OCR A Level Chemistry A: Spectroscopy (H432) – Revision Notes

These notes condense 6.3.2 Spectroscopy for final revision. For full explanations and five worked examples, use the Spectroscopy study guide.

They follow the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017. NMR outcomes 6.3.2(a) to (e), part of Module 6 (Organic chemistry and analysis), are all here. In the specification’s assessment overview, Module 6 content appears in Components 02 and 03. There is no PAG link for this section.

Links: OCR A Level Chemistry hub, H432 checklist, practice questions, Analytical techniques notes for IR and mass spectra, and free diagnostics.

Key words

• Environment: a set of atoms that are equivalent by symmetry. One environment gives one signal.
• Chemical shift, δ: signal position in ppm, measured from TMS at δ = 0.
• Proton decoupled: a carbon-13 spectrum in which each environment is one line, with no splitting. Every assessed carbon-13 spectrum is of this kind.
• High resolution: a proton spectrum showing splitting.
• Integration trace: a stepped line; step heights are in the ratio of the numbers of protons.
• Labile proton: an O–H or N–H proton that exchanges with D₂O.

Shift ranges for practice

The specification says chemical shift values will be provided on the Data Sheet in examinations, so use those in the exam. These typical ranges match the Marlbridge study guide and practice set.

Carbon	δ / ppm	Proton	δ / ppm
alkyl C–C	5–55	alkyl C–H	0.7–1.8
C–Cl, C–Br	30–70	H–C–C=O	2.0–2.7
C–N	30–65	H–C–N, H–C next to a benzene ring	2.3–3.0
C–O	50–90	H–C–Cl, H–C–Br	3.0–4.5
alkene C=C	115–150	H–C–O	3.3–4.5
aromatic C	110–160	alkene C=C–H	4.5–6.5
C≡N	110–125	aromatic H	6.5–8.2
acid, ester, amide C=O	160–185	aldehyde H	9.4–10.0
aldehyde, ketone C=O	190–220	carboxylic acid O–H	10.0–12.5
		alcohol O–H, amine N–H	0.5–5.0

Two pairs trip people up. A C=O carbon at δ 160–185 is an acid, ester or amide; at δ 190–220 it is an aldehyde or ketone. A proton at δ 9.4–10.0 is the aldehyde H; at δ 10.0–12.5 it is the acid O–H.

6.3.2(a): Carbon-13 method in steps

1. Count lines = number of carbon environments.
2. Fewer lines than carbons in the formula = symmetry.
3. Assign each line using the shift table.
4. List every structure that fits; say if more than one does.

Small reminder: benzene-1,4-dicarboxylic acid, HOOC–C₆H₄–COOH, has eight carbons but only three environments: the two C=O, the two ring carbons bonded to COOH, and the four ring C–H carbons.

6.3.2(b): Proton method in steps

1. Number of signals = number of proton environments.
2. Shift = type of environment. Aromatic protons are identified by shift only; you are not expected to analyse their splitting.
3. Area (peak area, integration trace or ratio numbers) = relative number of protons. Check the total against the formula.
4. Splitting: n adjacent non-equivalent protons give n + 1 peaks.
5. Structure: fit the fragments together and recheck every signal.

Group seen	Splitting pattern
CH₃CH₂–	3H triplet + 2H quartet
(CH₃)₂CH–	6H doublet + 1H septet
(CH₃)₃C–	9H singlet
CH₃ on C=O or on O	3H singlet
–CH₂CH₂– between groups with no H	two 2H triplets

Count protons on every adjacent carbon. A CH₂ between a CH₃ and a CH₂ has 3 + 2 = 5 neighbours and is a sextet. O–H and N–H protons usually show as broad singlets and do not split neighbouring signals.

Small reminder: in 1-chloropropan-2-one, ClCH₂COCH₃, both signals are singlets (2H and 3H), because the carbon between them is the C=O carbon, which has no hydrogen.

6.3.2(c): Predicting a spectrum

For each environment give three things: shift range, relative area, and splitting (with the number of adjacent H as the reason). For carbon-13, give the number of lines and a shift range for each. Do not predict splitting for carbon-13 or for aromatic protons.

6.3.2(d): TMS, solvents and D₂O

Item	What to write
TMS, Si(CH₃)₄	Standard, defined as δ = 0. 12 equivalent H (and 4 equivalent C) give one sharp signal, away from almost all sample signals. Inert and volatile, so easily removed.
Deuterated solvent, e.g. CDCl₃	Contains no ¹H, so the solvent gives no signal in the proton spectrum to swamp the sample.
D₂O shake	O–H and N–H protons exchange with deuterium: R–OH + D₂O ⇌ R–OD + HDO. Their signal disappears on rerunning. C–H signals stay.

6.3.2(e): Combined techniques in order

1. Elemental analysis: % ÷ Ar, divide by the smallest, scale to whole numbers = empirical formula.
2. Mass spectrum: molecular ion gives Mr; Mr ÷ empirical mass gives the multiplier. Fragments suggest parts of the structure. You are not expected to interpret mass spectra of organic halogen compounds.
3. IR: O–H (alcohol or acid), N–H, C=O.
4. Carbon-13: environments and carbon types.
5. Proton: environments, numbers, neighbours.
6. Check: every peak in every spectrum must be explained by the final structure.

Must-know distinctions

• Number of environments vs number of atoms: symmetry makes the first smaller.
• Proton decoupled carbon-13 vs high resolution proton: carbon-13 lines are never split in assessed spectra; proton signals are.
• Equivalent vs adjacent: equivalent protons do not split each other; non-equivalent protons on adjacent carbons do.
• Integration height vs signal height: use the step height of the trace, not how tall the peak looks.
• CH₃ on O vs CH₃ on C=O: both singlets, but at δ 3.3–4.5 and δ 2.0–2.7.
• D₂O vs CDCl₃: D₂O is shaken in to remove O–H and N–H signals; CDCl₃ is the solvent.

Quick self-test

1. Cyclohexanone: how many carbon-13 lines?
2. 1,3,5-trimethylbenzene: how many carbon-13 lines?
3. Give the relative peak areas in the proton spectrum of 3-methylbutanoic acid, (CH₃)₂CHCH₂COOH.
4. Predict the splitting of the CH₂ signal in propanal, CH₃CH₂CHO.
5. A proton signal appears at δ 9.7. What type of proton is it?
6. A carbon-13 line appears at δ 176. What can the carbon be?
7. Give the area ratio and splitting for each signal in ethyl methanoate, HCOOCH₂CH₃, ignoring the shift of the HCOO proton.
8. A compound C₃H₆O shows one singlet in its proton spectrum and two lines in its carbon-13 spectrum. Identify it.
9. What happens to the NH₂ signal of an amine when the sample is shaken with D₂O?
10. Why is TMS chosen as the reference? Give two reasons.
11. Why is CDCl₃, not CHCl₃, used as the solvent?

Answers

1. Four: C=O, the two carbons next to it, the next two, and the carbon opposite the C=O.
2. Three: CH₃ carbons, ring carbons carrying CH₃, and ring C–H carbons.
3. 6 : 1 : 2 : 1 (two CH₃, CH, CH₂, COOH), total 10 H.
4. Adjacent H = 3 (CH₃) + 1 (CHO) = 4, so a quintet.
5. An aldehyde H–C=O proton.
6. The C=O of a carboxylic acid, ester or amide.
7. 1 : 2 : 3. HCOO: singlet (its neighbour is O). OCH₂: quartet (3 adjacent H). CH₃: triplet (2 adjacent H).
8. Propanone, CH₃COCH₃: six equivalent H, and two carbon environments (C=O and CH₃).
9. It disappears, because N–H exchanges to N–D and deuterium gives no proton signal.
10. Any two: 12 equivalent H give one sharp signal; the signal is away from sample signals; it is inert; it is volatile and easily removed.
11. CDCl₃ has no ¹H, so it gives no solvent signal to swamp the sample.

Where marks are usually lost

• Counting the protons on only one adjacent carbon when both sides carry H.
• Splitting a signal by protons on oxygen or nitrogen, or by equivalent protons in the same group.
• Calling a 6-peak signal a “hexet” or a 4-peak signal a “quintet”: name the pattern from n + 1.
• Reading carbon-13 line heights as numbers of carbons.
• Giving “C=O” for a line at δ 176 without saying acid, ester or amide.
• Writing that TMS gives one peak “because it has 12 hydrogens” without saying they are equivalent.
• Saying D₂O makes the O–H signal move or split, instead of disappear.
• Stopping at one structure when the data fit two isomers.
• Forgetting to use the molecular ion to turn the empirical formula into the molecular formula.

Official syllabus

Source document: OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017, section 6.3.2 Spectroscopy, learning outcomes (a) to (e). Published by OCR.