Chemistry 9701 / 41 — May/June 2024

DifficultyEasy

Worked solution

Answer

pH increases from $\text{Mg(OH)}_2$ to $\text{Sr(OH)}_2$ . The more soluble the hydroxide, the greater the concentration of $\text{OH}^-$ ions in the saturated solution, so the higher the pH.

Final answer

pH increases from Mg(OH)2 to Sr(OH)2

Detailed explanation

Background Concept

The pH of a saturated solution of a metal hydroxide depends on the concentration of hydroxide ions it produces. A more soluble hydroxide dissolves to give a higher $[\text{OH}^-]$ , which means a lower $[\text{H}^+]$ (via $K_w = [\text{H}^+][\text{OH}^-]$ ) and therefore a higher pH.

Understanding the Question

This asks for the variation in pH of saturated solutions of $\text{Mg(OH)}_2$ , $\text{Ca(OH)}_2$ and $\text{Sr(OH)}_2$ , and an explanation. It builds directly on part (i): the more soluble the hydroxide, the higher the $[\text{OH}^-]$ and the higher the pH.

Approach

Link the solubility trend established in part (i) to the $[\text{OH}^-]$ in a saturated solution, then to pH.

Step-by-Step Reasoning

From part (i), solubility increases down the group: $\text{Sr(OH)}_2 > \text{Ca(OH)}_2 > \text{Mg(OH)}_2$ .
A saturated solution of a more soluble hydroxide contains a higher concentration of $\text{OH}^-$ ions.
A higher $[\text{OH}^-]$ means a lower $[\text{H}^+]$ (since $K_w = [\text{H}^+][\text{OH}^-]$ is constant at a given temperature), and therefore a higher pH.
So the pH of the saturated solutions increases from $\text{Mg(OH)}_2$ to $\text{Sr(OH)}_2$ .

Key Takeaways

pH of a saturated hydroxide solution is directly linked to its solubility.
More soluble hydroxide $\rightarrow$ higher $[\text{OH}^-]$ $\rightarrow$ higher pH.
This is a qualitative application of $K_w$ .

Common Mistakes

Saying pH decreases because "more soluble means more dissociated" — dissociation of a base produces $\text{OH}^-$ , so more solubility means higher pH, not lower.
Not linking the answer to the solubility trend from part (i).

Things to Be Careful About

The answer must mention both the direction (pH increases) and the reason (higher $[\text{OH}^-]$ ).
The mark scheme specifically wants " $[\text{OH}^-]$ ions increases" — use this exact phrasing.

Techniques used

relate solubility to hydroxide ion concentrationdeduce pH trend from [OH-]

(b)

Barium hydroxide, $\text{Ba(OH)}_2$ , is a strong base.

A $250.0\text{ cm}^3$ solution of $\text{Ba(OH)}_2$ with a pH of 12.2 is made by dissolving $\text{Ba(OH)}_2$ in distilled water.

Calculate the mass of $\text{Ba(OH)}_2$ required to make this solution.

Show your working.

[ $M_{\text{r}}$ : $\text{Ba(OH)}_2$ , 171.3]

mass of $\text{Ba(OH)}_2$ = .............................. g

4M Similar

DifficultyMedium

Worked solution

Working

$[\text{H}^+] = 10^{-12.2} = 6.31 \times 10^{-13} \text{ mol dm}^{-3}$

$[\text{OH}^-] = \frac{K_w}{[\text{H}^+]} = \frac{1.00 \times 10^{-14}}{6.31 \times 10^{-13}} = 0.0158 \text{ mol dm}^{-3}$

$[\text{Ba(OH)}_2] = \frac{[\text{OH}^-]}{2} = \frac{0.0158}{2} = 7.92 \times 10^{-3} \text{ mol dm}^{-3}$

$n(\text{Ba(OH)}_2) = 7.92 \times 10^{-3} \times 0.2500 = 1.98 \times 10^{-3} \text{ mol}$

$\text{mass} = 1.98 \times 10^{-3} \times 171.3 = 0.339 \text{ g}$

Answer

0.339 g

Final answer

0.339 g

Detailed explanation

Background Concept

pH is defined as $\text{pH} = -\log_{10}[\text{H}^+]$ . In aqueous solution at 298 K, the ionic product of water is $K_w = [\text{H}^+][\text{OH}^-] = 1.00 \times 10^{-14} \text{ mol}^2 \text{ dm}^{-6}$ . A strong base like $\text{Ba(OH)}_2$ fully dissociates in water:

\text{Ba(OH)}_2(\text{aq}) \rightarrow \text{Ba}^{2+}(\text{aq}) + 2\text{OH}^-(\text{aq})

So each mole of $\text{Ba(OH)}_2$ produces two moles of $\text{OH}^-$ .

Understanding the Question

We need to calculate the mass of $\text{Ba(OH)}_2$ ( $M_r = 171.3$ ) required to make $250.0 \text{ cm}^3$ of solution with pH 12.2. The working must be shown.

Approach

Work backwards from the pH: pH $\rightarrow$ $[\text{H}^+]$ $\rightarrow$ $[\text{OH}^-]$ (via $K_w$ ) $\rightarrow$ $[\text{Ba(OH)}_2]$ (via stoichiometry) $\rightarrow$ moles (via volume) $\rightarrow$ mass (via $M_r$ ). This is a standard chain of calculations.

Step-by-Step Reasoning

From pH to $[\text{H}^+]$ : $[\text{H}^+] = 10^{-12.2} = 6.31 \times 10^{-13} \text{ mol dm}^{-3}$ .
From $[\text{H}^+]$ to $[\text{OH}^-]$ : $[\text{OH}^-] = K_w/[\text{H}^+] = (1.00 \times 10^{-14})/(6.31 \times 10^{-13}) = 0.0158 \text{ mol dm}^{-3}$ . (Alternative: pOH = 14 − 12.2 = 1.8, so $[\text{OH}^-] = 10^{-1.8} = 0.0158 \text{ mol dm}^{-3}$ .)
From $[\text{OH}^-]$ to $[\text{Ba(OH)}_2]$ : Since each $\text{Ba(OH)}_2$ gives 2 $\text{OH}^-$ , $[\text{Ba(OH)}_2] = [\text{OH}^-]/2 = 0.0158/2 = 7.92 \times 10^{-3} \text{ mol dm}^{-3}$ .
From concentration to moles: $n = c \times V = 7.92 \times 10^{-3} \times 0.2500 = 1.98 \times 10^{-3} \text{ mol}$ . (Note: $250.0 \text{ cm}^3 = 0.2500 \text{ dm}^3$ .)
From moles to mass: $m = n \times M_r = 1.98 \times 10^{-3} \times 171.3 = 0.339 \text{ g}$ .

Key Takeaways

The pH $\rightarrow$ $[\text{H}^+]$ $\rightarrow$ $[\text{OH}^-]$ $\rightarrow$ concentration $\rightarrow$ moles $\rightarrow$ mass chain is a standard A-Level calculation.
Always divide $[\text{OH}^-]$ by 2 for $\text{Ba(OH)}_2$ (two $\text{OH}^-$ per formula unit).
Convert $\text{cm}^3$ to $\text{dm}^3$ before using in $c \times V$ .

Common Mistakes

Forgetting to divide $[\text{OH}^-]$ by 2 to get $[\text{Ba(OH)}_2]$ — this is the most common error.
Using 250 instead of 0.250 $\text{dm}^3$ (volume conversion error).
Using $[\text{H}^+]$ directly as $[\text{Ba(OH)}_2]$ without going through $[\text{OH}^-]$ .
Rounding intermediate values too aggressively, leading to a final answer off by a few percent.

Things to Be Careful About

$K_w = 1.00 \times 10^{-14}$ at 298 K — use this value.
The final answer should be given to 3 significant figures: 0.339 g.
The mark scheme uses "ecf" (error carried forward) — a wrong $[\text{OH}^-]$ can still earn later marks if used correctly.

Techniques used

convert pH to [H+]use Kw to find [OH-]apply stoichiometry of Ba(OH)2calculate mass from moles and molar mass

(c)

The solubility of iron(II) hydroxide, $\text{Fe(OH)}_2$ , is $5.85 \times 10^{-6}\text{ mol dm}^{-3}$ at $298\text{ K}$ .

3M

(i)

Write the expression for the solubility product, $K_{\text{sp}}$ , of $\text{Fe(OH)}_2$ .

$K_{\text{sp}} =$

DifficultyEasy

Worked solution

Answer

$K_{\text{sp}} = [\text{Fe}^{2+}][\text{OH}^-]^2$

Final answer

Ksp = [Fe2+][OH-]^2

Detailed explanation

Background Concept

The solubility product, $K_{\text{sp}}$ , is the equilibrium constant for the dissolution of a sparingly soluble ionic compound. For the dissolution of $\text{Fe(OH)}_2$ :

\text{Fe(OH)}_2(\text{s}) \rightleftharpoons \text{Fe}^{2+}(\text{aq}) + 2\text{OH}^-(\text{aq})

The equilibrium expression is:

K_{\text{sp}} = [\text{Fe}^{2+}][\text{OH}^-]^2

The solid $\text{Fe(OH)}_2$ is not included in the expression because its concentration (activity) is constant.

Understanding the Question

This is a recall task: write the $K_{\text{sp}}$ expression for $\text{Fe(OH)}_2$ . The mark scheme accepts exactly $K_{\text{sp}} = [\text{Fe}^{2+}][\text{OH}^-]^2$ .

Approach

Identify the ions formed on dissolution and their stoichiometric coefficients, then write the product of ion concentrations each raised to its stoichiometric coefficient.

Step-by-Step Reasoning

$\text{Fe(OH)}_2$ dissociates into $\text{Fe}^{2+}$ and $2\text{OH}^-$ .
The $K_{\text{sp}}$ expression is the product of the ion concentrations, each raised to the power of its stoichiometric coefficient: $K_{\text{sp}} = [\text{Fe}^{2+}][\text{OH}^-]^2$ .
The solid $\text{Fe(OH)}_2$ is omitted.

Key Takeaways

The solid phase is always omitted from the $K_{\text{sp}}$ expression.
The exponent on each ion equals its stoichiometric coefficient in the dissolution equation.

Common Mistakes

Including $\text{Fe(OH)}_2$ in the expression — the solid is always omitted.
Writing $[\text{OH}^-]$ without the power of 2.

Things to Be Careful About

The square on $[\text{OH}^-]$ is essential — it reflects the 2:1 stoichiometry.

Techniques used

write the solubility product expression

(ii)

Calculate the value of $K_{\text{sp}}$ of $\text{Fe(OH)}_2$ . Include its units.

$K_{\text{sp}} =$ ..............................

units = ..............................

DifficultyMedium-Easy

Worked solution

Working

Let $s$ = solubility of $\text{Fe(OH)}_2$ = $5.85 \times 10^{-6} \text{ mol dm}^{-3}$

$[\text{Fe}^{2+}] = s$ , $[\text{OH}^-] = 2s$

$K_{\text{sp}} = s(2s)^2 = 4s^3 = 4 \times (5.85 \times 10^{-6})^3 = 8.01 \times 10^{-16}$

Units: $\text{mol dm}^{-3} \times (\text{mol dm}^{-3})^2 = \text{mol}^3 \text{ dm}^{-9}$

Answer

$K_{\text{sp}} = 8.01 \times 10^{-16} \text{ mol}^3 \text{ dm}^{-9}$

Final answer

8.01 × 10^-16 mol^3 dm^-9

Detailed explanation

Background Concept

If the molar solubility of $\text{Fe(OH)}_2$ is $s \text{ mol dm}^{-3}$ , then at equilibrium:

$[\text{Fe}^{2+}] = s$ (one $\text{Fe}^{2+}$ per formula unit)
$[\text{OH}^-] = 2s$ (two $\text{OH}^-$ per formula unit)

Substituting into $K_{\text{sp}} = [\text{Fe}^{2+}][\text{OH}^-]^2$ gives:

K_{\text{sp}} = s \times (2s)^2 = 4s^3

The units of $K_{\text{sp}}$ depend on the stoichiometry: for a 1:2 salt, the units are $\text{mol}^3 \text{ dm}^{-9}$ .

Understanding the Question

Calculate $K_{\text{sp}}$ from the given solubility ( $5.85 \times 10^{-6} \text{ mol dm}^{-3}$ ) and state the units. Two marks: one for the value, one for the units.

Approach

Substitute $s$ into $K_{\text{sp}} = 4s^3$ , then determine the units from the expression.

Step-by-Step Reasoning

$s = 5.85 \times 10^{-6} \text{ mol dm}^{-3}$ .
$K_{\text{sp}} = 4s^3 = 4 \times (5.85 \times 10^{-6})^3$ .
Calculate $(5.85 \times 10^{-6})^3 = 5.85^3 \times 10^{-18} = 200.2 \times 10^{-18} = 2.002 \times 10^{-16}$ .
$K_{\text{sp}} = 4 \times 2.002 \times 10^{-16} = 8.01 \times 10^{-16}$ .
Units: $[\text{Fe}^{2+}]$ has units $\text{mol dm}^{-3}$ ; $[\text{OH}^-]^2$ has units $(\text{mol dm}^{-3})^2 = \text{mol}^2 \text{ dm}^{-6}$ . Product: $\text{mol}^3 \text{ dm}^{-9}$ .

Key Takeaways

For a 1:2 salt (like $\text{Fe(OH)}_2$ , $\text{Ca(OH)}_2$ , $\text{Ba(OH)}_2$ ), $K_{\text{sp}} = 4s^3$ .
The units of $K_{\text{sp}}$ depend on the stoichiometry of the dissolution — always derive them from the expression.

Common Mistakes

Forgetting the factor of 4 (using $s^3$ instead of $4s^3$ ).
Getting the units wrong — e.g. writing $\text{mol}^2 \text{ dm}^{-6}$ instead of $\text{mol}^3 \text{ dm}^{-9}$ .
Arithmetic errors in cubing $5.85 \times 10^{-6}$ .

Things to Be Careful About

The units must match the expression: $\text{mol}^3 \text{ dm}^{-9}$ .
Keep enough significant figures in intermediate steps (the final answer is $8.01 \times 10^{-16}$ , three significant figures).

Techniques used

relate ion concentrations to solubilitysubstitute into Ksp expressiondetermine units of Ksp

Q2MediumTransition ElementsElectrochemistry

(a)

2M

(i)

Define transition element.

DifficultyEasy

Worked solution

Answer

An element which forms one or more stable ion with incomplete d orbitals.

Final answer

An element which forms one or more stable ion with incomplete d orbitals.

Detailed explanation

Background Concept

Transition elements are defined not by their neutral atom configuration, but by the properties of the ions they form. While all transition metals have incomplete d subshells in their neutral atomic state (except Sc and Zn), the formal IUPAC definition requires that the element must form at least one stable ion with a partially filled d subshell. This is why Scandium (forms Sc³⁺ with empty d⁰) and Zinc (forms Zn²⁺ with full d¹⁰) are typically not classified as transition elements in this context.

Understanding the Question

The question asks for the standard definition of a transition element. This is a direct recall question testing precise terminology.

Approach

State the definition exactly as recognised by the examining body, ensuring the key phrases 'stable ion' and 'incomplete d orbitals' are included.

Step-by-Step Reasoning

The core of the definition is that the d-orbitals must be incomplete.
This incompleteness must occur in a 'stable ion', not just the neutral atom.
Combining these gives the mark-scheme answer: 'element which forms one or more stable ion with incomplete d orbitals'.

Key Takeaways

Always define transition elements in terms of their ions, not their neutral atoms. The phrase 'one or more stable ion' is crucial because some elements might have incomplete d-orbitals in a transient or unstable oxidation state.

Common Mistakes

Defining them as 'elements with incomplete d orbitals' (missing the 'stable ion' condition).
Saying 'd-block elements' (the d-block is a block of the periodic table; not all d-block elements are transition elements, e.g., Zn, Sc).

Things to Be Careful About

Ensure the wording matches the mark scheme: 'stable ion' and 'incomplete d orbitals' are the exact creditable phrases.

Techniques used

recall definition of transition element

(ii)

Explain why transition elements can form complex ions.

DifficultyEasy

Worked solution

Answer

They have vacant d orbitals that are energetically accessible (to accept lone pairs from ligands).

Final answer

They have vacant d orbitals that are energetically accessible.

Detailed explanation

Background Concept

Complex ions form when a central metal ion is surrounded by ligands. Ligands are molecules or ions that possess at least one lone pair of electrons. The bond formed between the ligand and the metal ion is a coordinate (dative covalent) bond, where both electrons in the bond come from the ligand.

Understanding the Question

The question asks why transition elements, in particular, are capable of forming complex ions. We need to identify the structural feature of transition metal ions that allows them to accept electron pairs.

Approach

Recall the requirements for coordinate bond formation: a lone pair donor (ligand) and an electron pair acceptor (central ion). Transition metal ions have accessible empty orbitals that can accept these lone pairs.

Step-by-Step Reasoning

Ligands need a vacant orbital to accept the lone pair of electrons.
Transition metal ions have (n-1)d orbitals that are partially filled or empty and are close in energy to the s and p orbitals.
These vacant d orbitals are energetically accessible to accept lone pairs from incoming ligands.
Therefore, the presence of 'vacant d orbitals that are energetically accessible' is the correct explanation.

Key Takeaways

The ability to form complexes is a direct consequence of having energetically accessible vacant orbitals (specifically d orbitals in transition metals) to accept electron pairs from ligands.

Common Mistakes

Saying 'they have d orbitals' (all d-block elements have d orbitals; the key is that they are vacant and energetically accessible).
Saying 'they have high charge density' (while true for small highly charged ions, it's not the primary reason for complex formation; the orbital availability is the fundamental reason).

Things to Be Careful About

Use the exact phrasing from the mark scheme: 'vacant d orbitals' and 'energetically accessible'.

Techniques used

identify vacant orbitals for ligand donation

(b)

The 3d orbitals in an isolated $\text{Ag}^+$ ion are degenerate.

2M

(i)

Define degenerate d orbitals.

(ii)

Sketch the shape of a $3\text{d}_{xy}$ orbital in Fig. 2.1.

Fig. 2.1

(c)

Tollens’ reagent can be used to distinguish between aldehydes and ketones. Tollens’ reagent contains $[\text{Ag(NH}_3)_2]\text{OH}$ , which can be prepared in a two-step process.

step 1 Aqueous $\text{NaOH}$ is added dropwise to aqueous $\text{AgNO}_3$ to form $\text{Ag}_2\text{O}$ as a brown precipitate.

step 2 Aqueous $\text{NH}_3$ is added dropwise to $\text{Ag}_2\text{O}$ to form a colourless solution containing $[\text{Ag(NH}_3)_2]\text{OH}$ .

Construct equations for each of the steps in the preparation of $[\text{Ag(NH}_3)_2]\text{OH}$ .

step 1 ........................................................................................................................................

step 2 ........................................................................................................................................

DifficultyMedium-Easy

Worked solution

Answer

step 1:
$2\text{AgNO}_3 + 2\text{NaOH} \rightarrow \text{Ag}_2\text{O} + 2\text{NaNO}_3 + \text{H}_2\text{O}$
or $2\text{Ag}^+ + 2\text{OH}^- \rightarrow \text{Ag}_2\text{O} + \text{H}_2\text{O}$

step 2:
$\text{Ag}_2\text{O} + 4\text{NH}_3 + \text{H}_2\text{O} \rightarrow 2[\text{Ag}(\text{NH}_3)_2]\text{OH}$
or $\text{Ag}^+ + 2\text{NH}_3 + \text{OH}^- \rightarrow [\text{Ag}(\text{NH}_3)_2]\text{OH}$

Final answer

step 1: 2AgNO3 + 2NaOH -> Ag2O + 2NaNO3 + H2O (or 2Ag+ + 2OH- -> Ag2O + H2O); step 2: Ag2O + 4NH3 + H2O -> 2[Ag(NH3)2]OH (or Ag+ + 2NH3 + OH- -> [Ag(NH3)2]OH)

Detailed explanation

Background Concept

Tollens' reagent is the diamminesilver(I) complex, [Ag(NH3)2]+. It is prepared by first precipitating silver oxide from silver nitrate using sodium hydroxide, and then dissolving the oxide in aqueous ammonia. Silver(I) ions react with hydroxide ions to form a brown precipitate of silver oxide (Ag2O), not the soluble silver hydroxide (AgOH is unstable and dehydrates immediately). When ammonia is added, it acts as a ligand, replacing the oxide/hydroxide to form the soluble complex ion.

Understanding the Question

The question asks for equations for the two steps in preparing Tollens' reagent. Molecular and ionic equations are both acceptable where specified.

Approach

Write the balanced equation for the precipitation of Ag2O (step 1) and the dissolution of Ag2O in ammonia to form the complex (step 2). Provide both molecular and ionic forms where the mark scheme allows.

Step-by-Step Reasoning

Step 1:

Reactants: AgNO3(aq) and NaOH(aq).
Products: Ag2O(s) is the brown precipitate. NaNO3 remains in solution. Water is also formed from the combination of H+ and OH- (or from the dehydration of AgOH).
Balanced molecular equation: $2\text{AgNO}_3 + 2\text{NaOH} \rightarrow \text{Ag}_2\text{O} + 2\text{NaNO}_3 + \text{H}_2\text{O}$ .
Net ionic equation: $2\text{Ag}^+ + 2\text{OH}^- \rightarrow \text{Ag}_2\text{O} + \text{H}_2\text{O}$ .

Step 2:

Reactants: Ag2O(s) and NH3(aq). Water is also involved to provide OH- for the final product.
Products: [Ag(NH3)2]OH(aq), which is colourless.
Balanced molecular equation: $\text{Ag}_2\text{O} + 4\text{NH}_3 + \text{H}_2\text{O} \rightarrow 2[\text{Ag}(\text{NH}_3)_2]\text{OH}$ .
Ionic equation: $\text{Ag}^+ + 2\text{NH}_3 + \text{OH}^- \rightarrow [\text{Ag}(\text{NH}_3)_2]\text{OH}$ (Note: this is a simplified ionic form; the full ionic would involve Ag2O dissolving, but the mark scheme accepts this net form showing complex formation).

Key Takeaways

Silver oxide is the key intermediate. It is insoluble in water but soluble in excess ammonia due to complex ion formation. Always balance carefully; Ag2O requires 2 Ag atoms, so 4 NH3 are needed for 2 complex ions.

Common Mistakes

Writing AgOH as the product in step 1 (AgOH is unstable and immediately forms Ag2O).
Forgetting the H2O in step 2 (needed to provide the OH- ion in the final product and to balance oxygen/hydrogen).
Writing the complex as [Ag(NH3)2]+ without the OH- counter ion in the molecular equation.

Things to Be Careful About

Ensure equations are fully balanced. The mark scheme explicitly allows ionic equations, which are often cleaner, but if writing molecular equations, ensure all products (including NaNO3 and H2O) are included and balanced.

Techniques used

write balanced molecular equationswrite net ionic equations

(d)

Name the shape of the complex ion $[\text{Ag(NH}_3)_2]^+$ .

State the bond angle for H-N-Ag and for N-Ag-N.

shape ........................................................................................................................................

bond angle for H-N-Ag = .............. °

bond angle for N-Ag-N = .............. °

DifficultyEasy

Worked solution

Answer

shape: linear
bond angle for H–N–Ag: 109.5 °
bond angle for N–Ag–N: 180 °

Final answer

shape: linear; bond angle for H-N-Ag = 109.5; bond angle for N-Ag-N = 180

Detailed explanation

Background Concept

The shape of a complex ion is determined by its coordination number (the number of ligand donor atoms bonded to the central metal ion). A coordination number of 2 always results in a linear geometry with a bond angle of 180°. For the ligand itself (ammonia, NH3), the nitrogen atom has 3 bonding pairs and 1 lone pair, giving a tetrahedral electron geometry and a trigonal pyramidal molecular geometry, with bond angles close to the tetrahedral angle of 109.5°.

Understanding the Question

The question asks for the shape of [Ag(NH3)2]+, the N-Ag-N bond angle, and the H-N-Ag bond angle. Note the trick: the H-N-Ag angle depends on the geometry around the nitrogen atom in the ammonia ligand, not the silver ion.

Approach

Determine the coordination number of Ag+ in the complex (it's 2, from two NH3 ligands).
State the shape and N-Ag-N angle for the complex.
Analyze the local geometry around the nitrogen atom in NH3 to find the H-N-Ag angle.

Step-by-Step Reasoning

Complex shape: Ag+ is bonded to 2 nitrogen atoms (from 2 NH3 ligands). Coordination number = 2. The shape is linear. The bond angle N–Ag–N is 180°.
Ligand geometry: In the NH3 ligand, the nitrogen atom is bonded to 3 hydrogen atoms and 1 silver atom (via a coordinate bond). This gives 4 electron domains (3 N-H bonds + 1 N-Ag coordinate bond). The electron geometry is tetrahedral. The bond angle around a tetrahedral center is 109.5°. Therefore, the H–N–Ag bond angle is 109.5°.

Key Takeaways

Always check which atom's geometry the question is asking about. The N-Ag-N angle is about the central metal (linear, 180°). The H-N-Ag angle is about the ligand atom (tetrahedral electron geometry, ~109.5°).

Common Mistakes

Saying the shape is 'tetrahedral' (confusing coordination number 2 with 4).
Saying the H-N-Ag angle is 107° (the lone pair on nitrogen is now involved in the coordinate bond to Ag, so there are 4 bonding pairs and 0 lone pairs around N in terms of domains, making it effectively tetrahedral 109.5°. Even if considered as 3 bonds + 1 lone pair originally, the coordinate bond uses the lone pair, so the 4 domains are all bonding: 3 N-H and 1 N-Ag. Thus 109.5° is correct).
Forgetting that the question asks for H-N-Ag, not just N-Ag-N.

Things to Be Careful About

The mark scheme specifically awards a mark for 109.5° for H-N-Ag. This is a common trap question. Ensure you distinguish between the geometry around the metal and the geometry around the ligand atom.

Techniques used

determine molecular geometry from coordination numberidentify bond angles in ligands and complex

(e)

An electrochemical cell uses $\text{Ag}_2\text{O}$ as the positive electrode and $\text{Zn}$ as the negative electrode immersed in an alkaline electrolyte.

The overall cell reaction is shown.

\text{Ag}_2\text{O} + \text{Zn} + \text{H}_2\text{O} \rightarrow 2\text{Ag} + \text{Zn(OH)}_2

Complete the half-equation for the reaction at each electrode.

at the positive electrode $\text{Ag}_2\text{O} +$ .....................................................................................

at the negative electrode $\text{Zn} +$ ........................................................................................

DifficultyMedium-Easy

Worked solution

Answer

at the positive electrode (reduction):
$\text{Ag}_2\text{O} + \text{H}_2\text{O} + 2\text{e}^- \rightarrow 2\text{Ag} + 2\text{OH}^-$

at the negative electrode (oxidation):
$\text{Zn} + 2\text{OH}^- \rightarrow \text{Zn(OH)}_2 + 2\text{e}^-$

Final answer

positive: Ag2O + H2O + 2e- -> 2Ag + 2OH-; negative: Zn + 2OH- -> Zn(OH)2 + 2e-

Detailed explanation

Background Concept

In an electrochemical cell, the positive electrode is the cathode (reduction occurs here) and the negative electrode is the anode (oxidation occurs here). The overall reaction is given: $\text{Ag}_2\text{O} + \text{Zn} + \text{H}_2\text{O} \rightarrow 2\text{Ag} + \text{Zn(OH)}_2$ . We need to split this into two half-equations. The electrolyte is alkaline, so we must balance oxygen and hydrogen using H2O and OH- ions, not H+.

Understanding the Question

The question asks to complete the half-equations for the positive and negative electrodes based on the overall cell reaction. Ag2O is reduced to Ag, and Zn is oxidised to Zn(OH)2.

Approach

Identify the reduction half-reaction (positive electrode): Ag2O -> Ag. Balance Ag, then O using H2O, then H using OH- (or vice versa), then balance charge with electrons.
Identify the oxidation half-reaction (negative electrode): Zn -> Zn(OH)2. Balance Zn, then O and H using OH-, then balance charge with electrons.

Step-by-Step Reasoning

Positive electrode (Cathode, Reduction):

Reactant: Ag2O. Product: Ag.
Balance Ag: $\text{Ag}_2\text{O} \rightarrow 2\text{Ag}$ .
Balance O: Add H2O to the right? No, add H2O to the left and OH- to the right (alkaline conditions).
$\text{Ag}_2\text{O} + \text{H}_2\text{O} \rightarrow 2\text{Ag} + 2\text{OH}^-$ .
Balance charge: Left is 0, right is -2. Add 2e- to the left.
Final: $\text{Ag}_2\text{O} + \text{H}_2\text{O} + 2\text{e}^- \rightarrow 2\text{Ag} + 2\text{OH}^-$ .

Negative electrode (Anode, Oxidation):

Reactant: Zn. Product: Zn(OH)2.
Balance Zn: $\text{Zn} \rightarrow \text{Zn(OH)}_2$ .
Balance O and H: Add 2OH- to the left.
$\text{Zn} + 2\text{OH}^- \rightarrow \text{Zn(OH)}_2$ .
Balance charge: Left is -2, right is 0. Add 2e- to the right.
Final: $\text{Zn} + 2\text{OH}^- \rightarrow \text{Zn(OH)}_2 + 2\text{e}^-$ .
Check: Add the two half-equations: $\text{Ag}_2\text{O} + \text{H}_2\text{O} + 2\text{e}^- + \text{Zn} + 2\text{OH}^- \rightarrow 2\text{Ag} + 2\text{OH}^- + \text{Zn(OH)}_2 + 2\text{e}^-$ . Cancel 2e- and 2OH-: $\text{Ag}_2\text{O} + \text{H}_2\text{O} + \text{Zn} \rightarrow 2\text{Ag} + \text{Zn(OH)}_2$ . Matches the overall reaction perfectly.

Key Takeaways

When balancing half-equations in alkaline solution, use H2O and OH- to balance oxygen and hydrogen. Always check that the electrons lost at the anode equal the electrons gained at the cathode.

Common Mistakes

Using H+ to balance hydrogen (wrong for alkaline electrolyte).
Forgetting to balance the charge with electrons.
Swapping the electrodes (positive is always reduction, negative is always oxidation).

Things to Be Careful About

Ensure the number of electrons is the same in both half-equations so they can be added to give the overall reaction. Here, both involve 2 electrons, which is convenient.

Techniques used

balance half-equations in alkaline mediumidentify oxidation and reduction sites

(f)

Coordination polymers are made when a bidentate ligand acts as a bridge between different metal ions.

Under certain conditions $\text{Ru}^{3+}\text{(aq)}$ and the bidentate ligand dps can form a coordination polymer containing $([\text{Ru(dps)Cl}_4]^-)_n$ chains.

Fig. 2.2

The bidentate ligand dps uses each of the nitrogen atoms to bond to a different $\text{Ru}^{3+}$ .

Complete Fig. 2.3 by drawing the structure for the coordination polymer $([\text{Ru(dps)Cl}_4]^-)_n$ .
Show two repeat units.

The dps ligand can be represented using N⌒N.

Fig. 2.3

DifficultyMedium

Worked solution

Answer

Final answer

Two Ru centers, each octahedrally coordinated to 4 Cl ligands and 2 N atoms from dps ligands bridging to adjacent Ru centers.

Detailed explanation

Background Concept

Coordination polymers are extended structures where metal ions are linked by bridging ligands. The ligand 'dps' (4,4'-dipyridyl sulfide, though the mark scheme image description mentions 1,3-di(4-pyridyl)propane, we follow the mark scheme's structural representation: two pyridine rings connected by a chain, with N atoms acting as donors). The dps ligand is bidentate, using its two nitrogen atoms to coordinate to different metal ions, thus bridging them.

The complex ion is $[\text{Ru(dps)Cl}_4]^-$ . Ru is in a +3 oxidation state (given as Ru³⁺). Cl is -1 (4 x -1 = -4). dps is neutral. Overall charge = +3 - 4 = -1. This matches the formula.

Each Ru³⁺ has a coordination number of 6 (octahedral geometry). It is bonded to 4 Cl ligands and 2 N atoms from dps ligands.

Understanding the Question

The question asks to complete the diagram showing two repeat units of the coordination polymer $([\text{Ru(dps)Cl}_4]^-)_n$ . The dps ligand is represented as N⌒N (a bidentate bridge). We need to show how Ru centers are linked by dps ligands, with the remaining coordination sites filled by Cl ligands.

Approach

Draw two Ru atoms.
Around each Ru, draw 4 Cl ligands (typically in a square planar or octahedral arrangement; the mark scheme shows 4 Cl in a roughly square planar arrangement around Ru, with the N atoms axial).
Draw the dps ligand bridging the two Ru atoms: Ru-N-CH2-CH2-CH2-N-Ru (or as per the mark scheme image: Ru-N-(chain)-N-Ru).
Extend the chain with dashed lines to indicate the polymer continues.

Step-by-Step Reasoning

Central atoms: Draw two Ru atoms.
Ligands on Ru: Each Ru needs 4 Cl atoms and 2 N atoms to complete its octahedral coordination (coordination number 6).
Bridging: The dps ligand connects two Ru atoms. One N atom bonds to the left Ru, the other N atom bonds to the right Ru. The chain between the N atoms (the sulfide bridge or propane chain as per the mark scheme's Fig 2f representation) is drawn between them.
Repeat units: Draw a second dps ligand bridging the right Ru to a third (implied) Ru on the right, using a dashed line. Similarly, draw a dashed line from the left N of the first dps to the left.
Structure:
Left Ru: bonded to 4 Cl (up, down, left-down, right-down roughly) and one N from the central dps ligand.
Central dps: N-CH2-CH2-CH2-N (or as drawn in mark scheme: a chain connecting two N's).
Right Ru: bonded to the other N of the central dps, 4 Cl, and another N from the next dps ligand (dashed line to the right).
The mark scheme figure shows: --- Ru(Cl)4 - N - (chain) - N - Ru(Cl)4 - N - (chain) - N ---

Key Takeaways

In coordination polymers, bidentate ligands bridge metal centers. Each metal's coordination sphere must be completed. Here, Ru is octahedral (CN=6), using 2 sites for bridging dps and 4 sites for terminal Cl ligands.

Common Mistakes

Drawing dps as a monodentate ligand (it must bridge two Ru atoms).
Giving Ru the wrong coordination number (must be 6, so 4 Cl + 2 N).
Forgetting the dashed lines to indicate the polymer chain continues.
Drawing the wrong chain between the N atoms (follow the mark scheme's representation of the dps backbone).

Things to Be Careful About

The mark scheme awards marks for: (1) presence of dps ligand bonded to two Ru, and (2) the rest of the structure correct (i.e., 4 Cl on each Ru, correct connectivity). Ensure the Cl ligands are clearly drawn attached to Ru and not to the dps ligand.

Techniques used

draw coordination polymer structureidentify bridging ligand connectivity

Q3MediumGroup 2Transition ElementsEquilibriaElectrochemistry

(a)

When a sample of hydrated lithium ethanedioate, $\text{Li}_2\text{C}_2\text{O}_4\cdot\text{H}_2\text{O}$ , is gently heated, two gaseous products are formed and a white solid residue remains.

The residue is added to $\text{HNO}_3\text{(aq)}$ . A gas is produced that turns limewater milky.

Complete the equation for the decomposition of $\text{Li}_2\text{C}_2\text{O}_4\cdot\text{H}_2\text{O}$ .

\text{Li}_2\text{C}_2\text{O}_4\cdot\text{H}_2\text{O} \rightarrow \text{....................................} + \text{....................................} + \text{....................................}

DifficultyMedium-Easy

Worked solution

Answer

\text{Li}_2\text{C}_2\text{O}_4\cdot\text{H}_2\text{O} \rightarrow \text{Li}_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}

Final answer

Li2C2O4.H2O -> Li2CO3 + H2O + CO

Detailed explanation

Background Concept

When hydrated metal ethanedioates (oxalates) are gently heated, they undergo thermal decomposition. The water of crystallisation is released as steam, and the anhydrous ethanedioate decomposes into a metal carbonate and carbon monoxide. This is analogous to the decomposition of Group 2 ethanedioates, where the smaller, more highly charged cations polarise the large ethanedioate ion ( $\text{C}_2\text{O}_4^{2-}$ ), weakening the C–C bond and causing it to break into $\text{CO}_3^{2-}$ and $\text{CO}$ .

Understanding the Question

The question states that heating hydrated lithium ethanedioate produces two gases and a white solid. The solid residue reacts with nitric acid to produce a gas that turns limewater milky, which is the standard test for carbon dioxide ( $\text{CO}_2$ ). This confirms the white solid is a carbonate, specifically lithium carbonate ( $\text{Li}_2\text{CO}_3$ ). The two gases must therefore be water vapour (from the hydrate) and carbon monoxide (from the decomposition of the ethanedioate ion).

Approach

Identify the white solid residue as $\text{Li}_2\text{CO}_3$ based on the $\text{CO}_2$ test with $\text{HNO}_3$ .
Recognise that the water of crystallisation is released as a gas ( $\text{H}_2\text{O}$ ).
Balance the remaining atoms: the $\text{C}_2\text{O}_4^{2-}$ ion splits into $\text{CO}_3^{2-}$ (in the solid) and $\text{CO}$ (gas).
Write the balanced equation with state symbols if required, though the mark scheme focuses on the correct formulae.

Step-by-Step Reasoning

Residue identification: The reaction $\text{Li}_2\text{CO}_3 + 2\text{HNO}_3 \rightarrow 2\text{LiNO}_3 + \text{H}_2\text{O} + \text{CO}_2$ produces $\text{CO}_2$ , which turns limewater milky. Thus, the residue is $\text{Li}_2\text{CO}_3$ .
Gaseous products: The hydrate loses water: $\text{H}_2\text{O(g)}$ . The ethanedioate ion decomposes: $\text{C}_2\text{O}_4^{2-} \rightarrow \text{CO}_3^{2-} + \text{CO(g)}$ . The second gas is $\text{CO}$ .
Balancing: One mole of $\text{Li}_2\text{C}_2\text{O}_4\cdot\text{H}_2\text{O}$ yields one mole each of $\text{Li}_2\text{CO}_3$ , $\text{H}_2\text{O}$ , and $\text{CO}$ . The equation is already balanced with coefficients of 1.

\text{Li}_2\text{C}_2\text{O}_4\cdot\text{H}_2\text{O} \rightarrow \text{Li}_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}

Key Takeaways

Hydrated ethanedioates decompose to give the metal carbonate, water, and carbon monoxide.
Carbonates react with acids to produce $\text{CO}_2$ , which is identified by turning limewater milky.

Common Mistakes

Writing $\text{CO}_2$ as a decomposition product of the ethanedioate ion itself (it is produced when the carbonate reacts with acid, not during the initial thermal decomposition of the ethanedioate).
Forgetting to include the water of crystallisation as a product.
Balancing the equation incorrectly (e.g., producing $\text{CO}_2$ and $\text{C}$ instead of $\text{CO}$ ).

Things to Be Careful About

Ensure the formula for the ethanedioate ion is $\text{C}_2\text{O}_4^{2-}$ and correctly splits into $\text{CO}_3^{2-}$ and $\text{CO}$ .
State symbols are not strictly required by this specific mark scheme entry, but adding (s), (g), (l) correctly is good practice: $\text{Li}_2\text{C}_2\text{O}_4\cdot\text{H}_2\text{O}(\text{s}) \rightarrow \text{Li}_2\text{CO}_3(\text{s}) + \text{H}_2\text{O}(\text{g}) + \text{CO}(\text{g})$ .

Techniques used

deduce decomposition products from residue testsbalance thermal decomposition equation

(b)

The trend in the decomposition temperatures of the Group 2 ethanedioates is similar to that of the Group 2 nitrates.

Suggest which of $\text{CaC}_2\text{O}_4$ and $\text{BaC}_2\text{O}_4$ will decompose at the lower temperature. Explain your answer.

DifficultyMedium-Easy

Worked solution

Answer

$\text{CaC}_2\text{O}_4$ will decompose at the lower temperature.

$\text{Ca}^{2+}$ has a smaller ionic radius than $\text{Ba}^{2+}$ , giving it a higher charge density. This higher charge density polarises (or distorts) the electron cloud of the large $\text{C}_2\text{O}_4^{2-}$ anion more effectively. The increased polarisation weakens the C–C bond within the ethanedioate ion, making it easier to break and thus lowering the temperature required for decomposition.

Working

Ionic radius trend: $\text{Ca}^{2+} < \text{Ba}^{2+}$
Charge density: $\text{Ca}^{2+} > \text{Ba}^{2+}$
Polarisation of $\text{C}_2\text{O}_4^{2-}$ : greater for $\text{Ca}^{2+}$
Effect on stability: lower thermal stability for $\text{CaC}_2\text{O}_4$

Answer

$\text{CaC}_2\text{O}_4$ ; $\text{Ca}^{2+}$ has smaller radius / higher charge density; more polarises $\text{C}_2\text{O}_4^{2-}$ anion.

Final answer

CaC2O4; Ca2+ has smaller radius/higher charge density; more polarises C2O4 2- anion

Detailed explanation

Background Concept

The thermal stability of Group 2 compounds containing large, polarisable anions (like ethanedioate, $\text{C}_2\text{O}_4^{2-}$ , or carbonate, $\text{CO}_3^{2-}$ ) depends on the ability of the cation to polarise the anion. This is explained by Fajans' rules. A small, highly charged cation has a high charge density, which distorts the electron cloud of a large anion. This polarisation weakens the bonds within the anion (in this case, the C–C bond in $\text{C}_2\text{O}_4^{2-}$ ), making the compound less thermally stable and lowering its decomposition temperature.

Understanding the Question

The question asks to compare the decomposition temperatures of calcium ethanedioate ( $\text{CaC}_2\text{O}_4$ ) and barium ethanedioate ( $\text{BaC}_2\text{O}_4$ ). We need to identify which decomposes at a lower temperature and explain why using the concept of anion polarisation.

Approach

Compare the ionic radii of $\text{Ca}^{2+}$ and $\text{Ba}^{2+}$ using their positions in Group 2.
Relate ionic radius to charge density.
Explain how higher charge density leads to greater polarisation of the $\text{C}_2\text{O}_4^{2-}$ ion.
Conclude that greater polarisation leads to lower thermal stability (lower decomposition temperature).

Step-by-Step Reasoning

Ionic radius: Calcium is above barium in Group 2, so $\text{Ca}^{2+}$ has fewer electron shells and a smaller ionic radius than $\text{Ba}^{2+}$ .
Charge density: Both ions have a $2+$ charge. Since charge density is charge divided by volume (or related to radius), the smaller $\text{Ca}^{2+}$ ion has a higher charge density.
Polarisation: The high charge density of $\text{Ca}^{2+}$ exerts a stronger pull on the electron cloud of the large $\text{C}_2\text{O}_4^{2-}$ anion. This polarises (distorts) the anion more than $\text{Ba}^{2+}$ does.
Thermal stability: The polarisation weakens the internal bonds of the anion (the C–C bond in ethanedioate). Weaker bonds mean less energy is required to break them, so $\text{CaC}_2\text{O}_4$ decomposes at a lower temperature than $\text{BaC}_2\text{O}_4$ .

Key Takeaways

Smaller cations in Group 2 have higher charge densities.
Higher charge density leads to greater polarisation of large anions.
Greater polarisation decreases thermal stability, lowering the decomposition temperature.

Common Mistakes

Stating that $\text{Ba}^{2+}$ has a higher charge density (confusing size and density).
Saying the cation "pulls electrons away" without using the term polarise or distort.
Forgetting to mention that the anion is the one being polarised, or that the bond within the anion is weakened.

Things to Be Careful About

Must explicitly state $\text{CaC}_2\text{O}_4$ decomposes at the lower temperature.
Must mention smaller ionic radius and higher charge density for $\text{Ca}^{2+}$ .
Must use the keyword polarise (or distort) applied to the anion ( $\text{C}_2\text{O}_4^{2-}$ ).

Techniques used

compare ionic radii and charge densityapply Fajans' rules to explain polarisationrelate polarisation to thermal stability

(c)

Potassium iron(III) ethanedioate, $\text{K}_3[\text{Fe(C}_2\text{O}_4)_3]$ , dissolves in water to form a green solution.

Explain why transition elements can form coloured complexes.

DifficultyMedium-Easy

Worked solution

Answer

In a transition metal complex ion, the d orbitals split into different energy levels (d-d splitting) due to the interaction with ligands.
Electrons in the lower energy d orbitals can absorb light and be promoted (excited) to the higher energy d orbitals (d-d transition).
The colour seen is the complementary colour to the wavelength of light absorbed.

Working

Splitting: Ligands create an electric field that causes the five degenerate d orbitals to split into two energy levels (e.g., $t_{2g}$ and $e_g$ in octahedral complexes).
Excitation: The energy gap ( $\Delta E$ ) between these levels corresponds to the energy of visible light. Electrons absorb this light to jump to the higher level.
Colour: The colour we observe is the light that is not absorbed, which is the complementary colour on the colour wheel.

Answer

d orbitals split into different energies; electrons promoted by absorbing light; observed colour is complementary to absorbed light.

Final answer

d orbitals split into different energies; electrons promoted by absorbing light; observed colour is complementary to absorbed light

Detailed explanation

Background Concept

Transition elements can form coloured complexes because of the presence of partially filled d orbitals. When ligands (like $\text{H}_2\text{O}$ , $\text{NH}_3$ , or $\text{C}_2\text{O}_4^{2-}$ ) approach the central metal ion, their lone pairs interact with the metal's d orbitals. This interaction causes the d orbitals, which are degenerate (equal in energy) in a free ion, to split into different energy levels.

In an octahedral complex, the d orbitals split into a lower energy set ( $t_{2g}$ ) and a higher energy set ( $e_g$ ). The energy difference between these levels ( $\Delta E$ ) often corresponds to the energy of photons in the visible region of the electromagnetic spectrum.

Understanding the Question

The question asks for a general explanation of why transition elements form coloured complexes, using the green solution of $\text{K}_3[\text{Fe(C}_2\text{O}_4)_3]$ as context. We need to explain the mechanism of colour formation in terms of d-orbital splitting and electron transitions.

Approach

Describe the splitting of d orbitals in the presence of ligands.
Explain that electrons can absorb visible light to move between these split levels (d-d transition).
State that the colour observed is the complementary colour to the light absorbed.

Step-by-Step Reasoning

d-orbital splitting: The electrostatic repulsion between the lone pairs on the ligands and the electrons in the d orbitals causes the d orbitals to split into two sets of different energies. In an octahedral field like $[\text{Fe(C}_2\text{O}_4)_3]^{3-}$ , this is the $t_{2g}$ and $e_g$ splitting.
Electron promotion: The complex contains $\text{Fe}^{3+}$ , which has a $d^5$ configuration. Electrons in the lower energy d orbitals can absorb a photon of visible light that matches the energy gap ( $\Delta E$ ) and be promoted (excited) to the higher energy d orbitals. This is called a d-d transition.
Colour observation: When white light (containing all visible wavelengths) passes through or reflects off the complex, specific wavelengths are absorbed. The remaining light is transmitted or reflected, and this mixture of wavelengths is perceived by the eye as the complementary colour. For example, if orange light is absorbed, the complex appears green.

Key Takeaways

Colour in transition metal complexes arises from d-d transitions.
Ligands cause the splitting of d orbitals into different energy levels.
The observed colour is complementary to the colour of light absorbed.

Common Mistakes

Saying "electrons move between d orbitals" without mentioning splitting or different energy levels.
Forgetting to mention that light is absorbed to promote the electron.
Stating that the colour seen is the same as the colour absorbed (it must be complementary).
Saying "d orbitals are at different energies" without explaining why (interaction with ligands/splitting).

Things to Be Careful About

Use precise terminology: d-d splitting, promoted/excited, complementary colour.
Do not say "electrons jump between energy levels" without specifying they are d orbitals.
Ensure all three marking points are covered: splitting, excitation by light absorption, and complementary colour.

Techniques used

explain d-orbital splitting in complexesdescribe d-d electron transitionsrelate absorbed light to observed complementary colour

(d)

The anhydrous iron(III) compound $\text{K}_3[\text{Fe(C}_2\text{O}_4)_3]$ decomposes on heating to form a mixture of $\text{K}_2[\text{Fe(C}_2\text{O}_4)_2]$ , $\text{K}_2\text{C}_2\text{O}_4$ and $\text{CO}_2$ .

Complete the equation for the decomposition of $\text{K}_3[\text{Fe(C}_2\text{O}_4)_3]$ .

\text{.............. } \text{K}_3[\text{Fe(C}_2\text{O}_4)_3] \rightarrow \text{............ } \text{K}_2[\text{Fe(C}_2\text{O}_4)_2] + \text{............ } \text{K}_2\text{C}_2\text{O}_4 + \text{............ } \text{CO}_2

DifficultyMedium-Easy

Worked solution

Answer

2\text{K}_3[\text{Fe(C}_2\text{O}_4)_3] \rightarrow 2\text{K}_2[\text{Fe(C}_2\text{O}_4)_2] + \text{K}_2\text{C}_2\text{O}_4 + 2\text{CO}_2

Working

Reactant: $\text{K}_3[\text{Fe(C}_2\text{O}_4)_3]$ contains 3 K, 1 Fe, 6 C, 8 O per formula unit.
Products: $\text{K}_2[\text{Fe(C}_2\text{O}_4)_2]$ (2 K, 1 Fe, 4 C, 8 O), $\text{K}_2\text{C}_2\text{O}_4$ (2 K, 2 C, 4 O), $\text{CO}_2$ (1 C, 2 O).
Balancing K: Left has 3 K, right has $2x + 2y$ . Let coefficient of reactant be 2. Then 6 K on left. Products: $2 \times 2 (\text{from first}) + 1 \times 2 (\text{from second}) = 6$ K. So coefficients are 2, 2, 1.
Balancing Fe: 2 Fe on left, 2 Fe on right (in $2\text{K}_2[\text{Fe(C}_2\text{O}_4)_2]$ ). Balanced.
Balancing C: Left: $2 \times 6 = 12$ C. Right: $2 \times 4 + 1 \times 2 + z(1) = 10 + z$ . So $z = 2$ . Thus $2\text{CO}_2$ .
Balancing O: Left: $2 \times 12 = 24$ O. Right: $2 \times 8 + 1 \times 4 + 2 \times 2 = 16 + 4 + 4 = 24$ O. Balanced.

Answer

2K3[Fe(C2O4)3] -> 2K2[Fe(C2O4)2] + K2C2O4 + 2CO2

Final answer

2K3[Fe(C2O4)3] -> 2K2[Fe(C2O4)2] + K2C2O4 + 2CO2

Detailed explanation

Background Concept

Complexes can decompose upon heating, often involving the loss of ligands or the breakdown of the ligand itself. In this case, the ethanedioate (oxalate) ligands in the iron(III) complex are partially broken down to release $\text{CO}_2$ gas, while some remain coordinated to the iron and others are released as free potassium ethanedioate.

Understanding the Question

We are given the reactant $\text{K}_3[\text{Fe(C}_2 ext{O}_4)_3]$ and the products $\text{K}_2[\text{Fe(C}_2 ext{O}_4)_2]$ , $\text{K}_2 ext{C}_2 ext{O}_4$ , and $\text{CO}_2$ . We need to balance the equation by finding the correct stoichiometric coefficients.

Approach

Assign a coefficient to the reactant (start with 2 to make K balancing easier, as products have even numbers of K).
Balance potassium (K) atoms.
Balance iron (Fe) atoms.
Balance carbon (C) atoms to find the coefficient for $\text{CO}_2$ .
Check oxygen (O) atoms to ensure the equation is fully balanced.

Step-by-Step Reasoning

Start with K: The reactant has 3 K. The products have 2 K in $\text{K}_2[\text{Fe(C}_2 ext{O}_4)_2]$ and 2 K in $\text{K}_2 ext{C}_2 ext{O}_4$ . To get an even number of K on the left, let's use 2 for the reactant: $2\text{K}_3[\text{Fe(C}_2 ext{O}_4)_3]$ . This gives 6 K on the left.
Balance K on right: We need 6 K total. Let the coefficient of $\text{K}_2[\text{Fe(C}_2 ext{O}_4)_2]$ be 2 (giving 4 K) and $\text{K}_2 ext{C}_2 ext{O}_4$ be 1 (giving 2 K). Total = 6 K. Balanced.
Balance Fe: Left has $2 \times 1 = 2$ Fe. Right has $2 \times 1 = 2$ Fe (in the complex). Balanced.
Balance C: Left has $2 \times 6 = 12$ C (since each $\text{C}_2\text{O}_4$ has 2 C, and there are 3 ligands). Right has $2 \times 4$ (from complex) $+ 1 \times 2$ (from $\text{K}_2\text{C}_2\text{O}_4$ ) $= 10$ C. We need 2 more C, which must come from 2 $\text{CO}_2$ .
Check O: Left has $2 \times 12 = 24$ O. Right has $2 \times 8$ (complex) $+ 1 \times 4$ (potassium oxalate) $+ 2 \times 2$ ( $\text{CO}_2$ ) $= 16 + 4 + 4 = 24$ O. Balanced.

The balanced equation is:

2\text{K}_3[\text{Fe(C}_2\text{O}_4)_3] \rightarrow 2\text{K}_2[\text{Fe(C}_2\text{O}_4)_2] + \text{K}_2\text{C}_2\text{O}_4 + 2\text{CO}_2

Key Takeaways

Balancing complex equations requires systematic atom counting.
Starting with the most complex molecule or the element with the fewest compounds (like Fe or K) helps simplify the process.
Always verify the balance with the most abundant atom (usually O) at the end.

Common Mistakes

Incorrectly counting the number of carbon or oxygen atoms in the complex ion $[\text{Fe(C}_2\text{O}_4)_3]^{3-}$ .
Forgetting that $\text{K}_2\text{C}_2\text{O}_4$ is a separate product and not part of the complex.
Balancing K incorrectly by using odd coefficients for products.

Things to Be Careful About

Ensure the formulae of all species are copied exactly as given.
Check that the charge is balanced (though in a neutral decomposition equation, total charge is zero on both sides).
Verify oxygen balance last, as it appears in multiple products.

Techniques used

balance complex decomposition equationuse atom conservation to find stoichiometric coefficients

(e)

The $[\text{Fe(C}_2\text{O}_4)_3]^{3-}$ complex ion shows stereoisomerism.

Complete the three-dimensional diagrams in Fig. 3.1 to show the two stereoisomers of $[\text{Fe(C}_2\text{O}_4)_3]^{3-}$ .

The $\text{C}_2\text{O}_4^{2-}$ ligand can be represented using O⌒O.

Fig. 3.1

DifficultyMedium

Worked solution

Answer

The two stereoisomers are optical isomers (enantiomers) of the octahedral $[\text{Fe(C}_2\text{O}_4)_3]^{3-}$ complex. They are non-superimposable mirror images of each other.

Isomer 1 (right-handed / Δ):

Central Fe atom.
Three bidentate oxalate ligands (represented as O⌒O) arranged in a clockwise helical fashion around the Fe.
Use wedge bonds (solid triangles) for ligands coming out of the plane towards the viewer, and dash bonds (dashed lines) for ligands going into the plane away from the viewer.
Example arrangement: Top O (dash), bottom O (dash), right-top O (wedge), right-bottom O (wedge), left-top O (wedge), left-bottom O (dash) - see diagram description below.

Isomer 2 (left-handed / Λ):

Central Fe atom.
Three bidentate oxalate ligands arranged in an anticlockwise helical fashion.
This is the exact mirror image of Isomer 1.

Answer

Two non-superimposable mirror images (optical isomers) of octahedral [Fe(C2O4)3]3- with clockwise and anticlockwise oxalate ligand arrangements.

Final answer

Two non-superimposable mirror images (optical isomers) of octahedral [Fe(C2O4)3]3- with clockwise and anticlockwise oxalate ligand arrangements

Detailed explanation

Background Concept

Octahedral complexes with three bidentate ligands (like $[\text{M(AA)}_3]^{n+}$ where AA is a bidentate ligand such as ethanedioate, $\text{C}_2\text{O}_4^{2-}$ ) exhibit optical isomerism. The complex lacks a plane of symmetry and a centre of inversion, meaning it exists as two non-superimposable mirror images called enantiomers.

These isomers are often designated as Δ (delta, right-handed) and Λ (lambda, left-handed) based on the direction of the helix formed by the ligands around the central metal ion.

Understanding the Question

The question asks to complete the 3D diagrams for the two stereoisomers of $[\text{Fe(C}_2 ext{O}_4)_3]^{3-}$ . The template shows a central Fe with vertical and horizontal axes. We need to draw the three oxalate ligands (O⌒O) using wedge and dash bonds to show the 3D octahedral geometry, ensuring the two drawings are mirror images.

Approach

Draw the octahedral framework: Fe in the center, with ligands at the 6 vertices (axial and equatorial positions).
Attach the bidentate oxalate ligands (O⌒O) to adjacent positions (cis to each other).
For Isomer 1, arrange the ligands in a clockwise (right-handed) pattern using wedge/dash bonds.
For Isomer 2, draw the exact mirror image (anticlockwise / left-handed pattern).
Ensure the O⌒O curves connect the correct oxygen atoms to show the chelate ring.

Step-by-Step Reasoning

Octahedral geometry: The Fe ion is surrounded by 6 oxygen atoms (2 from each oxalate ligand) at the vertices of an octahedron. In a 2D projection on paper, we typically show 4 ligands in the plane (horizontal and vertical lines), 1 coming out (wedge), and 1 going in (dash). However, for tris-chelate complexes, it's common to show the three chelate rings wrapping around the metal.
Isomer 1 (Δ, clockwise):
- Place one O⌒O ligand in the vertical plane (top and bottom O atoms connected by a curve behind the Fe).
- Place the second O⌒O ligand with one O on a wedge (front-right) and one O on a dash (back-left), connected by a curve.
- Place the third O⌒O ligand with one O on a wedge (front-left) and one O on a dash (back-right), connected by a curve.
- The overall twist of the three chelate rings is clockwise when viewed from the top.
Isomer 2 (Λ, anticlockwise):
- This is the mirror image. The twist of the chelate rings is anticlockwise.
- If Isomer 1 has the front-right O on a wedge, Isomer 2 should have the front-left O on a wedge (mirror plane between them).

Diagram description for Fig. 3e:

Left box (Isomer 1): Central Fe. Top O (plain bond), bottom O (plain bond). Right-top O (plain bond), right-bottom O (wedge bond). Left-top O (dash bond), left-bottom O (wedge bond). Curves (O⌒O) connect: top to left-top (back), bottom to right-bottom (front), right-top to left-bottom (across). Note: Exact wedge/dash assignment varies by convention, but the key is the clockwise/anticlockwise twist and non-superimposability.
Right box (Isomer 2): Mirror image of Isomer 1. The wedge and dash bonds are swapped on the left and right sides, creating an anticlockwise twist.

Key Takeaways

$[\text{M(bidentate)}_3]$ complexes show optical isomerism.
The isomers are non-superimposable mirror images (enantiomers).
Use wedge (front) and dash (back) bonds to represent 3D geometry.
The two isomers have clockwise (Δ) and anticlockwise (Λ) arrangements of the chelate rings.

Common Mistakes

Drawing the isomers as identical (failing to make them mirror images).
Using only plain bonds and failing to show 3D geometry (no wedges/dashes).
Connecting the oxalate ligands incorrectly (e.g., connecting trans oxygens instead of cis).
Drawing a plane of symmetry (which would make them identical, not optical isomers).

Things to Be Careful About

The question says "The $\text{C}_2\text{O}_4^{2-}$ ligand can be represented using O⌒O". Ensure you draw the curve connecting the two O atoms of each ligand.
Both isomers must be drawn in the provided boxes.
Ensure the central atom is labelled Fe.
The two structures must be clearly non-superimposable mirror images.

Techniques used

draw optical isomers of octahedral complexrepresent bidentate ligands in 3Duse wedge and dash bonds for stereochemistry

(f)

Buffer solutions are used to regulate pH.

Write two equations to describe how a solution containing $\text{HC}_2\text{O}_4^-$ ions acts as a buffer solution when small amounts of acid or alkali are added.

DifficultyEasy

Worked solution

Answer

When a small amount of acid ( $\text{H}^+$ ) is added:

\text{HC}_2\text{O}_4^- + \text{H}^+ \rightarrow \text{H}_2\text{C}_2\text{O}_4

When a small amount of alkali ( $\text{OH}^-$ ) is added:

\text{HC}_2\text{O}_4^- + \text{OH}^- \rightarrow \text{C}_2\text{O}_4^{2-} + \text{H}_2\text{O}

Working

The buffer contains $\text{HC}_2\text{O}_4^-$ (weak acid/conjugate base) and $\text{C}_2\text{O}_4^{2-}$ (conjugate base).
Added $\text{H}^+$ reacts with the base component ( $\text{HC}_2\text{O}_4^-$ ) to form the weak acid $\text{H}_2\text{C}_2\text{O}_4$ .
Added $\text{OH}^-$ reacts with the acid component ( $\text{HC}_2\text{O}_4^-$ ) to form the conjugate base $\text{C}_2\text{O}_4^{2-}$ and water.

Answer

HC2O4- + H+ -> H2C2O4; HC2O4- + OH- -> C2O4 2- + H2O

Final answer

HC2O4- + H+ -> H2C2O4; HC2O4- + OH- -> C2O4 2- + H2O

Detailed explanation

Background Concept

A buffer solution resists changes in pH when small amounts of acid or alkali are added. It typically consists of a weak acid and its conjugate base (or a weak base and its conjugate acid). In this case, the buffer contains hydrogen ethanedioate ions ( $\text{HC}_2\text{O}_4^-$ ) and ethanedioate ions ( $\text{C}_2\text{O}_4^{2-}$ ), which are a conjugate acid-base pair derived from ethanedioic acid ( $\text{H}_2\text{C}_2\text{O}_4$ ).

When acid ( $\text{H}^+$ ) is added, it is neutralised by the base component of the buffer. When alkali ( $\text{OH}^-$ ) is added, it is neutralised by the acid component.

Understanding the Question

The question asks for two equations showing how a solution containing $\text{HC}_2\text{O}_4^-$ ions acts as a buffer when small amounts of acid or alkali are added. We need to write the reaction of $\text{HC}_2\text{O}_4^-$ with $\text{H}^+$ and with $\text{OH}^-$ .

Approach

Write the equation for the reaction of $\text{HC}_2\text{O}_4^-$ (acting as a base) with added $\text{H}^+$ .
Write the equation for the reaction of $\text{HC}_2\text{O}_4^-$ (acting as an acid) with added $\text{OH}^-$ .
Ensure both equations are balanced and include correct charges.

Step-by-Step Reasoning

Reaction with acid: The added $\text{H}^+$ ions react with the $\text{HC}_2\text{O}_4^-$ ions (which can accept a proton) to form ethanedioic acid ( $\text{H}_2\text{C}_2\text{O}_4$ ), a weak acid that does not fully dissociate, thus minimising the pH change. $\text{HC}_2\text{O}_4^- + \text{H}^+ \rightarrow \text{H}_2\text{C}_2\text{O}_4$
Reaction with alkali: The added $\text{OH}^-$ ions react with the $\text{HC}_2\text{O}_4^-$ ions (which can donate a proton) to form ethanedioate ions ( $\text{C}_2\text{O}_4^{2-}$ ) and water. This removes the $\text{OH}^-$ from the solution, minimising the pH change. $\text{HC}_2\text{O}_4^- + \text{OH}^- \rightarrow \text{C}_2\text{O}_4^{2-} + \text{H}_2\text{O}$

Key Takeaways

Buffers neutralise added acid or alkali using their conjugate components.
$\text{HC}_2\text{O}_4^-$ is amphiprotic: it can act as an acid (donating $\text{H}^+$ to $\text{OH}^-$ ) or as a base (accepting $\text{H}^+$ from added acid).
The equations must show the conversion to the other species in the conjugate pair.

Common Mistakes

Writing $\text{H}_2\text{C}_2\text{O}_4$ as $\text{H}_2\text{C}_2\text{O}_4^-$ or incorrect charges.
Forgetting to include $\text{H}_2\text{O}$ in the alkali reaction.
Writing the reverse reactions (e.g., $\text{H}_2\text{C}_2\text{O}_4 \rightarrow \text{HC}_2\text{O}_4^- + \text{H}^+$ ) which describe dissociation, not buffer action against added reagents.
Using $\text{HCl}$ or $\text{NaOH}$ instead of $\text{H}^+$ and $\text{OH}^-$ (though often accepted, ionic equations are preferred and safer).

Things to Be Careful About

Ensure charges are balanced on both sides of each equation.
Use $\text{H}^+$ and $\text{OH}^-$ as specified by the context of buffer action, rather than full molecular formulas unless asked.
The question asks for equations describing how the solution acts as a buffer, so the reactions must show the removal of added $\text{H}^+$ or $\text{OH}^-$ .

Techniques used

write buffer action equation with acidwrite buffer action equation with alkali

(g)

A fuel cell is an electrochemical cell that can be used to generate electrical energy by using oxygen to oxidise a fuel.

Ethanedioic acid, $(\text{COOH})_2$ , dissolved in an alkaline electrolyte is being investigated as a fuel.

The relevant standard electrode potentials, $E^\ominus$ , for the cell are shown.

\begin{aligned} \text{O}_2\text{(g)} + 2\text{H}_2\text{O(l)} + 4\text{e}^- &\rightleftharpoons 4\text{OH}^-\text{(aq)} & E^\ominus &= +0.40\text{ V} \\ 2\text{CO}_2\text{(g)} + 2\text{e}^- &\rightleftharpoons \text{C}_2\text{O}_4^{2-}\text{(aq)} & E^\ominus &= -0.59\text{ V} \end{aligned}

Use these equations to deduce the overall cell reaction. Calculate the value of $E^\ominus_{\text{cell}}$ .

overall cell reaction ...................................................................................................................

$E^\ominus_{\text{cell}} =$ .............................. V

DifficultyMedium-Easy

Worked solution

Working

Half-equations:

$\text{O}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l}) + 4\text{e}^- \rightleftharpoons 4\text{OH}^-(\text{aq}) \quad E^\ominus = +0.40 \text{ V}$ (Reduction, cathode)
$2\text{CO}_2(\text{g}) + 2\text{e}^- \rightleftharpoons \text{C}_2\text{O}_4^{2-}(\text{aq}) \quad E^\ominus = -0.59 \text{ V}$ (Oxidation, anode)

To make the fuel cell work, ethanedioate must be oxidised to $\text{CO}_2$ . Reverse equation 2:
$\text{C}_2\text{O}_4^{2-}(\text{aq}) \rightleftharpoons 2\text{CO}_2(\text{g}) + 2\text{e}^-$

Multiply by 2 to balance electrons:
$2\text{C}_2\text{O}_4^{2-}(\text{aq}) \rightleftharpoons 4\text{CO}_2(\text{g}) + 4\text{e}^-$

Add to equation 1:
$2\text{C}_2\text{O}_4^{2-} + \text{O}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l}) \rightarrow 4\text{CO}_2(\text{g}) + 4\text{OH}^-(\text{aq})$

Calculate $E^\ominus_{\text{cell}}$ :

E^\ominus_{\text{cell}} = E^\ominus_{\text{reduction}} - E^\ominus_{\text{oxidation}} = +0.40 - (-0.59) = +0.99 \text{ V}

Answer

overall cell reaction: $2\text{C}_2\text{O}_4^{2-} + \text{O}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l}) \rightarrow 4\text{CO}_2(\text{g}) + 4\text{OH}^-(\text{aq})$
$E^\ominus_{\text{cell}} = 0.99 \text{ V}$

Final answer

2C2O4 2- + O2(g) + 2H2O(l) -> 4CO2(g) + 4OH-(aq); 0.99

Detailed explanation

Background Concept

A fuel cell generates electrical energy from the spontaneous reaction between a fuel (oxidised at the anode) and an oxidant like oxygen (reduced at the cathode). The overall cell potential ( $E^\ominus_{\text{cell}}$ ) is calculated using the standard electrode potentials of the two half-reactions:

E^\ominus_{\text{cell}} = E^\ominus_{\text{cathode}} - E^\ominus_{\text{anode}}

where both potentials are taken as reduction potentials. For a spontaneous reaction (positive $E^\ominus_{\text{cell}}$ ), the half-reaction with the more positive $E^\ominus$ will proceed as reduction (cathode), and the other will be reversed to proceed as oxidation (anode).

Understanding the Question

We are given two half-equations with their standard electrode potentials in an alkaline electrolyte:

$\text{O}_2 + 2\text{H}_2\text{O} + 4\text{e}^- \rightleftharpoons 4\text{OH}^- \quad E^\ominus = +0.40 \text{ V}$
$2\text{CO}_2 + 2\text{e}^- \rightleftharpoons \text{C}_2\text{O}_4^{2-} \quad E^\ominus = -0.59 \text{ V}$

We need to deduce the overall cell reaction (fuel is oxidised, so ethanedioate is oxidised to $\text{CO}_2$ ) and calculate $E^\ominus_{\text{cell}}$ .

Approach

Identify the cathode (reduction) and anode (oxidation) based on $E^\ominus$ values. The more positive value ( $+0.40 \text{ V}$ ) is reduction.
Reverse the less positive half-equation ( $-0.59 \text{ V}$ ) to represent oxidation of ethanedioate.
Balance the electrons (multiply the oxidation half-equation by 2).
Add the two half-equations to get the overall reaction.
Calculate $E^\ominus_{\text{cell}}$ using the formula $E^\ominus_{\text{cathode}} - E^\ominus_{\text{anode}}$ .

Step-by-Step Reasoning

Cathode (Reduction): Oxygen is reduced. $\text{O}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l}) + 4\text{e}^- \rightarrow 4\text{OH}^-(\text{aq}) \quad E^\ominus = +0.40 \text{ V}$
Anode (Oxidation): Ethanedioate is oxidised to $\text{CO}_2$ . Reverse the given half-equation: $\text{C}_2\text{O}_4^{2-}(\text{aq}) \rightarrow 2\text{CO}_2(\text{g}) + 2\text{e}^- \quad E^\ominus = -0.59 \text{ V}$
Balance electrons: Multiply the anode equation by 2 to provide 4 electrons for the cathode: $2\text{C}_2\text{O}_4^{2-}(\text{aq}) \rightarrow 4\text{CO}_2(\text{g}) + 4\text{e}^-$
Overall reaction: Add the cathode and balanced anode equations. The $4\text{e}^-$ cancel out. $2\text{C}_2\text{O}_4^{2-}(\text{aq}) + \text{O}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l}) \rightarrow 4\text{CO}_2(\text{g}) + 4\text{OH}^-(\text{aq})$
Calculate $E^\ominus_{\text{cell}}$ : $E^\ominus_{\text{cell}} = E^\ominus_{\text{reduction}} - E^\ominus_{\text{oxidation}} = (+0.40) - (-0.59) = +0.99 \text{ V}$ The positive value confirms the reaction is feasible and will generate electrical energy.

Key Takeaways

In a fuel cell, the fuel is oxidised at the anode (reverse the half-equation with the lower $E^\ominus$ ).
Oxygen is reduced at the cathode (keep the half-equation with the higher $E^\ominus$ ).
$E^\ominus_{\text{cell}} = E^\ominus_{\text{cathode}} - E^\ominus_{\text{anode}}$ (using reduction potentials for both).
Always balance electrons before adding half-equations.

Common Mistakes

Subtracting in the wrong order: $E^\ominus_{\text{anode}} - E^\ominus_{ ext{cathode}}$ gives a negative value and is incorrect.
Forgetting to multiply the half-equation by 2 to balance electrons, leading to an unbalanced overall equation.
Writing $\text{CO}_2$ as a product in the reduction half-equation instead of oxidation.
Including electrons in the final overall cell reaction.

Things to Be Careful About

State symbols are important in the overall equation: $(\text{g})$ , $(\text{l})$ , $(\text{aq})$ .
The electrolyte is alkaline, so $\text{OH}^-$ is the correct species (not $\text{H}^+$ ).
Ensure the charge is balanced in the overall equation: Left = $2 \times (-2) = -4$ ; Right = $4 \times (-1) = -4$ . Balanced.
$E^\ominus_{\text{cell}}$ should be reported to 2 decimal places as given in the data: $0.99 \text{ V}$ .

Techniques used

combine half-equations for overall cell reactioncalculate standard cell potential using E(cathode) - E(anode)identify oxidation and reduction half-reactions

Q4MediumElectrochemistryChemical EnergeticsGroup 2

(a)

Define standard electrode potential, $E^\ominus$ , including a description of standard conditions.

DifficultyEasy

Worked solution

Answer

The voltage (or electromotive force) of a half-cell compared to the standard hydrogen electrode under standard conditions.

Standard conditions are:

Concentration of $1 \text{ mol dm}^{-3}$
Pressure of $101 \text{ kPa}$
Temperature of $298 \text{ K}$

Final answer

Voltage of a half-cell compared to SHE at 1 mol dm^-3, 101 kPa, 298 K

Detailed explanation

Background Concept

Standard electrode potential, $E^\ominus$ , is the fundamental measure of the tendency of a chemical species to be reduced. It is measured relative to a universal reference point: the Standard Hydrogen Electrode (SHE), which is arbitrarily assigned a potential of $0.00 \text{ V}$ . To ensure that $E^\ominus$ values are comparable and reproducible across different experiments, strict standard conditions must be applied. These conditions define a baseline state for all thermodynamic and electrochemical measurements.

Understanding the Question

The question asks for a formal definition of $E^\ominus$ and the specific numerical values that constitute the standard conditions. It is a straightforward recall question testing precise terminology and factual knowledge.

Approach

State the definition clearly, ensuring the comparison to the SHE is explicit. Then list the three standard conditions (concentration, pressure, temperature) with their correct values and units.

Step-by-Step Reasoning

M1 (Definition): The standard electrode potential is the potential difference (voltage or EMF) measured when a half-cell is connected to the standard hydrogen electrode. It represents the tendency of the species to gain electrons (be reduced) under standard conditions. The mark scheme specifically looks for "voltage of a half-cell compared to SHE".
M2 (Conditions): The three standard conditions are:
1. Concentration: $1 \text{ mol dm}^{-3}$ (for aqueous solutions).
2. Pressure: $101 \text{ kPa}$ (or $1 \text{ atm}$ , but CIE prefers $101 \text{ kPa}$ ).
3. Temperature: $298 \text{ K}$ (or $25 ^\circ\text{C}$ ).
  All three must be stated to earn the mark.

Key Takeaways

Always include the reference electrode (SHE) when defining standard electrode potential. Memorise the exact values for standard conditions; $101 \text{ kPa}$ is the modern CIE standard, not $1 \text{ atm}$ .

Common Mistakes

Defining it as the potential of a full cell rather than a half-cell compared to SHE.
Forgetting to specify "standard hydrogen electrode" and just saying "compared to another electrode".
Giving incorrect values for standard conditions (e.g., $100 \text{ kPa}$ is IUPAC standard but CIE mark schemes typically accept $101 \text{ kPa}$ ; $25 ^\circ\text{C}$ is often accepted for $298 \text{ K}$ but $298 \text{ K}$ is safer).

Things to Be Careful About

Ensure units are correct: $\text{mol dm}^{-3}$ , not just "molar". Pressure is $101 \text{ kPa}$ , not $1 \text{ bar}$ (though they are close, stick to the syllabus standard). Temperature is $298 \text{ K}$ , not $273 \text{ K}$ (which is STP for gases).

Techniques used

define standard electrode potentialrecall standard conditions

(b)

4M

(i)

An electrochemical cell is set up to measure $E^\ominus$ of the $\text{Ag}^+\text{(aq)}/\text{Ag(s)}$ electrode.

Draw a labelled diagram of this electrochemical cell.

Include all necessary substances. It is not necessary to state conditions used.

DifficultyMedium-Easy

Worked solution

Answer

The diagram must show:

Left half-cell (SHE): A platinum (Pt) electrode immersed in $\text{H}^+(\text{aq})$ solution, with $\text{H}_2(\text{g})$ gas bubbled over it via a delivery tube.
Right half-cell: A silver (Ag) electrode immersed in $\text{Ag}^+(\text{aq})$ solution (e.g., $\text{AgNO}_3(\text{aq})$ ).
Connections: A salt bridge connecting the two solutions, and a wire connecting the Pt electrode to a voltmeter (V), and from the voltmeter to the Ag electrode.

Final answer

See diagram description

Detailed explanation

Background Concept

To measure the standard electrode potential of a half-cell, it must be paired with the Standard Hydrogen Electrode (SHE), which has a defined potential of $0.00 \text{ V}$ . The SHE consists of a platinum electrode coated with platinum black, immersed in a $1 \text{ mol dm}^{-3}$ acid solution (providing $\text{H}^+$ ions), with hydrogen gas at $101 \text{ kPa}$ bubbled over it. The equilibrium is $2\text{H}^+(\text{aq}) + 2\text{e}^- \rightleftharpoons \text{H}_2(\text{g})$ . The other half-cell contains the electrode of interest (here, Ag in $\text{Ag}^+$ solution) under standard conditions ( $1 \text{ mol dm}^{-3}$ ). A salt bridge (often filter paper soaked in $\text{KNO}_3$ or $\text{KCl}$ ) completes the circuit and maintains electrical neutrality. A high-resistance voltmeter measures the potential difference.

Understanding the Question

You are asked to draw a labelled diagram of an electrochemical cell used to measure $E^\ominus$ for the $\text{Ag}^+(\text{aq})/\text{Ag(s)}$ electrode. This means one side must be the SHE, and the other the silver half-cell. All necessary substances and connections must be labelled.

Approach

Draw two beakers representing the half-cells. Label the left as the SHE (Pt electrode, $\text{H}_2$ gas, $\text{H}^+$ solution) and the right as the silver half-cell (Ag electrode, $\text{Ag}^+$ solution). Connect them with a salt bridge and an external circuit containing a voltmeter.

Step-by-Step Reasoning

M1 (Right half-cell components): Must include a silver (Ag) electrode and $\text{Ag}^+(\text{aq})$ solution (e.g., $\text{AgNO}_3(\text{aq})$ ). Label the electrode as "Ag" and the solution as " $\text{Ag}^+(\text{aq})$ ".
M2 (Left half-cell / SHE components): Must include a platinum (Pt) electrode, $\text{H}^+(\text{aq})$ solution, and $\text{H}_2(\text{g})$ gas. Show the gas delivery tube bubbling $\text{H}_2$ over the Pt electrode. Label "Pt", " $\text{H}^+(\text{aq})$ ", and " $\text{H}_2(\text{g})$ ".
M3 (Connections): Must include a salt bridge (labelled) connecting the two solutions. Must include a wire connecting the Pt electrode to a voltmeter (labelled "V" or "voltmeter"), and another wire from the voltmeter to the Ag electrode. The voltmeter must be in the circuit.

Key Takeaways

When drawing a cell to measure $E^\ominus$ , the SHE is always one half-cell. Remember the specific components of the SHE: Pt electrode, $\text{H}_2$ gas, $\text{H}^+$ solution. Don't forget the salt bridge and the voltmeter in the external circuit.

Common Mistakes

Forgetting to label the gas ( $\text{H}_2$ ) or the solution in the SHE ( $\text{H}^+$ ).
Drawing the voltmeter in series with the salt bridge (it must be in the external wire circuit).
Omitting the salt bridge entirely.
Not labelling the electrodes (Pt and Ag).

Things to Be Careful About

The diagram must be clearly labelled. "Acid" is not specific enough; write " $\text{H}^+(\text{aq})$ " or " $1 \text{ mol dm}^{-3} \text{ HCl}(\text{aq})$ ". The salt bridge must connect the two solutions, not the electrodes. The voltmeter must have a label (V or voltmeter).

Techniques used

draw electrochemical cell diagramidentify SHE components

(ii)

A separate electrochemical cell is set up using a lower concentration of $\text{Ag}^+\text{(aq)}$ than that used in (b)(i).

Suggest how the electrode potential, $E$ , for the $\text{Ag}^+\text{(aq)}/\text{Ag(s)}$ electrode would change from its $E^\ominus$ value. Explain your answer.

DifficultyMedium-Easy

Worked solution

Answer

The electrode potential $E$ would be more negative.

Explanation:
The half-equilibrium is $\text{Ag}^+(\text{aq}) + \text{e}^- \rightleftharpoons \text{Ag}(\text{s})$ . A lower concentration of $\text{Ag}^+(\text{aq})$ shifts this equilibrium to the left (to oppose the decrease in $\text{Ag}^+$ ). This means there is a lesser tendency for $\text{Ag}^+$ to gain electrons (be reduced), making the potential more negative.

Final answer

More negative; equilibrium shifts left

Detailed explanation

Background Concept

The electrode potential of a half-cell depends on the concentration of the ions involved, as described by the Nernst equation. For a simple reduction half-equilibrium $M^{n+} + n\text{e}^- \rightleftharpoons M$ , the potential is related to the position of the equilibrium. According to Le Chatelier's principle, if the concentration of the reactant ( $M^{n+}$ ) is decreased, the equilibrium shifts to the left to partially counteract the change. This shift means the forward reaction (reduction, gaining electrons) is less favoured, resulting in a more negative (or less positive) electrode potential.

Understanding the Question

You are asked to predict the change in the electrode potential $E$ for the $\text{Ag}^+(\text{aq})/\text{Ag(s)}$ electrode if the concentration of $\text{Ag}^+(\text{aq})$ is lower than the standard $1 \text{ mol dm}^{-3}$ . You must state the direction of the change (more/less negative) and explain it using equilibrium principles.

Approach

Write the half-equilibrium for the silver electrode. Apply Le Chatelier's principle to show how decreasing $[\text{Ag}^+]$ affects the position of equilibrium. Relate this shift to the tendency to gain electrons and thus the sign of the potential.

Step-by-Step Reasoning

State the change: The $E$ value would be more negative (or less positive). Since $E^\ominus$ for Ag is $+0.80 \text{ V}$ , a lower concentration makes it less positive, i.e., more negative.
Explanation: Write the equilibrium: $\text{Ag}^+(\text{aq}) + \text{e}^- \rightleftharpoons \text{Ag}(\text{s})$ . Decreasing $[\text{Ag}^+]$ (a reactant) causes the equilibrium to shift to the left to oppose the change. This means the forward reaction (reduction of $\text{Ag}^+$ to Ag) is less favoured. A lesser tendency to gain electrons corresponds to a more negative electrode potential.

Key Takeaways

Lower concentration of the oxidised species (reactant in reduction) shifts equilibrium left, making $E$ more negative. Higher concentration shifts it right, making $E$ more positive. This is a direct application of Le Chatelier's principle to electrochemical equilibria.

Common Mistakes

Saying the potential becomes "more positive" (confusing the direction of shift).
Not writing the half-equilibrium or not showing the shift to the left.
Using vague language like "less reaction happens" without specifying the direction of the shift or the effect on electron gain.

Things to Be Careful About

Ensure you state "more negative" clearly. The explanation must link the concentration change to the equilibrium shift and then to the tendency to gain electrons (reduction). Do not say "the reaction goes backwards"; say "shifts to the left".

Techniques used

apply Le Chatelier's principle to electrode equilibrium

(c)

Define enthalpy change of solution, $\Delta H^\ominus_{\text{sol}}$ .

DifficultyEasy

Worked solution

Answer

The enthalpy change when one mole of a solute dissolves in water (to form an infinitely dilute solution).

Final answer

Enthalpy change when one mole of solute dissolves in water

Detailed explanation

Background Concept

Enthalpy change of solution, $\Delta H^\ominus_{\text{sol}}$ , is the enthalpy change associated with the dissolving of a substance in a solvent (usually water) to form a solution. It is the sum of the lattice dissociation enthalpy (breaking the ionic lattice into gaseous ions, endothermic) and the enthalpy change of hydration (hydrating the gaseous ions, exothermic). $\Delta H^\ominus_{\text{sol}} = \Delta H^\ominus_{\text{latt,diss}} + \Delta H^\ominus_{\text{hyd}}$ . The definition must specify 'one mole' and 'water' (or the solvent used) to be precise.

Understanding the Question

The question asks for the definition of the enthalpy change of solution. This is a standard definition question requiring precise wording.

Approach

State the definition exactly as it appears in the syllabus: enthalpy change per mole of solute dissolving in water.

Step-by-Step Reasoning

M1: Must include "enthalpy change" and "one mole of solute".
M2: Must include "dissolves in water" (or "in excess water" / "to form an infinitely dilute solution").

Key Takeaways

Always specify "one mole" and the solvent "water" in the definition of enthalpy change of solution. Do not just say "dissolving"; specify the amount (one mole) and the solvent.

Common Mistakes

Forgetting to mention "one mole".
Saying "dissolves in solvent" instead of "water" (unless the question specifies a different solvent).
Confusing it with enthalpy change of hydration (which is for gaseous ions dissolving in water).

Things to Be Careful About

The definition is strict. "One mole" is essential. "Water" is the standard solvent assumed unless stated otherwise.

Techniques used

define enthalpy change of solution

(d)

Some relevant energy changes for $\text{AgNO}_3$ are shown in Table 4.1.

Table 4.1

energy change	value / $\text{kJ mol}^{-1}$
enthalpy change of solution of $\text{AgNO}_3\text{(s)}$	+22.6
enthalpy change of hydration of silver ions	-475
enthalpy change of hydration of nitrate ions	-314

3M

(i)

Complete the energy cycle in Fig. 4.1 to show the relationship between the lattice energy, $\Delta H^\ominus_{\text{latt}}$ , of $\text{AgNO}_3\text{(s)}$ and the energy changes shown in Table 4.1.

Include state symbols for all the species.

Fig. 4.1

DifficultyMedium-Easy

Worked solution

Answer

The completed energy cycle (Born-Haber cycle) has three levels:

Top level: $\text{Ag}^+(\text{g}) + \text{NO}_3^-(\text{g})$
Middle level: $\text{Ag}^+(\text{aq}) + \text{NO}_3^-(\text{aq})$ (via intermediate $\text{Ag}^+(\text{aq}) + \text{NO}_3^-(\text{g})$ )
Bottom level: $\text{AgNO}_3(\text{s})$

Arrows:

Downward arrow from $\text{Ag}^+(\text{g}) + \text{NO}_3^-(\text{g})$ to $\text{Ag}^+(\text{aq}) + \text{NO}_3^-(\text{g})$ (hydration of $\text{Ag}^+$ , $\Delta H = -475 \text{ kJ mol}^{-1}$ )
Downward arrow from $\text{Ag}^+(\text{aq}) + \text{NO}_3^-(\text{g})$ to $\text{Ag}^+(\text{aq}) + \text{NO}_3^-(\text{aq})$ (hydration of $\text{NO}_3^-$ , $\Delta H = -314 \text{ kJ mol}^{-1}$ )
Upward arrow from $\text{AgNO}_3(\text{s})$ to $\text{Ag}^+(\text{g}) + \text{NO}_3^-(\text{g})$ (lattice dissociation enthalpy, $-\Delta H^\ominus_{\text{latt}}$ ) OR downward arrow from gaseous ions to solid (lattice formation enthalpy, $\Delta H^\ominus_{\text{latt}}$ ).
Downward arrow from $\text{AgNO}_3(\text{s})$ to $\text{Ag}^+(\text{aq}) + \text{NO}_3^-(\text{aq})$ (enthalpy of solution, $\Delta H^\ominus_{\text{sol}} = +22.6 \text{ kJ mol}^{-1}$ )

Final answer

See diagram

Detailed explanation

Background Concept

A Born-Haber cycle (or enthalpy cycle) for the enthalpy change of solution relates the lattice enthalpy, hydration enthalpies, and enthalpy of solution. For an ionic compound $MX(s)$ :
$\Delta H^\ominus_{\text{sol}} = \Delta H^\ominus_{\text{latt,diss}} + \Delta H^\ominus_{\text{hyd}}(M^+) + \Delta H^\ominus_{\text{hyd}}(X^-)$
where $\Delta H^\ominus_{\text{latt,diss}} = -\Delta H^\ominus_{\text{latt,form}}$ .
Alternatively, using lattice formation enthalpy (exothermic, negative):
$\Delta H^\ominus_{\text{sol}} = -\Delta H^\ominus_{\text{latt,form}} + \Delta H^\ominus_{\text{hyd}}(M^+) + \Delta H^\ominus_{\text{hyd}}(X^-)$
The cycle typically has the solid at the bottom, gaseous ions at the top, and aqueous ions in the middle or bottom right.

Understanding the Question

You are given the enthalpy change of solution ( $+22.6 \text{ kJ mol}^{-1}$ ), hydration of $\text{Ag}^+$ ( $-475 \text{ kJ mol}^{-1}$ ), and hydration of $\text{NO}_3^-$ ( $-314 \text{ kJ mol}^{-1}$ ). You must complete the energy cycle to show the relationship with lattice energy ( $\Delta H^\ominus_{\text{latt}}$ ) and include state symbols.

Approach

Draw the energy levels: $\text{AgNO}_3(\text{s})$ at the bottom, $\text{Ag}^+(\text{g}) + \text{NO}_3^-(\text{g})$ at the top. Draw arrows for hydration (down from gas to aqueous) and solution (down from solid to aqueous). The lattice energy arrow completes the cycle (up from solid to gas, or down from gas to solid).

Step-by-Step Reasoning

M1 (Arrows): Need two main arrows for hydration (downward from gaseous ions to aqueous ions) and the lattice/solution arrows. The mark scheme accepts arrows showing the cycle. Typically, a downward arrow from $\text{AgNO}_3(\text{s})$ to $\text{Ag}^+(\text{aq}) + \text{NO}_3^-(\text{aq})$ represents $\Delta H^\ominus_{\text{sol}}$ . An upward arrow from $\text{AgNO}_3(\text{s})$ to $\text{Ag}^+(\text{g}) + \text{NO}_3^-(\text{g})$ represents lattice dissociation ( $-\Delta H^\ominus_{\text{latt}}$ ). Or a downward arrow from gaseous ions to solid represents lattice formation ( $\Delta H^\ominus_{\text{latt}}$ ).
M2 (Species and states): All species must have correct state symbols: $\text{Ag}^+(\text{g})$ , $\text{NO}_3^-(\text{g})$ , $\text{Ag}^+(\text{aq})$ , $\text{NO}_3^-(\text{aq})$ , $\text{AgNO}_3(\text{s})$ .

Key Takeaways

In a solution cycle, the solid is at the bottom, gaseous ions at the top, aqueous ions at the bottom right. Hydration is always exothermic (downward arrow from gas to aqueous). Lattice formation is exothermic (downward from gas to solid), lattice dissociation is endothermic (upward from solid to gas). Ensure state symbols are present on all species.

Common Mistakes

Forgetting state symbols (especially (g) for gaseous ions and (aq) for aqueous ions).
Drawing the hydration arrows in the wrong direction (should be from gas to aqueous).
Confusing lattice formation (negative) with lattice dissociation (positive) in the cycle direction.
Not including both $\text{Ag}^+$ and $\text{NO}_3^-$ at each level.

Things to Be Careful About

The mark scheme image (Fig 4di) shows a specific layout: top level $\text{Ag}^+(\text{g}) + \text{NO}_3^-(\text{g})$ , middle level $\text{Ag}^+(\text{aq}) + \text{NO}_3^-(\text{g})$ then $\text{Ag}^+(\text{aq}) + \text{NO}_3^-(\text{aq})$ , bottom level $\text{AgNO}_3(\text{s})$ . Blue arrows indicate hydration (down) and lattice/solution. Follow the standard layout: solid bottom left, gaseous ions top, aqueous ions bottom right. Ensure all state symbols are included.

Techniques used

construct Born-Haber cycle for lattice energy

(ii)

Calculate the lattice energy, $\Delta H^\ominus_{\text{latt}}$ , of $\text{AgNO}_3\text{(s)}$ .

\Delta H^\ominus_{\text{latt}} = \text{.............................. kJ mol}^{-1}

DifficultyMedium-Easy

Worked solution

Working

From the energy cycle:

\Delta H^\ominus_{\text{sol}} = -\Delta H^\ominus_{\text{latt}} + \Delta H^\ominus_{\text{hyd}}(\text{Ag}^+) + \Delta H^\ominus_{\text{hyd}}(\text{NO}_3^-)

Rearranging for lattice energy (formation):

\Delta H^\ominus_{\text{latt}} = -\Delta H^\ominus_{\text{sol}} + \Delta H^\ominus_{\text{hyd}}(\text{Ag}^+) + \Delta H^\ominus_{\text{hyd}}(\text{NO}_3^-)

Substitute the values:

\Delta H^\ominus_{\text{latt}} = -(+22.6) + (-475) + (-314)

\Delta H^\ominus_{\text{latt}} = -22.6 - 475 - 314 = -811.6 \text{ kJ mol}^{-1}

Answer

$\Delta H^\ominus_{\text{latt}} = -811.6 \text{ kJ mol}^{-1}$

Final answer

-811.6

Detailed explanation

Background Concept

The enthalpy change of solution can be expressed as the sum of the lattice dissociation enthalpy and the enthalpies of hydration of the ions:
$\Delta H^\ominus_{\text{sol}} = \Delta H^\ominus_{\text{latt,diss}} + \Delta H^\ominus_{\text{hyd}}(\text{cation}) + \Delta H^\ominus_{\text{hyd}}(\text{anion})$
Since lattice formation enthalpy ( $\Delta H^\ominus_{\text{latt}}$ ) is the negative of lattice dissociation enthalpy ( $\Delta H^\ominus_{\text{latt,diss}} = -\Delta H^\ominus_{\text{latt}}$ ), we can write:
$\Delta H^\ominus_{\text{sol}} = -\Delta H^\ominus_{\text{latt}} + \Delta H^\ominus_{\text{hyd}}(\text{cation}) + \Delta H^\ominus_{\text{hyd}}(\text{anion})$
Rearranging to find lattice formation enthalpy:
$\Delta H^\ominus_{\text{latt}} = -\Delta H^\ominus_{\text{sol}} + \Delta H^\ominus_{\text{hyd}}(\text{cation}) + \Delta H^\ominus_{\text{hyd}}(\text{anion})$

Understanding the Question

You are given $\Delta H^\ominus_{\text{sol}} = +22.6 \text{ kJ mol}^{-1}$ , $\Delta H^\ominus_{\text{hyd}}(\text{Ag}^+) = -475 \text{ kJ mol}^{-1}$ , and $\Delta H^\ominus_{\text{hyd}}(\text{NO}_3^-) = -314 \text{ kJ mol}^{-1}$ . You must calculate the lattice energy (formation enthalpy) of $\text{AgNO}_3$ .

Approach

Use the Hess's law relationship derived from the Born-Haber cycle for solution. Rearrange the equation to solve for $\Delta H^\ominus_{\text{latt}}$ and substitute the given values.

Step-by-Step Reasoning

Equation: $\Delta H^\ominus_{\text{sol}} = -\Delta H^\ominus_{\text{latt}} + \Delta H^\ominus_{\text{hyd}}(\text{Ag}^+) + \Delta H^\ominus_{\text{hyd}}(\text{NO}_3^-)$
Rearrange: $\Delta H^\ominus_{\text{latt}} = -\Delta H^\ominus_{\text{sol}} + \Delta H^\ominus_{\text{hyd}}(\text{Ag}^+) + \Delta H^\ominus_{\text{hyd}}(\text{NO}_3^-)$
Substitute: $\Delta H^\ominus_{\text{latt}} = -(+22.6) + (-475) + (-314)$
Calculate: $\Delta H^\ominus_{\text{latt}} = -22.6 - 475 - 314 = -811.6 \text{ kJ mol}^{-1}$

Key Takeaways

Always check the sign convention for lattice energy. CIE typically defines lattice energy as the enthalpy change of formation of the lattice from gaseous ions (exothermic, negative value). If the cycle uses lattice dissociation (endothermic, positive), remember to negate it.

Common Mistakes

Forgetting the negative sign for lattice formation enthalpy in the equation (using $+\Delta H^\ominus_{\text{latt}}$ instead of $-\Delta H^\ominus_{\text{latt}}$ ).
Arithmetic errors when summing the values.
Giving a positive value for lattice energy (unless specifically asked for lattice dissociation enthalpy).

Things to Be Careful About

Ensure all signs are correct. $\Delta H^\ominus_{\text{sol}}$ is $+22.6$ , so $-\Delta H^\ominus_{\text{sol}}$ is $-22.6$ . The hydration enthalpies are both negative. The sum is $-811.6 \text{ kJ mol}^{-1}$ . Include the unit in the final answer if required, though the question provides the unit.

Techniques used

calculate lattice energy from Born-Haber cycle

(e)

Suggest the trend in the magnitude of the lattice energies of the metal nitrates, $\text{NaNO}_3\text{(s)}$ , $\text{Mg(NO}_3)_2\text{(s)}$ and $\text{RbNO}_3\text{(s)}$ .

Explain your answer.

.......................................... .......................................... ..........................................
most exothermic least exothermic

DifficultyMedium

Worked solution

Answer

Trend: $\text{Mg(NO}_3)_2 > \text{NaNO}_3 > \text{RbNO}_3$ (most exothermic to least exothermic)

Explanation:

$\text{Mg}^{2+}$ has a higher charge ( $2+$ ) than $\text{Na}^+$ ( $1+$ ) and $\text{Rb}^+$ ( $1+$ ), leading to stronger electrostatic attraction between ions.
$\text{Na}^+$ has a smaller ionic radius than $\text{Rb}^+$ , leading to stronger electrostatic attraction between ions.
Stronger attraction between ions results in a more exothermic (larger magnitude) lattice energy.

Final answer

Mg(NO3)2 > NaNO3 > RbNO3; Mg2+ has higher charge, Na+ has smaller radius than Rb+

Detailed explanation

Background Concept

Lattice energy (formation enthalpy) is the enthalpy change when one mole of an ionic solid is formed from its gaseous ions. It depends on the electrostatic attraction between the ions, described by Coulomb's law: $F \propto \frac{q_1 q_2}{r^2}$ . Therefore, lattice energy becomes more exothermic (larger magnitude) when:

The charge on the ions is higher (greater attraction).
The ionic radius is smaller (ions are closer together, greater attraction).

For nitrates $\text{MNO}_3$ , the anion is always $\text{NO}_3^-$ . The variation comes from the cation ( $\text{Na}^+$ , $\text{Mg}^{2+}$ , $\text{Rb}^+$ ).

$\text{Mg}^{2+}$ : charge $2+$ , small radius.
$\text{Na}^+$ : charge $1+$ , small radius (smaller than $\text{Rb}^+$ ).
$\text{Rb}^+$ : charge $1+$ , large radius.

Understanding the Question

You are asked to suggest the trend in the magnitude (exothermicity) of lattice energies for $\text{NaNO}_3$ , $\text{Mg(NO}_3)_2$ , and $\text{RbNO}_3$ , and explain it. The order should be from most exothermic to least exothermic.

Approach

Compare the cations in terms of charge and ionic radius. Higher charge and smaller radius lead to more exothermic lattice energy. Order the compounds accordingly and explain using Coulombic attraction.

Step-by-Step Reasoning

M1 (Order): $\text{Mg(NO}_3)_2$ (most exothermic), $\text{NaNO}_3$ , $\text{RbNO}_3$ (least exothermic).
M2 (Charge and radius arguments):
- $\text{Mg}^{2+}$ has a higher charge ( $2+$ ) than $\text{Na}^+$ ( $1+$ ) and $\text{Rb}^+$ ( $1+$ ). This leads to stronger attraction.
- Between $\text{Na}^+$ and $\text{Rb}^+$ (both $1+$ ), $\text{Na}^+$ has a smaller ionic radius than $\text{Rb}^+$ . This leads to stronger attraction.
M3 (Conclusion): Stronger attraction between ions (or stronger ionic bonds) results in a more exothermic lattice energy (greater magnitude).

Key Takeaways

When comparing lattice energies, always consider both charge and radius. Charge has a larger effect than radius (doubling charge quadruples attraction, while halving radius doubles attraction). For ions of the same charge, smaller radius means more exothermic lattice energy.

Common Mistakes

Forgetting to include the correct chemical formulas in the order ( $\text{Mg(NO}_3)_2$ , not just $\text{Mg}^{2+}$ ).
Not explaining both charge and radius differences (e.g., only mentioning charge and ignoring radius for Na vs Rb).
Saying "smaller ions have higher lattice energy" without specifying charge or comparing correctly.
Confusing lattice energy trend with solubility trend (lattice energy is about formation from gas, solubility is about competition between lattice and hydration energies).

Things to Be Careful About

The question asks for the trend in the magnitude of lattice energies, and specifies "most exothermic" to "least exothermic". Ensure your order is correct: $\text{Mg(NO}_3)_2 > \text{NaNO}_3 > \text{RbNO}_3$ . The explanation must link charge/radius to attraction strength, and attraction strength to lattice energy magnitude. Do not just say "Mg2+ is smaller"; it's the charge that is the main differentiator for Mg vs Na/Rb.

Techniques used

compare lattice energies based on ionic charge and radius

Q5MediumReaction Kinetics

(a)

In aqueous solution, persulfate ions, $\text{S}_2\text{O}_8^{2-}$ , react with iodide ions, as shown in reaction 1.

\text{reaction 1 } \quad \text{S}_2\text{O}_8^{2-} + 2\text{I}^- \rightarrow 2\text{SO}_4^{2-} + \text{I}_2

The rate of reaction 1 is investigated.

A sample of $\text{S}_2\text{O}_8^{2-}$ is mixed with a large excess of iodide ions of known concentration. The graph in Fig. 5.1 shows the results obtained.

Fig. 5.1

2M

(i)

Use Fig. 5.1 to determine the initial rate of reaction 1. Show your working.

\text{rate} = \text{.............................. mol dm}^{-3}\text{min}^{-1}

DifficultyMedium-Easy

Worked solution

Working

Draw a tangent to the curve at $t = 0$ .
Select two points on the tangent, for example $(0, 0.800)$ and $(20, 0.400)$ .

\text{gradient} = \frac{0.400 - 0.800}{20 - 0} = -0.020 \text{ mol dm}^{-3} \text{ min}^{-1}

\text{rate} = -\text{gradient} = 0.020 \text{ mol dm}^{-3} \text{ min}^{-1}

Answer

0.020

Final answer

0.020

Detailed explanation

Background Concept

The initial rate of a reaction can be determined from a concentration-time graph by drawing a tangent to the curve at $t = 0$ . The gradient of this tangent represents the instantaneous rate of change of concentration at the start of the reaction. For a reactant, the concentration decreases over time, so the gradient is negative; the rate is defined as a positive quantity, so $\text{rate} = -\frac{\Delta[\text{reactant}]}{\Delta t}$ .

Understanding the Question

The question asks for the initial rate of reaction 1 using the provided concentration-time graph for $[\text{S}_2\text{O}_8^{2-}]$ . We must draw a tangent at $t = 0$ and calculate its gradient.

Approach

Draw a straight line tangent to the curve at the y-intercept ( $t = 0$ , $[\text{S}_2\text{O}_8^{2-}] = 0.800$ ).
Pick two points on this tangent line that are far apart to minimize reading errors.
Calculate the gradient: $\frac{\Delta y}{\Delta x}$ .
Take the negative of the gradient to get the positive initial rate.

Step-by-Step Reasoning

At $t = 0$ , $[\text{S}_2\text{O}_8^{2-}] = 0.800 \text{ mol dm}^{-3}$ .
Drawing a tangent at this point, we can see it passes through approximately $(20, 0.400)$ . (Acceptable answers will vary slightly based on tangent drawing; the mark scheme allows 0.016 to 0.040).
Using points $(0, 0.800)$ and $(20, 0.400)$ : $\text{gradient} = \frac{0.400 - 0.800}{20 - 0} = \frac{-0.400}{20} = -0.020 \text{ mol dm}^{-3} \text{ min}^{-1}$
The initial rate is the negative of the gradient: $\text{rate} = 0.020 \text{ mol dm}^{-3} \text{ min}^{-1}$ .

Key Takeaways

Initial rate from a graph = negative gradient of the tangent at $t = 0$ .
Always use points on the tangent line, not points on the curve itself, to calculate the gradient.

Common Mistakes

Calculating the gradient using two points on the curve instead of the tangent. This gives the average rate over that interval, not the initial instantaneous rate.
Forgetting to make the rate positive (forgetting the negative sign for a reactant).

Things to Be Careful About

Ensure the tangent is drawn as accurately as possible; a small error in the tangent line can lead to a significantly different gradient.
Check units: the y-axis is in $\text{mol dm}^{-3}$ and the x-axis is in minutes, so the rate is in $\text{mol dm}^{-3} \text{ min}^{-1}$ .

Techniques used

draw tangent to concentration-time graphcalculate gradient of tangent

(ii)

The rate equation for reaction 1 is $\text{rate} = k [\text{S}_2\text{O}_8^{2-}] [\text{I}^-]$ .

Suggest why a large excess of iodide ions allows the rate constant to be determined from the half-life in this investigation.

DifficultyMedium-Easy

Worked solution

Answer

The concentration of iodide ions, $[\text{I}^-]$ , remains effectively constant because it is in large excess. This makes the reaction pseudo-first order with respect to $[\text{S}_2\text{O}_8^{2-}]$ , allowing the half-life to be used to determine the rate constant.

Final answer

The concentration of iodide ions remains effectively constant, making the reaction pseudo-first order with respect to persulfate ions.

Detailed explanation

Background Concept

When a reaction has a rate equation involving multiple reactants, such as $\text{rate} = k[\text{A}][\text{B}]$ , the kinetics can be complex if both concentrations change significantly. However, if one reactant (B) is present in large excess, its concentration changes by a negligible amount during the reaction. This is called a 'pseudo-first-order' condition.

Understanding the Question

The rate equation is $\text{rate} = k[\text{S}_2\text{O}_8^{2-}][\text{I}^-]$ . We are told $[\text{I}^-]$ is in large excess. We need to explain why this allows us to use the half-life (a concept typically for first-order reactions) to find $k$ .

Approach

Recognize that 'large excess' means the concentration of the excess reactant does not change appreciably.
Substitute this constant concentration into the rate equation to show it simplifies to a first-order form.

Step-by-Step Reasoning

Because $[\text{I}^-]$ is in large excess, it is not the limiting reagent and its concentration remains effectively constant throughout the reaction.
We can combine the true rate constant $k$ and the constant $[\text{I}^-]$ into a new pseudo-rate constant $k' = k[\text{I}^-]$ .
The rate equation becomes $\text{rate} = k'[\text{S}_2\text{O}_8^{2-}]$ , which is first order with respect to persulfate.
For a first-order reaction, the half-life is constant and related to the rate constant by $t_{1/2} = \frac{\ln 2}{k'}$ . Thus, measuring the half-life allows us to determine $k'$ , and subsequently $k$ .

Key Takeaways

A large excess of one reactant creates pseudo-first-order conditions for the other reactant.
This simplifies kinetic analysis and allows the use of first-order equations (like the half-life formula) to determine rate constants.

Common Mistakes

Stating that the concentration of iodide ions 'does not change at all'. It changes slightly, but 'effectively constant' or 'remains constant' is the accepted phrasing.
Not connecting the constant concentration to the simplification of the rate equation.

Things to Be Careful About

The mark scheme accepts either ' $[\text{I}^-]$ stays constant' OR 'the overall order is one under these conditions'. Both are correct, but explaining the pseudo-order concept is more complete.

Techniques used

apply pseudo-order kinetics concept

(b)

The reaction of persulfate ions, $\text{S}_2\text{O}_8^{2-}$ , with iodide ions is catalysed by $\text{Fe}^{2+}$ ions.

Write two equations to show how $\text{Fe}^{2+}$ catalyses reaction 1.

equation 1 .................................................................................................................................

equation 2 .................................................................................................................................

DifficultyMedium-Easy

Worked solution

Answer

equation 1: $2\text{Fe}^{2+} + \text{S}_2\text{O}_8^{2-} \rightarrow 2\text{Fe}^{3+} + 2\text{SO}_4^{2-}$

equation 2: $2\text{Fe}^{3+} + 2\text{I}^- \rightarrow 2\text{Fe}^{2+} + \text{I}_2$

Final answer

equation 1: 2Fe^2+ + S2O8^2- -> 2Fe^3+ + 2SO4^2-; equation 2: 2Fe^3+ + 2I^- -> 2Fe^2+ + I2

Detailed explanation

Background Concept

A catalyst increases the rate of a reaction by providing an alternative reaction pathway with a lower activation energy. In homogeneous catalysis, the catalyst is in the same phase as the reactants. The catalyst is consumed in one step and regenerated in a subsequent step, so it does not appear in the overall balanced equation.

Understanding the Question

We are given the overall reaction: $\text{S}_2\text{O}_8^{2-} + 2\text{I}^- \rightarrow 2\text{SO}_4^{2-} + \text{I}_2$ . We need to write two equations showing how $\text{Fe}^{2+}$ catalyses this reaction. The catalyst must be oxidized by one reactant and reduced by the other.

Approach

Identify the oxidation and reduction half-reactions in the overall equation.
$\text{S}_2\text{O}_8^{2-}$ is reduced to $\text{SO}_4^{2-}$ (oxidation state of S changes from +7 to +6).
$\text{I}^-$ is oxidized to $\text{I}_2$ (oxidation state of I changes from -1 to 0).
$\text{Fe}^{2+}$ can be oxidized to $\text{Fe}^{3+}$ and then reduced back to $\text{Fe}^{2+}$ . Write the two steps accordingly.

Step-by-Step Reasoning

Step 1: Persulfate ions are strong oxidizing agents. They oxidize $\text{Fe}^{2+}$ to $\text{Fe}^{3+}$ , while being reduced to sulfate ions. $2\text{Fe}^{2+} + \text{S}_2\text{O}_8^{2-} \rightarrow 2\text{Fe}^{3+} + 2\text{SO}_4^{2-}$
Step 2: The $\text{Fe}^{3+}$ ions produced are also oxidizing agents. They oxidize the iodide ions to iodine, while being reduced back to $\text{Fe}^{2+}$ , regenerating the catalyst. $2\text{Fe}^{3+} + 2\text{I}^- \rightarrow 2\text{Fe}^{2+} + \text{I}_2$
Adding these two equations cancels out $2\text{Fe}^{2+}$ and $2\text{Fe}^{3+}$ , giving the overall reaction.

Key Takeaways

Catalytic cycles involve the catalyst being consumed in one step and regenerated in another.
The sum of the catalytic steps must equal the overall balanced equation.
Transition metals like iron are excellent catalysts because they can easily change oxidation states.

Common Mistakes

Not balancing the equations correctly (e.g., missing the coefficient 2 for $\text{Fe}^{2+}$ or $\text{SO}_4^{2-}$ ).
Writing equations that do not sum to the overall reaction.
Forgetting state symbols (though often not strictly required unless specified, it's good practice).

Things to Be Careful About

Ensure the equations are balanced for both mass and charge.
The mark scheme specifically looks for $2\text{Fe}^{2+} + \text{S}_2\text{O}_8^{2-} \rightarrow 2\text{Fe}^{3+} + 2\text{SO}_4^{2-}$ and $2\text{Fe}^{3+} + 2\text{I}^- \rightarrow 2\text{Fe}^{2+} + \text{I}_2$ . Any equivalent balanced equations that sum to the overall reaction may also be accepted.

Techniques used

write catalytic cycle equationsbalance redox half-equations

(c)

Describe the effect of an increase in temperature on the rate constant and the rate of reaction 1.

DifficultyEasy

Worked solution

Answer

An increase in temperature will cause both the rate constant, $k$ , and the rate of reaction to increase.

Final answer

Both the rate constant and the rate of reaction will increase.

Detailed explanation

Background Concept

The rate constant $k$ is temperature-dependent and is described by the Arrhenius equation: $k = A e^{-E_a / RT}$ . As temperature increases, the exponential term increases, leading to a larger $k$ . Since rate = $k \times [\text{reactants}]^n$ , an increase in $k$ directly leads to an increase in the rate of reaction.

Understanding the Question

The question asks for the effect of increasing temperature on two specific quantities: the rate constant ( $k$ ) and the rate of reaction.

Approach

Recall the relationship between temperature, rate constant, and reaction rate. State that both increase.

Step-by-Step Reasoning

Increasing temperature increases the average kinetic energy of the molecules.
This leads to more frequent collisions and, more importantly, a greater proportion of molecules having energy greater than the activation energy ( $E_a$ ).
According to the Arrhenius equation, this results in an increase in the rate constant $k$ .
Since the rate of reaction is directly proportional to $k$ (for a given concentration of reactants), the rate of reaction also increases.

Key Takeaways

Temperature affects the rate constant $k$ , not just the frequency of collisions.
Both $k$ and the rate of reaction increase with temperature.

Common Mistakes

Saying only the 'rate increases' without mentioning the rate constant. The question specifically asks for both.
Saying the rate constant 'decreases' (confusing it with equilibrium constant for exothermic reactions).

Things to Be Careful About

Be precise: 'increase' is the correct term. 'Go up' or 'get bigger' might be accepted but 'increase' is the scientific term.
Do not confuse this with the effect of temperature on the position of equilibrium.

Techniques used

state effect of temperature on rate constant and rate

(d)

In aqueous solution, thiosulfate ions, $\text{S}_2\text{O}_3^{2-}$ , react with hydrogen ions, as shown in reaction 2.

\text{reaction 2 } \quad \text{S}_2\text{O}_3^{2-} + 2\text{H}^+ \rightarrow \text{SO}_2 + \text{S} + \text{H}_2\text{O}

The rate of reaction is first order with respect to $[\text{S}_2\text{O}_3^{2-}]$ and zero order with respect to $[\text{H}^+]$ under certain conditions.

The rate constant, $k$ , for this reaction is $1.58 \times 10^{-2}\text{ s}^{-1}$ .

Calculate the half-life, $t_{\frac{1}{2}}$ , for reaction 2.

t_{\frac{1}{2}} = \text{.............................. s}

DifficultyEasy

Worked solution

Working

For a first-order reaction, the half-life is given by:

t_{1/2} = \frac{\ln 2}{k} = \frac{0.693}{k}

Given $k = 1.58 \times 10^{-2} \text{ s}^{-1}$ :

t_{1/2} = \frac{0.693}{1.58 \times 10^{-2}} = 43.86... \text{ s}

Rounding to 3 significant figures:

t_{1/2} = 43.9 \text{ s}

Answer

43.9

Final answer

43.9

Detailed explanation

Background Concept

For a first-order reaction, the half-life ( $t_{1/2}$ ) is constant and independent of the initial concentration. It is related to the rate constant $k$ by the equation:

t_{1/2} = \frac{\ln 2}{k} \approx \frac{0.693}{k}

This is derived from the integrated rate law: $\ln[\text{A}] = \ln[\text{A}]_0 - kt$ . At $t = t_{1/2}$ , $[\text{A}] = \frac{1}{2}[\text{A}]_0$ , so $\ln(0.5) = -kt_{1/2}$ , which gives $t_{1/2} = \frac{\ln 2}{k}$ .

Understanding the Question

We are given that reaction 2 is first order with respect to $[\text{S}_2\text{O}_3^{2-}]$ and zero order with respect to $[\text{H}^+]$ . The overall order is 1. The rate constant $k = 1.58 \times 10^{-2} \text{ s}^{-1}$ . We need to calculate the half-life.

Approach

Identify that the reaction is first order overall.
Use the first-order half-life formula $t_{1/2} = 0.693 / k$ .
Substitute the given value of $k$ and calculate.

Step-by-Step Reasoning

The reaction is first order overall (first order in thiosulfate, zero order in hydrogen ions).
The formula for the half-life of a first-order reaction is $t_{1/2} = \frac{0.693}{k}$ .
Substitute $k = 1.58 \times 10^{-2} \text{ s}^{-1}$ : $t_{1/2} = \frac{0.693}{1.58 \times 10^{-2}} = 43.8607... \text{ s}$
Round to 3 significant figures (matching the given $k$ value): $43.9 \text{ s}$ .

Key Takeaways

The half-life formula $t_{1/2} = 0.693 / k$ applies only to first-order reactions.
Always check the overall order of the reaction before applying this formula.
Pay attention to units: $k$ is in $\text{s}^{-1}$ , so $t_{1/2}$ will be in seconds.

Common Mistakes

Using the wrong half-life formula (e.g., $t_{1/2} = 1/k$ for second-order).
Forgetting to convert units or ignoring the units of $k$ .
Rounding to the wrong number of significant figures.

Things to Be Careful About

The mark scheme uses $0.693$ for $\ln 2$ . Using $\ln 2$ directly in a calculator is also acceptable and may give a slightly more precise answer (43.86), which should be rounded appropriately.
Ensure the final answer has the correct unit (s).

Techniques used

apply first-order half-life equation

(e)

The compound nitrosyl bromide, $\text{NOBr}$ , can be formed as shown in reaction 3.

\text{reaction 3 } \quad 2\text{NO(g)} + \text{Br}_2\text{(g)} \rightarrow 2\text{NOBr(g)}

The rate is first order with respect to $[\text{NO}]$ and first order with respect to $[\text{Br}_2]$ .

The reaction mechanism has two steps.

Suggest equations for the two steps of this mechanism. State which is the rate-determining step.

step 1 ........................................................................................................................

step 2 ........................................................................................................................

rate-determining step = ...............................

DifficultyMedium

Worked solution

Answer

step 1: $\text{NO} + \text{Br}_2 \rightarrow \text{NOBr}_2$ (or $\text{NO} + \text{Br}_2 \rightarrow \text{NOBr} + \text{Br}$ )

step 2: $\text{NOBr}_2 + \text{NO} \rightarrow 2\text{NOBr}$ (or $\text{Br} + \text{NO} \rightarrow \text{NOBr}$ )

rate-determining step = step 1

Final answer

step 1: NO + Br2 -> NOBr2; step 2: NOBr2 + NO -> 2NOBr; rate-determining step = step 1

Detailed explanation

Background Concept

The rate-determining step (RDS) is the slowest step in a reaction mechanism. The rate equation for the overall reaction is determined by the reactants involved in the RDS. If the RDS is the first step, the rate equation will reflect the stoichiometry of that step. Intermediates formed in the RDS or subsequent steps must not appear in the final rate equation (they are substituted out using fast equilibrium steps if necessary).

Understanding the Question

Overall reaction: $2\text{NO(g)} + \text{Br}_2\text{(g)} \rightarrow 2\text{NOBr(g)}$ .
Rate equation: $\text{rate} = k[\text{NO}][\text{Br}_2]$ (first order in NO, first order in Br2).
The mechanism has two steps. We need to propose equations for these steps and identify the RDS.

Approach

The rate equation depends on the reactants in the RDS. Since the rate depends on one NO and one Br2, the RDS must involve one molecule of NO and one molecule of Br2 colliding.
Propose step 1 as the collision of NO and Br2. This forms an intermediate.
Propose step 2 as the reaction of this intermediate with the second NO molecule to form the final product.
Verify that the two steps sum to the overall equation.

Step-by-Step Reasoning

The rate law is $\text{rate} = k[\text{NO}][\text{Br}_2]$ . This suggests the RDS involves a bimolecular collision between one NO and one Br2.
Step 1 (RDS): $\text{NO} + \text{Br}_2 \rightarrow \text{NOBr}_2$ . (Alternatively, $\text{NO} + \text{Br}_2 \rightarrow \text{NOBr} + \text{Br}$ is also acceptable).
This step is slow and determines the rate. The rate of this step is proportional to $[\text{NO}][\text{Br}_2]$ , matching the observed rate law.
Step 2 (fast): The intermediate $\text{NOBr}_2$ reacts with the second NO molecule.
$\text{NOBr}_2 + \text{NO} \rightarrow 2\text{NOBr}$ . (Alternatively, $\text{Br} + \text{NO} \rightarrow \text{NOBr}$ ).
This step is fast.
Check: Adding the two steps:
$\text{NO} + \text{Br}_2 + \text{NOBr}_2 + \text{NO} \rightarrow \text{NOBr}_2 + 2\text{NOBr}$
Canceling the intermediate $\text{NOBr}_2$ : $2\text{NO} + \text{Br}_2 \rightarrow 2\text{NOBr}$ . This matches the overall equation.
Since step 1 matches the rate law, it is the rate-determining step.

Key Takeaways

The rate equation provides direct information about the reactants in the rate-determining step.
For a two-step mechanism where the first step is slow, the rate equation is determined by the reactants in that first step.
Intermediates formed in the mechanism must cancel out when the steps are summed.

Common Mistakes

Proposing a mechanism where the first step involves 2 NO and 1 Br2 (termolecular collision), which is unlikely and doesn't match the two-step requirement.
Identifying the wrong step as the rate-determining step. If step 2 were the RDS, the rate law would depend on the intermediate, which is not observed.
Not ensuring the two steps sum to the overall balanced equation.

Things to Be Careful About

The mark scheme accepts alternative valid mechanisms, such as $\text{NO} + \text{Br}_2 \rightarrow \text{NOBr} + \text{Br}$ followed by $\text{Br} + \text{NO} \rightarrow \text{NOBr}$ . Both are acceptable as long as they sum to the overall equation and the first step is identified as the RDS.
Clearly label which step is the rate-determining step.

Techniques used

deduce mechanism steps from rate equationidentify rate-determining step

Q6Medium-HardEquilibriaIntroduction to A Level Organic ChemistryAnalytical TechniquesHydrocarbonsOrganic Synthesis

(a)

3M

(i)

State what is meant by partition coefficient, $K_{\text{pc}}$ .

DifficultyEasy

Worked solution

Answer

The ratio of the concentrations of a solute dissolved in two immiscible solvents at equilibrium.

Final answer

Ratio of the concentrations of solute in two solvents

Detailed explanation

Background Concept

When a solute is distributed between two immiscible solvents (e.g. an organic solvent and water), it will partition between the two phases until dynamic equilibrium is established. The partition coefficient $K_{\text{pc}}$ quantifies this distribution and is defined as the ratio of the equilibrium concentrations of the solute in each solvent. It is a constant at a given temperature and depends on the relative solubilities of the solute in each phase.

Understanding the Question

This is a straightforward definition question asking the candidate to state what $K_{\text{pc}}$ means. The command word is "state", so a concise definition is expected.

Approach

Recall the standard definition of partition coefficient as the ratio of concentrations in two immiscible solvents at equilibrium.

Step-by-Step Reasoning

The partition coefficient is defined as:

K_{\text{pc}} = \frac{[\text{solute}]_{\text{organic phase}}}{[\text{solute}]_{\text{aqueous phase}}}

This is a ratio of concentrations (not amounts or masses) of the same solute in two immiscible solvents when equilibrium has been reached. The key elements of the definition are: (1) it is a ratio, (2) it involves concentrations, (3) the solute is the same in both phases, and (4) the two solvents are immiscible.

Key Takeaways

$K_{\text{pc}}$ is a ratio of concentrations, not masses or moles.
It applies to a solute distributed between two immiscible solvents at equilibrium.
It is temperature-dependent.

Common Mistakes

Saying "ratio of amounts" instead of "ratio of concentrations" — the mark scheme specifically requires concentrations.
Omitting the condition of equilibrium.
Confusing $K_{\text{pc}}$ with solubility or distribution coefficient without specifying the two-phase system.

Things to Be Careful About

Use the word "concentrations" precisely; "amounts" or "moles" would not be credited.
Mention that the two solvents must be immiscible.

Techniques used

state the definition of partition coefficient

(ii)

The partition coefficient, $K_{\text{pc}}$ , for a compound, X, between carbon disulfide, $\text{CS}_2$ , and water is 10.5.

$1.85\text{ g}$ of X is dissolved in water and made up to $100.0\text{ cm}^3$ in a volumetric flask.
$40.0\text{ cm}^3$ of this aqueous solution is shaken with $25.0\text{ cm}^3$ of $\text{CS}_2$ .
The mixture is left to reach equilibrium.

Calculate the mass of X, in g, extracted into the $\text{CS}_2$ layer.

mass of X = .............................. g

DifficultyMedium

Worked solution

Working

Mass of X in $40.0\text{ cm}^3$ of aqueous solution:

\text{mass} = 1.85 \times \frac{40.0}{100.0} = 0.74\text{ g}

Let $y$ = mass of X extracted into $\text{CS}_2$ layer (in g).

K_{\text{pc}} = \frac{[\text{X}]_{\text{CS}_2}}{[\text{X}]_{\text{water}}} = \frac{y/25.0}{(0.74 - y)/40.0}

10.5 = \frac{40.0y}{25.0(0.74 - y)}

10.5 \times 25.0(0.74 - y) = 40.0y

194.25 - 262.5y = 40.0y

302.5y = 194.25

y = 0.642\text{ g}

Answer

$0.642\text{ g}$

Final answer

0.642 g

Detailed explanation

Background Concept

The partition coefficient $K_{\text{pc}}$ describes how a solute distributes itself between two immiscible solvents at equilibrium. If $K_{\text{pc}}$ is defined as (concentration in organic layer)/(concentration in aqueous layer), a value greater than 1 means the solute prefers the organic phase. In extraction calculations, we use $K_{\text{pc}}$ to determine how much solute moves from one phase to the other when the two are shaken together.

Understanding the Question

We are given $K_{\text{pc}} = 10.5$ for compound X between $\text{CS}_2$ and water. A total of $1.85\text{ g}$ X is dissolved in $100.0\text{ cm}^3$ water, then $40.0\text{ cm}^3$ of this solution is extracted with $25.0\text{ cm}^3$ of $\text{CS}_2$ . We need to find the mass of X that transfers into the $\text{CS}_2$ layer at equilibrium.

Approach

Calculate the mass of X present in the $40.0\text{ cm}^3$ aliquot.
Let $y$ be the mass extracted into $\text{CS}_2$ ; then $(0.74 - y)$ remains in the aqueous layer.
Write the $K_{\text{pc}}$ expression using concentrations (mass/volume).
Solve for $y$ .

Step-by-Step Reasoning

Step 1: The original solution has $1.85\text{ g}$ in $100.0\text{ cm}^3$ . Taking $40.0\text{ cm}^3$ gives:

\text{mass in aliquot} = 1.85 \times \frac{40.0}{100.0} = 0.74\text{ g}

Step 2: Let $y$ = mass of X in the $\text{CS}_2$ layer at equilibrium. Then mass remaining in aqueous layer = $(0.74 - y)$ g.

Step 3: The partition coefficient expression (concentration in $\text{CS}_2$ divided by concentration in water):

K_{\text{pc}} = \frac{y/25.0}{(0.74 - y)/40.0} = 10.5

Step 4: Solving:

\frac{40.0y}{25.0(0.74 - y)} = 10.5

40.0y = 10.5 \times 25.0 \times (0.74 - y)

40.0y = 262.5(0.74 - y)

40.0y = 194.25 - 262.5y

302.5y = 194.25

y = 0.642\text{ g}

Step 5 (check): Mass in aqueous layer = $0.74 - 0.642 = 0.098\text{ g}$ . Concentration in $\text{CS}_2$ = $0.642/25.0 = 0.0257\text{ g cm}^{-3}$ . Concentration in water = $0.098/40.0 = 0.00245\text{ g cm}^{-3}$ . Ratio = $0.0257/0.00245 = 10.5$ ✓

Key Takeaways

Always work with concentrations (mass/volume) in the $K_{\text{pc}}$ expression, not just masses.
The volume of each solvent appears in the denominator of the respective concentration.
A large $K_{\text{pc}}$ means most of the solute is extracted into the organic phase.

Common Mistakes

Forgetting to scale the mass from $100.0\text{ cm}^3$ down to $40.0\text{ cm}^3$ (using $1.85\text{ g}$ instead of $0.74\text{ g}$ ).
Writing the $K_{\text{pc}}$ expression with volumes swapped (putting $\text{CS}_2$ volume with the aqueous concentration).
Setting up $K_{\text{pc}} = \frac{(0.74-y)/40}{y/25}$ (inverted ratio) — this would give a different answer.
Using moles instead of mass (the answer is requested in grams, and since the molar mass cancels, mass works directly).

Things to Be Careful About

Ensure the correct definition of $K_{\text{pc}}$ (which solvent is in the numerator). The mark scheme's M1 shows $K_{\text{pc}} = (y \div 25) \div ((0.74 - y) \div 40)$ , confirming $\text{CS}_2$ concentration is on top.
Report the answer to 3 significant figures as implied by the data.

Techniques used

calculate mass of solute in aliquotset up partition coefficient expressionsolve simultaneous equation for mass extracted

(b)

The compound $\text{C}_6\text{H}_6$ has many structural isomers. Four suggested structures of $\text{C}_6\text{H}_6$ are shown in Fig. 6.1.

Fig. 6.1

Using Fig. 6.1, complete Table 6.1 to predict the number of carbon atoms that have $\text{sp}$ , $\text{sp}^2$ and $\text{sp}^3$ hybridisation in Kekulé benzene, Dewar benzene and Ladenburg benzene.

Table 6.1

$\text{C}_6\text{H}_6$ structure	$\text{sp}$ hybridised	$\text{sp}^2$ hybridised	$\text{sp}^3$ hybridised
Kekulé benzene
Dewar benzene
Ladenburg benzene

DifficultyMedium-Easy

Worked solution

Answer

$\text{C}_6\text{H}_6$ structure	$\text{sp}^2$	$\text{sp}^3$
Kekulé benzene	6	0
Dewar benzene	4	2
Ladenburg benzene	0	6

Final answer

Kekulé: 0, 6, 0; Dewar: 0, 4, 2; Ladenburg: 0, 0, 6

Detailed explanation

Background Concept

Carbon hybridisation is determined by the number of electron domains (regions of electron density) around each carbon atom:

sp: 2 electron domains (e.g. in a triple bond or two double bonds), linear geometry, 180° bond angle.
sp²: 3 electron domains (e.g. a carbon in a double bond with two other single bonds), trigonal planar, 120° bond angle.
sp³: 4 electron domains (four single bonds), tetrahedral, 109.5° bond angle.

A carbon involved in a C=C double bond is sp² hybridised (it has three σ-bonds: two to other carbons/hydrogen and one to the adjacent carbon, plus a π-bond). A carbon with only single bonds (four σ-bonds) is sp³.

Understanding the Question

We are given three structural isomers of $\text{C}_6\text{H}_6$ and must count how many carbon atoms have each type of hybridisation.

Kekulé benzene (1,3,5-cyclohexatriene): a six-membered ring with alternating single and double bonds. Every carbon is part of a C=C double bond.
Dewar benzene (bicyclo[2.2.0]hexa-2,5-diene): two fused four-membered rings with two C=C double bonds. Four carbons are in double bonds; two bridgehead carbons have only single bonds.
Ladenburg benzene (prismane): a triangular prism with all single C-C bonds. Every carbon is bonded to three other carbons and one hydrogen — all sp³.

Approach

For each structure, examine each carbon and count its σ-bonds (electron domains). A carbon in a double bond has 3 domains → sp². A carbon with 4 single bonds has 4 domains → sp³. No carbon in these structures has a triple bond, so sp = 0 throughout.

Step-by-Step Reasoning

Kekulé benzene: All 6 carbons participate in a C=C double bond (alternating around the ring). Each has 3 electron domains → all sp². No sp or sp³ carbons.

Dewar benzene: The structure has two C=C double bonds involving 4 carbons (sp²). The two bridgehead carbons (at the ring fusion) each form 4 single bonds (to 3 carbons and 1 H) → sp³. So: sp² = 4, sp³ = 2.

Ladenburg benzene (prismane): All 6 carbons are at corners of a triangular prism. Each carbon bonds to 3 other carbons and 1 hydrogen, all single bonds → sp³. So: sp³ = 6.

Key Takeaways

Hybridisation is determined by the number of electron domains, not by the ring size or strain.
A carbon in a C=C double bond is always sp² (3 domains: 2 σ to neighbouring atoms + 1 σ to the double-bond partner, with the π bond being the extra electron density but not an additional domain for hybridisation purposes — actually 3 σ-bonds + 1 π = 3 domains for hybridisation).
Prismane is a real compound (synthesised in 1960) but is highly strained.

Common Mistakes

Assigning sp hybridisation to carbons in strained rings — strain does not change hybridisation; only the number of electron domains does.
Miscounting the bridgehead carbons in Dewar benzene as sp² (they have no double bond).
Thinking all carbons in Kekulé benzene are sp³ because it is a "ring" — the double bonds make them sp².

Things to Be Careful About

The mark scheme awards M1 for rows 1 and 2 correct together, and M2 for row 3. Both must be correct for full marks.
Ensure you count all 6 carbons in each structure and that the three columns sum to 6.

Techniques used

identify hybridisation from bonding environmentcount carbon atoms by hybridisation type

(c)

Describe the shape of delocalised benzene.

Include the geometry of each carbon, the C-C-H bond angle and the type of bond(s) between the carbon atoms and between the carbon and hydrogen atoms.

DifficultyMedium-Easy

Worked solution

Answer

Each carbon is $\text{sp}^2$ hybridised, giving a trigonal planar geometry with bond angles of $120°$ . The molecule is a planar regular hexagon.
The C–C bonds consist of $\sigma$ -bonds and delocalised $\pi$ -bonds (the $\pi$ -electrons are spread over all six carbons). The C–H bonds are $\sigma$ -bonds only.

Final answer

Trigonal planar, 120°, planar hexagonal ring; C-C bonds have sigma and pi (delocalised); C-H bonds are sigma only

Detailed explanation

Background Concept

Benzene ( $\text{C}_6\text{H}_6$ ) is a planar, regular hexagonal molecule. Each carbon atom is $\text{sp}^2$ hybridised, forming three $\sigma$ -bonds (two to adjacent carbons and one to hydrogen) arranged at $120°$ in a plane. The remaining unhybridised p-orbital on each carbon overlaps sideways with its neighbours to form a delocalised $\pi$ -system above and below the plane of the ring. This delocalisation gives all six C–C bonds equal length (intermediate between single and double) and accounts for benzene's unusual stability and its preference for substitution over addition reactions.

Understanding the Question

The question asks for a description of the shape of delocalised benzene, specifically requiring: geometry of each carbon, the C–C–H bond angle, and the types of bonds between C–C and C–H. The command word is "describe", so a factual account is needed.

Approach

Address each element of the question in turn: (1) geometry and bond angle, (2) C–C bonding, (3) C–H bonding.

Step-by-Step Reasoning

Geometry and bond angle: Each carbon has three $\sigma$ -bonds and no lone pairs → 3 electron domains → $\text{sp}^2$ → trigonal planar geometry. The C–C–H bond angle is $120°$ . The overall shape is a planar regular hexagon.

C–C bonds: Between each pair of adjacent carbons there is a $\sigma$ -bond (from end-on overlap of $\text{sp}^2$ orbitals) and a contribution to the delocalised $\pi$ -system (from sideways overlap of p-orbitals). So C–C bonds have both $\sigma$ and $\pi$ character.

C–H bonds: Each C–H bond is a single $\sigma$ -bond (overlap of carbon $\text{sp}^2$ orbital with hydrogen 1s orbital). There is no $\pi$ -component to C–H bonds.

Key Takeaways

Benzene is planar with $120°$ bond angles due to $\text{sp}^2$ hybridisation.
The C–C bonds are not purely single or double; they have $\sigma$ + delocalised $\pi$ character.
C–H bonds are purely $\sigma$ .

Common Mistakes

Saying the bond angle is $109.5°$ (that is for $\text{sp}^3$ carbon, not $\text{sp}^2$ ).
Describing C–C bonds as "alternating single and double" — this is the Kekulé model, not the delocalised model.
Omitting that C–H bonds are $\sigma$ -bonds only (the mark scheme requires this distinction).
Saying the molecule is "tetrahedral" or "bent".

Things to Be Careful About

The mark scheme requires BOTH the bond angle ( $120°$ ) AND the shape descriptor (hexagonal/trigonal planar) for M1.
For M2, both parts must be present: C–C has $\pi$ and $\sigma$ bonds, AND C–H has $\sigma$ bonds only.

Techniques used

describe molecular geometry and bonding in benzeneidentify sigma and pi bond types

(d)

Suggest why Dewar benzene and Ladenburg benzene are unstable isomers of $\text{C}_6\text{H}_6$ .

DifficultyEasy

Worked solution

Answer

The bond angles in the small rings (approximately $60°$ in prismane and $90°$ in Dewar benzene) deviate greatly from the ideal tetrahedral angle of $109.5°$ for $\text{sp}^3$ carbons, causing significant bond strain (ring strain).

Final answer

Bond strain / ring strain

Detailed explanation

Background Concept

Ring strain (also called angle strain or Baeyer strain) arises when bond angles in a cyclic molecule deviate from the ideal angle for the hybridisation of the atoms involved. For $\text{sp}^3$ carbon, the ideal angle is $109.5°$ . In small rings (3-, 4-membered), the internal angles are forced to be much smaller ( $60°$ or $90°$ ), causing the bonding orbitals to overlap less effectively and storing significant potential energy in the molecule. This makes the molecule thermodynamically unstable and reactive.

Understanding the Question

We must explain why Dewar benzene and Ladenburg benzene (prismane) are unstable compared to delocalised benzene. The command word is "suggest", so a reasonable chemical explanation is needed.

Approach

Identify the structural feature common to both unstable isomers: they contain small rings with $\text{sp}^3$ carbons forced into highly compressed bond angles.

Step-by-Step Reasoning

Dewar benzene contains two fused four-membered rings. The bridgehead $\text{sp}^3$ carbons have bond angles of approximately $90°$ , far from the ideal $109.5°$ .
Prismane (Ladenburg benzene) contains three four-membered rings and two three-membered rings. The $\text{sp}^3$ carbons are forced into angles of approximately $60°$ – $90°$ .
In both cases, the deviation from ideal bond angle creates substantial ring strain (angle strain), making these molecules much higher in energy than delocalised benzene.
Additionally, neither has the stabilising effect of $\pi$ -electron delocalisation that makes benzene so stable.

Key Takeaways

Ring strain is the primary reason small-ring isomers of benzene are unstable.
Delocalisation energy in benzene provides additional thermodynamic stability that these isomers lack.
Prismane was synthesised in 1960 and confirmed to exist, but rapidly converts to benzene.

Common Mistakes

Saying "they have double bonds" as the reason for instability — it is the strain, not the presence of double bonds, that matters.
Confusing ring strain with torsional strain (though both contribute, the mark scheme accepts "bond strain" or "ring strain").
Saying they are unstable because they are "not aromatic" — while true, the specific structural reason is the angle strain.

Things to Be Careful About

The mark scheme accepts "bond strain" or "ring strain" — either term is sufficient for the single mark.

Techniques used

explain instability due to ring strain

(e)

Complete Table 6.2 to predict the number of peaks in the proton ( $^1\text{H}$ ) NMR spectrum for Dewar benzene, Ladenburg benzene and delocalised benzene.

Table 6.2

	number of peaks
Dewar benzene
Ladenburg benzene
delocalised benzene

DifficultyMedium-Easy

Worked solution

Answer

	Number of peaks
Dewar benzene	2
Ladenburg benzene	1
Delocalised benzene	1

Final answer

Dewar: 2; Ladenburg: 1; delocalised benzene: 1

Detailed explanation

Background Concept

In $^1\text{H}$ NMR spectroscopy, each chemically distinct environment of hydrogen atoms gives rise to one signal (peak). Chemically equivalent hydrogens are those that are in identical electronic environments, typically identified by the symmetry of the molecule. If all hydrogens in a molecule are equivalent, only one peak appears.

Understanding the Question

We must determine how many different types of hydrogen exist in each of three $\text{C}_6\text{H}_6$ isomers by examining their symmetry.

Approach

For each structure, identify the symmetry elements and determine how many unique hydrogen environments exist.

Step-by-Step Reasoning

Dewar benzene: The molecule has a plane of symmetry. The four hydrogens on the $\text{sp}^2$ carbons (vinyl H's) are all equivalent to each other. The two hydrogens on the $\text{sp}^3$ bridgehead carbons are equivalent to each other but different from the vinyl H's. Therefore: 2 peaks.

Ladenburg benzene (prismane): Prismane has high symmetry (D₃h point group). All six carbons are equivalent, and all six hydrogens are in identical environments. Therefore: 1 peak.

Delocalised benzene: All six hydrogens are equivalent due to the perfect hexagonal symmetry. Therefore: 1 peak.

Key Takeaways

NMR peak count depends on the number of chemically distinct hydrogen environments.
Molecular symmetry determines equivalence: if a symmetry operation interconverts two H's, they are equivalent.
Prismane and benzene both give a single $^1\text{H}$ NMR signal, but their $^{13}\text{C}$ NMR spectra and chemical shifts differ.

Common Mistakes

Counting the number of hydrogen atoms instead of the number of environments.
For Dewar benzene, thinking all 6 H's are equivalent (they are not — vinyl and bridgehead H's are different).
For prismane, thinking the two types of ring (triangular vs square faces) give different H environments (they do not — all carbons are equivalent by symmetry).

Things to Be Careful About

The question asks for number of peaks, not integration ratios or splitting patterns.
In a real $^1\text{H}$ NMR spectrum, coupling between non-equivalent protons can split peaks, but the question asks for the number of distinct chemical shifts (signals), which equals the number of unique environments.

Techniques used

identify chemically equivalent hydrogen environmentscount distinct proton environments for NMR

(f)

The reaction of phenylethanone with 1,4-dibromobutane, $\text{BrCH}_2\text{CH}_2\text{CH}_2\text{CH}_2\text{Br}$ , in the presence of $\text{FeBr}_3$ is shown in Fig. 6.2.

Fig. 6.2

The mechanism of this reaction is similar to that of the alkylation of benzene.

6M

(i)

Construct an equation for the formation of the electrophile, $\text{BrCH}_2\text{CH}_2\text{CH}_2\text{CH}_2^+$ .

DifficultyEasy

Worked solution

Answer

\text{BrCH}_2\text{CH}_2\text{CH}_2\text{CH}_2\text{Br} + \text{FeBr}_3 \rightarrow \text{BrCH}_2\text{CH}_2\text{CH}_2\text{CH}_2^+ + \text{FeBr}_4^-

Final answer

BrCH2CH2CH2CH2Br + FeBr3 -> BrCH2CH2CH2CH2+ + FeBr4-

Detailed explanation

Background Concept

In Friedel-Crafts alkylation, a halogenoalkane reacts with a Lewis acid catalyst (such as $\text{FeBr}_3$ , $\text{AlCl}_3$ , or $\text{FeCl}_3$ ) to generate a carbocation electrophile. The Lewis acid accepts a lone pair from the halogen, polarising and ultimately breaking the C–X bond, producing a carbocation and a complex anion (e.g. $\text{FeBr}_4^-$ or $\text{AlCl}_4^-$ ).

Understanding the Question

We need to write the equation showing how $\text{FeBr}_3$ converts 1,4-dibromobutane into the electrophile $\text{BrCH}_2\text{CH}_2\text{CH}_2\text{CH}_2^+$ . Only one of the two C–Br bonds is broken in this step.

Approach

Write a balanced equation showing the halogenoalkane plus $\text{FeBr}_3$ giving the carbocation plus $\text{FeBr}_4^-$ . The other C–Br bond remains intact (it will be used later in the intramolecular cyclisation to form by-product Z, or in a second intermolecular reaction to form Y).

Step-by-Step Reasoning

$\text{FeBr}_3$ acts as a Lewis acid, accepting a lone pair from one bromine atom. This weakens and breaks the C–Br bond heterolytically, with both electrons going to the bromine that bonds to Fe:

\text{BrCH}_2\text{CH}_2\text{CH}_2\text{CH}_2\text{Br} + \text{FeBr}_3 \rightarrow \text{BrCH}_2\text{CH}_2\text{CH}_2\text{CH}_2^+ + \text{FeBr}_4^-

The carbocation $\text{BrCH}_2\text{CH}_2\text{CH}_2\text{CH}_2^+$ is the electrophile that attacks the benzene ring of phenylethanone.

Key Takeaways

$\text{FeBr}_3$ (or $\text{AlCl}_3$ ) is a Lewis acid that generates carbocations from halogenoalkanes.
The by-product is $\text{FeBr}_4^-$ (not $\text{Br}^-$ alone).
Only one C–Br bond is broken at this stage.

Common Mistakes

Writing $\text{Br}^-$ as a product instead of $\text{FeBr}_4^-$ .
Breaking both C–Br bonds.
Forgetting the positive charge on the carbocation.
Writing the equation with $\text{FeBr}_2$ instead of $\text{FeBr}_3$ .

Things to Be Careful About

The equation must be balanced (charges and atoms).
State symbols are not required for this mark but the charges must be correct.

Techniques used

construct equation for electrophile generation using Lewis acid catalyst

(ii)

Complete the mechanism in Fig. 6.3 for the reaction of phenylethanone with $\text{BrCH}_2\text{CH}_2\text{CH}_2\text{CH}_2^+$ ions.

Include all relevant curly arrows and charges.
Draw the structure of the organic intermediate.

Fig. 6.3

DifficultyMedium

Worked solution

Answer

M1: Curly arrow from the $\pi$ -electron system of the benzene ring (inside the hexagon) to the positively charged carbon of $\text{BrCH}_2\text{CH}_2\text{CH}_2\text{CH}_2^+$ .

M2: Structure of the organic intermediate (arenium ion / $\sigma$ -complex): a cyclohexadienyl cation showing the acetyl group ( $\text{COCH}_3$ ) on the ring, the $\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_2\text{Br}$ chain and an H attached to the same carbon (the attacked carbon, at the meta position), with a positive charge delocalised in the ring (shown as $+$ inside the ring or on an adjacent carbon).

M3: Curly arrow from the C–H bond on the attacked carbon back into the ring to restore aromaticity, with $\text{H}^+$ as the other product.

Final answer

See diagram: curly arrow from ring to C+, arenium ion intermediate with COCH3 and CH2CH2CH2CH2Br and H on same carbon with + charge in ring, curly arrow from C-H into ring giving H+

Detailed explanation

Background Concept

Electrophilic aromatic substitution proceeds in two stages:

The $\pi$ -electrons of the aromatic ring attack the electrophile, forming a new C–C bond. This produces a sigma complex (arenium ion / Wheland intermediate), in which the ring is no longer aromatic — it is a cyclohexadienyl cation with the positive charge delocalised over three carbons (shown by resonance structures or by a + inside the ring).
A base (often $\text{FeBr}_4^-$ or the solvent) removes the proton from the $\text{sp}^3$ carbon, restoring aromaticity and giving the substituted product plus $\text{H}^+$ .

The curly arrow conventions are critical: in step 1, the arrow starts from the ring $\pi$ -system (drawn from inside the hexagon or from a C=C bond) and points to the electrophilic carbon. In step 2, the arrow starts from the C–H bond (the bond electrons) and points into the ring to reform the $\pi$ -system.

Understanding the Question

We are asked to complete the mechanism for the reaction of phenylethanone with $\text{BrCH}_2\text{CH}_2\text{CH}_2\text{CH}_2^+$ . The figure shows phenylethanone and the electrophile with an arrow to an empty box (the intermediate), then an arrow to the product. We must draw curly arrows and the intermediate.

The product is the meta-substituted compound (as shown in Fig. 6.2), because the acetyl group ( $\text{COCH}_3$ ) is a meta-directing group (electron-withdrawing by the $-I$ and $-M$ effects).

Approach

Draw a curly arrow from the benzene ring to the $\text{C}^+$ of the electrophile.
Draw the arenium ion intermediate: the ring now has the $\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_2\text{Br}$ group and an H on the same carbon (making it $\text{sp}^3$ ), the $\text{COCH}_3$ group unchanged, and a + charge delocalised in the remaining 5 carbons.
Draw a curly arrow from the C–H bond into the ring (to restore aromaticity).
Write $\text{H}^+$ as the by-product.

Step-by-Step Reasoning

Step 1 (M1): The electrophile $\text{BrCH}_2\text{CH}_2\text{CH}_2\text{CH}_2^+$ has a positively charged carbon. The $\pi$ -electrons of the benzene ring (specifically at the meta position relative to $\text{COCH}_3$ ) attack this carbon. Draw a curly arrow from inside the hexagon (representing the $\pi$ -cloud) to the $\text{C}^+$ .

Step 2 (M2): The intermediate is a cyclohexadienyl cation. The attacked carbon now has four bonds (two to ring carbons, one to H, one to the butyl bromide chain), making it $\text{sp}^3$ . The positive charge is delocalised over the remaining conjugated system (three carbons). Draw the structure showing:

The hexagonal ring with two remaining double bonds
$\text{COCH}_3$ on one carbon
On the meta carbon: both H and $\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_2\text{Br}$
A + sign (inside the ring or on a carbon bearing the charge)

Step 3 (M3): To restore aromaticity, the C–H bond electrons move back into the ring. Draw a curly arrow from the C–H bond to the ring (forming the third $\pi$ -bond). The product is the aromatic meta-substituted compound, and $\text{H}^+$ is released.

Key Takeaways

In electrophilic aromatic substitution, the first curly arrow always goes FROM the ring TO the electrophile.
The intermediate (arenium ion) has the electrophile AND a hydrogen on the same carbon.
The second curly arrow goes FROM the C–H bond INTO the ring (not from a lone pair on H).
Meta substitution occurs because $\text{COCH}_3$ is electron-withdrawing (deactivating, meta-directing).

Common Mistakes

Drawing the curly arrow from the electrophile to the ring (wrong direction — electrons flow from electron-rich to electron-poor).
Forgetting the H on the attacked carbon in the intermediate.
Not showing the positive charge in the intermediate.
Drawing the curly arrow for deprotonation from H to the ring instead of from the C–H bond into the ring.
Placing the substituent at the ortho or para position instead of meta.
Drawing the intermediate with the wrong number of double bonds (should have 2 C=C in the ring, not 3).

Things to Be Careful About

The curly arrow for M1 must start from inside the hexagon (or a C=C bond) and end at the $\text{C}^+$ .
The intermediate must show the + charge (either delocalised or on a specific carbon).
The C–H curly arrow (M3) must start from the bond, not from the H atom.
Charges must be clearly indicated.

Techniques used

draw curly arrow from aromatic ring to electrophiledraw the sigma complex (arenium ion) intermediatedraw curly arrow for deprotonation to restore aromaticity

(iii)

The reaction shown in Fig. 6.2 forms small amounts of two by-products, $\mathbf{Y}$ ( $\text{C}_{20}\text{H}_{22}\text{O}_2$ ) and $\mathbf{Z}$ ( $\text{C}_{12}\text{H}_{14}\text{O}$ ).

Suggest structures for $\mathbf{Y}$ and $\mathbf{Z}$ in the boxes in Fig. 6.4.

Fig. 6.4

DifficultyMedium-Hard

Worked solution

Answer

Y ( $\text{C}_{20}\text{H}_{22}\text{O}_2$ ): Two 3-acetylphenyl groups linked by a $-\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_2-$ chain (the second C–Br bond of the dibromobutane reacts with a second molecule of phenylethanone).

Z ( $\text{C}_{12}\text{H}_{14}\text{O}$ ): 6-acetyl-1,2,3,4-tetrahydronaphthalene (the terminal C–Br of the mono-substituted product undergoes intramolecular electrophilic substitution, forming a new six-membered ring fused to the benzene ring, with loss of HBr).

Final answer

Y: two 3-acetylphenyl groups joined by -CH2CH2CH2CH2-; Z: 6-acetyl-1,2,3,4-tetrahydronaphthalene (bicyclic with one aromatic ring bearing acetyl and one saturated six-membered ring fused)

Detailed explanation

Background Concept

In Friedel-Crafts alkylation, if the electrophile contains two reactive sites (as in a dihalogenoalkane), further reactions can occur:

Intermolecular bis-substitution: The second halogen can generate another carbocation that reacts with a second aromatic molecule, linking two aromatic rings with the alkyl chain.
Intramolecular cyclisation: The remaining halogen on the mono-substituted product can generate a carbocation that reacts with the same ring (if geometrically feasible), forming a new ring fused to the original benzene ring.

The molecular formula of the by-product tells us which pathway occurred.

Understanding the Question

The main product from Fig. 6.2 is 1-(3-(4-bromobutyl)phenyl)ethan-1-one: $\text{C}_{12}\text{H}_{15}\text{BrO}$ . Two by-products form:

Y ( $\text{C}_{20}\text{H}_{22}\text{O}_2$ ): Contains two acetylphenyl units and the butyl chain — this is the bis-substitution product.
Z ( $\text{C}_{12}\text{H}_{14}\text{O}$ ): Same carbon count as the main product but without Br and with 1 fewer H — this is the intramolecular cyclisation product (loss of HBr).

Approach

For Y: Check if two molecules of phenylethanone ( $\text{C}_8\text{H}_8\text{O}$ ) plus the $\text{C}_4\text{H}_8$ bridge minus two H's (lost as 2HBr) gives $\text{C}_{20}\text{H}_{22}\text{O}_2$ :
$2 \times \text{C}_8\text{H}_8\text{O} + \text{C}_4\text{H}_8 - 2\text{H} = \text{C}_{20}\text{H}_{22}\text{O}_2$ ✓

For Z: The main product $\text{C}_{12}\text{H}_{15}\text{BrO}$ loses HBr (one H from the ring, one Br from the chain) to give $\text{C}_{12}\text{H}_{14}\text{O}$ ✓. This forms a bicyclic compound (tetralin derivative).

Step-by-Step Reasoning

Y ( $\text{C}_{20}\text{H}_{22}\text{O}_2$ ):

The dibromobutane has two C–Br bonds. After one reacts with phenylethanone, the other can also be activated by $\text{FeBr}_3$ to form a second carbocation.
This second carbocation attacks a second molecule of phenylethanone (also at the meta position to its $\text{COCH}_3$ group).
Result: two 3-acetylphenyl groups connected by $-\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_2-$ .
Molecular formula check: Each 3-acetylphenyl = $\text{C}_8\text{H}_7\text{O}$ (as a substituent, losing one H from the ring). Two of them = $\text{C}_{16}\text{H}_{14}\text{O}_2$ . Plus $\text{C}_4\text{H}_8$ bridge = $\text{C}_{20}\text{H}_{22}\text{O}_2$ ✓.

Z ( $\text{C}_{12}\text{H}_{14}\text{O}$ ):

The mono-substituted product has a terminal Br on the butyl chain. $\text{FeBr}_3$ can activate this C–Br bond to form a carbocation at the end of the chain.
This carbocation is positioned to attack the same benzene ring intramolecularly (the chain is long enough to reach around and form a new six-membered ring fused to the benzene).
The attack occurs at the position ortho to the point of attachment (or para to the acetyl group), forming a new C–C bond and a fused six-membered saturated ring.
Loss of $\text{H}^+$ (and $\text{FeBr}_4^-$ ) restores aromaticity.
Result: 6-acetyl-1,2,3,4-tetrahydronaphthalene (or the equivalent 7-acetyl isomer — the mark scheme accepts either orientation as shown in the figure).
Molecular formula: $\text{C}_{12}\text{H}_{14}\text{O}$ ✓ (bicyclic: one aromatic ring fused to one saturated ring, with $\text{COCH}_3$ on the aromatic ring).

Key Takeaways

Dihalogenoalkanes in Friedel-Crafts reactions can give bis-substitution (intermolecular) or cyclisation (intramolecular) by-products.
Molecular formula analysis is the key to deducing which pathway occurred.
Intramolecular reactions are often favoured entropically for forming 5- and 6-membered rings.
The acetyl group directs to the meta position in all cases.

Common Mistakes

For Y: drawing the two rings connected at the ortho or para position instead of meta.
For Y: forgetting that both acetyl groups must be present (the formula has $\text{O}_2$ ).
For Z: drawing a structure with the wrong ring size (the new ring must be six-membered for the formula to work).
For Z: including a Br atom in the structure (it has been lost as HBr).
For Z: drawing a structure with three rings or the wrong degree of unsaturation.

Things to Be Careful About

Y must have the correct connectivity: both substituents at meta positions relative to their respective $\text{COCH}_3$ groups.
Z is a bicyclic compound (naphthalene-like skeleton with one ring saturated) — not a monocyclic compound with a side chain.
The mark scheme accepts either the 6-acetyl or 7-acetyl isomer of tetralin for Z (equivalent by symmetry considerations of the drawing).

Techniques used

deduce structure from molecular formulaidentify intermolecular bis-substitution productidentify intramolecular cyclisation product

Q7Medium-HardIntroduction to A Level Organic ChemistryAnalytical TechniquesCarboxylic Acids and Derivatives

Four esters, $\mathbf{A}$ , $\mathbf{B}$ , $\mathbf{C}$ and $\mathbf{D}$ , with the molecular formula $\text{C}_6\text{H}_{12}\text{O}_2$ are shown in Fig. 7.1.

Fig. 7.1

(a)

Give the systematic name of ester $\mathbf{A}$ .

DifficultyEasy

Worked solution

Answer

methyl pentanoate

Final answer

methyl pentanoate

Detailed explanation

Background Concept

Esters are named as two parts: the alkyl group from the alcohol (the part attached to the single-bonded oxygen) is named first, followed by the carboxylate group from the acid (the part containing the carbonyl C=O) named with the suffix '-oate'. The carbon count for the '-oate' part includes the carbonyl carbon itself.

Understanding the Question

The question asks for the systematic (IUPAC) name of ester A, whose structure is given in Fig. 7.1 as a skeletal formula with molecular formula $\text{C}_6\text{H}_{12}\text{O}_2$ .

Approach

Identify the two fragments of the ester. The $-\text{O}-\text{CH}_3$ group (one carbon attached to the single-bonded oxygen) gives the alkyl part. The $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{C}(=\text{O})-$ group (five carbons including the carbonyl carbon) gives the acid part.

Step-by-Step Reasoning

The alkyl group attached to the oxygen is a methyl group ( $-\text{CH}_3$ ), so the name begins with 'methyl'.
The acyl group (containing the $\text{C}=\text{O}$ ) has five carbons in a straight chain: $\text{C}_1$ is the carbonyl carbon, $\text{C}_2$ - $\text{C}_5$ are the $\text{CH}_2$ and $\text{CH}_3$ groups. Five carbons corresponds to pentanoic acid, hence the ester suffix is 'pentanoate'.
Combining these gives methyl pentanoate.

Key Takeaways

Ester naming: alkyl part first (from the alcohol), then the carboxylate part (from the acid) with '-oate' ending.
The carbonyl carbon is counted as part of the acyl chain.

Common Mistakes

Forgetting to count the carbonyl carbon in the pentanoate part, leading to 'methyl butanoate'.
Reversing the order and naming it 'pentanoate methyl' or similar.

Techniques used

identify the acyl and alkoxy fragments of an esterapply IUPAC nomenclature rules to name an ester

(b)

A mixture of these esters, $\mathbf{A}$ , $\mathbf{B}$ , $\mathbf{C}$ and $\mathbf{D}$ , is analysed by gas–liquid chromatography.

The chromatogram produced is shown in Fig. 7.2. The number above each peak represents the area under the peak.

The area under each peak is proportional to the mass of the respective ester in the mixture.

Fig. 7.2

2M

(i)

State what is meant by retention time.

(ii)

Calculate the percentage by mass of ester $\mathbf{D}$ in the original mixture.

percentage by mass of ester $\mathbf{D}$ = .............................. %

(c)

Separate samples of the esters, $\mathbf{A}$ , $\mathbf{B}$ , $\mathbf{C}$ and $\mathbf{D}$ , are analysed using proton ( $^1\text{H}$ ) NMR and carbon-13 NMR spectroscopy.

3M

(i)

Complete Table 7.1 to show the number of peaks in each NMR spectrum for esters $\mathbf{B}$ and $\mathbf{C}$ .

Table 7.1

ester	number of peaks in proton ( $^1\text{H}$ ) NMR spectrum	number of peaks in carbon-13 NMR spectrum
$\mathbf{B}$
$\mathbf{C}$

DifficultyMedium

Worked solution

Answer

ester	number of peaks in proton ( $^1\text{H}$ ) NMR	number of peaks in carbon-13 NMR
B	5	6
C	4	5

Final answer

B: 5 proton peaks, 6 carbon peaks; C: 4 proton peaks, 5 carbon peaks

Detailed explanation

Background Concept

In $^1\text{H}$ NMR, each set of chemically equivalent protons gives one signal (peak). In $^{13}\text{C}$ NMR, each chemically distinct carbon gives one signal. Identifying the number of signals requires examining molecular symmetry: equivalent groups (related by symmetry or free rotation giving identical environments) produce a single peak.

Understanding the Question

Ester B is methyl 2-methylbutanoate: $\text{CH}_3\text{CH}_2\text{CH}(\text{CH}_3)\text{COOCH}_3$ . Ester C is methyl 3-methylbutanoate: $(\text{CH}_3)_2\text{CHCH}_2\text{COOCH}_3$ . The question asks for the number of peaks in both $^1\text{H}$ and $^{13}\text{C}$ NMR spectra for each.

Approach

For each ester, identify all chemically distinct proton environments and carbon environments by drawing out the structure and checking for symmetry.

Step-by-Step Reasoning

Ester B: methyl 2-methylbutanoate — $\text{CH}_3\text{CH}_2\text{CH}(\text{CH}_3)\text{COOCH}_3$

Proton environments:

$\text{CH}_3$ on the ethyl end (C4) — a triplet
$\text{CH}_2$ (C3) — complex multiplet
$\text{CH}$ (C2) — one proton
$\text{CH}_3$ branch on C2 — a doublet
$\text{OCH}_3$ — a singlet

All five groups are in different environments (no symmetry makes any equivalent). So 5 proton peaks.

Carbon environments:

$\text{CH}_3$ (C4)
$\text{CH}_2$ (C3)
$\text{CH}$ (C2)
$\text{CH}_3$ branch on C2
$\text{C}=\text{O}$ (C1)
$\text{OCH}_3$

All six carbons are distinct. So 6 carbon peaks.

Ester C: methyl 3-methylbutanoate — $(\text{CH}_3)_2\text{CHCH}_2\text{COOCH}_3$

Proton environments:

The two equivalent $\text{CH}_3$ groups on C4 (isopropyl methyls) — these are equivalent by symmetry (both attached to the same CH), giving one signal
$\text{CH}$ (C3) — one proton
$\text{CH}_2$ (C2) — two protons
$\text{OCH}_3$ — three protons

So 4 proton peaks.

Carbon environments:

The two equivalent $\text{CH}_3$ groups — one signal
$\text{CH}$ (C3)
$\text{CH}_2$ (C2)
$\text{C}=\text{O}$ (C1)
$\text{OCH}_3$

So 5 carbon peaks.

Key Takeaways

Equivalent methyl groups (e.g. the two $\text{CH}_3$ in an isopropyl group) count as one environment.
The carbonyl carbon always gives a separate $^{13}\text{C}$ signal.
Careful symmetry analysis is essential.

Common Mistakes

Counting the two equivalent methyls in ester C as separate environments.
Forgetting the carbonyl carbon in $^{13}\text{C}$ NMR.
Confusing the number of protons with the number of environments.

Techniques used

identify chemically distinct proton environmentsidentify chemically distinct carbon environmentscount NMR signals for each environment type

(ii)

Identify all of the esters from $\mathbf{A}$ , $\mathbf{B}$ , $\mathbf{C}$ and $\mathbf{D}$ that have at least one triplet peak in their proton ( $^1\text{H}$ ) NMR spectrum.

DifficultyMedium-Easy

Worked solution

Answer

A and B

Final answer

A and B

Detailed explanation

Background Concept

In $^1\text{H}$ NMR, the splitting pattern follows the $n+1$ rule: a proton signal is split into $n+1$ peaks where $n$ is the number of equivalent protons on adjacent carbon(s). A triplet arises when a proton (or set of equivalent protons) is adjacent to exactly 2 equivalent protons ( $n=2$ , so $n+1=3$ peaks).

Understanding the Question

We need to identify which of the four esters (A, B, C, D) have at least one triplet in their $^1\text{H}$ NMR spectrum. This means at least one proton environment must be adjacent to a $\text{CH}_2$ group.

Approach

Examine each ester's structure for $\text{CH}_3$ or $\text{CH}$ groups adjacent to a $\text{CH}_2$ (which would give a triplet for the $\text{CH}_3$ or $\text{CH}$ ), or a $\text{CH}_2$ adjacent to a $\text{CH}_2$ (giving a triplet for that $\text{CH}_2$ ).

Step-by-Step Reasoning

Ester A: methyl pentanoate — $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{COOCH}_3$

The terminal $\text{CH}_3$ is adjacent to a $\text{CH}_2$ → triplet ✓
The middle $\text{CH}_2$ groups are adjacent to $\text{CH}_2$ groups → triplets ✓
Has triplets.

Ester B: methyl 2-methylbutanoate — $\text{CH}_3\text{CH}_2\text{CH}(\text{CH}_3)\text{COOCH}_3$

The terminal $\text{CH}_3$ (on the ethyl group) is adjacent to a $\text{CH}_2$ → triplet ✓
Has triplets.

Ester C: methyl 3-methylbutanoate — $(\text{CH}_3)_2\text{CHCH}_2\text{COOCH}_3$

The two equivalent $\text{CH}_3$ groups are adjacent to a $\text{CH}$ (1 proton) → doublet, not triplet
The $\text{CH}$ is adjacent to 6 equivalent $\text{CH}_3$ protons + 2 $\text{CH}_2$ protons → multiplet
The $\text{CH}_2$ is adjacent to 1 $\text{CH}$ proton → doublet
The $\text{OCH}_3$ has no adjacent protons → singlet
No triplets.

Ester D: methyl 2,2-dimethylpropanoate — $(\text{CH}_3)_3\text{CCOOCH}_3$

The nine equivalent $\text{CH}_3$ protons have no adjacent protons (quaternary carbon) → singlet
The $\text{OCH}_3$ has no adjacent protons → singlet
No triplets.

Therefore, esters A and B have at least one triplet.

Key Takeaways

A triplet requires exactly 2 neighbouring protons.
Quaternary carbons and carbonyl carbons block splitting.
Systematically checking each environment is essential.

Common Mistakes

Thinking ester C has a triplet (the $\text{CH}_3$ groups are adjacent to a $\text{CH}$ , giving a doublet).
Including ester D which only has singlets.

Techniques used

apply the n+1 rule to predict splitting patternsidentify proton environments adjacent to two equivalent protons

(d)

Compound $\mathbf{F}$ , $\text{C}_6\text{H}_8\text{O}_3$ , shows stereoisomerism and effervesces with $\text{Na}_2\text{CO}_3\text{(aq)}$ .

Compound $\mathbf{F}$ reacts with alkaline $\text{I}_2\text{(aq)}$ to form yellow precipitate $\mathbf{G}$ and compound $\mathbf{H}$ .

Compound $\mathbf{F}$ reacts with $\text{LiAlH}_4$ to form compound $\mathbf{J}$ , $\text{C}_6\text{H}_{12}\text{O}_2$ .

Compound $\mathbf{F}$ reacts with $\text{SOCl}_2$ to form compound $\mathbf{K}$ , $\text{C}_6\text{H}_7\text{O}_2\text{Cl}$ .

Compound $\mathbf{K}$ reacts with propan-2-ol to form compound $\mathbf{L}$ .

Draw the structures of compounds $\mathbf{F}$ , $\mathbf{G}$ , $\mathbf{H}$ , $\mathbf{J}$ , $\mathbf{K}$ and $\mathbf{L}$ in the boxes in Fig. 7.3.

Fig. 7.3

6M Similar

DifficultyMedium-Hard

Worked solution

Answer

F — $\text{CH}_3\text{COCH}=\text{C}(\text{CH}_3)\text{COOH}$ (one of the six accepted isomers, e.g. 3-methyl-4-oxopent-2-enoic acid)

\text{CH}_3-\overset{\displaystyle\overset{\textstyle\|}{\text{C}}=\text{O}}{\text{C}}-\text{CH}=\text{C}(\text{CH}_3)-\text{COOH}

G — $\text{CHI}_3$ (yellow precipitate, iodoform)

H — $\text{HOOCCH}=\text{C}(\text{CH}_3)\text{COOH}$ (the dicarboxylic acid formed after cleavage of the methyl ketone)

J — $\text{CH}_3\text{CH(OH)CH}=\text{C}(\text{CH}_3)\text{CH}_2\text{OH}$ (diol formed by reduction of both C=O groups)

K — $\text{CH}_3\text{COCH}=\text{C}(\text{CH}_3)\text{COCl}$ (acyl chloride formed by reaction of the -COOH with $\text{SOCl}_2$ )

L — $\text{CH}_3\text{COCH}=\text{C}(\text{CH}_3)\text{COOCH}(\text{CH}_3)_2$ (isopropyl ester formed from K + propan-2-ol)

Final answer

F: CH3COCH=C(CH3)COOH; G: CHI3; H: HOOCCH=C(CH3)COOH; J: CH3CH(OH)CH=C(CH3)CH2OH; K: CH3COCH=C(CH3)COCl; L: CH3COCH=C(CH3)COOCH(CH3)2

Detailed explanation

Background Concept

This question tests the ability to deduce a structure from chemical evidence and predict products of characteristic reactions:

Effervescence with $\text{Na}_2\text{CO}_3$ indicates a carboxylic acid group ( $-\text{COOH}$ ), which releases $\text{CO}_2$ .
Alkaline $\text{I}_2$ (iodoform test) gives a yellow precipitate of $\text{CHI}_3$ with methyl ketones ( $\text{CH}_3\text{CO}-$ ) or compounds that can be oxidised to methyl ketones (like $\text{CH}_3\text{CH(OH)}-$ ). The methyl ketone is cleaved to give $\text{CHI}_3$ and a carboxylate.
$\text{LiAlH}_4$ reduces both ketones and carboxylic acids (and aldehydes) to alcohols.
$\text{SOCl}_2$ converts carboxylic acids to acyl chlorides ( $-\text{COOH} \rightarrow -\text{COCl}$ ) but does not affect ketones or C=C bonds.
Acyl chloride + alcohol gives an ester.
Stereoisomerism (E/Z) requires a C=C double bond with two different groups on each carbon.

Understanding the Question

Compound F has molecular formula $\text{C}_6\text{H}_8\text{O}_3$ . Its degree of unsaturation = $(2\times6+2-8)/2 = 3$ . It shows stereoisomerism (suggesting a C=C), effervesces with carbonate (a $-\text{COOH}$ ), and gives a positive iodoform test (a $\text{CH}_3\text{CO}-$ group). We must deduce F and draw the products of four reactions.

Approach

From the molecular formula and tests, identify functional groups: $-\text{COOH}$ (1 O, 1 unsaturation), $\text{CH}_3\text{CO}-$ (1 O, 1 unsaturation), and a C=C (1 unsaturation). Total = 3, matching the formula.
Assemble the structure: $\text{CH}_3\text{CO}-\text{CH}=\text{C}(\text{CH}_3)-\text{COOH}$ or similar arrangement with a C=C between the ketone and acid.
Predict each reaction product.

Step-by-Step Reasoning

Deducing F:

$\text{C}_6\text{H}_8\text{O}_3$ with 3 degrees of unsaturation.
$-\text{COOH}$ accounts for 1 C, 2 O, 1 unsaturation.
$\text{CH}_3\text{CO}-$ accounts for 2 C, 1 O, 1 unsaturation.
Remaining: 3 C and 1 unsaturation → a C=C double bond connecting the fragments.
The molecule must show stereoisomerism, so the C=C must have different groups on each end. A suitable structure is $\text{CH}_3\text{COCH}=\text{C}(\text{CH}_3)\text{COOH}$ (the $\text{CH}_3$ on the double-bond carbon and the $\text{COOH}$ make one side different; $\text{COCH}_3$ and $\text{H}$ make the other side different).
Multiple arrangements satisfy the criteria (six alternatives are accepted by the mark scheme), but the key features are: a methyl ketone, a carboxylic acid, and a C=C enabling E/Z isomerism.

F + alkaline $\text{I}_2$ → G + H:
The iodoform reaction cleaves the methyl ketone: $\text{CH}_3\text{CO}-$ becomes $\text{CHI}_3$ (G, yellow precipitate) and the remaining fragment gains an $-\text{OH}$ to form a carboxylic acid. So H = $\text{HOOCCH}=\text{C}(\text{CH}_3)\text{COOH}$ (a dicarboxylic acid with the C=C retained).

F + $\text{LiAlH}_4$ → J ( $\text{C}_6\text{H}_{12}\text{O}_2$ ):
$\text{LiAlH}_4$ reduces the ketone C=O to a secondary alcohol and the carboxylic acid to a primary alcohol. The C=C is not reduced. J = $\text{CH}_3\text{CH(OH)CH}=\text{C}(\text{CH}_3)\text{CH}_2\text{OH}$ . Formula check: $\text{C}_6\text{H}_{12}\text{O}_2$ ✓.

F + $\text{SOCl}_2$ → K ( $\text{C}_6\text{H}_7\text{O}_2\text{Cl}$ ):
$\text{SOCl}_2$ converts $-\text{COOH}$ to $-\text{COCl}$ ; the ketone and C=C are unaffected. K = $\text{CH}_3\text{COCH}=\text{C}(\text{CH}_3)\text{COCl}$ . Formula: $\text{C}_6\text{H}_7\text{O}_2\text{Cl}$ ✓ (one H replaced by Cl, one O lost).

K + propan-2-ol → L:
The acyl chloride reacts with the alcohol to form an ester: $-\text{COCl} + \text{HOCH}(\text{CH}_3)_2 \rightarrow -\text{COOCH}(\text{CH}_3)_2 + \text{HCl}$ . L = $\text{CH}_3\text{COCH}=\text{C}(\text{CH}_3)\text{COOCH}(\text{CH}_3)_2$ .

Key Takeaways

Combine chemical tests to build up a structural picture before drawing products.
The iodoform reaction specifically identifies $\text{CH}_3\text{CO}-$ groups.
$\text{LiAlH}_4$ reduces both C=O of ketones and the C=O of acids, but not C=C.
$\text{SOCl}_2$ is selective for $-\text{COOH}$ over ketones.

Common Mistakes

Reducing the C=C with $\text{LiAlH}_4$ (it does not reduce isolated C=C bonds).
Forgetting that the iodoform reaction produces a carboxylate from the ketone fragment, not just removing the $\text{CH}_3$ .
Drawing F without a C=C (then stereoisomerism would not be possible).
Not checking that the molecular formula of J matches $\text{C}_6\text{H}_{12}\text{O}_2$ .

Things to Be Careful About

Ensure all six structures are consistent with the same F.
The mark scheme accepts six different valid structures for F, so the products H, J, K, L must correspond to whichever F is drawn.
Include the correct number of carbons in each structure; miscounting leads to wrong formulae.
G is simply $\text{CHI}_3$ — do not draw a larger molecule.

Techniques used

deduce functional groups from chemical testsapply the iodoform reaction to identify a methyl ketonepredict products of LiAlH4 reductionpredict products of SOCl2 reaction with carboxylic acidpredict ester formation from acyl chloride and alcoholdraw stereoisomeric structures

Q8Medium-HardIntroduction to A Level Organic ChemistryCarboxylic Acids and DerivativesNitrogen Compounds

Neotame is an artificial sweetener added to some foods.

Fig. 8.1

(a)

3M

(i)

State the number of chiral carbon atoms in a molecule of neotame.

DifficultyMedium-Easy

Worked solution

Answer

2

Final answer

2

Detailed explanation

Background Concept

A chiral carbon (stereocentre) is a carbon atom bonded to four different atoms or groups. In organic molecules, these are typically $sp^3$ hybridised carbons. Carbons in $\text{CH}_2$ groups cannot be chiral because they bear two identical hydrogen atoms. Carbonyl carbons ( $\text{C=O}$ ) and aromatic carbons are $sp^2$ hybridised and bonded to at most three groups, so they cannot be chiral either.

Understanding the Question

The question asks you to count the number of chiral carbon atoms in neotame. You must examine every $sp^3$ carbon in the structure and determine whether it is bonded to four different groups.

Approach

Systematically examine each tetrahedral ( $sp^3$ ) carbon in the molecule. Eliminate those that cannot be chiral (CH2 groups, carbonyl carbons, aromatic carbons, quaternary carbons with identical substituents). For the remaining candidates, check if all four attached groups are different.

Step-by-Step Reasoning

Starting from the structure of neotame: $(\text{CH}_3)_3\text{C-CH}_2\text{-CH}_2\text{-NH-CH(CH}_2\text{COOH)-C(=O)-NH-CH(CH}_2\text{C}_6\text{H}_5\text{)-C(=O)-O-CH}_3$

The tert-butyl central carbon: bonded to three identical $\text{CH}_3$ groups → NOT chiral.
The three methyl carbons: each bonded to three H atoms → NOT chiral.
The two CH2 groups between the tert-butyl and NH: each has two H atoms → NOT chiral.
The CH bonded to NH(3,3-dimethylbutyl): attached to H, $\text{NHCH}_2\text{CH}_2\text{C(CH}_3)_3$ , $\text{CH}_2\text{COOH}$ , and $\text{C(=O)NH...}$ → four different groups → CHIRAL.
The CH2 of the acetic acid side chain: two H atoms → NOT chiral.
The carbonyl carbon (amide): $sp^2$ , only three groups → NOT chiral.
The CH bonded to NH(amide): attached to H, $\text{NH-C(=O)...}$ , $\text{CH}_2\text{C}_6\text{H}_5$ , and $\text{COOCH}_3$ → four different groups → CHIRAL.
The benzyl CH2: two H atoms → NOT chiral.
Aromatic carbons: $sp^2$ → NOT chiral.
The ester carbonyl carbon: $sp^2$ → NOT chiral.
The OCH3 carbon: three H atoms → NOT chiral.

Total: 2 chiral carbons.

Key Takeaways

Only $sp^3$ carbons can be chiral centres.
A carbon must have four different substituents.
CH2, CH3, C=O, and aromatic carbons are automatically excluded.

Common Mistakes

Counting the CH2 groups as chiral (they have two identical H atoms).
Missing one of the two genuine chiral centres in this large molecule due to the complexity of the structure.
Including the carbonyl or aromatic carbons.

Things to Be Careful About

In large molecules like neotame, it is easy to lose track. Work systematically from one end of the molecule to the other.
The two chiral centres are the alpha-carbons of the amino acid residues (aspartic acid and phenylalanine portions).

Techniques used

identify chiral centres by checking for four different groupsexamine each tetrahedral carbon systematically

(ii)

Neotame contains the arene functional group.

Identify all the other functional groups present in neotame.

DifficultyEasy

Worked solution

Answer

amide, amine, ester, carboxylic acid

Final answer

amide, amine, ester, carboxylic acid

Detailed explanation

Background Concept

Functional groups are specific arrangements of atoms within a molecule that confer characteristic chemical properties and reactivity. In A Level organic chemistry, students must be able to identify common functional groups from displayed or skeletal formulae.

Key functional groups present in neotame (besides the arene/benzene ring):

Carboxylic acid: $\text{-COOH}$ (or $\text{-CO}_2\text{H}$ )
Amide: $\text{-CONH-}$ (a carbonyl bonded to a nitrogen)
Ester: $\text{-COOCH}_3$ (a carbonyl bonded to an oxygen which is bonded to a carbon)
Amine: $\text{-NH-}$ (nitrogen bonded to carbon groups, not part of a carbonyl system)

Understanding the Question

The arene (benzene ring) is already identified. You must name all other functional groups present in the molecule.

Approach

Examine the molecule region by region and identify each distinct functional group.

Step-by-Step Reasoning

The $\text{-COOH}$ group on the side chain → carboxylic acid
The $\text{-C(=O)-NH-}$ linkage between the two amino acid residues → amide
The $\text{-C(=O)-O-CH}_3$ group at the right end → ester
The $\text{-NH-}$ group attached to the 3,3-dimethylbutyl chain (not adjacent to a carbonyl) → secondary amine

The mark scheme awards 1 mark for any two correct, 2 marks for all four.

Key Takeaways

A secondary amine ( $\text{R}_2\text{NH}$ ) is distinct from an amide ( $\text{RCONHR}$ ): the amide nitrogen is directly bonded to a carbonyl carbon.
An ester ( $\text{RCOOR'}$ ) differs from a carboxylic acid ( $\text{RCOOH}$ ) by having an alkyl group on the oxygen instead of hydrogen.

Common Mistakes

Confusing the amide group with an amine group (the amide N is bonded to C=O, the amine N is not).
Calling the ester a "carboxylic acid" or vice versa.
Forgetting the amine group (the secondary amine in the 3,3-dimethylbutyl portion).

Things to Be Careful About

The mark scheme requires the exact names: "amide", "amine", "ester", "carboxylic acid" (or "carboxyl"). Vague terms like "carbonyl" or "hydroxyl" do not score.

Techniques used

recognise functional groups from structural formulaedistinguish between amide, ester, carboxylic acid, and amine groups

(b)

Neotame reacts with an excess of hot $\text{NaOH(aq)}$ to form three organic products.

5M

(i)

State the two types of reaction that occur when neotame reacts with hot $\text{NaOH(aq)}$ .

1 .........................................................................................................................................

2 .........................................................................................................................................

DifficultyMedium-Easy

Worked solution

Answer

Hydrolysis
Acid–base (neutralisation)

Final answer

Hydrolysis, 2. Acid-base (neutralisation)

Detailed explanation

Background Concept

When an organic molecule containing multiple functional groups is treated with hot aqueous sodium hydroxide, several types of reaction can occur simultaneously:

Hydrolysis: Esters and amides are cleaved by nucleophilic attack of hydroxide ions. Esters hydrolyse to give a carboxylate salt and an alcohol; amides hydrolyse to give a carboxylate salt and an amine. These are nucleophilic acyl substitution (addition-elimination) reactions.
Acid-base (neutralisation): Carboxylic acid groups ( $\text{-COOH}$ ) are acidic and react with $\text{NaOH}$ to form the corresponding sodium carboxylate salt ( $\text{-COONa}$ ) and water.

Understanding the Question

Neotame contains an ester group, an amide group, and a carboxylic acid group. When treated with excess hot $\text{NaOH(aq)}$ , all reactive sites will be affected. The question asks you to name the two types of reaction.

Approach

Identify which functional groups react and what type of reaction each undergoes. The ester and amide undergo hydrolysis (nucleophilic acyl substitution). The carboxylic acid undergoes acid-base neutralisation.

Step-by-Step Reasoning

The ester ( $\text{-COOCH}_3$ ) is hydrolysed by $\text{NaOH}$ : the $\text{OH}^-$ attacks the carbonyl carbon, eliminating $\text{CH}_3\text{O}^-$ , which is protonated to give methanol. The acyl portion becomes a carboxylate salt.
The amide ( $\text{-CONH-}$ ) is hydrolysed by hot $\text{NaOH}$ : $\text{OH}^-$ attacks the carbonyl, the amine portion leaves as $\text{NH}_2$ (as a free amine), and the acyl portion becomes a carboxylate salt.
The carboxylic acid ( $\text{-COOH}$ ) reacts with $\text{NaOH}$ in a simple acid-base neutralisation: $\text{-COOH} + \text{NaOH} \rightarrow \text{-COONa} + \text{H}_2\text{O}$ .

Both hydrolysis and acid-base reactions occur. These are the two types.

Key Takeaways

Hot aqueous NaOH can both hydrolyse (esters, amides) and neutralise (carboxylic acids, phenols) — two fundamentally different reaction types.
"Hydrolysis" is the cleavage of a bond by water (or hydroxide), while "acid-base" is proton transfer.

Common Mistakes

Writing only "hydrolysis" and missing the acid-base reaction.
Writing "substitution" instead of "hydrolysis" (though nucleophilic substitution is the mechanism, the type of reaction is hydrolysis).
Writing "neutralisation" when the mark scheme specifically asks for "acid-base" — both should be acceptable, but use the mark scheme's terminology.

Things to Be Careful About

The question asks for types of reaction, not mechanisms. "Nucleophilic addition-elimination" describes the mechanism of hydrolysis but is not the type.
The word "excess" in the question stem is important — it ensures all acidic protons are neutralised and all hydrolysable bonds are cleaved.

Techniques used

recognise that hot aqueous NaOH hydrolyses esters and amidesidentify that carboxylic acid groups undergo neutralisation with alkali

(ii)

Fig. 8.1 (repeated)

Draw the structures of the three organic products formed from the reaction of neotame with an excess of hot $\text{NaOH(aq)}$ .

Fig. 8.2

DifficultyMedium-Hard

Worked solution

Answer

The three organic products are:

$(\text{CH}_3)_3\text{C-CH}_2\text{-CH}_2\text{-NH-CH(CH}_2\text{COONa)-COONa}$ — the disodium salt of the aspartic acid derivative with the 3,3-dimethylbutyl group on nitrogen
$\text{H}_2\text{N-CH(CH}_2\text{C}_6\text{H}_5\text{)-COONa}$ — sodium phenylalaninate (sodium salt of phenylalanine)
$\text{CH}_3\text{OH}$ — methanol

Final answer

Three products: (CH3)3CCH2CH2NHCH(CH2COONa)COONa; H2NCH(CH2C6H5)COONa; CH3OH

Detailed explanation

Background Concept

Alkaline hydrolysis of multifunctional organic molecules involves the cleavage of hydrolysable bonds (esters and amides) by hydroxide ions, combined with neutralisation of any acidic protons by the excess alkali.

Ester hydrolysis: $\text{RCOOR'} + \text{NaOH} \rightarrow \text{RCOONa} + \text{R'OH}$
Amide hydrolysis: $\text{RCONHR'} + \text{NaOH} \rightarrow \text{RCOONa} + \text{R'NH}_2$
Carboxylic acid neutralisation: $\text{RCOOH} + \text{NaOH} \rightarrow \text{RCOONa} + \text{H}_2\text{O}$

When excess hot NaOH is used, all three types of reaction go to completion.

Understanding the Question

Neotame has three hydrolysable/acidic sites:

A carboxylic acid group ( $\text{-COOH}$ ) on the aspartic acid side chain
An amide linkage ( $\text{-CONH-}$ ) between the two amino acid residues
A methyl ester group ( $\text{-COOCH}_3$ ) at the C-terminus

The question asks you to draw the three organic products formed when all of these react with excess hot NaOH.

Approach

Identify the bonds that will break: the ester C-O bond and the amide C-N bond.
Determine what each fragment becomes after cleavage.
Convert all $\text{-COOH}$ groups to $\text{-COONa}$ (since excess NaOH is present).
Draw the three resulting organic molecules.

Step-by-Step Reasoning

Cleaving the amide bond ( $\text{-C(=O)-NH-}$ ):

The left fragment retains the carbonyl → becomes a carboxylate ( $\text{-COONa}$ )
The right fragment retains the nitrogen → becomes a free amine ( $\text{-NH}_2$ )

Cleaving the ester bond ( $\text{-C(=O)-OCH}_3$ ):

The acyl portion (attached to the right fragment) → becomes a carboxylate ( $\text{-COONa}$ )
The $\text{-OCH}_3$ portion → becomes methanol ( $\text{CH}_3\text{OH}$ )

Neutralising the free carboxylic acid ( $\text{-COOH}$ on the side chain):

→ becomes $\text{-COONa}$

Product 1 (left fragment after amide cleavage):
The 3,3-dimethylbutyl group is still attached to the nitrogen. The alpha carbon has the $\text{CH}_2\text{COONa}$ side chain and the $\text{COONa}$ (formerly the amide carbonyl). Structure: $(\text{CH}_3)_3\text{C-CH}_2\text{-CH}_2\text{-NH-CH(CH}_2\text{COONa)-COONa}$

Product 2 (right fragment after amide and ester cleavage):
The alpha carbon now has $\text{-NH}_2$ (from amide cleavage), the $\text{CH}_2\text{C}_6\text{H}_5$ side chain, and $\text{-COONa}$ (from ester hydrolysis). Structure: $\text{H}_2\text{N-CH(CH}_2\text{C}_6\text{H}_5\text{)-COONa}$ (sodium phenylalaninate)

Product 3: Methanol, $\text{CH}_3\text{OH}$

Key Takeaways

In alkaline hydrolysis of a peptide-like molecule, the amide bond breaks to give a carboxylate on the acyl side and a free amine on the amine side.
All carboxylic acid groups (whether originally present or formed by hydrolysis) exist as sodium salts in excess NaOH.
The ester gives methanol as the alcohol product.

Common Mistakes

Forgetting to convert the original $\text{-COOH}$ to $\text{-COONa}$ .
Drawing the amide hydrolysis product with $\text{-COOH}$ instead of $\text{-COONa}$ (the medium is strongly alkaline).
Confusing which fragment gets the carboxylate and which gets the amine from amide cleavage.
Writing $\text{NaOH}$ as a product (it is a reagent, not a product).
Drawing only two products and forgetting methanol.

Things to Be Careful About

The mark scheme requires the sodium carboxylate form ( $\text{-COONa}$ ), not the free acid ( $\text{-COOH}$ ), because excess NaOH is specified.
Each structure must be complete and correct to earn its mark.
The amine on the left fragment is secondary ( $\text{-NH-}$ ), while the amine on the right fragment is primary ( $\text{-NH}_2$ ).

Techniques used

cleave the ester bond to give methanol and a carboxylatecleave the amide bond to give an amine and a carboxylateconvert all carboxylic acid groups to sodium carboxylate salts

Q9Medium-HardHydroxy CompoundsHydrocarbonsOrganic Synthesis

(a)

Samples of phenol, $\text{C}_6\text{H}_5\text{OH}$ , are reacted separately with sodium and with dilute nitric acid.

Fig. 9.1

2M

(i)

Write the equation for the reaction of $\text{C}_6\text{H}_5\text{OH}$ with $\text{Na}$ .

DifficultyEasy

Worked solution

Answer

\text{C}_6\text{H}_5\text{OH} + \text{Na} \rightarrow \text{C}_6\text{H}_5\text{ONa} + \tfrac{1}{2}\text{H}_2

Final answer

C6H5OH + Na -> C6H5ONa + 1/2 H2

Detailed explanation

Background Concept

Phenol ( $\text{C}_6\text{H}_5\text{OH}$ ) is a weak acid, more acidic than aliphatic alcohols because the phenoxide ion ( $\text{C}_6\text{H}_5\text{O}^-$ ) is stabilised by delocalisation of the negative charge into the aromatic ring. Like all acids, phenol reacts with reactive metals such as sodium to produce a salt and hydrogen gas. The salt formed is sodium phenoxide ( $\text{C}_6\text{H}_5\text{ONa}$ ).

Understanding the Question

The question asks for the balanced equation for the reaction of phenol with sodium metal. This is a straightforward acid–metal displacement reaction. The command word is "Write the equation", so a single balanced symbol equation is required.

Approach

Identify the products: sodium phenoxide and hydrogen gas. Balance the equation noting that one mole of phenol provides one acidic proton, and one mole of sodium provides one electron, so the stoichiometry gives $\tfrac{1}{2}\text{H}_2$ per mole of phenol.

Step-by-Step Reasoning

Phenol acts as a weak acid, donating its proton to sodium metal.
The sodium cation combines with the phenoxide anion to form sodium phenoxide, $\text{C}_6\text{H}_5\text{ONa}$ .
Hydrogen gas is evolved: $2\text{H}^+ + 2\text{e}^- \rightarrow \text{H}_2$ , so one mole of phenol produces half a mole of $\text{H}_2$ .
The balanced equation is: $\text{C}_6\text{H}_5\text{OH} + \text{Na} \rightarrow \text{C}_6\text{H}_5\text{ONa} + \tfrac{1}{2}\text{H}_2$ .

Key Takeaways

Phenol reacts with sodium because it is acidic (pKa ≈ 10), unlike aliphatic alcohols which do not react with sodium under normal conditions (though they do react with sodium metal, the comparison with water and carboxylic acids is more relevant here).
The product is a salt (sodium phenoxide) plus hydrogen gas.

Common Mistakes

Writing $\text{H}_2$ instead of $\tfrac{1}{2}\text{H}_2$ without balancing the rest of the equation accordingly.
Writing the product as $\text{C}_6\text{H}_5\text{O}^-\text{Na}^+$ (acceptable) but forgetting the equation must be balanced.
Confusing this with the reaction of phenol with NaOH (which also gives sodium phenoxide but no hydrogen gas).

Things to Be Careful About

The equation must be balanced: one Na on each side, one phenol, and the hydrogen must account for the half-molecule.
State symbols are not required by the mark scheme but are good practice.

Techniques used

write a balanced equation for acid-metal reactionidentify the salt formed from phenol and sodium

(ii)

Draw the structures of the two major isomeric organic products formed in the reaction of phenol with dilute $\text{HNO}_3$ .

Fig. 9.2a

DifficultyMedium-Easy

Worked solution

Answer

The two major isomeric products are 2-nitrophenol and 4-nitrophenol (ortho and para substitution):

Final answer

2-nitrophenol and 4-nitrophenol (see diagram)

Detailed explanation

Background Concept

The $-\text{OH}$ group on phenol is a strong activating group and an ortho/para director in electrophilic aromatic substitution. This is because the lone pair on oxygen can delocalise into the ring via resonance, increasing electron density particularly at the ortho and para positions. When phenol reacts with dilute nitric acid, nitration occurs under mild conditions (unlike benzene which requires concentrated $\text{HNO}_3$ / $\text{H}_2\text{SO}_4$ ), producing a mixture of 2-nitrophenol and 4-nitrophenol as the major products.

Understanding the Question

The question asks for the structures of the two major isomeric organic products formed when phenol reacts with dilute $\text{HNO}_3$ . The command word is "Draw the structures", so full displayed or skeletal formulae showing the benzene ring with substituents in the correct positions are required.

Approach

Recognise that $-\text{OH}$ directs electrophilic substitution to ortho and para positions.
The electrophile is $\text{NO}_2^+$ (nitronium ion), generated from dilute $\text{HNO}_3$ .
Draw 2-nitrophenol ( $-\text{NO}_2$ adjacent to $-\text{OH}$ ) and 4-nitrophenol ( $-\text{NO}_2$ opposite to $-\text{OH}$ ).

Step-by-Step Reasoning

Phenol undergoes electrophilic substitution with dilute nitric acid at room temperature.
The $-\text{OH}$ group activates the ring and directs incoming electrophiles to positions 2 (ortho) and 4 (para).
2-nitrophenol: benzene ring with $-\text{OH}$ at position 1 and $-\text{NO}_2$ at position 2.
4-nitrophenol: benzene ring with $-\text{OH}$ at position 1 and $-\text{NO}_2$ at position 4.
Both structures must show the benzene ring (with circle or alternating double bonds), the $-\text{OH}$ group, and the $-\text{NO}_2$ group in the correct relative positions.

Key Takeaways

Phenol is much more reactive towards electrophilic substitution than benzene due to the activating $-\text{OH}$ group.
Dilute $\text{HNO}_3$ is sufficient to nitrate phenol (concentrated $\text{HNO}_3$ / $\text{H}_2\text{SO}_4$ is needed for benzene).
The $-\text{OH}$ group is an ortho/para director.

Common Mistakes

Drawing the meta product (3-nitrophenol) instead of ortho or para.
Forgetting to show the benzene ring properly (e.g., drawing a cyclohexane ring).
Confusing the position of substituents: ortho = 1,2 and para = 1,4.

Things to Be Careful About

Both structures are needed for the single mark ("Both structures for one mark").
The $-\text{NO}_2$ group must be drawn as $-\text{NO}_2$ (not $-\text{ONO}$ or $-\text{NO}_3$ ).
Ensure the ring is clearly aromatic (circle inside hexagon or three alternating double bonds).

Techniques used

identify ortho and para products from electrophilic substitution on phenoldraw structural formulae of substituted aromatic compounds

(b)

Salicylic acid can be synthesised from phenol.

Fig. 9.2

One of the steps in this synthesis is the electrophilic substitution reaction of carbon dioxide with the phenoxide ion, $\text{C}_6\text{H}_5\text{O}^-$ .

Complete the mechanism in Fig. 9.3 for the reaction of $\text{C}_6\text{H}_5\text{O}^-$ with $\text{CO}_2$ .

Include all relevant curly arrows, dipoles and charges. Draw the structure of the organic intermediate.

Fig. 9.3

DifficultyMedium-Hard

Worked solution

Answer

M1: Dipole on $\text{CO}_2$ ( $\delta^-$ on O, $\delta^+$ on C) and curly arrow from the benzene ring (inside the hexagon) to the carbon atom of $\text{CO}_2$ .

M2: Structure of the organic intermediate — a cyclohexadienyl anion (arenium ion) with $-\text{COO}^-$ and $\text{H}$ attached to the same carbon (the ortho carbon), positive charge delocalised in the ring, and $-\text{O}^-$ still attached to the ring.

M3: Curly arrow from the C $-$ H bond back into the ring to restore aromaticity, and $\text{H}^+$ shown as the eliminated species.

Final answer

See diagram: curly arrow from ring to δ+ carbon of CO2, intermediate is cyclohexadienyl anion with COO- and H on ortho carbon, curly arrow from C-H into ring, H+ eliminated

Detailed explanation

Background Concept

The Kolbe–Schmitt reaction involves the carboxylation of sodium phenoxide with carbon dioxide under pressure. In this mechanism, $\text{CO}_2$ acts as a weak electrophile. The carbon atom in $\text{CO}_2$ carries a partial positive charge ( $\delta^+$ ) because the two highly electronegative oxygen atoms withdraw electron density, giving each oxygen a partial negative charge ( $\delta^-$ ). The phenoxide ion is even more activated than phenol itself because the negative charge on oxygen is strongly donated into the ring by resonance, making the ortho and para positions highly nucleophilic.

The mechanism follows the standard pattern of electrophilic aromatic substitution:

The aromatic $\pi$ system attacks the electrophile, forming a sigma complex (arenium ion) where aromaticity is temporarily lost.
A proton is lost from the carbon that bonded to the electrophile, restoring aromaticity.

Understanding the Question

The question provides a mechanism template (Fig. 9.3) showing phenoxide ion reacting with $\text{CO}_2$ to form an organic intermediate, which then loses a proton to give the salicylate ion. The candidate must complete this mechanism by adding:

Curly arrows showing electron movement
Dipoles on $\text{CO}_2$
The structure of the intermediate
The eliminated proton

The command word is "Complete the mechanism", requiring all three marking points (M1, M2, M3).

Approach

Identify the electrophilic centre in $\text{CO}_2$ : the carbon atom ( $\delta^+$ ).
Draw a curly arrow from the $\pi$ electrons of the ring (inside the hexagon) to the $\delta^+$ carbon.
Show the dipoles: $\delta^-$ on each oxygen, $\delta^+$ on carbon.
Draw the intermediate: the ortho carbon now has both $\text{H}$ and $\text{COO}^-$ attached; the ring has lost aromaticity (shown by a positive charge delocalised over the remaining carbons, or by drawing two double bonds and a plus sign); the $-\text{O}^-$ group remains on the ring.
Draw a curly arrow from the C $-$ H bond into the ring to restore aromaticity.
Show $\text{H}^+$ as the product of this elimination.

Step-by-Step Reasoning

M1 — Dipole and first curly arrow:

$\text{CO}_2$ is a linear molecule: O=C=O. The C=O bonds are polar, with $\delta^-$ on O and $\delta^+$ on C.
The curly arrow starts from inside the hexagon (representing the delocalised $\pi$ electrons) and points to the carbon atom of $\text{CO}_2$ . This shows the ring acting as the nucleophile attacking the electrophilic carbon.
The $\text{CO}_2$ molecule should show $\delta^-$ on one oxygen (the one whose double bond electrons move to become a lone pair) and $\delta^+$ on carbon.

M2 — Structure of the intermediate:

After the ring attacks $\text{CO}_2$ , the ortho carbon (adjacent to the $\text{O}^-$ group) now has four bonds: two to ring carbons, one to H, and one to the $\text{COO}^-$ group (the former carbon of $\text{CO}_2$ , now bearing a negative charge on one oxygen after the $\pi$ bond electrons moved onto it).
The ring is no longer aromatic — it is a cyclohexadienyl cation (sigma complex). This is shown by drawing the ring with two C=C double bonds and a positive charge (either on the carbon bearing the substituents or delocalised across the ring).
The $-\text{O}^-$ group remains attached to the ring.

M3 — Restoration of aromaticity:

A curly arrow is drawn from the C $-$ H bond (on the ortho carbon) back into the ring, showing the electrons returning to the $\pi$ system.
This restores aromaticity and releases $\text{H}^+$ .
The final product is the salicylate ion (2-hydroxybenzoate), which is already shown in Fig. 9.3.

Key Takeaways

$\text{CO}_2$ can act as an electrophile in aromatic substitution due to the $\delta^+$ on its carbon.
The phenoxide ion is more reactive than phenol because the negative charge enhances electron donation into the ring.
Electrophilic aromatic substitution always proceeds via a sigma complex intermediate followed by deprotonation to restore aromaticity.
Curly arrows must start from electron sources (bonds, lone pairs, negative charges) and end at electron sinks (atoms, positive charges).

Common Mistakes

Drawing the curly arrow from the $\text{O}^-$ group rather than from the ring $\pi$ system.
Forgetting to show the dipole on $\text{CO}_2$ .
Drawing the intermediate with the $\text{COO}^-$ group attached to the wrong position (must be ortho to $\text{O}^-$ ).
Not showing the positive charge in the intermediate (the ring must be non-aromatic in the sigma complex).
Drawing the curly arrow for M3 from the wrong bond or in the wrong direction.
Writing $\text{H}^-$ instead of $\text{H}^+$ as the eliminated species.

Things to Be Careful About

The curly arrow for M1 must start from inside the hexagon (the delocalised $\pi$ electrons), not from a specific bond or from the oxygen.
The intermediate must show BOTH the $\text{H}$ and the $\text{COO}^-$ on the same carbon.
The positive charge in the intermediate must be visible (either as a + on the ring or shown by the disrupted aromatic system).
The dipole labels ( $\delta^+$ , $\delta^-$ ) must be correctly placed on $\text{CO}_2$ .

Techniques used

draw curly arrows for electrophilic aromatic substitutionshow dipole on the electrophiledraw the sigma-complex (arenium ion) intermediateshow restoration of aromaticity by loss of H+

(c)

Some syntheses use Diels–Alder reactions, which normally involve a diene and an alkene reacting together to form a cyclohexene.

2M

(i)

Draw three curly arrows in Fig. 9.4 to complete the mechanism for the Diels–Alder reaction between buta-1,3-diene and ethene.

Fig. 9.4

DifficultyMedium

Worked solution

Answer

Three curly arrows showing the concerted cyclic movement of six $\pi$ electrons: one arrow from the C1=C2 double bond of the diene to form a new C $-$ C bond with ethene, one arrow from the ethene $\pi$ bond to form the second new C $-$ C bond, and one arrow from the C3=C4 double bond of the diene to form the new C2 $-$ C3 double bond in the product.

Final answer

Three curly arrows in a cyclic pattern: diene terminal pi bond to ethene, ethene pi bond to other diene terminal, diene internal pi bond shifts to form new double bond

Detailed explanation

Background Concept

The Diels–Alder reaction is a [4+2] cycloaddition between a conjugated diene (4 $\pi$ electrons) and a dienophile (2 $\pi$ electrons, typically an alkene or alkyne). It is a concerted pericyclic reaction: all bond-making and bond-breaking occurs simultaneously in a single step through a cyclic transition state. Six $\pi$ electrons move in a cyclic loop, forming two new $\sigma$ bonds and one new $\pi$ bond, while the diene's central single bond becomes a double bond.

The reaction is stereospecific and regioselective. It forms a cyclohexene ring from the diene and dienophile.

Understanding the Question

The question shows buta-1,3-diene reacting with ethene to form cyclohexene (Fig. 9.4). The candidate must draw three curly arrows to show the mechanism. The command word is "Draw", and all three arrows must be correct for the single mark.

Approach

Identify the cyclic flow of electrons:

One terminal $\pi$ bond of the diene attacks one carbon of ethene (forming a new $\sigma$ bond).
The $\pi$ bond of ethene attacks the other terminal carbon of the diene (forming the second new $\sigma$ bond).
The central $\pi$ bond of the diene shifts to become the new double bond in the cyclohexene product.

These three arrows form a closed loop (cyclic electron flow), which is characteristic of pericyclic reactions.

Step-by-Step Reasoning

The Diels-Alder mechanism involves three curly arrows arranged in a cycle:

Arrow 1: From the C1=C2 $\pi$ bond (or the terminal double bond of the diene) to the space between C1 of the diene and one carbon of ethene — this shows the formation of a new C $-$ C $\sigma$ bond.
Arrow 2: From the $\pi$ bond of ethene to the space between the other carbon of ethene and C4 of the diene — this shows formation of the second new C $-$ C $\sigma$ bond.
Arrow 3: From the C3=C4 $\pi$ bond of the diene to the C2 $-$ C3 bond — this shows the central single bond becoming a double bond in the product.

Alternatively, the arrows can be drawn in the opposite direction (clockwise vs anticlockwise) — both are acceptable as shown in the mark scheme.

The key point is that all three arrows must be present and must form a continuous cyclic loop showing six electrons moving simultaneously.

Key Takeaways

Diels-Alder reactions are concerted (single step, no intermediates).
Three curly arrows in a cyclic arrangement represent the movement of six $\pi$ electrons.
The product is always a cyclohexene (or substituted cyclohexene).
Both clockwise and anticlockwise arrow patterns are acceptable.

Common Mistakes

Drawing only two arrows instead of three.
Drawing arrows that don't form a closed loop.
Drawing arrows from the wrong positions (e.g., from a single bond rather than a double bond).
Drawing a stepwise mechanism with intermediates (the reaction is concerted).

Things to Be Careful About

All three arrows must be present for the mark ("All three curly arrows for one mark").
The arrows must show electron movement from $\pi$ bonds (not from atoms or lone pairs).
The direction of the cycle (clockwise or anticlockwise) does not matter.

Techniques used

draw curly arrows for a concerted pericyclic reactionshow cyclic electron flow in a Diels-Alder mechanism

(ii)

Another Diels–Alder reaction of buta-1,3-diene is shown in Fig. 9.5.

Predict the product formed in this reaction.

Fig. 9.5