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CHEMISTRY, M7 2025 HSC 24

65.0 g of ethyne gas reacts with an excess of gaseous hydrogen chloride to produce chloroethene.

  1. Draw the full structural formula of ethyne and identify the shape of the molecule.   (2 marks)

\begin{array}{|c|c|}
\hline
\rule{0pt}{2.5ex}\quad \quad \text{Structural formula } \quad \quad \rule[-1ex]{0pt}{0pt}& \quad \quad\text{ Shape of molecule } \quad \quad\\
\hline
\rule{0pt}{2.5ex}\rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\rule[-1ex]{0pt}{0pt}& \\
\hline
\end{array}

  1. The molar masses of the compounds in the reaction are provided.

\begin{array}{|l|c|}
\hline \rule{0pt}{2.5ex}\text{Compound} \rule[-1ex]{0pt}{0pt}& \text{ Molar mass } \\
\hline \rule{0pt}{2.5ex}\text{Ethyne} \rule[-1ex]{0pt}{0pt}& 26.04 \\
\hline \rule{0pt}{2.5ex}\text{Hydrogen chloride} \rule[-1ex]{0pt}{0pt}& 36.46 \\
\hline \rule{0pt}{2.5ex}\text{Chloroethene} \rule[-1ex]{0pt}{0pt}& 62.50 \\
\hline
\end{array}

  1. Calculate the mass of chloroethene produced, using the molar masses provided. Include a relevant chemical equation in your answer.   (3 marks)

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a.    Structural formula:

   

The shape of ethyne is linear.

b.    156 grams

Show Worked Solution

a.    Structural formula:

   

The shape of ethyne is linear.
 

b.    The chemical equation for the reacton occuring is shown below:

\(\ce{C2H2(g) + HCl(g) -> C2H3Cl(g)}\)

  • \(n\ce{(C2H2)} = \dfrac{m}{MM} = \dfrac{65.0}{26.04} = 2.496\ \text{mol}\)
  • As the reactants and products are in a \(1:1:1\) ratio, \(n\ce{(C2H2)}_{\text{reacted}} = n\ce{(C2H3Cl)}_{\text{produced}}\)
  • \(m\ce{(C2H3Cl)} = n \times MM = 2.496 \times 62.50 = 156\ \text{grams (3 sig.fig.)}\)

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, M7 EQ-Bank 26

This flow chart shows reactions involving six different organic compounds (A to F).
 

 

Draw the structures of compounds A to F, justifying your diagrams with reference to the information provided.   (7 marks)

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A:  
       

B:  
       

C:  
           

D:  
       

E:  
       

F:  
       

Show Worked Solution
  • A combines with dilute  \(\ce{H2SO4}\)  to produce two alcohols.
  • C doesn’t react with strong oxidant (i.e. it is a tertiary alcohol).

C:  
           

  • B must therefore be 2-methylbutan-1-ol or 3-methylbutan-2-ol.
  • Since B gives two products when heated with concentrated \(\ce{H2SO4}\), it must be 3-methylbutan-2-ol, as 2-methylbutan-1-ol will only produce one product.

B:  
       

  • D is a ketone, produced by B reacting with a strong oxidant.

D:  
       

  • A can be dehydrated using concentrated  \(\ce{H2SO4}\)  on either B or C.

A:  
       

  • B dehydrates to A or E. Given A‘s structure above,

E:  
       

  • C dehydrates to A or F. Again, given A’s structure above,

F:  
       

CHEMISTRY, M7 2020 HSC 13 MC

Which of the following conversions results in the formation of a different shape around the carbon atom?

  1. Methanoic acid to methanal
  2. Methanoic acid to methanol
  3. Methanoic acid to methanamide
  4. Methanoic acid to sodium methanoate
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`B`

Show Worked Solution
  • Methanoic acid is a trigonal planar shape around the carbon atom, whereas methanol is a tetrahedral shape.

`=> B`


♦ Mean mark 45%.

CHEMISTRY, M7 2021 HSC 7 MC

Methanol undergoes a substitution reaction using hydrogen bromide.

Compared to methanol, the product of this reaction has a

  1. lower boiling point.
  2. lower molecular mass.
  3. greater solubility in water.
  4. different molecular geometry at the carbon atom.
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`A`

Show Worked Solution
  • The product of the substitution reaction between methanol and hydrogen bromide is bromomethane.
  • Methanol contains an OH functional group and thus can form strong hydrogen bonds.
  • Bromomethane can only form dipole-dipole forces which are weaker than hydrogen bonds. As a result, bromomethane requires less energy to break these intermolecular forces, resulting in a lower boiling point than methanol.

`=> A`