OxfordAQA A-Level Chemistry: Organic analysis (9620) – Revision Notes

Here is 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, in condensed form. The notes summarise functional group tests and Required practical 5 (3.3.6.1), molecular formulae from precise masses (3.3.6.2) and IR spectra (3.3.6.3). Section 3.3.6 is International AS content and is examined in Unit 2; the one exception, the carbonate test, is labelled below. For full explanations and worked examples, use the organic analysis study guide.

When you can recall these notes, try the practice questions. Other links: 9620 chemistry hub, printable checklist and free 10-minute diagnostics.

3.3.6.1 Test tube reactions (Required practical 5)

Required practical 5: “Carry out tests for alcohols, aldehydes, alkenes and carboxylic acids”. The reactions come from other 9620 sections.

Looking for	Reagent and condition	Positive	Negative
Alkene, C=C	Shake with bromine water	Orange → colourless	Remains orange
1° or 2° alcohol	Warm with K₂Cr₂O₇ and acid	Orange → green	Remains orange (3° alcohol, ketone)
Aldehyde	Fehling’s solution, warm	Blue → red precipitate (Cu₂O)	Stays blue
Aldehyde	Tollens’ reagent, warm	Silver mirror	No mirror
Carboxylic acid	Na₂CO₃(aq)	Fizzing; CO₂ turns limewater milky	No fizzing

Stage label: the carbonate reaction appears in 3.3.9.1, which is International A-level only. The test is still needed for Required practical 5.

What changes in each test

Test	Species before	Species after	Type of change
Dichromate(VI)	Cr₂O₇²⁻ (orange)	Cr³⁺ (green)	Reduction of Cr; alcohol oxidised
Fehling’s	Cu(II) complex (blue)	Cu₂O (red solid)	Reduction of Cu; aldehyde oxidised
Tollens’	[Ag(NH₃)₂]⁺ (colourless)	Ag (mirror)	Reduction of Ag; aldehyde oxidised
Bromine water	Br₂ (orange)	Dibromoalkane (colourless)	Electrophilic addition
Carbonate	CO₃²⁻	CO₂ gas	Acid–carbonate reaction

Method in steps: sorting a set of unknowns

1. Write the functional groups present in each candidate.
2. Pick a first test that only one candidate passes. The carbonate test is often a good start.
3. Remove aldehydes with Fehling’s or Tollens’ before any dichromate(VI) test.
4. Use bromine water for C=C.
5. Use dichromate(VI) to split 1°/2° alcohols from everything that stays orange.
6. If a ketone and a 3° alcohol remain, no AS test tube reaction separates them: switch to IR.

Practical points

• Heat with a water bath; most of these liquids are flammable.
• Record the starting colour and the final colour.
• Run a blank with water for comparison.

3.3.6.2 Mass spectrometry

Key idea. Two formulae can share an integer Mr yet differ in precise mass. Measuring the molecular ion’s m/z to 4 decimal places pins down which formula is right. The instrument itself (ionisation, acceleration, drift, detection) is in the atomic structure guide.

Precise masses used here: C = 12.0000, H = 1.0078, O = 15.9949, N = 14.0031.

Method in steps

1. Write every candidate formula.
2. For each, add (number of atoms × precise mass) for every element.
3. Keep four decimal places.
4. Match one total to the measured m/z.
5. Show the totals for the losing candidates as well.

Worked reminder: m/z = 90.0678

C4H10O2  : 4(12.0000) + 10(1.0078) + 2(15.9949)               = 90.0678   match
C3H6O3   : 3(12.0000) + 6(1.0078) + 3(15.9949)                = 90.0315
C3H10N2O : 3(12.0000) + 10(1.0078) + 15.9949 + 2(14.0031)     = 90.0791

The molecular formula is C₄H₁₀O₂. It might be a diol such as butane-1,4-diol, or many other isomers. Only IR, test tube results or fingerprint comparison can go further.

3.3.6.3 Infrared spectroscopy

Key idea. Each type of bond absorbs IR radiation in its own wavenumber range (cm⁻¹). A spectrum shows transmittance against wavenumber; absorptions point downwards. You use the spectrum together with the Chemistry data booklet. In the exam, take the ranges from the booklet you are given; these are the ones used here.

Bond	Range / cm⁻¹	Shape clue
O–H (alcohols)	3230–3550	Broad (hydrogen bonding)
N–H	3300–3500	Narrower than alcohol O–H
C–H	2850–3300	Almost always present
O–H (acids)	2500–3000	Very broad, merges with C–H
C≡N	2220–2260	Sharp
C=O	1680–1750	Intense and narrow
C=C	1620–1680	Weaker than C=O
C–O	1000–1300	In the fingerprint region

Method in steps: reading a spectrum

1. Broad band at 3230–3550? Alcohol O–H.
2. Very broad band at 2500–3000? Acid O–H.
3. Strong band at 1680–1750? C=O.
4. Band at 1620–1680? C=C. Band at 2220–2260? C≡N.
5. List what is missing. Each absence rules structures out.
6. Do not use C–H as evidence for a functional group.

Fingerprint region

Below about 1500 cm⁻¹. The pattern belongs to the whole molecule, so it is characteristic of one compound. Match it against spectra of known compounds to identify the substance. It distinguishes isomers that share the same bonds.

Impurities

An absorption the expected product cannot produce signals an impurity. Typical cases:

Expected product	Extra band seen	Likely impurity
Aldehyde	Very broad 2500–3000	Carboxylic acid (over-oxidation)
Ketone or alkene	Broad 3230–3550	Unreacted alcohol, or water
Nitrile	Broad 3230–3550	Water or an alcohol side product

Worked reminder: reading clues together

A compound shows a band at 1660 cm⁻¹ and a broad band at 3310 cm⁻¹, with nothing at 1680–1750 cm⁻¹. It turns bromine water from orange to colourless and acidified dichromate(VI) from orange to green.

• 1660 cm⁻¹: C=C. Bromine water result agrees.
• 3310 cm⁻¹, broad: alcohol O–H.
• No C=O band, so the dichromate result is not from an aldehyde.
• Conclusion: an unsaturated primary or secondary alcohol, such as but-3-en-1-ol.

Global warming link

The Earth’s surface emits infrared. C=O bonds in CO₂, C–H bonds in methane and O–H bonds in water vapour absorb it. The energy is passed on or re-emitted in all directions, some back to the surface. Higher concentrations of these gases mean more infrared absorbed, which contributes to global warming.

Must-know distinctions

Pair	How to tell them apart
Fehling’s vs Tollens’	Blue → red precipitate vs silver mirror
Aldehyde vs ketone	Both C=O in IR; only the aldehyde reacts with Fehling’s or Tollens’
1°/2° alcohol vs 3° alcohol	Orange → green vs stays orange
Ketone vs 3° alcohol	No test tube difference at AS; IR shows C=O or O–H
Acid O–H vs alcohol O–H	Very broad at 2500–3000 vs broad at 3230–3550
Molecular formula vs structure	Mass spectrometry gives the first; IR, tests and fingerprinting give the second

Quick self-test

1. Write the formula of the silver complex in Tollens’ reagent.
2. Hex-3-ene is shaken with bromine water. Give the full observation.
3. Which one of these does not change the colour of warm acidified dichromate(VI): pentan-2-ol, 2-methylbutan-2-ol, 3-methylbutan-1-ol?
4. Name the red solid formed when an aldehyde reacts with Fehling’s solution.
5. A molecular ion appears at m/z = 56.0624. Is the compound C₄H₈, C₃H₄O or C₂H₄N₂? Show the sums.
6. A spectrum has a sharp absorption at 2240 cm⁻¹. Which bond is present?
7. Why is the O–H absorption of an alcohol broad?
8. A spectrum shows a very broad band from 2500 to 3000 cm⁻¹ and a strong band at 1705 cm⁻¹. Which functional group is present?
9. Where is the fingerprint region, and how is it used to identify a compound?
10. Which gas is released when a carboxylic acid reacts with sodium carbonate, and how do you test for it?
11. Why should aldehydes be ruled out before using dichromate(VI) to test for an alcohol?
12. Name the bond in each of carbon dioxide, methane and water vapour that absorbs infrared radiation.

Answers

1. [Ag(NH₃)₂]⁺.
2. The bromine water changes from orange to colourless.
3. 2-methylbutan-2-ol. It is a tertiary alcohol, so it is not easily oxidised and the solution stays orange.
4. Copper(I) oxide, Cu₂O.
5. C₄H₈: 4(12.0000) + 8(1.0078) = 56.0624. C₃H₄O: 3(12.0000) + 4(1.0078) + 15.9949 = 56.0261. C₂H₄N₂: 2(12.0000) + 4(1.0078) + 2(14.0031) = 56.0374. The compound is C₄H₈.
6. C≡N (2220–2260 cm⁻¹), a nitrile.
7. Hydrogen bonding between molecules gives a spread of O–H bond strengths, so absorption covers a range.
8. A carboxylic acid: acid O–H (2500–3000) plus C=O (1680–1750).
9. Below about 1500 cm⁻¹. Compare it with spectra of known compounds; an exact match identifies it.
10. Carbon dioxide; it turns limewater milky.
11. Aldehydes are oxidised by dichromate(VI) as well, so they also turn it from orange to green.
12. C=O in carbon dioxide, C–H in methane, O–H in water vapour.

Where marks are usually lost

• Describing bromine water as turning “clear”, or reporting just the end colour of a test.
• Swapping Fehling’s and Tollens’ observations, or calling the Fehling’s product “copper”.
• Concluding “alcohol” from a dichromate(VI) result while an aldehyde is still possible.
• Claiming a test tube reaction separates a ketone from a tertiary alcohol at AS.
• Choosing a molecular formula without showing the precise-mass total for every candidate.
• Rounding precise masses so that two candidates appear identical.
• Assigning a very broad 2500–3000 cm⁻¹ band to an alcohol.
• Using C–H absorptions as evidence for a particular functional group.
• Describing fingerprinting without saying the spectrum is compared with known spectra.
• Leaving bonds absorbing infrared out of the global warming explanation.

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.