The \({ }^{13}\text{C NMR}\) spectrum of an organic compound is shown below.
The organic compound could be
CHEMISTRY, M8 2023 VCE 16 MC
Consider the following molecule.
How many peaks will be observed in a \({ }^{13} \text{C NMR}\) spectrum of this molecule
- 5
- 6
- 7
- 8
CHEMISTRY, M8 2023 HSC 36
An organic reaction pathway involving compounds \(\text{A, B,}\) and \(\text{C}\) is shown in the flow chart.
The molar mass of \(\text{A}\) is 84.156 g mol\(^{-1}\).
A chemist obtained some spectral data for the compounds as shown.
| \( \text{Data from} \ ^{1} \text{H NMR spectrum of compound C} \) | ||
| \( Chemical \ Shift \ \text{(ppm)} \) | \( Relative \ peak \ area \) | \( Splitting \ pattern \) |
| \(1.01\) | \(3\) | \(\text{Triplet}\) |
| \(1.05\) | \(3\) | \(\text{Triplet}\) |
| \(1.65\) | \(2\) | \(\text{Multiplet}\) |
| \(2.42\) | \(2\) | \(\text{Triplet}\) |
| \(2.46\) | \(2\) | \(\text{Quartet}\) |
| \( ^{1} \text{H NMR chemical shift data}\) | |
| \( Type \ of \ proton \) | \( \text{δ/ppm} \) |
| \( \ce{R - C\textbf{H}3,R - C\textbf{H}2 - R}\) | \(0.7-1.7\) |
| \( \left.\begin{array}{l}\ce{\textbf{H}3C - CO - \\-C\textbf{H}2 - CO -}\end{array}\right\} \begin{aligned} & \text { (aldehydes, ketones,} \\ &\text{carboxylic acids or esters) }\end{aligned}\) | \(2.0-2.6\) |
| \( \ce{R - C\textbf{H}O} \) | \(9.4-10.00\) |
| \( \ce{R - COO\textbf{H}} \) | \(9.0-13.0\) |
Identify the functional group present in each of compounds \(\text{A}\) to \(\text{C}\) and draw the structure of each compound. Justify your answer with reference to the information provided. (9 marks)
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Compound \(\text{A}\): Alkene
Compound \(\text{B}\): Secondary alcohol
Compound \(\text{C}\): Ketone
Reasoning as follows:
→ Compound \(\text{A}\) is able to undergo an addition reaction to add water across a \(\ce{C=C}\) bond \(\Rightarrow \) Alkene
→ Compound \(\text{B}\) is the product of the above hydration reaction and is therefore an alcohol.
→ The \(\ce{^{13}C\ NMR}\) spectrum of Compound \(\text{A}\) confirms it is an alkene (132 ppm peak corresponding to the \(\ce{C=C}\) atoms). 3 spectrum peaks indicate 3 carbon environments. The molar mass of compound \(\text{A}\) is 84.156 g mol\(^{-1}\) which suggests symmetry within the molecule.
→ The Infrared Spectrum of Compound \(\text{B}\) has a broad peak at approximately 3400 cm\(^{-1}\). This indicates the presence of an hydroxyl group and confirms \(\text{B}\) is an alcohol.
→ Compound \(\text{C}\) is produced by the oxidation of Compound \(\text{B}\) with acidified potassium permanganate.
→ Compound \(\text{C}\) is a carboxylic acid if \(\text{B}\) is a primary alcohol or a ketone if \(\text{B}\) is a secondary alcohol.
→ Since the \(\ce{^{1}H NMR}\) spectrum of \(\text{C}\) does not show any peaks between 9.0 − 13.0 ppm, it cannot be a carboxylic acid. Compound \(\text{C}\) is therefore a ketone and Compound \(\text{B}\) is a secondary alcohol.
→ The \(\ce{^{1}H NMR}\) spectrum shows 5 peaks \(\Rightarrow \) 5 hydrogen environments.
→ Chemical shift and splitting patterns information indicate:
1.01 ppm – 1.05 ppm: \(\ce{CH3}\) (next to a \(\ce{CH2}\))
1.65 ppm: \(\ce{CH2}\) (with multiple neighbouring hydrogens)
2.42 ppm: \(\ce{CH2}\) (next to the ketone \(\ce{C=O}\) and a \(\ce{CH2}\))
2.46 ppm: \(\ce{CH2}\) (next to the ketone \(\ce{C=O}\) and a \(\ce{CH3}\))
Show Worked Solution
Compound \(\text{A}\): Alkene
Compound \(\text{B}\): Secondary alcohol
Compound \(\text{C}\): Ketone
Reasoning as follows:
→ Compound \(\text{A}\) is able to undergo an addition reaction to add water across a \(\ce{C=C}\) bond \(\Rightarrow \) Alkene
→ Compound \(\text{B}\) is the product of the above hydration reaction and is therefore an alcohol.
→ The \(\ce{^{13}C\ NMR}\) spectrum of Compound \(\text{A}\) confirms it is an alkene (132 ppm peak corresponding to the \(\ce{C=C}\) atoms). 3 spectrum peaks indicate 3 carbon environments. The molar mass of compound \(\text{A}\) is 84.156 g mol\(^{-1}\) which suggests symmetry within the molecule.
→ The Infrared Spectrum of Compound \(\text{B}\) has a broad peak at approximately 3400 cm\(^{-1}\). This indicates the presence of an hydroxyl group and confirms \(\text{B}\) is an alcohol.
→ Compound \(\text{C}\) is produced by the oxidation of Compound \(\text{B}\) with acidified potassium permanganate.
→ Compound \(\text{C}\) is a carboxylic acid if \(\text{B}\) is a primary alcohol or a ketone if \(\text{B}\) is a secondary alcohol.
→ Since the \(\ce{^{1}H NMR}\) spectrum of \(\text{C}\) does not show any peaks between 9.0 − 13.0 ppm, it cannot be a carboxylic acid. Compound \(\text{C}\) is therefore a ketone and Compound \(\text{B}\) is a secondary alcohol.
→ The \(\ce{^{1}H NMR}\) spectrum shows 5 peaks \(\Rightarrow \) 5 hydrogen environments.
→ Chemical shift and splitting patterns information indicate:
1.01 ppm – 1.05 ppm: \(\ce{CH3}\) (next to a \(\ce{CH2}\))
1.65 ppm: \(\ce{CH2}\) (with multiple neighbouring hydrogens)
2.42 ppm: \(\ce{CH2}\) (next to the ketone \(\ce{C=O}\) and a \(\ce{CH2}\))
2.46 ppm: \(\ce{CH2}\) (next to the ketone \(\ce{C=O}\) and a \(\ce{CH3}\))
CHEMISTRY, M8 2019 HSC 19 MC
Compound `text{X}` shows three signals in its \(\ce{^{13}C NMR}\) spectrum.
Treatment of `text{X}` with hot acidified potassium permanganate produces a compound `text{Y}`. Compound `text{Y}` turns blue litmus red.
Compound `text{X}` produces compound `text{Z}` upon reaction with hot concentrated sulfuric acid.
Which of the following correctly identifies compounds `text{X}`, `text{Y}` and `text{Z}`?
CHEMISTRY, M8 2020 HSC 5 MC
A \( \ce{^13C NMR} \) spectrum is shown.
Which compound gives rise to this spectrum?
- chloroethane
- 1-chloropropane
- 1, 2-dichloroethane
- 1, 2-dichloropropane
CHEMISTRY, M8 2022 HSC 27
A bottle labelled 'propanol' contains one of two isomers of propanol.
- Draw the TWO isomers of propanol. (2 marks)
- Describe how \( \ce{^13C NMR}\) spectroscopy might be used to identify which isomer is in the bottle. (2 marks)
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- Each isomer produces a different product when oxidised.
- Write equations to represent the oxidation reactions of the two isomers. Include reaction conditions. (3 marks)
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a. Isomer 1:
Isomer 2:
b. Identifying isomers with \( \ce{^13C NMR} \) spectroscopy:
→ this can be used to identify the isomers in the bottle because they show a different number of signals which helps deduce the carbon environment.
→ Propan-1-ol contains 3 \( \ce{C}\) environments so it would have 3 peaks on a \( \ce{^13C NMR}\) spectrum whereas propan-2-ol only contains 2 \( \ce{C}\) environments (due to symmetry), so it would only have 2 signals on a \( \ce{^13C NMR}\) spectrum.
c.
Show Worked Solution
a. Isomer 1:
Isomer 2:
b. Identifying isomers with \( \ce{^13C NMR} \) spectroscopy:
→ this can be used to identify the isomers in the bottle because they show a different number of signals which helps deduce the carbon environment.
→ Propan-1-ol contains 3 \( \ce{C}\) environments so it would have 3 peaks on a \( \ce{^13C NMR}\) spectrum whereas propan-2-ol only contains 2 \( \ce{C}\) environments (due to symmetry), so it would only have 2 signals on a \( \ce{^13C NMR}\) spectrum.
c.
♦ Mean mark (c) 52%.
CHEMISTRY, M8 2022 HSC 12 MC
Which isomer of \(\ce{C6H14}\) would have the fewest signals in \(\ce{^13C NMR}\)?
