Organic synthesis: heating, separation and purification
Key idea: Choose each purification step to remove a specific impurity. Reflux returns vapour to the reaction; distillation collects a volatile fraction, and recrystallisation separates a solid as it cools.
Offline pages
Saved on this device only. Connect at least once every 30 days to confirm Premium. Sign-out or a change of account removes saved copies.
Loading guide controls…
Guided lessons · coming soon
Guided lesson
Guided steps
- Choose where the solvent should return.
- Trace vapour up and condensate back into the flask.
- Compare the receiver, thermometer and cooling-water arrangement.
- Match each treatment to what it removes.
- Explain why a wet mass gives an inflated yield.
- Calculate the yield from the dried sample mass.
- Explain which impurities drying may leave behind.
- Choose the layer and explain washing, drying and purity.
On this page
You stopped at:
ContinueMaking an organic product and obtaining a clean sample are different jobs. A reaction mixture can contain product, unreacted reagents, solvent, water and by-products. Each work-up step should remove something identified, rather than simply follow a memorised apparatus list.
Use the reaction’s stated reagents and conditions from your organic chemistry lessons. Here the focus is why you heat, separate and purify the resulting mixture.
Check your understanding 1 Choose where the condensed liquid goes
You need to keep a volatile solvent in a boiling reaction mixture rather than collect it. Which arrangement fits?
Show answer
Use reflux: condensed solvent returns to the same flask. Distillation separates a fraction into a receiver; sealing hot apparatus can cause pressure to build.
Reflux retains volatile material
Heating under reflux lets a mixture react at its boiling temperature while a condenser returns much of the escaping solvent vapour as liquid. The vertical arrangement returns condensate to the same flask; you are not collecting a separate product fraction.
Cooling water enters the lower port and leaves the upper port, keeping the jacket filled. The upper end is vented as specified; heating a sealed system can build dangerous pressure. Clamp the apparatus securely and use the heating method specified for the solvent. A water bath is suitable only when its temperature range and the procedure make it appropriate.
The diagram shows vapour rising and condensed liquid returning. Reflux limits loss of volatile material; it does not separate the product from everything else in the flask.
Distillation collects a volatile fraction
In simple distillation, vapour passes into a sloping condenser and the condensed liquid runs into a separate receiver. The thermometer bulb measures vapour near the side-arm entrance, rather than the liquid in the flask.
Swipe or scroll sideways to inspect the complete overview.
Use simple distillation when the volatility difference makes the separation adequate, for example separating a volatile liquid from a non-volatile impurity. Close-boiling liquid mixtures may need fractional distillation; even then, a boiling range is evidence about a sample, insufficient to confirm its identity or complete purity.
Many organic solvents are flammable. Use the approved electric heater or suitable bath rather than a naked flame where flammable liquids are present. Keep the system vented, secure the glassware, and follow the instructed procedure to avoid boiling to dryness. Add anti-bumping granules before heating where specified, never to an already hot boiling liquid.
Practise assembling reflux and distillation apparatus only under laboratory supervision.
Match the work-up step to the impurity
Let the mixture cool as directed before a work-up involving transfer or a separating funnel. The method and the product’s properties determine which steps are appropriate.
| Step | What it can do | What else to check |
|---|---|---|
| Separate two immiscible liquid layers | Isolate the layer containing most of the desired product | The organic layer is not always uppermost; use the stated densities or an instructed identification test |
| Wash the product-containing layer | Transfer a selected soluble impurity into a suitable wash solution | One wash does not necessarily remove every impurity; the product may also have some solubility in the wash |
| Use a suitable drying agent | Remove residual water from an organic liquid, then separate the drying agent as directed | It does not remove all dissolved organic impurities |
| Distil a liquid | Collect a fraction according to volatility | A collected fraction is not automatically pure |
| Recrystallise a solid | Exploit greater product solubility in hot solvent than cold solvent | It is not suitable for every product–solvent pair and usually sacrifices some yield |
For a separating funnel, vent as instructed, directed away from people. Washing with carbonate can release carbon dioxide when acidic material is present. Release pressure safely; do not shake a sealed funnel without venting. Remove the stopper before draining through the tap and collect the required layer according to the method.
For recrystallisation, dissolve the crude solid in the minimum appropriate volume of hot solvent. Allow it to cool so crystals form, then separate them using the specified filtration method. A small amount of cold suitable solvent can wash away adhering mother liquor while limiting product loss. Dry the crystals before measuring yield or melting range. The remaining solvent contains both impurities and some product, which explains why purification can lower isolated yield.
Yield and purity answer different questions
Percentage yield compares the isolated product amount with the theoretical amount from stoichiometry. A weighed sample must be appropriately isolated and dried before its mass is treated as product.
Worked example 1
Interpret an apparent yield above 100%
Problem
The theoretical product mass is 5.00 g. A freshly filtered, still-wet sample weighs 5.30 g. Calculate the apparent percentage yield and explain whether it demonstrates more than 100% conversion.
Show full solution
Compare the masses
Method
Divide the measured sample mass by the theoretical product mass and multiply by 100.Reason
The theoretical mass is the reference amount for complete conversion of the limiting reagent.Working
Answer
Apparent percentage yield = (5.30/5.00) × 100 = 106%.Check what was weighed
Method
Separate wet sample mass and product mass.Reason
Retained water or solvent adds to the measured mass without adding product. Other impurities or measurement errors can also inflate it.Working
Answer
The 5.30 g sample cannot be treated as 5.30 g of dry product.State the conclusion
Working
Answer
The apparent yield is 106%, but it cannot show conversion above 100%. Isolate and dry the product appropriately before reporting its yield.
Check your understanding 2 Separate yield from purity
A synthesis has a theoretical product mass of 5.00 g. The isolated, dried sample weighs 3.80 g. Calculate the percentage yield if this mass is treated as product. Consider whether drying makes it pure in the following question.
Show answer
The yield is (3.80/5.00) × 100 = 76.0%.
Check your understanding 3 Does drying make the sample pure?
A dried sample weighs 3.80 g when the theoretical product mass is 5.00 g. Explain whether drying alone makes the sample pure and how that affects the interpretation of the 76.0% apparent yield.
Show answer
The yield is (3.80/5.00) × 100 = 76.0%. Drying removes retained water or solvent, but other impurities may remain. Yield measures how much product is recovered under the stated assumption; purity needs separate evidence.
A sharp melting range close to a reliable reference supports the purity and identity of a solid. It cannot, on its own, prove either. Impurities often depress and broaden the range, while some different substances can have similar melting points.
Check your understanding 4 Choose the layer and explain each treatment
A reaction gives an organic liquid in an immiscible solvent. The method states that this organic layer is denser than the aqueous layer and contains the desired product. A water-soluble acidic impurity is removed by an appropriate wash. The isolated organic layer is then treated with a drying agent. Which layer should be retained? Explain the different purposes of washing and drying. Would a high isolated mass alone prove a pure product?
Show answer
Retain the lower layer because this specified organic layer is denser. Washing transfers the selected acidic impurity to the wash layer; drying removes residual water from the retained organic liquid. Other dissolved impurities can remain after both treatments. A high mass can include solvent, water or other impurities, so check purity separately.
The H2 syllabus includes simple organic synthesis and purification, water-bath heating, reflux and distillation. Conducting a large-scale organic synthesis involving reflux, distillation and purification is not expected in the practical examination. Knowing the purpose of the steps is different from demonstrating that you can carry them out safely and effectively.
Before choosing a procedure, name the product, the impurity to remove and the physical or chemical property that makes separation possible.
Spotted an error? Report a correction
Syllabus and review details
- GCE A-Level H2 Chemistry 2027 · 9476-2027
9476 (2027), complete syllabus
Last reviewed: