Identify halogenoalkanes with ethanolic silver nitrate
Use aqueous silver nitrate in ethanol to detect halide ions released from a halogenoalkane.
AS Level; Cambridge International Chemistry 9701, 2025–2027 syllabus. Use the named reagents, solvents, temperatures and pressure conditions; begin curly arrows at a bond or lone pair and distinguish typical mechanisms from absolute rules. Written practice is self-checked
Before you begin
- Use covalent formulas, nucleophiles, electrophiles, homolytic and heterolytic fission, and the alkene and alcohol pathways in Topics 13–14.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What color is silver bromide precipitate?
What you will learn
- State the ethanolic silver nitrate test
- Match silver halide precipitate colors
- Use the result as evidence of halogen identity
Halide released forms a precipitate
Warm a halogenoalkane with aqueous silver nitrate in ethanol. Reaction with the solvent mixture releases halide ions; Ag⁺ reacts with X⁻ to form a silver halide precipitate.
AgCl is white, AgBr is cream and AgI is yellow. With bromoethane, a cream AgBr precipitate is expected.
Interpret color as a test result
The precipitation reaction is Ag⁺(aq) + X⁻(aq) → AgX(s). The observed color indicates which silver halide formed.
Color supports a halogen identification when the conditions and controls are suitable. Avoid claiming that any color alone identifies an unknown without considering mixtures or contamination.
Pause and explain
What observation indicates iodide in this silver-ion test?
Worked example
Describe the observation when bromoethane is warmed with aqueous silver nitrate in ethanol and explain it.
Show the worked solution
- The halogenoalkane releases Br⁻ into the mixture.
- Ag⁺ combines with Br⁻.
- The silver bromide precipitate is cream.
Answer A cream precipitate forms because Ag⁺(aq) + Br⁻(aq) → AgBr(s).
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Predict the precipitate color from chlorobutane with aqueous silver nitrate in ethanol.
Give me a hint
Identify the halide ion and corresponding AgX solid.
Compare my reasoning
- Chlorobutane can release Cl⁻.
- Ag⁺ combines with Cl⁻ to form AgCl(s).
- Silver chloride is white.
A white AgCl precipitate forms.
Look for these in your work
- Halide identified
- Precipitate formula correct
- Color correct
A cream precipitate forms after heating a halogenoalkane with ethanolic silver nitrate. What halide is supported, and what equation represents the precipitate step?
Give me a hint
Map cream to the silver halide and write the net ionic equation.
Compare my reasoning
- Cream corresponds to AgBr.
- The halide is Br⁻.
- Ag⁺ + Br⁻ → AgBr(s).
Bromide is supported: Ag⁺(aq) + Br⁻(aq) → AgBr(s), a cream precipitate.
Look for these in your work
- Halide inferred
- Ionic equation stated
- Color linked
Why should a precipitate color be interpreted with a clean apparatus and suitable comparison?
Give me a hint
Think about contamination and evidence quality.
Compare my reasoning
- Another halide could contaminate the sample or apparatus.
- Contamination may produce a second silver halide color or mixed appearance.
- A clean apparatus and known comparison improve the inference.
Contamination can produce misleading precipitates, so clean apparatus and suitable comparisons strengthen the identification.
Look for these in your work
- Confounder named
- Effect explained
- Control justified
Common mistakes
- Calling AgBr white
- Confusing silver nitrate with the silver-ion halide test using ammonia
- Reporting a color without naming the silver halide
What can you explain now?
What color is silver bromide precipitate?
Compare with the explanation
Cream
AgBr forms a cream precipitate in this test.
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Make it stick
What three precipitate colors distinguish AgCl, AgBr and AgI?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.