OCR A-Level Chemistry Synthesis and analytical techniques (02), June 2017: Question 16

20 marks · Hard difficulty · Structured Questions

Answer questions on unsaturated hydrocarbons involving isomers, stereoisomerism, pi-bonds, electrophilic addition mechanisms, and calculations involving amounts of substance.

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Question text

16 This question is about unsaturated hydrocarbons.

(a) Compound A and compound B are isomers.

compound A compound B

Compound A has a lower melting point than compound B.

Suggest why.

… [2]

(b) Compound C, CH3CH2CH=CHCH2CH2OH, exists as cis and trans stereoisomers.

(i) Name compound C.

… [1]

(ii) Define the term stereoisomers.

… [1]

(iii) Draw the structures of the cis and trans stereoisomers of compound C.

cis trans

[2]

Complete the diagram below to show how p-orbitals are involved in the formation of a π-bond.

[1]

(d) Compound D, shown below, reacts with hydrogen bromide by electrophilic addition. A mixture

of two organic compounds, E and F, is formed.

CH3 CH2CH3

HBr

C C Mixture of organic compounds E and F

CH3 H

compound D

(i) Suggest how an HBr molecule can act as an electrophile.

… [1]

(ii) Draw the structures of the two organic compounds E and F.

E F

[2]

(iii) Outline the mechanism of the reaction between compound D and hydrogen bromide to

form either compound E or compound F.

Include curly arrows and relevant dipoles.

[3]

(iv) Which of E or F is the major organic product?

Explain your answer.

Major organic product …

Explanation …

… [1]

(e) Myrcene, C10H16, is a naturally occurring hydrocarbon containing more than one

carbon-carbon double bond.

myrcene

(i) Reaction of 204 mg of myrcene with hydrogen gas produces a saturated alkane.

Calculate the volume of hydrogen gas, in cm3 and measured at RTP, needed for this reaction.

Show your working.

volume = … cm3 [2]

(ii) β-Carotene is a naturally occurring unsaturated hydrocarbon found in carrots.

A β-carotene molecule contains 40 carbon atoms, has two rings, and a branched chain.

0.0200 mol of β-carotene reacts with 5.28 dm3 of hydrogen gas to form a saturated

hydrocarbon.

Using molecular formulae, construct a balanced equation for this reaction.

Include relevant calculations and reasoning.

Equation … [4]

Mark scheme

Show the mark scheme

Question Answer Marks Guidance

16 (a) Compound A (is branched so) has less points of contact / Both answers need to be comparisons

less surface interaction between molecules 2 ALLOW ORA throughout

DO NOT ALLOW ‘more contact between atoms’

IGNORE van der Waals’ forces/VDW for induced

dipole–dipole interactions (ambiguous as this

term refers to both permanent dipole – dipole

and induced dipole–dipole forces)

Induced dipole–dipole interactions / London (dispersion) ALLOW fewer induced dipole-dipole interactions.

forces are weaker.

AND IGNORE it is easier to break the induced dipole-

Require less energy to break (these interactions / forces) dipole / London forces. (reference to energy

required)

IGNORE less energy required to separate

molecules

IGNORE less energy is needed to break the

bonds.

(b) (i) Hex-3-en-1-ol 1 ALLOW Hex-3-ene-1-ol

ALLOW 1-hydroxyhex-3-ene as this is

unambiguous

Hex-3-enol is not sufficient

IGNORE lack of hyphens, or addition of commas

Question Answer 7 Marks Guidance

(ii) Same structural formula 1 ALLOW have the same structure/displayed

AND formula/skeletal formula

Different arrangement (of atoms) in space OR different

spatial arrangement (of atoms) DO NOT ALLOW same empirical formula OR

same general formula

IGNORE same molecular formula

Reference to E/Z isomerism or optical isomerism

is not sufficient

(iii) 2 ALLOW any combination of skeletal OR

structural OR displayed formula as long as

CH3CH2 CH2CH2OH H CH2CH2OH unambiguous

C C C C ALLOW one mark if both stereoisomers of

compound C are shown but in the incorrect

H H CH3CH2 H columns

cis trans

ALLOW one mark for correct stereoisomers of

compound C in correct columns where –

CH2CH2OH is represented as -C2H5O or –

C2H4OH

DO NOT ALLOW incorrect connectivity e.g. –

CH3CH2 on first occasion but allow ECF in

second structure.

DO NOT ALLOW overlapping p orbitals on left

hand side in the diagram.

DO NOT ALLOW a diagram that contains four

lobes on the right hand side.

Two p-orbitals shown as a “dumb-bell” added to structure

on left. e.g.

AND

-bond on structure on right IGNORE any atoms joined to the bonds

Note: labels are not required

ALLOW the following diagram to show the -

bond

( -bond)

(d) (i) (The H atom of HBr) accepts a pair of electrons 1

(ii) CH3 CH2CH3 CH3 CH2CH3 2 ALLOW correct structural OR displayed OR

skeletal formulae OR a combination of above as

H C C C H long as unambiguous

3 H3C C C H

H Br Br H ALLOW in either order

(iii) Curly arrow from C=C bond to H of H–Br 3 ANNOTATE ANSWER WITH TICKS AND

CROSSES

Correct dipole shown on H–Br 9 ALLOW any combination of skeletal OR

AND curly arrow showing the breaking of H–Br bond structural OR displayed formula as long as

unambiguous

CH3 CH2CH3

C C DO NOT ALLOW partial charges shown on C=C

double bond ( the second marking point)

CH3 H

H +

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

Correct carbocation

AND

curly arrow from Br– to C+ of carbocation

CH CH CH CH3 CH2CH3

32 3

DO NOT ALLOW + on C of carbocation

CH3 C C H CH3 C C H

+ OR +

Curly arrow must come from a lone pair on Br–

H H –

- - OR from the negative sign of Br ion

Br –

(then lone pair on Br ion does not need to be

shown)

(iv) CH CH CH 1 Note: the correct product and explanation are

32 3 both required for the mark

H3C C C H 10 The major product may be identified by its

corresponding letter (E or F) from the table in

Br H (d)(ii)

correct structure

2-bromo-2-methylpentane correct name

AND

(the) carbocation intermediate (in the formation of 2- DO NOT ALLOW product comes from the more

bromo-2-methylpentane) is more stable (than the stable secondary or primary carbocation

carbocation in the formation of the other product) IGNORE explanations based on Markownikoff’s

rule.

(e) (i) 2

204 × 10–3 Correct working required for the first marking

n(myrcene) = = 1.5(0) × 10–3 (mol)

136.0 point.

Volume of H = 3 × 1.5(0) × 10–3 × 24000

= 108 (cm3) ALLOW ECF from incorrect moles of myrcene

i.e. n(myrcene) × 3 × 24000

Common incorrect answers

108000 cm3 = 1 mark (not converted to g)

12cm3 = 1 mark ( divided by 3)

36 cm3 = 1 mark ( not multiplied by 3)

IGNORE Calculations based on pV = nRT

(ii) Amount of hydrogen 4

5.28 5.28

n(H2) = 24.0 = 0.22(0) (mol) ALLOW Evidence of n(H2) = 24.0 if 0.22 is not

seen

Number of double bonds

0.220

= = 11 Evidence for 11 double bonds could come from

0.0200

11H2 in equation

Formula of saturated product

C40H78

Formula could be shown as the product of an

Equation equation

C40H56 + 11H2 C40H78

ALLOW ECF from C40H82 and C40H80 only

i.e. C40H60 + 11H2 C40H82

C40H58 + 11H2 C40H80

Total 20

How to answer it

Unsaturated Hydrocarbons Study Guide

OCR A-Level Chemistry

What this question tests

This comprehensive question assesses core organic chemistry concepts including intermolecular forces and branching, E/Z (cis/trans) stereoisomerism nomenclature, orbital overlap in pi-bonds, electrophilic addition mechanisms with curly arrows and dipoles, carbocation stability for major/minor products, and multi-step stoichiometric calculations involving gas volumes at RTP.

Part (a): Intermolecular Forces and Branching

Question 16(a): Melting point comparison between isomers A and B

✅ Correct Answer

Compound A is branched, resulting in fewer points of contact between molecules and weaker induced dipole-dipole interactions (London forces). Less energy is required to break these forces.

💡 Key Knowledge

Branching reduces surface contact area between molecules.
Smaller contact area leads to fewer induced dipole-dipole interactions.
Always use comparative language (e.g., "fewer", "weaker").

❌ Common Errors

Using ambiguous terms like "van der Waals' forces" without specifying induced dipole-dipole interactions.
Stating that covalent bonds within the molecules are broken (molecules separate, bonds do not break).
Failing to make direct comparative statements for both compounds.

Marks: 2 marks. 1 mark for branching / surface contact comparison; 1 mark for weaker forces / less energy required.

Part (b): Stereoisomerism and Naming

Questions 16(b)(i), (ii), and (iii): Naming, Defining, and Drawing Stereoisomers

✅ Correct Answers

(i) Name: Hex-3-en-1-ol (or 1-hydroxyhex-3-ene )

(ii) Definition: Same structural formula but different arrangement of atoms in space (or different spatial arrangement).

(iii) Structures:
cis (Z): Alkyl chains on the same side of the C=C bond.
trans (E): Alkyl chains on opposite sides of the C=C bond.

🧠 Exam Technique

When drawing stereoisomers, ensure the planar geometry around the carbon-carbon double bond (trigonal planar, 120° bond angles) is clearly visible so the spatial differences are unambiguous.

❌ Common Errors

Omission of hyphens or numbers in IUPAC nomenclature (e.g., hex3enol is penalised).
Mixing up structural and stereoisomer definitions.

Marks: 4 marks total (1 for naming, 1 for definition, 2 for cis/trans structures).

Part (c): Bonding in Alkenes

Question 16(c): Pi-bond Formation

✅ Correct Answer

Complete the diagram by drawing two p-orbitals as dumbbell shapes perpendicular to the sigma bond axis, merging sideways above and below the plane to show the formed pi-bond.

💡 Key Knowledge

A pi-bond is formed by the sideways overlap of two adjacent p-orbitals, one from each carbon of the double bond, resulting in electron density concentrated above and below the sigma bond axis.

❌ Common Errors

Overlapping p-orbitals incorrectly on only one side of the reaction arrow.
Drawing four lobes on the right-hand side instead of a unified sideways overlap cloud.

Marks: 1 mark.

Part (d): Electrophilic Addition Mechanisms

Questions 16(d)(i) to (iv): Electrophiles, Mechanisms, and Major Products

✅ Correct Answers

(i) Electrophile action: The H atom of HBr accepts a pair of electrons.

(ii) Structures E & F: The structural isomers formed via addition across the double bond (e.g., 2-bromo-2-methylpentane and 2-bromo-3-methylpentane variations depending on structure D).

(iii) Mechanism: Curly arrow from C=C bond to H; dipole on H-Br with curly arrow breaking H-Br bond; intermediate carbocation attacked by Br⁻ with a curly arrow from lone pair to C⁺.

(iv) Major Product: 2-bromo-2-methylpentane. Reason: The carbocation intermediate formed in its pathway is tertiary and therefore more stable than the alternative secondary carbocation.

🧠 Exam Technique

Make sure curly arrows start precisely from bonds or lone pairs and point exactly to where the new bond is forming. Examiners strictly penalize floating arrows.

❌ Common Errors

Omitting the partial charges (δ+ on H, δ- on Br) in the electrophile.
Referring to "Markovnikov's rule" in explanations without explicitly naming or explaining carbocation stability (Examiners require stability reasoning for the mark).

Marks: 7 marks total across parts (i) to (iv).

Part (e): Stoichiometric Calculations

Questions 16(e)(i) and (ii): Moles, Gas Volumes, and Balanced Equations

📐 Calculation Step-by-Step: Part (i)

Step 1: Find Mr of Myrcene (C₁₀H₁₆)
(10 × 12.0) + (16 × 1.0) = 136.0 g mol⁻¹

Step 2: Calculate moles of myrcene
Mass = 204 mg = 204 × 10⁻³ g = 0.204 g
Moles = 0.204 / 136.0 = 1.50 × 10⁻³ mol

Step 3: Relate to H₂ moles (myrcene has 2 double bonds, requiring 3 H₂ for total saturation based on structure/formula)
Moles of H₂ = 3 × 1.50 × 10⁻³ = 4.50 × 10⁻³ mol

Step 4: Calculate volume of H₂ at RTP
Volume = Moles × 24000 cm³
Volume = 4.50 × 10⁻³ × 24000 = 108 cm³

📐 Calculation Step-by-Step: Part (ii)

Step 1: Moles of H₂ reacted
Volume = 5.28 dm³
Moles H₂ = 5.28 / 24.0 = 0.220 mol

Step 2: Ratio of H₂ to β-carotene
Moles β-carotene = 0.0200 mol
Ratio = 0.220 / 0.0200 = 11 moles of H₂ per 1 mole of β-carotene. Thus, 11 double bonds / rings.

Step 3: Molecular formula of saturated product
Given: C₄₀H₁₆ originally unsaturated with 11 double bonds/rings. Adding 11 H₂ (22 hydrogens) gives C₄₀H₃₈ or similar based on starting unsaturation.

Step 4: Balanced Equation
C₄₀H₅₆ + 11H₂ → C₄₀H₇₈ (or equivalent fully saturated formula matched to input data).

❌ Common Calculation Traps

Forgetting to convert milligrams (mg) to grams (g) using ×10⁻³.
Using 24 dm³ instead of 24000 cm³ when calculating volume in cm³.
Incorrectly counting the number of double bonds/rings from the unsaturation index.

Marks: 6 marks total (2 for part i, 4 for part ii).

Topics

Module 2: Foundations in chemistry · Module 4: Core organic chemistry · 2.1 Atoms and reactions · 4.1 Basic concepts and hydrocarbons · 4.2 Alcohols, haloalkanes and analysis

Question and mark scheme from the OCR A-Level Chemistry examination, Synthesis and analytical techniques (02), June 2017. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.