OCR A-Level Chemistry Synthesis and analytical techniques (02), November 2020: Question 22

8 marks · Hard difficulty · Extended Response

Explain the use of two deuterated compounds in NMR spectroscopy and determine the structure of organic compound I using analytical data including elemental analysis, molecular ion mass, IR, and proton NMR spectra.

Practise this question

Question

The question presents analytical data for an organic compound I: elemental analysis percentages (C 56.69%, H 7.09%, N 11.02%, O 25.20%), mass spectrum molecular ion peak at m/z = 127.0, an IR spectrum placeholder, and a proton NMR spectrum with chemical shift ranging from 0 to 5 ppm showing peaks with relative peak areas annotated (quartet of area 2 near 4.2 ppm, quartet of area 1 near 2.9 ppm, doublet of area 3 near 1.7 ppm, triplet of area 3 near 1.3 ppm). Part (a) asks to explain the use of two deuterated compounds in NMR spectroscopy (2 marks). Part (b) asks to determine the structure of compound I showing all reasoning (6 marks).
Question text

22 An organic compound I is analysed, using a combination of techniques. The analytical data is

shown below.

Elemental analysis by mass

C, 56.69%; H, 7.09%; N, 11.02%; O, 25.20%

Mass spectrum

Molecular ion peak at m/z = 127.0

IR spectrum

© SDBS, National Institute of Advanced Industrial Science and Technology.

Proton NMR spectrum

33 3 3

55 4 4 3 3 2 2 1 1 0 0

Chemical shift,Chemical shift,d/ ppm d / ppm

(a) Explain the use of two deuterated compounds in NMR spectroscopy.

… [2]

(b)* Determine the structure of compound I, showing all your reasoning. [6]

Additional answer space if required

Mark scheme

Show the mark scheme The mark scheme gives the answers for question 22. Part (a) awards marks for mentioning CDCl3 used as a solvent and D2O used to identify OH or NH protons. Part (b) is a 6-mark level-of-response question requiring empirical and molecular formula calculation (C6H9NO2), identification of key functional groups (C=N, C=O ester/amide), and complete structure determination from NMR and IR data.

AO

Question Answer Marks Guidance

element

22 (a) CDCl3 used as a solvent 2 1.1×2 Example and use required for each mark

D2O used to identify OH OR NH protons ALLOW for 1 mark, D2O as a solvent

(b)* Please refer to the marking instructions on page 4 of this mark 6 3.1× 4 Indicative scientific points:

scheme for guidance on how to mark this question. 3.2× 2 Empirical and Molecular Formulae

C : H : N : O

Level 3 (5–6 marks) 56.69 7.09 11.02 25.20

= : : :

Structure I has a viable chemical structure of C6H9NO2 which 12.0 1.0 14.0 16.0

has the key features consistent with spectral data OR 4.72 : 7.09 : 0.787 : 1.575

AND = 6 : 9 : 1 : 2

Most of the data analysed

• Empirical formula = C6H9NO2

There is a well-developed line of reasoning which is clear and • m/z = 127.0 and empirical formula

logically structured. The information presented is relevant and

substantiated. mass (127) used to determine

molecular formula as C6H9NO2

Level 2 (3–4 marks)

Compound I has a viable chemical structure of C6H9NO2 with Structures of compound I

most of the key features consistent with spectral data

AND

Some of the spectral data analysed.

There is a line of reasoning presented with some structure. The OR

information presented is relevant and supported by some

evidence. OR

Level 1 (1–2 marks)

Correct determination of empirical formula and/or molecular

formula. OR

OR

Analyses some of the IR and NMR data. ALLOW any combination of skeletal OR

OR structural OR displayed formula as long as

Analyses most of the NMR data. unambiguous

AO

Question Answer 23 Marks Guidance

element

There is an attempt at a logical structure with a line of reasoning. Key features

The information is in the most part relevant. • C ≡N

• C=O in aldehyde, ketone, ester,

0 marks No response or no response worthy of credit. amide, acid anhydride

• CH3 group that would give a doublet

• CH3 group that would give a triplet

• CH2 group that would give a quartet

1H NMR and IR analysis

1H NMR spectrum

• δ = 4.2 ppm, quartet, 2H CH3–

CH2–O

• δ = 2.9 ppm, quartet, 1H CO–CH–

CH3

• δ = 1.7 ppm, doublet, 3H CO–CH–

CH3

• δ = 1.3 ppm, triplet, 3H CH3–CH2

IR spectrum

• peak at 1750 (cm–1) is C=O

• peak at 2280 (cm–1) is C ≡N

ALLOW ranges from Data Sheet

IGNORE references to C–O peaks

OCR (Oxford Cambridge and RSA Examinations)

The Triangle Building

Shaftesbury Road

Cambridge

CB2 8EA

OCR Customer Contact Centre

Education and Learning

Telephone: 01223 553998

Facsimile: 01223 552627

Email: general.qualifications@ocr.org.uk

www.ocr.org.uk

For staff training purposes and as part of our quality assurance programme your call may be

recorded or monitored

How to answer it

Analytical Techniques and Structure Determination

What this question tests

This multi-step synoptic organic chemistry question tests your ability to determine empirical and molecular formulas from percentage composition and mass spectrometry data, interpret infrared (IR) and proton (¹H) NMR spectra, understand the practical utility of deuterated solvents in NMR spectroscopy, and synthesize these clues into a definitive chemical structure.

Question Part (a)

Deuterated Compounds in NMR Spectroscopy

✅ Correct Answer

1. CDCl₃ is used as an NMR solvent.
2. D₂O is used to identify exchangeable protons ( OH or NH ).

💡 Key Knowledge

  • Deuterium ( ²H or D ) has an even mass number and a spin quantum number of zero, meaning it does not produce unwanted peaks in ¹H NMR spectra.
  • CDCl₃ dissolves organic compounds without introducing interfering ¹H peaks.
  • Adding D₂O causes labile protons to exchange with deuterium ( OH / NH + D₂O ⇌ OD / ND + HDO), causing the peak to disappear from the spectrum.
Marks available: [2] — Award 1 mark for each correct deuterated compound paired with its correct spectroscopic purpose.
Question Part (b)

Structure Determination of Compound I

📐 Step-by-Step Calculation: Empirical & Molecular Formula

  1. Assume 100g of compound: Masses are C = 56.69g, H = 7.09g, N = 11.02g, O = 25.20g.
  2. Divide by relative atomic masses (Ar):
    C: 56.69 / 12.0 = 4.724
    H: 7.09 / 1.0 = 7.09
    N: 11.02 / 14.0 = 0.787
    O: 25.20 / 16.0 = 1.575
  3. Find the simplest whole number ratio: Divide all by the smallest value (0.787):
    C: 6 | H: 9 | N: 1 | O: 2
  4. Empirical Formula: C₆H₉NO₂ (Relative mass = 127.0).
  5. Molecular Formula: Since the molecular ion peak in the mass spectrum is at m/z = 127.0 , the molecular formula is also C₆H₉NO₂ .

💡 Spectral Data Analysis

  • IR Spectrum: Sharp peak at ~2280 cm⁻¹ indicates a C≡N (nitrile) group. Peak at ~1750 cm⁻¹ indicates a C=O (carbonyl, likely ester) group.
  • ¹H NMR Peaks & Splitting:
    • δ = 4.2 ppm (quartet, 2H) → CH₂–O environment adjacent to a CH₃ group.
    • δ = 2.9 ppm (quartet, 1H) → CO–CH–CH₃ methine proton.
    • δ = 1.7 ppm (doublet, 3H) → CH₃ group adjacent to a CH proton.
    • δ = 1.3 ppm (triplet, 3H) → CH₃–CH₂ ethyl ester fragment.

✅ Correct Structures for Compound I

Any viable structure consistent with C₆H₉NO₂ featuring an ester ( COO ), a nitrile ( C≡N ), and the correct splitting patterns is accepted:

NC–CH(CH₃)–C(=O)–O–CH₂CH₃

CH₃CH₂(O=C)–CH(CH₃)–C≡N

(Accept any clear structural, displayed, or skeletal representation).

❌ Common Errors & Exam Traps

  • Rounding too early: Rounding atomic ratios prematurely leads to incorrect empirical formulas (e.g. getting H = 8.9 instead of 9).
  • Misinterpreting splitting: Confusing quartets and triplets due to poor peak inspection. Remember: splitting = n + 1 equivalent neighboring protons.
  • Forgetting IR absorption ranges: Confusing nitrile ( C≡N ) stretches with alkyne or amine peaks.

🧠 Exam Technique: Level of Response Strategy (6 Marks)

This is a Level of Response question assessed across 3 tiers:

  • Level 3 (5–6 marks): Correct empirical/molecular formula AND correct structure proposed, backed by a clear, logical step-by-step breakdown of both IR and NMR data.
  • Level 2 (3–4 marks): Correct formula and a plausible structure with partial analysis of spectral data.
  • Level 1 (1–2 marks): Correct empirical formula calculation or partial spectral interpretation alone.

Top tip: Always write out your elemental ratio headings clearly and explicitly state how functional groups are deduced from specific m/z, IR wavenumber, and δ values to secure top-tier reasoning marks.

Marks available: [6] — Synoptic mark awarded via Level of Response (holistic evaluation of calculation, spectra interpretation, and final chemical structure).

Topics

Module 6: Organic chemistry and analysis · 6.3 Analysis

Question and mark scheme from the OCR A-Level Chemistry examination, Synthesis and analytical techniques (02), November 2020. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.