A chemist uses spectroscopy to identify an unknown organic molecule, Molecule \(\text{J}\), that contains chlorine. The \({}^{13}\text{C NMR}\) spectrum of Molecule \(\text{J}\) is shown below. The infra-red (IR) spectrum of Molecule \(\text{J}\) is shown below. --- 2 WORK AREA LINES (style=lined) --- The mass spectrum of Molecule \(\text{J}\) is shown below --- 2 WORK AREA LINES (style=lined) --- The \({ }^1 \text{H NMR}\) spectrum of Molecule \(\text{J}\) is shown below. --- 3 WORK AREA LINES (style=lined) --- --- 5 WORK AREA LINES (style=blank) --- a. → Absence of a very broad \(\ce{OH}\) acid peak between 2500–3000. → Molecule \(\text{J}\) must be an ester. b. The peak at 110 m/z: → due to the Chlorine-37 isotope which is slightly heavier than the more abundant Chlorine-35. c. The two singlet peaks indicate: → two different hydrogen environments within the molecule. → there are no adjacent hydrogen environments. → The relative heights of the peaks show the ratios of the hydrogens in the environments are 2 : 3. d. Either of the two molecules show below are correct: a. → Absence of a very broad \(\ce{OH}\) acid peak between 2500–3000. → Molecule \(\text{J}\) must be an ester. b. The peak at 110 m/z: → due to the Chlorine-37 isotope which is slightly heavier than the more abundant Chlorine-35. c. The two singlet peaks indicate: → two different hydrogen environments within the molecule. → there are no adjacent hydrogen environments. → The relative heights of the peaks show the ratios of the hydrogens in the environments are 2 : 3. d. Either of the two molecules show below are correct:
COMMENT: Know the masses of common isotopes.
CHEMISTRY, M8 2021 VCE 7*
Two students are given a homework assignment that involves analysing a set of spectra and identifying an unknown compound. The unknown compound is one of the molecules shown below. The \(^{13}\text{C NMR}\) spectrum of the unknown compound is shown below. --- 1 WORK AREA LINES (style=lined) --- --- 6 WORK AREA LINES (style=lined) --- --- 1 WORK AREA LINES (style=lined) --- --- 4 WORK AREA LINES (style=lined) ---
The infra-red (IR) spectrum of the unknown compound is shown below.
CHEMISTRY, M8 2021 VCE 11 MC
The spectroscopy information for an organic molecule is given below.
| number of peaks in \({}^{13}\text{C NMR}\) | 2 |
| number of sets of peaks in \({}^{1}\text{H NMR}\) | 3 |
| m/z of the last peak in the mass spectrum | 60 |
| infra-red (IR) spectrum | an absorption peak appears at 3350 cm\(^{-1}\) |
The organic molecule is
CHEMISTRY, M8 2023 VCE 7-1*
Molecule \(\text{V}\) contains only carbon atoms, hydrogen atoms and one oxygen atom. The mass spectrum of molecule \(\text{V}\) is shown below. --- 4 WORK AREA LINES (style=lined) --- --- 1 WORK AREA LINES (style=lined) --- The \({ }^1 \text{H NMR}\) spectrum of molecule \(\text{V}\) is shown below. --- 2 WORK AREA LINES (style=lined) --- The \({ }^{13} \text{C NMR}\) spectrum of molecule \(\text{V}\) is shown below. --- 8 WORK AREA LINES (style=blank) --- a.i. Molar mass of \(\text{V}\ = 86\ \text{g mol}^{-1}\) → Indicated by the parent ion peak at 86 m/z. → Molecular formula: \(\ce{C5H10O}\) a.ii. Small peak at m/z = 87: → A carbon-13 isotope being present in the molecule. b. The doublet peak at 1.1 ppm: → Indicates there is a single hydrogen with a different hydrogen environment bonded to an adjacent carbon atom. c. a.i. Molar mass of \(\text{V}\ = 86\ \text{g mol}^{-1}\) → Indicated by the parent ion peak at 86 m/z. → Molecular formula: \(\ce{C5H10O}\) a.ii. Small peak at m/z = 87: → A carbon-13 isotope being present in the molecule. b. The doublet peak at 1.1 ppm: → Indicates there is a single hydrogen with a different hydrogen environment bonded to an adjacent carbon atom. c. From the carbon NMR graph: → There are 4 carbon environments, one shifted above 200 ppm indicating a ketone or aldehyde. → As there are 5 carbons, 2 of the carbons must have the same environment. From the hydrogen NMR graph: → There are 3 hydrogen environments. → The septet peak indicates there are 6 hydrogens with the same chemical environment on adjacent carbon atoms.
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♦ Mean mark (b) 50%.
♦ Mean mark (c) 43%.
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 26b
Explain why a chemist should use more than one spectroscopic technique to identify an organic compound. Use TWO spectroscopic techniques to support your answer. (3 marks)
CHEMISTRY, M8 2019 HSC 26a
The following data were obtained for an organic compound containing carbon, hydrogen and oxygen. The compound is a colourless liquid that reacts with sodium carbonate powder to produce bubbles.
What is the structural formula of this compound? Justify your answer with reference to the information given on its reactivity and to at least THREE of the provided spectra. (5 marks)
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→ The compound exhibits characteristics that suggest it is an organic acid.
→ This is demonstrated by its reaction with sodium carbonate, which produces carbon dioxide bubbles, as well as the presence of a strong absorption around 1700 cm ¯1 in the IR spectrum, which is characteristic of a \(\ce{CO}\) bond, and another broad absorption in the region 2500-3500 cm ¯1, which is characteristic of an \(\ce{OH}\) bond in acids.
→ The mass spectrum of the compound has a parent peak at m/z = 74, which is consistent with its molecular formula \(\ce{C3H6O2}\) (molar mass = 74 g mol ¯1).
→ The \(\ce{^13C NMR}\) spectrum shows 3 signals, including one around 180 ppm, which is characteristic of a carbonyl carbon, and the \(\ce{^1H NMR}\) spectrum shows 3 signals, including a quartet with an integration of 2 and a triplet with an integration of 3, indicating the presence of a \(\ce{CH3}\) group and \(\ce{CH2}\) group respectively.
→ These observations confirm the presence of a \(\ce{COOH}\) group in the compound.
Show Worked Solution
→ The compound exhibits characteristics that suggest it is an organic acid.
→ This is demonstrated by its reaction with sodium carbonate, which produces carbon dioxide bubbles, as well as the presence of a strong absorption around 1700 cm ¯1 in the IR spectrum, which is characteristic of a \(\ce{CO}\) bond, and another broad absorption in the region 2500-3500 cm ¯1, which is characteristic of an \(\ce{OH}\) bond in acids.
→ The mass spectrum of the compound has a parent peak at m/z = 74, which is consistent with its molecular formula \(\ce{C3H6O2}\) (molar mass = 74 g mol ¯1).
→ The \(\ce{^13C NMR}\) spectrum shows 3 signals, including one around 180 ppm, which is characteristic of a carbonyl carbon, and the \(\ce{^1H NMR}\) spectrum shows 3 signals, including a quartet with an integration of 2 and a triplet with an integration of 3, indicating the presence of a \(\ce{CH3}\) group and \(\ce{CH2}\) group respectively.
→ These observations confirm the presence of a \(\ce{COOH}\) group in the compound.
CHEMISTRY, M8 2020 HSC 30
A chemist discovered a bottle simply labelled \( '\ce{C5H10O2}' \).
To confirm the molecular structure of the contents of the bottle, a sample was submitted for analysis by infrared spectroscopy and \( \ce{^1H} \) and \( \ce{^13C NMR} \) spectroscopy. The resulting spectra are shown.
Draw a structural formula for the unknown compound that is consistent with all of the information provided. Justify your answer with reference to the information provided. (7 marks)
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Infrared Spectrum
→ There’s a signal between 1680 – 1750 cm ¯1, indicating the presence of a \( \ce{C=O} \) bond (consistent with carbonyl group).
→ No absorption at 3230 − 3550 cm ¯1 indicates the absence of the \( \ce{O-H} \) bond (rules out carboxylic acid\( \ce{-COOH} \)).
\( \ce{^13C NMR} \):
→ The \( \ce{^13C NMR} \) spectrum has 4 peaks, indicating that there are 4 unique carbon environments in the compound.
→ This is consistent with the proposed structure of the compound.
→ The peak at 170 ppm is characteristic of the carbonyl carbon, and the peak at 70 ppm corresponds to a carbon nucleus adjacent to an oxygen, which confirms the presence of an ester group.
The \( \ce{^1H NMR} \) Spectrum:
→ A septet in the \( \ce{^1H NMR} \) spectrum is consistent with the presence of six neighbouring hydrogen atoms on two \( \ce{CH3} \) groups.
→ A doublet in the spectrum is consistent with one neighbouring hydrogen atom.
→ The combination of a septet and a doublet is consistent with the presence of a \( \ce{-CH(CH3)2} \) group in the compound.
→ A singlet in the spectrum is consistent with the absence of neighbouring hydrogen atoms, which would be produced by an isolated methyl group.
Show Worked Solution
Infrared Spectrum
→ There’s a signal between 1680 – 1750 cm ¯1, indicating the presence of a \( \ce{C=O} \) bond (consistent with carbonyl group).
→ No absorption at 3230 − 3550 cm ¯1 indicates the absence of the \( \ce{O-H} \) bond (rules out carboxylic acid\( \ce{-COOH} \)).
\( \ce{^13C NMR} \):
→ The \( \ce{^13C NMR} \) spectrum has 4 peaks, indicating that there are 4 unique carbon environments in the compound.
→ This is consistent with the proposed structure of the compound.
→ The peak at 170 ppm is characteristic of the carbonyl carbon, and the peak at 70 ppm corresponds to a carbon nucleus adjacent to an oxygen, which confirms the presence of an ester group.
The \( \ce{^1H NMR} \) Spectrum:
→ A septet in the \( \ce{^1H NMR} \) spectrum is consistent with the presence of six neighbouring hydrogen atoms on two \( \ce{CH3} \) groups.
→ A doublet in the spectrum is consistent with one neighbouring hydrogen atom.
→ The combination of a septet and a doublet is consistent with the presence of a \( \ce{-CH(CH3)2} \) group in the compound.
→ A singlet in the spectrum is consistent with the absence of neighbouring hydrogen atoms, which would be produced by an isolated methyl group.
Mean mark 55%.
CHEMISTRY, M8 2022 HSC 30
The following spectra were obtained for an unknown organic compound.
In the space provided, draw and name the unknown compound that is consistent with all the information provided. Justify your answer with reference to the information provided. (7 marks)
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Structure: 3-methylbutan-2-one
Mass spectrum:
→ There is a parent ion peak at m/z = 86, indicating that the compound has a molar mass of 86 g mol¯1.
IR spectrum:
→ There is a strong absorption at 1680 – 1750 cm¯1, indicating a carbonyl functional group.
→ There isn’t a strong absorption at 2500 – 3000 cm¯1, indicating an absence of hydroxyl group (consistent with a ketone).
Proton NMR spectrum:
→ There are 3 signals, indicating 3 unique hydrogen environments:
→ The signal at 1.1 ppm, consists of 6 hydrogens and is a doublet.
→ The signal at 2 ppm, consists of 3 hydrogens and is a singlet.
→ The signal at 2.6 ppm, consists of 1 hydrogen and is a septet.
Carbon-13 NMR spectrum:
→ There are 4 signals, indicating 4 unique carbon environments.
→ A signal at 19 ppm indicates a carbon atom single bonded to another adjacent carbon atom.
→ A signal at 28 ppm indicates a carbon adjacent to a carbonyl carbon.
→ A signal at 41 ppm indicates a carbon adjacent to a carbonyl carbon and methyl groups.
→ A signal at 210 indicates a ketone carbon.
Show Worked Solution
Structure: 3-methylbutan-2-one
Mass spectrum:
→ There is a parent ion peak at m/z = 86, indicating that the compound has a molar mass of 86 g mol¯1.
IR spectrum:
→ There is a strong absorption at 1680 – 1750 cm¯1, indicating a carbonyl functional group.
→ There isn’t a strong absorption at 2500 – 3000 cm¯1, indicating an absence of hydroxyl group (consistent with a ketone).
Proton NMR spectrum:
→ There are 3 signals, indicating 3 unique hydrogen environments:
→ The signal at 1.1 ppm, consists of 6 hydrogens and is a doublet.
→ The signal at 2 ppm, consists of 3 hydrogens and is a singlet.
→ The signal at 2.6 ppm, consists of 1 hydrogen and is a septet.
Carbon-13 NMR spectrum:
→ There are 4 signals, indicating 4 unique carbon environments.
→ A signal at 19 ppm indicates a carbon atom single bonded to another adjacent carbon atom.
→ A signal at 28 ppm indicates a carbon adjacent to a carbonyl carbon.
→ A signal at 41 ppm indicates a carbon adjacent to a carbonyl carbon and methyl groups.
→ A signal at 210 indicates a ketone carbon.
♦ Mean mark 51%.
CHEMISTRY, M8 2021 HSC 29
A chemist obtained spectral data of pentane-1,5-diamine \(\ce{(C5H14N2)}\).
Relate the highlighted features of the spectra to the structure of pentane-1,5-diamine. (7 marks)
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