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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`