OCR A-Level Chemistry Synthesis and analytical techniques (02), June 2022: Question 16

17 marks · Hard difficulty · Structured Questions

Name and outline the mechanism for the reaction of an unsaturated hydrocarbon with bromine, define structural and stereoisomers, draw cis/trans and optical isomers, deduce structures from NMR and IR spectral data, and draw products and mechanisms of ozonolysis.

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Question text

16 This question is about unsaturated hydrocarbons.

(a) The unsaturated hydrocarbon A, shown below, is reacted with bromine.

H3C CH3

C C

CH3CH2CH2 H

Hydrocarbon A

(i) What is the systematic name of hydrocarbon A?

… [1]

(ii) Outline the mechanism for the reaction of hydrocarbon A with bromine.

The structure of hydrocarbon A has been provided.

Include curly arrows and relevant dipoles.

H3C CH3

C C

CH3CH2CH2 H

[3]

(b) Compounds B and C are branched hydrocarbons that are structural isomers of C6H12.

Compounds B and C both have stereoisomers.

• Compound B has cis and trans isomers but does not have optical isomers.

• Compound C has optical isomers but does not have cis and trans isomers.

(i) What is meant by the term structural isomers?

… [1]

(ii) What is meant by the term stereoisomers?

… [1]

(iii) Draw structures for the cis and trans isomers of the branched hydrocarbon B.

cis isomer trans isomer

[2]

(iv) Draw 3D structures for the optical isomers of compound C.

Optical isomers

[2]

(v) Compounds D and E are two more structural isomers of C6H12.

Compounds D and E do not show stereoisomerism.

Table 16.1 shows NMR and infrared (IR) spectral data for D and E.

Number of peaks in Number of peaks in IR peak at

1H NMR spectrum 13C NMR spectrum 1620–1680 cm–1

D 1 1 No

E 1 2 Yes

Table 16.1

Draw the structures of D and E and explain how the spectral data in Table 16.1

provides evidence for the structures.

D E

… [4]

compounds.

For example, the complete ozonolysis of methylbut-2-ene is shown below.

H3C H H3C H

C C C O + O C

H3C CH3 H3C CH3

(i) Draw the structures of the products you would expect from the ozonolysis of the

two compounds below.

[2]

(ii) The mechanism for ozonolysis takes place in several steps.

The curly arrows in the first step in the ozonolysis of methylbut-2-ene are shown below.

In the box, draw the structure(s) for the product(s) of this step.

O –O..

H3C H

C C

H3C CH3

[1]

Mark scheme

Show the mark scheme

Question Answer Marks Guidance

element

16 (a) (i) 3-methylhex-2-ene 1 AO1.2 IGNORE lack of hyphens, or addition of commas

DO NOT ALLOW 3-methyhex-2-ene

OR 3-methhex-2-ene

OR 3-methlyhex-2-ene

OR 3-methylhexan-2-ene

IGNORE references to E/Z or cis/trans

16 (a) (ii) ANNOTATE ANSWER WITH TICKS AND CROSSES 3 AO1.2 ALLOW any combination of skeletal OR structural

×1 OR displayed formula as long as unambiguous

AO2.5 IGNORE connectivity of CH3CH2CH2 and CH3

×2 groups in carbocation and product

ALLOW C3H7 for CH3CH2CH2

DO NOT ALLOW half headed or double headed

arrows but allow ECF if seen more than once

DO NOT ALLOW use of HBr but ECF for

subsequent use

Curly arrow from C=C bond to Brδ+ of Br–Br For curly arrows, ALLOW straight or snake-like

arrows and small gaps (see examples):

AND ------------------------------------------------------------

Correct dipole on Br–Br DO NOT ALLOW partial charge on C=C

AND

curly arrow for breaking of Br–Br bond 1st curly arrow must

• go to a Br atom of Br–Br

AND

start from, OR be traced back to any point across

width of C=C

element

2nd curly arrow must

• start from, OR be traced back to, any part

of δ+Br–Brδ– bond

• AND go to Br δ–

Correct carbocation to match mechanism 3rd curly arrow must

• go to the C+ of carbocation

AND curly arrow from Br– to C+ of carbocation

AND

• start from, OR be traced back to any point

across width of lone pair on :Br–

• OR start from – charge on Br– ion

(Lone pair NOT needed if curly arrow shown

i.e. ALLOW carbonium + on either C atom –

from – charge on Br )

ALLOW bromonium ion (Contact TL)

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

element

Correct product to match mechanism

16 (b) (i) Same molecular formula 1 AO1.1 Same formula is not sufficient

AND (no reference to molecular)

Different structural formulae Different arrangement of atoms is not sufficient

(no reference to structure/structural)

OR For ‘structural formulae’,

ALLOW structure/displayed/skeletal formulae/

Both have the molecular formula C6H12 functional groups

AND

Different structural formulae DO NOT ALLOW any reference to spatial/space

16 (b) (ii) Same structural formula 1 AO1.1 ALLOW structure/displayed/skeletal formula

AND

Different arrangement (of atoms) in space DO NOT ALLOW same empirical formula

OR different spatial arrangement (of atoms) OR same general formula

IGNORE same molecular formula

Reference to E/Z isomerism or optical isomerism is

not sufficient

element

16 (b) (iii) Correct identification of cis AND trans isomers of 2 ALLOW any combination of skeletal OR structural

4-methylpent-2-ene AO1.2 OR displayed formula as long as unambiguous

H H H3C H

AO2.5

C C C C C3H7 is not sufficient (could be unbranched)

H3C CH(CH3)2 H CH(CH ) ALLOW one mark if cis AND trans isomers of

4-methylpent-2-ene are in the wrong boxes

cis isomer trans isomer

ALLOW the isomers of 3-methylpent-2-ene in either

Identification of 3-methylpent-2-ene as cis AND trans

isomers

cis isomer trans isomer

Ambiguity with cis/trans identification system

ALLOW one mark for correct identification of cis

AND trans isomers of unbranched C6H12

e.g.

cis isomer trans isomer

13 AO

element

16 (b) (iv) Correct groups attached to chiral carbon of compound C 2 AO2.5 ALLOW any combination of skeletal OR structural

seen once e.g. ×2 OR displayed formula as long as unambiguous

For C2H5–, ALLOW CH3CH2–

For –CH=CH2, ALLOW –C2H3 OR –CHCH2

For bond into paper accept:

Two 3D structures of compound C that are mirror images ALLOW two 3D structures with 2 groups swapped

with correct connectivity in both e.g.

DO NOT ALLOW a bond angle of 180°

e.g.

element

16 (b) (v) 4

ALLOW 1 mark for structures if shown in wrong

boxes.

AO2.5

×2

D E

Two of the following for D CHECK table 16.1 for annotations that may be

• All H are equivalent/in the same chemical AO2.2 worthy of credit

environment/ the same type ×2

• All C are equivalent/ in the same chemical

environment/ the same type

• No C=C present

Two of the following for E

• All H are equivalent/ in the same chemical

environment/ the same type

• 2 C environments

• C=C present

15 AO

element

16 (c) (i) 2 AO3.1 ALLOW any combination of skeletal OR structural

×1 OR displayed formula as long as unambiguous

BOTH structures required for

AO3.2

×1

16 (c) (ii) 1 AO3.2 ALLOW any combination of skeletal OR structural

OR displayed formula as long as unambiguous

Total 17

How to answer it

Organic Chemistry: Alkenes, Isomerism, Spectroscopy & Ozonolysis

What this question tests

This comprehensive question assesses your mastery of alkene nomenclature, electrophilic addition mechanisms with curly arrows and dipoles, definitions of structural and stereoisomerism, drawing geometric (cis/trans) and optical isomers (including 3D tetrahedral representations), interpreting spectroscopic data (¹H NMR, ¹³C NMR, and IR) to deduce structures, and applying ozonolysis cleavage patterns to cyclic and branched alkenes.

Question 16(a): Alkene Nomenclature & Electrophilic Addition

✅ Correct Answers

(a)(i): 3-methylhex-2-ene
(a)(ii): Electrophilic addition mechanism showing:
- Curly arrow from C=C bond to the Br (δ+) atom.
- Dipole ( Br δ+— Br δ-) on the bromine molecule.
- Curly arrow for the breaking of the Br—Br bond.
- Correct carbocation intermediate matching the mechanism.
- Curly arrow from the lone pair on the bromide ion ( Br⁻ ) to the positively charged carbon atom.
- Correct halogenoalkane product: 3-bromo-3-methylhexane .

💡 Key Knowledge

Electrophilic Addition: Alkenes are electron-rich due to the pi bond, making them vulnerable to attack by electrophiles (electron pair acceptors).
Markovnikov's Rule: The major carbocation intermediate forms via the more stable intermediate (usually tertiary > secondary > primary), which dictates the final substitution position of the halogen.

🧠 Exam Technique

Curly Arrows: Must start precisely from a bond (or lone pair) and point precisely to where electrons are moving (an atom or forming a bond). The first arrow starts from the C=C π-bond and goes to the Br atom.
Nomenclature: Count the longest carbon chain containing the functional group carefully. Use hyphens between numbers and letters (e.g., 3-methylhex-2-ene ).

❌ Common Errors

Drawing partial charges directly on the C=C bond (incorrect; only polar bonds like Br—Br have dipoles).
Starting the bromide attack arrow from the negative charge rather than directly from the lone pair of electrons.
Failing to include the carbocation intermediate step in the mechanism.

Question 16(b): Isomerism & Spectroscopy of C₆H₁₂ Isomers

✅ Correct Answers

(b)(i) Structural isomers: Compounds with the same molecular formula but different structural formulae.
(b)(ii) Stereoisomers: Compounds with the same structural formula, but with a different arrangement of atoms in space.
(b)(iii) Cis/trans isomers of B: Structures of 4-methylpent-2-ene showing either cis (methyl groups on same side) or trans (methyl groups on opposite sides) arrangements across the double bond.
(b)(iv) Optical isomers of C: 3D tetrahedral representations of enantiomers showing non-superimposable mirror images around a chiral carbon bonded to four different groups ( -H , -CH₃ , -C₂H₅ , -CH=CH₂ ). Use wedge-dash bonds.
(b)(v) Structures & Explanations:
- D: Cyclohexane ( C₆H₁₂ ). 1 peak in ¹H NMR (all equivalent H), 1 peak in ¹³C NMR (all equivalent C), no IR peak at 1620–1680 cm⁻¹ (no C=C).
- E: 2,3-dimethylbut-2-ene. 1 peak in ¹H NMR, 2 peaks in ¹³C NMR, has an IR peak at 1620–1680 cm⁻¹ (C=C present).

💡 Key Knowledge

Chirality: A chiral center requires a carbon atom bonded to four completely different groups. Enantiomers rotate plane-polarised light in opposite directions.
NMR Environments: The number of peaks corresponds to the number of unique chemical environments for hydrogen (¹H) or carbon (¹³C) atoms.
IR Spectroscopy: Absorption in the 1620–1680 cm⁻¹ region corresponds to alkene C=C stretching vibrations.

🧠 Exam Technique

When drawing 3D optical isomers, ensure you use wedge-dash notation clearly to show 3D spatial orientation around the central chiral carbon.
For spectroscopic deduction questions, systematically eliminate structures that conflict with the number of peaks or presence/absence of specific IR absorption bands.

❌ Common Errors

Using vague definitions like "same molecular formula, different arrangement of atoms" without specifying structural vs spatial differences.
Drawing flat 90-degree bond angles for optical isomers instead of tetrahedral arrangement.

Question 16(c): Ozonolysis Reactions

✅ Correct Answers

(c)(i) Products of ozonolysis:
- For the acyclic alkene ( pent-2-ene derivative): Cleaves the C=C bond to form ethanal and propanal (or equivalent carbonyl fragments).
- For the cyclic alkene: Ring opens up to form a dicarbonyl compound (e.g., a hexane-1,6-dial derivative or keto-aldehyde structure).
(c)(ii) Ozonolysis intermediate step: Shows the molozonide/cyclic ozonide intermediate structure formed when ozone adds across the carbon-carbon double bond.

💡 Key Knowledge

Ozonolysis Mechanism Summary: Ozone ( O₃ ) reacts across alkene C=C bonds, cleaving the carbon skeleton completely into two separate carbonyl compounds (aldehydes or ketones depending on original alkyl substitution).
Cyclic Alkenes: Cleaving a ring double bond does not yield two separate molecules, but rather a single chain containing two carbonyl groups at the cleavage sites.

🧠 Exam Technique

To predict ozonolysis products quickly: erase the C=C double bond, replace each doubly-bonded carbon with a double-bonded oxygen ( =O ), and draw the resulting fragments.

❌ Common Errors

Forgetting that cyclic alkenes open into chain structures rather than splitting into two separate molecules.
Miscounting carbon atoms across the ring cleavage boundary.

Topics

Module 4: Core organic chemistry · Module 6: Organic chemistry and analysis · 6.2 Nitrogen compounds, polymers and synthesis · 6.3 Analysis · 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 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.