IB DP Chemistry Mock Exam HL Paper 1A Set 5 - 2025 Syllabus
IB DP Chemistry Mock Exam HL Paper 1A Set 5
Prepare for the IB DP Chemistry Exam with our comprehensive IB DP Chemistry Exam Mock Exam HL Paper 1A Set 5. 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
Question
Which pairs of reactants could produce the following intermediate?
A. I and II only
B. I and III only
C. II and III only
D. I, II and III
Answer/Explanation
Markscheme: A
Question
Which terms describe the nitronium ion in the nitration of benzene?
▶️Answer/Explanation
Markscheme: D
In the nitration of benzene, the nitronium ion (\(\rm NO_2^+\)) acts as an electrophile. Electrophiles are species that are electron-deficient and seek electrons from other molecules. The nitronium ion is attracted to the electron-rich benzene ring and acts as an electrophile in the reaction.
So, the correct terms describing the nitronium ion in the nitration of benzene are:
Type of reactant: Electrophile
Acid-base nature: Lewis acid
Question
What is the percentage error if the enthalpy of combustion of a substance is determined experimentally to be -2100kJ \(mol^{-1}\) , but the literature value is -3500kJ \(mol^{-1}\)?
A. 80%
B. 60%
C. 40%
D. 20%
▶️Answer/Explanation
Markscheme: C
The percentage error is calculated using the formula:
\[ \text{Percentage Error} = \left| \frac{\text{Experimental Value} – \text{Literature Value}}{\text{Literature Value}} \right| \times 100\]
Given:
– Experimental Value = -2100 kJ/mol
– Literature Value = -3500 kJ/mol
\[ \text{Percentage Error} = \left| \frac{-2100 – (-3500)}{-3500} \right| \times 100\]
\[ \text{Percentage Error} = \left| \frac{1400}{3500} \right| \times 100\]
\[ \text{Percentage Error} = \frac{2}{5} \times 100\]
\[ \text{Percentage Error} = 40\%\]
Question
Which molecule produces this \(^1H-NMR\) spectrum?
A. \(CH_3COOCH_3\)
B. \(CH_3COCH_3\)
C. \(CH_3CHO\)
D. \(CH_3CH_2CH_3\)
Answer/Explanation
Markscheme: B
Question
What is the index of hydrogen deficiency of adenine?
A. 3
B. 4
C. 5
D. 6
▶️Answer/Explanation
Markscheme: D
The index of hydrogen deficiency (IHD), also known as the degree of unsaturation or double bond equivalent, can be calculated using the formula:
\[IHD = \frac{{2C + 2 + N – H – X}}{2}\]
where:
\(C\) is the number of carbon atoms,
\(N\) is the number of nitrogen atoms,
\(H\) is the number of hydrogen atoms, and
\(X\) is the number of halogen atoms (Cl, Br, I).
For adenine, the molecular formula is \(C_5H_5N_5\). Applying these values to the formula:
\[IHD = \frac{{2(5) + 2 + 5 – 5 – 0}}{2} = \frac{{12}}{2} = 6\]
So, the index of hydrogen deficiency (IHD) for adenine is 6.