📋 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 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 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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