OCR A-Level Chemistry Synthesis and analytical techniques (02), November 2021: Question 11

1 mark · Medium difficulty · Multiple Choice

Identify the compound that produced the given infrared spectrum from four options.

Practise this question

Question

Multiple choice question 11 asks to identify which compound could have produced the IR spectrum shown. The spectrum displays transmittance (%) from 0 to 100 on the y-axis against wavenumber in cm⁻¹ from 4000 to 500 on the x-axis. Notable peaks include a broad absorption around 3300 cm⁻¹, strong peaks around 3000 cm⁻¹, and a strong peak around 1700 cm⁻¹. Below the spectrum are four options: A, (CH3)2CHCHO; B, (CH3)2CHCOOH; C, CH3CH(OH)CH=CH2; D, CH3COCH(OH)CH3, followed by a box for the answer.
Question text

11 Which compound could have produced the IR spectrum below?

transmittance (%) 50

4000 3000 2000 1500 1000 500

wavenumber / cm–1

A (CH3)2CHCHO

B (CH3)2CHCOOH

C CH3CH(OH)CH=CH2

D CH3COCH(OH)CH3

Your answer

[1]

Mark scheme

Show the mark scheme The mark scheme indicates the correct answer for question 11 is D, awarding 1 mark.

11 D 1 AO2.5

How to answer it

Identifying Functional Groups from an Infrared (IR) Spectrum

📌 What this question tests

This question assesses your ability to interpret infrared (IR) spectra by matching characteristic absorption wavenumbers to specific bonds and functional groups, allowing you to deduce the correct molecular structure from a given list of isomers and related compounds.

Question 11: Spectral Analysis

Identifying the correct compound from IR absorption data

✅ Correct Answer: D

CH₃COCH(OH)CH₃

Mark Awarded: 1 / 1 (AO2.5)

💡 Key Knowledge

  • O-H stretch (alcohol): Broad absorption peak around 3200 - 3600 cm⁻¹ .
  • C=O stretch (carbonyl): Strong, sharp absorption peak around 1680 - 1750 cm⁻¹ .
  • C-H aliphatic stretch: Sharp peaks just below 3000 cm⁻¹ .

🧠 Exam Technique

  • Scan for key peaks first: Look at the spectrum and immediately note the very broad peak around 3400 cm⁻¹ (Alcohol O-H) and the sharp peak near 1715 cm⁻¹ (Carbonyl C=O).
  • Eliminate options systematically: Check each option against the identified functional groups to narrow down your choices rapidly.

❌ Common Errors

  • Confusing carboxylic acid and alcohol O-H: Carboxylic acids ( -COOH ) show an extremely broad O-H stretch that merges with the C-H region down to 2500 cm⁻¹ . The peak here is a standard alcohol O-H.
  • Missing the carbonyl peak: Failing to notice the strong dip around 1700 cm⁻¹ , leading students to mistakenly select unsaturated alcohols like option C.

🔬 Step-by-Step Examiner Breakdown

To arrive at the correct answer D, analyse the spectrum step-by-step:

  1. Identify the O-H peak: There is a very broad, strong absorption spanning from roughly 3200 cm⁻¹ to 3550 cm⁻¹ . This confirms the presence of an alcohol ( -OH ) group.
  2. Identify the C=O peak: There is a sharp, deep trough right around 1715 cm⁻¹ . This confirms the presence of a carbonyl ( C=O ) group (ketone or aldehyde).
  3. Evaluate the Options:
    • A ((CH₃)₂CHCHO): Has a C=O, but no O-H group. (Incorrect)
    • B ((CH₃)₂CHCOOH): Has a C=O and an O-H, but the O-H is a carboxylic acid (very broad, extending down to 2500 cm⁻¹ ). (Incorrect)
    • C (CH₃CH(OH)CH=CH₂): Has an alcohol O-H and a C=C alkene bond, but no carbonyl C=O peak around 1700 cm⁻¹ . (Incorrect)
    • D (CH₃COCH(OH)CH₃): Contains both a ketone carbonyl ( C=O ) giving the peak at ~1715 cm⁻¹ and a secondary alcohol ( -OH ) giving the broad peak at ~3400 cm⁻¹ . (Correct)

Topics

Module 6: Organic chemistry and analysis · Module 4: Core organic chemistry · 6.3 Analysis · 4.2 Alcohols, haloalkanes and 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.