May/June 2026 Paper 22

2026 · 6 questions · 40 parts · 60 marks

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6(a)(i)Structural Geometric IsomerismEasy1 mark
Figure 6.1 shows the structure of compound P.
Figure 6.1, the skeletal structure of compound P: a four-carbon chain with a C=C double bond drawn trans (zig-zag) across the middle, a −COOH group (C=O above, OH below) at the left end, and a −COOH group (C=O below, OH above) at the right end — but-2-enedioic acid drawn in its trans (E) form.
Draw the structure of the stereoisomer of P.
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6(a)(ii)Structural Geometric IsomerismMedium2 marks
Explain why P shows stereoisomerism.

Answer

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6(b)(i)AlkenesEasy2 marks
P reacts to produce compound Q.
Figure 6.2, P reacting to compound Q: on the left, the same trans-butenedioic-acid structure of P (a C=C double bond flanked by two −COOH groups); a plain right-pointing arrow leads to Q on the right, a four-carbon chain with a −COOH group at each end and an −OH group on the second carbon (the C=C of P has become a CH(OH)–CH₂ chain in Q), the structure of malic acid, HOOC–CH₂–CH(OH)–COOH.
Identify the reagent and conditions required to convert P to Q.
Reagent
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Conditions
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6(b)(ii)Optical IsomerismMedium2 marks
Complete Figure 6.3 to show the pair of stereoisomers of Q.
Diagram to annotate
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6(c)(i)Mass Spectrometry Infrared SpectroscopyMedium2 marks
An unlabelled sample contains one of compounds P, R or S.
Figure 6.4, the skeletal structures of three compounds, labelled P, R and S. P (left) is the same trans-butenedioic-acid structure as Figure 6.1/6.2 (C=C flanked by two −COOH groups). R (centre) is a four-carbon diacid with no double bond, HOOC–CH₂–CH₂–COOH (butanedioic acid). S (right) is a four-carbon diol with a central C=C double bond, HO–CH₂–CH=CH–CH₂–OH (but-2-ene-1,4-diol).
The mass spectrum of the sample is shown in Figure 6.5.
Figure 6.5, a mass spectrum with m/e on the x-axis (10 to 120) and relative abundance (0 to 100) on the y-axis. Small peaks appear near m/e = 13, 18 and 32; a cluster around m/e = 26–29 (peaking near 27); a cluster around m/e = 41–46 with a tall peak at 45 (height about 76); peaks around m/e = 53–55 (peaking near 53, height about 48); small peaks at 60 and 63; a cluster around m/e = 69–71; peaks around m/e = 81–82; a peak at m/e = 88; the tallest peak (the base peak) at m/e = 98 (height 100) with a smaller peak at 99 and a trace at 100; and a peak at m/e = 116, the highest-mass peak shown.
Identify the one peak on Figure 6.5 which confirms that the sample is P. Explain your reasoning.
Explain your reasoning
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6(c)(ii)Mass Spectrometry Infrared SpectroscopyMedium2 marks
Suggest the structural formulae of the fragments responsible for the peaks at m/ and m/.
Structural formula of the fragment at m/
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Structural formula of the fragment at m/
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6(c)(iii)Organic Functional Group TestsMedium2 marks
Suggest one reagent that is used to distinguish between samples of R and S. Include appropriate observations.

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