Edexcel International A Level Chemistry: Organic nitrogen compounds: amines, amides, amino acids and proteins (YCH11) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

This set covers Topic 19: Organic Nitrogen Compounds: Amines, Amides, Amino Acids and Proteins of the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry specification (YCH11), Issue 1, September 2017, outcomes 19.1 to 19.12 including Core Practical 15. Unit 5: Transition Metals and Organic Nitrogen Chemistry (IA2) contains Topic 19. Before attempting it, learn the topic from the study guide and revision notes; see also a topic-by-topic printable checklist, the course page for Edexcel IAL Chemistry plus a free diagnostic.

Use these relative atomic masses: N = 14.0, C = 12.0, Cl = 35.5, H = 1.0, O = 16.0.

Questions

1. This question is about naming nitrogen compounds.

(a) Name H₂NCH(CH₂CH₂CH₃)COOH. [1] (b) Name CH₃CH₂CH₂CONHCH₂CH₃. [1] (c) Give the structural formula of propylamine and classify it as a primary, secondary or tertiary amine. [2]

2. Explain the order of base strength of ammonia, pentylamine and phenylamine. [4]

3. This question is about butylamine, C₄H₉NH₂.

(a) Write an equation to show why aqueous butylamine is alkaline. [1] (b) Write an equation for the reaction of excess butylamine with ethanoyl chloride, and name the organic product containing the amide group. [2] (c) Butylamine is added a drop at a time, until in excess, to copper(II) sulfate solution. Describe what you would see. [2]

4. Pentylamine, CH₃(CH₂)₄NH₂, can be made in two ways.

(a) Give the reagent and conditions to make pentylamine from 1-chloropentane. [2] (b) Explain why one reagent is used in excess. [1] (c) Give the formula of one secondary amine that could also form in (a). [1] (d) Give the reagent needed to make pentylamine from a nitrile. [1]

5. 0.261 g of a primary aliphatic amine, RNH₂, reacts exactly with 30.00 cm³ of 0.100 mol dm⁻³ hydrochloric acid. Calculate the Mr of the amine and deduce its molecular formula. [4]

6. Phenylamine is made from nitrobenzene.

(a) Give the reagents and conditions for the reduction. [2] (b) Explain why sodium hydroxide solution is added before the phenylamine is separated. [1] (c) 6.15 g of nitrobenzene (Mr 123.0) gives 3.26 g of phenylamine (Mr 93.0). Calculate the percentage yield. [2]

7. A student makes an azo dye from phenylamine and phenol.

(a) Give the reagents and the temperature for converting phenylamine into benzenediazonium chloride. [2] (b) Write an equation for this reaction. [1] (c) State the conditions for the coupling step and give the structural formula of the dye. [2] (d) 2.79 g of phenylamine (Mr 93.0) gives 5.05 g of dye (Mr 198.0). Assuming a 1 : 1 ratio throughout, calculate the percentage yield. [2]

8. This question is about polymers in Topic 19.

(a) Draw the repeat unit of poly(propenamide). [1] (b) Draw the repeat unit of poly(ethenol). [1] (c) Nylon-6,10 is made from hexane-1,6-diamine and decanedioyl dichloride, ClOC(CH₂)₈COCl. Draw the repeat unit and name the other product. [2] (d) 2.32 g of hexane-1,6-diamine (Mr 116.0) reacts with 4.78 g of decanedioyl dichloride (Mr 239.0). Calculate the maximum mass of nylon-6,10 that can form (repeat unit Mr 282.0). [2]

9. (a) Explain why nylon fibres are strong. [2] (b) Liquid-detergent capsules have a poly(ethenol) film. Explain why the film is useful in a washing machine. [2]

10. This question is about 2-amino-3-methylbutanoic acid (valine), (CH₃)₂CHCH(NH₂)COOH.

(a) Draw the zwitterion. [1] (b) Give the structures of the main species in strongly acidic solution and in strongly alkaline solution. [2] (c) 0.597 g of an amino acid H₂NCH(R)COOH reacts exactly with 25.50 cm³ of 0.200 mol dm⁻³ hydrochloric acid (1 : 1). Calculate its Mr and show that it could be valine. [3] (d) Describe an experiment to show that a solution of naturally occurring valine is optically active, and give the expected result. [2]

11. (a) Glycine and serine, HOCH₂CH(NH₂)COOH, join to form a dipeptide. Give the structural formulae of the two possible dipeptides. [2] (b) Calculate the Mr of the tripeptide formed from glycine (Mr 75.0), alanine (Mr 89.0) and serine (Mr 105.0). [2]

12. Four unlabelled samples, A to D, are tested.

• A: colourless liquid, fishy smell; turns red litmus blue; with copper(II) sulfate solution gives a pale blue precipitate that dissolves in excess A to a deep blue solution.
• B: colourless liquid; fizzes with sodium hydrogencarbonate solution; the gas turns limewater milky.
• C: colourless liquid, C₄H₈O; yellow-orange precipitate with 2,4-DNPH; no change with Fehling’s solution; pale yellow precipitate with iodine in alkali.
• D: white solid; lilac flame; with dilute nitric acid then silver nitrate solution gives a cream precipitate.

(a) Identify the functional group in A and suggest why its solution is alkaline. [2] (b) Identify the functional group in B. [1] (c) Deduce the structure and name of C, using each observation. [3] (d) Identify D. [2]

Answers

1. (a) 2-aminopentanoic acid [1] (b) N-ethylbutanamide [1] (c) CH₃CH₂CH₂NH₂ [1]; primary amine [1]. Examiner insight: In N-substituted amides the N- locant is part of the name, so “ethylbutanamide” alone describes a different compound.

2. Bases accept a proton using the lone pair on nitrogen, so strength depends on how available that lone pair is [1]. Pentylamine is the strongest: the alkyl group releases electron density onto nitrogen (positive inductive effect) [1]. Phenylamine is the weakest: the nitrogen lone pair is delocalised into the π system of the benzene ring [1]. Order: phenylamine < ammonia < pentylamine [1]. Examiner insight: Each explanation must mention the lone pair; “the ring makes it weaker” alone is unlikely to score.

3. (a) C₄H₉NH₂(aq) + H₂O(l) ⇌ C₄H₉NH₃⁺(aq) + OH⁻(aq) [1] (b) CH₃COCl + 2C₄H₉NH₂ → CH₃CONHC₄H₉ + C₄H₉NH₃⁺Cl⁻ [1]; N-butylethanamide [1]. (c) A pale blue precipitate forms [1]; with more butylamine the precipitate disappears, leaving a deep blue solution [1]. Examiner insight: Give both stages and both colours; naming the complex ion is not an observation.

4. (a) Concentrated ammonia, in excess, dissolved in ethanol [1]; heated under pressure in a sealed container [1]. (b) Excess ammonia makes further substitution less likely [1]. (c) (C₅H₁₁)₂NH [1] (d) LiAlH₄ (dry ether solvent), or H₂ and nickel, to reduce pentanenitrile [1]. Examiner insight: “Ammonia” on its own is a weak answer here: the conditions (excess, ethanol, sealed tube) carry the credit.

5. n(HCl) = 0.100 × 30.00 ÷ 1000 = 0.00300 mol [1]; ratio 1 : 1 so n(RNH₂) = 0.00300 mol [1]; Mr = 0.261 ÷ 0.00300 = 87.0 [1]; CₙH₂ₙ₊₁NH₂: 14n + 17 = 87, so n = 5 and the formula is C₅H₁₃N [1]. Examiner insight: Show the general formula step, so the working can still earn credit if the final formula slips.

6. (a) Sn with concentrated HCl [1]; reflux [1]. (b) The product is the phenylammonium salt; NaOH removes the proton to release free phenylamine [1]. (c) n(nitrobenzene) = 6.15 ÷ 123.0 = 0.0500 mol, so theoretical mass = 0.0500 × 93.0 = 4.65 g [1]; yield = 3.26 ÷ 4.65 × 100 = 70.1% [1]. Examiner insight: Work through moles; dividing the two masses directly is a common slip.

7. (a) Sodium nitrite and hydrochloric acid (to make nitrous acid) [1]; below 10 °C [1]. (b) C₆H₅NH₂ + HCl + HNO₂ → C₆H₅N₂⁺Cl⁻ + 2H₂O [1] (c) Phenol dissolved in sodium hydroxide solution, kept cold [1]; C₆H₅N=NC₆H₄OH [1]. (d) n(phenylamine) = 2.79 ÷ 93.0 = 0.0300 mol, theoretical dye = 0.0300 × 198.0 = 5.94 g [1]; yield = 5.05 ÷ 5.94 × 100 = 85.0% [1]. Examiner insight: A temperature range such as “0–5 °C” or “below 10 °C” is safer than “cold”, which is vague.

8. (a) –[CH₂–CH(CONH₂)]– with extension bonds [1] (b) –[CH₂–CH(OH)]– with extension bonds [1] (c) –[NH(CH₂)₆NHCO(CH₂)₈CO]– [1]; hydrogen chloride [1]. (d) n(diamine) = 2.32 ÷ 116.0 = 0.0200 mol and n(dichloride) = 4.78 ÷ 239.0 = 0.0200 mol, so they react 1 : 1 with neither in excess [1]; mass = 0.0200 × 282.0 = 5.64 g [1]. Examiner insight: Check each amide link carefully: an –OH or –Cl left on the end of a repeat unit, or a C=C left in an addition repeat unit, loses the mark.

9. (a) Hydrogen bonds form between N–H on one chain and C=O on a neighbouring chain [1]; with London forces along the long chains, these hold the chains firmly together [1]. (b) Its many –OH groups hydrogen bond with water [1], so the film dissolves in the wash and releases the detergent [1]. Examiner insight: Say which atoms form the hydrogen bonds; “strong intermolecular forces” alone is too vague for both marks.

10. (a) (CH₃)₂CHCH(NH₃⁺)COO⁻ [1] (b) Acid: (CH₃)₂CHCH(NH₃⁺)COOH [1]; alkali: (CH₃)₂CHCH(NH₂)COO⁻ [1]. (c) n(HCl) = 0.200 × 25.50 ÷ 1000 = 0.00510 mol = n(amino acid) [1]; Mr = 0.597 ÷ 0.00510 = 117 [1]; valine, C₅H₁₁NO₂, has Mr = 60.0 + 11.0 + 14.0 + 32.0 = 117.0, which matches [1]. (d) Pass plane-polarised monochromatic light through the solution in a polarimeter [1]; the plane of polarisation is rotated [1]. Examiner insight: In (b) the charges must be on the correct groups; a positive charge on the carboxylic acid group earns nothing.

11. (a) H₂NCH₂CONHCH(CH₂OH)COOH [1]; H₂NCH(CH₂OH)CONHCH₂COOH [1]. (b) Two water molecules are lost when three amino acids join [1]; Mr = 75.0 + 89.0 + 105.0 − 2 × 18.0 = 233.0 [1]. Examiner insight: Count the peptide links, not the amino acids: a chain of three has two links, so two waters are removed.

12. (a) Amine (–NH₂) [1]; the nitrogen lone pair accepts a proton from water, forming OH⁻ ions [1]. (b) Carboxylic acid (–COOH) [1] (c) 2,4-DNPH shows a C=O group, and no change with Fehling’s means it is a ketone, not an aldehyde [1]; the iodoform result shows a CH₃CO– group [1]; with C₄H₈O this is CH₃COCH₂CH₃, butanone [1]. (d) Lilac flame shows potassium ions [1]; cream precipitate shows bromide ions, so D is potassium bromide, KBr [1]. Examiner insight: Link each deduction to its observation; a correct name with no reasoning earns little.

Where marks are usually lost

• Explaining base strength without the nitrogen lone pair.
• Using one mole of amine, not two, in the reaction with ethanoyl chloride.
• Giving “ammonia” without excess, ethanol and a sealed tube.
• Forgetting the NaOH step after tin and hydrochloric acid.
• Writing “cold” rather than a temperature for diazotisation.
• Repeat units with a C=C, or –OH or –Cl on the ends.
• Hydrogen bonding without naming the groups involved.
• Putting the charges on the wrong groups in amino acid ions.
• Removing the wrong number of water molecules from a peptide.
• Giving test results without a matching deduction.

Next steps

• Topic 19 revision notes
• Topic 19 study guide
• Edexcel International A Level Chemistry hub
• Printable Edexcel Chemistry checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017, Pearson Education Limited – Unit 5, Topic 19: Organic Nitrogen Compounds: Amines, Amides, Amino Acids and Proteins, outcomes 19.1 to 19.12 (Core Practical 15).