OCR A-Level Chemistry Synthesis and analytical techniques (02), November 2020: Question 16

9 marks · Medium difficulty · Structured Questions

Name the halogenated product of hydrocarbon A, explain stereoisomerism, draw 3D diagrams of stereoisomers, complete the radical substitution mechanism table, and state limitations of radical substitution.

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Question

Question text

16 The structure of hydrocarbon A is shown below.

(a) Hydrocarbon A can be reacted with bromine in the presence of ultraviolet radiation to prepare

(CH3)3CCHBrCH3.

What is the systematic name for (CH3)3CCHBrCH3?

… [1]

(b) (CH3)3CCHBrCH3has stereoisomers.

(i) Explain the term stereoisomers and name this type of stereoisomerism.

Explanation: …

Type of stereoisomerism: … [1]

(ii) Draw 3D diagrams for the stereoisomers of (CH3)3CCHBrCH3.

[2]

(CH3)3CCHBrCH3.

Use skeletal formulae for all organic compounds.

Use ‘dots’ (●) to show the position of unpaired electrons.

Initiation

Propagation

+ … + …

… + … + …

[3]

(d) State two limitations of using radical substitution in organic synthesis.

1 …

2 …

[2]

Mark scheme

Show the mark scheme

Question Answer Marks Guidance

element

16 (a) (i) 2-bromo-3,3-dimethylbutane 1 1.2 IGNORE lack of hyphens or addition of commas

ALLOW 3,3-dimethyl-2-bromobutane

DO NOT ALLOW

2-bromo-3-dimethylbutane

methy for methyl

methly for methyl

brom for bromo

(b) (i) Stereoisomers 1 1.2 ALLOW structure/displayed/skeletal formula

Same structural formula

AND DO NOT ALLOW same empirical formula

Different arrangement (of atoms) in space OR same general formula

OR different spatial arrangement (of atoms)

AND IGNORE same molecular formula

Type: Optical IGNORE references to chiral

molecules/compounds

(ii) One 3D structure with correct groups attached to the 2 2.5 ALLOW small slip in one of the groups OR use of

chiral C C4H9

3D structures must have four central bonds with at

1.2 least two wedges.

Two 3D structures of (CH3)3CCHBrCH3 that are mirror For bond into paper accept:

images AND correct connectivity in both

Br Br

C C ALLOW two 3D structures with 2 groups swapped

CH3 H3C e.g.

(CH3)3C C(CH3)3

H H

element

Br Br

C C

CH3 C(CH3)3

(CH3)3C H3C

H H

Initiation IGNORE dots for initiation step,

Br2 → 2Br• 1.2 i.e. ALLOW Br2 → Br + Br OR Br2 → 2Br

Propagation

DOT REQUIRED at correct position on chain.

+ Br• + HBr

ALLOW 1 mark if both propagation equations are

2.5×2 correct by atom but dot(s) missing or on incorrect

C in chain

ALLOW 1 mark if both propagation equations are

correct including position of dot(s) but structures

are not shown using skeletal formula

ALLOW ECF from incorrect intermediate

7 element

(d) further substitution/s 2 1.1×2 ALLOW dibromo/multibromo compounds formed

OR OR an example of a further substitution product

produces different termination products OR an example of a different termination product

OR ALLOW more than one hydrogen (atom) can be

More than one termination step replaced

OR ALLOW radicals react with each other to form

Mixture of products are formed other products

IGNORE references to separation of products

IGNORE references to atom economy or yield

substitution at different positions along chain ALLOW a hydrogen (atom) on a different carbon

(atom) can be replaced

How to answer it

Free Radical Substitution & Optical Isomerism Study Guide

What this question tests

This question assesses your mastery of alkane chemistry, specifically free radical substitution mechanisms involving ultraviolet radiation, IUPAC organic nomenclature rules, defining and drawing 3D representations of optical isomers (stereoisomerism), and understanding the synthetic limitations of free-radical mechanisms in organic chemistry.

Question Part (a)

IUPAC Nomenclature

✅ Correct Answer

2-bromo-3,3-dimethylbutane

Awarded 1 mark for correct spelling, hyphenation, and numerical locants (AO1.2).

💡 Key Knowledge

Find the longest continuous carbon chain containing the principal functional group (4 carbons = butane).
Number the chain from the end that gives the substituents the lowest possible locants (from right to left gives the bromine at carbon 2 rather than carbon 3).
Alphabetise substituents ( bromo before methyl ).

❌ Common Errors

Writing 3,3-dimethyl-2-bromobutane is allowed, but mixing up locants like 2-bromo-3-dimethylbutane loses the mark.
Failing to use hyphens between numbers and letters (e.g., 2 bromo 3 3 dimethylbutane ).

Question Part (b)(i)

Explaining Stereoisomerism and Optical Isomers

✅ Correct Answer

Explanation: Same structural formula AND different arrangement (of atoms) in space.

Type of stereoisomerism: Optical (isomerism)

Awarded 2 marks total for the correct definition pairing and identifying optical isomerism (AO1.2).

🧠 Exam Technique

When defining stereoisomerism, always include two key phrases: "same structural formula" and "different spatial arrangement / different arrangement of atoms in space". Do not use vague terms like "same molecular formula" as that describes structural isomers.

Question Part (b)(ii)

Drawing 3D Diagrams of Optical Isomers

✅ Correct Answer

Two 3D tetrahedral structures drawn as non-superimposable mirror images centred around the chiral carbon (C2), clearly showing standard 3D bonds (wedges and dashes).

Awarded 2 marks: 1 for correct groups attached to the chiral centre, 1 for correct non-superimposable mirror image 3D representation (AO2.5, AO1.2).

🧠 Exam Technique

To draw clear 3D representations:

Draw two standard tetrahedral bonds in the plane of the paper (lines).
Draw one wedge (coming out of the page) and one dash (going into the page).
Ensure the mirror image relationship is obvious by placing a dashed mirror line between the two structures or reflecting the groups precisely.

Question Part (c)

Free Radical Substitution Mechanism

✅ Correct Answer

Initiation: Br₂ → 2Br• (dot required on Br)

Propagation Step 1: Alkane skeleton + Br• → Alkyl radical + HBr

Propagation Step 2: Alkyl radical + Br₂ → Halogenoalkane product + Br•

Awarded 3 marks: 1 for initiation, 2 for correct propagation steps including radical dots (•) and correct skeletal structures (AO1.2, AO2.5).

❌ Common Errors

Missing radical dots: Omitting the unpaired electron dot (•) on bromine or on the carbon chain in propagation equations loses marks.
Incorrect structural format: Failing to use skeletal formulas for the organic molecules as explicitly requested in the stem.
Placing the radical dot on the wrong carbon atom of the chain during propagation.

Question Part (d)

Limitations of Radical Substitution

✅ Correct Answer

Any two of the following:

Further substitution (substitution of more than one hydrogen atom / formation of multibromo compounds).
Substitution at different positions along the carbon chain (yielding a mixture of structural isomers).
Formation of different termination products (side reactions where radicals combine).

Awarded 2 marks (1 mark per valid limitation) (AO1.1).

💡 Key Knowledge

Radical substitution is rarely used in industrial or laboratory organic synthesis for preparing specific halogenoalkanes because it is unselective. The reaction produces an uncontrollable mixture of products, lowering the percentage yield of the desired single compound.

Topics

Module 4: Core organic chemistry · Module 6: Organic chemistry and analysis · 4.1 Basic concepts and hydrocarbons · 6.2 Nitrogen compounds, polymers and synthesis

Question and mark scheme from the OCR A-Level Chemistry examination, Synthesis and analytical techniques (02), November 2020. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.