May/June 2026 Paper 43

2026 · 9 questions · 63 parts · 100 marks

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8(a)(i)Acyl Chlorides AmidesMedium1 mark
Secondary amine M can be made by the three-step synthesis shown in Figure 8.1.
Figure 8.1, a three-step synthesis: CH3COOH reacts in step 1 to give CH3COCl, which reacts in step 2 to give amide L (shown as a blank box), which reacts in step 3 to give secondary amine M, drawn as CH3CH2-NH-CH(CH3)2
Write an equation for the reaction of CHCOOH with SOCl in step 1.

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8(a)(ii)Acyl Chlorides AmidesEasy1 mark
State an alternative reagent that reacts with CHCOOH to produce CHCOCl in step 1.

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8(a)(iii)Acyl Chlorides AmidesMedium1 mark
Compound L is an amide. Draw the structure of L in Figure 8.1.
Diagram to annotate
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8(a)(iv)Acyl Chlorides AmidesMedium1 mark
Draw the structure of the reagent added to CHCOCl in step 2.
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8(a)(v)Acyl Chlorides AmidesMedium1 mark
Name the mechanism of the reaction forming L in step 2.
8(a)(vi)Acyl Chlorides AmidesMedium1 mark
Suggest a reagent for step 3.
8(b)(i)Nmr Spectroscopy Structure ElucidationMedium2 marks
The proton (H) NMR spectrum of M dissolved in DO is obtained. Four of the proton environments, r, s, t and u, are labelled on the structure of M in Figure 8.2.
Figure 8.2, the structure CH3CH2-NH-CH(CH3)2 with proton environments labelled: r on the CH3 of the ethyl group, s on the CH2 of the ethyl group, t on the two equivalent CH3 groups of the isopropyl group, and u on the CH of the isopropyl group
Identify the splitting pattern for each of the proton environments r, s, t and u.
Proton environment r
Proton environment s
Proton environment t
Proton environment u
8(b)(ii)Nmr Spectroscopy Structure ElucidationMedium2 marks
The proton (H) NMR spectrum of M dissolved in CDCl is obtained. State and explain how this spectrum differs from the proton (H) NMR spectrum of M dissolved in DO.

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