Home / AS & A Level Chemistry 30.1 Arenes: Exam Style Questions Paper 4

AS & A Level Chemistry 30.1 Arenes: Exam Style Questions Paper 4

A Level Chemistry Paper 4 - All Chapters
Question

(a) State the relative basicities of phenylamine, C₆H₅NH₂, benzylamine, C₆H₅CH₂NH₂, and ammonia, NH₃, in aqueous solution. Explain your answer.

Structures of phenylamine, benzylamine, and ammonia

(b) An excess of \(Br_2(aq)\) is added to separate samples of \(C_6H_5NH_2\) and benzene, \(C_6H_6\).
(i) \(C_6H_5NH_2\) reacts readily with \(Br_2 (aq)\) to form organic product M. State the expected observations for this reaction. Draw the structure of M.

(ii) \(C_6H_6\) does not react with \(Br_2 (aq)\). Suggest why \(Br_2 (aq)\) reacts with \(C_6H_5NH_2\) but not with \(C_6H_6\).

(c) Explain why benzamide, C₆H₅CONH₂, is a much weaker base than ammonia, NH₃.

(d) C₆H₅CONH₂ is formed by reacting benzoyl chloride, \(C_6H_5COCl\), with \(NH_3\). Complete the mechanism in Fig. 8.1 for the reaction of \(C_6H_5COCl\) with \(NH_3\). Include all relevant lone pairs of electrons, curly arrows, charges and dipoles. Draw the structure of the organic intermediate.

Fig. 8.1: Reaction mechanism

(e) Phenylalanine, C₆H₅CH₂CH(NH₂)COOH, is an amino acid with an isoelectric point of 5.5.
(i) State what is meant by isoelectric point.

(ii) Draw the structure of C₆H₅CH₂CH(NH₂)COOH at pH 10.
(f) C₆H₅CH₂CH(NH₂)COOH and alanine, \(CH_3CH(NH_2)COOH\), react to form a dipeptide containing both amino acid residues. Draw the structure of this dipeptide. The peptide functional group formed should be displayed.

▶️ Answer/Explanation
Solution

(a) Benzylamine > Ammonia > Phenylamine

Explanation:

  • Benzylamine’s alkyl group (\(CH_2\)) donates electrons via the +I effect, increasing basicity.
  • Ammonia has no electron-withdrawing or donating groups.
  • Phenylamine’s lone pair delocalizes into the benzene ring, reducing availability for protonation.

(b)(i) Observations: White precipitate (2,4,6-tribromoaniline) forms; brown bromine color decolorizes.

Structure of M (2,4,6-tribromoaniline)

Explanation: The \(-NH_2\) group activates the benzene ring, enabling electrophilic substitution at the 2, 4, and 6 positions.

(b)(ii) \(C_6H_5NH_2\) reacts because the \(-NH_2\) group donates electron density to the ring, polarizing \(Br_2\). Benzene lacks an activating group and requires a catalyst (e.g., FeBr₃) for bromination.

Explanation: The lone pair on \(-NH_2\) increases the ring’s electron density, facilitating electrophilic attack.

(c) Benzamide’s lone pair on nitrogen is delocalized into the carbonyl (\(C=O\)) group, reducing its availability to accept protons.

Explanation: The adjacent carbonyl withdraws electron density via resonance, making the nitrogen less basic than in NH₃.

(d) Nucleophilic acyl substitution mechanism:

Mechanism of benzoyl chloride with NH₃

Explanation: NH₃ attacks the electrophilic carbonyl carbon, forming a tetrahedral intermediate. Chloride leaves, yielding benzamide.

(e)(i) Isoelectric point (pI): The pH at which an amino acid has no net charge (zwitterion form).

Explanation: At pI (5.5 for phenylalanine), the \(-NH_3^+\) and \(-COO^-\) groups are balanced.

(e)(ii) At pH 10, phenylalanine is deprotonated: \(C_6H_5CH_2CH(NH_2)COO^-\).

Explanation: The \(-COOH\) group loses \(H^+\) (pKa ~2.2), and \(-NH_3^+\) deprotonates to \(-NH_2\) (pKa ~9.1).

(f) Dipeptide structure (phenylalanine-alanine or alanine-phenylalanine):

Dipeptide structure

Explanation: Condensation reaction forms a peptide bond (\(-CONH-\)) between the carboxyl and amino groups of the two amino acids.

More A Level Chemistry Paper 4 Question.....
Scroll to Top