OCR A-Level Chemistry Synthesis and analytical techniques (02), June 2018: Question 19

10 marks · Hard difficulty · Structured Questions

State and explain the effect of the halogen on the rate of haloalkane hydrolysis, outline the mechanism for the hydrolysis of chlorocyclohexane, draw a reflux apparatus diagram, and identify unknown haloalkane E, product F, and precipitate G from analytical data.

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Question

Question text

19 This question is about the hydrolysis of haloalkanes.

(a) The rate of hydrolysis of a haloalkane depends on the halogen present.

State and explain how the halogen in the haloalkane affects the rate of hydrolysis.

… [2]

(b) Chlorocyclohexane is hydrolysed with aqueous sodium hydroxide.

Outline the mechanism for this reaction.

Show curly arrows, relevant dipoles and the products.

[3]

• 0.0100 mol of haloalkane E is refluxed with excess NaOH(aq) to form a reaction mixture

containing an organic product F.

• The reaction mixture is neutralised with dilute nitric acid.

• Excess AgNO3(aq) is added to the reaction mixture. 1.88 g of a precipitate G forms.

Organic product, F, has a molar mass of 74.0 g mol−1 and has a chiral carbon atom.

(i) Draw a labelled diagram to show how the student would carry out the hydrolysis of

haloalkane E.

[2]

(ii) Analyse the information to identify E, F and G.

Show your working.

[3]

Mark scheme

Show the mark scheme

Question Answer Marks Guidance

19 (a) 2

Links rate of reaction to strength of bond/bond enthalpy Each marking point must be a comparison

e.g.

the weaker the bond the faster the reaction

stronger bond takes longer to break

lower bond enthalpy reacts faster

IGNORE references to halogens as elements:

i.e. chlorine is less reactive than bromine etc.

Correct comparison of rate of reaction for at least two C–

Hal bonds

e.g.

C–F bond is hydrolysed slowest

C–I bond is hydrolysed faster than C–Br DO NOT ALLOW chloride, bromide and iodide

C–Br has shorter reaction time than C–Cl

Correct comparison of C–Hal bond strength/enthalpy of IGNORE references to bond length, polarity and

at least two of C–Hal bonds electronegativity

e.g.

C–I bond is the weakest

C–I has lower bond enthalpy than C–Br

C–Br is broken more easily/readily than C–Cl

C–Hal bond strength decreases down group (7)

Question Answer 17 Marks Guidance

(b) 3 ANNOTATE ANSWER TICKS AND CROSSES

NOTE: curly arrows can be straight, snake-like, etc.

Curly arrow from HO– to carbon atom of C−Cl bond but NOT double headed or half headed arrows

Dipole shown on C–Cl bond, C + and Cl – 1st curly arrow must

AND go to the C of C–Cl

curly arrow from C−Cl bond to Cl atom AND

start from, OR be traced back to any point across

width of lone pair on O of OH–

+ -

- OR start from – charge on O of –OH ion

IGNORE presence of Na+ but OH– needed

i.e. Na+OH–can be allowed if criteria met

------------------------------------------------------------------------- –

(Lone pair NOT needed if curly arrow shown from O )

Correct organic product AND Cl–

2nd curly arrow must start from, OR be traced back to,

any part of C–Cl bond and go to Cl

OH -

+ Cl

IGNORE presence of Na+ but Cl– needed

i.e. Na+Cl–can be allowed

BUT NaCl does NOT show Cl–

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

ALLOW SN1 mechanism

First mark

Dipole shown on C–Cl bond, C + and Cl −,

AND curly arrow from C−Cl bond to Cl atom

+ -

Cl + + Cl-

Second mark

Correct carbocation AND curly arrow from HO– to

carbocation

+ OH

Curly arrow must come from lone pair on O of HO– OR

OH–

OR from minus on O of HO– ion (no need to show lone

pair if curly came from negative charge)

Third mark

Correct organic product AND Cl–

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

Diagram showing round bottom/pear shaped flask AND

upright condenser

DO NOT ALLOW conical flask, volumetric flask, beaker

in place of round bottom/pear shaped flask

DO NOT ALLOW distillation

DO NOT ALLOW stopper/bung on top of condenser

IGNORE a thermometer in condenser

IGNORE a small gap between flask and condenser

Labels

(Round-bottom/pear-shaped) flask

AND condenser

AND water in at bottom and out at top

AND heat (source) ALLOW diagram of heating apparatus as an alternative

to heat label

Precipitate G 1 mark displayed formula as long as unambiguous

silver bromide/AgBr

AND Note: working is required for first mark

M = 1.88/0.01 = 188 (g mol–1)

188 – 107.9 = 80.1 (so halide is Br–) ALLOW use of 108 as A of Ag

Alcohol F and Haloalkane E 2 marks

E and F clearly identified Note: E and F can be identified by correct name or

structure BUT IGNORE incorrect names

F/alcohol: butan-2-ol

H OH

H3C C C CH3

H H

E/haloalkane:

E is haloalkane of C4H9X with

same halogen as G

AND

same carbon chain as F

Total 10

How to answer it

Chemistry Study Guide: Hydrolysis of Haloalkanes

What this question tests

This question assesses your understanding of nucleophilic substitution reactions of haloalkanes, carbon-halogen bond enthalpies, reaction mechanisms using curly arrows and dipoles, practical organic chemistry setup (refluxing apparatus), and quantitative analytical chemistry involving molar mass and stoichiometry calculations.

Question Part (a)

Bond Enthalpy and Rates of Hydrolysis

✅ Correct Answer

Rates of hydrolysis depend directly on carbon-halogen bond enthalpy. As you go down Group 7, the C-Hal bond becomes weaker (lower bond enthalpy), meaning less energy is required to break it. Therefore, iodoalkanes hydrolyse the fastest and fluoroalkanes the slowest.

💡 Key Knowledge

Bond enthalpy decreases down Group 7 (C-F > C-Cl > C-Br > C-I).
Comparisons must be explicit (e.g., C-I bond is weaker and hydrolysed faster than C-Cl).
Polarity and electronegativity are distractors here; rate is controlled entirely by bond strength, not bond polarity.

❌ Common Errors

Students frequently lose marks by discussing electronegativity or dipole moments instead of bond enthalpy. Examiners also penalize loose answers that do not explicitly compare at least two specific halogens.

🎯 Marks available: 2 marks. Awarded for linking rate to bond strength/enthalpy and making a correct comparison of at least two specific carbon-halogen bonds.

Question Part (b)

Mechanism of Hydrolysis (Chlorocyclohexane)

✅ Correct Answer

An SN2 mechanism (or concerted pathway) featuring:

Partial positive charge ( δ+ ) on the carbon bonded to Cl, and partial negative ( δ- ) on Cl.
A curly arrow starting from the lone pair on the oxygen of the hydroxide ion ( :OH⁻ ) to the δ+ carbon.
A curly arrow from the C-Cl bond going directly to the chlorine atom.
Organic product shown as cyclohexanol plus a chloride ion ( Cl⁻ ).

🧠 Exam Technique

Ensure your first curly arrow originates clearly from a lone pair on the oxygen atom of the hydroxide ion, not from the negative charge floating in space. The second curly arrow must point straight at the Cl atom.

❌ Common Errors

Writing NaCl instead of free Cl⁻ as a product will lose you the product mark. Double-headed or half-headed arrows used incorrectly will also be rejected.

🎯 Marks available: 3 marks. Breakdown: (1) Dipole and first curly arrow from OH⁻, (2) Second curly arrow from C-Cl bond to Cl, (3) Correct organic product (cyclohexanol) and Cl⁻ ion.

Question Part (c)(i)

Practical Technique: Reflux Apparatus

✅ Correct Answer

A properly drawn and fully labelled reflux setup featuring:

A round-bottomed or pear-shaped flask containing the reaction mixture.
An upright vertical Liebig condenser connected directly to the flask.
Clear water jackets with water entering at the bottom and leaving at the top.
A heat source shown underneath the flask (no stopper or bung at the top of the condenser!).

🧠 Exam Technique

Never seal a reflux apparatus with a bung or stopper, as this creates a dangerous closed system susceptible to explosion. Keep the top of the condenser completely open.

🎯 Marks available: 2 marks. Awarded for correct apparatus drawing (pear/round-bottom flask with upright condenser) and complete clear labels (flask, condenser, water in/out, heat).

Question Part (c)(ii)

Quantitative Analysis & Structure Identification

📐 Step-by-Step Calculation & Deduction

Find moles of precipitate G:
Mass of precipitate G = 1.88 g. Given 0.0100 mol of haloalkane E is used, and haloalkanes react in a 1:1 ratio with silver nitrate to form 1 mol of silver halide precipitate:
Molar mass of G = Mass / Moles = 1.88 g / 0.0100 mol = 188.0 g mol⁻¹.
Identify Precipitate G:
Molar mass of Ag = 107.9 g mol⁻¹. Halide mass = 188.0 - 107.9 = 80.1 g mol⁻¹. This corresponds to bromide ( Br⁻ ), identifying G as silver bromide (AgBr) and confirming haloalkane E contains a bromine atom.
Identify Organic Product F:
Product F has a molar mass of 74.0 g mol⁻¹ and a chiral carbon atom. Since F is an alcohol formed from the hydrolysis of a haloalkane, let's test general formulas for saturated aliphatic alcohols ( CₙH₂ₙ₊₂O ):
C₄H₁₀O: (4 × 12.0) + (10 × 1.0) + 16.0 = 74.0 g mol⁻¹.
An alcohol with 4 carbons and a chiral centre is butan-2-ol ( CH₃CH(OH)CH₂CH₃ ).
Identify Haloalkane E:
Haloalkane E must share the same carbon chain as F and the same halogen as G, making E 2-bromobutane ( C₄H₉Br ).

✅ Summary of Identifications

G: Silver bromide / AgBr
F: butan-2-ol
E: 2-bromobutane (or bromobutane)

❌ Common Errors

Forgetting to show working for the calculation of G will lose you credit, even if the final formula is correct. Ensure structural or displayed formulas for F clearly demonstrate the chiral center.

🎯 Marks available: 3 marks. Breakdown: (1) Identifying precipitate G as AgBr via molar mass calculation, (2) Identifying alcohol F (butan-2-ol) and its structure, (3) Identifying haloalkane E based on matching carbon chain and halogen.

Topics

Module 4: Core organic chemistry · Practical Activity Groups · PAG 5: Synthesis of an organic liquid · 4.2 Alcohols, haloalkanes and analysis

Question and mark scheme from the OCR A-Level Chemistry examination, Synthesis and analytical techniques (02), June 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.