AQA A-Level Chemistry (7405)

## Learning Outcomes

- Identify the correct reagent, conditions and positive observation for major organic functional group tests
- Explain how functional group tests are used in sequence to identify an unknown organic compound
- Distinguish between tests that confirm a group directly and tests that only narrow down possibilities
- Recall the exact observations for alkenes, alcohols, aldehydes, ketones, carboxylic acids and halogenoalkanes
- Apply synoptic reasoning to choose between overlapping test results, such as aldehyde versus ketone or alkene versus phenol
- Avoid common exam errors involving vague observations, missing conditions and incorrect conclusions

## AQA A-Level Chemistry (7405) Syllabus

For the AQA A-Level Chemistry (7405), you are expected to understand organic functional group tests, with a focus on the following syllabus points:

- Know the tests for alcohols, aldehydes, alkenes and carboxylic acids from Required Practical 6
- Identify carbonyl compounds using **2,4-DNPH** and then distinguish aldehydes from ketones using **Tollens' reagent** or **Fehling's solution**
- Explain how **acidified potassium dichromate(VI)** is used to test oxidisable compounds and why it does not uniquely identify alcohols
- Understand that halogenoalkanes are tested by **hydrolysis first**, followed by testing the halide ions with **nitric acid and silver nitrate**
- Recall the colours of **silver halide precipitates** and their behaviour with ammonia
- Recognise that **bromine water** can be decolourised by more than one type of compound and use the observation carefully
- Know the **iodoform test** and what structural feature gives a positive result

## Test Your Knowledge

_Attempt these questions before reading this article. If you find some difficult or cannot remember the answers, look more closely at that area during your revision._

1. What reagent is used to test for an alkene, and what is the positive result?
2. Why does acidified potassium dichromate(VI) not prove on its own that an unknown compound is an alcohol?
3. How can you distinguish an aldehyde from a ketone using test-tube reactions?
4. Why must a halogenoalkane be hydrolysed before using silver nitrate?
5. Which compounds give a positive iodoform test?

## Introduction

Organic analysis often looks simple in revision tables, but exam questions are rarely just a memory test. You usually need to choose the right reagent, state the conditions, give the exact observation, and then explain what the result means and what it does not mean.

In synoptic questions, one test is often not enough. A positive result may only narrow the possibilities, so you need a sequence of tests and a clear chain of logic. That is why precise wording matters so much in this topic.

> **Key Term: functional group**
> A specific atom or group of atoms in an organic molecule that causes the molecule to have characteristic chemical reactions.

### Core Required Practical 6 tests

The highest-yield tests are the ones from **Required Practical 6**: alkene, alcohol, aldehyde and carboxylic acid tests. These are the tests most likely to appear directly or as part of a longer identification problem.

For an **alkene**, add **bromine water** at room temperature. A positive result is that the **orange/brown bromine water decolourises to colourless**. This happens because bromine adds across the **C=C** double bond. The key chemical idea is addition: the double bond is electron-rich and reacts readily.

> **Key Term: bromine water test**
> A test for unsaturation in which bromine water is decolourised from orange or brown to colourless by a compound containing a carbon-carbon double bond.

This test is useful, but it is not fully unique. **Phenol** and **phenylamine** also decolourise bromine water, but they also produce a **white precipitate**. A simple alkene gives decolourisation only. That distinction is a favourite exam trap.

For an **alcohol**, warm with **acidified potassium dichromate(VI)**. Primary and secondary alcohols change the solution from **orange to green** because dichromate(VI) ions are reduced to chromium(III) ions while the alcohol is oxidised. Tertiary alcohols do not react under these conditions, so the solution stays orange.

> **Key Term: acidified potassium dichromate(VI) test**
> A test in which acidified potassium dichromate(VI) changes from orange to green when it oxidises a primary alcohol, a secondary alcohol or an aldehyde.

The weakness of this test is easy to miss: **aldehydes also turn dichromate from orange to green**. So if the question asks you to identify an alcohol, this reagent on its own is not enough unless you already know the unknown cannot be an aldehyde.

For an **aldehyde**, two classic tests are used. Warm with **Tollens' reagent** to get a **silver mirror**, or sometimes a grey silver precipitate. Warm with **Fehling's solution** to get a **brick-red precipitate**. Ketones do not give these results.

> **Key Term: Tollens' reagent**
> An alkaline solution containing diamminesilver(I) ions, $[Ag(NH_3)_2]^+$, which is reduced by an aldehyde to form a silver mirror.
>
> **Key Term: Fehling's solution**
> An alkaline blue solution of complexed copper(II) ions that gives a brick-red precipitate with an aldehyde on warming.

The reason aldehydes react and ketones do not is structural. An aldehyde has a hydrogen attached to the carbonyl carbon, so it is readily oxidised to a carboxylic acid. Oxidising a ketone further would require breaking a carbon-carbon bond, which does not happen easily under these mild conditions.

For a **carboxylic acid**, add **sodium carbonate** or **sodium hydrogencarbonate**. A positive result is **effervescence** as **carbon dioxide** is produced. If the gas is bubbled through limewater, the limewater turns cloudy.

Equation with hydrogencarbonate:

$$
RCOOH(aq) + NaHCO_3(aq) \rightarrow RCOONa(aq) + H_2O(l) + CO_2(g)
$$

Equation with carbonate:

$$
2RCOOH(aq) + Na_2CO_3(aq) \rightarrow 2RCOONa(aq) + H_2O(l) + CO_2(g)
$$

> **Key Term: carbonate test**
> A test for acids in which a carbonate or hydrogencarbonate reacts to produce carbon dioxide, seen as effervescence.

Be careful with the conclusion. This test shows the presence of an **acid**, not specifically a carboxylic acid, unless the context tells you the unknown is organic.

> **Test Tip:** In AQA questions, a full-mark answer usually needs three parts: **reagent**, **conditions** if relevant, and **exact observation**. "It changes colour" is too vague. Write "orange acidified potassium dichromate(VI) turns green".

### Carbonyl compounds and oxidation logic

A common exam sequence starts with a compound that contains oxygen. You then need to decide whether it is an alcohol, aldehyde, ketone or carboxylic acid. The hardest part is dealing with overlapping results.

The fastest first step for a possible aldehyde or ketone is **2,4-DNPH**, also called **Brady's reagent**. A positive result is an **orange or yellow precipitate**, showing the presence of a **carbonyl compound**.

> **Key Term: 2,4-DNPH test**
> A test in which aldehydes and ketones react with 2,4-dinitrophenylhydrazine to form an orange or yellow precipitate.

This test does not distinguish between aldehydes and ketones. It only tells you that a carbonyl group is present. You then need a second test. If **Tollens' reagent** gives a silver mirror, or **Fehling's solution** gives a brick-red precipitate, the compound is an **aldehyde**. If there is no reaction in those tests after a positive 2,4-DNPH result, it is a **ketone**.

This is a classic two-step logic chain:

1. **2,4-DNPH positive** means aldehyde or ketone.
2. **Tollens' or Fehling's positive** means aldehyde.
3. **Tollens' and Fehling's negative** means ketone.

Worked example:

An unknown liquid gives:

- orange precipitate with 2,4-DNPH
- no reaction with Tollens' reagent
- no change with Fehling's solution

Reasoning:

- The 2,4-DNPH result shows a carbonyl compound is present.
- The negative Tollens' and Fehling's tests show it is not an aldehyde.
- Therefore, the compound is a **ketone**.

Students often confuse this with dichromate. A ketone does not react with acidified dichromate(VI), but a negative dichromate result alone does not prove ketone, because **tertiary alcohols** also give no reaction.

Now consider alcohols in more detail. AQA often expects you to distinguish **primary**, **secondary** and **tertiary** alcohols by using oxidation products.

- Primary alcohol + acidified dichromate(VI) gives orange to green and can form an **aldehyde** on distillation or a **carboxylic acid** under reflux.
- Secondary alcohol + acidified dichromate(VI) gives orange to green and forms a **ketone**.
- Tertiary alcohol gives **no reaction**.

Why does distillation matter for a primary alcohol? Because the aldehyde product is removed as it forms, so it is less likely to be oxidised further to the carboxylic acid. Under reflux, the aldehyde stays in the flask and is oxidised further.

Worked distinction:

You have two unknown alcohols, one primary and one secondary.

Step 1: Warm each with acidified potassium dichromate(VI).

- Both turn from orange to green, so both are oxidisable alcohols.

Step 2: Distil the oxidation products and test with Tollens' reagent.

- If the distillate gives a silver mirror, the product is an aldehyde, so the original alcohol was **primary**.
- If the distillate does not react with Tollens', the product is a ketone, so the original alcohol was **secondary**.

Common student errors here:

- saying dichromate identifies any alcohol
- forgetting that aldehydes also reduce dichromate
- confusing reflux with distillation and therefore misidentifying the product
  > **Test Tip:** If the question says "confirm the presence of an aldehyde", give a **positive aldehyde test** such as Tollens' or Fehling's. Do not answer with "it is not a ketone because..." unless the question specifically asks for elimination logic.

### Halogenoalkanes, bromine water overlaps and the iodoform test

The test for a **halogenoalkane** is often described badly by students. Silver nitrate does **not** test the halogenoalkane directly. First, the compound must be **hydrolysed** so that the halogen leaves as a halide ion.

General hydrolysis equation:

$$
R-X + OH^- \rightarrow R-OH + X^-
$$

After hydrolysis, the halide ion is tested with **nitric acid and silver nitrate**. The precipitate colour identifies the halide:

- **AgCl**: white
- **AgBr**: cream
- **AgI**: yellow
  > **Key Term: halogenoalkane hydrolysis**
  > The reaction of a halogenoalkane with hydroxide ions to produce an alcohol and a halide ion before testing the halide ion.
  >
  > **Key Term: silver halide precipitate**
  > An insoluble silver compound formed when silver ions react with halide ions, used to identify chloride, bromide or iodide ions.

Why is nitric acid added? To remove interfering ions such as carbonate or hydroxide that might also react with silver ions and give misleading precipitates. Nitric acid is used because nitrate ions do not form a precipitate with silver ions under these conditions.

A further check with ammonia can confirm the identity:

- **AgCl** dissolves in **dilute ammonia**
- **AgBr** dissolves in **concentrated ammonia**
- **AgI** does **not** dissolve in ammonia

This gives stronger evidence when the white, cream and yellow shades are hard to tell apart.

Another synoptic overlap comes with **bromine water**. If bromine water decolourises:

- **alkene**: decolourisation only
- **phenol or phenylamine**: decolourisation and **white precipitate**

That extra precipitate matters. Without it, you could make the wrong conclusion.

The **iodoform test** is also useful in Paper 3 because it is more structural than simply naming a functional group. Warm with **iodine in alkaline solution**, usually **I₂/NaOH**. A positive result is a **yellow precipitate of iodoform, CHI₃**.

> **Key Term: iodoform test**
> A test using iodine in alkaline solution that gives a yellow precipitate of CHI₃ for compounds containing the CH₃CO– group or the CH₃CH(OH)– group.

This test is positive for:

- **methyl ketones**, containing **CH₃CO–**
- secondary alcohols containing **CH₃CH(OH)–**
- **ethanol**, because it is oxidised under the test conditions to ethanal

This is a good example of why pattern recognition matters. Not all alcohols give the iodoform test. Only ethanol and alcohols with the right structural fragment do.

A useful unknown-identification sequence is:

1. Test with **bromine water**
   - decolourises only: alkene possible
   - decolourises and white precipitate: phenol or phenylamine possible

2. Test with **2,4-DNPH**
   - orange precipitate: aldehyde or ketone

3. Test with **Tollens' or Fehling's**
   - positive: aldehyde
   - negative after positive 2,4-DNPH: ketone

4. Test with **acidified dichromate(VI)**
   - orange to green: primary alcohol, secondary alcohol or aldehyde

5. Test with **carbonate or hydrogencarbonate**
   - effervescence of CO₂: carboxylic acid

6. For suspected halogenoalkane:
   - hydrolyse, then acidify and add **AgNO₃**
   - identify halide by precipitate colour

A final practical point: many organic liquids are flammable, so gentle warming in a **water bath** is safer than using a naked flame. That matters in practical-method questions.

## Key Point Checklist

_This article has covered the following key knowledge points:_

- Bromine water is decolourised from orange or brown to colourless by an alkene
- Phenol and phenylamine also decolourise bromine water, but they give a white precipitate as well
- Acidified potassium dichromate(VI) changes from orange to green with primary alcohols, secondary alcohols and aldehydes
- Tertiary alcohols and ketones do not react with acidified potassium dichromate(VI) under these conditions
- 2,4-DNPH gives an orange or yellow precipitate with aldehydes and ketones
- Tollens' reagent gives a silver mirror, or grey silver precipitate, with an aldehyde
- Fehling's solution gives a brick-red precipitate with an aldehyde
- Carboxylic acids react with sodium carbonate or sodium hydrogencarbonate to produce effervescence of carbon dioxide
- Halogenoalkanes must be hydrolysed first before the halide ions are tested with nitric acid and silver nitrate
- Silver halide precipitates are white for chloride, cream for bromide and yellow for iodide
- The iodoform test gives a yellow precipitate of CHI₃ for compounds containing CH₃CO– or CH₃CH(OH)–
- Strong exam answers state the reagent, any needed conditions and the exact observation

## Key Terms and Concepts

- functional group
- bromine water test
- acidified potassium dichromate(VI) test
- Tollens' reagent
- Fehling's solution
- carbonate test
- 2,4-DNPH test
- halogenoalkane hydrolysis
- silver halide precipitate
- iodoform test


Organic analysis - Functional group tests