HSC Exam Practice

Sample response. The carboxyl group (−COOH) is a functional group consisting of a carbonyl group (C=O) and a hydroxyl group (−OH) bonded to the same carbon atom. It is the characteristic functional group of carboxylic acids.

Marking notes. 1 mark for defining it as C=O and −OH on the same carbon; 1 mark for correctly identifying it as the functional group of carboxylic acids (or “the −COOH group”).

Sample response. IUPAC name: butanoic acid. The chain has 4 carbons (including the −COOH carbon, which is always C1). The prefix “but-” indicates 4 carbons; the suffix “−anoic acid” indicates a saturated carboxylic acid. The COOH carbon is always C1 and no locant is needed.

Marking notes. 1 mark for correct IUPAC name (butanoic acid); 1 mark for explaining that the −COOH carbon is always counted as C1 and determines the prefix.

Sample response. 2CH₃COOH + Na₂CO₃ → 2CH₃COONa + H₂O + CO₂(g). Products: sodium ethanoate, water, carbon dioxide. Observable: effervescence / bubbles of CO₂ gas; solid Na₂CO₃ dissolves.

Marking notes. 1 mark for correct balanced equation (condone state symbols absent if formula is correct); 1 mark for naming all three products; 1 mark for stating effervescence or CO₂ production as the observable.

Sample response. Both propanoic acid and propan-1-ol have one O−H bond and can form hydrogen bonds (1 mark). Propanoic acid molecules, however, form hydrogen-bonded dimers: two molecules align face-to-face and simultaneously form two H-bonds — O−H (mol 1) ··· O=C (mol 2) and O=C (mol 1) ··· H−O (mol 2) (1 mark). To vaporise propanoic acid, both H-bonds of a dimer must break at once, requiring significantly more energy than breaking the single O−H H-bond between two propan-1-ol molecules (1 mark).

Marking notes. 1 mark for identifying both have O−H H-bonding; 1 mark for naming and describing the cyclic dimer (2 simultaneous H-bonds); 1 mark for linking dimer disruption to higher energy / higher BP.

Sample response. A carboxylic acid (pKa ≈ 5) reacts with NaHCO₃ to produce CO₂ gas (effervescence): RCOOH + NaHCO₃ → RCOONa + H₂O + CO₂. A phenol (pKa ≈ 10) does not react with NaHCO₃ — no gas is produced. The chemical basis is acid strength: carboxylic acid (pKa ~5) is a stronger acid than carbonic acid (pKa1 ~6.35), so it can donate H⁺ to HCO₃⁻; phenol (pKa ~10) is weaker than carbonic acid, so it cannot donate H⁺ to HCO₃⁻ under standard conditions.

Marking notes. 1 mark for stating carboxylic acid produces CO₂/effervescence; 1 mark for stating phenol gives no reaction with NaHCO₃; 1 mark for the pKa-based explanation (carboxylic acid stronger than carbonic acid; phenol weaker than carbonic acid).

Sample response. When a carboxylic acid loses a proton, the carboxylate ion (RCOO⁻) is formed. In this ion, the negative charge is delocalised by resonance across both oxygen atoms: RCOO⁻ has two equivalent resonance structures, each placing the negative charge on a different oxygen (1 mark). This resonance stabilisation lowers the energy of the carboxylate ion relative to the undissociated acid (1 mark). A lower-energy conjugate base is a weaker base (less tendency to re-accept H⁺), which shifts the equilibrium further toward ionisation → stronger acid (pKa ~5) (1 mark). By contrast, when an alcohol loses a proton, the alkoxide (RO⁻) has the negative charge localised on a single oxygen — no resonance is possible. This makes the alkoxide a strong base and the equilibrium lies far left → alcohol is a very weak acid (pKa ~16) (1 mark).

Marking notes. 1 mark for identifying resonance delocalisation across both oxygens in RCOO⁻; 1 mark for linking resonance to lower energy / stabilised conjugate base; 1 mark for connecting weaker conjugate base to stronger acid; 1 mark for contrast with alkoxide (no resonance, stronger base, weaker acid).

Sample response. Chloroethanoic acid (Ka = 1.36×10⁻³) is approximately 77 times stronger than ethanoic acid (Ka = 1.76×10⁻⁵). The Cl atom bonded to the α-carbon is strongly electronegative and withdraws electron density inductively through the C−C bond toward the carboxyl group (inductive effect). This partial withdrawal of electron density reduces the electron density on the carboxylate oxygens, making the negative charge in the conjugate base more stable (less concentrated). A more stable conjugate base is a weaker base, shifting the equilibrium further toward ionisation → higher Ka. (1 mark for identifying the Cl inductive withdrawal; 1 mark for stabilisation of the conjugate base; 1 mark for linking to higher Ka / stronger acid.)

Sample response. The compound is a carboxylic acid. The production of CO₂ with NaHCO₃ confirms the compound has pKa < 6.35 (the pKa of carbonic acid) — from Table 2.1, only the two carboxylic acids (pKa 2.87 and 4.75) meet this criterion. The negative Fehling’s test rules out an aldehyde group. Therefore the compound contains a −COOH functional group. (1 mark for identification; 1 mark for pKa < 6.35 / Ka argument; 1 mark for using Fehling’s to exclude aldehyde.)

Sample response. Phenol (pKa = 10.0) reacts with NaOH because NaOH is a strong base and the phenoxide ion (C₆H₅O⁻) is stabilised by partial delocalisation of the negative charge into the benzene ring. However, phenol does not donate H⁺ to HCO₃⁻ because the pKa of phenol (10.0) is greater than the pKa of carbonic acid (pKa1 ≈ 6.35). An acid can only donate H⁺ to a base whose conjugate acid is weaker than the donor acid. Since phenol is weaker than carbonic acid, the reaction phenol + HCO₃⁻ → phenoxide + H₂CO₃ is thermodynamically unfavourable → no CO₂ produced. (1 mark for phenol reacts with NaOH due to strong base; 1 mark for pKa comparison — phenol weaker than carbonic acid; 1 mark for concluding the reaction with NaHCO₃ is unfavourable.)

Sample response. Compound A is a carboxylic acid (NaHCO₃ → effervescence; Na metal → H₂). Balanced equation with Na metal: 2RCOOH + 2Na → 2RCOONa + H₂(g). For butanoic acid specifically: 2CH₃CH₂CH₂COOH + 2Na → 2CH₃CH₂CH₂COONa + H₂. (1 mark for correct class identification; 1 mark for correct balanced equation.)

Sample response. Compound D is methanoic acid (HCOOH; molecular formula CH₂O₂; IUPAC name: methanoic acid). It is unique among carboxylic acids: its “R group” is simply H, giving the molecule a formyl-like (HCO–) component in addition to the –COOH group. The formyl group is structurally similar to an aldehyde and is oxidisable — hence the positive Tollens’ test (silver mirror). Simultaneously, HCOOH is a carboxylic acid (Ka = 1.77×10⁻⁴) and reacts with NaHCO₃ to give CO₂. No other simple carboxylic acid behaves as a reducing agent with Tollens’ reagent. (1 mark for IUPAC name methanoic acid; 1 mark for explaining both positive tests — the formyl group enables Tollens’ positive AND the –COOH group reacts with NaHCO₃.)

Sample response. The carboxyl group (−COOH) is one of the most chemically informative functional groups in organic chemistry because it simultaneously defines anomalous physical behaviour, weak-acid equilibria, and a set of diagnostic reactions that distinguish it from all other oxygen-containing functional groups.

Physically, carboxylic acids have the highest boiling points of any organic compounds with the same carbon chain length. Ethanoic acid (the active component of vinegar, used in Australian wine chemistry by the AWRI) boils at 118°C despite having a lower molecular mass (60 g/mol) than many comparable molecules. This arises because the −COOH group simultaneously donates (O−H) and accepts (C=O) hydrogen bonds. Two −COOH molecules can therefore form a cyclic dimer with two simultaneous H-bonds per pair. Vaporising the acid requires disrupting both H-bonds at once, demanding substantially more energy than breaking the single H-bond between two alcohol molecules — explaining the 40°C BP gap between ethanoic acid and ethanol at C2.

As a weak acid, −COOH partially ionises: RCOOH + H₂O ↔ RCOO⁻ + H₃O⁺; Ka « 1. Methanoic acid (Ka = 1.77×10⁻⁴), the component of Australian bull ant (Myrmecia) venom, has only ~4% ionisation at 0.1 mol/L — confirming the weak-acid classification. The structural basis for the group’s acidity lies in resonance stabilisation of the carboxylate anion (RCOO⁻): the negative charge is delocalised equally across both oxygens, making the conjugate base far more stable than the alkoxide (RO⁻) produced from an alcohol. This stability shifts the ionisation equilibrium further right for carboxylic acids (pKa ~5) than for alcohols (pKa ~16) or phenols (pKa ~10).

The four diagnostic reactions are: (1) with NaOH → carboxylate salt + H₂O (no gas); (2) with Na₂CO₃ → salt + H₂O + CO₂; (3) with NaHCO₃ → salt + H₂O + CO₂ — the key distinguishing reaction, as phenols and alcohols do not produce CO₂ with NaHCO₃; (4) with reactive metals (e.g. Mg) → metal carboxylate + H₂. The NaHCO₃ test reflects the pKa ordering: carboxylic acid (pKa ~5) is stronger than carbonic acid (pKa1 ~6.35), allowing H⁺ transfer; phenol (pKa ~10) cannot.

In summary, the −COOH group is a powerful dual classifier: its dimerisation capacity places carboxylic acids uniquely at the top of the BP ranking for any carbon chain length, while its resonance-stabilised conjugate base and Ka value allow precise placement on the pKa acid-strength scale and a three-level discrimination ladder (acid/phenol/alcohol) via NaHCO₃ reactivity.

Marking criteria. 1 mark — correctly describes the −COOH structure (C=O + O−H on same C) and its dual H-bond capacity. 1 mark — explains dimer formation and links it to anomalously high BP vs alcohols. 1 mark — defines weak acid / partial ionisation with Ka expression or Ka value. 1 mark — applies resonance to explain carboxylate stability and the pKa ordering relative to alcohols and/or phenols. 1 mark — accurately presents at least two of the four diagnostic reactions with correct products. 1 mark — uses the NaHCO₃ test as a discriminating tool between carboxylic acid, phenol and alcohol, with pKa justification. 1 mark — incorporates at least one named Australian example (e.g. vinegar/AWRI, bull ant/methanoic acid, citrus/citric acid, sourdough/lactic or propanoic acid) correctly applied to the chemistry discussed.