Structured Questions
Question 1: Analysis of Isomeric Esters
6 marks(a) Two isomeric esters have the molecular formula C₄H₈O₂. Ester A has 4 peaks in its Carbon-13 (¹³C) NMR spectrum. Ester B has only 3 peaks. Draw the structure of Ester B and explain why it shows only 3 peaks. [3]
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- Structure of Ester B: Isopropyl methanoate, HCOOCH(CH₃)₂ (or 1-methylethyl methanoate) [1]
- Equivalent groups: It contains two equivalent methyl groups (−CH₃) bonded to the same −CH− carbon [1]
- Explanation: Because of symmetry, these two methyl carbons are in the identical chemical environment, so they resonate at the same chemical shift and produce a single peak. Along with the C=O carbon and the −CH− carbon, this yields exactly 3 peaks [1]
(b) Ester A has 4 peaks in its Carbon-13 (¹³C) NMR spectrum. In ¹H NMR, Ester A shows a singlet at 2.0 ppm (integrating to 3H), a quartet at 4.1 ppm (2H), and a triplet at 1.2 ppm (3H). Identify Ester A and explain how the splitting patterns support your answer. [3]
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- Identity of Ester A: Ethyl ethanoate, CH₃COOCH₂CH₃ [1]
- Ethyl group coupling: The quartet at 4.1 ppm (2H) and triplet at 1.2 ppm (3H) indicate a neighbouring −CH₂-CH₃ (ethyl) group [1]
- Splitting explanation: The −CH₂− protons are split into a quartet (3+1) by the three adjacent protons on the −CH₃ group, and the −CH₃ protons are split into a triplet (2+1) by the two adjacent protons on the −CH₂− group [1] (Singlet at 2.0 ppm is the CH₃C=O group with no neighbouring protons)
Question 2: Proton NMR Comparison
6 marks(a) Contrast the number of peaks and the splitting patterns in the ¹H NMR spectrum of 1,2-dichloroethane (Cl-CH₂-CH₂-Cl) with those of 1,1-dichloroethane (CH₃-CH-Cl₂). [4]
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- 1,2-dichloroethane: Shows only 1 peak / environment (due to symmetry of the molecule) [1]
- 1,2-dichloroethane splitting: The peak is a singlet / has no splitting because all 4 protons are equivalent and do not couple with each other [1]
- 1,1-dichloroethane: Shows 2 peaks / environments (−CH₃ and −CH−) [1]
- 1,1-dichloroethane splitting: The −CH₃ peak is a doublet (split by the single neighbouring CH proton) and the −CH− peak is a quartet (split by the three neighbouring methyl protons) [1]
(b) Explain how a chemist could use deuterium oxide (D₂O) to confirm if a broad peak in a ¹H NMR spectrum is due to an O-H group. [2]
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- Add D₂O to the NMR tube, shake, and re-run the ¹H NMR spectrum [1]
- The peak due to the O-H group will disappear/vanish from the spectrum (because the acidic/labile O-H proton exchanges with deuterium to form O-D, which does not resonate in the standard ¹H range) [1]
Question 3: Reference Calibration
3 marks(a) State two physical properties and one chemical property of tetramethylsilane (TMS) that make it suitable as an internal reference standard. [3]
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Any three from:
- Physical: Highly volatile / has a low boiling point, meaning it can be easily evaporated off to recover the pure sample after analysis [1]
- Chemical: Chemically inert / unreactive, so it will not react with the sample compound being analysed [1]
- Spectral: Gives a single, intense, sharp peak because all 12 protons (and 4 carbons) are in completely equivalent environments [1]
- Positioning: Its protons/carbons are highly shielded, producing a peak far to the right (upfield, defined as 0 ppm) away from almost all other organic signals [1]
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