OCR A-Level Chemistry Synthesis and analytical techniques (02), June 2018: Question 20

18 marks · Hard difficulty · Extended Response

Analyze cinnamaldehyde and methylcinnamaldehyde regarding E/Z isomerism, functional group tests, reaction pathways including carbonyl and alkene chemistry, and electrophilic addition mechanisms with ICl.

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Question

Question text

20 Cinnamaldehyde and methylcinnamaldehyde are naturally occurring organic compounds.

CHO CHO

cinnamaldehyde methylcinnamaldehyde

(a) Methylcinnamaldehyde is an E stereoisomer.

Explain this statement in terms of the Cahn-Ingold-Prelog (CIP) rules.

… [2]

(b) A student plans to carry out some chemical tests on both cinnamaldehyde and

methylcinnamaldehyde.

CHO CHO

cinnamaldehyde methylcinnamaldehyde

(i) Suggest a suitable chemical test to confirm that both compounds contain an unsaturated

carbon chain.

Your answer should include the reagent and observations.

… [1]

(ii) Describe a chemical test to confirm that both compounds contain an aldehyde functional

group.

Your answer should include the reagent and observations.

… [1]

(iii) Describe a chemical test to confirm that cinnamaldehyde and methylcinnamaldehyde

contain a carbonyl group.

How could the products of this test be used to distinguish between the two compounds?

Your answer should not include spectroscopy.

… [3]

Draw the structures of the missing organic compounds in the boxes and add the missing

reagent(s) on the dotted line.

excessH2/Ni

CHO

reagent(s): …

cinnamaldehyde

NaBH4 H+(aq)

[5]

(d)* Methylcinnamaldehyde reacts with iodine monochloride, ICl, by electrophilic addition. The

reaction produces a mixture containing two different organic products.

CHO

methylcinnamaldehyde

The electronegativity values of chlorine and iodine are given in the table below.

Pauling electronegativity value

Cl 3.0

I 2.5

Outline the mechanism, using the ‘curly arrow’ model, for the formation of one of the organic

products and explain which of the two possible organic products is more likely to be formed.

In your mechanism, you can show the phenyl group as C6H5. [6]

… 23

Additional answer space if required.

Mark scheme

Show the mark scheme

Question Answer Marks Guidance

20 (a) 2

priority groups/atoms are on different/opposite sides ALLOW suitable alternatives to ‘priority’ e.g.

groups with highest atomic number or more

important groups etc.

ALLOW high priority groups are diagonal(ly

across)

IGNORE references to relative mass of groups,

Ar, Mr,

High(est) priority groups are C6H5 AND CHO ALLOW identification by name e.g

OR aldehyde for CHO

Lowest priority groups are H and CH3 phenyl/benzene group for C6H5

alkyl for CH3

ALLOW response in terms that O has higher

priority than H in context of –CH3 and –CHO

IF ‘priority’ is not mentioned ALLOW 1 mark for

‘C6H5 and CHO are on different sides’ OR H and

CH3 are on different sides

(b) (i) 1 Note: both reagent and observation are required

Bromine/ Br2

AND ALLOW bromine water/ Br2(aq)

goes colourless/decolourised

(ii) 1 Note: both reagent and observation are required

for the mark.

Tollens’ (reagent) +

AND ALLOW ammoniacal silver nitrate OR Ag /NH3

Silver (mirror/precipitate/ppt/solid) ALLOW black ppt OR grey ppt

Question Answer 22 Marks Guidance

(iii) (Add) 2,4-dinitrophenylhydrazine AND orange/yellow/red 3 ALLOW errors in spelling

precipitate ALLOW 2,4(-)DNP OR 2,4(-)DNPH

ALLOW Brady’s reagent or Brady’s Test

ALLOW solid OR crystals OR ppt as alternatives

for precipitate

Take melting point (of crystals) Mark second and third points independently

of response for first marking point

Compare to known values/database

DO NOT ALLOW 2nd and 3rd marks for taking

and comparing boiling points OR

chromatograms

ALLOW any combination of skeletal OR

CH2NH2 structural OR displayed formula as long as

unambiguous

reduction of nitrile to form amine For reaction with excess H2/Ni IGNORE

hydrogenation of benzene ring

hydrogenation of C=C

i.e. the following structure scores two marks

excess H2/Ni

CH2NH2

CHO

NaCN/H+

cinnamaldehyde

ALLOW KCN/H+

NaBH H+(aq)

4 ALLOW HCN

ALLOW H SO or HNO or HCl for H+

24 3

OH COOH

(d)* Please refer to marking instructions on page 5 of mark scheme for 6 Please check all of page 23 which is included

guidance on how to mark this question. with this response. If this page is blank

please annotate with SEEN

Level 3 (5–6 marks)

An outline of the mechanism for the formation of either product Throughout: ALLOW correct structural OR

which is mostly correct. displayed OR skeletal formulae OR a

AND combination of above if unambiguous

Major and minor products identified with a correct explanation of

which product is most/least likely to be formed. Indicative scientific points:

There is a well-developed line of reasoning which is clear and Mechanism for formation of either product.

logically structured. The information presented is relevant and Curly arrow from C=C to attack the I atom

substantiated. of the I-Cl

Correct dipole on I-Cl

Level 2 (3–4 marks) Curly arrow from I-Cl bond to Cl

An outline of the mechanism for the formation of either product Carbocation with full positive charge on

but with a few omissions/errors. carbon atom

AND –

Curly arrow from negative charge on Cl

Identifies major/minor product correctly OR Explanation of which –

or lone pair on Cl to carbon atom with

product is most/least likely to be formed.

positive charge

There is a line of reasoning presented with some structure. The H CHO H CHO H CHO

information presented is relevant and supported by some

C C C6H5 C C CH3 C6H5 C C CH3

evidence.

C6H5 CH3 I Cl I

Level 1 (1–2 marks)

A basic outline of the mechanism for the formation of either OR

product is attempted. H CHO H CHO H CHO

OR C C C6H5 C C CH3 C6H5 C C CH3

Basic explanation of which of the products is most/least likely to C6H5 CH3 Cl

I Cl I

be formed. I

There is an attempt at a logical structure with a line of reasoning.

The information is in the most part relevant.

0 marks Organic products

No response or no response worthy of credit.

Major/most likely product

H CHO

C6H5 C C CH3

I Cl

Minor/least likely product

H CHO

C6H5 C C CH3

Cl I

Major/most likely product is formed from

the most stable carbocation intermediate

OR – Cl is attached to carbon atom with

the least hydrogens attached

OR the carbon with the most –C atoms

attached

OR the – I is attached to the carbon atom

with most hydrogens attached

Total 18

How to answer it

Organic Synthesis, Stereoisomerism, and Electrophilic Addition

What this question tests

This comprehensive OCR A-Level question evaluates your mastery of several core organic chemistry modules: applying Cahn-Ingold-Prelog (CIP) priority rules for E/Z isomerism, recalling practical chemical tests for functional groups (alkenes, aldehydes, and carbonyls), predicting products of organic synthesis flowcharts (reduction, nucleophilic addition), and outlining complex electrophilic addition mechanisms involving unsymmetrical alkenes and polar reagents (ICI) linked to carbocation stability.

Part (a) — 2 Marks

Explaining E Isomerism via CIP Rules

✅ Correct Answer

Priority groups/atoms are on different / opposite sides across the C=C double bond.
Highest priority groups are C₆H₅ (phenyl) AND CHO (aldehyde) [OR lowest priority groups are H AND CH₃].

💡 Key Knowledge

To determine E / Z stereoisomerism, assign priority to the two groups attached to each carbon of the C=C double bond using atomic numbers (CIP rules). If the highest priority groups are on opposite sides, it is the E isomer.

❌ Common Errors

Students often lose the second mark by failing to explicitly state which groups have high priority on each carbon, writing vague statements like "the big groups are apart" instead of naming C₆H₅ and CHO.

Mark breakdown: 1 mark for stating priority groups are on opposite sides; 1 mark for correctly identifying the high priority groups (C₆H₅ and CHO).

Part (b) — 5 Marks Total

Chemical Tests for Functional Groups

(i) Unsaturated Carbon Chain (Alkenes) [1 mark]

✅ Correct Answer

Reagent: Bromine / Br₂ (or bromine water / Br₂(aq))

Observation: Goes colourless / decolourised (from orange/brown).

❌ Common Errors

Forgetting to state both the correct reagent and the specific visual observation. Saying "it turns clear" instead of "colourless" is a minor linguistic slip, but stick to "colourless" for precision.

(ii) Aldehyde Functional Group [1 mark]

✅ Correct Answer

Reagent: Tollens' reagent (or ammoniacal silver nitrate / Ag+/NH₃)

Observation: Silver mirror (or grey/black precipitate/solid).

🧠 Exam Technique

Fehling's or Benedict's solution (red precipitate) or acidified potassium dichromate(VI) (turns green) are also valid alternatives for aldehydes, but Tollens' is classic and reliable.

(iii) Distinguishing Carbonyl Compounds [3 marks]

✅ Correct Answer

Add 2,4-dinitrophenylhydrazine (2,4-DNPH / Brady's reagent) → Orange/yellow precipitate forms.
Take the melting point of the separated crystalline product.
Compare the measured melting point to a known database/literature values.

🧠 Exam Technique

Since both compounds contain a carbonyl group (aldehyde), they both react with 2,4-DNPH. To distinguish them physically without spectroscopy, derivatives must be purified, their melting points measured, and matched against standard data books.

Mark breakdown: 1 mark for 2,4-DNPH + orange/yellow ppt; 1 mark for taking melting point; 1 mark for comparing with database values.

Part (c) — 5 Marks

Organic Synthesis Flowchart

✅ Correct Answers & Reagents

Reagent for middle box (Cyanohydrin formation): NaCN / H⁺ (or KCN/H⁺ or HCN ).
Top-right box (Reduction of nitrile + hydrogenation of C=C): Structure with -CH₂NH₂ replacing -CN and a saturated single carbon-carbon bond in the chain ( -CH₂-CH(OH)-CH₂-CH₂NH₂ ).
Bottom-left box (Reduction of aldehyde): Structure where -CHO is reduced to primary alcohol -CH₂OH , keeping the C=C double bond intact.
Bottom-right box (Hydrolysis of nitrile): Structure where -CN is hydrolysed to carboxylic acid -COOH , keeping the C=C double bond intact.

❌ Common Errors

Over-reduction traps: NaBH₄ only reduces aldehydes/ketones, not alkenes or nitriles. excess H₂/Ni reduces both the nitrile to an amine and hydrogenates the alkene chain to single bonds (and can even affect the benzene ring under harsh conditions, though examiners accept benzene ring survival here).

Mark breakdown: 1 mark per correct box/reagent (total 5 marks). Ensure skeletal or structural formulas clearly show the functional group transformations.

Part (d) — 6 Marks (Level-of-Response)

Electrophilic Addition Mechanism & Regioselectivity

✅ Mechanism Requirements

Correct Dipole: Iodine is less electronegative (2.5) than Chlorine (3.0), giving I a delta plus ( δ⁺ ) and Cl a delta minus ( δ⁻ ).
Curly Arrow 1: From the C=C double bond to the δ⁺ Iodine atom.
Curly Arrow 2: From the I-Cl bond towards the Cl atom, showing heterolytic fission to form a Cl⁻ ion.
Carbocation intermediate: Must show a positive charge on the correct carbon atom with the I atom attached to the adjacent carbon.
Curly Arrow 3: From the lone pair/negative charge on the Cl⁻ ion attacking the positive carbon atom.

💡 Explaining Major vs. Minor Products

Major product: Chlorine bonds to the carbon with the least hydrogens attached (or most alkyl groups/stabilising influence), because the reaction proceeds via the most stable carbocation intermediate.

Iodine ( δ⁺ ) attaches first to the carbon with more hydrogens, generating a more stable secondary/tertiary carbocation.

❌ Common Errors

Reversing the dipole of I-Cl (putting δ⁺ on chlorine) collapses the entire mechanism marks. Missing partial charges, starting curly arrows away from bonds/electrons, or placing the carbocation on the wrong carbon atom will restrict answers to Level 1 or 2.

Mark breakdown: Level 3 (5–6 marks) requires a complete, accurate mechanism with all curly arrows, correct dipoles, intermediate, and a sound chemical explanation of carbocation stability determining the major product.

Topics

Module 4: Core organic chemistry · Module 6: Organic chemistry and analysis · 4.1 Basic concepts and hydrocarbons · 4.2 Alcohols, haloalkanes and analysis · 6.1 Aromatic compounds, carbonyls and acids · 6.2 Nitrogen compounds, polymers and synthesis

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.