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CHEMISTRY, M8 2025 HSC 36

Use the data sheet provided and the information in the table to answer this question.
 

 

Consider the molecule shown.

For each of the following instrumental techniques, predict the expected features of the spectra produced.

Refer to the structural features of the molecule in your answer.

  • Infrared (IR)    (Ignore any absorptions due to \(\ce{C - C}\) or \(\ce{C - H}\) )
  • Carbon-13 NMR
  • Proton NMR
  • Mass spectrometry   (7 marks)

Show Answers Only

Infrared (IR)

  • Strong broad \(\ce{OH}\) signal at approximately 2750 cm\(^{-1}\)
  • Strong sharp \(\ce{CO}\) signal at approximately 1700 cm\(^{-1}\)

\(^{13}\text{C NMR}\)

  • 3 distinct carbon environments 
  • 1 signal at 5−40 ppm \(\ce{(CH3)}\)
  • 1 signal at 20−50 ppm \(\ce{(CH2)}\)
  • 1 signal at 160−185 ppm \(\ce{(acid CO)}\)

\(^{1}\text{H NMR (Proton NMR)}\)

  • 3 distinct hydrogen environments
  • 1 signal at 0.7−2.1 ppm \(\ce{(CH3)}\), triplet splitting pattern, integration of 3
  • 1 signal at 2.1−4.5 ppm \(\ce{(CH2)}\), quartet splitting pattern, integration of 2
  • 1 signal at 9.0−13.0 ppm \(\ce{(COOH)}\), singlet, integration of 1

Mass Spectrometry

  • Molecular ion at 74 m/z (molar mass of propanoic acid ~74 g/mol)
Show Worked Solution

Infrared (IR)

  • Strong broad \(\ce{OH}\) signal at approximately 2750 cm\(^{-1}\)
  • Strong sharp \(\ce{CO}\) signal at approximately 1700 cm\(^{-1}\)
♦ Mean mark 62%.

\(^{13}\text{C NMR}\)

  • 3 distinct carbon environments 
  • 1 signal at 5−40 ppm \(\ce{(CH3)}\)
  • 1 signal at 20−50 ppm \(\ce{(CH2)}\)
  • 1 signal at 160−185 ppm \(\ce{(acid CO)}\)

\(^{1}\text{H NMR (Proton NMR)}\)

  • 3 distinct hydrogen environments
  • 1 signal at 0.7−2.1 ppm \(\ce{(CH3)}\), triplet splitting pattern, integration of 3
  • 1 signal at 2.1−4.5 ppm \(\ce{(CH2)}\), quartet splitting pattern, integration of 2
  • 1 signal at 9.0−13.0 ppm \(\ce{(COOH)}\), singlet, integration of 1

Mass Spectrometry

  • Molecular ion at 74 m/z (molar mass of propanoic acid ~74 g/mol)

CHEMISTRY, M8 2025 HSC 17 MC

The chemical environment of an atom depends on the species surrounding that atom within a molecule.

In which of the following compounds does the number of carbon chemical environments equal the number of proton chemical environments?
 

 

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\(A\)

Show Worked Solution
  • In the diagrams below, each colour represents a different carbon or proton environment (ignoring the oxygen atoms).

  • Only \(A\) has equal number of carbon and proton environments with two of each.

\(\Rightarrow A\)

CHEMISTRY, M8 2024 HSC 33

Acetone can be reduced, as shown.
 

  1. Identify the shape around the central carbon atom in each molecule.   (2 marks)

  2. Explain how \({ }^{13} \text{C NMR}\) spectroscopy could be used to monitor the progress of this reaction.   (3 marks)

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a.   Acetone:

  • Double bond and 2 single bonds coming off the central carbon atom  \(\Rightarrow\)  trigonal planar.

Product:

  • Contains single bonds coming off the central carbon atom (\Rightarrow\) tetrahedral. (Note: the hydrogen bonded to the central carbon atom in the product molecule is not shown due to the skeletal structure)

b.    \({ }^{13} \text{C NMR}\) Spectroscopy:

  • \({ }^{13} \text{C NMR}\) will differentiate between molecules with different carbon environments. This produces different signals on the \({ }^{13} \text{C NMR}\) spectrum.
  • The acetone would produce two signals on the \({ }^{13} \text{C NMR}\) spectrum. The first signal would be due to the \(\ce{CH3}\) groups either side of the central carbon between 20-50 ppm. The second signal would be from the carbonyl group between 190-220 ppm.
  • The product of the reduction would also produce two signals on the \({ }^{13} \text{C NMR}\) spectrum. The carbon with the hydroxyl group attached to it would produce a signal between 50-90 ppm and the \(\ce{CH3}\) groups either side would produce a signal between 5-40 ppm.
  • The reaction can be monitored by observing the disappearance of the carbonyl signal (190-220 ppm) and appearance of the hydroxyl signal (50-90 ppm) as acetone is reduced to the product.

Show Worked Solution

a.   Acetone:

  • Double bond and 2 single bonds coming off the central carbon atom  \(\Rightarrow\)  trigonal planar.

Product:

  • Contains single bonds coming off the central carbon atom (\Rightarrow\) tetrahedral. (Note: the hydrogen bonded to the central carbon atom in the product molecule is not shown due to the skeletal structure)
♦ Mean mark (a) 44%.

b.    \({ }^{13} \text{C NMR}\) Spectroscopy:

  • \({ }^{13} \text{C NMR}\) will differentiate between molecules with different carbon environments. This produces different signals on the \({ }^{13} \text{C NMR}\) spectrum.
  • The acetone would produce two signals on the \({ }^{13} \text{C NMR}\) spectrum. The first signal would be due to the \(\ce{CH3}\) groups either side of the central carbon between 20-50 ppm. The second signal would be from the carbonyl group between 190-220 ppm.
  • The product of the reduction would also produce two signals on the \({ }^{13} \text{C NMR}\) spectrum. The carbon with the hydroxyl group attached to it would produce a signal between 50-90 ppm and the \(\ce{CH3}\) groups either side would produce a signal between 5-40 ppm.
  • The reaction can be monitored by observing the disappearance of the carbonyl signal (190-220 ppm) and appearance of the hydroxyl signal (50-90 ppm) as acetone is reduced to the product.

CHEMISTRY, M8 2022 VCE 28 MC

The \({ }^{13}\text{C NMR}\) spectrum of an organic compound is shown below.

The organic compound could be

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\(D\)

Show Worked Solution
  • The \({ }^{13}\text{C NMR}\) has five peaks indicating 5 different carbon environments within the molecule.
  • The peak at 140 indicates the presence of the \(\ce{C=C}\).

\(\Rightarrow D\)

♦ Mean mark 42%.
COMMENT: Solving by elimination is an effective strategy here.

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

  1. 5
  2. 6
  3. 7
  4. 8
Show Answers Only

\(C\)

Show Worked Solution
  • There are 7 different carbon environments in the molecule:

\(\Rightarrow C\)

♦ Mean mark 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 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}`?
 

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`D`

Show Worked Solution
  • The given information suggests that compound `text{Y}` is a carboxylic acid because it is produced through the oxidation of compound `text{X}`, a primary alcohol, with an oxidising agent and turns blue litmus red.
  • The treatment of compound `text{X}` with hot concentrated sulfuric acid results in a dehydration reaction.
  • In summary, the information provided suggests that compound `text{Y}` is a carboxylic acid and compound `text{X}` is a primary alcohol, and that the treatment of compound `text{X}` with hot concentrated sulfuric acid results in a dehydration reaction.

`=>D`


♦♦♦ Mean mark 33%.

CHEMISTRY, M8 2020 HSC 5 MC

A \( \ce{^13C NMR} \) spectrum is shown.
 


 

Which compound gives rise to this spectrum?

  1. chloroethane
  2. 1-chloropropane
  3. 1, 2-dichloroethane
  4. 1, 2-dichloropropane
Show Answers Only

`A`

Show Worked Solution
  • The \( \ce{^13C NMR} \) spectrum shows two signals, indicating 2 unique carbon environments.
  • Chloroethane is the only compound with 2 unique carbon environments.

`=> A`

CHEMISTRY, M8 2022 HSC 27

A bottle labelled 'propanol' contains one of two isomers of propanol.

  1. Draw the TWO isomers of propanol.   (2 marks)
     

  1. Describe how \( \ce{^13C NMR}\) spectroscopy might be used to identify which isomer is in the bottle.   (2 marks)
  1. Each isomer produces a different product when oxidised.
  2. 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 \(\ce{^13C NMR}\)?
 

 

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`D`

Show Worked Solution
  • Structure D only contains two unique carbon environments, and thus would only contain two signals on a \(\ce{^13C NMR}\) spectrum.
  • Structure A contains 3 `text{C}` signals, structure B contains 5 `text{C}` signals, and structure C contains 4 `text{C}` signals.

`=> D`


♦ Mean mark 49%.