OCR A-Level Chemistry AS Breadth in chemistry (01), June 2022: Question 26

8 marks · Medium difficulty · Structured Questions

Outline the mechanism for the hydrolysis of 1-chloropropane with sodium hydroxide, and use ideal gas equation data to determine the molar mass and molecular formula of bromoalkane D.

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Question text

26 This question is about haloalkanes.

(a) 1-Chloropropane, C2H5CH2Cl, can be hydrolysed with aqueous sodium hydroxide, NaOH.

Outline the mechanism for this reaction.

The structure of 1-chloropropane has been provided.

Show curly arrows, relevant dipoles and product(s).

C2H5 C Cl

[3]

(b) A bromoalkane D is a liquid at room temperature and pressure but can easily be vaporised.

When vaporised, 0.330 g of D produces 74.0 cm3 of gas at 1.01 × 105 Pa and 100 °C.

Determine the molar mass and molecular formula of bromoalkane D.

molar mass = … g mol−1

molecular formula = …

[5]

Mark scheme

Show the mark scheme

Question Answer Marks Guidance

element

26 (a) 3 ANNOTATE ANSWER WITH TICKS AND

Curly arrow from HO– to C atom of C–Cl bond AO1.2 CROSSES

Dipole shown on C–Cl bond, Cδ+ and Clδ– NOTE: curly arrows can be straight, snake-like, etc.

AND but NOT double headed or half headed arrows

curly arrow from C−Cl bond to Cl atom AO1.2

1st curly arrow must

• go to C of C–Cl

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

DO NOT ALLOW H O instead of OH–

------------------------------------------------------------------------- AO2.5 –

(Lone pair NOT needed if curly arrow shown from O

×1

– )

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

ALLOW ECF NaCl– ONLY from NaOH–

IGNORE presence of Na+ but Cl– needed

i.e. Na+Cl– can be allowed

BUT NaCl does NOT show Cl–

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

23 AO

element

ALLOW SN1 mechanism

First mark

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

AND curly arrow from C−Cl bond to Cl atom

Second mark

Correct carbocation AND curly arrow from HO– to

carbocation

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–

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

24 element

(b) FIRST check the molar mass on answer line 5 ANNOTATE ANSWER WITH TICKS AND

MUST be derived from pV = nRT, CROSSES

Award 4 marks for calculation for: -------------------------------------------------------------------

• answer = 136.9 OR 137 If there is an alternative answer, check to see if

-------------------------------------------------------------------------- there is any ECF credit possible using working

Rearranging ideal gas equation to make n subject below

n = 1st mark may be implicit by direct substitution of

AO2.4 correct values below into rearranged equation.

Substituting all values including conversion to m3 and K ×4

5 –6 ALLOW use of 8.31 for R → 2.411 × 10–3

(1.01 × 10 ) × (74.0 × 10 )

n =

8.314 × 373

ONLY award this mark if n has been derived

n = 2.410095443 × 10–3 → 2.41 × 10–3 (mol) from correct rearranged ideal gas equation

unrounded rounded to 3 SF ALLOW 3 SF up to calculator value, correctly

rounded

Calculation of molar mass, M –3

m 0.330 2.41 × 10 OR 0.002411255537 → first 3 marks

M = = = 136.9.. (g mol–1)

n 2.410095443 × 10–3 → 136.868581616 → C4H9Br

0.330 –1

→ –3 = 136.9 (g mol ) NOTE: ALLOW 137 (i.e. to 3 SF)

2.41 × 10

ALLOW calculated M in range 136.9 – 137

ALLOW any unambiguous structure

ALLOW ECF provided that formula given is a

Molecular formula of D AO3.2

haloalkane and matches M calculated from

C4H9Br

0.330 g AND pV = nRT

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

IF candidate has failed to derive suitable value of n,

0.330 0.330

ALLOW value of M from 0.330 AND 24000 with M = OR –3

74.0/24000 3.0833.. × 10

haloalkane closest to calculated value for last 2 marks

See Guidance column. = 107 to 3 SF

From 107, ONLY ALLOW = C2H5Br (108.9)

How to answer it

Haloalkanes Study Guide

What this question tests

This question assesses your understanding of nucleophilic substitution mechanisms in haloalkanes (specifically S_N2 pathways using curly arrows and dipoles), alongside the application of the ideal gas equation ( pV = nRT ) to determine molar mass and deduce an unknown molecular formula.

Part (a) — Mechanism of Hydrolysis

Outline the mechanism for the hydrolysis of 1-chloropropane

💡 Key Knowledge

1-Chloropropane is a primary haloalkane, meaning it reacts predominantly via an S_N2 mechanism (though S_N1 is credited on the mark scheme if drawn consistently).
Hydroxide ions ( OH⁻ ) act as nucleophiles, donating an electron pair to the electron-deficient carbon.
Polar bonds are essential: Carbon carries a partial positive charge ( δ⁺ ) and chlorine carries a partial negative charge ( δ⁻ ) due to electronegativity differences.

🧠 Exam Technique (S_N2)

Dipole: Clearly label δ⁺ on C and δ⁻ on Cl.
First Curly Arrow: Must start from either the lone pair on the OH⁻ oxygen or the negative charge, and point directly to the C of the C–Cl bond.
Second Curly Arrow: Must start from the C–Cl bond and point directly to the Cl atom.
Product: Show the formation of propan-1-ol ( C₂H₅CH₂OH ) and a chloride ion ( Cl⁻ ).

❌ Common Errors

Starting the first curly arrow from the sodium ion ( Na⁺ ) or an incorrect part of the OH⁻ ion.
Drawing a half-headed (radical) arrow or double-headed fishhook incorrectly. Use standard curly arrows.
Failing to include the leaving group Cl⁻ alongside the organic product.
Using water ( H₂O ) instead of the hydroxide ion ( OH⁻ ) as the attacking nucleophile.

Marks available: [3] — 1 mark for dipoles and first curly arrow, 1 mark for second curly arrow, 1 mark for correct organic product and Cl⁻ .

Part (b) — Ideal Gas Calculation & Molecular Formula

Determine the molar mass and molecular formula of bromoalkane D

📐 Step-by-Step Calculation

Rearrange the ideal gas equation:
pV = nRT → n = pV / (RT)
Convert units carefully:
• Pressure ( p ) = 1.01 × 10⁵ Pa
• Volume ( V ) = 74.0 cm³ = 74.0 × 10⁻⁶ m³
• Temp ( T ) = 100 °C + 273.15 = 373 K
• Gas constant ( R ) = 8.31 J mol⁻¹ K⁻¹
Calculate moles ( n ):
n = (1.01 × 10⁵ × 74.0 × 10⁻⁶) / (8.31 × 373)
n = 2.4101 × 10⁻³ mol (unrounded)
Calculate molar mass ( M ):
M = m / n
M = 0.330 / (2.4101 × 10⁻³) = 136.9 g mol⁻¹
Deduce molecular formula:
Subtract the mass of Bromine ( Br = 79.9 ):
136.9 - 79.9 = 57.0
Divide by carbon mass ( 12.0 ) to find the alkyl chain ( C₄H₉ ).
Formula: C₄H₉Br

🧠 Exam Technique & Formatting

Significant Figures: Give your final molar mass to 3 significant figures ( 137 g mol⁻¹ or 136.9 g mol⁻¹ ).
Unit Conversions: Cubic centimeters ( cm³ ) must be multiplied by 10⁻⁶ to convert to cubic meters ( m³ ). Degrees Celsius must always be converted to Kelvin by adding 273 .
Error Carried Forward (ECF): If your calculated n is slightly off due to a minor arithmetic error, examiners will award ECF for calculating M and deducing a matching haloalkane formula.

❌ Common Calculation Traps

Forgetting to convert cm³ to m³ (the most common reason students lose marks on gas equations).
Forgetting to add 273 to convert Celsius to Kelvin.
Rounding intermediate values too early, which skews the final molar mass calculation away from standard atomic masses.

Marks available: [5] — 4 marks for calculating molar mass via pV = nRT (answering 136.9 or 137 ), and 1 mark for the correct molecular formula ( C₄H₉Br ).

Topics

Module 2: Foundations in chemistry · Module 4: Core organic chemistry · 2.1 Atoms and reactions · 4.2 Alcohols, haloalkanes and analysis

Question and mark scheme from the OCR A-Level Chemistry examination, AS Breadth in chemistry (01), June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.