OCR A-Level Chemistry Unified chemistry (03), June 2019: Question 4

18 marks · Medium difficulty · Structured Questions

Analyze the structure, acidity, spectra, and reactions of chloroxylenol and alpha-terpineol found in Dettol.

Practise this question

Question

Question 4 regarding Dettol antiseptics. Part (a) displays the chemical structure of chloroxylenol (a benzene ring with an OH at position 1, two methyl groups at positions 3 and 5, and a chlorine atom at position 4) and asks for its systematic name, the number of peaks in its 13C NMR spectrum, the acidic functional group and a test for it, and a multi-step calculation of the acid dissociation constant Ka from given pH and percentage composition data. Part (b) shows the skeletal structure of alpha-terpineol (a cyclohexene ring with a methyl group on the double bond and a 2-hydroxypropan-2-yl group at position 4), asking candidates to identify chiral centres, explain E/Z isomerism constraints in cyclic systems, and provide specific reagents and reaction products for both functional groups.
Question text

4 Dettol® is a disinfectant containing the antiseptic chloroxylenol, shown below.

OH

H3C CH3

Cl

chloroxylenol

(a) Chloroxylenol is a weak Brønsted–Lowry acid.

(i) What is the systematic name of chloroxylenol?

… [1]

(ii) Predict the number of peaks in a 13C NMR spectrum of chloroxylenol.

… [1]

(iii) Name the functional group responsible for the acidity of chloroxylenol and describe a

simple test which would confirm the presence of this group.

Functional group …

Test …

… [2]

(iv) A student measures the pH of the contents in a bottle of Dettol® as 5.14.

The label on the bottle shows the percentage of chloroxylenol in Dettol® as 4.80%

i.e. 100 cm3 of Dettol® contains 4.80 g of chloroxylenol.

Assume the following:

• Chloroxylenol is the only acidic component in Dettol®.

• Chloroxylenol is a weak monobasic acid.

• The density of Dettol® is 1.00 g cm−3.

Write the equation, using molecular formulae, for the acid dissociation of chloroxylenol.

Calculate the acid dissociation constant, Ka, for chloroxylenol.

K = … mol dm−3 [5]

a

(b) Dettol® contains other chemicals including α-terpineol, shown below.

(i) α-Terpineol is a chiral compound.

Show with an asterisk, (*), the chiral centre(s) in the structure of α-terpineol.

OH

α-terpineol

[1]

(ii) α-Terpineol meets the requirements for E/Z isomerism.

However, only one E/Z isomer of α-terpineol exists.

Explain

• why α-terpineol meets the requirements for E/Z isomerism

• whether α-terpineol is an E- or Z- isomer

• why only one E/Z isomer of α-terpineol exists.

… [4]

(iii) α-Terpineol contains two functional groups.

For each functional group, choose a reagent that reacts with that group only.

Draw the structures for the organic products of the reactions.

Show structures for organic compounds.

Reagent(s) …

Name of functional group that reacts …

Structure of organic product

Reagent(s) …

Name of functional group that reacts …

Structure of organic product

[4]

Mark scheme

Show the mark scheme Mark scheme for Question 4. Marks allocated: (a)(i) 4-chloro-3,5-dimethylphenol (1 mark); (a)(ii) 5 peaks (1 mark); (a)(iii) Phenol group, test using pH indicator/carbonate or bromine water/FeCl3 (2 marks); (a)(iv) Dissociation equation C8H9ClO = H+ + C8H8ClO-, molar mass 156.5 g/mol, [chloroxylenol] = 0.3067 mol dm^-3, [H+] = 7.24 x 10^-6 mol dm^-3, Ka = 1.71 x 10^-10 mol dm^-3 (5 marks); (b)(i) Asterisk on the C4 ring carbon (1 mark); (b)(ii) Explanation of C=C double bond criteria, high-priority groups on same side (Z-isomer), and ring strain preventing the E-isomer (4 marks); (b)(iii) Reagent and product choices for alkene addition and tertiary alcohol substitution/elimination/esterification (4 marks). Total: 18 marks.

AO

Question Answer Marks Guidance

element

4 (a) (i) 4-chloro-3,5-dimethylphenol 1 AO1.2 ALLOW 3,5-dimethyl-4-chlorophenol

CARE: Look for dimethyl ALLOW absence of hyphens or extra hyphen or

space, e.g. 4 chloro 3,5 dimethylphenol

ALLOW full stops or spaces between numbers

e.g. 4-chloro-3.5-dimethylphenol

ALLOW name based on benzene, if unambiguous

e.g.1-chloro-4-hydroxy-2,6-dimethylbenzene

DO NOT ALLOW meth OR methy

(ii) 5 1 AO2.5

(iii) Functional group 2

Phenol AO1.2 DO NOT ALLOW alcohol OR hydroxide

IGNORE hydroxyl OR hydroxy

IGNORE OH (name asked for)

Test

Indicator/pH paper turns red / orange OR pH < 7

OR pH meter < 7 ALLOW Add bromine AND white precipitate

AND

No reaction with Na CO /CO 2–/carbonate AO2.3 ALLOW FeCl AND violet/blue colour

23 3 3

AO

12 element

(iv) FIRST, CHECK THE ANSWER ON ANSWER LINE 5

IF answer = 1.71 × 10–10,

award FOUR calculation marks

CARE Separate mark for equation

--------------------------------------------------------------------

Equation (1 mark)

C H ClO ⇌ H+ + C H ClO– AO1.2 ALLOW → for ⇌

89 8 8

Molecular formulae required (atoms in any order) ×1

DO NOT ALLOW C8H8ClOH in equation

i.e. C H ClOH ⇌ H+ + C H ClO–

88 8 8

If equation is omitted,

ALLOW equation mark for a correct Ka expression

with molecular formula

[H+] [C H ClO–]

i.e.

[C8H9ClO]

[C8H9ClO] calculation (2 marks)

Molar mass C H ClO = 156.5 (g mol–1) AO2.8

89 NO ECF from an incorrect formula in equation

ONLY correct answer ×4

ALLOW ECF from incorrect molar mass

4.8 × 10 –3 ALLOW 0.307 up to calculator value: 0.306709265

[C8H9ClO] = OR 0.3067.…. (mol dm )

156.5 correctly rounded

Subsumes mark for molar mass = 156.5

ALLOW 7.24 × 10–6 up to calculator value:

K calculation (2 marks) 7.244359601 × 10–6 correctly rounded

a

[H+] = 10–5.14 = 7.244….. × 10–6 (mol dm–3)

ALLOW 2 SF (1.7…. × 10–10) up to calculator value,

(7.244…. × 10–6)2 correctly rounded (but take care from acceptable

K = = 1.71 × 10–10 (mol dm–3)

a 0.3067….. intermediate rounding)

COMMON ERRORS

2.36….. × 10–5 3/4 calculation marks

H432/03 Mark Scheme No squaring of 7.24 × 10–6 June 2019

AO

element

(b) (i) 1 AO2.5

DO NOT ALLOW more than one *

ALLOW a circle for *

(ii) MAXIMUM OF 4 MARKS FROM 5 MARKING POINTS 4

Requirement for E/Z isomerism 2 marks

C=C/double bond AO1.2 IGNORE no H attached to C=C

×2 IGNORE functional’,

Each C (in C=C) is attached to (two) different

groups/atoms i.e. ALLOW different functional groups

Identification as E- or Z- isomer 2 marks

E/Z isomerism linked to (high) priority groups AO2.5 ALLOW in context of groups with largest atomic number

×2 ORA

Z- isomer AND groups are on same side Award BOTH identification marks for:

OR the ring carbons Z- isomer AND (high) priority groups on same side

Reason why other E/Z isomer does not exist 1 mark Mark independently of previous part

ring would be strained

Response MUST be linked to the ring/cyclic structure

OR ring would break/deform

OR Cannot form ring if high priority groups are on

IGNORE just ‘E isomer is impossible’

opposite sides

OR ring locks groups on one side of C=C bond

IGNORE C=C bond cannot rotate

IGNORE Groups can’t swap sides

AO

14 element

(iii) First group: 4 AO3.2 CONTACT TEAM LEADER FOR OTHER REACTIONS

Reagent ×4 --------------------------------------------------------------------

AND ALLOW GROUPS EITHER WAY ROUND IN BOXES

Functional group: Alkene OR cycloalkene

Functional group MUST be named

Examples of reagents

Br2 or other halogen, HBr, H2 AND Ni (catalyst), DO NOT ALLOW UV with halogens

+ ALLOW H SO /H PO /acid for H+

H2O(g)/steam AND H (catalyst) 2 4 3 4

Organic product for reagent with C=C in α-terpineol

ALLOW product from H or H O if H+ catalyst has been ALLOW addition of HBr/ H2O either way across C=C

omitted from reagent.

---------------------------------------

Second group

Reagent

AND

Functional group: (Tertiary) alcohol

Examples of reagents – – –

NaBr/KBr/Br– AND acid/H+ (substitution), ALLOW ANY HALIDE, i.e. Cl , Br , I

ALLOW H SO /H PO /acid for H+

OR HBr 2 4 3 4

ALLOW HBr for H+ and Br–

Acid/H+ (catalyst) (elimination),

CH COOH AND acid/H+ (catalyst) (esterification) ALLOW name or formula of any carboxylic acid or

CH3COOCOCH3 (esterification) acyl chloride for esterification

CH3COCl (esterification)

ALLOW Na → product with –ONa OR –O–

Organic product for reagent with OH in α-terpineol

ALLOW product if catalyst omitted from reagent 2– +

DO NOT ALLOW Cr2O7 /H (tertiary alcohol)

Total 18

How to answer it

Aromatic Chemistry, Acidity & Stereochemistry: Chloroxylenol & α-Terpineol

WHAT THIS QUESTION TESTS
  • Aromatic IUPAC Nomenclature & Carbon-13 NMR Symmetry: Numbering substituted benzenes and identifying planes of symmetry.
  • Phenol Acidity & Distinguishing Tests: Differentiating weak phenol acidity from carboxylic acids and primary/secondary/tertiary alcohols.
  • Weak Acid Quantitative Chemistry: Converting percentage mass/volume concentration into mol dm⁻³, calculating [H⁺] from pH, and determining Ka.
  • Stereoisomerism in Rings: Locating chiral carbon centres and applying CIP E/Z priority rules to cyclic alkenes with ring-strain constraints.
  • Selective Functional Group Reactions: Distinguishing chemical reactivity between cycloalkenes and tertiary alcohols.
PART (a)(i) - (a)(iii)

Structure, Symmetry & Functional Group Properties of Chloroxylenol

✅ Correct Answers

(i) Systematic Name [1 mark]:
4-chloro-3,5-dimethylphenol
(Also allowed: 3,5-dimethyl-4-chlorophenol)

(ii) Number of ¹³C NMR Peaks [1 mark]:
5 peaks

(iii) Functional Group & Test [2 marks]:
• Functional Group [1]: Phenol
• Simple Test [1]: Add indicator paper / pH meter to show pH < 7 AND add Na₂CO₃ (or carbonate) to show no effervescence / no reaction.
Alternative tests accepted: Add bromine water → forms a white precipitate; OR add neutral FeCl₃ → purple/violet solution formed.

💡 Key Knowledge

  • Numbering priority: The principal functional group (-OH) takes carbon 1. Symmetrical substitution gives positions 3 and 5 for methyl groups, leaving chlorine at position 4. Alphabetical ordering: 'c' before 'm' gives 4-chloro-3,5-dimethylphenol.
  • ¹³C Symmetry: Chloroxylenol possesses a vertical mirror plane running straight through C1 (-OH) and C4 (-Cl). Carbon environments:
    1. C-OH (C1)
    2. C-H (C2 & C6 equivalent)
    3. C-CH₃ (C3 & C5 equivalent)
    4. C-Cl (C4)
    5. -CH₃ carbons (both equivalent)
    Total = 5 peaks.
  • Phenol vs Alcohol Acidity: Phenols are weakly acidic (pH ~5–6), so they react with strong bases (NaOH) but are too weak to react with carbonates (no CO₂ bubbles). Regular aliphatic alcohols do not show acidic pH in aqueous solution.

🧠 Exam Technique

  • In (a)(iii), the prompt asks for the name of the functional group. Writing the formula " -OH " or "hydroxyl" scores 0. Always write phenol.
  • When using acidity to identify a phenol, stating only "pH < 7" could also imply a carboxylic acid. To be conclusive, the mark scheme specifically requires you to show it is acidic and does not react with Na₂CO₃.

❌ Common Errors

  • Writing "alcohol" or "hydroxide" in part (iii) — fatal error; -OH directly bonded to an aromatic ring is strictly a phenol.
  • Miscounting ¹³C peaks by treating the two methyl groups or the two meta carbons (C2 and C6) as non-equivalent, giving 6 or 8 peaks.
  • Forgetting "di" in nomenclature or writing incorrect hyphenation (e.g., 4-chlorodimethylphenol).
PART (a)(iv)

Quantitative Acid Dissociation Constant (Ka) Calculation

5 Marks: Molecular Equation + Concentration + [H⁺] + Ka

📐 Step-by-Step Calculation

  1. Step 1: Write the molecular dissociation equation [1 mark]
    Count atoms from structure: 8 Carbons, 9 Hydrogens, 1 Chlorine, 1 Oxygen.
    C₈H₉ClO ⇌ H⁺ + C₈H₈ClO⁻
    (Care: The question demands molecular formulae, not displayed or structural formulae).
  2. Step 2: Calculate molar mass of chloroxylenol [1 mark]
    M(C₈H₉ClO) = (8 × 12.0) + (9 × 1.0) + 35.5 + 16.0 = 156.5 g mol⁻¹
  3. Step 3: Calculate initial concentration [HA] [1 mark]
    Label states 4.80% w/v = 4.80 g in 100 cm³ of Dettol.
    Mass in 1000 cm³ (1 dm³) = 4.80 × 10 = 48.0 g
    [C₈H₉ClO] = 48.0 / 156.5 = 0.3067... mol dm⁻³
  4. Step 4: Calculate [H⁺] from pH [1 mark]
    [H⁺] = 10-pH = 10-5.14 = 7.24436 × 10⁻⁶ mol dm⁻³
  5. Step 5: Calculate Ka [1 mark]
    For a weak monobasic acid, assuming [H⁺] ≈ [A⁻] and negligible dissociation:
    Ka = [H⁺]² / [C₈H₉ClO]
    Ka = (7.24436 × 10⁻⁶)² / 0.3067... = 1.71 × 10⁻¹⁰ mol dm⁻³
    (Allow 2 SF: 1.7 × 10⁻¹⁰ mol dm⁻³)

❌ Common Calculation Traps

  • Forgetting to square [H⁺]: Doing 7.24 × 10⁻⁶ / 0.3067 = 2.36 × 10⁻⁵. Always check your expression: numerator is [H⁺]².
  • Writing condensed/structural formula in equation: Writing C₆H₂(CH₃)₂Cl(OH) was heavily penalised because the prompt specifically stated "using molecular formulae".
  • Incorrect units conversion: Forgetting that 4.80 g is in 100 cm³, not 1 dm³. Dividing 4.80 g directly by molar mass without multiplying by 10 leads to 0.0307 mol dm⁻³.
Mark Allocation Summary:
• 1 mark: Reversible molecular dissociation equation.
• 1 mark: Correct molar mass (156.5 g mol⁻¹).
• 1 mark: Calculating [HA] = 0.307 mol dm⁻³.
• 1 mark: Calculating [H⁺] = 7.24 × 10⁻⁶ mol dm⁻³.
• 1 mark: Final value for Ka = 1.71 × 10⁻¹⁰ mol dm⁻³.
PART (b)(i) & (b)(ii)

Stereochemistry of α-Terpineol: Chirality & E/Z Isomerism

✅ Part (b)(i) Chiral Centre [1 mark]

There is exactly one chiral centre in α-terpineol.

Position: Put an asterisk (*) on C4 of the cyclohexene ring (the ring carbon bonded directly to the bulky -C(CH₃)₂OH group).

Why? That carbon is bonded to 4 different groups: -H, the -C(CH₃)₂OH group, the -CH₂-CH= branch, and the -CH₂-CH₂- branch.

✅ Part (b)(ii) E/Z Explanation [4 marks max]

Requirements for E/Z isomerism [2 marks]:

  • Contains a C=C double bond (restricted rotation) [1].
  • Each carbon of the C=C double bond is attached to two different groups or atoms [1].

Identification of isomer [1 mark]:

  • It is the Z-isomer because the highest priority CIP groups on each double-bond carbon are on the same side (the ring carbons form the backbone on one side) [1].

Why only one isomer exists [1 mark]:

  • The ring would be too strained / cannot span across if the priority groups were on opposite sides (an E-isomer in a 6-membered ring would cause extreme ring strain/break the ring) [1].

🧠 CIP Priority Breakdown on the C=C Bond

  • Ring carbon with methyl (C1): Attached to -CH₃ (priority 2, C bonded to H,H,H) and ring carbon C2/C6 (priority 1, C bonded to C,H,H). Ring carbon has higher priority.
  • Ring carbon with H (C2): Attached to -H (priority 2) and ring carbon C3 (priority 1). Ring carbon has higher priority.
  • Both high priority ring groups are linked together on the same side of the double bond, making it the Z-isomer.

❌ Common Errors in Ring Stereochemistry

  • Marking the carbon with the -OH group as chiral: it has two identical -CH₃ groups attached to it, so it is achiral.
  • Vague phrasing: saying "different groups attached to double bond" rather than specifying that each carbon atom of the C=C must have two different groups.
  • Merely stating "E is impossible" without linking to ring strain / cyclic structure geometry.
PART (b)(iii)

Selective Functional Group Reactions of α-Terpineol [4 Marks]

Reaction targeting ONE group while leaving the other unaffected

Group 1: Alkene (C=C Double Bond)

Name of group: Alkene (or cycloalkene) [1]

Reagent options:

  • Br₂ (or Cl₂, I₂) in organic solvent / water
  • H₂ / Ni catalyst (electrophilic addition/reduction)
  • HBr or HCl
  • Steam / H⁺ (acid catalyst)

Structure of organic product [1]:

The C=C bond is saturated while the tertiary alcohol group -C(CH₃)₂OH remains completely intact! (e.g. addition of Br₂ across the C=C to form a 1,2-dibromo derivative).

Group 2: Alcohol (Tertiary -OH)

Name of group: Alcohol (or tertiary alcohol) [1]

Reagent options:

  • Substitution: NaBr + conc. H₂SO₄ (or HBr / conc. HCl)
  • Esterification: CH₃COOH + conc. H₂SO₄ or CH₃COCl
  • Acid-catalysed Elimination: Conc. H₂SO₄ / H₃PO₄ + heat (dehydration)
  • Metal reaction: Na (forms alkoxide)

Structure of organic product [1]:

The ring C=C remains unchanged, while the -OH group is replaced (e.g. converted to -C(CH₃)₂Br , -C(CH₃)₂Cl , or ester -C(CH₃)₂OCOCH₃ ).

❌ Major Reaction Traps to Avoid

  • Attempting oxidation with K₂Cr₂O₇ / H⁺: The alcohol is tertiary! Tertiary alcohols cannot be oxidised by acidified dichromate. Proposing this scores 0.
  • Using UV light with halogens: Adding UV light causes free radical substitution of C-H bonds throughout the ring rather than clean, selective functional group reaction.
Mark Allocation for Part (b)(iii):
• 1 mark: Reagent & correct name for alkene reaction.
• 1 mark: Correct structure of product with reacted C=C and unchanged tertiary alcohol.
• 1 mark: Reagent & correct name for alcohol reaction.
• 1 mark: Correct structure of product with reacted alcohol and intact ring C=C.

Topics

Module 4: Core organic chemistry · Module 5: Physical chemistry and transition elements · Module 6: Organic chemistry and analysis · 4.1 Basic concepts and hydrocarbons · 4.2 Alcohols, haloalkanes and analysis · 5.1 Rates, equilibrium and pH · 6.1 Aromatic compounds, carbonyls and acids · 6.2 Nitrogen compounds, polymers and synthesis · 6.3 Analysis

Question and mark scheme from the OCR A-Level Chemistry examination, Unified chemistry (03), June 2019. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.