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Mock Exam IB DP Chemistry HL Paper 2 Set 3

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 (\(\text{Cl}\)).

Question

(iv) But-2-ene can be polymerized. Draw a section of the resulting polymer showing two repeating units.

▶️Answer/Explanation

Ans:

x

Question

(b) Chloroethane can be converted into ethanoic acid in a two-step process.

Identify reagents for each step.

Step 1:
Step 2:

▶️Answer/Explanation

Ans:

Step 1:
In the first step, chloroethane (\(\text{C}_2\text{H}_5\text{Cl}\)) can be converted into ethanol (\(\text{C}_2\text{H}_5\text{OH}\)) by a nucleophilic substitution reaction using a strong nucleophile such as hydroxide ion (\(\text{OH}^-\)) or water (\(\text{H}_2\text{O}\)) in the presence of a suitable solvent and possibly heat. The reaction proceeds via an SN2 mechanism, where the hydroxide ion attacks the carbon atom bonded to chlorine, leading to the displacement of the chloride ion and the formation of ethanol.

Step 2:
In the second step, ethanol (\(\text{C}_2\text{H}_5\text{OH}\)) can be oxidized to ethanoic acid (\(\text{CH}_3\text{COOH}\)) by an oxidizing agent such as potassium dichromate (\(\text{K}_2\text{Cr}_2\text{O}_7\)) in acidic medium (\(\text{H}^+\)). The reaction involves the conversion of the primary alcohol group (\(\text{-OH}\)) in ethanol to a carboxylic acid group (\(\text{-COOH}\)).

Therefore, the reagents for each step are as follows:
Step 1: Hydroxide ion (\(\text{OH}^-\)) or water (\(\text{H}_2\text{O}\)) in the presence of a suitable solvent and possibly heat.
Step 2: Potassium dichromate (\(\text{K}_2\text{Cr}_2\text{O}_7\)) in acidic medium (\(\text{H}^+\)).

(c) (i) Identify the type of reaction that takes place in step 1 of part (b).

Ans:

The type of reaction that takes place in step 1 of part (b) is a nucleophilic substitution reaction, specifically an SN2 (substitution nucleophilic bimolecular) reaction.

Question

(ii) Sketch the mechanism of the reaction for step 1 in part (b), using curly arrows to show the movement of electron pairs.

▶️Answer/Explanation

Ans:

curly arrow going from lone pair/negative charge on O in –OH to C  curly arrow showing Cl leaving
representation of transition state showing negative charge, square brackets and partial bonds

Question

(iii) Identify the products formed from the reaction of ethanol and ethanoic acid in the presence of an acid catalyst.

▶️Answer/Explanation

Ans:

In the presence of an acid catalyst, ethanol (\(\text{C}_2\text{H}_5\text{OH}\)) and ethanoic acid (\(\text{CH}_3\text{COOH}\)) can undergo esterification to form an ester and water.

The general reaction is:

\[ \text{Ethanol} + \text{Ethanoic acid} \xrightarrow{\text{Acid catalyst}} \text{Ester} + \text{Water} \]

The specific ester formed depends on the reaction conditions and the specific acid catalyst used. However, a common ester formed from this reaction is ethyl ethanoate (ethyl acetate), which has the structural formula \(\text{CH}_3\text{COOCH}_2\text{CH}_3\).

Therefore, the products formed from the reaction of ethanol and ethanoic acid in the presence of an acid catalyst are an ester (such as ethyl ethanoate) and water.

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