OCR A-Level Chemistry Unified chemistry (03), June 2025: Question 5

16 marks · Medium difficulty · Structured Questions

Calculate the mass of reducing agent and percentage yield for an ester reduction, outline the mechanism with curly arrows, draw a dot-and-cross diagram for NaBH4, and determine cell potential and half-equation for a borohydride fuel cell.

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Question text

5 This question is about two metal hydrides, lithium aluminium hydride, LiAlH4, and sodium

borohydride, NaBH4. Both hydrides are used as reducing agents in organic chemistry.

(a) LiAlH4, is a stronger reducing agent than NaBH4, and can reduce esters to alcohols.

A student plans to use LiAlH4 to reduce the liquid ester, ethyl phenylmethanoate, C6H5COOC2H5

(density = 1.05 g cm–3) to the alcohol, C H CH OH (M = 108).

65 2 r

The student uses C6H5COOC2H5 and LiAlH4 (an excess) in a 1.0 : 2.5 molar ratio.

The student’s method is outlined below.

Step 1 7.50 cm3 of C H COOC H is dissolved in an organic solvent.

65 2 5

Step 2 LiAlH4 is added to the solution from Step 1. The reduction takes place.

Step 3 The mixture from Step 2 is heated with dilute aqueous acid.

Step 4 The impure C6H5CH2OH is purified.

The student carries out this preparation and obtains 4.40 g of pure C6H5CH2OH.

(i) What is the systematic name of the alcohol C6H5CH2OH?

… [1]

(ii) Calculate the mass of LiAlH4 that the student should use in Step 2 and the percentage yield of

pure C6H5CH2OH obtained in Step 4.

Give the mass of LiAlH4 and the percentage yield of pure C6H5CH2OH to 3 significant figures.

Mass of LiAlH4 = … g

Percentage yield of pure C6H5CH2OH = … %

[4]

(iii) In the preparation, LiAlH4 behaves as a nucleophile.

What is meant by the term nucleophile?

… [1]

(iv) The mechanism for Step 2 in the reduction of C6H5COOC2H5 is shown below.

In the mechanism, LiAl H can be represented as H–.

Complete the mechanism by adding curly arrows to the three boxes.

C6H5 C C6H5 C OC2H5

OC2H5

H– H

C6H5 C + OC H

H–

C6H5 C H + OC2H5

[4]

(v) Suggest why the student heated the mixture from Step 2 with dilute aqueous acid in Step 3.

… [1]

(b) NaBH contains Na+ and BH – ions.

Draw a ‘dot‑and‑cross’ diagram for NaBH4.

Use a different symbol for the electrons from each element.

Show outer electrons only.

[2]

• The positive electrode contains O2.

• The negative electrode contains NaBH4.

The reaction at the positive electrode and the overall cell reaction for a NaBH4/O2 fuel cell are

shown below.

Reaction at positive electrode: O + 2H O + 4e− 4OH− E o = +0.40 V

Overall cell reaction: NaBH + 2O NaBO + 2H O E o = 1.64 V

42 2 2 cell

(i) What is the standard electrode potential of the negative electrode in this cell?

Standard electrode potential = … V [1]

(ii) Construct the equation for the reaction at the negative electrode of this cell.

… [2]

Mark scheme

Show the mark scheme

Question Answer Marks Guidance

5 (a) (i) phenylmethanol 1 ALLOW 1-phenylmethanol,

Phenylmethan-1-ol, 1-phenylmethan-1-ol

DO NOT ALLOW

Hydroxymethylbenzene OR methylphenol

(a) (ii) FIRST CHECK THE ANSWER ON ANSWER LINE 4 ALLOW ECF for each step

If mass LiAlH4 = 4.96 OR 4.97 g

AND % yield = 77.6 OR 77.5 (%) award 4 marks

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

Percentage yield 3 marks

Theoretical moles

n(C6H5COOC2H5) OR n(C6H5CH2OH)

7.50 1.05

= 150.0 OR 0.0525 (mol) ✓ NOTE: 7.50 1.05 = 7.875 g

Actual moles

4.40 Alternative method using mass

n(C6H5CH2OH) = 108.0 OR 0.0407…. (mol) ✓ 1. Theoretical moles = 0.0525 mol

Calculator = 0.04074074074

2. Mass = 0.0525 108.0 OR 5.67 g

% yield

0.0407…. 4.40

% yield = 100 = 77.6% (3 sig fig) ✓ 3. % yield = 5.67 100 = 77.6%

0.0525

ALLOW 77.5 – 77.6 % Common errors

taking into account rounding to 0.0407 7.50

No density: = 0.0500 mol x

150.0

Mass of LiAlH4 1 mark • % yield = 81.4 – 81.5% ✓✓ ECF

• LiAlH4 mass = 4.74 g ✓ ECF

mass LiAlH4 = 0.0525 2.5 37.9 = 4.97 ……. g ✓

Calculator = 4.974375 g 3SF 7.50/1.05

÷ density: 150.0 = 0.0476.. x

ALLOW 4.96 – 4.97 g

3 SF • % yield = 85.5 – 85.6 % ✓✓ ECF

taking into account rounding to 0.131 • LiAlH4 mass = 4.46 g ✓ ECF

(a) (iii) Donates an electron pair (to another species) 1 ALLOW donates a lone pair

(a) (iv) 4 IGNORE any added charges OR dipoles.

Marks solely for curly arrows

O O

For each mark, an extra curly arrow is a

C6H5 C C H C OC H

65 2 5 CON

OC2H5

H– H :

ALLOW curly arrow from

OR from – charge on O– or H–

6 curly arrows correct Curly arrow from a bond MUST go to O

✓✓✓✓ in C=O and in –OC2H5

5 curly arrows correct

✓✓✓ O IGNORE any curly arrows on bottom

structures (not in boxes):

C6H5 C + OC H

4 curly arrows correct 2 5 i.e.

✓✓ H

H–

3 curly arrows correct

✓

Annotate incorrect or absent curly arrows first with x or ^

Helps to account for marks.

(a) (v) H+ adds to –ve ion/anion/O– 1 ALLOW O– + H+ → OH

–ve ions are protonated

Protonates the product (from Step 2) IGNORE ‘to form an OH group/alcohol

OR Must include idea of H+ involved.

Provides H+ to form the alcohol

Needs idea of H+ being gained/donated

Question Answer 20 Marks Guidance

(b) 2 NOT REQUIRED

• Brackets

• Circles

• B and H symbols

IGNORE inner shells

DO NOT ALLOW ANY marks for a

covalent NaBH4

IF 8 electrons around Na, the ‘extra’

8 electrons around B electron around B must match Na electrons

AND 4 covalent bonds between B and 4 H atoms

AND Na with no electrons OR 8 electrons ALLOW other symbols than x and o

provided that the origin of the electron is

3 covalent bonds containing x or o AND clear, e.g.

1 covalent bond containing x OR o AND extra Na electron

shown differently AND charges

ALLOW dot and cross labels swapped:

i.e. • for B electrons and for H electrons

44 2

AND NaBO2 + H2O on right-hand side ✓

IGNORE O2 for the 1st mark only

ALLOW multiples

Correct balanced equation with electrons ✓ IGNORE state symbols

NaBH + 8OH– → NaBO + 6H O + 8e–

42 2

For 2nd mark, any common species on

ALLOW 1 mark for ‘correct equation’ but wrong way both sides of equation needs to be

round, i.e. NaBO + 6H O + 8e– → NaBH + 8OH– cancelled

22 4

How to answer it

Metal Hydrides: Organic Reduction, Bonding & Fuel Cells

📌 What this question tests

This synoptic question spans across Year 1 and Year 2 chemistry, assessing your ability to:

Apply IUPAC nomenclature to aromatic alcohols.
Perform multi-step reacting mass and percentage yield calculations using density and molar ratios.
Understand nucleophilic addition-elimination mechanisms for ester reduction by LiAlH₄.
Draw complex dot-and-cross diagrams for polyatomic ionic species showing origins of electrons.
Calculate electrode potentials and derive half-cell oxidation equations from fuel cell data.

Part (a)(i) — 1 Mark

Systematic Naming of C₆H₅CH₂OH

✅ Correct Answer

phenylmethanol

Also allowed: 1-phenylmethanol, phenylmethan-1-ol, 1-phenylmethan-1-ol

❌ Common Errors

Do NOT write:

hydroxymethylbenzene (not systematic IUPAC)
methylphenol / cresol (this is an isomer where -OH is directly attached to the benzene ring)
benzyl alcohol (common name, not systematic)

Mark scheme: [1 mark] for correctly identifying phenylmethanol.

Part (a)(ii) — 4 Marks

Mass of LiAlH₄ and Percentage Yield Calculation

📐 Step-by-Step Calculation

Find mass and moles of starting ester (ethyl phenylmethanoate, C₆H₅COOC₂H₅):
Mass = Volume × Density = 7.50 cm³ × 1.05 g cm⁻³ = 7.875 g
Molar mass (Mᵣ) of C₆H₅COOC₂H₅ = (8 × 12.0) + (10 × 1.0) + (2 × 16.0) = 150.0 g mol⁻¹
Moles of ester = 7.875 / 150.0 = 0.0525 mol
Calculate mass of LiAlH₄ required (1.0 : 2.5 molar ratio):
Moles of LiAlH₄ = 0.0525 mol × 2.5 = 0.13125 mol
Mᵣ of LiAlH₄ = 6.9 + 27.0 + (4 × 1.0) = 37.9 g mol⁻¹
Mass of LiAlH₄ = 0.13125 mol × 37.9 g mol⁻¹ = 4.9744 g → 4.97 g (3 sig figs; 4.96 to 4.97 g allowed)
Calculate theoretical yield and actual moles of C₆H₅CH₂OH:
Theoretical moles of C₆H₅CH₂OH = 0.0525 mol (1:1 stoichiometric conversion from ester)
Actual moles of C₆H₅CH₂OH = 4.40 g / 108.0 g mol⁻¹ = 0.04074 mol
Calculate percentage yield:
Percentage Yield = (0.04074 / 0.0525) × 100 = 77.6% (3 sig figs; 77.5 – 77.6% allowed)

✅ Final Answers (to 3 sig figs)

Mass of LiAlH₄ = 4.97 g

Percentage yield = 77.6%

❌ Common Calculation Traps

Forgetting density: Using 7.50 g directly gives 0.0500 mol (yield = 81.5%, mass = 4.74 g).
Inverting density: Dividing by density (7.50 / 1.05) gives 0.0476 mol (yield = 85.6%, mass = 4.46 g).
Significant figures: Forgetting to round both final answers to 3 sig figs costs marks.

Mark scheme: [1] for theoretical moles (0.0525 mol); [1] for actual moles (0.0407 mol); [1] for % yield (77.6%); [1] for mass of LiAlH₄ (4.97 g). Full 4 marks awarded if final answers are correct. ECF applies throughout.

Part (a)(iii) — 1 Mark

Definition of a Nucleophile

✅ Correct Definition

An electron pair donor (or donates a lone pair to another species).

🧠 Exam Technique

Always specify an electron pair. Stating "electron donor" is incomplete and will score zero because that defines a reducing agent, not a nucleophile!

Mark scheme: [1 mark] for "donates an electron pair" or "donates a lone pair".

Part (a)(iv) — 4 Marks

Mechanism: Reduction of Ester to Alcohol by Hydride (H⁻)

💡 Detailed Curly Arrow Descriptions (6 Arrows Required)

LiAlH₄ reduces the ester via a two-stage nucleophilic addition-elimination followed by a second nucleophilic addition:

Box 1 (First nucleophilic attack):
Arrow 1: From the lone pair or negative charge on :H⁻ to the carbonyl carbon atom (C=O).
Arrow 2: From the C=O double bond to the carbonyl oxygen atom.
Box 2 (Collapse of tetrahedral intermediate & loss of ethoxide):
Arrow 3: From the lone pair or negative charge on :O⁻ back into the C–O bond to reform the C=O double bond.
Arrow 4: From the C–OC₂H₅ single bond onto the oxygen of the –OC₂H₅ leaving group.
Box 3 (Second nucleophilic attack on aldehyde intermediate):
Arrow 5: From the lone pair or negative charge on the second :H⁻ to the carbonyl carbon atom (C=O) of the aldehyde.
Arrow 6: From the C=O double bond onto the carbonyl oxygen atom.

🧠 Marking Rules for Mechanisms

6 correct arrows = 4 marks
5 correct arrows = 3 marks
4 correct arrows = 2 marks
3 correct arrows = 1 mark
Every extra (incorrect) curly arrow acts as a contradiction (CON) and deducts 1 mark!

❌ Common Errors

Arrow starting from H rather than the lone pair or negative charge on :H⁻ .
Drawing an arrow pointing to the wrong atom (e.g. into the oxygen instead of the C–O bond in Box 2).
Drawing arrows on the final product outside the boxes (these are ignored).

Part (a)(v) — 1 Mark

Role of Dilute Aqueous Acid (Step 3)

✅ Correct Answer

Protonates the intermediate alkoxide ion ( C₆H₅CH₂O⁻ / negative ion) OR H⁺ adds to the negative ion/O⁻ to form the alcohol.

C₆H₅CH₂O⁻ + H⁺ → C₆H₅CH₂OH

❌ Insufficient Answers

Do not simply write "to form an alcohol" or "to neutralise the mixture". You must explicitly communicate the transfer or gain of a proton / H⁺ by the intermediate.

Mark scheme: [1 mark] for idea of H⁺ being gained/donated or alkoxide/negative ion protonated.

Part (b) — 2 Marks

'Dot-and-Cross' Diagram for NaBH₄

💡 Structure Breakdown

NaBH₄ is an ionic compound made of [Na]⁺ and [BH₄]⁻ ions:

Sodium ion: Draw [Na]⁺ with no outer electrons (or full shell of 8).
Borohydride ion: Draw [BH₄]⁻ with central Boron surrounded by 4 single covalent bonds to 4 H atoms (total 8 electrons in 4 shared pairs).
Electron symbols requirement: The question strictly states: "Use a different symbol for the electrons from each element." You need 3 distinct symbols!
• e.g. Crosses (×) for B electrons
• Dots (•) for H electrons
• Triangles (▲) or stars for the electron coming from Na
3 of the B–H bonds contain a pair from B and H. The 4th bond has the extra electron donated by Na paired with an electron from H (or B).

❌ Major Examiner Trap

Never draw NaBH₄ as a covalent molecule!

If you draw covalent bonds between Na and B or Na and H, the mark scheme states: DO NOT ALLOW ANY marks for a covalent structure.

Mark scheme:
[1 mark] 8 electrons around B, 4 covalent bonds to 4 H, and [Na]⁺ with no outer electrons (or 8).
[1 mark] 3 covalent bonds containing B & H electrons AND 1 covalent bond containing the extra electron from Na shown with a 3rd distinct symbol, with overall brackets and charges [+ and -].

Part (c)(i) — 1 Mark

Standard Electrode Potential of Negative Electrode

📐 Calculation

Formula: E°cell = E°(positive electrode) − E°(negative electrode)

Given data:

E°cell = +1.64 V
E°(positive electrode) = +0.40 V

Rearranging:

E°(negative) = E°(pos) − E°cell = +0.40 − 1.64 = −1.24 V

✅ Correct Answer

−1.24 V

Note: The negative sign (−) is essential!

Mark scheme: [1 mark] for −1.24 V (negative sign required).

Part (c)(ii) — 2 Marks

Half-Equation at the Negative Electrode

📐 Deriving the Negative Electrode Equation

Overall cell reaction:
NaBH₄ + 2O₂ → NaBO₂ + 2H₂O
Positive electrode (reduction) equation:
O₂ + 2H₂O + 4e⁻ → 4OH⁻
Multiply positive electrode reaction by 2 so O₂ cancels out:
2O₂ + 4H₂O + 8e⁻ → 8OH⁻
Subtract the positive reaction from the overall cell reaction (or recall that Oxidation occurs at the negative terminal: Reactants → Products + e⁻):
Reactants: NaBH₄ + 8OH⁻
Products: NaBO₂ + (2H₂O + 4H₂O) + 8e⁻ = NaBO₂ + 6H₂O + 8e⁻

✅ Balanced Half-Equation

NaBH₄ + 8OH⁻ → NaBO₂ + 6H₂O + 8e⁻

Also allowed: Ions separated ( BH₄⁻ + 8OH⁻ → BO₂⁻ + 6H₂O + 8e⁻ )

🧠 Mark Breakdown

Mark 1: Correct species identified ( NaBH₄ + OH⁻ on LHS and NaBO₂ + H₂O on RHS).
Mark 2: Fully balanced equation including 8e⁻ .
If written in reverse as a reduction ( NaBO₂ + 6H₂O + 8e⁻ → NaBH₄ + 8OH⁻ ), only 1 mark is awarded.

Mark scheme: [2 marks] for fully correct oxidation equation. Multiples allowed. State symbols ignored.

Topics

Module 2: Foundations in chemistry · Module 4: Core organic chemistry · Module 5: Physical chemistry and transition elements · Module 6: Organic chemistry and analysis · 2.1 Atoms and reactions · 2.2 Electrons, bonding and structure · 4.1 Basic concepts and hydrocarbons · 4.2 Alcohols, haloalkanes and analysis · 5.2 Energy · 6.1 Aromatic compounds, carbonyls and acids

Question and mark scheme from the OCR A-Level Chemistry examination, Unified chemistry (03), June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.