Edexcel A-Level Chemistry Paper 2, June 2024: Question 6

7 marks · Medium difficulty · Short Open Response

Identify NMR region, explain TMS use, define secondary amine, deduce amine structure, and distinguish cyclic alcohol isomers using 13C NMR spectra.

Practise this question

Question

Question text

6 This question is about nuclear magnetic resonance (NMR) spectroscopy.

(a) Which part of the electromagnetic spectrum is used in NMR?

(1)

A infrared

B radio waves

C ultraviolet

D X‑ray

(b) Explain why tetramethylsilane, TMS, is used in NMR spectroscopy.

(2)

The low resolution 1H NMR spectrum of Q has three peaks.

(i) State what is meant by the term ‘secondary amine’.

(1)

(ii) Deduce the structure of Q.

(1)

(d) Two cyclic alcohols have the structures shown.

OH OH

cyclohexane‑1,2‑diol cyclohexane‑1,3‑diol

Show that the 13C NMR spectra of these compounds can be used to distinguish

18 between the two alcohols, labelling the diagrams to justify your answer.

*P76896A01832* (2)

(Total for Question 6 = 7 marks)

Mark scheme

Show the mark scheme

Question

Answer Mark

Number

6(a) The only correct answer is B (radio waves) (1)

A is not correct because IR interacts with bonds

C is not correct because UV interacts with electrons

D is not correct because X-rays are diffracted by crystals in the solid state

Question

Answer Additional Guidance Mark

Number

6(b) An explanation that makes reference to the following points: (2)

• TMS acts as a standard / reference (at 0 ppm) (1) Allow acts as a control / comparison point /

baseline / to calibrate / to mark 0 ppm

• as it has (12 equivalent) hydrogens that produce a (1) Ignore just ‘has a shift at 0 ppm’

single / strong peak Allow produces a peak (well) away from those

or caused by other hydrogens / carbons

as it has (4 equivalent) carbons that produce a single / Allow (most organic) compounds result in shift

strong peak values (much) greater than TMS

Allow easy to separate from sample (after spectrum

is obtained)

Allow unreactive (so will not react with sample)

Question

Answer Additional Guidance Mark

Number

6(c)(i) An answer that makes reference to the following point: (1)

• the nitrogen (in the amine group) is attached to two Allow contains an NH group / two alkyl groups

carbon atoms / the nitrogen (in the amine group) is attached to the N / two R groups attached to the N /

attached to two carbons (atoms) N is adjacent to two carbons

Ignore amine group is attached to two carbons

Question

Answer Additional Guidance Mark

Number

6(c)(ii) Note – take care to check H is present on N in (1)

• skeletal formula

Allow structural, displayed or hybrid formulae, e.g.

Question

Answer Additional Guidance Mark

Number

6(d) An answer that makes reference to the If no marks scored allow 1 mark for idea that the spectrum of (2)

following points: cyclohexane-1,2-diol has fewer peaks / fewer environments than

cyclohexane-1,3-diol (even if any numbers quoted are incorrect)

• cyclohexane-1,2-diol has 3 peaks (in (1)

its 13C spectra)

and

3 environments shown on diagram (1)

• cyclohexane-1,3-diol has 4 peaks (in

its 13C spectra)

and

4 environments shown on diagram

Allow alternative labelling systems on diagrams e.g. use of letter

Ignore comments related to size of peaks / position of shift values /

splitting

(Total for Question 6 = 7 marks)

How to answer it

Nuclear Magnetic Resonance (NMR) Spectroscopy

What this question tests

This question assesses your understanding of NMR spectroscopy principles, including the electromagnetic region used, the role of tetramethylsilane (TMS) as a calibration standard, defining structural isomers/functional groups (secondary amines), deducing organic structures from molecular formulas and peak counts, and predicting carbon-13 (¹³C) environments in cyclic systems.

Question Part (a)

Electromagnetic Spectrum in NMR

✅ Correct Answer

B: radio waves

💡 Key Knowledge

NMR uses low-energy radio frequency radiation to cause nuclei with spin (like ¹H and ¹³C) to flip between energy states.
Infrared interacts with covalent bonds (stretching/bending), UV-visible interacts with electronic transitions, and X-rays interact with inner-shell electrons.

Marks: 1 / 1

Question Part (b)

Why use Tetramethylsilane (TMS)?

✅ Correct Answer

To score full marks, you must state both points:

TMS acts as a reference standard/calibrant set at 0 ppm .
It produces a single, intense peak because it has 12 equivalent hydrogens (in ¹H NMR) or 4 equivalent carbons (in ¹³C NMR).

🧠 Exam Technique

Examiners penalize vague answers like "it gives a peak at 0". You must explicitly link why it gives a single peak (chemical equivalence of its protons/carbons) and its role as a reference point.

❌ Common Errors

Students often lose the second mark by forgetting to mention symmetry/equivalence, simply stating "it has many hydrogens" instead of specifying that they are equivalent and produce a single peak.

Marks: 2 / 2

Question Part (c)(i)

Defining a Secondary Amine

✅ Correct Answer

The nitrogen atom in the amine group is attached directly to two carbon atoms (or two alkyl groups).

💡 Key Knowledge

Classification of amines depends on how many carbon chains are attached to the nitrogen atom: Primary (1), Secondary (2), Tertiary (3).

Marks: 1 / 1

Question Part (c)(ii)

Deducing the Structure of Q

✅ Correct Answer

Structure: HN(CH(CH₃)₂₂)₂ (Diisopropylamine), drawn as a skeletal or displayed formula showing a central -NH- connected to two -CH(CH₃)₂ groups.

🧠 Exam Technique

With molecular formula C₆H₁₅N and only 3 peaks in the ¹H NMR spectrum, the molecule must be highly symmetrical. A secondary amine with a central nitrogen splitting the 6 carbons equally into two identical isopropyl branches creates very few chemical environments.

Marks: 1 / 1

Question Part (d)

Distinguishing Cyclic Alcohols via ¹³C NMR

✅ Correct Answer

Cyclohexane-1,2-diol has 3 peaks in its ¹³C NMR spectrum (due to high symmetry down a plane of reflection passing between C1-C2).
Cyclohexane-1,3-diol has 4 peaks in its ¹³C NMR spectrum (lower symmetry resulting in four distinct carbon environments).

🧠 Exam Technique

To gain full marks, clearly label the carbon environments on the ring diagrams with numbers (1, 2, 3...) corresponding to the number of non-equivalent carbon atoms present in each isomer.

❌ Common Errors

Confusing ¹³C NMR environments with ¹H splitting patterns. Carbon NMR does not show spin-spin splitting (no multiplets); each peak corresponds strictly to a unique carbon environment.
Miscounting cyclic symmetry planes.

Marks: 2 / 2

Topics

Organic Chemistry · Topic 19: Modern Analytical Techniques II · Topic 18: Organic Chemistry III

Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 2, June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.