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CHEMISTRY, M7 2024 HSC 1 MC

Which two substances are members of the same homologous series?
 

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\(A\)

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→ A homologous series refers to compounds with the same or similar functional group and same general formula.

→ Both substances in \(A\) contain a hydroxyl group.

\(\Rightarrow A\)

CHEMISTRY, M7 2020 VCE 4 MC

What is the IUPAC name of the molecule shown above?

  1. 3-hydroxy-3-ethyl-propan-1-amine
  2. 3-amino-1-methylpropan-1-ol
  3. 3-hydroxypentan-1-amine
  4. 1-aminopentan-3-ol
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\(D\)

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→ Molecule has a 5-carbon chain with single bonds (“pentan”)

→ Alcohol functional group attached to 3rd carbon (“-3-ol”)

\(\Rightarrow D\)

CHEMISTRY, M7 EQ-Bank 24

Primary, unbranched alcohols and alkanes of the same carbon length have quite different boiling points.

Explain the difference in boiling point of these organic compounds, showing all intermolecular forces. Support your answer with diagrams.   (4 marks)

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→ Alkanes are saturated hydrocarbons – i.e. they are made up of carbon and hydrogen atoms only and all atoms are joined together by single covalent bonds which are non-polar.

→ Weak intermolecular forces (Van der Waals) are therefore the only forces holding alkane molecules together and as a result, alkanes have low boiling points.

→ In contrast, alcohols have an \(\ce{OH}\) functional group. The \(\ce{OH}\) bond is polar with an oxygen “pole” that is slightly negatively charged and hydrogen “pole” that is slightly positively charged.

→ The hydrogen atom on one molecule will form an electrostatic bond with the oxygen atom on another atom creating a hydrogen bond.

→ Since hydrogen bonds are much stronger intermolecular forces than dispersion forces, the boiling points of alcohols are significantly higher than those of alkanes with the same carbon lengths.

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→ Alkanes are saturated hydrocarbons – i.e. they are made up of carbon and hydrogen atoms only and all atoms are joined together by single covalent bonds which are non-polar.

→ Weak intermolecular forces (Van der Waals) are therefore the only forces holding alkane molecules together and as a result, alkanes have low boiling points.

→ In contrast, alcohols have an \(\ce{OH}\) functional group. The \(\ce{OH}\) bond is polar with an oxygen “pole” that is slightly negatively charged and hydrogen “pole” that is slightly positively charged.

→ The hydrogen atom on one molecule will form an electrostatic bond with the oxygen atom on another atom creating a hydrogen bond.

→ Since hydrogen bonds are much stronger intermolecular forces than dispersion forces, the boiling points of alcohols are significantly higher than those of alkanes with the same carbon lengths.

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.