, warm X with acidified dichromate: an orange → green change confirms a primary or secondary alcohol (no change suggests a tertiary alcohol or no –OH). Together these distinguish a simple alkene from an unsaturated alcohol.

Q2 (3 marks): (i) 2CH₃COOH + Na₂CO₃ → 2CH₃COONa + H₂O + CO₂, observation: brisk effervescence. (ii) CO₂ + Ca(OH)₂ → CaCO₃ + H₂O, observation: limewater turns milky/cloudy. (Any valid carboxylic acid may replace ethanoic acid.)

Q3 (5 marks): Qualitative tests can establish which functional groups are present or absent, a C=C (bromine water), an –OH (esterification/oxidation) and a –COOH (carbonate effervescence), and the oxidation test can further classify an alcohol as primary/secondary or tertiary. They are quick, cheap and use observable changes. However, they cannot give the molecular formula, the carbon skeleton, the position of a group, or distinguish isomers, and overlapping results (an unsaturated alcohol decolourises bromine water) limit certainty. They are therefore a first screen, combined with mass spectrometry, IR and NMR to determine a full structure.

Return to Think First

Return to the unknown liquid that decolourised bromine water but did not fizz with carbonate.

  • Which group is confirmed, and which is ruled out?
  • Why can you not yet decide between a simple alkene and an unsaturated alcohol, and what one test resolves it?

Review

What is the positive result of the bromine water test, and why must it be done in the dark?

How do you test for a carboxylic acid, and how do you confirm the gas?

How does acidified dichromate distinguish alcohol classes?

Why is bromine water alone insufficient to identify an unknown?

In what order should you test an unknown, and why?