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CHEMISTRY, M8 2024 HSC 38

Compounds \(\text{A}\) and \(\text{B}\) are isomers with formula \(\ce{C3H7X}\), where \(\ce{X}\) is a halogen. The mass spectrum for compound \(\text{A}\) is shown.
 

  

Compounds \(\text{A}\) and \(\text{B}\) undergo substitution reactions in the presence of hydroxide ions, producing alcohols \(\text{C}\) and \(\text{D}\). Compound \(\text{D}\) can be oxidised to a ketone; compound \(\text{C}\) can also be oxidised, but does not produce a ketone.
 

Compound \(\text{E}\) can be produced by refluxing 3-methylbutanoic acid, with one of the alcohols \(\text{C}\) or \(\text{D}\), in the presence of a catalyst.

The \({ }^1 \text{H NMR}\) spectrum for compound \(\text{E}\) contains the following features.
 

Draw the structure of compounds \(\text{A}\), \(\text{B}\) and \(\text{E}\). Explain your answer with reference to the information provided.   (7 marks)
 


Show Answers Only

 

  • The mass spectrum has two peaks to the far right of the spectrum of similar height at 122 m/z and 124 m/z. This is due to the halogen having two isotopes with the same relative abundance.
  • Isotope \(\ce{X}\) must be bromine, whose isotopes are \(\ce{Br-79}\) and \(\ce{Br-81}\). The two molecular ion peaks both correspond correctly to the molar mass of the parent molecule, \(\ce{C3H7Br}\), depending on which isomer is present in the compound.
  •    \(MM\ce{(C3H7Br-79)} = 3 \times 12 + 1 \times 7 + 79 = 122\)
  •    \(MM\ce{(C3H7Br-81)} = 3 \times 12 + 1 \times 7 + 81 = 124\)
  • The two isomers of \(\ce{C3H7Br}\) are 1-bromopropane and 2-bromopropane and when they undergo a substitution reaction with \(\ce{OH^{-}}\), they will produce propan-1-ol and propan-2-ol respectively. 
  • Only secondary alcohols will oxidise to produce a ketone. Hence compound \(\text{D}\) must be propan-2-ol and compound \(\text{C}\) must be propan-1-ol which can be oxidised to an aldehyde and the aldehyde can be oxidised to a carboxylic acid.
  • Therefore, compound \(\text{B}\) is 2-bromopropane and compound \(\text{A}\) is 1-bromopropane as during the substitution reaction the bromine atom is substituted with the hydroxide ion.
  • When 3-methylbutanoic acid is reacted with alcohol \(\text{C}\) (propan-1-ol) the following reaction takes place:

  • When 3-methylbutanoic acid is reacted with alcohol \(\text{D}\) (propan-2-ol) the following reaction takes place:

  • There are 6 unique hydrogen environments present in compound \(\text{E}\). Therefore compound \(\text{E}\) must be the ester produced in the reaction between 3-methylbutanoic acid and propan-1-ol (it has 6 hydrogen environments vs propan-2-ol ester which has 5).
  • Compound \(\text{E}\) is propyl 3-methylbutanoate.
  • This can be confirmed by the integration (ratio of hydrogens in each environment) and peak splitting columns (number of  splits = number of adjacent hydrogens + 1)
  • The shift at 0.96 is due to environment 1 which has six hydrogen atoms and has one neighbouring hydrogen atom (produces a doublet).
  • The shift at 2.1 is due to environment 2 which has one hydrogen 1 atom and has 8 neighbouring hydrogen atoms a (produces a mutliplet of 9)(.
  • The shift at 4.0 is due to environment 4. This \(\ce{CH2}\) group is bonded to an oxygen atom corresponding to a large chemical shift between 3.2–5.0. It also has 2 neighbouring hydrogens and so produces a triplet.
  • Other answers could have included a further explanation regarding the integration and peak splitting of all the hydrogen environments and their relative chemical shifts.

Show Worked Solution

 

  • The mass spectrum has two peaks to the far right of the spectrum of similar height at 122 m/z and 124 m/z. This is due to the halogen having two isotopes with the same relative abundance.
  • Isotope \(\ce{X}\) must be bromine, whose isotopes are \(\ce{Br-79}\) and \(\ce{Br-81}\). The two molecular ion peaks both correspond correctly to the molar mass of the parent molecule, \(\ce{C3H7Br}\), depending on which isomer is present in the compound.
  •    \(MM\ce{(C3H7Br-79)} = 3 \times 12 + 1 \times 7 + 79 = 122\)
  •    \(MM\ce{(C3H7Br-81)} = 3 \times 12 + 1 \times 7 + 81 = 124\)
  • The two isomers of \(\ce{C3H7Br}\) are 1-bromopropane and 2-bromopropane and when they undergo a substitution reaction with \(\ce{OH^{-}}\), they will produce propan-1-ol and propan-2-ol respectively. 
  • Only secondary alcohols will oxidise to produce a ketone. Hence compound \(\text{D}\) must be propan-2-ol and compound \(\text{C}\) must be propan-1-ol which can be oxidised to an aldehyde and the aldehyde can be oxidised to a carboxylic acid.
  • Therefore, compound \(\text{B}\) is 2-bromopropane and compound \(\text{A}\) is 1-bromopropane as during the substitution reaction the bromine atom is substituted with the hydroxide ion.
  • When 3-methylbutanoic acid is reacted with alcohol \(\text{C}\) (propan-1-ol) the following reaction takes place:

  • When 3-methylbutanoic acid is reacted with alcohol \(\text{D}\) (propan-2-ol) the following reaction takes place:

  • There are 6 unique hydrogen environments present in compound \(\text{E}\). Therefore compound \(\text{E}\) must be the ester produced in the reaction between 3-methylbutanoic acid and propan-1-ol (it has 6 hydrogen environments vs propan-2-ol ester which has 5).
  • Compound \(\text{E}\) is propyl 3-methylbutanoate.
  • This can be confirmed by the integration (ratio of hydrogens in each environment) and peak splitting columns (number of  splits = number of adjacent hydrogens + 1)
  • The shift at 0.96 is due to environment 1 which has six hydrogen atoms and has one neighbouring hydrogen atom (produces a doublet).
  • The shift at 2.1 is due to environment 2 which has one hydrogen 1 atom and has 8 neighbouring hydrogen atoms a (produces a mutliplet of 9)(.
  • The shift at 4.0 is due to environment 4. This \(\ce{CH2}\) group is bonded to an oxygen atom corresponding to a large chemical shift between 3.2–5.0. It also has 2 neighbouring hydrogens and so produces a triplet.
  • Other answers could have included a further explanation regarding the integration and peak splitting of all the hydrogen environments and their relative chemical shifts.
♦ Mean mark 52%.

CHEMISTRY, M8 2022 VCE 5*

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.
 

  1. Name the functional group that produces the peak at 168 ppm in the \({}^{13}\text{C NMR}\) spectrum on the first image, which is consistent with the IR spectrum shown above. Justify your answer with reference to the IR spectrum.   (2 marks)

The mass spectrum of Molecule \(\text{J}\) is shown below
 

  1. The molecular mass of Molecule \(\text{J}\) is 108.5
  2.  Explain the presence of the peak at 110 m/z.  (1 mark)

The \({ }^1 \text{H NMR}\) spectrum of Molecule \(\text{J}\) is shown below.
 

  1. The \({ }^1 \text{H NMR}\) spectrum consists of two singlet peaks.
  2. What information does this give about the molecule?   (2 marks)

  3. Draw a structural formula for Molecule \(\text{J}\) that is consistent with the information provided in parts a–c.   (2 marks)

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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 shown below are correct:

Show Worked Solution

a.    Absence of a very broad \(\ce{OH}\) acid peak between 2500–3000. 

  • Molecule \(\text{J}\) must be an ester. 
♦ Mean mark (a) 41%.

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. 
♦♦♦ Mean mark (b) 15%.
COMMENT: Know the masses of common isotopes.

d.    Either of the two molecules shown below are correct:
 

♦ Mean mark (d) 40%.

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. Based on the number of peaks in the \(^{13}\text{C NMR}\) spectrum above, which compound – \(\text{P, Q, R, S}\) or \(\text{T}\) – could be eliminated as the unknown compound?   (1 mark)

  2. The infra-red (IR) spectrum of the unknown compound is shown below.
     

  1. Identify which of the five compounds (1 or more) can be eliminated on the basis of the IR spectrum. Justify your answer using data from the IR spectrum.   (3 marks)

  2. The mass spectrum of the unknown compound is shown below.
     

  1. Write the chemical formula of the species that produces a peak at m/z = 43.   (1 mark)

  2. Explain why one molecule can produce multiple peaks on a mass spectrum.   (2 mark)

Show Answers Only

a.    The \(^{13}\text{C NMR}\) shows 4 different carbon environments.

  • Compound \(T\) has 5 unique carbon environments and can be eliminated. 

b.    Compounds \(\text{P, Q}\) and \(\text{S}\) can be eliminated.

  • Compounds \(\text{S}\) and \(\text{P}\) both don’t have an \(\ce{OH}\) alcohol group, however the IR spectrum clearly shows an \(\ce{OH}\) alcohol group with an absorbance at 3500 cm\(^{-1}\).
  • Compound \(\text{Q}\) contains an \(\ce{OH}\) acid group whereas the IR spectrum shows an \(\ce{OH}\) alcohol group at 3500 cm\(^{-1}\). There is no evidence of a broad \(\ce{OH}\) acid group between 2500–3000 cm\(^{-1}\). 

c.i.  The unknown compound is compound \(\text{R}\).

  • The chemical species are fragments of the original compound with a positive charge.
  • The chemical formulas could include \(\ce{[CH3CO]^+}\) or \(\ce{[C2H3O]^+}\) 

 ii.   Ions of that molecule must be formed to produce peaks on the mass spectrum.

  • Organic compounds can be split up into numerous different ions when producing fragment patterns leading to multiple different peaks on the mass spectrum.

Show Worked Solution

a.    The \(^{13}\text{C NMR}\) shows 4 different carbon environments.

  • Compound \(T\) has 5 unique carbon environments and can be eliminated.

b.    Compounds \(\text{P, Q}\) and \(\text{S}\) can be eliminated.

  • Compounds \(\text{S}\) and \(\text{P}\) both don’t have an \(\ce{OH}\) alcohol group, however the IR spectrum clearly shows an \(\ce{OH}\) alcohol group with an absorbance at 3500 cm\(^{-1}\).
  • Compound \(\text{Q}\) contains an \(\ce{OH}\) acid group whereas the IR spectrum shows an \(\ce{OH}\) alcohol group at 3500 cm\(^{-1}\). There is no evidence of a broad \(\ce{OH}\) acid group between 2500–3000 cm\(^{-1}\). 
♦ Mean mark (b) 43%.
COMMENT: Recognise the difference between OH alcohols and OH acid groups.

c.i.   The unknown compound is compound \(\text{R}\).

  • The chemical species are fragments of the original compound with a positive charge.
  • The chemical formulas could include \(\ce{[CH3CO]^+}\) or \(\ce{[C2H3O]^+}\)
♦ Mean mark (c.i.) 50%.
COMMENT: Ions require a positive charge.

 ii.    Ions of that molecule must be formed to produce peaks on the mass spectrum.

  • Organic compounds can be split up into numerous different ions when producing fragment patterns leading to multiple different peaks on the mass spectrum.
♦ Mean mark (c.ii.) 39%.

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 
 

Show Answers Only

\(B\)

Show Worked Solution
  • Only \(B\) and \(C\) have three different hydrogen environments (eliminate \(A\) and \(D\)). 
  • The molar mass of \(B\) is 60 g mol\(^{-1}\) which matches the last m/z peak in the mass spectrum.

\(\Rightarrow B\)

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.
 

  1.  i.  State the molecular formula of molecule \(\text{V}\). Justify your answer by using the information in the mass spectrum.   (2 marks)

  2. ii.  State why there is a small peak at \(\text{m/z} =87\).   (1 mark)

The \({ }^1 \text{H NMR}\) spectrum of molecule \(\text{V}\) is shown below.
 

  1. State what information the doublet at 1.1 ppm provides about the structure of the molecule.   (1 mark)

The \({ }^{13} \text{C NMR}\) spectrum of molecule \(\text{V}\) is shown below.
 

  1. In the space below, draw a structural formula of molecule \(\text{V}\) that is consistent with the information provided in parts a.-c.   (3 marks)

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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.   
       

Show Worked Solution

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.

♦ 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)

Show Answers Only
  • Different techniques in organic chemistry can be used to identify and characterise the structure of organic molecules.
  • These techniques such as \(\ce{^1H NMR}\) spectroscopy and mass spectrometry, provide different pieces of information about the molecule’s structure.
  • \(\ce{^1H NMR}\) spectroscopy can be used to identify functional groups and distinguish between isomers by providing information about the chemical environment and relative number of hydrogen nuclei.
  • Mass spectrometry, on the other hand, gives information about the molecular weight of a molecule and its characteristic fragments.
  • It is important to use a combination of these techniques in order to obtain a complete understanding of the structure of an organic compound.
Show Worked Solution
  • Different techniques in organic chemistry can be used to identify and characterise the structure of organic molecules.
  • These techniques such as \(\ce{^1H NMR}\) spectroscopy and mass spectrometry, provide different pieces of information about the molecule’s structure.
  • \(\ce{^1H NMR}\) spectroscopy can be used to identify functional groups and distinguish between isomers by providing information about the chemical environment and relative number of hydrogen nuclei.
  • Mass spectrometry, on the other hand, gives information about the molecular weight of a molecule and its characteristic fragments.
  • It is important to use a combination of these techniques in order to obtain a complete understanding of the structure of an organic compound.

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)
 

Show Answers Only

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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Infrared Spectrum:

  • IR spectrum allows us to find key functional groups in a molecule.
  • There is a peak signal at 3300 – 3400 cm¯1, indicating the presence of the amino group. 

Mass spectrum:

  • The mass spectrum represents the molecular mass and fragments of a molecule. The highlighted feature shows a 30 m/z signal, this is due to the fragmentation of the molecule.
  • The highlighted signal depicts the fragment `text{CH}_2 text{NH}_2 ^(\ +)` , which has a molecular mass of 30 (12 × 1 + 1.0 × 4 + 14 × 1 = 30). 

Carbon 13 NMR Spectrum:

  • The molecule has 5 carbon atoms, however, the spectrum only has 3 signals. This indicates symmetry, and that there are 3 different carbon environments in the molecule. 
  • Carbon 1 and 5 are in the same carbon environment, carbon 2 and 4 are in the same carbon environment, and carbon 3 is in a unique carbon environment.
  • The signals at 24 and 33 ppm are due to the `text{–CH}_2text{–CH}_2–` carbon atoms.
  • The signal at 42 ppm is due to the `text{C–N–H}` groups. 

Proton NMR:

  • The highlighted signal results from the middle three `text{CH}_2` groups. It is formed from similar chemical shifts of protons in two different environments, which creates an overlap of a `text{2H}` signal and a `text{4H}` signal, giving `text{6H}`.
  • The quintets arise because each `text{H}` atom has four `text{H}` atoms on neighbouring `text{C}` atoms.
Show Worked Solution

Infrared Spectrum:

  • IR spectrum allows us to find key functional groups in a molecule.
  • There is a peak signal at 3300 – 3400 cm¯1, indicating the presence of the amino group. 

Mass spectrum:

  • The mass spectrum represents the molecular mass and fragments of a molecule. The highlighted feature shows a 30 m/z signal, this is due to the fragmentation of the molecule.
  • The highlighted signal depicts the fragment `text{CH}_2 text{NH}_2 ^(\ +)` , which has a molecular mass of 30 (12 × 1 + 1.0 × 4 + 14 × 1 = 30). 

Carbon 13 NMR Spectrum:

  • The molecule has 5 carbon atoms, however, the spectrum only has 3 signals. This indicates symmetry, and that there are 3 different carbon environments in the molecule. 
  • Carbon 1 and 5 are in the same carbon environment, carbon 2 and 4 are in the same carbon environment, and carbon 3 is in a unique carbon environment.
  • The signals at 24 and 33 ppm are due to the `text{–CH}_2text{–CH}_2–` carbon atoms.
  • The signal at 42 ppm is due to the `text{C–N–H}` groups. 

Proton NMR:

  • The highlighted signal results from the middle three `text{CH}_2` groups. It is formed from similar chemical shifts of protons in two different environments, which creates an overlap of a `text{2H}` signal and a `text{4H}` signal, giving `text{6H}`.
  • The quintets arise because each `text{H}` atom has four `text{H}` atoms on neighbouring `text{C}` atoms.

♦ Mean mark 55%.

CHEMISTRY, M8 2021 HSC 12 MC

The mass spectrum and carbon-13 NMR for an organic compound are shown.
 

 

Which compound could produce the two spectra?
 

 

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`A`

Show Worked Solution
  • The mass spectrum indicates that the m/z peak is at 98, indicating that the molecular mass of the substance is approximately 98.
  • Additionally, the C-13 NMR spectrum shows 4 signals indicating that there are 4 unique carbon environments.
  • Thus, only A has a molecular mass of 98 and fits the 4 unique carbon environments.

`=> A`


♦♦ Mean mark 35%.