CAIE A-Level Chemistry A2 37.3 Carbon 13 Nmr Spectroscopy Questions
Practise counting carbon environments, assigning chemical shifts and testing possible molecular structures.
- Syllabus
- 2028–2030
- Course
- Chemistry 9701
- Level
- A2
Practise counting carbon environments, assigning chemical shifts and testing possible molecular structures.
Deduce the number of peaks that would be present in the carbon-13 NMR spectrum of benzophenone.
number of peaks
5 peaks
Identify two different environments of carbon atom that would result in different chemical shift ranges in this carbon-13 NMR spectrum of benzophenone.
| environment of carbon atom | chemical shift range (δ) |
|---|---|
| carbonyl / RCOR | 190–220 |
| arene / benzene | 110–160 |
Award one mark for each correct row.
Asparagine is an amino acid that contains a chiral carbon atom and displays stereoisomerism.
Separate samples of asparagine are dissolved in CDCl3 and analysed using carbon-13 and proton (1H) NMR spectroscopy.
Fig. 6.1
Predict the number of peaks seen in the carbon-13 and proton ( 1H ) NMR spectra of asparagine.
& carbon-13 NMR & proton NMR
number of peaks in CDCl3 & 4 & 5
A student analyses an aromatic compound, X,C8H8O3, using NMR spectroscopy.
Fig. 6.4 shows the carbon-13 NMR spectrum of a sample of \(\mathbf{X
Separate samples of X were analysed using proton (1H) NMR spectroscopy.
Table 6.1 gives information obtained from this analysis.
Table 6.1
Identify the number of different carbon environments present in X.
8 / eight
Aromatic compound X gives a yellow precipitate when it reacts with alkaline I2 (aq).
Use the information in (c) to suggest a structure for X.
Explain your reasoning.
Table 6.2
Molecule must not be symmetrical (as eight different C environments)
Viable structures of X
- methyl ketone / CH3CO group because (yellow) ppt / CHIX3 formed (with alkaline I2(aq))
any two linked statements:
- two - OH groups
because 1H NMR signals lost in D2O
- ketone / carbonyl / CH3CO
as (13C NMR) signal with Δ=205/190−220ppm
- methyl / carbon next to C=O because
(13CNMR) signal with Δ=30−65ppm