HSC Exam Practice

Sample response. A primary amine is an amine in which exactly one alkyl group is bonded to the nitrogen atom (R–NH₂); two N–H bonds remain. Amine classification (primary, secondary, tertiary) is determined by counting the number of alkyl groups bonded directly to the nitrogen: one = primary; two = secondary; three = tertiary.

Marking notes. 1 mark for correct definition (one alkyl group on N, or equivalent). 1 mark for correctly stating the classification rule (count alkyl groups on N, not carbons in the chain).

Sample response. (a) (CH₃)₂CHNH₂: one alkyl group (isopropyl) on N → primary amine. (b) (C₂H₅)₂NH: two ethyl groups on N → secondary amine. (c) (CH₃)₃N: three methyl groups on N, no H atoms on N → tertiary amine.

Marking notes. 1 mark per correctly classified compound with correct reason (max 3). Note: (a) is primary despite the N being on a secondary carbon — classification is by alkyl groups on N, not carbon substitution pattern.

Sample response. CH₃CH₂CH₂NH₂ + HCl → [CH₃CH₂CH₂NH₃⁺][Cl⁻] (propylammonium chloride). Reaction type: acid–base neutralisation (the amine lone pair accepts H⁺ from HCl). Two properties of the product: (1) ionic compound (composed of a propylammonium cation and a chloride anion); (2) non-volatile / solid at room temperature, and therefore odourless (no free lone pair, cannot reach olfactory receptors).

Marking notes. 1 mark — correct equation with ionic product notation ([RNH₃⁺][Cl⁻] or equivalent). 1 mark — type of reaction identified as acid–base (or neutralisation). 1 mark each (max 2) for any two valid properties: ionic; solid at RT; water-soluble; non-volatile; odourless; no free lone pair. Do NOT accept “water is a product” — this is a neutralisation without water production.

Sample response. An amine contains the functional group –NH₂ (or –NHR, –NR₂); in water, the lone pair on N freely accepts H⁺ → basic solution (pH > 7); Kb ∼ 10⁻⁴. An amide contains the functional group –CO–NH₂ (carbonyl bonded to N); the lone pair on N is delocalised into the C=O by resonance → not freely available to accept protons → neutral in water (pH ≈ 7); Kb ∼ 10⁻¹⁵.

Marking notes. 1 mark — amine functional group formula. 1 mark — amide functional group formula (must include C=O adjacent to N). 1 mark — amines produce basic solution (Kb or pH evidence). 1 mark — amides neutral because N lone pair is delocalised (resonance explanation required).

Sample response. RCOOH + H₂N–R’ → RCONH–R’ + H₂O. Type of reaction: condensation (two molecules combine with the loss of a small molecule, in this case water). The –OH from the carboxylic acid and one H from the amine –NH₂ combine as water; the amide bond –CO–NH– is formed.

Marking notes. 1 mark — correct general equation (reactants, products including water). 1 mark — correctly identifies as condensation reaction (accept “condensation polymerisation” if student extends to polymer context).

Sample response. Oxygen (EN = 3.5) is more electronegative than nitrogen (EN = 3.0) [1]. The O–H bond in propan-1-ol is therefore more polar than the N–H bond in propan-1-amine; the hydrogen in O–H is more strongly δ⁺ [1]. This makes O–H a stronger H-bond donor than N–H: the O–H···O hydrogen bonds in propan-1-ol require more energy to break than the N–H···N H-bonds in propan-1-amine, resulting in a higher boiling point for propan-1-ol [1].

Marking notes. 1 mark — correctly states EN values (O 3.5, N 3.0). 1 mark — links greater EN to greater bond polarity → stronger H-bond donation from O–H. 1 mark — concludes that stronger H-bonds require more energy to break → higher BP for propan-1-ol. Do not award marks for statements that only say “O–H bonds are stronger” without the electronegativity reasoning.

Sample response (a). Boiling point increases across the series from left to right: propane (−42°C) < trimethylamine (3°C) < propan-1-amine (48°C) < propan-1-ol (97°C) < propanamide (213°C). The trend reflects increasing strength of intermolecular forces: propane has only dispersion forces; trimethylamine adds dipole–dipole but no H-bond donation; propan-1-amine adds N–H H-bonding; propan-1-ol has the stronger O–H H-bonding; propanamide has the cooperative N–H donor + C=O acceptor H-bond network. [3 marks: 1 for overall direction of trend + at least one data value quoted; 1 for correctly linking each group to its IMF type; 1 for identifying that all five points follow this logic consistently.]

Sample response (b). Both propan-1-amine and propanamide have two N–H bonds per molecule (H-bond donors). However, propanamide additionally has a highly polar C=O group which is a strong H-bond acceptor. This creates a cooperative H-bond network: each propanamide molecule simultaneously donates H-bonds via N–H to a neighbour’s C=O, and accepts H-bonds into its own C=O from a neighbour’s N–H [1]. This networked, multi-directional H-bonding is far stronger than the N–H···N H-bonding in propan-1-amine, which lacks the strong C=O acceptor [1]. Far more energy is needed to disrupt the propanamide network than the amine network → BP 213°C vs 48°C [1].

Sample response (c). Trimethylamine bar would appear between propane and propan-1-amine (BP 3°C) — above propane but below propan-1-amine [1]. Although both are C₃ amines with similar MW (trimethylamine MW 59; propan-1-amine MW 59), trimethylamine is a tertiary amine with NO N–H bonds, so it cannot donate H-bonds between its own molecules. Intermolecular forces in trimethylamine are limited to dipole–dipole and dispersion. Propan-1-amine (primary) has two N–H bonds and forms N–H···N H-bonds, requiring more energy to vaporise → higher BP [1].

Sample response (a). (CH₃)₃N + HCl → [(CH₃)₃NH⁺][Cl⁻]   [1 mark]

Sample response (b). The product [(CH₃)₃NH⁺][Cl⁻] is an ionic ammonium salt [1]. The nitrogen’s lone pair has been used to form a bond with H⁺; it is no longer free to interact with olfactory receptors. The ionic salt is also non-volatile (high melting point, does not evaporate at room temperature), so no molecules reach the nose → odourless [1].

Sample response. The student’s claim is incorrect [1]. Ethanamide does NOT produce a basic solution in water; it gives pH ≈ 7 (neutral). The reason is structural: in ethanamine (CH₃CH₂NH₂), the lone pair on nitrogen sits in a free sp³ orbital with no adjacent carbonyl group, so it is freely available to accept H⁺ from water → CH₃CH₂NH₂ + H₂O ⇌ CH₃CH₂NH₃⁺ + OH⁻; Kb ∼ 4 × 10⁻⁴ (weak base; pH > 7) [1]. In ethanamide (CH₃CONH₂), the lone pair on N is delocalised into the adjacent C=O pi system by resonance: CH₃C(=O)NH₂ ↔ CH₃C(O⁻)=NH₂⁺ [1]. The lone pair participates in this pi system rather than sitting in a purely non-bonding sp³ orbital [1]. It therefore cannot be donated to accept H⁺ from water → Kb (ethanamide) ∼ 10⁻¹⁵, roughly 10¹¹ times smaller than Kb of ethanamine; pH ≈ 7; neutral [1]. Having N–H bonds does NOT confer basicity if the lone pair is delocalised. The same resonance that removes basicity also causes the amide bond to have partial C=N double-bond character, making the bond planar with restricted rotation [1]. In proteins, each amide bond is a peptide bond (–CO–NH–); because it cannot rotate freely, the protein backbone is constrained into repeating patterns: the alpha helix and beta sheet (secondary structures) [1]. These secondary structures define the three-dimensional shape of each protein — including the geometry of an enzyme’s active site. Without the planarity of the peptide bond, proteins would be random coils with no defined active site and no catalytic function [1].

Marking criteria.

• 1 mark — States the claim is incorrect; ethanamide does not produce a basic solution (pH ≈ 7).
• 1 mark — Ethanamine: lone pair in free sp³ orbital → accepts H⁺ from water → OH⁻ produced → basic; Kb ∼ 10⁻⁴.
• 1 mark — Draws or describes the resonance of ethanamide (CH₃C(=O)NH₂ ↔ CH₃C(O⁻)=NH₂⁺).
• 1 mark — Explains that resonance places the lone pair into the C=O pi system → not freely available to accept protons.
• 1 mark — Compares Kb values: ethanamine Kb ∼ 10⁻⁴ (measurably basic) vs ethanamide Kb ∼ 10⁻¹⁵ (effectively neutral) — must cite or compare actual values.
• 1 mark — Resonance also gives partial C=N double-bond character → restricted rotation → amide bond is planar.
• 1 mark — Links planarity to protein secondary structure (alpha helix or beta sheet named).
• 1 mark — Links protein secondary structure to 3D shape and/or biological function (active site, enzyme function, structural protein, etc.).