Inorganic & Organic Chemistry Core Practicals
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Inorganic & Organic Chemistry Core Practicals
Summary — What This Chapter Covers
- Core Practical 5 — Hydrolysis of Halogenoalkanes: reacting halogenoalkanes with warm aqueous silver nitrate to measure how fast the C–X bond breaks, based on bond enthalpy.
- Core Practical 6 — Chlorination of 2-Methylpropan-2-ol: converting an alcohol into a haloalkane using concentrated HCl, then purifying the product using a separating funnel, drying agent, and distillation.
- Core Practical 7 — Oxidation of Propan-1-ol: using acidified potassium dichromate(VI) under reflux (to make the acid) or distillation (to stop at the aldehyde), and understanding why the apparatus choice matters.
- Core Practical 8 — AS Qualitative Analysis: a toolkit of test-tube reactions to identify ammonium, carbonate/hydrogencarbonate, and sulfate ions, plus flame tests for metal ions.
Core Practical 5: Hydrolysis of Halogenoalkanes
Why halogenoalkanes hydrolyse
A halogenoalkane has a polar C–X bond (X = F, Cl, Br, or I), because the halogen is more electronegative than carbon. This makes the carbon slightly positive (δ+) — a perfect target for an electron-rich species (a nucleophile) to attack. When that nucleophile is hydroxide, OH⁻, the halogen atom is kicked out and replaced by an –OH group. This is called hydrolysis because water (or its more reactive cousin, hydroxide) is doing the breaking-apart.
(bromoethane) (ethanol) (bromide ion)
Notice this is very slow at room temperature — that's why the reaction mixture always needs warming. Warming gives the molecules enough kinetic energy to collide successfully and overcome the activation energy of breaking the C–X bond.
Bond enthalpy controls the rate
This is the part students most often get backwards, so let's slow down. The rate of hydrolysis depends on how strong the C–X bond is — not on how electronegative the halogen is, and not on the size of the halogen atom directly (though those two things are related to bond strength).
Rate of hydrolysis order: C–I > C–Br > C–Cl (C–F essentially doesn't react)
Think of it like this: imagine each C–X bond as a padlock holding the halogen in place. The C–F padlock is the strongest — practically un-pickable at these temperatures, so fluoroalkanes don't hydrolyse at all under normal lab conditions. The C–I padlock is the weakest and flimsiest, because iodine is a large atom and its bonding electrons are further from the nucleus and more weakly held — so it's the easiest bond to break, and iodoalkanes react fastest.
The experiment: measuring the rate
Here's the clever trick of this practical: you can't easily "watch" a C–X bond break directly, so instead you use a reaction that turns the invisible halide ion product into something you see — a precipitate.
- Set up three test tubes containing a mixture of ethanol (as a solvent so everything mixes — halogenoalkanes don't dissolve well in water) and acidified silver nitrate solution, all sitting in a 50 °C water bath.
- Add a few drops of a chloroalkane, a bromoalkane, and an iodoalkane to separate tubes, and start a stopwatch the moment each is added.
- Time how long each tube takes to form a visible precipitate.
As hydrolysis proceeds, halide ions (Cl⁻, Br⁻, I⁻) are released into solution. These react instantly with the Ag⁺ ions already present to form an insoluble silver halide precipitate:
Because this silver halide precipitation step is essentially instantaneous compared to the hydrolysis step, the time you measure is really a stand-in for how long the hydrolysis (bond-breaking) step took — that's the whole logic of the experiment.
| Halogenoalkane | Precipitate colour | Speed |
|---|---|---|
| Chloroalkane | White | Slowest |
| Bromoalkane | Cream | Medium |
| Iodoalkane | Yellow | Fastest |
Explain, in terms of bond enthalpy, why 1-chlorobutane hydrolyses more slowly than 1-iodobutane. Your answer should reference the strength of the relevant bonds.
Why is ethanol included in the reaction mixture along with the halogenoalkane and silver nitrate solution?
Core Practical 6: Chlorination of 2-Methylpropan-2-ol
The idea: running hydrolysis in reverse
Core Practical 5 turned a halogenoalkane into an alcohol using OH⁻ as the nucleophile. This practical does the opposite — it turns an alcohol, 2-methylpropan-2-ol, into a haloalkane, 2-chloro-2-methylpropane, by reacting it with concentrated hydrochloric acid. The –OH group is swapped out for a –Cl.
This is a genuinely useful synthesis technique, but the tricky part of this practical isn't really the chemistry — it's the purification procedure. You're going to make a small amount of product mixed in with leftover acid, water, and other impurities, and you need a systematic way to isolate a pure, dry product.
The purification sequence — step by step
Think of this as a five-stage clean-up chain. Each stage removes one specific type of impurity, and the order matters:
Why distil, and why that exact range?
Distillation is the final purification step because it separates compounds by boiling point. 2-chloro-2-methylpropane boils at around 51 °C — so collecting the fraction distilling over between 47–53 °C ensures you are collecting essentially pure product and leaving behind any higher- or lower-boiling impurities still in the flask.
Sodium hydrogencarbonate solution is added to the organic layer during purification, and the funnel is shaken with the pressure released at regular intervals. Explain why releasing the pressure is necessary at this stage.
Describe how you would separate the organic and aqueous layers using a separating funnel, and identify which layer contains the product.
Core Practical 7: Oxidation of Propan-1-ol
Two products, one starting material
This is the single most important idea in this practical: a primary alcohol like propan-1-ol can be oxidised to two different products depending on how far you push the reaction and what apparatus you use.
The oxidising agent throughout is acidified potassium dichromate(VI), K₂Cr₂O₇(aq)/H⁺. It's written as "[O]" in equations as shorthand for "an oxygen atom supplied by the oxidising agent."
C₂H₅CHO + [O] → C₂H₅COOH (aldehyde → acid)
Reflux vs. distillation — the apparatus IS the answer
This is where the practical really tests understanding, not just memorisation. The choice between reflux and distillation apparatus directly determines which product you end up with, and you need to be able to explain why.
| Goal | Apparatus | Why |
|---|---|---|
| Make propanoic acid (full oxidation) | Reflux (vertical condenser, vapour condenses and drips back down) | Keeps the alcohol and intermediate aldehyde in contact with the oxidising agent for a long time (here, 20 minutes of heating), so oxidation goes all the way to the carboxylic acid |
| Make propanal (stop at aldehyde) | Distillation (angled condenser, vapour escapes and is collected) | Propanal has a lower boiling point than propan-1-ol, so as soon as it forms it evaporates and distils off it can be oxidised further to the acid |
The step-by-step method
- Add 20 cm³ acidified K₂Cr₂O₇(aq) to a pear-shaped flask and cool it in an ice bath.
- Set up the reflux apparatus while keeping the flask cool.
- Add anti-bumping granules to the flask.
- Slowly add 1 cm³ of propan-1-ol dropwise into the reflux condenser using a pipette.
- Once added, remove the ice bath and let the mixture warm to room temperature.
- Heat gently (electric heater or water bath — never a naked flame) for 20 minutes.
- Purify the product by distillation.
Hazards, risks and precautions
- Alcohols (propan-1-ol, butan-1-ol, pentan-1-ol) are flammable and often harmful — keep away from naked flames and avoid skin contact/inhalation. Use a fume cupboard for harmful alcohols.
- Potassium dichromate is a strong oxidising agent (and is also toxic/carcinogenic) — handle with care and avoid contact.
- Any spillages should be mopped up immediately with plenty of water.
- Because ethanol/propanol vapours are flammable, heating must be done with an electric heater or water bath — never a Bunsen burner flame directly.
A student wants to prepare propanal from propan-1-ol. Explain why distillation apparatus, rather than reflux apparatus, should be used, and explain the role of boiling point in your answer.
Why should the reaction mixture be heated using an electric heater or a water bath rather than a Bunsen burner with a naked flame?
Core Practical 8: AS Qualitative Analysis
This practical is essentially a toolkit of ion "fingerprint" tests. Each test tube reaction has a distinct, unambiguous observation that lets you identify an ion is present. If the sample is a solid, dissolve it in deionised water first to make an aqueous solution before testing.
Testing for ammonium ions (NH₄⁺)
Method: Add ~10 drops of the ammonium-containing solution to a clean test tube, then add ~10 drops of sodium hydroxide solution and swirl to mix. Warm gently in a water bath. Hold a piece of damp red litmus paper near the mouth of the tube (don't let it touch the sides) using tongs.
Ionic: NH₄⁺(aq) + OH⁻(aq) → NH₃(g) + H₂O(l)
Testing for carbonate & hydrogencarbonate ions
Method: Add a small amount (~1 cm³) of dilute hydrochloric acid to a test tube, then add an equal amount of the sample solution. Immediately attach a bung with a delivery tube, directing the gas produced into a second test tube containing limewater.
Ionic: 2H⁺(aq) + CO₃²⁻(aq) → CO₂(g) + H₂O(l)
Ionic: H⁺(aq) + HCO₃⁻(aq) → CO₂(g) + H₂O(l)
CO₂(g) + Ca(OH)₂(aq) → CaCO₃(s) + H₂O(l)
Note that carbonate and hydrogencarbonate give the same observation — both produce CO₂ that turns limewater milky. So this test alone can't distinguish between the two; it only confirms "a carbonate-type species is present."
Testing for sulfate ions (SO₄²⁻)
Method: Acidify the sample with dilute hydrochloric acid, then add a few drops of aqueous barium chloride.
Flame tests for metal ions
Method: Dip a loop of unreactive wire (nichrome or platinum) in concentrated acid, then hold it in the blue part of a Bunsen flame until it produces no colour (this cleans the wire and prevents contamination from previous tests). Dip the clean loop into the solid sample and hold it at the edge of the blue flame. Observe the colour produced.
| Ion | Flame colour |
|---|---|
| Li⁺ | Scarlet red |
| Na⁺ | Yellow |
| K⁺ | Lilac |
| Rb⁺ | Red |
| Cs⁺ | Blue |
| Mg²⁺ | No flame colour |
| Ca²⁺ | Brick red |
| Sr²⁺ | Red |
| Ba²⁺ | Apple green |
A white solid is thought to contain sulfate ions. Describe a test you could carry out to confirm this, including the expected positive result and the reason for any preliminary step.
Explain why the nichrome wire loop is dipped in acid and heated in a flame until it shows no colour, before it is used to test an unknown solid sample.
What to Memorise
Bond enthalpy & precipitate colours (CP5)
| Bond | Relative strength | Precipitate colour | Rate |
|---|---|---|---|
| C–Cl | Strongest | White | Slowest |
| C–Br | Medium | Cream | Medium |
| C–I | Weakest | Yellow | Fastest |
Purification stages (CP6)
React → Separate (discard aqueous) → Wash with NaHCO₃ (removes acid) → Dry with MgSO₄ (removes water) → Distil (collect 47–53 °C fraction)
Reflux vs distillation (CP7)
Reflux traps vapour → full oxidation → carboxylic acid. Distillation removes vapour early → stops at aldehyde.
Ion tests (CP8)
| Ion | Reagent | Positive result |
|---|---|---|
| NH₄⁺ | NaOH, warm, damp red litmus | Litmus turns blue |
| CO₃²⁻ / HCO₃⁻ | Dilute HCl, gas into limewater | Limewater turns milky |
| SO₄²⁻ | Dilute HCl then BaCl₂(aq) | White precipitate |
Flame colours: Li⁺ scarlet red · Na⁺ yellow · K⁺ lilac · Ca²⁺ brick red · Ba²⁺ apple green · Cu²⁺ not listed here but commonly blue-green in wider spec.
Concepts Checklist
Exam Tips — Traps & What Examiners Look For
- Exam Tips — Traps & What Examiners Look For
- Testing for carbonate & hydrogencarbonate ions
- Bond enthalpy & precipitate colours (CP5)
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