Chromatography and qualitative analysis

In a nutshell

Chromatography separates the components of a mixture so you can identify them (R_f values, retention times) and, with gas chromatography, find how much of each is there.

Test-tube tests then tell you which functional groups an unknown organic compound contains. Each test is a reaction you have met before; here you use them together to identify an unknown.

Assumed knowledge: Alkenes, Alcohols, Haloalkanes, Carbonyl compounds.

Core content

Types of chromatography: the two phases

Every type of chromatography has two phases:

• Stationary phase: stays still. It is a solid, or a liquid held on a solid support.
• Mobile phase: moves through or over the stationary phase, carrying the mixture with it. It is a liquid (TLC) or a gas (GC).

Components separate because each one is held back by the stationary phase to a different extent. A component that is held back more travels more slowly.

Still don't get it? · why the components separate

Picture a crowd walking along a high street full of shops. Everyone walks at the same speed, but some people keep stopping to look in windows. The people who stop most often get furthest behind. After ten minutes the crowd has spread out into groups.

Now step by step:

1. The mobile phase is the walking crowd: it keeps everything moving forward at the same rate.
2. The stationary phase is the shops. Each component of the mixture keeps "stopping": in TLC it sticks to the solid surface (adsorption); in GC it dissolves in the liquid stationary phase.
3. Each component spends a different fraction of its time stuck to (or dissolved in) the stationary phase.
4. The component held back most moves slowest. In TLC it travels the shortest distance (small R_f). In GC it comes out last (long retention time).

Exam wording: in TLC, components separate by adsorption onto the solid stationary phase; in GC, components separate by their relative solubility in the liquid stationary phase. The more strongly a component is held by the stationary phase, the lower its R_f or the longer its retention time.

Types of chromatography: TLC and R_f values

A one-way TLC plate is spotted on a pencil baseline, stood in a shallow solvent, and the solvent rises up the plate. Each component stops as a separate spot.

$R_f = \frac{\text{distance moved by the component}}{\text{distance moved by the solvent front}}$

• Measure both distances from the baseline, not from the bottom of the plate.
• Measure to the centre of the spot.
• R_f has no units and is always between 0 and 1.

How to interpret a TLC chromatogram:

• Number of spots = the minimum number of components. Two components with very similar R_f values can sit in one spot, so one spot does not prove a substance is pure.
• Identify a component by comparing its R_f with the R_f of a known (standard) substance.
• R_f depends on the solvent and the stationary phase (and temperature). So compare only with standards run under the same conditions, ideally on the same plate alongside the mixture.
• Most organic compounds are colourless. Show the spots under a UV lamp (or with a chemical locating agent such as iodine) and draw round them in pencil.

Types of chromatography: gas chromatography and retention times

The sample is injected into the carrier gas, which carries it through a heated column to a detector. The detector output is a gas chromatogram: one peak per separated component.

Retention time is the time taken for a component to pass from the column inlet (injection) to the detector.

• Each peak's position on the time axis gives the retention time, which identifies the component.
• Identify it by comparing with the retention time of a known substance run under the same conditions (same column, temperature and carrier-gas flow rate).
• Two compounds with very similar retention times may give a single peak and not be distinguished.

Types of chromatography: amounts and proportions from a gas chromatogram

The area under a peak is proportional to the amount of that component. The area is called the peak integration value, and in exam questions it is supplied: you do not measure it.

To find the proportions (percentage composition) of a mixture:

$\text{percentage of component} = \frac{\text{peak integration value of the component}}{\text{sum of all peak integration values}} \times 100$

• Use peak area, never peak height.
• Add up every peak in the chromatogram for the total.
• Strictly, this assumes the detector responds equally to each component. To find an actual concentration, use a calibration curve.

Types of chromatography: external calibration curves for concentration

A detector does not give the same peak area for the same concentration of every compound. So to find the concentration of a component, you calibrate the GC with that component.

1. Make standard solutions of the pure component at several known concentrations.
2. Run each standard through the GC under the same conditions as the sample.
3. Record the peak integration value at the component's retention time for each standard.
4. Plot peak area (y-axis) against concentration (x-axis) and draw the line of best fit: this is the external calibration curve.
5. Run the sample, find the component's peak by its retention time, and read its concentration off the curve from its peak area.

The curve is only valid within the range of the standards; do not extrapolate far beyond them.

Still don't get it? · what a calibration curve does

Imagine a kitchen scale that has lost its markings. You can still use it: put on a 100 g weight and mark where the needle points, then 200 g, then 300 g. Now any unknown object can be weighed by seeing where the needle lands between your marks.

Now step by step:

1. The GC detector gives a peak area, not a concentration. That area is the "needle position".
2. Different compounds move the needle by different amounts, so you must mark the scale for the compound you are measuring.
3. You run known concentrations of the pure component (the known weights) and plot peak area against concentration.
4. You run the sample, take the peak area of that component, and read across to the line and down to the concentration axis.

Exam wording: an external calibration curve is a graph of peak area (integration value) against concentration for standard solutions of known concentration of the pure component, run under the same conditions. The concentration of the component in the sample is read from the curve using its peak area.

PAG6: using TLC in the synthesis of an organic solid

In PAG6 you make an organic solid (for example aspirin) and use TLC to check its purity. This method gives the reason for each step, because the reasons are what earn marks.

Interpreting a purity check:

• Pure product: one spot at the same R_f as the pure substance.
• Impure product: an extra spot. If it has the same R_f as the starting material, unreacted starting material is present. Purify (e.g. by recrystallisation) and run the TLC again.

Tests for organic functional groups: the seven tests

Each test identifies a functional group by an observation you can see. Learn the reagent, the conditions and the exact observation for each.

The sections below give the chemistry behind each test, in spec order.

Tests for organic functional groups: alkenes with bromine

Bromine adds across the C=C to form a colourless dibromo compound, so the orange colour disappears.

CH₂=CH₂ + Br₂ → CH₂BrCH₂Br    (1,2-dibromoethane)

• Positive: orange to colourless.
• Negative (no C=C, e.g. an alkane): stays orange.
• Watch out: phenol also decolourises bromine water, and gives a white precipitate (2,4,6-tribromophenol). So decolourising alone does not prove an alkene.

Tests for organic functional groups: haloalkanes with silver nitrate in ethanol

Water hydrolyses the C-X bond, releasing a halide ion, which then forms a silver halide precipitate:

CH₃CH₂CH₂CH₂Br + H₂O → CH₃CH₂CH₂CH₂OH + H⁺ + Br^-

Ag⁺(aq) + Br^-(aq) → AgBr(s)

• Ethanol is there as a solvent so the haloalkane (insoluble in water) and the aqueous silver nitrate mix.
• Warm in a water bath, not with a flame: ethanol and haloalkanes are flammable.
• The colour of the precipitate identifies the halogen:

The precipitate forms fastest with an iodoalkane and slowest with a chloroalkane, because C-I has the lowest bond enthalpy. (The rate comparison is covered in Haloalkanes.)

Tests for organic functional groups: phenols by weak acidity

A phenol is a weak acid. It is acidic enough to neutralise NaOH, but too weak to react with carbonates.

C₆H₅OH + NaOH → C₆H₅ONa + H₂O    (sodium phenoxide)

• With aqueous sodium carbonate: no effervescence. This is what separates a phenol from a carboxylic acid.
• With aqueous sodium hydroxide: phenol reacts; phenol is only slightly soluble in water, but it dissolves in NaOH(aq) because sodium phenoxide is soluble.
• An alcohol reacts with neither.

Tests for organic functional groups: carbonyl compounds with 2,4-DNP

2,4-DNP (2,4-dinitrophenylhydrazine) gives an orange precipitate with aldehydes and ketones.

• It shows a carbonyl group (C=O) is present, but it cannot tell an aldehyde from a ketone.
• The C=O in carboxylic acids and esters does not give a precipitate.
• You do not need the equation or the structure of the orange product.
• The orange solid (the 2,4-DNP derivative) can be filtered, recrystallised and its melting point measured. Comparing it with known values identifies the specific aldehyde or ketone.

Tests for organic functional groups: aldehydes with Tollens' reagent

Tollens' reagent (ammoniacal silver nitrate) gives a silver mirror with an aldehyde on warming. A ketone gives no change.

The explanation needs both halves:

• the aldehyde is oxidised to a carboxylic acid;
• the silver ions are reduced to silver: Ag⁺ + e^- → Ag.

CH₃CHO + [O] → CH₃COOH

Ketones cannot be oxidised, so no silver forms.

Tests for organic functional groups: alcohols and aldehydes with acidified dichromate

Acidified potassium dichromate(VI) is an oxidising agent. Warm the unknown with it in a water bath.

• Orange to green: Cr₂O₇^2- (Cr +6, orange) is reduced to Cr³⁺ (Cr +3, green).
• Positive for primary alcohols, secondary alcohols and aldehydes, which are oxidised.
• Stays orange for tertiary alcohols and ketones, which are not oxidised.

CH₃CH₂OH + [O] → CH₃CHO + H₂O    (primary alcohol to aldehyde)

CH₃CH(OH)CH₃ + [O] → CH₃COCH₃ + H₂O    (secondary alcohol to ketone)

Because aldehydes also turn it green, dichromate alone cannot separate an aldehyde from an alcohol: use 2,4-DNP as well.

Tests for organic functional groups: carboxylic acids with carbonates

Carboxylic acids react with carbonates, giving a salt, water and carbon dioxide:

2CH₃COOH + Na₂CO₃ → 2CH₃COONa + H₂O + CO₂

• Observation: effervescence (fizzing).
• Confirm the gas is CO₂: bubble it through limewater, which turns milky.
• Phenols and alcohols give no effervescence with carbonate.

PAG7: qualitative analysis of organic functional groups

In PAG7 you identify the functional groups in a series of unknown organic compounds. Work in a logical order, and use a fresh sample for each test.

1. Sodium carbonate(aq). Effervescence: carboxylic acid.
2. Sodium hydroxide(aq) (if no effervescence). Reacts or dissolves: phenol.
3. Bromine water. Orange to colourless: alkene (unless a white precipitate also forms, which points to phenol).
4. 2,4-DNP. Orange precipitate: aldehyde or ketone. Then:
   • Tollens' reagent: silver mirror = aldehyde; no change = ketone.
5. Acidified dichromate, warm. Orange to green with no precipitate in step 4: primary or secondary alcohol. Stays orange: not a primary or secondary alcohol (a tertiary alcohol stays orange).
6. Silver nitrate in ethanol, warm. Precipitate: haloalkane; the colour gives the halogen.

Practical points that earn marks:

• Heat with a water bath or electric heater, never a naked flame: most organic compounds are flammable.
• Use a clean test tube for Tollens' reagent, otherwise the silver forms a grey-black precipitate instead of a mirror.
• Record observations (what you see, with colours before and after) separately from conclusions (the functional group).

Worked examples

Example 1: calculating R_f values and identifying components (TLC)

A mixture (lane 1) and three pure standards P, Q and R (lanes 2 to 4) were run on the same TLC plate. The solvent front moved 8.0 cm from the baseline.

Step 1: read the distances (to the centres of the spots, from the baseline). The mixture has spots at 2.4 cm and 5.6 cm.

Step 2: calculate each R_f.

$R_f(\text{lower spot}) = \frac{2.4\ \text{cm}}{8.0\ \text{cm}}$

$R_f(\text{lower spot}) = 0.30$

$R_f(\text{upper spot}) = \frac{5.6\ \text{cm}}{8.0\ \text{cm}}$

$R_f(\text{upper spot}) = 0.70$

The units cancel, so R_f has no units. Both data values have 2 significant figures, so give R_f to 2 s.f.

Step 3: compare with the standards on the same plate.

$R_f(\text{P}) = \frac{5.6\ \text{cm}}{8.0\ \text{cm}} = 0.70$

$R_f(\text{Q}) = \frac{4.0\ \text{cm}}{8.0\ \text{cm}} = 0.50$

$R_f(\text{R}) = \frac{2.4\ \text{cm}}{8.0\ \text{cm}} = 0.30$

Answer: the mixture contains P (R_f 0.70) and R (R_f 0.30). It does not contain Q.

Where marks are lost: measuring from the bottom edge of the plate instead of the baseline (that adds the baseline height, about 1.5 cm, to every distance and gives the wrong R_f), or measuring to the top of a spot rather than its centre.

Example 2: percentage composition from a gas chromatogram

The gas chromatogram drawn in Core content (peaks at 1.8, 3.2 and 4.6 min) was recorded for a mixture. Under the same conditions, the pure standards gave these retention times: P 1.8 min, Q 2.6 min, R 3.2 min, S 4.6 min. The peak integration values were supplied:

Step 1: identify each component by matching retention times. 1.8 min = P, 3.2 min = R, 4.6 min = S. There is no peak at 2.6 min, so Q is absent.

Step 2: add up all the peak integration values.

$\text{total} = 2.42 + 5.15 + 1.83$

$\text{total} = 9.40$

Step 3: find the percentage of each component.

$\%\,\text{P} = \frac{2.42}{9.40} \times 100 = 25.7\,\%$

$\%\,\text{R} = \frac{5.15}{9.40} \times 100 = 54.8\,\%$

$\%\,\text{S} = \frac{1.83}{9.40} \times 100 = 19.5\,\%$

Check: 25.7 + 54.8 + 19.5 = 100.0 %. Answers to 3 s.f., matching the integration values.

Where marks are lost: dividing by the wrong total (missing a peak), or using peak heights read off the graph instead of the supplied areas.

Example 3: concentration from an external calibration curve

Standard solutions of pure compound R were run through the GC under the same conditions as the mixture.

A sample was then run. The peak at R's retention time (3.2 min) had an integration value of 1435. Find the concentration of R in the sample.

Step 1: plot the calibration curve (peak area against concentration) and draw the line of best fit.

Step 2: read across from 1435 to the line, then down. The line meets 1435 at a concentration of 0.070 mol dm^-3.

Step 3 (check by calculation): the line passes through the origin, so find its gradient and divide.

$\text{gradient} = \frac{2050}{0.100\ \text{mol dm}^{-3}} = 20\,500\ \text{dm}^{3}\ \text{mol}^{-1}$

$\text{concentration} = \frac{1435}{20\,500\ \text{dm}^{3}\ \text{mol}^{-1}}$

$\text{concentration} = 0.070\ \text{mol dm}^{-3}$

Answer: 0.070 mol dm^-3 (2 s.f., matching the concentration data), with units.

Where marks are lost: reading the area off the wrong component's peak (match the retention time first), plotting the axes the wrong way round, or leaving the units off.

Example 4: identifying an unknown with test-tube tests

Three unlabelled liquids are isomers of C₃H₆O: propanal, propanone and prop-2-en-1-ol (CH₂=CHCH₂OH). Plan how to identify each, with observations.

1. Add 2,4-DNP to each. Propanal and propanone give an orange precipitate (both are carbonyls). Prop-2-en-1-ol gives no precipitate, so it is identified.

2. Confirm prop-2-en-1-ol with bromine water: orange to colourless, because it has a C=C.

   CH₂=CHCH₂OH + Br₂ → CH₂BrCHBrCH₂OH

3. Warm the other two with Tollens' reagent in a water bath. Propanal gives a silver mirror (it is an aldehyde, oxidised to propanoic acid). Propanone gives no change (a ketone is not oxidised).

   CH₃CH₂CHO + [O] → CH₃CH₂COOH

Notice the structure of the answer: reagent, observation for each compound, conclusion. Each identification needs the observation, not just the name of the test.

Common exam mistakes

• Measuring R_f distances from the bottom of the plate. Both distances are measured from the baseline, and to the centre of the spot.
• Giving R_f units, or a value above 1. R_f is a ratio of two lengths: no units, always between 0 and 1. A value above 1 means the fraction is upside down.
• Saying "one spot, so it is pure". Compounds with similar R_f values can share a spot. Compare with a standard run on the same plate.
• Comparing R_f values or retention times obtained under different conditions. Both depend on the conditions (solvent and plate; column, temperature and flow rate), so standards must be run under the same conditions.
• Baseline drawn in ink, or solvent above the baseline. Ink runs with the solvent; solvent above the line dissolves the spots off the plate.
• Calling retention time "the time the component spends in the column" without the endpoints. Say from injection (column inlet) to the detector.
• Using peak height for amounts. The area (integration value) is proportional to the amount.
• Saying the percentage from peak areas is a concentration. Peak areas give proportions; a calibration curve of the pure component is needed for a concentration.
• Bromine water "goes clear". Clear is not a colour change: write orange to colourless (decolourised).
• 2,4-DNP "turns orange". The observation is an orange precipitate, not an orange solution.
• Using 2,4-DNP to tell an aldehyde from a ketone. Both give the precipitate; only Tollens' distinguishes them.
• Half the Tollens' explanation. You need aldehyde oxidised to a carboxylic acid and Ag⁺ reduced to Ag.
• Dichromate colour backwards, or a tertiary alcohol turning green. It goes orange to green, and tertiary alcohols stay orange.
• Thinking phenol fizzes with sodium carbonate. Phenol is too weak an acid; only carboxylic acids give effervescence.
• "CO₂ is produced" as an observation. The observation is effervescence (fizzing); CO₂ is the conclusion.
• Saying every haloalkane gives a white precipitate. White = chloro, cream = bromo, yellow = iodo.
• Forgetting the ethanol, or heating with a Bunsen. Ethanol makes the haloalkane and aqueous silver nitrate mix; warm in a water bath because the compounds are flammable.

Key definitions

• Stationary phase: the phase that does not move; a solid, or a liquid on a solid support.
• Mobile phase: the phase that moves through or over the stationary phase, carrying the components; a liquid (TLC) or a gas (GC).
• Adsorption: the process by which a component is held on the surface of the solid stationary phase (the basis of separation in TLC).
• R_f value: distance moved by the component ÷ distance moved by the solvent front, both measured from the baseline.
• Retention time: the time taken for a component to pass from the column inlet (injection) to the detector.
• Peak integration value: the area under a peak in a gas chromatogram; it is proportional to the amount of that component.
• External calibration curve: a graph of peak area against concentration for standard solutions of the pure component of known concentration, run under the same conditions, used to find the concentration of that component in a sample.
• Tollens' reagent: ammoniacal silver nitrate; gives a silver mirror with aldehydes because the aldehyde is oxidised to a carboxylic acid and silver ions are reduced to silver.
• 2,4-DNP (2,4-dinitrophenylhydrazine): a reagent that gives an orange precipitate with aldehydes and ketones, detecting a carbonyl group.

Specification

• I can interpret a one-way TLC chromatogram: calculate R_f values and identify components by comparison with standards. (6.3.1 a, PAG6)
• I can describe how to run a TLC plate, with the reason for each step, and use it to check the purity of an organic solid. (6.3.1 a, PAG6)
• I can interpret a gas chromatogram using retention times to identify components. (6.3.1 b i)
• I can use supplied peak integration values to find the amounts and proportions of the components in a mixture. (6.3.1 b ii)
• I can create and use an external calibration curve to find the concentration of a component. (6.3.1 b ii, additional guidance)
• I can identify an alkene by its reaction with bromine. (6.3.1 c i)
• I can identify a haloalkane by its reaction with aqueous silver nitrate in ethanol, and the halogen from the precipitate colour. (6.3.1 c ii)
• I can identify a phenol by its weak acidity: reaction with NaOH but no reaction with CO₃^2-. (6.3.1 c iii)
• I can identify a carbonyl compound by its reaction with 2,4-DNP. (6.3.1 c iv)
• I can identify an aldehyde by its reaction with Tollens' reagent. (6.3.1 c v)
• I can identify primary and secondary alcohols and aldehydes by their reaction with acidified dichromate. (6.3.1 c vi)
• I can identify a carboxylic acid by its reaction with CO₃^2-. (6.3.1 c vii)
• I can plan a sequence of test-tube tests to identify the functional groups in an unknown compound, giving observations. (6.3.1 c, PAG7)

Related notes

• Alkenes: addition of bromine and the test for unsaturation
• Alcohols: oxidation with acidified dichromate
• Haloalkanes: hydrolysis with silver nitrate and the rate comparison
• Qualitative analysis (AS): silver halide precipitates and the carbonate test
• Analytical techniques: IR spectroscopy and mass spectrometry
• Aromatic compounds: the weak acidity of phenol and its reaction with bromine
• Carbonyl compounds: 2,4-DNP and Tollens' reagent in full
• Carboxylic acids and esters: reactions with carbonates
• Organic synthesis: recrystallisation and melting points for purifying an organic solid
• Spectroscopy: NMR, combined with other techniques to identify a structure