May/June 2026 Paper 42

2026 · 9 questions · 56 parts · 100 marks

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9(a)Amines Nitriles Azo DyesHard4 marks
Compound and compound are isomers.
Figure 9.1, compounds $P$ and $Q$: $P$ is a benzene ring with a –CH$_{3}$ group at one position and an –NH$_{2}$ group directly attached to the ring at the para position (4-methylaniline); $Q$ is a benzene ring with a –CH$_{2}$NH$_{2}$ group attached (benzylamine, i.e. the –NH$_{2}$ is on a side-chain carbon, not directly on the ring).
State the relative basicities of , and ammonia. Explain your answer.
Order (most basic to least basic)
Explanation
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9(b)(i)Organic SynthesisMedium2 marks
can be prepared from methylbenzene in two steps, as shown in Figure 9.2.
Figure 9.2, a two-step synthesis: methylbenzene (a benzene ring with –CH$_{3}$) reacts in step 1 to give a benzene ring with –CH$_{3}$ and, para to it, –NO$_{2}$; this reacts in step 2 to give $P$, a benzene ring with –CH$_{3}$ and, para to it, –NH$_{2}$.
Suggest reagents and conditions for steps 1 and 2 in Figure 9.2.
Step 1
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Step 2
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9(b)(ii)Amines Nitriles Azo DyesMedium1 mark
Complete the equation for step 2. Use [H] to represent an atom of hydrogen from the reducing agent.
Missing reagent and product
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9(c)(i)Amines Nitriles Azo DyesMedium1 mark
Figure 9.4 shows a two-step synthesis of azo dye .
Draw the structure of compound in the box in Figure 9.4.
Diagram to annotate
PenEraserUndoRedoClear
Draw with the mouse or a finger.
9(c)(ii)Amines Nitriles Azo DyesMedium2 marks
Figure 9.4, a two-step synthesis of azo dye $N$: $P$ reacts in step 1 to give diazonium salt $M$; $M$ reacts in step 2 (a coupling reaction) to give azo dye $N$ (shown: a benzene ring bearing –CH$_{3}$, joined through an –N=N– azo linkage to a second benzene ring that carries –SO$_{3}^{-}$Na$^{+}$ and –OH).
Give the reagents and conditions for steps 1 and 2 in Figure 9.4.
Step 1
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Step 2
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