OxfordAQA A-Level Chemistry: Organic analysis (9620)

This study guide works through Organic analysis, section 3.3.6 of the OxfordAQA International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards. Its three parts are functional groups identified by test tube reactions, with Required practical 5 (3.3.6.1); molecular formulae by mass spectrometry (3.3.6.2); and infrared spectra (3.3.6.3). The whole section is International AS content. It is examined in the Unit 2 paper, and the Unit 5 practical and synoptic paper can draw on any content.

Afterwards, use the revision notes and the practice questions. Course links: the 9620 chemistry hub, the printable checklist and the free 10-minute diagnostics.

Topic coverage at a glance

Section	Outcome in plain words	Stage
3.3.6.1 Test tube reactions	Name the functional group present from the results of reactions given elsewhere in the specification	International AS
Required practical 5	Carry out tests for alcohols, aldehydes, alkenes and carboxylic acids	International AS
3.3.6.2 Mass spectrometry	Work out a molecular formula from precise atomic masses and a measured precise molecular mass	International AS
3.3.6.3 Infrared spectroscopy	Read IR spectra with the Chemistry data booklet to find bonds, functional groups and impurities; identify a compound by its fingerprint; explain how infrared absorption by CO₂, CH₄ and H₂O links to global warming	International AS

A stage label you need. The acid–carbonate reaction of carboxylic acids is stated in section 3.3.9.1, which is International A-level only (A2) content. Required practical 5 is an AS practical and still lists a test for carboxylic acids, so the carbonate test is taught here.

3.3.6.1 Identifying functional groups with test tube reactions

This outcome has no list of its own. It relies on reactions taught in other sections, so you gather the tests from the AS organic sections and use them as a toolkit.

The toolkit

Group	Reagent	What you see if the group is present	Source in 9620
C=C	Bromine water, shaken	Orange solution turns colourless	3.3.4.2 (“use of bromine to test for unsaturation”)
Primary or secondary alcohol	Acidified potassium dichromate(VI), warmed	Orange solution turns green	3.3.5.1
Aldehyde	Fehling’s solution, warmed	Blue solution, then a red precipitate	3.3.5.1
Aldehyde	Tollens’ reagent, warmed	A silver mirror coats the tube	3.3.5.1
Carboxylic acid	Sodium carbonate solution	Fizzing; the gas turns limewater milky	3.3.9.1 (International A-level only)

Negative results matter too. A tertiary alcohol leaves dichromate(VI) orange, as it is not easily oxidised. A ketone leaves Fehling’s solution blue and gives no silver mirror.

The chemistry behind each colour change

• Bromine. Br₂ adds across the double bond (electrophilic addition), so the orange bromine is used up. For but-2-ene: CH₃CH=CHCH₃ + Br₂ → CH₃CHBrCHBrCH₃.
• Dichromate(VI). Orange Cr₂O₇²⁻ ions gain electrons and become green Cr³⁺ ions while the alcohol is oxidised. With [O] as the oxidant, octan-2-ol (secondary) gives octan-2-one: CH₃CH(OH)(CH₂)₅CH₃ + [O] → CH₃CO(CH₂)₅CH₃ + H₂O.
• Fehling’s solution. An aldehyde reduces the blue copper(II) complex to red copper(I) oxide, Cu₂O, and is itself oxidised.
• Tollens’ reagent. The active species is the linear complex [Ag(NH₃)₂]⁺ (the specification names it in 3.2.5). An aldehyde reduces it to silver metal, which forms the mirror.
• Carbonate. The acid is strong enough to release carbon dioxide: 2HCOOH + Na₂CO₃ → 2HCOONa + CO₂ + H₂O (methanoic acid shown).

Required practical 5: doing the tests well

• Use a few drops of each liquid. Many are flammable, so warm them in a hot water bath, not a flame.
• Note the colour before and after every test. An observation names both.
• Run a blank (reagent plus water) alongside, so you can judge whether a colour has really changed.

Planning a sequence

Two tests overlap. An aldehyde also turns acidified dichromate(VI) green, because aldehydes are oxidised too. So identify any aldehyde first, and use dichromate(VI) later, once aldehydes are out of the running.

Some pairs cannot be split by any AS test tube reaction. A ketone and a tertiary alcohol both give no change with every reagent in the toolkit. You then need infrared evidence (below): C=O against O–H.

Worked example 1: five unlabelled liquids

Five bottles hold pentanoic acid, 3-methylbutanal, oct-1-ene, pentan-3-one and hexan-1-ol. Design a scheme that identifies each one.

Step	Test (on the liquids still unidentified)	Result that identifies a bottle
1	Add sodium carbonate solution	Fizzing, gas turns limewater milky: pentanoic acid
2	Warm with Fehling’s solution	Blue to red precipitate: 3-methylbutanal
3	Shake with bromine water	Orange to colourless: oct-1-ene
4	Heat gently with acidified potassium dichromate(VI)	Turns from orange to green: hexan-1-ol; no change: pentan-3-one

Hexan-1-ol, a primary alcohol, is oxidised; pentan-3-one, a ketone, is not. Using dichromate(VI) before Fehling’s solution would have turned 3-methylbutanal green as well.

Had the fifth bottle held 2-methylbutan-2-ol (tertiary) instead, both liquids at step 4 would stay orange, and only IR could separate them.

3.3.6.2 Mass spectrometry: from precise mass to molecular formula

The time of flight mass spectrometer, and its use to find relative molecular mass, belong to 3.1.1.2 and are taught in our atomic structure guide. Here the point is precision.

Many different molecular formulae share the same whole-number Mr. Their precise masses are not the same, because the relative mass of each isotope is not an exact whole number, except for carbon-12, which is 12.0000 by definition. A high-resolution instrument measures the molecular ion’s m/z to 4 decimal places. For a 1+ ion, that m/z is the precise molecular mass.

Precise relative masses used on this page:

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

Method

1. List each candidate formula. Check they all have the same whole-number Mr.
2. Multiply the number of each atom by its precise mass and add the products.
3. Keep four decimal places throughout.
4. Choose the candidate whose total equals the measured m/z, and show the sums for the others.

Worked example 2: liquid L, m/z = 128.1197

A colourless liquid L gives a molecular ion at m/z = 128.1197. The possible formulae are C₈H₁₆O, C₇H₁₂O₂, C₉H₂₀ and C₇H₁₆N₂. All four have Mr = 128.

C8H16O  : 8(12.0000) + 16(1.0078) + 15.9949              = 128.1197
C7H12O2 : 7(12.0000) + 12(1.0078) + 2(15.9949)           = 128.0834
C9H20   : 9(12.0000) + 20(1.0078)                        = 128.1560
C7H16N2 : 7(12.0000) + 16(1.0078) + 2(14.0031)           = 128.1310

Only C₈H₁₆O matches, so L is C₈H₁₆O. This is a molecular formula, not a structure. C₈H₁₆O could be an aldehyde, a ketone, an unsaturated alcohol or a cyclic alcohol. Worked example 4 narrows it down.

3.3.6.3 Infrared spectroscopy

Why bonds absorb

Covalent bonds vibrate. Infrared radiation of a matching frequency is absorbed and makes the bond vibrate more strongly. Different bonds absorb at different, characteristic wavenumbers (measured in cm⁻¹). An IR spectrum plots transmittance (%) on the vertical axis against wavenumber, with wavenumber decreasing from left to right. Each absorption appears as a downward dip.

Ranges to work with

You identify bonds using the spectrum and the Chemistry data booklet. The ranges below are the ones used on this page. In an exam, read the values from the booklet you are given.

Bond	Wavenumber / cm⁻¹	Typical appearance
O–H in alcohols	3230–3550	Broad band
N–H	3300–3500	Medium
C–H	2850–3300	Present in nearly every organic compound
O–H in carboxylic acids	2500–3000	Very broad, often merged with C–H
C≡N	2220–2260	Sharp, in a region that is otherwise quiet
C=O	1680–1750	Strong and sharp
C=C	1620–1680	Medium to weak
C–O	1000–1300	Several bands, inside the fingerprint region

O–H bands are broad because hydrogen bonding spreads the O–H bond strengths over a range.

Fingerprinting

Below about 1500 cm⁻¹ the spectrum holds many overlapping absorptions. They come from vibrations of the molecule as a whole, so the pattern is characteristic of one compound. Comparing this region with spectra of known compounds identifies a molecule, even when two candidates contain exactly the same bonds. The comparison only works if the correct compound is among the reference spectra you check.

Identifying impurities

Any band the expected product cannot produce means another substance is present. Match the band to a bond, then suggest its source: unreacted starting material, a side product or water.

Worked example 3: an aldehyde that went too far

A student oxidises heptan-1-ol with acidified potassium dichromate(VI) to make heptanal, CH₃(CH₂)₅CHO. The IR spectrum of the product shows:

• a strong, sharp absorption at 1725 cm⁻¹
• a very broad absorption across roughly 2500–3000 cm⁻¹, merged with the C–H bands
• no separate broad band in the 3230–3550 cm⁻¹ region.

Interpretation:

• 1725 cm⁻¹ lies in the C=O range (1680–1750). Heptanal has a C=O bond, so this is expected.
• The very broad 2500–3000 cm⁻¹ band is O–H in a carboxylic acid. Heptanal has no O–H.
• No band at 3230–3550 cm⁻¹, so there is no unreacted heptan-1-ol.
• Conclusion: the sample contains heptanoic acid. The aldehyde stayed in contact with the oxidising agent and was oxidised further. Distilling the aldehyde off as it forms avoids this (section 3.3.5.1).

Worked example 4: identifying liquid L

From worked example 2, L is C₈H₁₆O. Its IR spectrum shows a strong absorption at 1718 cm⁻¹, C–H bands just below 3000 cm⁻¹, nothing in the 3230–3550 cm⁻¹ region and nothing at 1620–1680 cm⁻¹. L leaves Fehling’s solution blue.

• 1718 cm⁻¹ is C=O. L is an aldehyde or a ketone.
• No O–H (alcohol) band and no C=C band, so unsaturated or cyclic alcohols are ruled out.
• Fehling’s solution stays blue, so L is not an aldehyde. L is a ketone.

Several ketones are C₈H₁₆O, including octan-2-one, octan-3-one and octan-4-one, with near-identical group absorptions. The fingerprint region of L matches the reference spectrum of octan-4-one exactly. L is octan-4-one, CH₃CH₂CH₂CH₂COCH₂CH₂CH₃.

Infrared absorption and global warming

Sunlight warms the Earth’s surface, which then radiates energy as infrared. The C=O bonds in carbon dioxide, the C–H bonds in methane and the O–H bonds in water vapour absorb some of this outgoing infrared. The molecules pass this energy on or re-emit it in all directions, so part returns towards the surface instead of escaping to space. As the concentrations of these gases rise, more infrared is absorbed, and this adds to global warming.

Common errors

• Describing bromine water as “going clear”. Colourless is the observation; clear only means transparent.
• Giving a silver mirror for Fehling’s solution or a red precipitate for Tollens’ reagent.
• Running dichromate(VI) before ruling out aldehydes, so an aldehyde is mistaken for an alcohol.
• Rounding precise masses so two candidates look equal.
• Using the C=O band alone to tell an aldehyde from a ketone. Both show it.
• Saying greenhouse gases “trap heat” without bonds absorbing infrared.

Where to go next

Next, try the practice questions. For Mr and molecular formulae, see amount of substance.

Official syllabus

OxfordAQA International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards (first teaching 2019). Topic covered: Organic analysis, section 3.3.6, subsections 3.3.6.1, 3.3.6.2 and 3.3.6.3 (International AS), and Required practical 5.