AQA A-Level Chemistry: Organic analysis (7405)

This guide teaches section 3.3.6 Organic analysis of the AQA AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams from June 2016 onwards. It covers all three parts: 3.3.6.1 identification of functional groups by test-tube reactions (with Required practical 6), 3.3.6.2 mass spectrometry and 3.3.6.3 infrared spectroscopy. Section 3.3.6 is AS content. The specification places it on Paper 2 at AS (sections 3.3.1 to 3.3.6) and at A-level (section 3.3), and A-level Paper 3 can test any content and any practical skill.

For quick recall, use the revision notes. To test yourself, use the practice questions. The course hub is AQA A-Level Chemistry, the printable checklist lists every specification point, and the free 10-minute diagnostics show where your gaps are.

What this unit covers

Spec point	What you must be able to do	Status
3.3.6.1 Test-tube reactions	Identify functional groups using the reactions in the specification	AS and A-level
Required practical 6	Tests for alcohol, aldehyde, alkene and carboxylic acid	AS and A-level
3.3.6.2 Mass spectrometry	Use precise atomic masses and the precise molecular mass to find a molecular formula	AS and A-level
3.3.6.3 Infrared spectroscopy	Use IR spectra and the Chemistry Data Sheet or Booklet to identify bonds, functional groups and impurities; fingerprinting; the link between IR absorption by CO₂, methane and water vapour and global warming	AS and A-level

One detail needs a label. The statement that carboxylic acids “will liberate CO₂ from carbonates” sits in 3.3.9.1, which is A-level only. Required practical 6 still asks for a carboxylic acid test, and the carbonate reaction is the one used below.

3.3.6.1 Functional groups by test-tube reactions

You identify functional groups using “the reactions of functional groups listed in the specification”. So this section collects tests from 3.3.4, 3.3.5 and 3.3.9 and uses them to tell compounds apart.

The tests

Functional group	Reagent and conditions	Positive result	Where it is in the spec
Alkene (C=C)	Bromine water, shake at room temperature	Orange to colourless	3.3.4.2
Primary or secondary alcohol	Acidified potassium dichromate(VI), warm	Orange to green	3.3.5.2
Tertiary alcohol	Acidified potassium dichromate(VI), warm	Stays orange (not easily oxidised)	3.3.5.2
Aldehyde	Tollens’ reagent, warm in a water bath	Silver mirror (grey-black precipitate of silver)	3.3.5.2
Aldehyde	Fehling’s solution, warm in a water bath	Blue solution gives a brick-red precipitate	3.3.5.2
Ketone	Tollens’ reagent or Fehling’s solution	No change	3.3.5.2
Carboxylic acid	Sodium carbonate or sodium hydrogencarbonate	Effervescence; the gas turns limewater cloudy	3.3.9.1 (A-level only statement)

What is happening in each test

• Bromine water. Bromine adds across the C=C bond by electrophilic addition, so the orange colour of bromine disappears. For ethene: CH₂=CH₂ + Br₂ → CH₂BrCH₂Br.
• Acidified dichromate(VI). The orange Cr₂O₇²⁻ ion is reduced to green Cr³⁺ as the alcohol is oxidised. A primary alcohol gives an aldehyde (and then a carboxylic acid); a secondary alcohol gives a ketone. Using [O]: CH₃CH₂OH + [O] → CH₃CHO + H₂O.
• Tollens’ reagent. This contains [Ag(NH₃)₂]⁺. An aldehyde reduces the silver(I) to silver metal: [Ag(NH₃)₂]⁺ + e⁻ → Ag + 2NH₃. The aldehyde is oxidised to a carboxylic acid: CH₃CHO + [O] → CH₃COOH. The complex ion itself is covered in the transition metals guide.
• Fehling’s solution. The blue copper(II) complex is reduced by an aldehyde to copper(I) oxide, Cu₂O, which is the red precipitate.
• Carbonate. The acid releases carbon dioxide: CH₃COOH + NaHCO₃ → CH₃COONa + CO₂ + H₂O.

Required practical 6: running the tests safely

• Use small volumes of the organic liquid in clean test tubes.
• Warm with a water bath, not a Bunsen flame. Most of these organic liquids are flammable.
• Record the colour before and after. “Goes green” is incomplete; “orange to green” is an observation.
• Run a control (reagent with water) to judge a change.

Order matters

Some tests give positive results with more than one group. An aldehyde also turns acidified dichromate(VI) green, because it is oxidised to a carboxylic acid. So test for an aldehyde before you use dichromate to look for an alcohol. A good key uses one feature per step.

Worked example 1: four unlabelled liquids

Four bottles contain hex-1-ene, propanoic acid, butanal and butan-2-ol. Plan tests to identify each one.

1  Add sodium hydrogencarbonate solution.
   Effervescence, gas turns limewater cloudy ...... propanoic acid
   No effervescence ................................ go to 2
2  Warm with Tollens' reagent.
   Silver mirror ................................... butanal
   No change ....................................... go to 3
3  Add bromine water.
   Orange to colourless ............................ hex-1-ene
   Stays orange .................................... go to 4
4  Warm with acidified potassium dichromate(VI).
   Orange to green ................................. butan-2-ol

Why this order works: the aldehyde is removed from the list at step 2, so at step 4 the only compound left is the alcohol. If you had used dichromate first, both butanal and butan-2-ol would have turned it green.

3.3.6.2 Mass spectrometry: finding a molecular formula

How a mass spectrometer works, and how it gives the relative molecular mass from the molecular ion peak, is part of 3.1.1.2. See the atomic structure guide for that. Here you use the spectrometer’s high precision.

Propan-1-ol (C₃H₈O) and ethanoic acid (C₂H₄O₂) both have Mr = 60 to the nearest whole number. A high-resolution mass spectrometer measures m/z of the molecular ion to four decimal places, and the precise masses differ because only ¹²C has a mass that is exactly a whole number.

You will be given precise atomic masses in the question. The values used on this page are:

Atom	Precise relative mass
¹H	1.0078
¹²C	12.0000
¹⁴N	14.0031
¹⁶O	15.9949

Method

1. Write down each candidate molecular formula with the right whole-number Mr.
2. Calculate the precise mass of each one from the given atomic masses.
3. Compare each with the measured value and pick the one that matches.
4. Show every calculation.

Worked example 2: molecular ion at m/z = 60.0210

A compound gives a molecular ion peak at m/z = 60.0210. Which of C₂H₄O₂, C₃H₈O, CH₄N₂O and C₂H₈N₂ is it?

C2H4O2 : 2(12.0000) + 4(1.0078) + 2(15.9949) = 60.0210
C3H8O  : 3(12.0000) + 8(1.0078) + 15.9949    = 60.0573
CH4N2O : 12.0000 + 4(1.0078) + 2(14.0031) + 15.9949 = 60.0323
C2H8N2 : 2(12.0000) + 8(1.0078) + 2(14.0031) = 60.0686

Only C₂H₄O₂ gives 60.0210, so the molecular formula is C₂H₄O₂. Mass spectrometry tells you the formula, not the structure. C₂H₄O₂ could be ethanoic acid, methyl methanoate or hydroxyethanal. You need infrared data or a test-tube reaction to decide, which is worked example 4.

3.3.6.3 Infrared spectroscopy

Why bonds absorb infrared

Covalent bonds vibrate. A bond absorbs infrared radiation when the frequency of the radiation matches a frequency at which the bond vibrates, and different bonds absorb at different, characteristic wavenumbers. The spectrum plots transmittance (%) against wavenumber in cm⁻¹. An absorption shows as a dip (a trough) in transmittance, so the spectrum reads “upside down” compared with most graphs.

The data you use

The specification expects you to use the Chemistry Data Sheet or Booklet. The ranges used on this page are below; always read them from the copy you are given.

Bond	Wavenumber / cm⁻¹
N–H (amines)	3300–3500
O–H (alcohols)	3230–3550
C–H	2850–3300
O–H (acids)	2500–3000
C≡N	2220–2260
C=O	1680–1750
C=C	1620–1680
C–O	1000–1300
C–C	750–1100

Points to read from the shape of the spectrum, not just the position:

• An alcohol O–H gives a broad absorption, because of hydrogen bonding.
• A carboxylic acid O–H is very broad and lower, at 2500–3000, and usually overlaps the C–H absorptions.
• A C=O absorption is strong and sharp, around 1700.
• Almost every organic molecule has C–H, so a C–H absorption on its own tells you very little.

Fingerprinting

The region at lower wavenumbers (below about 1500 cm⁻¹) contains many overlapping absorptions from the whole molecule. This pattern is unique to each compound, like a fingerprint. To identify a compound, you compare its spectrum with the spectrum of a known sample (for example from a database). An exact match across the fingerprint region confirms the identity. This also lets you tell apart two compounds with the same bonds, such as propanal and propanone, which both show C=O.

Worked example 3: identifying an impurity

A student oxidises propan-1-ol and distils off propanal. The IR spectrum of the distillate has a strong sharp absorption at 1730 cm⁻¹ and a broad absorption at 3400 cm⁻¹.

• 1730 cm⁻¹ is in the C=O range (1680–1750), as expected for propanal.
• 3400 cm⁻¹ is in the O–H (alcohol) range (3230–3550). Propanal has no O–H bond.
• Conclusion: the product contains unreacted propan-1-ol as an impurity.

A very broad absorption across 2500–3000 cm⁻¹ would instead point to propanoic acid, from further oxidation.

Worked example 4: putting the techniques together

Return to C₂H₄O₂ from worked example 2. Its IR spectrum shows a very broad absorption from 2500 to 3000 cm⁻¹ and a strong absorption at 1715 cm⁻¹. There is no separate broad absorption at 3230–3550 cm⁻¹.

Candidate	O–H (acid) 2500–3000	O–H (alcohol) 3230–3550	C=O 1680–1750	Fits?
Ethanoic acid, CH₃COOH	Yes	No	Yes	Yes
Methyl methanoate, HCOOCH₃	No	No	Yes	No
Hydroxyethanal, HOCH₂CHO	No	Yes	Yes	No

The compound is ethanoic acid. A test-tube check: it should effervesce with sodium hydrogencarbonate, giving a gas that turns limewater cloudy.

Infrared absorption and global warming

The Earth’s surface absorbs radiation from the Sun and re-emits energy as infrared. Bonds in carbon dioxide (C=O), methane (C–H) and water vapour (O–H) absorb some of this infrared radiation, for the same reason the bonds in an organic sample absorb in a spectrometer. The molecules then re-emit energy in all directions, including back towards the surface, so energy is kept in the lower atmosphere. Rising concentrations of these gases mean more infrared is absorbed, and this contributes to global warming.

Common errors

• Writing “bromine water goes clear”. Clear is not a colour; the change is orange to colourless.
• Using acidified dichromate(VI) to identify an alcohol when an aldehyde may still be present.
• Saying a tertiary alcohol “does not react” without stating the observation: the solution stays orange.
• Giving “red precipitate” for Tollens’ reagent. That is Fehling’s solution; Tollens’ gives a silver mirror.
• Picking a molecular formula from precise masses without showing the calculation for each candidate.
• Rounding precise masses to fewer than four decimal places, so two candidates look the same.
• Assigning a broad absorption at 2500–3000 cm⁻¹ to an alcohol. That range is the acid O–H.
• Claiming that functional-group absorptions alone tell propanal from propanone. Both show C=O; use the fingerprint region or Tollens’.
• Saying greenhouse gases “trap” heat without mentioning absorption of infrared by bonds.

Where to go next

Work through the practice questions, then see relative mass and the mole for Mr and molecular formulae and the exam preparation guide. At A-level, NMR spectroscopy (A-level only) builds on this evidence.

Official syllabus

AQA, AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams June 2016 onwards: section 3.3.6 Organic analysis (3.3.6.1 to 3.3.6.3), with Required practical 6.