AQA A-Level Chemistry Paper 2, 2018: Question 10

10 marks · Medium difficulty · State/Explain/Describe

Identify compounds from infrared spectra, draw skeletal isomers of secondary amines, compare amine synthesis routes, and give equations and explanations for reactions of amines and phenylamine.

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Question text

10 This question is about amines.

10.1 The infrared spectra A, B and C are those of a primary amine, a tertiary amine and a

nitrile, but not necessarily in that order.

Give the letter of each compound in the correct box.

[1 mark]

primary amine tertiary amine nitrile

10.2 There are three secondary amines that contain four carbon atoms per molecule.

*24* Draw the skeletal formulas of these three secondary amines.

[2 marks]

10.3 Primary amines can be prepared by the reaction of halogenoalkanes with ammonia or

by the reduction of nitriles.

Justify the statement that it is better to prepare primary amines from nitriles rather

than from halogenoalkanes.

[2 marks]

10.4 Draw the structure of a primary amine with four carbon atoms that cannot be

formed from a nitrile.

26 [1 mark]

A student dissolves a few drops of propylamine in 1 cm3 of water in a test tube.

10.5

Give an equation for the reaction that occurs.

Describe what is observed when Universal Indicator is added to this solution.

[2 marks]

Equation

Observation

10.6 Phenylamine can be prepared by a process involving the reduction of nitrobenzene

using tin and an excess of hydrochloric acid.

Give an equation for the reduction of nitrobenzene to form phenylamine. Use [H] to

represent the reducing agent.

Explain why an aqueous solution is obtained in this reduction even though

phenylamine is insoluble in water.

[2 marks]

Equation

Explanation

Mark scheme

Show the mark scheme

Question Answers Mark Additional Comments/Guidance

A 10.1 C B A this order only 1

Must be skeletal – allow with or without H on N

Any three from

All 3 correct score 2 (or one if not skeletal)

or H Any two correct score 1 (or zero if not skeletal)

2 Allow cyclic IIo amines but NOT amines also

containing other functional groups

OR H

10.2 N H N N

N H

OR H

N H

N H N H

– – –

With halogenoalkane:

1 Ignore bi-product / yield

further reaction (of primary amines)

OR 25

Impure product/mixture of products/lower atom economy

10.3

With nitriles

No further reaction 1

Single product / higher atom economy

Allow cyclic Io amines but NOT amines also

1 containing other functional groups

or NH2

or H3C

NH2

10.4

NH2

NH2 CH3

or NH2

CH CH CH NH + H O ⇌ CH CH CH NH + + OH– 1 Allow simple arrow

32 2 2 2 3 2 2 3

10.5 Not C3H7

(green) turns blue 1

–Allow blue-green, blue-purple– –

C6H5NO2 + 6[H] C6H5NH2 + 2H2O 1 Not H2

26 Not molecular formulae

NO2 + 6[H] NH2 + 2H2O

10.6

+ – 1

C6H5NH2 present as ionic salt OR C6H5NH3 (Cl ) OR phenyl Allow present as an ion

ammonium (chloride) But not phenylammonium hydroxide

Total 10

How to answer it

Amines: Spectroscopy, Synthesis, and Basic Properties

📋 What this question tests

This 10-mark question evaluates core organic chemistry across Year 2 topics:

Infrared Spectroscopy: Identifying distinguishing absorption bands for nitriles (C≡N), primary amines (N–H doublet), and tertiary amines (no N–H bond).
Structural Isomerism & Representation: Drawing skeletal formulas of secondary amines with a specific molecular formula (C₄H₁₁N).
Synthetic Pathways: Evaluating halogenoalkane substitution vs. nitrile reduction, and deducing structural constraints imposed by reduction mechanisms.
Acid-Base Behaviour: Writing weak-base ionisation equilibria in water and predicting pH observations.
Aromatic Chemistry: Formulating reduction equations for nitrobenzene and accounting for the water solubility of phenylammonium salts.

Question 10.1: Identifying Compounds from Infrared Spectra

1 Mark

✅ Correct Answer

Primary Amine	Tertiary Amine	Nitrile
C	B	A

Order required: C, B, A

💡 Key Knowledge

Nitrile (A): Characteristic sharp, distinct peak at ~2220–2260 cm⁻¹ corresponding to the C≡N triple bond.
Primary Amine (C): Characteristic broad N–H stretch at 3300–3500 cm⁻¹ appearing as a twin peak ("fork") due to symmetric and asymmetric stretching of the –NH₂ group.
Tertiary Amine (B): Has nitrogen bonded to three alkyl groups (R₃N). Lacks any N–H bonds, so there is no absorption in the 3300–3500 cm⁻¹ region. Only C–H stretching (~2950 cm⁻¹) is observed.

❌ Common Errors

Confusing primary and tertiary amines: students often forget that tertiary amines have no hydrogen atoms directly bonded to nitrogen, meaning they produce zero N–H absorption above 3000 cm⁻¹.

[1 mark] All three letters must be placed correctly in order (C, B, A).

Question 10.2: Skeletal Formulas of C₄ Secondary Amines

2 Marks

✅ Correct Answers (Any 3)

Diethylamine: Two-carbon chain on each side of N.
Skeletal: A "V" shape connected to N, connected to an inverted "V": /\–NH–/\ or simply /\–N–/\
N-methylpropylamine: A single methyl carbon on one side and a 3-carbon propyl chain on the other.
Skeletal: –NH–/\_
N-methylpropan-2-amine: A methyl group on one side and an isopropyl (branched) group on the other.
Skeletal: –NH–CH(\)– (branched Y-shape off the nitrogen)

Note: The H atom on nitrogen may be shown explicitly (–NH–) or omitted (–N–) in skeletal representations. Cyclic secondary amines containing 4 carbons (e.g. pyrrolidine) are also accepted.

🧠 Exam Technique

Definition check: A secondary amine has exactly two carbon groups attached to the nitrogen atom (R–NH–R').
Count carbons carefully: Ensure every drawn structure has exactly four carbon vertices/ends.
Strict skeletal rules: Do not write out carbon chains as CH₃–CH₂– . If drawn as structural or displayed formulas instead of skeletal, you lose 1 mark even if all structures are correct!

[2 marks] All 3 correct skeletal structures = 2 marks (1 mark if drawn non-skeletal). Any 2 correct skeletal structures = 1 mark.

Question 10.3: Preparation from Nitriles vs. Halogenoalkanes

2 Marks

✅ Mark Scheme Breakdown

Point 1 (Halogenoalkane): Reaction with halogenoalkane leads to further reaction (nucleophilic substitution) of the primary amine product / produces a mixture of products (secondary/tertiary amines, quaternary salt) / has a lower atom economy. [1 mark]
Point 2 (Nitrile): Reduction of nitrile gives no further reaction / produces a single product / has a higher atom economy. [1 mark]

🧠 Why Nitrile Reduction is Superior

When ammonia attacks a halogenoalkane, the primary amine formed is itself a nucleophile (often even more nucleophilic than ammonia due to the positive inductive effect of the alkyl group). It attacks unreacted halogenoalkane, inevitably producing secondary and tertiary amines. In contrast, catalytic hydrogenation of a nitrile specifically reduces the –C≡N bond to –CH₂NH₂ with no alkylating agent present to cause further substitution.

❌ Common Errors

Stating vague answers like "gives a better yield" or "fewer by-products" without specifying that halogenoalkanes give a mixture of amine products or undergo further reaction. The mark scheme explicitly states: Ignore bi-product / yield.

[2 marks] 1 mark for the drawback of halogenoalkane route; 1 mark for the advantage of nitrile reduction.

Question 10.4: C₄ Primary Amine That Cannot Be Formed From a Nitrile

1 Mark

✅ Acceptable Structures

Either of the following primary amines where the –NH₂ group is bonded to a secondary or tertiary carbon:

Butan-2-amine (sec-butylamine):
CH₃CH(NH₂)CH₂CH₃ or CH₃–CH(NH₂)–CH₂CH₃
2-Methylpropan-2-amine (tert-butylamine):
(CH₃)₃CNH₂ or (CH₃)₃C–NH₂

(Displayed, structural, or skeletal forms are all accepted).

💡 The Chemical Logic

Reducing a nitrile always converts the functional group as follows:

R–C≡N + 4[H] → R–CH₂–NH₂

Notice that the carbon attached to the –NH₂ group must always have at least two hydrogen atoms attached (it must be a –CH₂–NH₂ group). Therefore, any primary amine where the –NH₂ is on a –CH– (secondary carbon) or –C– (tertiary carbon) can never be made by reduction of a nitrile!

[1 mark] Correct structure of butan-2-amine or 2-methylpropan-2-amine.

Question 10.5: Propylamine in Water & Indicator Observation

2 Marks

✅ Correct Answer

Equation:

CH₃CH₂CH₂NH₂ + H₂O ⇌ CH₃CH₂CH₂NH₃⁺ + OH⁻

(A single forward arrow → is also accepted).

Observation:

Turns blue (also allow blue-green or blue-purple; or green turns blue).

❌ Common Errors

Molecular formula penalised: Writing C₃H₇NH₂ is not accepted by the mark scheme because it doesn't clearly show propylamine (could be 1-methylethylamine). Write CH₃CH₂CH₂NH₂ .
Missing charge: Forgetting the positive charge on the propylammonium ion ( CH₃CH₂CH₂NH₃⁺ ) or the negative charge on the hydroxide ion ( OH⁻ ).
Incorrect indicator colour: Stating purple alone without blue, or stating red (confusing base with acid).

[2 marks] 1 mark for balanced equation with correct ionic products; 1 mark for the alkaline colour observation (blue).

Question 10.6: Reduction of Nitrobenzene & Aqueous Solubility

2 Marks

✅ Correct Answers

Equation:

C₆H₅NO₂ + 6[H] → C₆H₅NH₂ + 2H₂O

(Can also be shown using skeletal/aromatic ring diagrams).

Explanation:

Phenylamine is present as an ionic salt / phenylammonium ion / phenylammonium chloride / C₆H₅NH₃⁺ (Cl⁻) .

💡 Why the Product is Soluble

The reduction takes place in the presence of excess concentrated hydrochloric acid (HCl). Phenylamine has a lone pair on nitrogen and acts as a Brønsted-Lowry base, reacting immediately with the acid:

C₆H₅NH₂ + HCl → C₆H₅NH₃⁺Cl⁻

Because the product exists as an ionic compound (phenylammonium chloride), it readily dissolves in water via strong ion-dipole attractions, even though molecular phenylamine itself is insoluble due to the large non-polar benzene ring.

❌ Common Errors

Writing H₂ instead of [H] in the reduction equation. The question explicitly states: "Use [H] to represent the reducing agent." Writing 3H₂ scores 0 marks.
Forgetting that 2 molecules of water are formed as a by-product, needing a total of 6[H].
Writing molecular formula ( C₆H₇N ) instead of structural formula ( C₆H₅NH₂ ).
Claiming the product is soluble as phenylammonium hydroxide (which makes no sense in acidic HCl medium!).

[2 marks] 1 mark for balanced equation using 6[H]; 1 mark for identifying the salt/ion formed ( C₆H₅NH₃⁺ ).

Topics

Organic Chemistry · 3.3.1 Introduction to Organic Chemistry · 3.3.3 Halogenoalkanes · 3.3.6 Organic Analysis · 3.3.10 Aromatic Chemistry · 3.3.11 Amines

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 2, 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.