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CHEMISTRY, M7 2024 HSC 24

The boiling points for two series of compounds are listed.

 

  1. Plot the boiling points for each series of compounds against the number of carbon atoms per molecule.   (3 marks)

  1. With reference to hydrogen bonding and dispersion forces, explain the trends in the boiling point data of these compounds, within each series and between the series.  (4 marks)

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a.     

b.     The alcohols have higher boiling points than amines of the same chain length.

→ Both alcohols and amines have polar hydrogen bonding as a result of the \(\ce{OH}\) and \(\ce{NH2}\) functional groups respectively.

→ Oxygen molecules have higher electronegativity than nitrogen molecules making the hydrogen bonding in the alcohols significantly stronger than the hydrogen bonding in amines.

→ Since the dispersion forces in amines and alcohols of the same chain length are very similar, the difference in the strength of the intermolecular bonding is dependent on the strength of the hydrogen bonding.

→ Therefore, a larger amount of thermal energy is required to separate the alcohol molecules resulting in higher boiling points than amines.
 

As the chain length of the alcohols and amines increase, the boiling points also increase.

→ When the chain length increases, the number of electrons in the molecules also increase which corresponds to a larger number of dispersion forces between neighbouring molecules.

→ The stronger dispersion forces between the molecules increase the overall strength of the intermolecular forces in both the alcohols and amines therefore leading to higher boiling points as chain length increases.

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a.   

       
 

b.     The alcohols have higher boiling points than amines of the same chain length.

→ Both alcohols and amines have polar hydrogen bonding as a result of the \(\ce{OH}\) and \(\ce{NH2}\) functional groups respectively.

→ Oxygen molecules have higher electronegativity than nitrogen molecules making the hydrogen bonding in the alcohols significantly stronger than the hydrogen bonding in amines.

→ Since the dispersion forces in amines and alcohols of the same chain length are very similar, the difference in the strength of the intermolecular bonding is dependent on the strength of the hydrogen bonding.

→ Therefore, a larger amount of thermal energy is required to separate the alcohol molecules resulting in higher boiling points than amines.
 

As the chain length of the alcohols and amines increase, the boiling points also increase.

→ When the chain length increases, the number of electrons in the molecules also increase which corresponds to a larger number of dispersion forces between neighbouring molecules.

→ The stronger dispersion forces between the molecules increase the overall strength of the intermolecular forces in both the alcohols and amines therefore leading to higher boiling points as chain length increases.

CHEMISTRY, M7 2020 VCE 16 MC

The following table provides information about three organic compounds, \(\text{X, Y}\) and \(\text{Z}\).

Which one of the following is the best estimate for the boiling point of Compound \(\text{Z}\)?

  1. 31 °C
  2. 101 °C
  3. 114 °C
  4. 156 °C
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\(A\)

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→ Boiling points of molecular compounds are related to the strength of their intermolecular bonding.

→ All given molecules are polar molecules. Each intermolecular bonding will have contributions from both dispersion forces and dipole-dipole bonding.

→ Since the molar mass of each compound is the same, the contribution from dispersion forces will be similar for all.

→ Compounds \(\text{X}\) and \(\text{Y}\) will both have hydrogen bonding, due to the presence of the \(\ce{O-H}\) functional group.

→ Compound \(\text{Z}\) does not have intermolecular hydrogen bonding and therefore would have the lowest boiling point.

\(\Rightarrow A\)

♦♦♦ Mean mark 20%.

CHEMISTRY, M7 2023 HSC 10 MC

Which of the following correctly lists the compounds in order of increasing boiling point?

  1. Heptane < heptan-2-one < heptan-1-o1 < heptanoic acid
  2. Heptane < heptan-1-o1 < heptan-2-one < heptanoic acid
  3. Heptanoic acid < heptan-2-one < heptan-1-o1 < heptane
  4. Heptanoic acid < heptan-1-o1 < heptan-2-one < heptane
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\(A\)

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→ Compounds with functional groups capable of hydrogen bonding have higher boiling points (due to stronger bonds, more energy is required to break)

\(\Rightarrow A\)

CHEMISTRY, M7 2018 HSC 13 MC

Pentanol, propyl acetate, pentanoic acid and ethyl propanoate all contain five carbon atoms. These four compounds are mixed in a flask and then separated by fractional distillation.

Which compound would be most likely to remain in the flask?

  1. Pentanol
  2. Propyl acetate
  3. Pentanoic acid
  4. Ethyl propanoate
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`C`

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→ Fractional distillation is the separating of a mixture into its component parts by heating.

→ The compound with the highest boiling point is pentanoic acid which will remain after all other compounds have vaporised. 

→ Pentanoic acid has stronger hydrogen bonding than pentanol and hence stronger intermolecular forces.

`=>C`


♦ Mean mark 46%.

CHEMISTRY, M7 2017 HSC 27

The boiling points and molar masses of three compounds are shown in the table.
 

Acetic acid, butan-1-ol and butyl acetate have very different molar masses but similar boiling points. Explain why in terms of the structure and bonding of the three compounds.   (5 marks)

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→ Although the three listed compounds different molar masses, they have similar boiling points due to their different structures and resulting intermolecular forces.

→ Butyl acetate has the largest molar mass and therefore greatest dispersion forces but it is only slightly polar and has no hydrogen bonding.

→ Butan-1-ol has lower molar mass than butyl acetate and therefore smaller dispersion forces but it is polar and contains a hydrogen bound to an oxygen. Therefore, it exhibits hydrogen bonding resulting in strong intermolecular forces and a boiling point in the middle of the three compounds.

→ Acetic acid has the lowest molar mass and hence the weakest dispersion forces. It is however highly polar due to the presence of the carboxyl group \(\ce{(COOH)}\) and contains a hydrogen bound to an oxygen allowing the formation of hydrogen bonds between molecules.

→ The presence of a second oxygen in acetic acid increases the hydrogen bonding compared with butan-1-ol. 

→ These factors lead to acetic acid possessing the highest boiling point despite its molar mass being the lowest.

→ In summary, the totality of the intermolecular forces of all three molecules is similar and therefore similar boiling points.

Show Worked Solution

→ Although the three listed compounds different molar masses, they have similar boiling points due to their different structures and resulting intermolecular forces.

→ Butyl acetate has the largest molar mass and therefore greatest dispersion forces but it is only slightly polar and has no hydrogen bonding.

→ Butan-1-ol has lower molar mass than butyl acetate and therefore smaller dispersion forces but it is polar and contains a hydrogen bound to an oxygen. Therefore, it exhibits hydrogen bonding resulting in strong intermolecular forces and a boiling point in the middle of the three compounds.

→ Acetic acid has the lowest molar mass and hence the weakest dispersion forces. It is however highly polar due to the presence of the carboxyl group \(\ce{(COOH)}\) and contains a hydrogen bound to an oxygen allowing the formation of hydrogen bonds between molecules.

→ The presence of a second oxygen in acetic acid increases the hydrogen bonding compared with butan-1-ol. 

→ These factors lead to acetic acid possessing the highest boiling point despite its molar mass being the lowest.

→ In summary, the totality of the intermolecular forces of all three molecules is similar and therefore similar boiling points.


♦♦ Mean mark 41%.

CHEMISTRY, M7 2019 HSC 9 MC

All of the following compounds have similar molar masses.

Which has the highest boiling point?

  1. Butane
  2. Ethanoic acid
  3. Propan-1-ol
  4. Propane
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`B`

Show Worked Solution

→ Carboxyllic acids have a high affinity for hydrogen bonding, the strongest molecular force.

→ They therefore require more heat to break the intermolecular forces to convert liquid to gas versus other substances.

`=>B`

CHEMISTRY, M7 2020 HSC 32

The table shows three compounds and their boiling points.
 

An ester does not always have a lower boiling point than both the alcohol and the alkanoic acid from which it is produced.

Using the information in the table, account for this observation.   (4 marks)

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→ The boiling points of all compounds are a function of the strength of their intermolecular forces.

→ All three compounds are polar and have dispersion forces between molecules and dipole-dipole interactions.

→ Methanol and propanoic acid can also form hydrogen bonds (strongest type of intermolecular force).

→ Methyl propanoate’s larger size gives it stronger dispersion forces than methanol and propanoic acid, but it cannot form hydrogen bonds.

→ Despite having weaker dispersion forces, propanoic acid can form two hydrogen bonds per molecule, which makes up for its weaker dispersion forces and results in stronger overall intermolecular forces than methyl propanoate.

→ Methanol is a polar molecule that can form strong hydrogen bonds due to its hydroxyl group. However, it has the lowest boiling point due to its small molar mass, resulting in weaker dispersion forces.

Show Worked Solution

→ The boiling points of all compounds are a function of the strength of their intermolecular forces.

→ All three compounds are polar and have dispersion forces between molecules and dipole-dipole interactions.

→ Methanol and propanoic acid can also form hydrogen bonds (strongest type of intermolecular force).

→ Methyl propanoate’s larger size gives it stronger dispersion forces than methanol and propanoic acid, but it cannot form hydrogen bonds.

→ Despite having weaker dispersion forces, propanoic acid can form two hydrogen bonds per molecule, which makes up for its weaker dispersion forces and results in stronger overall intermolecular forces than methyl propanoate.

→ Methanol is a polar molecule that can form strong hydrogen bonds due to its hydroxyl group. However, it has the lowest boiling point due to its small molar mass, resulting in weaker dispersion forces.