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Chemistry (IAL)

Inorganic & Organic Chemistry Core Practicals

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Edexcel IAL Chemistry · Core Practicals 5–8

Inorganic & Organic Chemistry Core Practicals

The big idea: Every core practical here is really the same story told four ways — you take a molecule or ion, force a change on it (break a bond, add an oxidiser, swap a functional group), and then use a simple, observable signal (a precipitate colour, a gas turning litmus blue, a flame colour) as your evidence for what actually happened at the molecular level.

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.

CH₃CH₂Br + OH⁻ → CH₃CH₂OH + Br⁻
(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.

Why hydroxide beats water as a nucleophile Water's oxygen only carries a negative charge (δ−) because it's still bonded to two hydrogens pulling electron density away. Hydroxide's oxygen carries a formal negative charge, because one of those O–H bonds has been replaced by a lone pair. A fuller negative charge means a stronger pull towards the δ+ carbon, so OH⁻ attacks faster than H₂O — that's why we use aqueous NaOH or KOH rather than just water to hydrolyse halogenoalkanes at a reasonable rate.

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).

The Rule Bond enthalpy order: C–F > C–Cl > C–Br > C–I
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.

Common mistake Students often say "iodine is the most reactive halogen so it reacts fastest" and leave it there — but on the exam, you must explain it through bond enthalpy, not just "reactivity." The examiner wants: "the C–I bond has the lowest bond enthalpy, so it requires the least energy to break, so the rate of hydrolysis is fastest." Vague reactivity language without the bond-enthalpy reasoning loses marks.

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.

  1. 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.
  2. Add a few drops of a chloroalkane, a bromoalkane, and an iodoalkane to separate tubes, and start a stopwatch the moment each is added.
  3. 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:

Ag⁺(aq) + X⁻(aq) → AgX(s)

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.

HalogenoalkanePrecipitate colourSpeed
ChloroalkaneWhiteSlowest
BromoalkaneCreamMedium
IodoalkaneYellowFastest
Memory hook "White, Cream, Yellow" — the colours get progressively deeper/darker as you go down the halogen group (Cl → Br → I), and conveniently that's also the order from slowest to fastest reaction. If you remember the colour sequence, you get the rate sequence for free.
Practice Question 1

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.

Practice Question 2

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:

1. React: 2-methylpropan-2-ol + conc. HCl (added in small portions, in a fume hood, releasing pressure each time) → left to stand 20 min, shaken at 2-min intervals 2. Separate: Pour into separating funnel → two layers form → run off (discard) the LOWER aqueous layer 3. Neutralise: Add NaHCO₃ solution to organic layer, shake, release pressure → removes acidic impurities (HCl residue) → run off (discard) the lower aqueous layer again 4. Dry: Pour organic layer into a dry conical flask, add MgSO₄ (anhydrous) → removes remaining water → decant off the clear liquid 5. Purify: Distil the liquid, collect fraction at 47–53 °C → this is the pure 2-chloro-2-methylpropane
Key point In a separating funnel, the organic layer sits on top because 2-chloro-2-methylpropane is less dense than water and immiscible with it. The aqueous layer (denser) sits below and is run off through the tap first.
Why the fume hood and slow addition of HCl? Concentrated hydrochloric acid is highly volatile at room temperature and constantly releases HCl gas, which is toxic and corrosive to breathe in. Adding it in small portions (2–3 cm³ at a time) and releasing the pressure through the stopper after each addition prevents a dangerous build-up of gas pressure inside the sealed separating funnel, which could otherwise force the stopper out or cause the funnel to leak.
Common mistake Students frequently mix up the order of "which layer do I keep." Remember: you keep the organic (top) layer at every single separation step in this practical, because that's where your product (2-chloro-2-methylpropane) has dissolved. You always discard the lower aqueous layer. Also — don't confuse the sodium hydrogen carbonate wash with the drying step; NaHCO₃ removes , while MgSO₄ removes . They are not interchangeable and both steps are needed.

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.

Practice Question 3

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.

Practice Question 4

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.

propan-1-ol --[O], mild/distil--> propanal --[O], reflux--> propanoic acid

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."

C₂H₅CH₂OH + [O] → C₂H₅CHO + H₂O (to aldehyde)
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.

GoalApparatusWhy
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 core logic Reflux = trap the vapour so the reaction has time to go further (→ acid). Distillation = remove the product as soon as it forms so the reaction can't go further (→ aldehyde).
REFLUX (vertical condenser) DISTILLATION (angled condenser) ┌─ open top ┌─ thermometer at side-arm │ vapour rises, │ vapour rises, │ condenses, │ travels sideways, │ drips BACK DOWN │ condenses, │ into the flask │ LEAVES the system │ (nothing escapes) │ (collected separately) └─ heat └─ heat → stays in reaction mixture → removed from reaction mixture → oxidation continues → oxidation stops → CARBOXYLIC ACID forms → ALDEHYDE forms

The step-by-step method

  1. Add 20 cm³ acidified K₂Cr₂O₇(aq) to a pear-shaped flask and cool it in an ice bath.
  2. Set up the reflux apparatus while keeping the flask cool.
  3. Add anti-bumping granules to the flask.
  4. Slowly add 1 cm³ of propan-1-ol dropwise into the reflux condenser using a pipette.
  5. Once added, remove the ice bath and let the mixture warm to room temperature.
  6. Heat gently (electric heater or water bath — never a naked flame) for 20 minutes.
  7. Purify the product by distillation.
Why cool the mixture first, before adding the alcohol? The oxidation of alcohols by dichromate is exothermic. Adding propan-1-ol to a warm, already-reactive mixture could cause the reaction to proceed too violently, and since propan-1-ol is flammable, a sudden release of heat and vapour is a real fire and safety hazard. Cooling first, and adding the alcohol slowly dropwise, keeps the reaction controlled.

Hazards, risks and precautions

🔥 Flammable ☣ Harmful to health ⚗ Oxidising
  • 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.
Practice Question 5

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.

Practice Question 6

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.

Overall: NH₄Cl(aq) + NaOH(aq) → NH₃(g) + H₂O(l) + NaCl(aq)
Ionic: NH₄⁺(aq) + OH⁻(aq) → NH₃(g) + H₂O(l)
Positive result Damp red litmus paper turns blue — because ammonia gas (NH₃) is alkaline and dissolves in the water on the paper to form a basic solution.
Common mistake Students often say "the gas turns litmus paper blue" without saying , or forget the litmus paper needs to be damp (not dry) — the gas needs to dissolve into the moisture on the paper to react and change the colour. A dry piece of litmus paper won't work nearly as reliably.

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.

Carbonate — Overall: 2HCl(aq) + Na₂CO₃(aq) → 2NaCl(aq) + CO₂(g) + H₂O(l)
Ionic: 2H⁺(aq) + CO₃²⁻(aq) → CO₂(g) + H₂O(l)
Hydrogencarbonate — Overall: HCl(aq) + NaHCO₃(aq) → NaCl(aq) + CO₂(g) + H₂O(l)
Ionic: H⁺(aq) + HCO₃⁻(aq) → CO₂(g) + H₂O(l)
Positive result Limewater (calcium hydroxide solution) turns from clear to milky/cloudy, because insoluble calcium carbonate (CaCO₃) precipitate forms:
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.

Ba²⁺(aq) + SO₄²⁻(aq) → BaSO₄(s)
Positive result A white precipitate of barium sulfate forms.
Why acidify with HCl first? Carbonate ions, if present, would also react with barium ions to form a white precipitate (BaCO₃), which would give a false positive for sulfate. Adding dilute HCl first destroys any carbonate ions (releasing them as CO₂ gas, which escapes), so that if a white precipitate still forms after adding BaCl₂, you can be confident it's genuinely from sulfate ions, not carbonate.

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.

IonFlame colour
Li⁺Scarlet red
Na⁺Yellow
K⁺Lilac
Rb⁺Red
Cs⁺Blue
Mg²⁺No flame colour
Ca²⁺Brick red
Sr²⁺Red
Ba²⁺Apple green
Why the loop must be cleaned between tests The test only works reliably if just one type of metal ion is present on the wire. If the loop is contaminated with a leftover ion from the previous test, two flame colours mix together and become impossible to identify correctly — this is a genuine cause of error, not just "bad technique."
Common mistake Letting the wire get so hot it glows red-hot itself (from the metal of the wire, not the sample) can be mistaken for a flame colour result. Also, students sometimes confuse Ca²⁺ (brick red) with Sr²⁺ or Li⁺ (both simply "red") — these are genuinely hard to distinguish by eye alone in an exam scenario, which is worth acknowledging as a real limitation of flame tests.
Practice Question 7

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.

Practice Question 8

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)

BondRelative strengthPrecipitate colourRate
C–ClStrongestWhiteSlowest
C–BrMediumCreamMedium
C–IWeakestYellowFastest

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)

IonReagentPositive result
NH₄⁺NaOH, warm, damp red litmusLitmus turns blue
CO₃²⁻ / HCO₃⁻Dilute HCl, gas into limewaterLimewater 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

Trap 1: "Reactivity" without mechanism Never just say a halogenoalkane is "more reactive" — examiners want the bond enthalpy explanation every time: weaker bond → less energy needed to break it → faster rate. This applies across the whole spec, not just this chapter.
Trap 2: Mixing up drying and neutralising In CP6, students often say MgSO₄ "removes impurities" vaguely, or say NaHCO₃ "dries" the product. Be precise: NaHCO₃ removes impurities (and you'll see effervescence/CO₂ release as evidence), while anhydrous MgSO₄ removes (it will look damp/cloudy when wet, and clear once dry).
Trap 3: Forgetting to justify apparatus choice in CP7 A common exam question gives you a target product (aldehyde or carboxylic acid) and asks you to justify the apparatus. Always connect it back to boiling point and whether the vapour is retained (reflux) or removed (distillation) — don't just name the apparatus without the reasoning.
Trap 4: Sulfate/carbonate test order If you add barium chloride before acidifying with HCl in the sulfate test, any carbonate present will also give a white precipitate — a false positive. Examiners frequently ask "why is HCl added first" — the answer is always to eliminate this false-positive risk from carbonate ions.
Trap 5: Ionic vs overall equations Practice writing both the full (overall) equation and the ionic equation for each test-tube reaction in CP8 — mark schemes often accept either, but ask specifically for the ionic form (spectator ions like Na⁺ and Cl⁻ removed) to test real understanding of what's chemically happening.
What examiners are really checking for across this whole chapter Can you connect an (colour change, precipitate, gas turning litmus blue) to the correct (which bond broke, which ion is present, why this apparatus was chosen)? That link is worth more marks than simply describing what you saw.
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  • Exam Tips — Traps & What Examiners Look For
  • Testing for carbonate & hydrogencarbonate ions
  • Bond enthalpy & precipitate colours (CP5)
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