IB DP Chemistry Mock Exam HL Paper 1A Set 3 - 2025 Syllabus

IB DP Chemistry Mock Exam HL Paper 1A Set 3

Prepare for the IB DP Chemistry Exam with our comprehensive IB DP Chemistry Exam Mock Exam HL Paper 1A Set 3. Test your knowledge and understanding of key concepts with challenging questions covering all essential topics. Identify areas for improvement and boost your confidence for the real exam

IB DP Chemistry Mock Tests -All Sets

Question

Ammonia (NH₃) and nitrous acid (HNO₂) are nitrogen compounds. What are the oxidation states of nitrogen in these compounds?

	NH₃	HNO₂
(A)	+3	+3
(B)	–3	–3
(C)	+3	–3
(D)	–3	+3

▶️ Answer/Explanation

Detailed solution

NH₃: H is +1 each, so \(x + 3(+1) = 0\) ⇒ \(x = -3\).
HNO₂: H is +1, O is –2 each, so \(1 + x + 2(-2) = 0\) ⇒ \(x – 3 = 0\) ⇒ \(x = +3\).
✅ Answer: (D)

Question

Which choice of alcohol and reaction conditions would give the greatest yield of methylpropanoic acid?

Option	Alcohol	Conditions
A	\(2\)-methylpropan-\(2\)-ol	Reflux
B	\(2\)-methylpropan-\(1\)-ol	Reflux
C	\(2\)-methylpropan-\(2\)-ol	Distillation
D	\(2\)-methylpropan-\(1\)-ol	Distillation

▶️ Answer/Explanation

Detailed solution

Methylpropanoic acid is a carboxylic acid. To obtain a high yield of a carboxylic acid from an alcohol, a primary alcohol must be fully oxidized under reflux with an oxidizing agent.

• \(2\)-methylpropan-\(1\)-ol is a primary alcohol, which can be oxidized first to an aldehyde and then to the carboxylic acid under reflux.
• \(2\)-methylpropan-\(2\)-ol is a tertiary alcohol, which is not readily oxidized under normal conditions, so it will not give a good yield of the acid.
• Distillation is used to stop oxidation early (at the aldehyde stage), so it does not give the highest yield of the acid.

Therefore, the correct combination is the primary alcohol \(2\)-methylpropan-\(1\)-ol under reflux.

✅ Answer: (B)

Question

What is the role of the CN^– ion in the reaction of 1-chloropropane with excess KCN in ethanol?

\[ \text{CH}_3\text{CH}_2\text{CH}_2\text{Cl} + \text{KCN} \rightarrow \text{CH}_3\text{CH}_2\text{CH}_2\text{CN} + \text{KCl} \]

(A) Electrophile and Lewis base
(B) Nucleophile and Lewis acid
(C) Electrophile and Lewis acid
(D) Nucleophile and Lewis base

▶️ Answer/Explanation

Detailed solution

CN^– attacks the electrophilic carbon in 1-chloropropane, so it acts as a nucleophile.
A nucleophile is a species that donates an electron pair — this is also the definition of a Lewis base.
✅ Answer: (D)

IB DP Chemistry Mock Exam HL Paper 1A Set 3 - 2025 Syllabus