OxfordAQA A-Level Chemistry: Nuclear magnetic resonance spectroscopy (9620)

Here you learn section 3.3.15, Nuclear magnetic resonance spectroscopy (sub-section 3.3.15.1), 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. The whole section is International A-level only (International A2 content). Sections 3.3.7 to 3.3.16 are examined in Unit 4, and Unit 5 (practical and synoptic) covers all content. OxfordAQA stresses solving problems with analytical data over spectroscopic theory.

After this guide, the revision notes give the condensed version and the practice questions let you test yourself. NMR is added to the mass spectrometry and infrared work in Organic analysis. See also the course hub, the checklist to print and the free diagnostics.

Topic coverage at a glance

9620 reference	What you must be able to do
3.3.15.1	Appreciate that several analytical methods are combined to confirm a new compound’s structure
3.3.15.1	State that NMR shows where ¹³C or ¹H atoms sit within a molecule
3.3.15.1	Know that ¹³C NMR gives simpler spectra than ¹H NMR
3.3.15.1	Use the δ scale for chemical shift; know that shift depends on the molecular environment
3.3.15.1	Read an integrated ¹H spectrum for the ratio of ¹H atoms between environments
3.3.15.1	Know that ¹H samples are dissolved in deuterated solvents or CCl₄
3.3.15.1	Explain why TMS is a suitable standard
3.3.15.1	Use ¹H and ¹³C spectra with data-booklet shifts to suggest structures or part structures
3.3.15.1	Apply the n+1 rule to non-equivalent protons on neighbouring carbons (doublet, triplet, quartet; aliphatic only)

Every row is International A-level only. The section lists no practical activity.

Why one technique is never enough

New compounds are confirmed by a set of techniques that each answer a different question:

• Mass spectrometry: the molecular ion gives Mr, and a precise mass fixes the molecular formula.
• Infrared: the bonds present, so the functional group (C=O, O–H and so on).
• ¹³C NMR: how many different carbon environments there are, and what type each one is.
• ¹H NMR: how many hydrogen environments there are, how many H are in each, and how many H sit on the neighbouring carbons.

A structure is only confirmed when it agrees with every piece of data at once.

What NMR measures

Placed in a powerful magnet, ¹H and ¹³C nuclei can take in energy from radio waves. The exact frequency depends on the atoms around the nucleus, its environment. Nuclei in the same environment absorb at the same place, so each signal locates a group of ¹H or ¹³C atoms in the molecule.

Atoms are equivalent when the molecule’s symmetry makes them interchangeable. Mark the symmetry on a displayed formula before you count anything.

Chemical shift and the δ scale

The horizontal axis of a spectrum is chemical shift, δ, in parts per million (ppm), measured relative to a standard set at δ = 0. Zero is on the right and δ increases to the left. Chemical shift depends on the molecular environment: an electronegative atom (O, Cl, Br) or a C=O group nearby draws electron density away from the nucleus and moves its signal to a larger δ. Two halogen atoms on the same carbon move it further still.

You are expected to take shift values from the Chemistry data booklet. The figures in these tables are rounded ones prepared for this guide; in the exam, use the booklet you are given.

¹H environment	δ / ppm
H on a carbon in an alkyl chain	0.7–1.7
H on a carbon next to C=O	2.0–3.0
H on a carbon bonded to Cl or Br	3.0–4.5
H on a carbon bonded to O (alcohol, ether, ester)	3.1–4.3
H in O–H of an alcohol (variable)	1.0–5.0
H on C=C	4.5–6.0
Aldehyde H	9.0–10.0
Carboxylic acid H	10.0–12.0

¹³C environment	δ / ppm
Alkyl carbon	5–45
Carbon next to C=O	20–55
Carbon bonded to Cl or Br	30–75
Carbon bonded to O	50–90
C=C	100–150
C=O in an acid or ester	160–185
C=O in an aldehyde or ketone	190–220

Why TMS is the standard

Tetramethylsilane, Si(CH₃)₄, is added as the reference and its signal is defined as δ = 0. You must be able to explain why it is suitable:

• Its 12 H atoms are all equivalent, as are its 4 C atoms, so it gives one sharp, intense signal.
• Its signal sits at the far right (low δ), clear of nearly every signal from organic compounds, so the two never clash.
• It is chemically inert: the sample is left unchanged.
• It is volatile: it boils off easily, so the sample can be recovered.

Why deuterated solvents or CCl₄

¹H NMR samples are run in solution. A solvent such as CH₂Cl₂ contains ¹H atoms and is present in far larger amounts than the sample, so its own ¹H signal would dwarf the sample’s. A deuterated solvent (for example CDCl₃) has its ¹H replaced by deuterium, ²H, which gives no signal in a ¹H spectrum. Tetrachloromethane, CCl₄, is also used: it has no hydrogen atoms.

¹³C NMR

In the ¹³C spectra you meet, each carbon environment gives one line:

• number of lines = number of carbon environments;
• each line’s δ shows the type of carbon.

¹³C spectra are simpler than ¹H spectra because their lines are never split, and their δ range (about 0–220 ppm) is far wider than for ¹H (about 0–12 ppm), so lines rarely overlap.

Worked example 1: four bromoalkanes, C₄H₉Br

Predict the number of lines in the ¹³C spectrum of each isomer.

1-bromobutane            CH3–CH2–CH2–CH2Br
2-bromobutane            CH3–CHBr–CH2–CH3
1-bromo-2-methylpropane  (CH3)2CH–CH2Br
2-bromo-2-methylpropane  (CH3)3CBr

1. 1-bromobutane: no symmetry, four different carbons: 4 lines.
2. 2-bromobutane: the two CH₃ groups differ (one is on the C–Br carbon, one on the CH₂): 4 lines.
3. 1-bromo-2-methylpropane: the two CH₃ groups on the CH are equivalent, so CH₃ (×2), CH and CH₂Br: 3 lines.
4. 2-bromo-2-methylpropane: three equivalent CH₃ and one C–Br carbon: 2 lines.

The count separates the branched isomers but not the two straight-chain ones; δ values or a ¹H spectrum are needed for those.

Worked example 2: an acid or an ester? C₄H₈O₂

Compound D, C₄H₈O₂, gives ¹³C lines at δ 184, 34 and 19. Suggest a structure.

1. Three lines for four carbons, so two carbons are equivalent.
2. δ 184 is a C=O in an acid or ester (160–185).
3. There is no line between 50 and 90, so no carbon is bonded to the single-bonded O of an ester. That rules out propan-2-yl methanoate, HCOOCH(CH₃)₂, which also has three environments.
4. δ 34 fits a carbon next to C=O and δ 19 an alkyl carbon; two equivalent CH₃ groups on one CH fit.
5. D is 2-methylpropanoic acid, (CH₃)₂CHCOOH. Butanoic acid would give four lines.

¹H NMR

A ¹H spectrum gives three kinds of information:

1. Number of signals: the number of hydrogen environments.
2. Chemical shift: what each set of H is near.
3. Integration: the relative number of H in each environment.

Integration

The integration trace rises in steps over each signal. Step heights are proportional to the numbers of ¹H atoms. Divide every height by the smallest, make whole numbers, then scale to the formula’s H count.

Worked example 3: reading an integration trace

Compound E, C₅H₁₂O, shows two singlets: δ 3.2 with a trace height of 14 mm and δ 1.2 with a trace height of 42 mm. Deduce its structure.

1. Ratio = 14 : 42 = 1 : 3.
2. The formula has 12 H and 1 + 3 = 4, so multiply by 3: 3 H at δ 3.2 and 9 H at δ 1.2.
3. δ 3.2 is H on a carbon bonded to O, and a 3H singlet is a CH₃ with no H on its neighbour: CH₃–O–.
4. A 9H singlet at alkyl δ is a (CH₃)₃C– unit: three identical methyls on a carbon carrying no H.
5. E is 2-methoxy-2-methylpropane, (CH₃)₃COCH₃. The atoms total C₅H₁₂O.

Spin–spin splitting: the n+1 rule

Non-equivalent H atoms on neighbouring carbons split a ¹H signal:

n such neighbouring H turn the signal into n + 1 peaks.

Pattern	Peaks	H on neighbouring carbons
singlet	one	none
doublet	two	one
triplet	three	two
quartet	four	three

You only need doublets, triplets and quartets, and only in aliphatic compounds. Remember:

• The splitting describes the neighbours, not the group itself.
• Equivalent H do not split each other.
• An H on a carbon whose neighbours carry no H (next to O, C=O or a carbon with no H) is a singlet.

Worked example 4: two trichloroethanes

Compound F, C₂H₃Cl₃, has two ¹H signals: δ 5.8 (triplet, 1H) and δ 3.9 (doublet, 2H). Its isomer G has one signal, a singlet at δ 2.7 (3H). Identify F and G.

1. In F, the 1H signal is a triplet, so its carbon is next to a carbon carrying 2 H.
2. The 2H signal is a doublet, so its carbon is next to a carbon carrying 1 H.
3. A CH next to a CH₂: F is 1,1,2-trichloroethane, CHCl₂–CH₂Cl. The CHCl₂ proton has the larger δ because two Cl atoms are on its carbon.
4. G’s three H are equivalent and their neighbour (CCl₃) has no H: G is 1,1,1-trichloroethane, CH₃–CCl₃.

Worked example 5: combining the evidence

Compound H has the molecular formula C₄H₇BrO₂.

• Infrared: strong absorption near 1745 cm⁻¹; no broad O–H absorption.
• ¹³C NMR: four lines, at δ 171, 53, 40 and 22.
• ¹H NMR: δ 4.4 (quartet, 1H), δ 3.8 (singlet, 3H), δ 1.7 (doublet, 3H).

Deduce the structure.

1. The IR absorption is a C=O. With no O–H and two O atoms, an ester is likely.
2. ¹³C: four lines for four carbons, so no symmetry. δ 171 is an ester C=O; δ 53 is a C–O carbon; δ 40 fits a carbon bonded to Br; δ 22 is an alkyl CH₃.
3. ¹H: the 3H singlet at δ 3.8 is CH₃–O with no H neighbours, the ester’s methyl group.
4. The 1H quartet is next to a CH₃, and the 3H doublet is next to a CH. Together they form CH₃–CH<.
5. The CH at δ 4.4 carries the Br.
6. H is methyl 2-bromopropanoate, CH₃CHBrCOOCH₃. Every signal, ratio and splitting fits.

Common errors

• Counting atoms, not environments: look for equivalent CH₃ groups before counting.
• Reading a quartet as “a CH₃ group”. A quartet is next to a CH₃.
• Treating the integration ratio as the actual H count when the formula has more H.
• Splitting a CH₃ that sits next to O or C=O. With no H on its neighbour it is a singlet.
• Saying CDCl₃ is used “because it is inert”. The point is that it gives no ¹H signal.
• Giving “its signal is at zero” as a reason for TMS. Zero is a definition; the reasons are one intense signal, no overlap, inertness and volatility.

Where to go next

Next, use the revision notes, the practice questions, the Organic analysis revision notes and the diagnostics.

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, published by OxfordAQA. This guide covers section 3.3.15 Nuclear magnetic resonance spectroscopy (3.3.15.1), International A-level only.