Edexcel International A Level Chemistry: Spectroscopy and chromatography (YCH11)

This study guide teaches Topic 15E: Spectroscopy and chromatography from the Pearson Edexcel International Advanced Level in Chemistry (YCH11) specification, Issue 1, September 2017. It works through every outcome from 15.17 to 15.23. Being IA2 (International A2) material, it belongs in Unit 4: Rates, Equilibria and Further Organic Chemistry, closing Topic 15 (Organic Chemistry: Carbonyls, Carboxylic Acids and Chirality). The specification says that Unit 2, Topic 10D (mass spectra and IR) is assumed and extended here, so revise Topic 10D on mass spectra and IR first if fragmentation or IR absorptions feel shaky.

Then use the Topic 15E revision notes and Topic 15E practice set. The Edexcel IAL Chemistry course hub lists every topic, the Edexcel chemistry checklist tracks outcomes, and a free diagnostic finds weak spots.

What Topic 15E asks of you

Outcome	You must be able to
15.17	Calculate an accurate relative molecular mass from atomic masses given to four decimal places; use accurate masses to pick a molecular formula and suggest structures
15.18	Explain that ¹³C NMR shows where the ¹³C atoms sit in a molecule
15.19	Predict carbon environments using chemical shift values, δ, and justify the number of ¹³C peaks
15.20	Use low and high resolution proton NMR: proton types from δ, peak areas, the (n+1) rule, and predicting a spectrum
15.21	State that chromatography separates a mixture using a mobile phase and a stationary phase
15.22	Calculate Rf values from one-way paper and TLC chromatograms and explain why Rf values differ
15.23	Describe HPLC and GC as column methods that separate by retention time, used with mass spectrometry in forensics and drug testing in sport

Shift values used on this page

The ranges below are approximate values for the examples on this page. In a question, use the shift values that question gives you.

Carbon environment	Approximate ¹³C δ / ppm
Alkyl carbon (C–C, C–H only)	5–45
Carbon bonded to O (alcohols, ethers, esters)	50–90
C=C in alkenes; carbons in a benzene ring	110–160
Carbonyl carbon of an acid or ester	160–185
Carbonyl carbon of an aldehyde or ketone	190–220

Proton environment	Approximate ¹H δ / ppm
Alkyl C–H (CH₃, CH₂, CH away from functional groups)	0.7–1.7
H on the carbon beside a C=O (for example CH₃CO–)	2.0–2.7
H on a carbon bonded to O or Cl	3.3–4.5
O–H in alcohols (varies)	1.0–5.0
Alkene C–H	4.5–6.5
Aldehyde –CHO	9.0–10.0
Carboxylic acid –COOH	10.0–12.0

Chemical shift is measured in ppm from a reference compound, tetramethylsilane (TMS), which is set at δ = 0. The more electronegative atoms near a nucleus, the higher its δ value tends to be.

15.17 Accurate relative molecular mass

In Topic 10D the molecular ion gave a whole-number Mr, which many formulae share. A high resolution mass spectrometer measures m/z to four decimal places. Atomic masses are not exact whole numbers, so each formula has its own accurate mass.

On the ¹²C scale, carbon-12 is exactly 12.0000. The other atoms differ slightly from whole numbers. On this page use: H 1.0078, C 12.0000, O 15.9949, N 14.0031.

Worked example 1: calculating an accurate Mr

Calculate the accurate relative molecular mass of C₅H₁₁NO₂.

1. Carbon: 5 × 12.0000 = 60.0000
2. Hydrogen: 11 × 1.0078 = 11.0858
3. Nitrogen: 1 × 14.0031 = 14.0031
4. Oxygen: 2 × 15.9949 = 31.9898
5. Total: 117.0787

Keep four decimal places in every term.

Worked example 2: from accurate mass to structure

A neutral organic liquid gives a molecular ion at m/z = 114.1043. Four formulae have whole-number mass 114: C₇H₁₄O, C₆H₁₀O₂, C₆H₁₄N₂ and C₅H₁₀N₂O. Its IR spectrum has a strong absorption near 1715 cm⁻¹ and no broad O–H absorption.

Formula	Accurate Mr	Difference from 114.1043
C₇H₁₄O	114.1041	0.0002
C₆H₁₀O₂	114.0678	0.0365
C₆H₁₄N₂	114.1154	0.0111
C₅H₁₀N₂O	114.0791	0.0252

1. The closest match is C₇H₁₄O.
2. C₇H₁₄O has one C=O or one C=C or one ring. The IR band near 1715 cm⁻¹ points to C=O, so it is an aldehyde or ketone.
3. Possible structures include heptanal, heptan-2-one, heptan-3-one and heptan-4-one. Isomers share an accurate mass, so NMR is needed to decide (worked example 4).

15.18 What carbon-13 NMR tells you

About 1% of carbon atoms are ¹³C. In a strong magnetic field, ¹³C nuclei absorb radio-frequency radiation at a frequency that depends on the surrounding atoms, so carbons in different positions give signals at different δ values.

A ¹³C NMR spectrum therefore shows:

• how many different carbon environments the molecule has (one peak for each)
• what kind of environment each one is (from its δ value), for example an alkyl carbon, a carbon bonded to oxygen or a C=O carbon.

15.19 Using ¹³C NMR data

Carbons related by symmetry are in the same environment and give one peak between them. So the number of peaks equals the number of different carbon environments, which can be fewer than the number of carbons.

Worked example 3: 2-methylbutan-2-ol

Structure: (CH₃)₂C(OH)CH₂CH₃, five carbons. Predict the ¹³C spectrum.

1. The two CH₃ groups on C2 are equivalent: one environment.
2. C2 carries the OH: one environment.
3. The CH₂: one environment.
4. The CH₃ at the end of the ethyl group: one environment.
5. Four peaks for five carbons.
6. Shifts from the table: C2 (bonded to O) falls in 50–90 ppm; the other three are alkyl carbons in 5–45 ppm.

To “justify” a number of peaks, name each environment and say which carbons are equivalent.

15.20 Proton NMR

Low resolution: environments and peak areas

In a low resolution ¹H spectrum, each proton environment gives one peak. Its δ value tells you the type of proton (use the table). The relative peak area (or the ratio given on the spectrum) is proportional to the number of ¹H atoms in that environment.

High resolution: the (n+1) rule

At high resolution, a peak splits because of protons on the adjacent carbon. If there are n non-equivalent protons on neighbouring carbons, the peak splits into n + 1 lines.

H atoms on adjacent carbons	Pattern you see
none	singlet (one line)
one	doublet (two lines)
two	triplet (three lines)
three	quartet (four lines)
six, from two CH₃ groups	septet (seven lines)

• Equivalent protons do not split each other. The six H atoms in (CH₃)₂ give one peak.
• Count H on the neighbouring carbons, not on the carbon you are looking at.
• An O–H proton normally appears as a singlet and does not split its neighbours.
• A carbon with no H, such as a C=O carbon, passes on no splitting. A CH₃ next to C=O is a singlet.

Worked example 4: predicting the spectrum of heptan-4-one

Heptan-4-one is CH₃CH₂CH₂COCH₂CH₂CH₃. It is symmetrical about the C=O.

Environment	Number of H	Neighbouring H	Splitting	Approximate δ / ppm
CH₃ (two equivalent)	6	2	triplet	0.9
middle CH₂ (two equivalent)	4	3 + 2 = 5	sextet (6 lines)	1.6
CH₂ next to C=O (two equivalent)	4	2	triplet	2.4

Peak area ratio 6 : 4 : 4, which simplifies to 3 : 2 : 2. The ¹³C spectrum would show four peaks: C=O (190–220 ppm) and three alkyl carbons.

This separates the isomers from worked example 2: heptan-2-one, for instance, would show a 3H singlet near 2.1 ppm from CH₃CO.

Worked example 5: deducing a structure

Compound P is a ketone, C₅H₁₀O. Its high resolution ¹H spectrum shows:

• δ 1.1, doublet, area 6
• δ 2.1, singlet, area 3
• δ 2.6, septet, area 1

1. Area 6 as a doublet: two equivalent CH₃ groups next to one H, so (CH₃)₂CH–.
2. Area 1 as a septet: one H with six neighbours, the CH of that same group. δ 2.6 places it next to C=O.
3. Area 3 as a singlet at δ 2.1: a CH₃ next to C=O with no H on the neighbouring carbon, so CH₃CO–.
4. Join the pieces: CH₃COCH(CH₃)₂, 3-methylbutan-2-one.
5. Check: 6 + 3 + 1 = 10 H and five carbons, matching C₅H₁₀O.

15.21 What chromatography does

Chromatography separates the components of a mixture using a mobile phase, which moves, and a stationary phase, which does not. Each component is distributed between the two phases to a different extent: the more strongly it is held by the stationary phase, the more slowly it moves.

Method	Stationary phase	Mobile phase
Paper	Water held in the paper fibres	Solvent soaking up the paper
Thin-layer (TLC)	Thin layer of solid such as silica or alumina on a plate	Solvent moving up the plate

15.22 Rf values

For a one-way chromatogram:

Rf = (baseline to centre of spot) ÷ (baseline to solvent front)

As a ratio of two lengths, Rf has no units; its value is from 0 to 1.

Worked example 6: a TLC plate

A TLC plate is run until the solvent front is 7.4 cm above the baseline. Three spots appear at 1.6 cm, 4.2 cm and 6.5 cm.

1. Spot 1: 1.6 ÷ 7.4 = 0.22
2. Spot 2: 4.2 ÷ 7.4 = 0.57
3. Spot 3: 6.5 ÷ 7.4 = 0.88

Spot 1 travels least: polar silica holds it most strongly, so it is likely the most polar component.

Why Rf values differ

• Substances differ in how strongly the stationary phase holds them (for example by hydrogen bonds to silica or to water in paper) and in how soluble they are in the mobile phase.
• The same substance gives a different Rf if you change the solvent, the stationary phase or the temperature. To compare an unknown with a known substance, run both on the same plate under the same conditions.
• In any one solvent, two different compounds may give the same Rf. A matching Rf suggests, but does not prove, that two spots are the same compound.

15.23 HPLC and GC

High-performance liquid chromatography (HPLC) and gas chromatography (GC) are both column chromatography. The stationary phase is packed into or coats the inside of a column, and the mobile phase carries the mixture through it. Components leave the column at different times. The time from injection to detection is the retention time, and components separate because their retention times differ.

	HPLC	GC
Mobile phase	Liquid solvent pumped under high pressure	Unreactive carrier gas
Stationary phase	Small solid particles packed in the column	Liquid coating on a solid support or on the column wall
Sample	Dissolved in the solvent	Must vaporise in the heated column

Retention time depends on the column, temperature and flow rate, so compare it with a standard run under the same conditions. A match is evidence, not proof, of identity.

With mass spectrometry. The column output can feed straight into a mass spectrometer (GC–MS or HPLC–MS). Each separated component gives its own mass spectrum, which is matched against a database of known spectra. This is far more reliable than retention time alone. The specification names two uses:

• Forensics: identifying drugs, poisons or fire accelerants in samples from a scene.
• Drug testing in sport: separating the substances in an athlete’s urine sample and identifying any banned compound.

Common errors

• Choosing a formula by whole-number mass when the question gives accurate masses.
• Counting carbon atoms instead of carbon environments for the number of ¹³C peaks.
• Measuring Rf from the bottom of the plate instead of from the baseline.
• Calling the GC mobile phase “air” rather than an unreactive carrier gas.

Official syllabus

Specification: Pearson Edexcel YCH11, International Advanced Subsidiary/Advanced Level in Chemistry, Issue 1 of September 2017 (Pearson Education Limited). Covered here: Unit 4, Topic 15E, outcomes 15.17 to 15.23.