OCR A-Level Chemistry Synthesis and analytical techniques (02), November 2020: Question 18

✅ Correct Answer

Reagents and conditions: Acidified potassium dichromate ( K₂Cr₂O₇ / H⁺ ) AND reflux.

Equation: HO(CH₂)₄OH + 4[O] → HOOC(CH₂)₂COOH + 2H₂O (or showing [O] and H₂O products separately).

💡 Key Knowledge

A diol with primary alcohol groups ( HO(CH₂)₄OH ) requires complete oxidation to form a dicarboxylic acid ( HOOC(CH₂)₂COOH ). This requires excess oxidising agent and heating under reflux to ensure complete conversion without losing volatile intermediates.

🧠 Exam Technique

When writing equations with [O] , count your carbon atoms carefully! Notice how the carbon chain length changes or remains consistent depending on the reactant. Balance the oxygen atoms using water ( H₂O ) on the product side.

❌ Common Errors

Forgetting to specify reflux will lose you the condition mark. Writing just "dichromate" without stating it must be acidified will also be penalised.

✅ Correct Answer

A clear diagram showing a hydrogen bond (dotted line) between a lone pair on an oxygen atom of the dicarboxylic acid and the δ+ hydrogen atom of a water molecule (or vice versa), complete with clear partial charges ( δ+ / δ- ) and labels.

💡 Key Knowledge

Carboxylic acids are soluble in water because their polar -COOH and -OH groups can form strong hydrogen bonds with polar water molecules.

🧠 Exam Technique

To secure full marks: 1. Draw a clear hydrogen bond with a dotted line between O and H . 2. Label the partial charges ( δ+ and δ- ) correctly on the bond involved. 3. Explicitly state that hydrogen bonding occurs between the molecules.

❌ Common Errors

Placing a δ+ charge directly onto a hydrogen atom belonging to a non-polar -CH₂- group will instantly lose you credit. Ensure partial charges are placed exclusively on polar bonds ( O-H or C=O ).

✅ Correct Answer

--C(=O)-(CH₂)₂-C(=O)-O-(CH₂)₄-O-- (Showing the ester linkage clearly displayed with continuation bonds passing through brackets or extending out).

💡 Key Knowledge

Condensation polymerisation between a dicarboxylic acid and a diol eliminates water and forms an ester link ( -COO- ). A repeat unit must show the structural units derived from both monomers joined together.

🧠 Exam Technique

Always show open-ended continuation bonds ( -- ) at both ends of your repeat unit. Make sure the ester linkage is fully displayed as -C(=O)-O- rather than condensed as -CO₂- .

❌ Common Errors

Drawing more than one repeat unit or forgetting to include the continuation bonds at the ends will cause marks to be dropped.

✅ Correct Answer

The ester bond / ester group / polyester can be broken down (or hydrolysed).

💡 Key Knowledge

Polyesters contain polar ester linkages in their polymer backbone. These bonds are susceptible to nucleophilic attack by water (hydrolysis), allowing biological catalysts (enzymes) in the environment to break the polymer chains down.

🧠 Exam Technique

Keep your answer concise and targeted. Examiners require explicit reference to the ester bond or hydrolysis. Vague statements like "the bonds break" are insufficient.

✅ Correct Answer

Equation: HOOC(CH₂)₂COOH + 2SOCl₂ → ClOC(CH₂)₂COCl + 2SO₂ + 2HCl (with correct structural/skeletal representation for the diacyl chloride product).

💡 Key Knowledge

Carboxylic acids react with thionyl chloride ( SOCl₂ ), phosphorus(V) chloride ( PCl₅ ), or phosphorus(III) chloride ( PCl₃ ) to form acyl chlorides, releasing acidic gas byproducts.

🧠 Exam Technique

Check stoichiometry carefully! Since the starting dicarboxylic acid contains two -COOH groups, you need 2 moles of SOCl₂ to convert both groups, producing 2SO₂ and 2HCl alongside the diacyl chloride.

❌ Common Errors

Missing coefficients in balanced equations is the most frequent place students lose easy marks. Balance your inorganic byproducts ( SO₂ and HCl ) relative to the number of acid groups reacting.