Mock Exam IB DP Chemistry SL Paper 2 Set 4

Question

Alkanes form a homologous series.
(a) (i) Outline the meaning of homologous series.

▶️Answer/Explanation

Ans:

A homologous series is a group of organic compounds with similar chemical properties, exhibiting a gradual change in physical properties and molecular structures as the number of carbon atoms increases. In a homologous series:

  1. Similar Chemical Properties: All members of the series share the same functional group, which determines their chemical behavior. They undergo similar types of chemical reactions and exhibit analogous chemical properties.

  2. Gradual Change in Physical Properties: As the number of carbon atoms increases, there is a gradual change in physical properties such as boiling point, melting point, density, and solubility. This change occurs due to increasing molecular size and surface area, leading to differences in intermolecular forces.

  3. General Formula: Members of a homologous series can be represented by a general formula, which shows the relationship between the number of carbon atoms and other elements or functional groups present in the molecule.

  4. Common Structural Feature: Despite variations in molecular structure, all members of the series share a common structural feature or backbone due to the presence of the same functional group.

  5. Regular Increase in Molecular Mass: Each member of the series differs from the next by a constant molecular mass increment, usually the molecular mass of a CH2 unit.

  6. Isomerism: Members of a homologous series may exhibit structural isomerism, where different compounds have the same molecular formula but different structural arrangements.

Examples of homologous series include alkanes, alkenes, alkynes, alcohols, carboxylic acids, and ethers. In the case of alkanes, the general formula is \(\mathrm{C}_n \mathrm{H}_{2 n+2}\), where \(n\) represents the number of carbon atoms in the molecule.

Question

(ii) State the preferred IUPAC name for the following compounds.

▶️Answer/Explanation

Ans:

Question

(iii) Identify one chiral carbon atom present in one of the following structures with an asterisk (*).

▶️Answer/Explanation

Ans:

In order to identify a chiral carbon atom in the structures provided, we need to understand that a chiral carbon is a carbon atom that is bonded to four different atoms or groups. Let’s examine the structures:

1. 2-Chlorobutane:

\[ \text{CH}_3 – \text{CH}_2 – \text{CH}(\text{Cl}) – \text{CH}_3 \]

In this structure, the carbon labeled as (\(\text{Cl}\)) is bonded to four different groups: three hydrogen atoms and a chlorine atom. Therefore, the carbon labeled (\(\text{Cl}\)) is a chiral carbon.

2. 1-Chloro-2-methylpropane:

\[ \text{CH}_3 – \text{C}(\text{H})(\text{Cl}) – \text{CH}_3 \]

In this structure, the carbon labeled as (\(\text{Cl}\)) is bonded to three hydrogen atoms and a methyl group. Since it’s not bonded to four different groups, it is not a chiral carbon.

So, the chiral carbon atom is present in 2-chlorobutane, and it is the carbon labeled as (\(\text{Cl}\)).

Scroll to Top