Edexcel A-Level Chemistry: Analytical Techniques
6 exam-style questions with full mark schemes and model answers. Write your own answer and the AI examiner marks it against the mark scheme.
A liquid compound W contains only carbon, hydrogen and oxygen and has the molecular formula C₄H₈O (Mr=72). It does not decolourise bromine water and gives no reaction with Tollens reagent. The following spectroscopic data were recorded.
Infrared spectrum (selected absorptions)
| Wavenumber / cm⁻¹ | Appearance |
|---|---|
| 2960 | medium |
| 1715 | strong, sharp |
There is no broad absorption between 2500 and 3600 cm⁻¹.
¹H NMR spectrum
| δ / ppm | Integration ratio | Splitting |
|---|---|---|
| 2.4 | 2 | quartet |
| 2.1 | 3 | singlet |
| 1.0 | 3 | triplet |
Mass spectrum (selected peaks)
| m/z | Relative abundance / % |
|---|---|
| 72 | 24 |
| 57 | 30 |
| 43 | 100 |
| 29 | 55 |
Deduce the structure of W, explaining how each piece of evidence supports your answer. (6 marks)
A volatile organic compound Y contains carbon, hydrogen and one other element. Its low-resolution mass spectrum shows the peaks below. (Use ³⁵Cl = 35 and ⁷⁹Br = 79.)
| m/z | Relative abundance / % | Note |
|---|---|---|
| 78 | 30 | |
| 80 | 10 | |
| 43 | 100 | base peak |
| 27 | 35 |
A small peak is also seen at m/z 79, with an abundance of about 3.3 % of the m/z 78 peak.
(a) Use the peaks at m/z 78 and 80 to identify the other element present, and state the species responsible for the m/z 78 peak. (2 marks)
(b) Use the m/z 79 peak to determine the number of carbon atoms in Y. (1 mark)
(c) Deduce the molecular formula of Y, calculate its degree of unsaturation, and hence identify the fragment ions at m/z 43 and m/z 27. (3 marks)
A halogenoalkane T has the molecular formula C₃H₇Br. Its high-resolution ¹H NMR spectrum, recorded in CDCl₃, is summarised below.
| δ / ppm | Integration ratio | Splitting |
|---|---|---|
| 3.4 | 2 | triplet |
| 1.9 | 2 | sextet |
| 1.0 | 3 | triplet |
Interpret this spectrum to deduce the structure of T. Assign each signal and explain how the integration ratio and the splitting patterns establish the connectivity. (5 marks)
Propan-1-ol (CH₃CH₂CH₂OH) and propan-2-ol ((CH₃)₂CHOH) are isomers with the molecular formula C₃H₈O. Both show a broad O–H absorption near 3350 cm⁻¹ in their infrared spectra, so IR alone cannot distinguish them.
Explain how their ¹³C NMR spectra would allow you to tell the two isomers apart. In your answer, state the number of peaks each isomer produces and justify your reasoning. (5 marks)
A student uses thin-layer chromatography (TLC) on a silica plate to analyse a mixture of two organic compounds, P and R, using a non-polar solvent as the mobile phase. After development, the solvent front had travelled 8.0 cm from the origin. Compound P had travelled 6.0 cm and compound R had travelled 2.0 cm.
| Spot | Distance moved / cm |
|---|---|
| P | 6.0 |
| R | 2.0 |
| Solvent front | 8.0 |
(a) Calculate the Rf value of each compound. (2 marks)
(b) State and explain which compound, P or R, is the more polar. (2 marks)
2-Methylpropan-2-ol has the structure (CH₃)₃COH.
State how many peaks would appear in its ¹³C NMR spectrum, and explain your answer. (3 marks)