OCR A-Level Chemistry Synthesis and analytical techniques (02), June 2018: Question 12

1 mark · Medium difficulty · Multiple Choice

Identify the ion responsible for the peak with the greatest relative intensity in the mass spectrum of (CH3)2CHCH2OH.

Practise this question

Question

A multiple-choice question showing the mass spectrum of (CH3)2CHCH2OH with relative intensity on the y-axis (0 to 100) and m/z on the x-axis (10 to 95). The base peak (greatest relative intensity of 100) occurs around m/z = 43. Four options are provided: A, CHCH2OH+; B, CH3CH2CH+; C, (CH3)2CH+; D, CH3CO+.
Question text

12 The mass spectrum of (CH3)2CHCH2OH is shown below.

relative 60

intensity

10 20 30 40 50 60 70 80 90

m/z

Which ion is responsible for the peak with the greatest relative intensity?

A CHCH OH+

B CH CH CH+

C (CH ) CH+

D CH CO+

Your answer

[1]

Mark scheme

Show the mark scheme The mark scheme table shows question number 12 with the correct answer option C and 1 mark allocated.

12 C 1

How to answer it

Mass Spectrometry of 3-methylbutan-1-ol

What this question tests

This question assesses your understanding of mass spectrometry, specifically how molecular ions fragment upon electron bombardment. You must be able to link peaks on a mass spectrum (m/z values) to specific fragment ions generated from a branched alcohol and identify the base peak (the peak with the greatest relative intensity).

Question Analysis & Answer

Question 12 [1 Mark]

✅ Correct Answer: C ((CH₃)₂CH⁺)

The peak at 100% relative intensity (the base peak) occurs at an m/z value of 43. Calculating the relative molecular mass (Mr) of the isopropyl cation (CH₃)₂CH⁺ gives (2 × 15) + 13 = 43, which matches the base peak on the spectrum.

💡 Key Knowledge

  • Base Peak: The tallest peak in a mass spectrum, representing the most stable (and therefore most abundant) fragment ion formed. It is assigned a relative intensity of 100%.
  • Fragmentation: When high-energy electrons strike a molecule in a mass spectrometer, it forms a molecular ion ( M⁺ ) which can break apart into a positive fragment ion and a neutral radical.
  • Stable Carbocations: Branched carbocations (like secondary carbocations) are more stable than primary ones, making them more likely to form and survive in high abundances.

🧠 Exam Technique

When tackling multiple-choice questions involving mass spectra:

  1. Locate the base peak (height = 100%) on the vertical axis.
  2. Drop down to the horizontal axis to read off its m/z value (here, m/z = 43 ).
  3. Calculate the m/z (formula mass) of each option to find a match.

❌ Common Errors

  • Confusing the molecular ion peak ( M⁺ ) with the base peak. The molecular ion shows the Mr of the whole molecule, whereas the base peak is simply the most abundant fragment.
  • Making arithmetic errors when summing up atomic masses ( C = 12, H = 1, O = 16 ) for the given ionic options.

📐 Step-by-Step Mass Verification

Let's check the mass-to-charge ratios for the options provided:

  • Option A: CHCH₂OH⁺ → 12 + 1 + (2×1) + 16 + 1 = 33
  • Option B: CH₃CH₂CH⁺ → (3×12) + (7×1) = 43 (Note: While this has an m/z of 43, the branched structure in option C is the more stable and logical carbocation formed via cleavage next to the branching point of (CH₃)₂CHCH₂OH).
  • Option C: (CH₃)₂CH⁺ → (3×12) + (7×1) = 43. Matches the base peak at m/z = 43 perfectly.
  • Option D: CH₃CO⁺ → (2×12) + (3×1) + 16 = 43 (Not structurally derivable from the carbon skeleton without major rearrangement).
Mark Scheme Allocation: 1 mark awarded for selecting C.

Topics

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

Question and mark scheme from the OCR A-Level Chemistry examination, Synthesis and analytical techniques (02), June 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.