OCR A-Level Chemistry Synthesis and analytical techniques (02), June 2017: Question 21

6 marks · Hard difficulty · Extended Response

Deduce the structure of an organic compound L containing carbon, hydrogen, and oxygen only using its 1H NMR and infrared spectra after reflux with hydrochloric acid.

Practise this question

Question

An exam question showing the 1H NMR spectrum of compound L with chemical shifts from 0 to 5 ppm and relative peak areas of 2, 2, 9, and 3. Below are infrared spectra for products M (partially redacted due to copyright) and N, with transmittance percentage against wavenumber from 4000 to 500 cm-1. The question asks to use the information provided to suggest a structure for compound L and show all reasoning, worth 6 marks.
Question text

Compound L is an organic compound containing carbon, hydrogen and oxygen only.

The 1H NMR spectrum of compound L is shown below.

The numbers by the peaks are the relative peak areas.

54 3 2 1 0

chemical shift, δ/ppm

Compound L is refluxed with aqueous hydrochloric acid, forming two organic compounds M

and N. The infrared spectra of M and N are shown below.

Infrared spectrum of M

Item removed due to third party copyright restrictions.

Infrared spectrum of N

transmittance

(%) 50

4000 3000 2000 1500 1000 500

wavenumber / cm–1

Use the information provided to suggest a structure for compound L.

Show all of your reasoning.

… [6]

Mark scheme

Show the mark scheme Mark scheme showing a level-based response grid from Level 1 to Level 3 (up to 6 marks). Indicative scientific points include details on 1H NMR chemical shifts, splitting patterns and relative peak areas for L, infrared spectra analysis for M (carboxylic acid) and N (alcohol), and the identification of L as an ester.

Question Answer Marks Guidance

Please refer to the marking instructions on page 5 of this mark 6 Indicative scientific points may include:

scheme for guidance on how to mark this question. 1. 1H NMR spectrum

= 1.1 ppm, triplet, 3H CH3–CH2–

Level 3 (5–6 marks)

Structure of L is CH3CH2COOCH2C(CH3)3 OR = 1.3 ppm, singlet, 9H (CH3)3C–

(CH3)3CCH2COOCH2CH3

AND = 2.3 ppm, quartet, 2H CH –CH –C=O

A comprehensive explanation with most of the spectral data

analysed and few omissions. = 4.0 ppm, singlet, 2H –CH –O–

There is a well-developed line of reasoning which is clear and

logically structured. The information presented is relevant and

substantiated. ALLOW approximate values for chemical

shifts.

Splitting patterns used to deduce the correct structure of L.

2. Infrared spectra

Level 2 (3–4 marks)

IR spectrum of M

Attempts all three scientific points but explanations may be –1

incomplete and/or structure of L incorrect. peak at 2300–3700 (cm ) is O–H

OR –1

Explains two scientific points thoroughly with few omissions. peak at ~1720 (cm ) is C=O

There is a line of reasoning presented with some structure. The M is a carboxylic acid

information presented in the most part relevant and supported

IR spectrum of N

by some evidence. –1

peak at 3100-3700 (cm ) is O–H

The analysis is clear and includes some interpretation of

NMR/IR peaks. N is an alcohol

Level 1 (1–2 marks)

A simple explanation based on at least two of the main scientific ALLOW ranges from Data Sheet

points. IGNORE references to C–O peaks

OR

3. Structure of L

Explains one scientific point thoroughly with few omissions.

L is an ester (as it reacts with HCl(aq) to

There is an attempt at a logical structure with a line of reasoning. form carboxylic acid and alcohol)

The information is in the most part relevant.

The analysis is communicated in an unstructured way and Correct structure

includes interpretation of a few peaks from the NMR/IR spectra. H H O H CH

0 marks

H C C C O C C CH3

No response or no response worthy of credit.

H H H CH3

ALLOW any combination of skeletal OR

structural OR displayed formula as long as

unambiguous

Total 6

How to answer it

Elucidating Ester Structure from NMR and IR Spectra

What this question tests

This 6-mark Level-of-Response question assesses your ability to combine spectroscopic data from two different sources (¹H NMR and Infrared spectroscopy) with chemical reaction knowledge (ester hydrolysis) to deduce an unknown organic structure. You must interpret chemical shifts, splitting patterns, relative peak areas, and characteristic IR absorption bands to construct a logical, step-by-step argument.

Question 21 • 6 Marks

Structure Determination of Compound L

✅ Correct Answer

Compound L is an ester:

CH₃CH₂COOCH₂C(CH₃)₃

(Alternatively accepted: (CH₃)₃CCH₂COOCH₂CH₃ or any unambiguous structural/skeletal/displayed formula).

💡 Key Knowledge

  • Reflux with HCl(aq): Esters undergo acid hydrolysis to form a carboxylic acid ( M ) and an alcohol ( N ).
  • ¹H NMR Integration: Relative peak areas give the ratio of hydrogen environments (9:3:2:2 simplifies to total protons).
  • Splitting Patterns (n+1 rule): Triangles, quartets, and singlets reveal neighbouring proton counts.
  • IR Spectroscopy: Broad O-H stretches identify acids and alcohols; sharp C=O stretches confirm carbonyl groups.

🧠 Exam Technique (Level 3 Strategy)

To secure a Level 3 (5–6 marks), you must systematically break down both the NMR spectrum and the IR spectra, tying them together to justify the functional groups and carbon skeleton before proposing the final structure.

❌ Common Errors

  • Ignoring peak integration ratios and assigning incorrect numbers of equivalent protons.
  • Failing to deduce that L is an ester despite the reaction with aqueous acid producing two separate organic products.
  • Confusing splitting patterns (e.g., misinterpreting a singlet as a triplet due to poor peak resolution reading).
Mark Scheme Breakdown:
• Level 3 (5–6 marks): Correct structure of L given WITH a comprehensive explanation analysing most spectral data and splitting patterns.
• Level 2 (3–4 marks): Attempts all scientific points with minor omissions or an incorrect final structure but clear partial reasoning.
• Level 1 (1–2 marks): Simple explanation based on at least two main scientific points with unstructured communication.

Step-by-Step Examiner Analysis

Step 1: Interpreting the Reaction Clue

Compound L is refluxed with aqueous hydrochloric acid to form two organic compounds, M and N. This tells us immediately that L contains an ester functional group ( -COO- ), which hydrolyses into a carboxylic acid and an alcohol.

Step 2: Analysing the ¹H NMR Spectrum Data

  • δ = 1.1 ppm (triplet, 3H): Corresponds to a CH₃–CH₂– fragment (methyl group adjacent to a CH₂ ).
  • δ = 1.3 ppm (singlet, 9H): Corresponds to a (CH₃)₃C– (tert-butyl) group. The singlet indicates zero adjacent protons ( n+1 = 1 ).
  • δ = 2.3 ppm (quartet, 2H): Corresponds to a CH₃–CH₂–C=O fragment (methylene group adjacent to a methyl and next to a carbonyl).
  • δ = 4.0 ppm (singlet, 2H) or region near 3.5–4.1 ppm: Corresponds to a –CH₂–O– group attached to oxygen in the ester linkage.

Step 3: Analysing the Infrared (IR) Spectra

  • IR Spectrum of M (Carboxylic Acid): Shows a characteristic broad absorption peak at 2300–3700 cm⁻¹ for the O-H bond in carboxylic acids, alongside a sharp peak at ~1720 cm⁻¹ for the carbonyl ( C=O ).
  • IR Spectrum of N (Alcohol): Shows a broad absorption peak at 3100–3700 cm⁻¹ for the alcoholic O-H stretch.

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), June 2017. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.