OCR A Level Chemistry A: Spectroscopy (H432)

This guide teaches 6.3.2 Spectroscopy from the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017. NMR-centred outcomes 6.3.2(a) to (e) are all taught on this page, which sits within Module 6 (Organic chemistry and analysis); because H432 is the full A level, none of it carries an AS or A2 label. The specification’s assessment overview puts Module 6 in two papers: Component 02, Synthesis and analytical techniques, plus Component 03, Unified chemistry, which spans every module. No PAG is linked to 6.3.2.

Browse every topic on the OCR A Level Chemistry hub, record progress on the printable H432 checklist and test yourself with the free diagnostics. Infrared and mass spectra are taught in Analytical techniques (4.2.4); this guide adds NMR and then combines all of them.

Outcomes in 6.3.2

Statement	You must be able to	Specification guidance
6.3.2(a)	From a carbon-13 spectrum, predict (i) the number of carbon environments, (ii) the types of environment from chemical shifts, (iii) possible structures	M3.1; spectra proton decoupled; shifts on the Data Sheet; specification functional groups; HSW3,5
6.3.2(b)	From a high resolution proton spectrum, predict (i) the number of proton environments, (ii) their types, (iii) relative numbers from peak areas, integration traces or ratio numbers, (iv) adjacent non-equivalent protons by the n + 1 rule, (v) possible structures	M3.1; shifts on the Data Sheet; aromatic protons by shift only, no splitting analysis; HSW3,5
6.3.2(c)	Predict a carbon-13 or proton spectrum for a given molecule	M3.1
6.3.2(d)	Explain (i) TMS as the chemical shift standard, (ii) deuterated solvents such as CDCl₃, (iii) identifying O–H and N–H protons by exchange with D₂O	None
6.3.2(e)	Work out organic structures by combining (i) elemental analysis, (ii) mass spectra, (iii) IR spectra and (iv) NMR spectra	M3.1; reference data on the Data Sheet; no mass spectra of organic halogen compounds; HSW3,5,6

What an NMR spectrum shows

Placed in a strong magnetic field, ¹H and ¹³C nuclei take in radio-frequency energy, and the frequency each one absorbs depends on its environment. Atoms related by symmetry are equivalent and share one signal. A signal’s position is its chemical shift, δ, in ppm from a reference at δ = 0. Nearby O, N, Cl or C=O moves signals to higher δ.

Chemical shift values used on this page

The specification says NMR chemical shift values will be provided on the Data Sheet in examinations; use those in the exam. On this page, use these typical ranges.

Carbon environment	δ / ppm	Proton environment	δ / ppm
C–C (alkyl)	5–55	R–CH₃, R–CH₂–R, R₃CH	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
C=C (alkene)	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
C=O in acids, esters, amides	160–185	aldehyde H–C=O	9.4–10.0
C=O in aldehydes, ketones	190–220	carboxylic acid O–H	10.0–12.5
		alcohol O–H, amine N–H	0.5–5.0 (variable)

6.3.2(a): Carbon-13 NMR

Assessed carbon-13 spectra are proton decoupled: each carbon environment gives one line, with no splitting. Line heights do not show how many carbons share an environment, so count lines, not heights.

Method: count the lines (environments); fewer lines than carbons means symmetry; assign each line a type; write every structure that fits.

Worked example 1: an ester or an acid?

Compound A, C₅H₁₀O₂, gives carbon-13 lines at δ 170.6, 67.6, 21.9 and 21.4.

Count. Four lines for five carbons: two carbons are equivalent.

Types. δ 170.6 is a C=O in an acid, ester or amide. δ 67.6 is a carbon bonded to oxygen. δ 21.9 and 21.4 are two different alkyl carbons.

Rule out. An acid such as 3-methylbutanoic acid also has four environments, but it has no line at δ 50–90. So A is an ester.

Possible structures. Three C₅H₁₀O₂ esters have four environments: propan-2-yl ethanoate, CH₃COOCH(CH₃)₂; methyl 2-methylpropanoate, (CH₃)₂CHCOOCH₃; and 2-methylpropyl methanoate, HCOOCH₂CH(CH₃)₂.

Carbon-13 data alone cannot separate them. If A’s proton spectrum has a 3H singlet at δ 2.0, that CH₃ is next to C=O, so A is propan-2-yl ethanoate. Methyl 2-methylpropanoate would show its 3H singlet (OCH₃) at δ 3.3–4.5, and 2-methylpropyl methanoate has no 3H singlet.

6.3.2(b): High resolution proton NMR

(i) Number of signals = number of proton environments.

(ii) Chemical shift gives the type of environment. Recognise aromatic protons (δ 6.5–8.2) from their shift; you are not expected to analyse their splitting.

(iii) Relative peak area is proportional to the number of protons. It may be shown as an integration trace (divide each step height by the smallest) or as ratio numbers.

(iv) Splitting. A signal is split by the non-equivalent protons on the adjacent carbon atoms. By the n + 1 rule, n adjacent non-equivalent protons split a signal into n + 1 peaks.

Pattern	Peaks (n + 1)	Neighbouring non-equivalent H (n)
singlet	1	none
doublet	2	one
triplet	3	two
quartet	4	three
quintet	5	four
sextet	6	five
septet	7	six

Add up the protons on all adjacent carbons. Equivalent protons do not split each other. O–H and N–H protons usually give broad singlets and do not split neighbours.

(v) Structure. Combine the evidence and check the total proton count against the formula.

Worked example 2: reading the splitting

Compound B, C₆H₁₂O₂, gives this proton spectrum.

δ / ppm	Relative area	Splitting
0.93	3	triplet
1.38	2	sextet
1.60	2	quintet
2.04	3	singlet
4.06	2	triplet

Environments. Five signals; areas total 12, matching the formula.

Shifts. δ 2.04 (3H, singlet) is CH₃ next to C=O with no neighbouring H, so CH₃CO–. δ 4.06 (2H) is a CH₂ on oxygen.

Splitting. The δ 4.06 triplet has 2 adjacent H. The δ 1.60 quintet has 4 adjacent H, so it sits between two CH₂ groups. The δ 1.38 sextet has 5 adjacent H: a CH₂ on one side and CH₃ on the other. The δ 0.93 triplet is a CH₃ next to a CH₂.

Structure. The chain is O–CH₂–CH₂–CH₂–CH₃ joined to CH₃CO–. B is butyl ethanoate, CH₃COOCH₂CH₂CH₂CH₃.

6.3.2(c): Predicting a spectrum

Reverse the method: for each environment give the shift range, relative area and splitting.

Worked example 3: methyl 3-chloropropanoate, ClCH₂CH₂COOCH₃

Carbon-13. Four carbons, all different, so four lines: C=O at δ 160–185, OCH₃ at δ 50–90, CH₂Cl at δ 30–70 and CH₂ next to C=O at δ 5–55.

Proton. Three environments, area ratio 2 : 2 : 3.

Protons	δ range	Area	Splitting
Cl–CH₂–	3.0–4.5	2	triplet (2 adjacent H)
–CH₂–C=O	2.0–2.7	2	triplet (2 adjacent H)
–O–CH₃	3.3–4.5	3	singlet (no adjacent H)

The OCH₃ is a singlet: its neighbouring carbon (C=O) has no hydrogen.

6.3.2(d): TMS, deuterated solvents and D₂O

(i) Tetramethylsilane, TMS, Si(CH₃)₄, is the standard. Its signal is defined as δ = 0. It is used because:

• all 12 hydrogens are equivalent (and all four carbons are), so it gives one sharp signal
• its signal lies away from almost all organic signals
• it is inert and volatile, so it does not react with the sample and is easily removed afterwards.

(ii) Deuterated solvents such as CDCl₃ are used because deuterium, ²H, does not absorb in the region recorded for ¹H. A solvent containing ordinary hydrogen would give a huge signal that swamps the sample.

(iii) D₂O exchange. O–H and N–H shifts vary, so run the spectrum, shake the sample with D₂O and run it again. These protons swap for deuterium:

R–OH + D₂O ⇌ R–OD + HDO

R–NH₂ + 2D₂O ⇌ R–ND₂ + 2HDO

The O–H or N–H signal disappears from the second spectrum, which identifies it. C–H protons do not exchange.

Worked example 4: confirming an amine

2-methylpropan-2-amine, (CH₃)₃CNH₂, gives a 9H singlet at δ 1.1 and a broad 2H singlet near δ 1.3. After shaking with D₂O only the 9H singlet remains, so the 2H signal was NH₂. The 9H signal is a singlet because the central carbon has no hydrogen.

6.3.2(e): Combining analytical data

Each technique adds evidence:

• elemental analysis gives the empirical formula (2.1.3(c))
• the mass spectrum gives Mr from the molecular ion, and fragments (4.2.4(f) and (g))
• the IR spectrum gives functional groups, especially O–H, N–H and C=O
• carbon-13 NMR gives the carbon environments
• proton NMR gives the proton environments, their numbers and their neighbours.

You will not be asked to interpret mass spectra of organic halogen compounds.

Worked example 5: from percentages to a structure

Compound C contains 50.8% C, 8.5% H and 40.7% O by mass. Relative atomic masses: O 16.0, C 12.0, H 1.0. The data are:

• mass spectrum: m/z 118 (very small), 73, 45 (tallest) and 29
• IR: broad absorption at 3450 cm⁻¹, strong absorption at 1735 cm⁻¹
• carbon-13: five lines, at δ 175.7, 66.8, 61.7, 20.4 and 14.1
• proton: δ 1.29 (3H, triplet), 1.41 (3H, doublet), 2.9 (1H, broad singlet, gone after D₂O), 4.23 (2H, quartet), 4.28 (1H, quartet).

Empirical formula. C: 50.8 ÷ 12.0 = 4.233. H: 8.5 ÷ 1.0 = 8.5. O: 40.7 ÷ 16.0 = 2.544. Dividing by 2.544 gives 1.664 : 3.342 : 1. Multiplying by 3 gives 4.99 : 10.02 : 3, so C₅H₁₀O₃, with Mr = 118.0.

Molecular formula. The molecular ion at m/z 118 matches: C₅H₁₀O₃.

IR. 3450 cm⁻¹ (broad) is an alcohol O–H, not the very broad acid band. 1735 cm⁻¹ is C=O.

Carbon-13. δ 175.7 is an ester C=O. δ 66.8 and 61.7 are two different C–O carbons. δ 20.4 and 14.1 are two alkyl carbons.

Proton. The δ 1.29 triplet (3H) and δ 4.23 quartet (2H) are an ethyl group on oxygen, –OCH₂CH₃. The δ 1.41 doublet (3H) and δ 4.28 quartet (1H) are CH₃–CH< with the CH on oxygen. The δ 2.9 signal vanishes with D₂O, so it is the O–H.

Fragments. m/z 45 fits CH₃CH(OH)⁺, m/z 73 fits COOCH₂CH₃⁺ and m/z 29 fits CH₂CH₃⁺.

Structure. C is ethyl 2-hydroxypropanoate, CH₃CH(OH)COOCH₂CH₃.

Common errors

• Counting carbon-13 line heights as numbers of carbons.
• Using n instead of n + 1, so a CH₂ next to a CH₃ is called a triplet.
• Forgetting to add the protons on both adjacent carbons.
• Splitting an OCH₃ signal because “the oxygen is next to something”: count H on adjacent carbons only.
• Trying to work out splitting for aromatic protons, which you are not expected to analyse.
• Saying D₂O makes the O–H peak “move” rather than disappear.

Next steps

• Condensed tables and a quick self-test: Spectroscopy revision notes
• Original questions with mark-by-mark answers: Spectroscopy practice questions
• IR, mass spectra and elemental analysis first: Analytical techniques
• Ten-minute course checks: Marlbridge diagnostics

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.