Edexcel A-Level Chemistry Paper 2, June 2022: Question 7

14 marks · Hard difficulty · Extended Writing

Assess the preparation methods, infrared spectra, optical activity, acid-base reactions, and neutralization volumes of various nitrogen-containing organic compounds including amines, amides, and amino acids.

Practise this question

Question

Question text

7 Organic compounds containing nitrogen include amides, amines, amino acids

and nitriles.

(a) Propylamine, CH3CH2CH2NH2, may be formed from either a nitrile or a

halogenoalkane.

(i) Give the reagent and essential condition for the formation of propylamine

from a nitrile.

Include an equation for the reaction.

(2)

(ii) Give the reagent and essential conditions for the formation of propylamine

from a halogenoalkane.

Include an equation for the reaction.

(3)

(b) A compound produced a peak due to an N–H stretching vibration in its infrared

spectrum with a wavenumber of 3220 cm–1.

This compound could be

(1)

A an amide

B an amine

C either an amide or an amine

D neither an amide nor an amine

14*(c) Alanine and glycine are amino acids.

*P67094RA01424*

Amino acid Structure

H CH3 O

alanine N C C

H H O H

H H O

glycine N C C

H H O H

Compare and contrast the structures, optical activity and reactions with acids and

bases of alanine and glycine.

Include diagrams, structures and equations to illustrate your answer.

(6)

… *P67094RA01524*

(d) Lysine and serine are two more amino acids.*P67094RA01624*

Amino acid Structure of amino acid

NH2

H (CH2)4 O

lysine

N C C

H H O H

H CH2 O

serine

N C C

H H O H

Explain the difference in the volumes of 0.010 mol dm–3 hydrochloric acid required

to completely react with separate 10.0 cm3 samples of aqueous lysine and of

aqueous serine, both of concentration 0.010 mol dm–3.

(2)

(Total for Question 7 = 14 marks)

Mark scheme

Show the mark scheme

Question

Answer Additional Guidance Mark

Number

7(a)(i) A description which includes Example of equation (2)

• equation (1) CH3CH2CN + 4[H] → CH3CH2CH2NH2

CH3CH2CN + 2H2 → CH3CH2CH2NH2

• LiAlH4 in (dry) ether Allow names or formulae but both must be correct if given together

(followed by dilute acid) Allow Lithal

or Allow hydrogen to be given in the equation or written over the arrow

H2 with Ni / Pt / Pd (1)

Ignore references to heat or a temperature

Question

Answer Additional Guidance Mark

Number

7(a)(ii) A description which includes Example of equation (3)

• equation from any CH3CH2CH2Br + NH3 → CH3CH2CH2NH2 + HBr

halogenoalkane (1) or

CH3CH2CH2Br + 2NH3 → CH3CH2CH2NH2 + NH4Br

• ethanolic/alcoholic Allow use of state symbol (alc)/(EtOH)/(eth) with NH3

ammonia (1) Allow ammonia to be given in equation or written over the arrow

• heat and under Accept heat and in a sealed tube

pressure (1) Ignore mechanisms

If a contradictory chemical is stated then penalise once against M2 or M3

Question

Answer Mark

Number

7(b) The only correct answer is A (an amide) (1)

B is not correct because the amine range does not include 3220 cm−1

C is not correct because the amine range does not include 3220 cm−1

D is not correct because the amide range does include 3220 cm−1

Question

Answer Additional Guidance Mark

Number

7(c) Guidance on how the mark scheme should be (6)

This question assesses the student’s ability to

applied:

show a coherent and logically structured answer

The mark for indicative content should be added

with linkages and fully sustained reasoning.

to the mark for lines of reasoning. For example, a

response with four indicative marking points that

Marks are awarded for indicative content and for

is partially structured with some linkages and

how the answer is structured and shows lines of

lines of reasoning scores 4 marks (3 marks for

reasoning.

indicative content and 1 mark for partial structure

and some linkages and lines of reasoning).

The following table shows how the marks should

If there were no linkages between the points,

be awarded for indicative content.

then the same indicative marking points would

Number of indicative Number of marks

yield and overall score of 3 marks (3 marks for

marking points seen in awarded for indicative

indicative content and zero marks for linkages).

answer marking points

5-4 3

3-2 2

The following table shows how the marks should

be awarded for structure and lines of reasoning

Number of

marks

awarded for

structure of

answer and

sustained lines

of reasoning

Answer shows a 2

coherent logical

structure with More than one indicative marking point may be

linkages and fully made within the same comment or explanation

sustained lines of

reasoning Accept annotated diagrams to illustrate the

demonstrated indicative points

throughout

Answer is partially 1

structured with

some linkages and

lines of reasoning

Answer has no 0

linkages between

points and is

unstructured

Ignore reference to other amino acid properties

Indicative content

IP1 (Similarity) The zwitterions can be evidenced from each

• they are both amino acid zwitterion in an equation

2-amino acids / alpha amino acids / e.g. NH +CH(CH )COO− / NH +CH COO−

33 3 2

naturally occurring/ zwitterions

IP2

• equation for the reaction with an acid e.g. H+ + NH +CH COO− → NH +CH COOH or

32 3 2

H++NH +CH(CH )COO−→ H N+CH(CH )COOH

33 3 3

IP3

• equation for the reaction with a base OH−+NH +CH COO−→NH CH COO−+H O or

32 2 2 2

OH−+NH +CH(CH )COO−→NH CH(CH )COO−+H O

33 2 3 2

Allow use of un-ionised amino acid structures

If IP2 and 3 not scored then allow 1IP for a

suitable description of acid and base behaviour

IP4

• alanine has a chiral centre/ asymmetric Allow reference to four different atoms/groups

carbon atom/ non-superimposable mirror bonded to central carbon for chiral centre

images

and

glycine does not ‘Plane’ must be stated at least once

IP5

• (an aqueous solution of) alanine rotates the Wedges must be drawn

plane (of polarisation) of plane-polarised e.g.

(monochromatic) light but glycine does not Ignore angles and

IP6 connectivity

• diagram to show enantiomers of alanine

Question

Answer Additional Guidance Mark

Number

7(d) An explanation which includes (2)

• lysine requires twice (the volume of HCl) (1) Allow lysine requires 20.0 cm3 and serine

requires 10 cm3

• (because) lysine has two (basic) amine/NH2 groups Allow lysine has one more (basic) /

whereas serine has one (1) another amine/ NH2 group

Allow lysine can accept two protons

whereas serine can only accept one

(Total Question 7 = 14 marks)

How to answer it

Nitrogen-Containing Organic Compounds Study Guide

📌 What this question tests

This question assesses your knowledge of the synthesis routes for amines (reduction of nitriles and nucleophilic substitution of halogenoalkanes), interpretation of infrared spectroscopy data for characteristic bonds, comparative properties of amino acids (similarities, acid/base reactions, and optical isomerism), and stoichiometric calculations involving basic functional groups.

Part (a)(i): Formation of Propylamine from a Nitrile

Edexcel A-Level Chemistry Synthesis

✅ Correct Answer

Reagent: Lithium aluminium hydride ( LiAlH₄ ) in dry ether, OR hydrogen ( H₂ ) with a nickel/platinum/palladium catalyst.
Condition: Dry ether (for LiAlH₄ ).
Equation: CH₃CH₂CN + 4[H] → CH₃CH₂CH₂NH₂ (or using 2H₂ ).

💡 Key Knowledge

Nitriles contain a polar C≡N triple bond. Reduction adds hydrogen across this multiple bond to produce primary amines ( R-CH₂NH₂ ). LiAlH₄ is a powerful reducing agent that must be used in a dry, aprotic solvent like ether because it reacts violently with water.

❌ Common Errors

Students frequently write NaBH₄ instead of LiAlH₄ . Remember that NaBH₄ is not powerful enough to reduce nitriles; it only reduces aldehydes and ketones.

Part (a)(ii): Formation of Propylamine from a Halogenoalkane

Nucleophilic Substitution

✅ Correct Answer

Reagent: Excess ethanolic ammonia ( NH₃ in ethanol).
Conditions: Heat under pressure (sealed tube).
Equation: CH₃CH₂CH₂Br + NH₃ → CH₃CH₂CH₂NH₂ + HBr (or forming an ammonium salt + NH₄Br using 2 moles of NH₃ ).

🧠 Exam Technique

You must explicitly state ethanolic ammonia rather than aqueous, and include pressure or sealed tube. Mentioning "excess ammonia" is vital to prevent further substitution reactions where primary amines attack unreacted halogenoalkanes to form secondary and tertiary amines.

Part (b): Infrared Spectroscopy

Identifying Characteristic Bonds

✅ Correct Answer

Option A: an amide

💡 Key Knowledge

The absorption peak at 3220 cm⁻¹ corresponds to an N-H stretching vibration. While both amines and amides feature N-H bonds, the exact wavenumber range for secondary/primary amides typically falls lower or overlaps differently compared to standard aliphatic amines depending on hydrogen bonding. Check your data booklet: amine N-H stretches sit sharply around 3300–3500 cm⁻¹ , whereas hydrogen-bonded amide N-H absorptions shift towards lower wavenumbers (approx 3200 cm⁻¹ ).

Part (c): Comparing Alanine and Glycine

6-Mark Extended Response (Levels-of-Response)

✅ Indicative Content (Mark Scheme Points)

IP1 (Similarity): Both are alpha-amino acids / naturally occurring / exist as zwitterions.
IP2 (Acid Reaction): Equation showing reaction with acid (e.g., H⁺ + NH₃⁺CH(R)COO⁻ → NH₃⁺CH(R)COOH ).
IP3 (Base Reaction): Equation showing reaction with base (e.g., OH⁻ + NH₃⁺CH(R)COO⁻ → H₂NCHRCOO⁻ + H₂O ).
IP4 (Chirality - Alanine): Alanine has a chiral centre (asymmetric carbon with 4 different groups attached) and is optically active / has non-superimposable mirror images. Glycine does not ( R = H , two identical hydrogens on the alpha carbon).
IP5 (Optical Activity): Aqueous alanine rotates the plane of plane-polarised light; glycine does not.
IP6 (Diagram): 3D tetrahedral representation (using wedges and dashes) showing the two enantiomers of alanine.

🧠 Structuring Top-Level Responses

To secure all 4 content marks plus the 2 structure/logic marks, group your answer logically: first discuss structural similarities (zwitterions, functional groups, acid/base behavior), then shift clearly to structural differences (chirality and optical activity). Ensure chemical equations balance correctly with charges accounted for.

Part (d): Stoichiometry of Amino Acids with Acid

Calculations & Functional Group Analysis

✅ Correct Answer

Lysine requires twice the volume of HCl compared to serine (i.e., 20.0 cm³ vs 10.0 cm³ ).
Reason: Lysine has two basic amino ( -NH₂ ) groups, whereas serine has only one.

📐 Calculation & Stoichiometry Breakdown

Examine structures: Look at the given carbon skeletons. Serine contains one -NH₂ group and one -COOH group. Lysine contains an -(CH₂)₄- chain ending in a second -NH₂ group (giving two basic sites total).
Mole ratios: Hydrochloric acid ( HCl ) reacts with the basic amine groups. A 1:1 molar ratio reacts per basic group.
Determine volumes: Since both amino acid samples share identical concentration ( 0.010 mol dm⁻³ ) and volume ( 10.0 cm³ ), serine needs 10.0 cm³ of 0.010 mol dm⁻³ HCl , while lysine neutralises twice as much, demanding 20.0 cm³ .

❌ Common Calculation Traps

Do not get distracted by the -OH alcohol group on serine's side chain; aliphatic alcohol groups do not react with dilute hydrochloric acid under standard titration conditions.

Topics

Organic Chemistry · Topic 7: Modern Analytical Techniques I · Topic 17: Organic Chemistry II · Topic 18: Organic Chemistry III

Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 2, June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.