OCR A-Level Chemistry Synthesis and analytical techniques (02), November 2020: Question 21

15 marks · Medium difficulty · Structured Questions

Synthesise an organic nitrogen salt from propanone, identify functional groups in aspartame, draw its acid hydrolysis products, and calculate the safe number of cans of diet drink that can be consumed.

Practise this question

Question

A three-part structured chemistry question about organic nitrogen compounds. Part (a) asks to complete a reaction flowchart starting from propanone to form salt H via reduction, halogenation, and amination steps. Part (b)(i) gives the skeletal structure of aspartame and asks to name its functional groups excluding the benzene ring. Part (b)(ii) asks to draw the three organic products of the complete acid hydrolysis of aspartame. Part (b)(iii) is a calculation question determining the safe number of cans of a diet drink containing aspartame a typical adult can consume in one day.
Question text

21 This question is about organic compounds containing nitrogen.

(a) Salt H, (CH3)2CHNH3Cl, is used in the manufacture of garden weedkillers.

The flowchart shows the synthesis of the salt H from propanone.

Complete the flowchart.

Show structures for organic compounds.

propanone

NaBH4

reagent: …

Br

CH

H C CH3

reagent: …

NH3Cl

reagent: …

CH

H C CH3

salt H

[5]

(b) Aspartame, shown below, is an artificial sweetener commonly used as a sugar substitute.

O NH2 O

H

N

O OH

O

aspartame

(i) Aspartame contains several functional groups.

Apart from the benzene ring, name the functional groups in aspartame.

… [3]

(ii) A sample of aspartame is hydrolysed with aqueous acid.

Draw the structures of the three organic products of the complete acid hydrolysis of

aspartame.

[4]

(iii) Some people are concerned that aspartame, C14H18N2O5, may have adverse health

effects.

Research shows that the safe maximum daily intake of aspartame is 1.7 × 10−4 mol kg−1.

• A typical UK adult has a mass of 75 kg.

• A can of a diet drink contains 167 mg of aspartame.

How many cans of this diet drink is it safe for a typical adult to drink in one day?

Number of cans = … [3]

Mark scheme

Show the mark scheme The mark scheme provides the expected answers for all parts of question 21. Part (a) shows the intermediate structures and reagents for the flowchart. Part (b)(i) lists ester, amide, amine, and carboxylic acid. Part (b)(ii) shows the structures of methanol and the two amino acid products from aspartame hydrolysis. Part (b)(iii) outlines the step-by-step stoichiometric calculation leading to the final answer of 22.4 or 22 cans.

AO

Question Answer Marks Guidance

element

21 (a) 5 2.5×5 ALLOW any combination of skeletal OR structural OR

displayed formula as long as unambiguous

ALLOW HBr

ALLOW for the bottom left structure

20 AO

element

(b) (i) Ester 3 1.2×3

Amide IGNORE amino acid

Amine

Carboxylic acid ALLOW carboxyl

4 groups correct

3 groups correct IGNORE attempt to classify amide, e.g. secondary

2 groups correct IGNORE formulae (question asks for names)

IF > 4 functional groups are shown,

• Count 4 groups max but incorrect groups first

IGNORE aryl OR alkyl group

e.g. benzene, phenyl, aryl, arene, methyl

(ii) Methanol 1 mark 4 2.5×4 ALLOW any combination of skeletal OR structural OR

displayed formula as long as unambiguous

Amino Acids 3 marks

ALLOW + charge on H of NH group, i.e.NH +

If BOTH amino acids are shown with NH3 groups

(without the + charge) OR as NH + groups,

award 2 of the 3 marks for the amino acids

If BOTH amino acids are shown as correctly balanced

salts, e.g NH3Cl, all marks can be awarded.

Both amino acids shown with NH +

AO

element

(iii) FIRST CHECK ANSWER ON THE ANSWER LINE 3 2.2×3 If there is an alternative answer, apply ECF and

If answer = 22.4 OR 22 OR 23 award 3 marks look for alternative methods

n(aspartame) in 1 can = 0.167 / 294 = 5.68 x 10-4 (mol) Alternative methods

n(aspartame) in 1 can = 0.167 / 294

n(aspartame) limit per day = 1.7x10-4 x 75 = 0.01275 = 5.68 x 10-4 (mol)

(mol)

n(aspartame) per kg = 5.68 x 10-4 / 75

number of cans = 0.01275 / 5.68 x 10-4 = 22.4 = 7.57 x 10-6 (mol)

number of cans = 1.7 x 10-4 / 7.57 x 10-6

= 22.4

OR

n(aspartame) limit per day = 1.7x10-4 x 75

=0.01275 (mol)

mass(aspartame) limit per day =0.01275 x 294

= 3.7485 (g)

number of cans = 3.7485 / 0.167

= 22.4

How to answer it

Organic Nitrogen Compounds, Functional Groups & Quantitative Analysis

What this question tests

This multi-part exam question assesses your core organic synthesis pathways (carbonyl reduction, halogenation, nucleophilic substitution, and salt formation), your ability to identify multiple functional groups within complex biomolecules, peptide/ester hydrolysis products under acidic conditions, and multi-step moles calculations involving concentration limits and mass constraints.

Question Part (a) - Synthesis Flowchart

Synthesising Salt H from Propanone

✅ Correct Answers & Reagents

  • Intermediate 1 (Alcohol): Propan-2-ol ( (CH₃)₂CHOH )
  • Reagent 1: NaBr / H₂SO₄ (or HBr )
  • Intermediate 2 (Amine): Propan-2-amine ( (CH₃)₂CHNH₂ )
  • Reagent 2: NH₃ in ethanol (or excess NH₃ )
  • Final Reagent: HCl (to form salt H )

💡 Key Knowledge

  • Reduction: NaBH₄ reduces ketones to secondary alcohols.
  • Substitution (OH to Br): Hydroxyl groups are substituted using sodium bromide and concentrated sulfuric acid in situ to generate HBr .
  • Amine Synthesis: Haloalkanes react with excess ethanolic ammonia via nucleophilic substitution to form primary amines.
  • Salt Formation: Amines act as weak bases and react with hydrochloric acid to form alkylammonium chloride salts.

❌ Common Errors

  • Omitting "ethanol" when specifying ammonia as a reagent for nucleophilic substitution, leading to substitution with water instead.
  • Forgetting to show the protonated ammonium group ( -NH₃Cl or -NH₃⁺ ) in the final salt structure.

🧠 Exam Technique

Check bond connectivity carefully when drawing skeletal or structural formulas. Ensure your intermediate boxes explicitly display carbon chain branching.

Total for part (a): 5 marks (1 mark per correct box/reagent step)
Question Part (b)(i) - Functional Groups

Identifying Functional Groups in Aspartame

✅ Correct Answers

Any three (or four) of the following groups (excluding the benzene ring):

  • Ester ( -COO- )
  • Amide (or peptide bond: -CONH- )
  • Amine (primary aliphatic amine: -NH₂ )
  • Carboxylic acid ( -COOH )

❌ Common Errors & Misconceptions

  • Writing "amino acid" as a functional group. Amino acid is a classification molecule containing multiple functional groups, not a single functional group name.
  • Classifying the amide as "secondary amide" when the mark scheme only requires "amide". Extra erroneous classifications are ignored unless incorrect.
Total for part (b)(i): 3 marks (Allocated based on number of correct functional groups identified: 4 correct = 3 marks, 3 correct = 2 marks, 2 correct = 1 mark)
Question Part (b)(ii) - Acid Hydrolysis

Acid Hydrolysis Products of Aspartame

✅ Correct Answers (3 Organic Products)

  1. Methanol: CH₃OH (from ester cleavage)
  2. Phenylalanine derivative: HOOC-CH(NH₂ or NH₃⁺)-CH₂-C₆H₅
  3. Aspartic acid derivative: HOOC-CH(NH₂ or NH₃⁺)-CH₂-COOH

💡 Key Knowledge

  • Acid hydrolysis of an ester yields a carboxylic acid and an alcohol.
  • Acid hydrolysis of an amide/peptide bond yields a carboxylic acid and an ammonium salt (or amine if neutral, but under acidic conditions, basic amine groups protonate to form -NH₃⁺ ).
Total for part (b)(ii): 4 marks (1 mark for methanol + 3 marks for the two amino acid residues, accepting either neutral or protonated -NH₃⁺ forms).
Question Part (b)(iii) - Quantitative Calculation

Safe Daily Intake of Diet Drink

📐 Step-by-Step Calculation

  1. Calculate Mr of Aspartame (C₁₄H₁₈N₂O₅):
    (14 × 12.0) + (18 × 1.0) + (2 × 14.0) + (5 × 16.0) = 294.0 g mol⁻¹
  2. Find moles of aspartame in one can:
    Mass = 167 mg = 0.167 g
    Moles = 0.167 / 294 = 5.68 × 10⁻⁴ mol
  3. Find maximum daily molar limit for a 75 kg adult:
    Limit = 1.7 × 10⁻⁴ mol kg⁻¹ day⁻¹ × 75 kg = 0.01275 mol
  4. Calculate maximum number of safe cans:
    Number of cans = 0.01275 / (5.68 × 10⁻⁴) = 22.4 cans

✅ Final Answer

Number of cans = 22 (Accept 22.4, 22, or truncated whole integer as specified by context; rounding down to whole cans 22 is standard for safe limits).

❌ Calculation Traps

  • Forgetting to convert milligrams ( mg ) to grams ( g ) before dividing by Mr .
  • Failing to multiply the safe daily limit per kg by the full body mass ( 75 kg ).
Total for part (b)(iii): 3 marks (Awarded for correct moles in a can, correct daily limit moles, and final correct division leading to 22.4 / 22).

Topics

Module 6: Organic chemistry and analysis · Module 2: Foundations in chemistry · 6.1 Aromatic compounds, carbonyls and acids · 6.2 Nitrogen compounds, polymers and synthesis · 2.1 Atoms and reactions

Question and mark scheme from the OCR A-Level Chemistry examination, Synthesis and analytical techniques (02), November 2020. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.