Edexcel A-Level Chemistry Paper 3, June 2023: Question 9

✅ Correct Answer

The hydrogen ions from the acid protonate the benzoate ion (or sodium benzoate) to form insoluble, unionised benzoic acid.

💡 Key Knowledge

Alkaline hydrolysis using NaOH produces a carboxylate salt ( C6H5COO⁻Na⁺ ) rather than the carboxylic acid directly because the alkaline conditions neutralise any acid formed.

❌ Common Errors

Students often state vaguely that acidification "neutralises the alkali" or "stops the reaction" without explicitly stating that it converts the benzoate salt into the benzoic acid product.

🧠 Exam Technique

Always name the specific species being created or altered. Mentioning protonation of the carboxylate/benzoate ion guarantees the mark.

✅ Correct Answer

Using structural or semi-structural formulae:

💡 Key Knowledge

Unlike alkaline hydrolysis, acid hydrolysis is a reversible equilibrium reaction. Therefore, a balance sign ( ⇌ ) or reverse arrow must be used instead of a straight arrow.

❌ Common Errors

Using molecular formulae (e.g., C8H8O2 ) instead of structural/displayed/skeletal formulae when requested, or omitting water as a reactant.

🧠 Exam Technique

Double-check formula requirements. Whenever the command paper asks for "structural formulae", ensure bonds or clear groupings (like COOCH3 ) are shown.

✅ Correct Answers & Structures

• W: Phenyl methanoate ( HCOOCH2C6H5 or ester derived from methanoic acid)
• X: Benzyl methanoate / Methyl benzoate isomer variant with an ester link showing methanoate origin.
• Y: Methylbenzoic acid / 4-methylbenzoic acid (or equivalent mono-substituted aromatic releasing CO₂ with carbonate).
• Z: 4-hydroxybenzoic acid (Disubstituted, 6 carbon environments in ¹³C NMR, forms a sweet-smelling ester with ethanol).

💡 Key Knowledge Clues

• Carbonate reaction ( Na2CO3 → CO2 ) proves the presence of a carboxylic acid group.
• Ester production from methanoic acid requires an HCOO- group.
• 6 peaks in ¹³C NMR for a disubstituted benzene ring points to a symmetrical 1,4-disubstituted (para) structure.

❌ Common Errors

Failing to link justifications explicitly to the clues in the stem text (e.g., ignoring why W must specifically originate from methanoic acid, or missing symmetry arguments for Z).

🧠 Exam Technique

Break down the multi-mark structural deduction systematically. Tick off each constraint: check mono- vs disubstituted, functional group tests, and NMR environments one by one.

✅ Mechanism Requirements

• Curly arrow starting from the benzene ring pi-system pointing directly to the C of the [CH3C=O]⁺ electrophile.
• Intermediate (horseshoe structure) with the positive charge bracketed inside a partial ring, and both H and the acyl group attached at the same ring carbon.
• Curly arrow originating from the C-H bond breaking to return electrons into the delocalised ring system.
• Regeneration of the aluminium chloride catalyst: AlCl4⁻ + H⁺ → AlCl3 + HCl .

💡 Key Knowledge

Phenol undergoes electrophilic substitution readily due to the oxygen lone pair interacting with the pi-system, directing incoming groups primarily to the 4- (para) position.

❌ Common Errors

Starting the initial curly arrow from inside the ring carbon rather than the delocalised electron cloud, or drawing the horseshoe intermediate pointing incorrectly/terminating at the wrong carbons.

🧠 Exam Technique

Make sure arrowheads are precise: curly arrows representing electron movement must start precisely at bonds or lone pairs and end where the new bond forms.

✅ Correct Reagents & Observations

Reagents: Alkaline iodine ( I2 / NaOH ) or NaOCl with KI (Triiodomethane / Iodoform test).

Observation: A pale yellow (or yellow) precipitate/solid forms (with a characteristic antiseptic smell).

💡 Key Knowledge

Piceol contains a methyl ketone group ( -COCH3 ). Methyl ketones (and secondary alcohols adjacent to a methyl group) undergo the iodoform reaction to form triiodomethane ( CHI3 ).

❌ Common Errors

Using standard carbonyl tests like Tollens' or Fehling's reagents. Both molecules contain carbonyls (one a ketone, one an aldehyde), so generic tests would give a positive result for both or fail to distinguish them effectively.

🧠 Exam Technique

Always scan structures for specific identifying structural units: methyl ketones ( CH3C=O ) immediately scream "triiodomethane test", while aldehydes point towards Tollens' or Fehling's.