A-LevelChemistryNitrogen and sulfur9701/42 · May/June 2024 · Q816 Marks

State the relative basicities of phenylamine, C₆H₅NH₂, benzylamine, C₆H₅CH₂NH₂, and ammonia, NH₃, in aqueous solution. Explain your answer. most basic > > least basic

(a)State the relative basicities of phenylamine, C₆H₅NH₂, benzylamine, C₆H₅CH₂NH₂, and ammonia, NH₃, in aqueous solution. Explain your answer. most basic > > least basic[3]
(b)(i)An excess of Br₂(aq) is added to separate samples of C₆H₅NH₂ and benzene, C₆H₆. C₆H₅NH₂ reacts readily with Br₂(aq) to form organic product M. State the expected observations for this reaction. Draw the structure of M. observations structure of M[2]
(b)(ii)C₆H₆ does not react with Br₂(aq). Suggest why Br₂(aq) reacts with C₆H₅NH₂ but not with C₆H₆.[2]
(c)Explain why benzamide, C₆H₅CONH₂, is a much weaker base than ammonia, NH₃.[1]
(d)C₆H₅CONH₂ is formed by reacting benzoyl chloride, C₆H₅COCl, with NH₃. Complete the mechanism in Fig. 8.1 for the reaction of C₆H₅COCl with NH₃. Include all relevant lone pairs of electrons, curly arrows, charges and dipoles. Draw the structure of the organic intermediate. [Figure 8.1 showing C₆H₅COCl reacting with NH₃, then an empty box for "organic intermediate", then this intermediate converting to C₆H₅CONH₂ + HCl][4]
(e)(i)Phenylalanine, C₆H₅CH₂CH(NH₂)COOH, is an amino acid with an isoelectric point of 5.5. State what is meant by isoelectric point.[1]
(e)(ii)Draw the structure of C₆H₅CH₂CH(NH₂)COOH at pH 10.[1]
(f)C₆H₅CH₂CH(NH₂)COOH and alanine, CH₃CH(NH₂)COOH, react to form a dipeptide containing both amino acid residues. Draw the structure of this dipeptide. The peptide functional group formed should be displayed.[2]

This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.

Mark scheme

Marking points from the official mark scheme (May/June 2024). The mark or mark type is shown in brackets.

[3] M1 benzylamine > ammonia > phenylamine
(basicity linked to) p orbital of N / lone pair on N AND being able to bond / accept / donate to / coordinate to a proton / H+
(benzylamine) R / alkyl / CH2 group AND is electron donating / positive inductive effect /has + I effect
(phenylamine) p orbital of N / lone pair on N is overlaps / incorporated / delocalised in the ring / π-bond system
[2] M1 white ppt.
M2 structure (drawing of 2,4,6-tribromoaniline)
[2] M1 lone pair/p-orbital / electrons on the nitrogen / NH2 AND overlap / delocalised / incorporated AND with the (π-electrons) ring / π system
M2 increasing its electron density (of the ring) OR it can polarise the electrophile / Br2 better
[1] lone pair/p-orbital on N is delocalised AND into C=O group / across the two electronegative O and N
[4] M1/M2 any two for one mark, all four for two marks:
lone pair on N
correct arrow from lone pair N to C (of C=O)
correct dipole on C=O
correct arrow from the C=O bond to O atom
M3 correct intermediate (drawing of tetrahedral intermediate with -O- (negatively charged), -Cl, -NH3+ (positively charged) attached to the central carbon, benzene ring also attached to central carbon)
M4 arrow from O(-) to C-O bond AND arrow from C-Cl bond to Cl
ALLOW arrow from anywhere on O- to C-O bond
[1] pH at which a molecule has no overall charge / is neutral / exist as a zwitterion
[1] C6H5CH2CH(NH2)COO-
[2] M1 correct peptide bond displayed (there must be a saturated carbon attached to either side of the peptide group)
M2 rest of the dipeptide correct
(drawing of dipeptide structure, two options for which amino acid is N-terminal)