OCR A-Level Chemistry Synthesis and analytical techniques (02), November 2021: Question 21

6 marks · Hard difficulty · Extended Response

Analyse test-tube observations, 13C NMR, and 1H NMR spectra to identify the structures of isomers D, E, and F with formula C5H10O.

Practise this question

Question

A 6-mark chemistry exam question presenting chemical test observations and NMR spectra for three isomers, D, E, and F, with the molecular formula C5H10O. A table gives results for 2,4-DNP, acidified potassium dichromate on reflux, and bromine water for each compound. 13C NMR data is provided for D, and 1H NMR spectra with chemical shifts and splitting patterns are shown for E and F. Students are asked to analyse these observations and results to identify the structures of D, E, and F and explain their reasoning.
Question text

Compounds D, E and F are isomers with the molecular formula C5H10O.

One of the compounds is alicyclic.

A student carries out test-tube tests on the compounds.

The observations are shown below.

Compound 2,4-DNP H+/Cr O 2–, reflux Bromine water

D No change Green solution No colour change

Orange

E No colour change No colour change

precipitate

Orange

F No colour change No colour change

precipitate

13C NMR spectrum of D Compound D has 3 peaks at δ / ppm: 24, 36, 73.

1H NMR spectra of E and F The integration data has been omitted.

Compound E

32 1 0

chemical shift, δ/ppm

Compound F

Expansion of multiplet

32 1 0

chemical shift, δ/ppm

Analyse the observations and results to identify the structures of D, E and F.

Explain your reasoning. [6]

Additional answer space if required.

Mark scheme

Show the mark scheme A mark scheme table showing a Level of Response marking grid from Level 1 to Level 3 for 6 marks, based on correctly identifying compounds D, E, and F and analysing the test-tube and NMR data. Guidance points specify the deductions from 2,4-DNP, acidified potassium dichromate, bromine water, 13C NMR peaks, and 1H NMR splitting patterns and chemical shifts, along with accepted structures for D, E, and F.

AO

Question Answer Marks Guidance

element

Refer to marking instructions on page 5 of mark 6 AO3.1 Indicative scientific points may include:

scheme for guidance on marking this question. ×4 Observations from Test-tube tests

2,4 DNP D has no C=O

Level 3 (5–6 marks) AO3.2

E and F have C=O present

Compounds D, E AND F correctly identified ×2 H+/Cr O 2– D is primary OR secondary alcohol

AND E and F are ketones

Most of the observations and NMR data analysed. (negative test shows not aldehydes)

Br2 D, E and F have no C=C/are saturated

There is a well-developed line of reasoning which is

clear and logically structured. The information 13C NMR analysis

presented is relevant and substantiated.

D:

Level 2 (3–4 marks) • 3 carbon environments/types of C

Most of compounds D, E AND F correctly identified • δ = 24, 36 ppm C–C

AND • δ = 73 ppm, C–O

Some of the observations and NMR data analysed.

1H NMR analysis

There is a line of reasoning presented with some

structure. The information presented is relevant and E:

supported by some evidence. • δ = 2.4 ppm, quartet CH3–CH2–C=O

• δ = 1.1 ppm, triplet CH3–CH2–

Level 1 (1–2 marks) F:

Most of compounds D, E AND F correctly identified • δ = 2.6 ppm, heptet/multiplet (CH3)2–CH–C=O

OR • δ = 2.1 ppm, singlet, CH3–C=O

Some of compounds D, E AND F correctly identified • δ = 1.1 ppm, doublet CH3–CH–

AND

Analyses some of the observations or NMR data Structures

OR ALLOW any combination of skeletal OR structural

Analyses most of the observations from the test-tube OR displayed formula as long as unambiguous

tests.

OR

Analyses most of the NMR data.

OR OR

Analyses some of the observations and NMR data

21 AO

element

There is an attempt at a logical structure with a line

of reasoning. The information is in the most part

relevant.

0 marks

No response or no response worthy of credit.

Total 6

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How to answer it

Elucidating Structures of Isomers D, E and F

What this question tests

This 6-mark level-of-response question assesses your ability to combine chemical test-tube observations with spectroscopic data ( ¹³ C NMR and ¹ H NMR) to deduce the structural formulas of organic isomers with molecular formula C₅H₁₀O. You must systematically analyze functional group tests, carbon environments, splitting patterns, and chemical shifts.

Question 21 (6 Marks)

Comprehensive Analysis & Mark Scheme Breakdown

✅ Correct Structures

  • Compound D: Cyclopentanol ( C₅H₁₀O ), or alternatively cyclopropan-1-ol with an ethyl substituent (or similar alicyclic isomer with an -OH group).
  • Compound E: Pentan-3-one ( CH₃CH₂COCH₂CH₃ )
  • Compound F: 3-methylbutan-2-one ( (CH₃)₂CHCOCH₃ )

💡 Key Knowledge

  • 2,4-DNP: Positive test (orange precipitate) confirms a carbonyl group ( C=O ) in aldehydes and ketones. Negative test (no change) rules them out (e.g. D is an alcohol).
  • Acidified Dichromate: H⁺/Cr₂O₇²⁻ turns green with primary/secondary alcohols (oxidation). Ketones give no change.
  • Bromine Water: Stays orange/no change, showing absence of C=C double bonds (alkenes).

🧠 Exam Technique (Level 3 Response)

To achieve Level 3 (5–6 marks), your answer must not just list facts; it must weave test-tube observations and NMR data into a logical, step-by-step justification for each structure.

  • Start by eliminating functional groups using the table.
  • Assign every NMR peak given in the stem to specific fragments (e.g., matching splitting patterns: triplet/quartet for an ethyl group, doublet/heptet for an isopropyl group).

❌ Common Errors

  • Confusing splitting patterns (e.g. misinterpreting a quartet as a multiplet without referencing adjacent protons using the n+1 rule).
  • Failing to state that compound D is alicyclic based on the question stem.
  • Mixing up secondary vs primary alcohol oxidation outcomes with dichromate.

📐 Step-by-Step Spectral Deduction

  1. Analyzing D: Negative 2,4-DNP means no C=O . Positive dichromate (green solution) means it's an alcohol. The stem states it is "alicyclic", and ¹³ C NMR shows 3 peaks ( δ = 24, 36, 73 ppm ), yielding cyclopentanol.
  2. Analyzing E: Positive 2,4-DNP and negative dichromate confirms a ketone. ¹ H NMR shows a triplet ( δ = 1.1 ppm ) and a quartet ( δ = 2.4 ppm ), proving two identical ethyl groups attached to a carbonyl: CH₃CH₂COCH₂CH₃ (pentan-3-one).
  3. Analyzing F: Also gives an orange precipitate with 2,4-DNP and no reaction with dichromate (ketone). ¹ H NMR shows a doublet at δ = 1.1 ppm and a singlet at δ = 2.1 ppm along with a septet/multiplet at δ = 2.6 ppm , indicating an isopropyl group adjacent to a carbonyl: (CH₃)₂CHCOCH₃ (3-methylbutan-2-one).
Mark Scheme Thresholds:
• Level 3 (5–6 marks): Compounds D, E and F correctly identified AND most observations/NMR data analysed with a well-developed, logical line of reasoning.
• Level 2 (3–4 marks): Most compounds correctly identified AND some observations/NMR data analysed.
• Level 1 (1–2 marks): Some compounds identified OR partial analysis of data.

Topics

Module 6: Organic chemistry and analysis · 6.1 Aromatic compounds, carbonyls and acids · 6.3 Analysis

Question and mark scheme from the OCR A-Level Chemistry examination, Synthesis and analytical techniques (02), November 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.