(a) Solid manganese(IV) oxide, $\text{MnO}_2$, catalyses the decomposition of hydrogen peroxide. $$ 2\text{H}_2\text{O}_2\text{(aq)} \rightarrow 2\text{H}_2\text{O(l)} + \text{O}_2\text{(g)} $$ State the type of catalysis for this reaction. Explain your answer. (b) Hydrogen peroxide reacts with iodide ions in acidic conditions as shown. $$ \text{H}_2\text{O}_2 + 2\text{I}^- + 2\text{H}^+ \rightarrow 2\text{H}_2\text{O} + \text{I}_2 $$ The initial rate of this reaction is…

(a) Heterogeneous catalysis; MnO2 is a solid and H2O2 is aqueous, so they are in different phases. · (b)(i) rate = k[H2O2][I-] · (b)(ii) 1.79 dm^3 mol^-1 s^-1 · (c)(i) Two constant half-lives of approximately 60 s. · (c)(ii) 0.0116 s^-1 · (d) Increases k and increases the rate of reaction.

(a) Define standard cell potential, $E^\ominus_{\text{cell}}$. Include a description of standard conditions. (b) The Daniell cell is an electrochemical cell consisting of a $\text{Cu}^{2+}\text{(aq)/Cu(s)}$ electrode and a $\text{Zn}^{2+}\text{(aq)/Zn(s)}$ electrode. (i) Draw a labelled diagram of this electrochemical cell. Include all necessary substances and relevant pieces of apparatus needed to measure the $E^\ominus_{\text{cell}}$. It is not necessary to state the…

(a) Potential difference between two half-cells under standard conditions (1 mol dm^-3, 1 atm/101 kPa, 298 K). · (b)(i) See diagram. · (b)(ii) Ions in solution; electrons in the wire. · (b)(iii) -0.78 V · (c)(i) +4 to +3 · (c)(ii) Zn + 2MnO2 -> ZnO + Mn2O3 · (c)(iii) +0.19 V

(a) State the relative acidities of bromoethanoic acid, $\text{BrCH}_2\text{COOH}$, chloroethanoic acid, $\text{ClCH}_2\text{COOH}$, ethanoic acid, $\text{CH}_3\text{COOH}$ and ethanol, $\text{CH}_3\text{CH}_2\text{OH}$. Explain your answer. (b) Fig. 7.1 shows the reaction of methylbenzene and ethanedioic acid with $\text{KMnO}_4$. Predict the major carbon-containing product for each of these reactions. (c) Polyamide X can be synthesised from ethanedioic acid and…

(a) ClCH2COOH > BrCH2COOH > CH3COOH > CH3CH2OH · (b) Benzoic acid and CO2 · (c)(i) Repeat unit: -NH-C6H4-NH-CO-CO- · (c)(ii) O2N-C6H4-NO2 + 12[H] -> H2N-C6H4-NH2 + 4H2O · (d)(i) V: 1,4-benzenediazonium dichloride; W: bis-azo dye (see diagram) · (d)(ii) NaNO2 and HCl at <= 10 C

9701/41 Oct/Nov 2025 — Chemistry Solved Past Paper

Q1MediumGroup 2Chemical EnergeticsEquilibria

(a)

Solutions of Group 2 hydrogencarbonates, $\text{M(HCO}_3)_2$ , decompose on heating to give the corresponding metal carbonate, carbon dioxide and water.

3M

(i)

Write an equation for the decomposition of strontium hydrogencarbonate, $\text{Sr(HCO}_3)_2$ .

1M Similar

(ii)

The thermal stability of Group 2 carbonates increases down the group.

Explain this trend.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Down the group the cation radius increases, so the charge density of the $\text{M}^{2+}$ ion decreases.
The cation therefore polarises/distorts the carbonate ion less, so the carbonate is more stable and decomposes less readily.

Final answer

Cation radius increases, charge density decreases, so less polarisation of the carbonate ion; carbonate more stable.

Detailed explanation

Background Concept

The thermal stability of Group 2 carbonates is controlled by the polarising power of the metal cation. A small, highly charged cation has a high charge density and strongly polarises (distorts) the electron cloud of the large carbonate ion. This distortion weakens the carbonate ion and makes it easier for it to decompose on heating.

Understanding the Question

The question states that thermal stability increases down the group and asks for an explanation. You need to link the change in cation size down the group to the extent of polarisation of the carbonate ion.

Approach

Start with the change in cation radius down the group, then connect it to charge density and polarising power, and finally to the stability of the carbonate ion.

Step-by-Step Reasoning

Down Group 2, the cation radius increases, so the charge density of the $\text{M}^{2+}$ ion decreases. A lower charge density means the cation has less polarising power. As a result, the carbonate ion is less polarised or distorted. A less distorted carbonate ion is more stable, so a higher temperature is needed to decompose it. This is why thermal stability increases down the group.

Key Takeaways

Thermal stability of carbonates depends on cation polarising power. Larger cations with lower charge density stabilise the carbonate ion.

Common Mistakes

Saying that charge density increases down the group.
Saying that the cation polarises the carbon dioxide or the metal oxide instead of the carbonate ion.
Confusing this trend with lattice energy arguments.

Things to Be Careful About

Both marking points are needed: the increase in cation radius/decrease in charge density, and the resulting decrease in polarisation/distortion of the carbonate ion. Use the term 'polarisation' or 'distortion' precisely.

Techniques used

relate cation radius to polarising powerexplain thermal stability trend

(b)

The hydroxides and fluorides of Group 2 elements show similar trends in solubility.

Describe the trend in the solubility of the fluorides of calcium, strontium and barium.

Explain your answer.

4M Similar

DifficultyMedium

Worked solution

Answer

Solubility increases down the group: $\text{CaF}_2 < \text{SrF}_2 < \text{BaF}_2$ .
As the cation radius increases, both the lattice energy and the enthalpy of hydration become less exothermic.
The lattice energy becomes less exothermic by a larger amount than the enthalpy of hydration.
Therefore $\Delta H_{\text{sol}}$ becomes more exothermic, favouring dissolution, so solubility increases.

Final answer

Solubility increases down the group: CaF2 < SrF2 < BaF2; lattice enthalpy decreases more than hydration enthalpy, so ΔH_sol becomes more exothermic.

Detailed explanation

Background Concept

For a sparingly soluble salt such as a Group 2 fluoride, solubility is related to the enthalpy change of solution, $\Delta H_{\text{sol}}$ . Dissolving $\text{MF}_2(\text{s})$ can be thought of in two steps: breaking the lattice to form gaseous ions (the reverse of lattice formation) and hydrating those ions. Thus $\Delta H_{\text{sol}} = \Delta H_{\text{hyd}}(\text{M}^{2+}) + 2\Delta H_{\text{hyd}}(\text{F}^-) - \Delta H_{\text{latt}}$ , where $\Delta H_{\text{latt}}$ is the exothermic lattice energy. A more exothermic (more negative) $\Delta H_{\text{sol}}$ favours dissolution.

Understanding the Question

This part asks you to describe the solubility trend for the fluorides of calcium, strontium and barium, and then explain it using lattice and hydration enthalpy changes.

Approach

State the trend first, then compare how lattice energy and hydration enthalpy change down the group. The key is which one changes by the greater amount, because that decides the sign of $\Delta H_{\text{sol}}$ .

Step-by-Step Reasoning

For $\text{CaF}_2$ , $\text{SrF}_2$ and $\text{BaF}_2$ , solubility increases down the group: $\text{CaF}_2 < \text{SrF}_2 < \text{BaF}_2$ . As the cation radius increases, both the lattice energy and the enthalpy of hydration become less exothermic. However, for fluorides, the lattice energy becomes less exothermic by a larger amount than the hydration enthalpy. Therefore $\Delta H_{\text{sol}}$ becomes more exothermic (more negative), which favours the dissolving process, so solubility increases.

Key Takeaways

Solubility trends down Group 2 depend on the relative changes in lattice and hydration enthalpy. For fluorides, lattice enthalpy changes more, so solubility increases down the group.

Common Mistakes

Confusing with Group 2 sulfates, whose solubility decreases down the group.
Saying that hydration enthalpy changes more than lattice enthalpy.
Stating the trend without giving the enthalpy explanation.

Things to Be Careful About

Use 'less exothermic' or 'less negative' consistently. The mark scheme requires the comparison of the extent of change: lattice enthalpy changes by a larger amount than hydration enthalpy for fluorides, making $\Delta H_{\text{sol}}$ more exothermic.

Techniques used

compare lattice and hydration enthalpy changesexplain solubility trend down Group 2

(c)

3M

(i)

Define enthalpy change of hydration, $\Delta H_{\text{hyd}}$ .

1M Similar

(ii)

State the main factors that affect the magnitude of enthalpy change of hydration.

Explain your answer.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

The two main factors are ionic radius and ionic charge.
As ionic radius increases, charge density decreases, so there is less attraction between the ion and water molecules and $\Delta H_{\text{hyd}}$ becomes less exothermic.
As ionic charge increases, charge density increases, so there is more attraction between the ion and water molecules and $\Delta H_{\text{hyd}}$ becomes more exothermic.

Final answer

Ionic radius and ionic charge; smaller radius/higher charge gives more exothermic hydration.

Detailed explanation

Background Concept

The magnitude of hydration enthalpy depends on the strength of attraction between an ion and water molecules. This attraction is stronger when the ion has a higher charge density, which occurs when the ion is smaller or has a higher charge.

Understanding the Question

The question asks for the main factors affecting $\Delta H_{\text{hyd}}$ and an explanation of how they affect it.

Approach

State the two factors, ionic radius and ionic charge, then explain each in terms of charge density and ion-water attraction.

Step-by-Step Reasoning

The two main factors are ionic radius and ionic charge. As ionic radius increases, charge density decreases, so there is less attraction between the ion and water molecules and $\Delta H_{\text{hyd}}$ becomes less exothermic. As ionic charge increases, charge density increases, so there is more attraction between the ion and water molecules and $\Delta H_{\text{hyd}}$ becomes more exothermic.

Key Takeaways

Higher charge density means stronger ion-dipole attraction and a more exothermic hydration enthalpy.

Common Mistakes

Giving only one factor.
Not explaining how the factor changes $\Delta H_{\text{hyd}}$ .
Confusing hydration enthalpy with lattice enthalpy.

Things to Be Careful About

Mention both ionic radius and ionic charge. Use 'charge density' to link the factor to the strength of attraction.

Techniques used

identify factors affecting hydration enthalpyrelate charge density to ion-water attraction

(d)

Table 1.1 shows various energy changes.

energy change	value / $\text{kJ mol}^{-1}$
lattice energy of $\text{MgF}_2$	$-2957$
enthalpy change of hydration, $\Delta H_{\text{hyd}}$ , of $\text{Mg}^{2+}$	$-1926$
enthalpy change of hydration, $\Delta H_{\text{hyd}}$ , of $\text{F}^-$	$-505$

Use data from Table 1.1 to calculate the enthalpy change of solution, $\Delta H_{\text{sol}}$ , for $\text{MgF}_2\text{(s)}$ .

It may be helpful to draw a labelled energy cycle. Show your working.

2M Similar

DifficultyMedium-Easy

Worked solution

Working

$\Delta H_{\text{sol}}(\text{MgF}_2) = \Delta H_{\text{hyd}}(\text{Mg}^{2+}) + 2\Delta H_{\text{hyd}}(\text{F}^-) - \Delta H_{\text{latt}}$
$= (-1926) + 2(-505) - (-2957)$
$= -1926 - 1010 + 2957 = +21\ \text{kJ mol}^{-1}$

Answer

$+21\ \text{kJ mol}^{-1}$

Final answer

+21 kJ mol^-1

Detailed explanation

Background Concept

The enthalpy change of solution of an ionic compound can be found using an energy cycle. For $\text{MgF}_2(\text{s})$ , the solid is first converted to gaseous ions, which requires energy equal to the reverse of the lattice energy. The gaseous ions are then hydrated, releasing energy. Therefore $\Delta H_{\text{sol}} = \Delta H_{\text{hyd}}(\text{Mg}^{2+}) + 2\Delta H_{\text{hyd}}(\text{F}^-) - \Delta H_{\text{latt}}$ .

Understanding the Question

Use the data in Table 1.1 to calculate $\Delta H_{\text{sol}}$ for $\text{MgF}_2$ . The factor 2 is needed because there are two fluoride ions per formula unit.

Approach

Add the hydration enthalpies of the ions, remembering to double the fluoride value, and add the energy needed to break the lattice (the positive reverse of the given lattice energy).

Step-by-Step Reasoning

Hydration of the ions: $(-1926) + 2(-505) = -2936\ \text{kJ mol}^{-1}$ . The reverse of lattice formation requires $+2957\ \text{kJ mol}^{-1}$ . Total: $-2936 + 2957 = +21\ \text{kJ mol}^{-1}$ . The enthalpy change of solution is therefore $+21\ \text{kJ mol}^{-1}$ .

Key Takeaways

For dissolution, use the hydration enthalpies of all ions and the reverse of the lattice energy. Watch the stoichiometry and signs.

Common Mistakes

Using $-2957$ instead of $+2957$ for the lattice-breaking step.
Forgetting to multiply the fluoride hydration enthalpy by 2.
Sign errors in the final sum.

Things to Be Careful About

The lattice energy is given as $-2957\ \text{kJ mol}^{-1}$ ; breaking the lattice requires $+2957\ \text{kJ mol}^{-1}$ . The final answer should include the sign and units.

Techniques used

apply enthalpy cycle for dissolutioncombine hydration and lattice enthalpy values

(e)

Mercury(I) fluoride, $\text{Hg}_2\text{F}_2$ , is sparingly soluble in water.

The cation in $\text{Hg}_2\text{F}_2$ exists as the diatomic ion $\text{Hg}_2^{2+}$ with a covalent $\text{Hg}-\text{Hg}$ bond.

3M

(i)

Write the expression for the solubility product, $K_{\text{sp}}$ , of $\text{Hg}_2\text{F}_2$ . Include the units.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

$\text{Hg}_2\text{F}_2(\text{s}) \rightleftharpoons \text{Hg}_2^{2+}(\text{aq}) + 2\text{F}^-(\text{aq})$
$K_{\text{sp}} = [\text{Hg}_2^{2+}][\text{F}^-]^2$
Units: $\text{mol dm}^{-3} \times (\text{mol dm}^{-3})^2 = \text{mol}^3\ \text{dm}^{-9}$

Final answer

Ksp = [Hg2^2+][F^-]^2; units mol^3 dm^-9

Detailed explanation

Background Concept

The solubility product, $K_{\text{sp}}$ , is the equilibrium constant for the dissolution of a sparingly soluble salt. The solid is not included in the expression. The concentration of each ion is raised to the power of its stoichiometric coefficient in the dissolution equation.

Understanding the Question

Write the $K_{\text{sp}}$ expression for $\text{Hg}_2\text{F}_2$ and include the units.

Approach

Write the dissolution equilibrium, then form the expression from the ion concentrations and deduce the units by multiplying the units of each concentration term.

Step-by-Step Reasoning

The dissolution is $\text{Hg}_2\text{F}_2(\text{s}) \rightleftharpoons \text{Hg}_2^{2+}(\text{aq}) + 2\text{F}^-(\text{aq})$ . Hence $K_{\text{sp}} = [\text{Hg}_2^{2+}][\text{F}^-]^2$ . Units: $\text{mol dm}^{-3} \times (\text{mol dm}^{-3})^2 = \text{mol}^3\ \text{dm}^{-9}$ .

Key Takeaways

The powers in the $K_{\text{sp}}$ expression match the stoichiometric coefficients, and the units depend on the total number of ions.

Common Mistakes

Forgetting the square on $[\text{F}^-]$ .
Including the solid in the expression.
Giving incorrect units such as $\text{mol dm}^{-3}$ .

Things to Be Careful About

For a 1:2 salt, the units are always $\text{mol}^3\ \text{dm}^{-9}$ . Write the expression exactly as $[\text{Hg}_2^{2+}][\text{F}^-]^2$ .

Techniques used

write solubility product expressionderive units from concentration powers

(ii)

The solubility of $\text{Hg}_2\text{F}_2$ is $9.20 \times 10^{-3}\text{ mol dm}^{-3}$ at $298\text{ K}$ .

Calculate the value of $K_{\text{sp}}$ of $\text{Hg}_2\text{F}_2$ at $298\text{ K}$ .

1M Similar

DifficultyMedium-Easy

Worked solution

Working

Let solubility $s = 9.20 \times 10^{-3}\ \text{mol dm}^{-3}$ .
$[\text{Hg}_2^{2+}] = s$ and $[\text{F}^-] = 2s$ .
$K_{\text{sp}} = s(2s)^2 = 4s^3 = 4(9.20 \times 10^{-3})^3 = 3.11 \times 10^{-6}$

Answer

$3.11 \times 10^{-6}\ \text{mol}^3\ \text{dm}^{-9}$

Final answer

3.11 × 10^-6 mol^3 dm^-9

Detailed explanation

Background Concept

If the molar solubility of a salt is $s$ , the concentrations of its ions can be written in terms of $s$ using the stoichiometry of the dissolution. For $\text{Hg}_2\text{F}_2$ , each formula unit gives one $\text{Hg}_2^{2+}$ and two $\text{F}^-$ .

Understanding the Question

Given the solubility $9.20 \times 10^{-3}\ \text{mol dm}^{-3}$ , calculate $K_{\text{sp}}$ .

Approach

Set $[\text{Hg}_2^{2+}] = s$ and $[\text{F}^-] = 2s$ , substitute into the $K_{\text{sp}}$ expression and evaluate.

Step-by-Step Reasoning

$K_{\text{sp}} = s(2s)^2 = 4s^3$ . With $s = 9.20 \times 10^{-3}$ , $K_{\text{sp}} = 4(9.20 \times 10^{-3})^3 = 3.11 \times 10^{-6}\ \text{mol}^3\ \text{dm}^{-9}$ .

Key Takeaways

The factor $2^2$ arises because $[\text{F}^-] = 2s$ . Always include the units from the expression.

Common Mistakes

Using $K_{\text{sp}} = s^2$ instead of $4s^3$ .
Forgetting the units.
Giving too many significant figures; 2 or 3 significant figures is appropriate.

Things to Be Careful About

The answer should be $3.11 \times 10^{-6}$ or $3.1 \times 10^{-6}$ , with units $\text{mol}^3\ \text{dm}^{-9}$ .

Techniques used

relate solubility to ion concentrationscalculate Ksp from molar solubility

Q2MediumTransition ElementsElectrochemistry

(a)

Iron can form stable ions in the $+2$ and $+3$ oxidation states.

Explain why transition elements have variable oxidation states.

1M Similar

DifficultyEasy

Worked solution

Answer

The (3)d and (4)s sub-shells / orbitals / electrons are close / similar in energy.

Final answer

The 3d and 4s subshells are close in energy

Detailed explanation

Background Concept

Transition elements are d-block elements that form at least one ion with a partially filled d subshell. A defining feature is their ability to exhibit variable oxidation states. This arises because the 3d and 4s subshells are very close in energy. In the neutral atom, the 4s subshell fills before the 3d (e.g., Fe: $[\text{Ar}]\,3d^6 4s^2$ ), but when ionising, the 4s electrons are removed first, and then 3d electrons can also be removed without a large energy penalty. Because the energy difference between 3d and 4s is small, removing different numbers of electrons costs similar energy, so several oxidation states are accessible.

Understanding the Question

The question asks why transition elements have variable oxidation states, using iron as an example ( $\text{Fe}^{2+}$ and $\text{Fe}^{3+}$ ). The answer must focus on the electronic structure — specifically the close energy of the 3d and 4s subshells.

Approach

Recall the electronic configuration of transition metals and the relative energies of 3d and 4s. The key idea is that these subshells are close in energy, so electrons can be removed from either without a huge energy cost, giving variable oxidation states.

Step-by-Step Reasoning

In a transition metal atom, the 3d and 4s subshells are close in energy.
When forming ions, electrons are removed from the 4s subshell first, then from the 3d.
Because the 3d and 4s energies are so similar, removing one, two, or more d electrons costs only a small amount of extra energy.
This allows the element to exist in multiple oxidation states, e.g., $\text{Fe}^{2+}$ ( $[\text{Ar}]\,3d^6$ ) and $\text{Fe}^{3+}$ ( $[\text{Ar}]\,3d^5$ ).
The mark scheme rewards the statement that the 3d and 4s subshells/orbitals/electrons are close/similar in energy.

Key Takeaways

Variable oxidation states are a direct consequence of the small energy gap between 3d and 4s subshells.

Common Mistakes

Saying "the d subshell is partially filled" without mentioning the energy similarity — this does not explain variable oxidation states.
Confusing the order of filling (4s fills before 3d) with the order of ionisation (4s lost first).
Not mentioning both 3d and 4s.

Things to Be Careful About

The mark scheme requires the idea that 3d and 4s are close/similar in energy. Use precise wording such as "close in energy" or "similar in energy".

Techniques used

explain the electronic-structure basis of variable oxidation states

(b)

Aqueous solutions of iron(II) salts contain the complex ion $[\text{Fe(H}_2\text{O)}_6]^{2+}$ .

Define complex ion.

1M Similar

(c)

$[\text{Fe(H}_2\text{O)}_6]^{2+}$ can be converted into $[\text{Fe(H}_2\text{O)}_4\text{(OH)}_2]$ .

4M

(i)

Suggest a suitable reagent for this conversion. State the type of reaction.

1M Similar

DifficultyEasy

Worked solution

Answer

NaOH(aq) (or OH⁻(aq)); precipitation / ligand exchange / deprotonation / acid–base reaction.

Final answer

NaOH(aq); precipitation/ligand exchange/deprotonation/acid–base

Detailed explanation

Background Concept

When a base such as $\text{NaOH}(\text{aq})$ is added to an aqueous solution of a transition metal aqua complex, the water ligands can be deprotonated. For $[\text{Fe(H}_2\text{O)}_6]^{2+}$ , adding $\text{OH}^-$ removes $\text{H}^+$ from two water ligands, giving $[\text{Fe(H}_2\text{O)}_4\text{(OH)}_2]$ . This is a ligand exchange (water replaced by hydroxide) and an acid–base reaction ( $\text{H}^+$ transferred to $\text{OH}^-$ ). The product is a neutral complex that is insoluble, so a precipitate forms.

Understanding the Question

Suggest a reagent to convert $[\text{Fe(H}_2\text{O)}_6]^{2+}$ to $[\text{Fe(H}_2\text{O)}_4\text{(OH)}_2]$ , and state the type of reaction.

Approach

Recognise that the product has two $\text{OH}^-$ ligands instead of two $\text{H}_2\text{O}$ ligands. This is achieved by adding a base, typically $\text{NaOH}(\text{aq})$ . The reaction type is precipitation / ligand exchange / deprotonation / acid–base.

Step-by-Step Reasoning

The conversion is:

[\text{Fe(H}_2\text{O)}_6]^{2+} + 2\text{OH}^- \rightarrow [\text{Fe(H}_2\text{O)}_4\text{(OH)}_2] + 2\text{H}_2\text{O}

Adding $\text{NaOH}(\text{aq})$ supplies $\text{OH}^-$ ions. Two water ligands lose $\text{H}^+$ to $\text{OH}^-$ , forming two hydroxide ligands. This is a deprotonation (acid–base) reaction and a ligand exchange. The resulting neutral complex is insoluble, so it precipitates.

Key Takeaways

Adding base to an aqua complex deprotonates water ligands, forming hydroxide complexes that often precipitate.

Common Mistakes

Suggesting "add water" — this does not change the complex.
Suggesting "add acid" — this would not deprotonate.
Not stating the reaction type.

Things to Be Careful About

The mark scheme accepts $\text{NaOH}$ / $\text{OH}^-(\text{aq})$ and any of precipitation / ligand exchange / deprotonation / acid–base.

Techniques used

identify the reagent for deprotonation of an aqua complexclassify the reaction type

(ii)

$[\text{Fe(H}_2\text{O)}_4\text{(OH)}_2]$ is a green precipitate that turns brown on standing in air.

Table 2.1 shows electrode potentials for some electrode reactions.

electrode reaction	$E^\ominus / \text{V}$
$\text{Fe(H}_2\text{O)}_3\text{(OH)}_3 + \text{H}_2\text{O} + \text{e}^- \rightleftharpoons \text{Fe(H}_2\text{O)}_4\text{(OH)}_2 + \text{OH}^-$	$-0.56$
$\text{O}_2 + 2\text{H}_2\text{O} + 4\text{e}^- \rightleftharpoons 4\text{OH}^-$	$+0.40$

Use the information in Table 2.1 to explain why $[\text{Fe(H}_2\text{O)}_4\text{(OH)}_2]$ turns brown on standing in air.

Include an equation for this reaction.

3M Similar

DifficultyMedium

Worked solution

Answer

$E^\ominus_{\text{cell}} = 0.40 - (-0.56) = +0.96\text{ V}$ . Since $E^\ominus_{\text{cell}}$ is positive, the reaction is feasible: $[\text{Fe(H}_2\text{O)}_4\text{(OH)}_2]$ is oxidised from $+2$ to $+3$ by oxygen in the air.

4\text{Fe(H}_2\text{O)}_4\text{(OH)}_2 + \text{O}_2 \rightarrow 4\text{Fe(H}_2\text{O)}_3\text{(OH)}_3 + 2\text{H}_2\text{O}

Final answer

Fe(II) complex is oxidised to Fe(III) by O2; 4Fe(H2O)4(OH)2 + O2 → 4Fe(H2O)3(OH)3 + 2H2O

Detailed explanation

Background Concept

Electrode potentials ( $E^\ominus$ values) predict the feasibility of redox reactions. For two half-reactions, the one with the more positive $E^\ominus$ is reduced (acts as the oxidising agent); the other is oxidised. The cell potential is $E^\ominus_{\text{cell}} = E^\ominus(\text{reduction}) - E^\ominus(\text{oxidation})$ . If $E^\ominus_{\text{cell}} > 0$ , the reaction is feasible.

Understanding the Question

$[\text{Fe(H}_2\text{O)}_4\text{(OH)}_2]$ is green but turns brown in air. Use the given $E^\ominus$ values to explain why, and write the balanced equation. The brown colour is the Fe(III) hydroxide complex.

Approach

Identify the two half-reactions: the Fe(III)/Fe(II) couple ( $E^\ominus = -0.56\text{ V}$ ) and the $\text{O}_2/\text{OH}^-$ couple ( $E^\ominus = +0.40\text{ V}$ ). Since $+0.40 > -0.56$ , $\text{O}_2$ is the oxidising agent. Calculate $E^\ominus_{\text{cell}}$ to confirm feasibility. Then combine the half-equations to write the overall equation.

Step-by-Step Reasoning

The Fe couple: $\text{Fe(H}_2\text{O)}_3\text{(OH)}_3 + \text{H}_2\text{O} + \text{e}^- \rightleftharpoons \text{Fe(H}_2\text{O)}_4\text{(OH)}_2 + \text{OH}^-$ , $E^\ominus = -0.56\text{ V}$ .
The $\text{O}_2$ couple: $\text{O}_2 + 2\text{H}_2\text{O} + 4\text{e}^- \rightleftharpoons 4\text{OH}^-$ , $E^\ominus = +0.40\text{ V}$ .
Since $+0.40\text{ V}$ is more positive than $-0.56\text{ V}$ , $\text{O}_2$ is reduced and the Fe(II) complex is oxidised.
$E^\ominus_{\text{cell}} = 0.40 - (-0.56) = +0.96\text{ V} > 0$ , so the reaction is feasible.
The Fe(II) complex is oxidised from $+2$ to $+3$ , forming the brown Fe(III) hydroxide complex.
Balanced equation: multiply the Fe half-reaction by 4 and add to the $\text{O}_2$ reduction:

4\text{Fe(H}_2\text{O)}_4\text{(OH)}_2 + \text{O}_2 \rightarrow 4\text{Fe(H}_2\text{O)}_3\text{(OH)}_3 + 2\text{H}_2\text{O}

Key Takeaways

Positive $E^\ominus_{\text{cell}}$ means the reaction is feasible; the species with the more positive $E^\ominus$ is reduced.

Common Mistakes

Writing an unbalanced equation.
Getting the direction of electron flow wrong (saying Fe is reduced).
Not stating that the Fe(II) complex is oxidised.

Things to Be Careful About

The equation must be balanced. The $E^\ominus$ comparison must show $\text{O}_2$ has the more positive $E^\ominus$ .

Techniques used

calculate cell potential from E° valuesdetermine redox feasibility from E°write a balanced redox equation

(d)

The complex $[\text{Co(NH}_3)_6]^{2+}$ reacts with hydrogen peroxide as shown.

2[\text{Co(NH}_3)_6]^{2+} + \text{H}_2\text{O}_2 \rightarrow 2[\text{Co(NH}_3)_6]^{3+} + 2\text{OH}^-

E^\ominus_{\text{cell}} = +1.67\text{ V}

Calculate $\Delta G^\ominus$ , in $\text{kJ mol}^{-1}$ , for reaction 1.

2M Similar

DifficultyMedium-Easy

Worked solution

Working

$n = 2$

\Delta G^\ominus = -nE^\ominus_{\text{cell}}F = -2 \times 1.67 \times 96\,500 = -322\,310\text{ J mol}^{-1}

Answer

\Delta G^\ominus = -322.3\text{ kJ mol}^{-1}

Final answer

-322.3 kJ mol^-1

Detailed explanation

Background Concept

The relationship $\Delta G^\ominus = -nFE^\ominus_{\text{cell}}$ connects Gibbs free energy change to cell potential. $n$ is the number of moles of electrons transferred per mole of reaction as written, $F$ is the Faraday constant ( $96\,500\text{ C mol}^{-1}$ ), and $E^\ominus_{\text{cell}}$ is in volts. The result is in $\text{J mol}^{-1}$ ; convert to $\text{kJ mol}^{-1}$ by dividing by 1000.

Understanding the Question

Given the reaction $2[\text{Co(NH}_3)_6]^{2+} + \text{H}_2\text{O}_2 \rightarrow 2[\text{Co(NH}_3)_6]^{3+} + 2\text{OH}^-$ and $E^\ominus_{\text{cell}} = +1.67\text{ V}$ , calculate $\Delta G^\ominus$ in $\text{kJ mol}^{-1}$ .

Approach

Determine $n$ from the stoichiometry: each $\text{Co}^{2+}$ loses one electron to become $\text{Co}^{3+}$ . Since there are two Co ions, $n = 2$ . Then substitute into $\Delta G^\ominus = -nFE^\ominus_{\text{cell}}$ .

Step-by-Step Reasoning

Oxidation state of Co in $[\text{Co(NH}_3)_6]^{2+}$ is $+2$ ; in $[\text{Co(NH}_3)_6]^{3+}$ it is $+3$ . Each Co loses one electron.
The equation shows $2[\text{Co(NH}_3)_6]^{2+}$ , so 2 electrons are transferred in total: $n = 2$ .
$\Delta G^\ominus = -2 \times 1.67 \times 96\,500 = -322\,310\text{ J mol}^{-1}$ .
Convert: $-322\,310 / 1000 = -322.3\text{ kJ mol}^{-1}$ .

Key Takeaways

$\Delta G^\ominus = -nFE^\ominus_{\text{cell}}$ ; $n$ is the number of electrons transferred per mole of reaction as written.

Common Mistakes

Using $n = 1$ (forgetting the coefficient 2 in front of Co).
Forgetting to convert J to kJ.
Sign error ( $\Delta G^\ominus$ is negative for a spontaneous reaction).

Things to Be Careful About

The answer must be in $\text{kJ mol}^{-1}$ . The mark scheme gives $-322.3\text{ kJ mol}^{-1}$ . Use $F = 96\,500\text{ C mol}^{-1}$ .

Techniques used

apply ΔG° = −nFE°celldetermine n from the redox equationconvert J mol⁻¹ to kJ mol⁻¹

Q3MediumReaction Kinetics

(a)

Solid manganese(IV) oxide, $\text{MnO}_2$ , catalyses the decomposition of hydrogen peroxide.

2\text{H}_2\text{O}_2\text{(aq)} \rightarrow 2\text{H}_2\text{O(l)} + \text{O}_2\text{(g)}

State the type of catalysis for this reaction. Explain your answer.

1M Similar

DifficultyEasy

Worked solution

Answer

Heterogeneous catalysis.
The catalyst, $\text{MnO}_2$ , is a solid, while the reactant $\text{H}_2\text{O}_2$ is in aqueous solution; they are in different phases (states of matter).

Final answer

Heterogeneous catalysis; MnO2 is a solid and H2O2 is aqueous, so they are in different phases.

Detailed explanation

Background Concept

Catalysts can be classified based on their phase relative to the reactants. In homogeneous catalysis, the catalyst and reactants are in the same phase (e.g., all in solution). In heterogeneous catalysis, the catalyst is in a different phase from the reactants (e.g., a solid catalyst with liquid or gaseous reactants). The reaction occurs at the surface of the solid catalyst where reactant molecules adsorb, bonds weaken, new bonds form, and products desorb.

Understanding the Question

The question asks for the type of catalysis when solid manganese(IV) oxide ( $\text{MnO}_2$ ) catalyses the decomposition of aqueous hydrogen peroxide ( $\text{H}_2\text{O}_2$ ). You must state the type and provide the reasoning based on the physical states of the substances involved.

Approach

Identify the physical state (phase) of the catalyst and the reactant from the state symbols in the chemical equation. Compare them to determine if they are in the same or different phases.

Step-by-Step Reasoning

Identify the states: The equation gives $\text{MnO}_2$ as a solid ( $\text{s}$ ) and $\text{H}_2\text{O}_2$ as aqueous ( $\text{aq}$ ).
Compare phases: A solid and an aqueous solution are different phases.
Conclude: Because the catalyst and reactants are in different phases, the catalysis is heterogeneous.

Key Takeaways

Always check the state symbols ( $\text{s}$ , $\text{l}$ , $\text{g}$ , $\text{aq}$ ) in the given equation to determine the phase of each substance. If they differ, it is heterogeneous catalysis; if they are the same, it is homogeneous.

Common Mistakes

Stating 'heterogeneous' without explaining why (the mark scheme requires the explanation that they are in different phases).
Confusing heterogeneous with homogeneous catalysis.
Forgetting to mention the specific states (solid vs. aqueous) and just saying 'different states'.

Things to Be Careful About

Ensure you explicitly state both the type ('heterogeneous') and the reason ('different phases/states'). The mark scheme awards marks for both the classification and the justification.

Techniques used

identify the phase of the catalyst relative to the reactantsclassify the type of catalysis

(b)

Hydrogen peroxide reacts with iodide ions in acidic conditions as shown.

\text{H}_2\text{O}_2 + 2\text{I}^- + 2\text{H}^+ \rightarrow 2\text{H}_2\text{O} + \text{I}_2

The initial rate of this reaction is investigated with different concentrations of $\text{H}_2\text{O}_2$ , $\text{I}^-$ and $\text{H}^+$ . The results obtained are shown in Table 3.1.

experiment	$[\text{H}_2\text{O}_2] / \text{mol dm}^{-3}$	$[\text{I}^-] / \text{mol dm}^{-3}$	$[\text{H}^+] / \text{mol dm}^{-3}$	initial rate / $\text{mol dm}^{-3}\text{ s}^{-1}$
1	0.0450	0.0300	0.0125	$2.42 \times 10^{-3}$
2	0.0225	0.0600	0.0125	$2.42 \times 10^{-3}$
3	0.0225	0.120	0.0125	$4.84 \times 10^{-3}$
4	0.0450	0.120	0.0500	$9.68 \times 10^{-3}$

6M

(i)

Use the information in Table 3.1 to deduce the rate equation for this reaction.

Explain your reasoning.

4M Similar

DifficultyMedium

Worked solution

Answer

Comparing experiments 2 and 3: $[\text{I}^-]$ is doubled ( $0.0600 \rightarrow 0.120$ ) while $[\text{H}_2\text{O}_2]$ and $[\text{H}^+]$ are constant; the rate doubles ( $2.42 \times 10^{-3} \rightarrow 4.84 \times 10^{-3}$ ). Thus, the reaction is first order with respect to $[\text{I}^-]$ .

Comparing experiments 1 and 2: $[\text{I}^-]$ is doubled and $[\text{H}_2\text{O}_2]$ is halved; the rate is unchanged. Since doubling $[\text{I}^-]$ doubles the rate, halving $[\text{H}_2\text{O}_2]$ must halve it to keep the rate constant. Thus, the reaction is first order with respect to $[\text{H}_2\text{O}_2]$ .

Comparing experiments 3 and 4: $[\text{H}_2\text{O}_2]$ is doubled and $[\text{H}^+]$ is quadrupled ( $0.0125 \rightarrow 0.0500$ ) while $[\text{I}^-]$ is constant; the rate doubles ( $4.84 \times 10^{-3} \rightarrow 9.68 \times 10^{-3}$ ). Since doubling $[\text{H}_2\text{O}_2]$ doubles the rate, changing $[\text{H}^+]$ has no effect. Thus, the reaction is zero order with respect to $[\text{H}^+]$ .

Rate equation:

\text{rate} = k[\text{H}_2\text{O}_2][\text{I}^-]

Final answer

rate = k[H2O2][I-]

Detailed explanation

Background Concept

The rate equation (or rate law) expresses the relationship between the rate of a reaction and the concentrations of the reactants: $\text{rate} = k[A]^m[B]^n$ . The exponents $m$ and $n$ are the reaction orders with respect to each reactant and must be determined experimentally; they cannot be deduced from the stoichiometry of the overall balanced equation. To find the order with respect to a specific reactant, we compare two experiments where the concentration of that reactant changes while all others are held constant.

Understanding the Question

You are given a table of initial rates for four experiments with varying concentrations of $\text{H}_2\text{O}_2$ , $\text{I}^-$ , and $\text{H}^+$ . You must deduce the order of reaction with respect to each species and write the complete rate equation, explaining your reasoning by referencing specific experiments from the table.

Approach

Find two experiments where only $[\text{I}^-]$ changes (Exp 2 and 3) to find its order.
Find two experiments where $[\text{H}_2\text{O}_2]$ changes while others are constant or cancel out (Exp 1 and 2) to find its order.
Find two experiments where $[\text{H}^+]$ changes (Exp 3 and 4) to find its order.
Combine the orders into the final rate equation.

Step-by-Step Reasoning

Order with respect to $[\text{I}^-]$ : Look at experiments 2 and 3. $[\text{H}_2\text{O}_2]$ is constant at $0.0225$ , $[\text{H}^+]$ is constant at $0.0125$ . $[\text{I}^-]$ doubles from $0.0600$ to $0.120$ . The rate doubles from $2.42 \times 10^{-3}$ to $4.84 \times 10^{-3}$ . Since $2^1 = 2$ , the order is 1 (first order).
Order with respect to $[\text{H}_2\text{O}_2]$ : Look at experiments 1 and 2. $[\text{H}^+]$ is constant. $[\text{I}^-]$ doubles ( $0.0300 \rightarrow 0.0600$ ) and $[\text{H}_2\text{O}_2]$ halves ( $0.0450 \rightarrow 0.0225$ ). The rate is unchanged ( $2.42 \times 10^{-3}$ ). We know doubling $[\text{I}^-]$ doubles the rate. For the overall rate to remain constant, halving $[\text{H}_2\text{O}_2]$ must halve the rate. Since $0.5^1 = 0.5$ , the order is 1 (first order).
Order with respect to $[\text{H}^+]$ : Look at experiments 3 and 4. $[\text{I}^-]$ is constant at $0.120$ . $[\text{H}_2\text{O}_2]$ doubles ( $0.0225 \rightarrow 0.0450$ ) and $[\text{H}^+]$ quadruples ( $0.0125 \rightarrow 0.0500$ ). The rate doubles ( $4.84 \times 10^{-3} \rightarrow 9.68 \times 10^{-3}$ ). We know doubling $[\text{H}_2\text{O}_2]$ doubles the rate. Since the rate only doubled in total, the quadrupling of $[\text{H}^+]$ must have had no effect ( $4^0 = 1$ ). Thus, the order is 0 (zero order).
Rate equation: Substitute the orders into the general form: $\text{rate} = k[\text{H}_2\text{O}_2]^1[\text{I}^-]^1[\text{H}^+]^0$ , which simplifies to $\text{rate} = k[\text{H}_2\text{O}_2][\text{I}^-]$ .

Key Takeaways

When deducing orders, look for pairs of experiments where only one concentration changes. If two concentrations change simultaneously, you can still deduce the orders if you already know the order of one of them (as in Exp 1 vs 2). Zero-order reactants do not appear in the final rate equation.

Common Mistakes

Assuming the orders match the stoichiometric coefficients in the balanced equation (e.g., assuming first order in $\text{H}^+$ because of the $2\text{H}^+$ in the equation). Orders must always be determined experimentally.
Failing to explain the reasoning clearly by explicitly stating which experiments are being compared and how the concentrations and rates change.

Things to Be Careful About

Ensure your comparisons are logically sound. For example, in Exp 1 vs 2, both $[\text{I}^-]$ and $[\text{H}_2\text{O}_2]$ change, so you must account for the known effect of $[\text{I}^-]$ to isolate the effect of $[\text{H}_2\text{O}_2]$ . The mark scheme accepts multiple valid pairs of experiments for the explanations.

Techniques used

compare experiments to determine order with respect to each reactantconstruct the rate equation from experimental orders

(ii)

Use your rate equation from (b)(i) and the data from Experiment 1 to calculate the rate constant, $k$ , for this reaction. Include the units of $k$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Rearranging the rate equation for $k$ :

k = \frac{\text{rate}}{[\text{H}_2\text{O}_2][\text{I}^-]}

Substituting the values from Experiment 1:

k = \frac{2.42 \times 10^{-3}}{0.0450 \times 0.0300} = 1.79

For the units:

\text{Units of } k = \frac{\text{mol dm}^{-3} \text{ s}^{-1}}{(\text{mol dm}^{-3})(\text{mol dm}^{-3})} = \text{dm}^3 \text{ mol}^{-1} \text{ s}^{-1}

Answer

$k = 1.79 \text{ dm}^3 \text{ mol}^{-1} \text{ s}^{-1}$

Final answer

1.79 dm^3 mol^-1 s^-1

Detailed explanation

Background Concept

The rate constant $k$ is a proportionality constant in the rate equation that is specific to a given reaction at a specific temperature. Its value depends on the overall order of the reaction. The units of $k$ vary with the overall order to ensure that the rate always has units of concentration per time (e.g., $\text{mol dm}^{-3} \text{ s}^{-1}$ ). For an overall order $n$ , the units of $k$ are $(\text{mol dm}^{-3})^{1-n} \text{ s}^{-1}$ .

Understanding the Question

Using the rate equation derived in part (b)(i) and the data from Experiment 1, calculate the numerical value of the rate constant $k$ and determine its correct units.

Approach

Rearrange the rate equation to isolate $k$ .
Substitute the concentration and rate values from Experiment 1 into the rearranged equation.
Derive the units by substituting the units of rate and concentration into the rearranged equation and simplifying.

Step-by-Step Reasoning

Calculation: From $\text{rate} = k[\text{H}_2\text{O}_2][\text{I}^-]$ , we get $k = \frac{\text{rate}}{[\text{H}_2\text{O}_2][\text{I}^-]}$ . Using Experiment 1: rate $= 2.42 \times 10^{-3}$ , $[\text{H}_2\text{O}_2] = 0.0450$ , $[\text{I}^-] = 0.0300$ .
$k = \frac{2.42 \times 10^{-3}}{0.0450 \times 0.0300} = \frac{2.42 \times 10^{-3}}{1.35 \times 10^{-3}} = 1.79259...$
Rounding to 3 significant figures gives $1.79$ .
Units: Rate is in $\text{mol dm}^{-3} \text{ s}^{-1}$ . Concentrations are in $\text{mol dm}^{-3}$ .
$\text{Units of } k = \frac{\text{mol dm}^{-3} \text{ s}^{-1}}{(\text{mol dm}^{-3}) \times (\text{mol dm}^{-3})} = \frac{1}{\text{mol dm}^{-3} \text{ s}^{-1}} \times \text{mol dm}^{-3} \text{ s}^{-1} \text{ (wait, simplify directly)} = \text{dm}^3 \text{ mol}^{-1} \text{ s}^{-1}$
Alternatively, for a second-order overall reaction (1+1=2), units are $\text{M}^{-1} \text{ s}^{-1}$ or $\text{dm}^3 \text{ mol}^{-1} \text{ s}^{-1}$ .

Key Takeaways

Always calculate $k$ using data from a single experiment where all values are known. Derive the units of $k$ systematically by substituting the units of all variables into the rearranged equation and cancelling terms.

Common Mistakes

Forgetting to include the units for $k$ (the mark scheme explicitly requires them).
Using data from the wrong experiment or misreading values from the table.
Incorrectly simplifying the units (e.g., writing $\text{mol}^{-1} \text{ dm}^3 \text{ s}^{-1}$ is fine, but $\text{dm}^3 \text{ mol}^{-1} \text{ s}^{-1}$ is the standard convention).

Things to Be Careful About

Ensure you use the correct number of significant figures. The mark scheme accepts a minimum of 2 significant figures (1.8 or 1.79). Using values from Experiment 1 is required, but using data from any other experiment with the correct rate equation will yield the same $k$ value (within rounding errors).

Techniques used

rearrange the rate equation to solve for kdetermine the units of the rate constant from the rate equation

(c)

The rate of the thermal decomposition of azomethane, $\text{CH}_3\text{N=NCH}_3$ , is investigated.

\text{CH}_3\text{N=NCH}_3 \rightarrow \text{N}_2 + \text{C}_2\text{H}_6

Fig. 3.1 shows the results obtained. The reaction is first order with respect to $\text{CH}_3\text{N=NCH}_3$ .

Fig. 3.1

3M

(i)

Use Fig. 3.1 to calculate two half-lives, $t_{1/2}$ , to show that the reaction is first order.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Initial concentration $[\text{CH}_3\text{N=NCH}_3]_0 = 0.050 \text{ mol dm}^{-3}$ .

First half-life ( $t_{1/2}$ ): Time for concentration to fall from $0.050$ to $0.025 \text{ mol dm}^{-3}$ .
From the graph, $[\text{CH}_3\text{N=NCH}_3] = 0.025 \text{ mol dm}^{-3}$ at $t \approx 60 \text{ s}$ . So, $t_{1/2} = 60 \text{ s}$ .

Second half-life: Time for concentration to fall from $0.025$ to $0.0125 \text{ mol dm}^{-3}$ .
From the graph, $[\text{CH}_3\text{N=NCH}_3] = 0.0125 \text{ mol dm}^{-3}$ at $t \approx 120 \text{ s}$ . The time taken is $120 - 60 = 60 \text{ s}$ .

Since the two half-lives are constant ( $60 \text{ s}$ ), the reaction is first order.

Final answer

Two constant half-lives of approximately 60 s.

Detailed explanation

Background Concept

For a first-order reaction, the half-life ( $t_{1/2}$ ) is constant and independent of the initial concentration. This is a defining characteristic used to identify first-order kinetics from a concentration-time graph. The half-life is the time taken for the concentration of a reactant to decrease to half of its current value. If you plot concentration against time, you can read off the time it takes to go from $C_0$ to $C_0/2$ , and then from $C_0/2$ to $C_0/4$ . If these time intervals are equal, the reaction is first order.

Understanding the Question

You are given a graph of concentration of azomethane against time for its thermal decomposition. You must calculate two half-lives from the graph and use the fact that they are constant to confirm the reaction is first order.

Approach

Identify the initial concentration from the y-intercept.
Find the time at which the concentration is half the initial value (first half-life).
Find the time at which the concentration is half of that value (second half-life start), and calculate the time interval to reach the next half (second half-life).
Compare the two intervals.

Step-by-Step Reasoning

Initial concentration: At $t = 0$ , $[\text{CH}_3\text{N=NCH}_3] = 0.050 \text{ mol dm}^{-3}$ .
First half-life: Half of $0.050$ is $0.025 \text{ mol dm}^{-3}$ . Locate $0.025$ on the y-axis, move across to the curve, and read down to the x-axis. The time is approximately $60 \text{ s}$ . So, $t_{1/2(1)} = 60 - 0 = 60 \text{ s}$ .
Second half-life: Half of $0.025$ is $0.0125 \text{ mol dm}^{-3}$ . Locate $0.0125$ on the y-axis, move across to the curve, and read down to the x-axis. The time is approximately $120 \text{ s}$ . The time taken for this second half-life is $120 - 60 = 60 \text{ s}$ . So, $t_{1/2(2)} = 60 \text{ s}$ .
Conclusion: Since $t_{1/2(1)} = t_{1/2(2)} = 60 \text{ s}$ , the half-life is constant, which confirms the reaction is first order.

Key Takeaways

For first-order reactions, the half-life is constant. Always measure half-lives from the curve by finding the time interval between consecutive halving of concentration, not just the time from $t=0$ .

Common Mistakes

Reading the time for the second half-life directly from the x-axis at $0.0125$ (which gives $120 \text{ s}$ ) and forgetting to subtract the first half-life time to get the actual interval ( $60 \text{ s}$ ).
Misreading the graph values due to grid line spacing. Ensure you interpolate correctly between grid lines.

Things to Be Careful About

The mark scheme allows a tolerance of $\pm 5 \text{ s}$ for the half-life value (so $55-65 \text{ s}$ is acceptable). Make sure you explicitly state that the two half-lives are constant/equal to earn the second mark.

Techniques used

read half-life values from a concentration-time graphverify first-order kinetics by checking for constant half-life

(ii)

Use your answer to (c)(i) to calculate the rate constant, $k$ , for the decomposition of azomethane.

1M Similar

DifficultyEasy

Worked solution

Answer

For a first-order reaction:

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

Using $t_{1/2} = 60 \text{ s}$ :

k = \frac{0.693}{60} = 0.0116 \text{ s}^{-1}

Final answer

0.0116 s^-1

Detailed explanation

Background Concept

For a first-order reaction, the relationship between the rate constant $k$ and the half-life $t_{1/2}$ is derived from the integrated rate law: $\ln[A] = \ln[A]_0 - kt$ . At $t = t_{1/2}$ , $[A] = [A]_0/2$ . Substituting this gives $\ln(1/2) = -kt_{1/2}$ , which simplifies to $k = \frac{\ln 2}{t_{1/2}}$ . Note that $\ln 2 \approx 0.693$ .

Understanding the Question

Using the constant half-life determined in part (c)(i), calculate the rate constant $k$ for the decomposition of azomethane.

Approach

Substitute the value of $t_{1/2}$ into the first-order half-life equation $k = \frac{\ln 2}{t_{1/2}}$ and calculate the result.

Step-by-Step Reasoning

Formula: $k = \frac{\ln 2}{t_{1/2}}$ or $k = \frac{0.693}{t_{1/2}}$ .
Substitution: $t_{1/2} = 60 \text{ s}$ .
$k = \frac{0.693}{60} = 0.01155 \approx 0.0116 \text{ s}^{-1}$
Units: For a first-order reaction, the units of $k$ are always $\text{time}^{-1}$ , so $\text{s}^{-1}$ .

Key Takeaways

The equation $k = \frac{\ln 2}{t_{1/2}}$ is specific to first-order reactions. For zero-order or second-order reactions, different equations apply (e.g., $t_{1/2} = \frac{[A]_0}{2k}$ for second order).

Common Mistakes

Using the wrong formula for $k$ (e.g., using a second-order formula).
Forgetting the units for $k$ (though the mark scheme here only asks for the value, it's good practice to include $\text{s}^{-1}$ ).
Rounding errors: using $0.69$ instead of $0.693$ or $\ln 2$ directly on a calculator.

Things to Be Careful About

Use the unrounded value of $t_{1/2}$ if possible, or use the value you calculated ( $60 \text{ s}$ ). The mark scheme accepts $0.0116$ (2 or 3 s.f.). Ensure you use $\ln 2$ (natural log) and not $\log_{10} 2$ .

Techniques used

calculate rate constant from half-life using the first-order formula

(d)

Describe the effect of increasing temperature on the rate constant and on the rate of a reaction.

1M Similar

DifficultyEasy

Worked solution

Answer

Increasing temperature increases the rate constant $k$ and therefore increases the rate of reaction.

Final answer

Increases k and increases the rate of reaction.

Detailed explanation

Background Concept

The rate constant $k$ is temperature-dependent, as described by the Arrhenius equation: $k = A e^{-E_a / RT}$ . As temperature increases, the exponential term $e^{-E_a / RT}$ increases because the denominator $RT$ increases, making the negative exponent less negative. This means a larger proportion of molecules have kinetic energy greater than or equal to the activation energy $E_a$ , leading to more successful collisions per unit time and a larger $k$ . Since rate $= k[\text{reactants}]^n$ , an increase in $k$ directly leads to an increase in the rate of reaction at any given concentration.

Understanding the Question

The question asks for the effect of increasing temperature on two specific quantities: the rate constant $k$ and the rate of the reaction. You need to state the direction of the change for both.

Approach

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

Step-by-Step Reasoning

Effect on $k$ : According to the Arrhenius equation, an increase in temperature increases the value of the rate constant $k$ .
Effect on rate: Since the rate is directly proportional to $k$ (for a given set of concentrations), an increase in $k$ results in an increase in the rate of reaction.

Key Takeaways

Temperature affects the rate constant $k$ , not the concentrations. An increase in temperature always increases $k$ and thus increases the rate of reaction (for both exothermic and endothermic reactions). Catalysts also increase $k$ by providing an alternative pathway with lower $E_a$ , but temperature increases $k$ by increasing the fraction of molecules with sufficient energy.

Common Mistakes

Stating that temperature affects the equilibrium position (that's thermodynamics, not kinetics; the question asks about rate and $k$ ).
Saying 'increases the rate' but forgetting to mention it also increases $k$ , or vice versa. The mark scheme requires both.
Confusing the effect of temperature on $k$ with the effect of concentration or pressure (which do not change $k$ ).

Things to Be Careful About

Be precise with terminology. Use 'rate constant' for $k$ and 'rate of reaction' for the overall speed. Both must increase.

Techniques used

describe the effect of temperature on the rate constantrelate rate constant change to reaction rate

Q4MediumElectrochemistryTransition Elements

(a)

Define standard cell potential, $E^\ominus_{\text{cell}}$ . Include a description of standard conditions.

2M Similar

DifficultyEasy

Worked solution

Answer

The standard cell potential, $E^\ominus_{\text{cell}}$ , is the potential difference (or voltage / EMF) between two half-cells (or two electrodes) when all reactants and products are in their standard states.

Standard conditions are:

concentration of $1 \text{ mol dm}^{-3}$
pressure of $1 \text{ atm}$ (or $101 \text{ kPa}$ )
temperature of $298 \text{ K}$ (or $25 \text{ }^\circ\text{C}$ )

Final answer

Potential difference between two half-cells under standard conditions (1 mol dm^-3, 1 atm/101 kPa, 298 K).

Detailed explanation

Background Concept

The standard electrode potential, $E^\ominus$ , of a half-cell is measured relative to the standard hydrogen electrode (SHE). When two half-cells are connected to form a complete electrochemical cell, the difference in their electrode potentials is the cell potential, $E_{\text{cell}}$ . To ensure that these potentials are comparable and reproducible, they are defined under a strict set of 'standard conditions'. Any deviation from these conditions will alter the measured potential, as described by the Nernst equation.

Understanding the Question

The question asks for a precise definition of $E^\ominus_{\text{cell}}$ and a list of the standard conditions that must be met for this value to be valid. The command word 'define' requires a concise, textbook-accurate statement, and the additional instruction 'Include a description of standard conditions' means the three specific parameters (concentration, pressure, temperature) must be stated.

Approach

Recall the formal IUPAC definition of standard cell potential. It is not just a voltage; it is the potential difference between two half-cells under standard states. Then, list the three quantitative values that constitute standard conditions.

Step-by-Step Reasoning

M1 (Definition): The cell potential is the potential difference (also acceptable: voltage or EMF) measured between the two half-cells (or two electrodes) that make up the cell. It is crucial to specify 'between two half-cells' rather than just 'voltage', as voltage is a general term.
M2 (Conditions): The three standard conditions are:
1. All aqueous solutions have a concentration of $1 \text{ mol dm}^{-3}$ .
2. All gases involved are at a pressure of $1 \text{ atm}$ (or $101 \text{ kPa}$ ).
3. The temperature is $298 \text{ K}$ (which is $25 \text{ }^\circ\text{C}$ ).
  All three must be stated to gain the mark.

Key Takeaways

A standard potential is always a relative measurement between two half-cells, and it is only valid when the system is at standard state: $1 \text{ mol dm}^{-3}$ for solutes, $1 \text{ atm}$ for gases, and $298 \text{ K}$ for temperature.

Common Mistakes

Stating 'voltage' without specifying it is the potential difference between two half-cells.
Forgetting one of the standard conditions, particularly pressure (which is often overlooked if no gases are involved in the specific half-cell, but is part of the general definition).
Writing $100 \text{ kPa}$ instead of $101 \text{ kPa}$ (though $1 \text{ bar}$ is now technically the IUPAC standard, CIE accepts $1 \text{ atm}$ or $101 \text{ kPa}$ ).

Things to Be Careful About

Ensure all three conditions (concentration, pressure, temperature) are explicitly listed. The temperature must be $298 \text{ K}$ or $25 \text{ }^\circ\text{C}$ ; $273 \text{ K}$ is incorrect as that is standard temperature for STP gas volume calculations, not standard electrode potentials.

Techniques used

define standard cell potentialstate standard conditions

(b)

The Daniell cell is an electrochemical cell consisting of a $\text{Cu}^{2+}\text{(aq)/Cu(s)}$ electrode and a $\text{Zn}^{2+}\text{(aq)/Zn(s)}$ electrode.

6M

(i)

Draw a labelled diagram of this electrochemical cell.

Include all necessary substances and relevant pieces of apparatus needed to measure the $E^\ominus_{\text{cell}}$ .

It is not necessary to state the conditions used.

3M Similar

DifficultyMedium-Easy

Worked solution

Answer

The diagram must show:

Two separate beakers (or half-cells).
Left half-cell: a copper electrode, $\text{Cu(s)}$ , dipping into a copper(II) solution, $\text{Cu}^{2+}(\text{aq})$ .
Right half-cell: a zinc electrode, $\text{Zn(s)}$ , dipping into a zinc(II) solution, $\text{Zn}^{2+}(\text{aq})$ .
A salt bridge (e.g., filter paper soaked in $\text{KNO}_3$ or a U-tube) connecting the two solutions, labelled 'salt bridge'.
Wires connecting each metal electrode to a voltmeter (labelled 'V' or 'voltmeter').
The liquid levels in both beakers and the salt bridge must be shown to indicate a complete circuit.

Final answer

See diagram.

Detailed explanation

Background Concept

A Daniell cell is a classic galvanic (voltaic) cell that converts chemical energy into electrical energy using the spontaneous redox reaction between zinc and copper(II) ions. To measure the cell potential without allowing the two solutions to mix directly (which would cause a short circuit and direct reaction), the cell is divided into two half-cells. Each half-cell contains a metal electrode immersed in a solution of its own ions. The two half-cells are connected electrically by a salt bridge and externally by a high-resistance voltmeter.

Understanding the Question

The question asks for a labelled diagram of a Daniell cell capable of measuring $E^\ominus_{\text{cell}}$ . This requires drawing the two half-cells, the connecting apparatus (salt bridge and external circuit), and correctly labelling all chemical species and apparatus.

Approach

Draw two separate containers. Place the appropriate metal and ion in each. Connect them with a salt bridge and a voltmeter with wires. Ensure all labels match the mark scheme requirements.

Step-by-Step Reasoning

M1 (Apparatus and circuit): Must include a salt bridge (labelled), a voltmeter (labelled 'V' or 'voltmeter'), and wires connecting the electrodes to the voltmeter. The liquid levels must be shown to indicate the salt bridge is immersed and the circuit is complete.
M2 (Copper half-cell): Must show $\text{Cu(s)}$ (the electrode) and $\text{Cu}^{2+}(\text{aq})$ (the solution, typically $\text{CuSO}_4$ or $\text{Cu(NO}_3)_2$ ).
M3 (Zinc half-cell): Must show $\text{Zn(s)}$ (the electrode) and $\text{Zn}^{2+}(\text{aq})$ (the solution, typically $\text{ZnSO}_4$ ).

Key Takeaways

A complete electrochemical cell diagram requires two half-cells, a salt bridge to maintain electrical neutrality, and an external circuit with a voltmeter. All chemical species must be labelled with their state symbols where appropriate.

Common Mistakes

Forgetting to label the salt bridge or the voltmeter.
Drawing a single beaker containing both ions (this would cause direct reaction, not a measurable cell potential).
Omitting state symbols for the aqueous ions, though the mark scheme primarily looks for the chemical formulas.
Forgetting to show the liquid levels or the wires connecting to the electrodes.

Things to Be Careful About

The diagram must represent a complete circuit. The salt bridge must connect the two solutions, and the wires must connect the solid metals to the voltmeter. Do not draw the salt bridge connecting to the wires; it must connect the liquid solutions.

Techniques used

draw electrochemical cell diagramlabel cell components

(ii)

State the charge carriers that transfer current through the solutions and through the wire.

1M Similar

(iii)

The standard electrode potential, $E^\ominus$ , for the $\text{Zn}^{2+}\text{(aq)/Zn(s)}$ electrode is $-0.76\text{ V}$ .

Water is added to a standard $\text{Zn}^{2+}\text{(aq)/Zn(s)}$ electrode.

The new concentration of $\text{Zn}^{2+}\text{(aq)}$ is $0.25\text{ mol dm}^{-3}$ .

Use the Nernst equation to calculate the electrode potential, $E$ , for this new $\text{Zn}^{2+}\text{(aq)/Zn(s)}$ electrode.

2M Similar

DifficultyMedium

Worked solution

Working

The Nernst equation for the $\text{Zn}^{2+}(\text{aq})/\text{Zn}(\text{s})$ half-cell is:
$E = E^\ominus + \frac{0.059}{z} \log[\text{Zn}^{2+}]$
where $z = 2$ (the number of electrons transferred in the half-equation $\text{Zn}^{2+} + 2\text{e}^- \rightarrow \text{Zn}$ ).

Substitute the given values ( $E^\ominus = -0.76 \text{ V}$ , $[\text{Zn}^{2+}] = 0.25 \text{ mol dm}^{-3}$ ):
$E = -0.76 + \frac{0.059}{2} \log(0.25)$
$E = -0.76 + 0.0295 \times (-0.6021)$
$E = -0.76 - 0.01776$
$E = -0.77776 \text{ V}$

Rounding to an appropriate number of significant figures (matching the data, typically 2 or 3 s.f.):
$E \approx -0.78 \text{ V} \text{ (or } -0.778 \text{ V})$

Answer

$E = -0.78 \text{ V}$

Final answer

-0.78 V

Detailed explanation

Background Concept

The Nernst equation relates the electrode potential ( $E$ ) under non-standard conditions to the standard electrode potential ( $E^\ominus$ ), the temperature, and the concentrations (or partial pressures) of the species involved. For a reduction half-reaction at $298 \text{ K}$ , the equation is commonly written as:
$E = E^\ominus + \frac{0.059}{z} \log[\text{Ox}]$
where $z$ is the number of moles of electrons transferred in the balanced half-equation, and $[\text{Ox}]$ is the concentration of the oxidised species. For the $\text{Zn}^{2+}/\text{Zn}$ half-cell, the half-equation is $\text{Zn}^{2+}(\text{aq}) + 2\text{e}^- \rightleftharpoons \text{Zn}(\text{s})$ . Since $\text{Zn}(\text{s})$ is a solid, its activity is 1 and does not appear in the expression.

Alternatively, using natural logarithm and fundamental constants:
$E = E^\ominus + \frac{RT}{zF} \ln[\text{Zn}^{2+}]$
where $R = 8.31 \text{ J K}^{-1} \text{mol}^{-1}$ , $T = 298 \text{ K}$ , and $F = 96500 \text{ C mol}^{-1}$ .

Understanding the Question

We are given the standard potential ( $E^\ominus = -0.76 \text{ V}$ ) and a new, non-standard concentration ( $0.25 \text{ mol dm}^{-3}$ ). We must calculate the new electrode potential $E$ using the Nernst equation.

Approach

Write down the Nernst equation for this specific half-cell.
Identify $z$ (number of electrons = 2).
Substitute the values and calculate.

Step-by-Step Reasoning

M1 (Equation): State the Nernst equation correctly: $E = E^\ominus + \frac{0.059}{z} \log[\text{Zn}^{2+}]$ . (Note: $\log$ here is base 10. If using $\ln$ , the constant is $\frac{RT}{F} \approx 0.0257 \text{ V}$ , so the term is $\frac{0.0257}{z} \ln[\text{Zn}^{2+}]$ or $\frac{8.31 \times 298}{96500 \times 2} \ln(0.25)$ ).
M2 (Calculation): Substitute $E^\ominus = -0.76$ , $z = 2$ , and $[\text{Zn}^{2+}] = 0.25$ :
$E = -0.76 + \frac{0.059}{2} \log(0.25)$
$\log(0.25) \approx -0.602$
$E = -0.76 + (0.0295 \times -0.602) = -0.76 - 0.0178 = -0.7778 \text{ V}$
The result is approximately $-0.78 \text{ V}$ (to 2 s.f.) or $-0.778 \text{ V}$ (to 3 s.f.). The mark scheme accepts $-0.778 \text{ V}$ min 2 s.f.

Key Takeaways

Decreasing the concentration of the oxidised species ( $\text{Zn}^{2+}$ ) makes the log term negative, which makes $E$ more negative (less positive) than $E^\ominus$ . This makes sense: lower concentration means less driving force for reduction.
Always check the sign of the log term. $\log(0.25)$ is negative, so the potential decreases.

Common Mistakes

Using the wrong value for $z$ (e.g., $z=1$ instead of $z=2$ ).
Forgetting the negative sign in front of $0.76$ .
Calculating $\log(0.25)$ incorrectly (it is negative, not positive).
Using the wrong constant in the Nernst equation (e.g., using $0.059$ with $\ln$ instead of $\log$ ).

Things to Be Careful About

Ensure you use $\log_{10}$ when using the $0.059/z$ form, or $\ln$ when using the $RT/zF$ form.
Significant figures: the given $E^\ominus$ is to 2 decimal places ( $-0.76$ ), so the answer should ideally be given to 2 or 3 decimal places. $-0.78 \text{ V}$ or $-0.778 \text{ V}$ are both acceptable.

Techniques used

apply Nernst equationcalculate electrode potential

(c)

An electrochemical cell consists of a $\text{ZnO/Zn}$ electrode and a $\text{MnO}_2/\text{Mn}_2\text{O}_3$ electrode in an alkaline electrolyte.

The standard cell potential, $E^\ominus_{\text{cell}}$ , for this cell is $+1.47\text{ V}$ .

The half-equation at each electrode when this cell is discharging is shown.

\text{Zn} + 2\text{OH}^- \rightarrow \text{ZnO} + \text{H}_2\text{O} + 2\text{e}^-

2\text{MnO}_2 + \text{H}_2\text{O} + 2\text{e}^- \rightarrow \text{Mn}_2\text{O}_3 + 2\text{OH}^-

3M

(i)

Use this information to determine the change in oxidation state of manganese when this cell is discharging.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

In $\text{MnO}_2$ , oxygen is $-2$ , so Mn is $+4$ .
In $\text{Mn}_2\text{O}_3$ , oxygen is $-2$ ( $3 \times -2 = -6$ ), so $2 \times \text{Mn} = +6$ , meaning Mn is $+3$ .

The change in oxidation state of manganese is from $+4$ to $+3$ .

Final answer

+4 to +3

Detailed explanation

Background Concept

Oxidation states (or oxidation numbers) are used to track electron transfer in redox reactions. In oxides, oxygen typically has an oxidation state of $-2$ . The sum of oxidation states in a neutral compound is zero. Manganese is a transition metal and can exhibit multiple oxidation states, commonly $+2$ , $+3$ , $+4$ , $+6$ , and $+7$ .

Understanding the Question

We are given the half-equation for the cathode (reduction) during discharge: $2\text{MnO}_2 + \text{H}_2\text{O} + 2\text{e}^- \rightarrow \text{Mn}_2\text{O}_3 + 2\text{OH}^-$ . We need to find the oxidation state of Mn in the reactant ( $\text{MnO}_2$ ) and the product ( $\text{Mn}_2\text{O}_3$ ) and state the change.

Approach

Calculate the oxidation state of Mn in $\text{MnO}_2$ and $\text{Mn}_2\text{O}_3$ using the rule that O is $-2$ and the compound is neutral.

Step-by-Step Reasoning

In $\text{MnO}_2$ : Let oxidation state of Mn be $x$ . $x + 2(-2) = 0 \Rightarrow x = +4$ .
In $\text{Mn}_2\text{O}_3$ : Let oxidation state of Mn be $y$ . $2y + 3(-2) = 0 \Rightarrow 2y = +6 \Rightarrow y = +3$ .
The manganese is reduced from $+4$ to $+3$ (it gains electrons, which is consistent with it being on the left side of a reduction half-equation).

Key Takeaways

Transition metals often change oxidation states in redox reactions. Always calculate the oxidation state for each element in the reactant and product to determine the change.

Common Mistakes

Assuming the oxidation state of Mn is the same as its group number or a fixed value.
Forgetting that there are two Mn atoms in $\text{Mn}_2\text{O}_3$ and dividing the total charge by 2.

Things to Be Careful About

The question asks for the change, so stating '+4 to +3' or 'decreases by 1' is correct. Simply stating '+3' is incomplete.

Techniques used

calculate oxidation states in compoundsdetermine change in oxidation state

(ii)

Write the equation for the overall reaction that occurs when this cell is discharging.

1M Similar

DifficultyEasy

Worked solution

Answer

Add the two half-equations together, cancelling the electrons and any species that appear on both sides:

Anode (oxidation): $\text{Zn} + 2\text{OH}^- \rightarrow \text{ZnO} + \text{H}_2\text{O} + 2\text{e}^-$
Cathode (reduction): $2\text{MnO}_2 + \text{H}_2\text{O} + 2\text{e}^- \rightarrow \text{Mn}_2\text{O}_3 + 2\text{OH}^-$

Overall: $\text{Zn} + 2\text{MnO}_2 \rightarrow \text{ZnO} + \text{Mn}_2\text{O}_3$

(Note: $2\text{OH}^-$ and $\text{H}_2\text{O}$ cancel out from both sides.)

Final answer

Zn + 2MnO2 -> ZnO + Mn2O3

Detailed explanation

Background Concept

The overall reaction in an electrochemical cell is the sum of the oxidation half-reaction (at the anode) and the reduction half-reaction (at the cathode). When adding them, the electrons must cancel out completely, and any other species (like $\text{OH}^-$ or $\text{H}_2\text{O}$ ) that appear on both sides of the arrow should also be cancelled.

Understanding the Question

We are given the two half-equations for the cell discharging. We need to combine them to write the overall balanced equation.

Approach

Add the left-hand sides together and the right-hand sides together. Cancel the $2\text{e}^-$ , the $2\text{OH}^-$ , and the $\text{H}_2\text{O}$ that appear on both sides.

Step-by-Step Reasoning

Reactants: $\text{Zn} + 2\text{OH}^- + 2\text{MnO}_2 + \text{H}_2\text{O} + 2\text{e}^-$
Products: $\text{ZnO} + \text{H}_2\text{O} + 2\text{e}^- + \text{Mn}_2\text{O}_3 + 2\text{OH}^-$
Cancel $2\text{e}^-$ , $2\text{OH}^-$ , and $\text{H}_2\text{O}$ from both sides.
Remaining: $\text{Zn} + 2\text{MnO}_2 \rightarrow \text{ZnO} + \text{Mn}_2\text{O}_3$

Key Takeaways

When combining half-equations, always check that electrons cancel and that spectator ions/molecules (like water or hydroxide in alkaline cells) are cancelled if they appear on both sides.

Common Mistakes

Forgetting to cancel the $\text{H}_2\text{O}$ and $\text{OH}^-$ .
Not balancing the electrons before adding (here, both have $2\text{e}^-$ , so they add directly).

Things to Be Careful About

The final equation should not contain any electrons. Check that the atoms and charges are balanced.

Techniques used

combine half-equations for overall cell reaction

(iii)

The $E^\ominus$ for the $\text{ZnO/Zn}$ electrode is $-1.28\text{ V}$ .

Calculate the standard electrode potential, $E^\ominus$ , for the $\text{MnO}_2/\text{Mn}_2\text{O}_3$ electrode.

1M Similar

DifficultyMedium-Easy

Worked solution

Working

The standard cell potential is given by:
$E^\ominus_{\text{cell}} = E^\ominus_{\text{cathode}} - E^\ominus_{\text{anode}}$

From the half-equations:

Oxidation (anode): $\text{Zn} + 2\text{OH}^- \rightarrow \text{ZnO} + \text{H}_2\text{O} + 2\text{e}^-$ (so $E^\ominus_{\text{anode}} = -1.28 \text{ V}$ )
Reduction (cathode): $2\text{MnO}_2 + \text{H}_2\text{O} + 2\text{e}^- \rightarrow \text{Mn}_2\text{O}_3 + 2\text{OH}^-$ (let this be $E^\ominus_{\text{cathode}}$ )

Given $E^\ominus_{\text{cell}} = +1.47 \text{ V}$ :
$1.47 = E^\ominus_{\text{cathode}} - (-1.28)$
$1.47 = E^\ominus_{\text{cathode}} + 1.28$
$E^\ominus_{\text{cathode}} = 1.47 - 1.28 = +0.19 \text{ V}$

Answer

$E^\ominus(\text{MnO}_2 / \text{Mn}_2\text{O}_3) = +0.19 \text{ V}$

Final answer

+0.19 V

Detailed explanation

Background Concept

The standard cell potential is the difference between the standard electrode potentials of the cathode (reduction) and the anode (oxidation):
$E^\ominus_{\text{cell}} = E^\ominus_{\text{reduction}} - E^\ominus_{\text{oxidation}}$
(or $E^\ominus_{\text{cathode}} - E^\ominus_{\text{anode}}$ ). Both values must be standard reduction potentials. If a half-equation is given as an oxidation (losing electrons), its standard reduction potential is still the value given in data booklets, but we use the formula above which handles the sign automatically.

Understanding the Question

We are given $E^\ominus_{\text{cell}} = +1.47 \text{ V}$ and $E^\ominus(\text{ZnO}/\text{Zn}) = -1.28 \text{ V}$ . We need to find $E^\ominus(\text{MnO}_2/\text{Mn}_2\text{O}_3)$ . We must first identify which half-cell is the cathode and which is the anode.

Approach

Identify the oxidation and reduction half-reactions from the given equations.
Assign them to anode and cathode.
Use the cell potential formula to solve for the unknown.

Step-by-Step Reasoning

The Zn half-equation shows loss of electrons (oxidation), so it is the anode. $E^\ominus_{\text{anode}} = -1.28 \text{ V}$ .
The Mn half-equation shows gain of electrons (reduction), so it is the cathode. Let $E^\ominus_{\text{cathode}} = x$ .
Apply the formula:
$E^\ominus_{\text{cell}} = E^\ominus_{\text{cathode}} - E^\ominus_{\text{anode}}$
$1.47 = x - (-1.28)$
$1.47 = x + 1.28$
$x = 1.47 - 1.28 = +0.19 \text{ V}$

Key Takeaways

Always use the formula $E^\ominus_{\text{cell}} = E^\ominus_{\text{cathode}} - E^\ominus_{\text{anode}}$ with standard reduction potentials. Do not change the sign of the anode potential before subtracting; the formula handles it.

Common Mistakes

Using $E^\ominus_{\text{cell}} = E^\ominus_{\text{oxidation}} + E^\ominus_{\text{reduction}}$ and getting the signs wrong.
Forgetting that the anode potential is subtracted, leading to $1.47 = x + 1.28 \Rightarrow x = 2.75 \text{ V}$ (incorrect).
Misidentifying the anode and cathode.

Things to Be Careful About

Ensure the final answer has the correct sign. $+0.19 \text{ V}$ is positive, which is consistent with a spontaneous cell ( $E^\ominus_{\text{cell}} > 0$ ).

Techniques used

calculate cell potential from electrode potentialsidentify cathode and anode from half-equations

Q5MediumTransition Elements

(a)

Copper shows typical properties of transition elements, including its behaviour as a catalyst.

Complete Table 5.1 to show the total number of unpaired electrons in the 3d and 4s orbitals of an isolated gaseous Cu atom and a $\text{Cu}^{2+}$ ion.

species	3d	4s
Cu
$\text{Cu}^{2+}$

1M Similar

DifficultyEasy

Worked solution

Answer

species	3d	4s
Cu	0	1
$\text{Cu}^{2+}$	1	0

Final answer

Cu: 3d=0, 4s=1; Cu $^{2+}$ : 3d=1, 4s=0

Detailed explanation

Background Concept

The electron configuration of transition elements follows the Aufbau principle, but there are notable exceptions in the first transition series due to the small energy difference between the 4s and 3d orbitals. For copper (Z=29), the expected configuration is $[\text{Ar}] 3\text{d}^9 4\text{s}^2$ , but the actual ground-state configuration of an isolated gaseous Cu atom is $[\text{Ar}] 3\text{d}^{10} 4\text{s}^1$ . This occurs because a completely filled 3d subshell and a half-filled 4s subshell provide extra stability (exchange energy and reduced electron-electron repulsion).

When a transition metal forms positive ions, electrons are removed from the 4s orbital before the 3d orbitals, because the 4s electrons are further from the nucleus on average and are higher in energy than the 3d electrons once the 3d subshell begins to fill.

Understanding the Question

The question asks for the number of unpaired electrons in the 3d and 4s orbitals for two species: an isolated gaseous Cu atom and a $\text{Cu}^{2+}$ ion. We must write their electron configurations, identify the occupancy of the 3d and 4s subshells, and count the unpaired electrons using Hund's rule (each orbital in a subshell is singly occupied before pairing occurs).

Approach

Write the full or abbreviated electron configuration for the neutral Cu atom.
Identify the 3d and 4s occupancies and count unpaired electrons.
Write the electron configuration for the $\text{Cu}^{2+}$ ion by removing two electrons from the Cu atom (first from 4s, then from 3d).
Identify the 3d and 4s occupancies and count unpaired electrons for the ion.

Step-by-Step Reasoning

Neutral Cu atom: The atomic number is 29. The noble gas core is Argon (18 electrons). The remaining 11 electrons go into 4s and 3d. Due to the stability of a full d-subshell, the configuration is $[\text{Ar}] 3\text{d}^{10} 4\text{s}^1$ .
- The 3d subshell has 5 orbitals, fully occupied with 10 electrons (all paired). Unpaired electrons in 3d = 0.
- The 4s subshell has 1 electron. Unpaired electrons in 4s = 1.
$\text{Cu}^{2+}$ ion: To form a 2+ ion, remove 2 electrons from the neutral atom. Remove the two 4s electrons first (but there is only one, so remove the 4s $^1$ electron and one 3d electron). The configuration becomes $[\text{Ar}] 3\text{d}^9$ .
- The 3d subshell has 9 electrons. According to Hund's rule, 5 orbitals get one electron each (5 unpaired), and the remaining 4 electrons pair up in 4 of the orbitals. This leaves 1 unpaired electron in the 3d subshell.
- The 4s subshell is now empty. Unpaired electrons in 4s = 0.

Key Takeaways

Copper is an exception to the Aufbau principle: $3\text{d}^{10} 4\text{s}^1$ rather than $3\text{d}^9 4\text{s}^2$ .
When forming cations, transition metals lose their 4s electrons before their 3d electrons.
A $3\text{d}^9$ configuration has exactly one unpaired electron.

Common Mistakes

Writing the Cu configuration as $3\text{d}^9 4\text{s}^2$ and then getting the unpaired electron counts wrong for both.
Removing electrons from the 3d subshell before the 4s subshell when forming $\text{Cu}^{2+}$ .
Miscounting unpaired electrons in a $3\text{d}^9$ configuration (it is 1, not 3 or 5).

Things to Be Careful About

Ensure you are counting unpaired electrons in separate columns for 3d and 4s, not the total. The table specifically asks for 3d and 4s separately.

Techniques used

write the electron configuration of atoms and ionscount unpaired electrons in d and s orbitals

(b)

The 3d orbitals in an isolated $\text{Cu}^{2+}$ ion are degenerate.

Complete the diagram to show the relative energies of the 3d orbitals in an isolated $\text{Cu}^{2+}$ ion and in $\text{Cu}^{2+}$ in a tetrahedral complex.

Fig. 5.1

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Isolated $\text{Cu}^{2+}$ ion (left side):
Draw five horizontal lines (or boxes) at the same energy level, representing the five degenerate 3d orbitals.

$\text{Cu}^{2+}$ in a tetrahedral complex (right side):
Draw two horizontal lines (or boxes) at a lower energy level, and three horizontal lines (or boxes) at a higher energy level. Ensure that all five lines on the right side are positioned at a higher energy level than the single line on the left side.

Final answer

Left: 5 degenerate lines. Right: 2 lower lines, 3 higher lines, all above the left side level.

Detailed explanation

Background Concept

In an isolated gaseous transition metal ion, the five 3d orbitals are degenerate (have the same energy). When ligands approach the central metal ion to form a complex, the electrostatic repulsion between the lone pairs on the ligands and the electrons in the d-orbitals causes the d-orbitals to split into sets of different energies. This is called d-orbital splitting.

The pattern of splitting depends on the geometry of the complex:

Octahedral: The orbitals split into a lower-energy $t_{2g}$ set (3 orbitals: $d_{xy}, d_{xz}, d_{yz}$ ) and a higher-energy $e_g$ set (2 orbitals: $d_{x^2-y^2}, d_{z^2}$ ). The $e_g$ orbitals point directly at the ligands, so they are higher in energy.
Tetrahedral: The splitting is reversed and smaller. The orbitals split into a lower-energy $e$ set (2 orbitals: $d_{z^2}, d_{x^2-y^2}$ ) and a higher-energy $t_2$ set (3 orbitals: $d_{xy}, d_{xz}, d_{yz}$ ). Because the ligands do not point directly at any of the d-orbitals in a tetrahedral arrangement, the overall repulsion is less than in an octahedral complex, but the average energy of all five d-orbitals in the complex is still higher than that of the isolated ion (this is a consequence of the barycentric rule: the center of gravity of the split levels must equal the original degenerate level plus the ligand-field stabilization energy contribution; practically, all split levels in a complex are higher than the isolated ion's level).

Understanding the Question

The question provides a blank energy level diagram with two sections: one for an isolated $\text{Cu}^{2+}$ ion and one for $\text{Cu}^{2+}$ in a tetrahedral complex. We must draw the correct number of energy levels (lines/boxes) in each section, show the correct relative splitting for a tetrahedral field, and ensure the complex levels are higher in energy than the isolated ion levels.

Approach

Draw five degenerate lines for the isolated ion on the left.
For the tetrahedral complex on the right, draw two lower lines ( $e$ set) and three higher lines ( $t_2$ set).
Shift the entire right-hand set upwards so that both the lower and upper lines are above the isolated ion's line.

Step-by-Step Reasoning

Isolated ion: Five 3d orbitals are degenerate. Draw five horizontal lines at the same height on the left side of the dashed line.
Tetrahedral complex: The d-orbitals split into two sets. The $e$ set (2 orbitals) is lower in energy, and the $t_2$ set (3 orbitals) is higher in energy. Draw two lines below and three lines above.
Relative energy: The electrostatic interaction with ligands raises the energy of the d-orbitals overall. Therefore, the lowest energy level in the tetrahedral complex must still be higher than the degenerate level of the isolated ion. Draw all five lines on the right side above the line on the left side.

Key Takeaways

Tetrahedral splitting: 2 lower ( $e$ ), 3 higher ( $t_2$ ).
All d-orbital energy levels in a complex are higher than those in the isolated gaseous ion.

Common Mistakes

Drawing the octahedral splitting pattern (3 lower, 2 higher) instead of tetrahedral (2 lower, 3 higher).
Placing the lower $e$ orbitals in the tetrahedral complex below the energy level of the isolated ion. (The splitting is relative to the new average, but the new average is higher than the isolated ion's level).

Things to Be Careful About

The mark scheme specifically requires that the energy of all five d orbitals in the complex is higher than all d orbitals in the isolated ion. Do not let the lower $e$ level dip below the isolated ion's line.

Techniques used

draw d-orbital splitting diagrams for tetrahedral complexescompare energy levels of isolated ions and complexes

(c)

Explain why transition elements behave as catalysts.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Transition elements can exist in more than one stable oxidation state (variable oxidation states).
They have vacant (or empty) d orbitals that are energetically accessible to form dative (coordinate) bonds with ligands (or to accept electron pairs from reactants).

These properties allow them to provide an alternative reaction pathway with a lower activation energy by forming intermediate complexes with reactants.

Final answer

Variable oxidation states and vacant d orbitals available to form dative bonds.

Detailed explanation

Background Concept

Catalysts increase the rate of a chemical reaction by providing an alternative reaction pathway with a lower activation energy, without being consumed in the overall reaction. Transition metals are excellent catalysts (both homogeneous and heterogeneous) due to their unique electronic structure.

Two key features enable this:

Variable oxidation states: Transition metals can easily gain or lose electrons from their d orbitals, allowing them to change oxidation states during a reaction. This enables them to participate in redox reactions by accepting electrons from one reactant and donating them to another.
Vacant d orbitals: Transition metals have partially filled or empty d orbitals that can accept lone pairs of electrons from ligands (reactants). This allows them to form temporary intermediate complexes (adsorption in heterogeneous catalysis, or coordination complexes in homogeneous catalysis), weakening the bonds within the reactant molecules and facilitating the reaction.

Understanding the Question

The question asks for a 2-mark explanation of why transition elements behave as catalysts. We need to provide two distinct, creditable points that link their electronic properties to their catalytic function.

Approach

Identify the two main electronic features of transition metals that enable catalysis: variable oxidation states and vacant d orbitals. State these clearly and link them to their function (forming intermediates or accepting/donating electrons).

Step-by-Step Reasoning

Point 1 (M1): State that transition elements exhibit more than one stable oxidation state (or variable oxidation states). This allows them to act as redox agents in intermediate steps.
Point 2 (M2): State that they have vacant (or empty) d orbitals that are energetically accessible. Explain that these can form dative (coordinate) bonds with ligands (or reactants), allowing the formation of intermediate complexes that lower the activation energy.

Key Takeaways

Catalysis in transition metals relies on variable oxidation states and accessible vacant d orbitals.
These features allow the metal to form intermediate complexes with reactants.

Common Mistakes

Saying 'they have low activation energy' (the catalyst provides a pathway with lower activation energy, it doesn't 'have' it intrinsically in this context).
Stating only one point (e.g., only variable oxidation states) and losing a mark.
Using vague terms like 'they can react easily' without specifying the electronic reason.

Things to Be Careful About

Ensure both points are distinct. 'Variable oxidation states' and 'vacant d orbitals' are the two standard creditable points.
Use precise terminology: 'dative bonds', 'ligands', 'energetically accessible'.

Techniques used

explain catalytic activity of transition metalslink variable oxidation states to catalyst function

(d)

$\text{CN}^-$ is a monodentate ligand.

Table 5.2 shows information about two complex ions that contain only $\text{CN}^-$ ions as ligands.

Complete Table 5.2.

metal ion	coordination number	formula of complex ion	charge of complex ion
$\text{Ag}^+$	2
$\text{Fe}^{2+}$			4–

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

metal ion	coordination number	formula of complex ion	charge of complex ion
$\text{Ag}^+$	2	$[\text{Ag(CN)}_2]^-$	$1-$
$\text{Fe}^{2+}$	6	$[\text{Fe(CN)}_6]^{4-}$	$4-$

Final answer

Ag+: [Ag(CN)2]-, charge 1-; Fe2+: coordination number 6, [Fe(CN)6]4-, charge 4-

Detailed explanation

Background Concept

A complex ion consists of a central metal ion surrounded by ligands. Ligands are ions or molecules that donate a lone pair of electrons to the central metal ion to form a dative (coordinate) bond.

Coordination number: The number of dative bonds formed between the ligands and the central metal ion. For monodentate ligands (like $\text{CN}^-$ ), the coordination number equals the number of ligands attached.
Formula of complex ion: Written as $[\text{M(L)}_n]^{z\pm}$ , where M is the metal, L is the ligand, n is the number of ligands, and $z\pm$ is the overall charge.
Overall charge: The sum of the charge of the central metal ion and the charges of all the ligands. $\text{Charge} = (\text{metal ion charge}) + n \times (\text{ligand charge})$ .

Understanding the Question

We are given two rows to complete. Both use $\text{CN}^-$ (cyanide, charge $1-$ ) as a monodentate ligand. We need to find the missing coordination number, formula, or charge for $\text{Ag}^+$ and $\text{Fe}^{2+}$ complexes.

Approach

Row 1 ( $\text{Ag}^+$ ): Coordination number is given as 2. Calculate the formula and overall charge.
Row 2 ( $\text{Fe}^{2+}$ ): Overall charge is given as $4-$ . Use the charge balance equation to find the number of ligands (coordination number), then write the formula.

Step-by-Step Reasoning

Row 1 ( $\text{Ag}^+$ ):
- Coordination number = 2, so there are 2 $\text{CN}^-$ ligands.
- Formula: $[\text{Ag(CN)}_2]$
- Charge calculation: $(+1) + 2 \times (-1) = +1 - 2 = -1$ . So the charge is $1-$ .
- Final: $[\text{Ag(CN)}_2]^-$ , charge $1-$ .
Row 2 ( $\text{Fe}^{2+}$ ):
- Metal ion charge = $+2$ . Ligand charge = $-1$ . Overall complex charge = $-4$ .
- Let $n$ be the number of ligands (coordination number).
- Equation: $(+2) + n(-1) = -4 \Rightarrow 2 - n = -4 \Rightarrow n = 6$ .
- Coordination number = 6.
- Formula: $[\text{Fe(CN)}_6]^{4-}$ .
- Final: coordination number 6, formula $[\text{Fe(CN)}_6]^{4-}$ , charge $4-$ .

Key Takeaways

For monodentate ligands, coordination number = number of ligands.
Overall complex charge = metal charge + (number of ligands × ligand charge).
You can work backwards from the overall charge to find the coordination number.

Common Mistakes

Forgetting to include the overall charge in the formula box (e.g., writing $\text{Ag(CN)}_2$ instead of $[\text{Ag(CN)}_2]^-$ ).
Calculating the charge incorrectly: $+1 + (-1) = 0$ instead of $+1 + 2(-1) = -1$ .
Assuming the coordination number for Fe is always 6 without checking (though it is common, it must be derived from the given charge here).

Things to Be Careful About

The question asks for the 'charge of complex ion', so write ' $1-$ ' or ' $4-$ ', not just the number.
Ensure the formula includes square brackets if required by your convention, though the mark scheme often accepts without if the charge is clear. Here, standard notation $[\text{Ag(CN)}_2]^-$ is best.

Techniques used

determine coordination number from ligand countcalculate complex ion charge from central metal and ligand chargeswrite complex ion formulae

(e)

The complex ion $[\text{Au(CN)}_2\text{Br}_2]^-$ displays geometrical (cis/trans) isomerism.

Draw the structure of trans- $[\text{Au(CN)}_2\text{Br}_2]^-$ . State its shape and the Br-Au-Br bond angle.

2M Similar

DifficultyMedium

Worked solution

Answer

Structure of trans- $[\text{Au(CN)}_2\text{Br}_2]^-$ :
Draw a square planar structure with Au in the center. Place the two Br ligands opposite each other (e.g., top and bottom) and the two CN ligands opposite each other (e.g., left and right). Alternatively, draw a 3D representation with wedges and dashes showing trans arrangement.

Shape: Square planar

Br-Au-Br bond angle: $180^\circ$

Final answer

Trans isomer drawn (Br opposite Br). Shape: square planar. Bond angle: 180°.

Detailed explanation

Background Concept

Geometrical (cis-trans) isomerism occurs in complex ions with coordination number 4 (square planar) or 6 (octahedral) when there are at least two different types of monodentate ligands.

Square planar complexes ( $MA_2B_2$ ): Can exist as cis (identical ligands adjacent, $90^\circ$ apart) or trans (identical ligands opposite, $180^\circ$ apart).
Bond angles: In a perfect square planar geometry, all bond angles are $90^\circ$ , but the angle between opposite ligands (e.g., Br-Au-Br in the trans isomer) is $180^\circ$ .
Gold(I) complexes: $\text{Au}^+$ has a $[\text{Kr}] 4\text{d}^{10}$ configuration. With a coordination number of 4 and $d^{10}$ , there is no crystal field stabilization energy preference for octahedral, and $sp^3$ or $dsp^2$ hybridization leads to square planar geometry (typically $dsp^2$ for $d^8$ or $d^{10}$ with strong field ligands, though Au(I) is often simply described as square planar due to $sp^2d$ or relativistic effects, but for A-Level, coordination number 4 with no unpaired electrons or specific ligand fields yielding square planar is the key). Actually, $[\text{Au(CN)}_2\text{Br}_2]^-$ has Au in +3 oxidation state here (2x-1 + 2x-1 = -4, overall -1, so Au is +3). Au(III) is $d^8$ , which strongly favors square planar geometry.

Understanding the Question

We are given the complex ion $[\text{Au(CN)}_2\text{Br}_2]^-$ . We must draw the trans isomer, state its shape, and give the Br-Au-Br bond angle.

Approach

Determine the coordination number: 2 CN + 2 Br = 4.
Recognize that $MA_2B_2$ with coordination number 4 and a $d^8$ metal (Au $^{3+}$ ) is square planar.
Draw the trans isomer: identical ligands (Br) opposite each other.
State the shape (square planar) and the angle between opposite ligands ( $180^\circ$ ).

Step-by-Step Reasoning

Coordination number and shape: There are 4 ligands. For Au(III) ( $d^8$ ), a coordination number of 4 results in a square planar geometry.
Trans isomer drawing: In the trans isomer, the two Br atoms are $180^\circ$ apart. Draw Au in the center, with Br at the top and bottom, and CN at the left and right. (Or use 3D wedges/dashes: Br top-left solid, Br bottom-right dashed, CN top-right dashed, CN bottom-left solid - wait, trans means opposite. Top and bottom are opposite. Left and right are opposite.)
Bond angle: The angle between the top Br, the central Au, and the bottom Br is $180^\circ$ .

Key Takeaways

$MA_2B_2$ square planar complexes show cis-trans isomerism.
Trans means identical ligands are opposite ( $180^\circ$ ).
Cis means identical ligands are adjacent ( $90^\circ$ ).

Common Mistakes

Drawing an octahedral or tetrahedral shape for a 4-coordinate complex (tetrahedral doesn't show cis-trans isomerism for $MA_2B_2$ ).
Drawing the cis isomer instead of trans (Br atoms adjacent at $90^\circ$ ).
Stating the bond angle as $90^\circ$ (this is the angle between Br and CN, not Br and Br in the trans isomer).
Forgetting to draw the Au atom in the center or omitting bonds.

Things to Be Careful About

The question specifically asks for the trans isomer. Ensure Br is opposite Br.
The bond angle asked for is Br-Au-Br. In trans, this is $180^\circ$ . In cis, it would be $90^\circ$ .
Use clear lines for bonds. 2D cross representation is usually acceptable for square planar.

Techniques used

draw geometrical isomers of square planar complexesidentify shape and bond angles from coordination number and ligand types

(f)

An impure sample of a vanadium(V) compound of mass $0.250\text{ g}$ is dissolved in aqueous acid. This solution contains $\text{VO}_3^-$ ions.

An excess of zinc is added to this solution. All the $\text{VO}_3^-$ ions are reduced to $\text{V}^{2+}$ ions and Zn atoms are oxidised to $\text{Zn}^{2+}$ ions.

The unreacted zinc is removed and the resulting solution is titrated with acidified $\text{MnO}_4^-$ .

The end-point is reached when $22.5\text{ cm}^3$ of $0.0750\text{ mol dm}^{-3}\text{ MnO}_4^-$ is added.

A redox reaction takes place and all the $\text{V}^{2+}$ reacts forming $\text{VO}_3^-$ .

3\text{MnO}_4^- + 5\text{V}^{2+} + 3\text{H}_2\text{O} \rightarrow 3\text{Mn}^{2+} + 5\text{VO}_3^- + 6\text{H}^+

5M

(i)

Calculate the percentage by mass of vanadium in the $0.250\text{ g}$ of impure sample. Assume the impurities do not contain any vanadium ions.

Show your working.

3M Similar

DifficultyMedium

Worked solution

Working

Moles of $\text{MnO}_4^-$ :

\text{moles} = \text{concentration} \times \text{volume} = 0.0750 \times \frac{22.5}{1000} = 1.6875 \times 10^{-3} \text{ mol}

From the balanced equation, the ratio of $\text{MnO}_4^-$ to $\text{V}^{2+}$ is $3 : 5$ .

Moles of $\text{V}^{2+}$ :

\text{moles } \text{V}^{2+} = \frac{5}{3} \times \text{moles } \text{MnO}_4^- = \frac{5}{3} \times 1.6875 \times 10^{-3} = 2.8125 \times 10^{-3} \text{ mol}

All $\text{V}^{2+}$ came from the vanadium in the sample, so moles of V = $2.8125 \times 10^{-3}$ mol.

Mass of V:

\text{mass} = \text{moles} \times A_r = 2.8125 \times 10^{-3} \times 50.9 = 0.143156 \text{ g}

Percentage by mass of V:

\% \text{V} = \frac{0.143156}{0.250} \times 100 = 57.26\% \approx 57.3\%

Answer

$57.3\%$ (or $57\%$ to 2 s.f.)

Final answer

57.3%

Detailed explanation

Background Concept

Redox titrations involve using a solution of known concentration (titrant) to determine the concentration or amount of an unknown substance (analyte) through a redox reaction. The key steps are:

Calculate moles of the titrant used.
Use the stoichiometric ratio from the balanced chemical equation to find moles of the analyte.
Convert moles of analyte to mass or concentration as required.

In this problem, vanadium is reduced to $\text{V}^{2+}$ , and then titrated with permanganate ( $\text{MnO}_4^-$ ). The balanced equation is provided:
$3\text{MnO}_4^- + 5\text{V}^{2+} + 3\text{H}_2\text{O} \rightarrow 3\text{Mn}^{2+} + 5\text{VO}_3^- + 6\text{H}^+$

Understanding the Question

We have a $0.250\text{ g}$ impure sample containing vanadium. All vanadium is converted to $\text{V}^{2+}$ and titrated with $0.0750\text{ mol dm}^{-3}$ $\text{MnO}_4^-$ . The volume used is $22.5\text{ cm}^3$ . We need to calculate the percentage by mass of vanadium in the original sample.

Approach

Calculate moles of $\text{MnO}_4^-$ from concentration and volume (convert cm $^3$ to dm $^3$ ).
Use the 3:5 mole ratio from the equation to find moles of $\text{V}^{2+}$ (which equals moles of V in the sample).
Calculate the mass of V using its relative atomic mass ( $A_r = 50.9$ ).
Calculate the percentage mass: $(\text{mass of V} / \text{mass of sample}) \times 100$ .

Step-by-Step Reasoning

Moles of $\text{MnO}_4^-$ :
$\text{Volume} = 22.5\text{ cm}^3 = 0.0225\text{ dm}^3$
$\text{Moles} = 0.0750 \times 0.0225 = 1.6875 \times 10^{-3}\text{ mol}$
Moles of $\text{V}^{2+}$ :
From the equation, 3 moles of $\text{MnO}_4^-$ react with 5 moles of $\text{V}^{2+}$ .
$\text{Moles } \text{V}^{2+} = \frac{5}{3} \times 1.6875 \times 10^{-3} = 2.8125 \times 10^{-3}\text{ mol}$
Mass of V:
$\text{Mass} = 2.8125 \times 10^{-3} \times 50.9 = 0.14315625\text{ g}$
Percentage mass:
$\% = \frac{0.14315625}{0.250} \times 100 = 57.2625\%$
Rounding to 3 significant figures (or 2, as $57\%$ is also acceptable per mark scheme): $57.3\%$ .

Key Takeaways

Always convert volume to dm $^3$ when using concentration in mol dm $^{-3}$ .
Carefully read the stoichiometric ratio from the balanced equation (it is not always 1:1).
The moles of titrant reactant determine the moles of analyte, which directly gives the mass of the element in the sample.

Common Mistakes

Forgetting to divide the volume by 1000 to convert cm $^3$ to dm $^3$ .
Using the wrong mole ratio (e.g., 5:3 instead of 5/3).
Calculating the percentage using the mass of the complex ion instead of the mass of the element V.
Significant figures: The mark scheme accepts 2 s.f. ( $57\%$ ) or 3 s.f. ( $57.3\%$ ). Using too many (e.g., 57.2625%) might be penalized or ignored depending on the examiner, but 3 s.f. is safe.

Things to Be Careful About

Ensure you use the correct $A_r$ for Vanadium (50.9).
The question states 'Assume the impurities do not contain any vanadium ions', so all V $^{2+}$ comes from the V in the sample.

Techniques used

calculate moles from concentration and volumeuse stoichiometric ratio from balanced equationcalculate mass and percentage composition

(ii)

Complete the equation for the reaction between acidified $\text{VO}_3^-$ ions and Zn metal.

\dots\dots \text{VO}_3^- + \dots\dots \text{Zn} + \dots\dots\dots\dots \rightarrow \dots\dots \text{V}^{2+} + \dots\dots \text{Zn}^{2+} + \dots\dots\dots\dots

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

2\text{VO}_3^- + 3\text{Zn} + 12\text{H}^+ \rightarrow 2\text{V}^{2+} + 3\text{Zn}^{2+} + 6\text{H}_2\text{O}

Final answer

2VO3- + 3Zn + 12H+ -> 2V2+ + 3Zn2+ + 6H2O

Detailed explanation

Background Concept

Balancing redox equations in acidic conditions involves ensuring that both the atoms and the charges are balanced. A reliable method is to use oxidation numbers or to write and balance half-equations.

Oxidation number method: Determine the change in oxidation number for the oxidized and reduced species. Multiply the species by coefficients to equalize the total electron transfer. Then balance the remaining atoms (usually O with H $_2$ O and H with H $^+$ ) and finally check the charge balance.
Half-equation method: Write the reduction half-equation and the oxidation half-equation. Balance atoms and charges in each, multiply by factors to equalize electrons, and add them together.

Understanding the Question

We need to complete the equation for the reaction between acidified $\text{VO}_3^-$ ions and Zn metal. The products are $\text{V}^{2+}$ , $\text{Zn}^{2+}$ , and we need to deduce the other product (likely H $_2$ O, since it's acidified and O needs to go somewhere).

Given skeleton: $\dots\text{VO}_3^- + \dots\text{Zn} + \dots\rightarrow \dots\text{V}^{2+} + \dots\text{Zn}^{2+} + \dots$

Approach

Determine oxidation states of V in $\text{VO}_3^-$ and in $\text{V}^{2+}$ .
Determine oxidation state change for Zn.
Find the ratio of $\text{VO}_3^-$ to Zn to balance electron transfer.
Balance O and H using H $_2$ O and H $^+$ .
Check charge balance.

Step-by-Step Reasoning

Oxidation states:
In $\text{VO}_3^-$ : O is $-2$ . Let V be $x$ . $x + 3(-2) = -1 \Rightarrow x - 6 = -1 \Rightarrow x = +5$ . So V is $+5$ .
In $\text{V}^{2+}$ : V is $+2$ .
Change for V: $+5 \rightarrow +2$ (reduction, gain of 3 electrons per V).

Zn: $0 \rightarrow +2$ (oxidation, loss of 2 electrons per Zn).
Electron balance:
To equalize electrons: V needs 3e $^-$ , Zn gives 2e $^-$ .
Least common multiple is 6.
Multiply V species by 2 (total 6e $^-$ gained).
Multiply Zn species by 3 (total 6e $^-$ lost).
So, $2\text{VO}_3^-$ and $3\text{Zn}$ on the left; $2\text{V}^{2+}$ and $3\text{Zn}^{2+}$ on the right.
Balance O and H:
Left side has $2 \times 3 = 6$ oxygen atoms (in $2\text{VO}_3^-$ ).
Right side needs 6 oxygen atoms. Add $6\text{H}_2\text{O}$ to the right.
Right side now has $6 \times 2 = 12$ hydrogen atoms.
Left side needs 12 hydrogen atoms. Add $12\text{H}^+$ to the left (acidified solution).
Check charge:
Left: $2(-1) + 3(0) + 12(+1) = -2 + 12 = +10$ .
Right: $2(+2) + 3(+2) + 6(0) = +4 + 6 = +10$ .
Charges are balanced.
Final equation:
$2\text{VO}_3^- + 3\text{Zn} + 12\text{H}^+ \rightarrow 2\text{V}^{2+} + 3\text{Zn}^{2+} + 6\text{H}_2\text{O}$

Key Takeaways

Always check oxidation numbers to find the electron transfer ratio.
In acidic conditions, balance oxygen with H $_2$ O and hydrogen with H $^+$ .
Always verify the final equation by checking both atom and charge balance.

Common Mistakes

Incorrect oxidation state for V in $\text{VO}_3^-$ (remember O is $-2$ and the overall ion is $-1$ , so V is $+5$ , not $+3$ ).
Forgetting to balance the final charge with H $^+$ ions.
Writing H $_2$ instead of H $_2$ O as the product (oxygen must go somewhere, and in acid, it forms water).
Incorrect stoichiometric coefficients (e.g., 1:1 ratio instead of 2:3).

Things to Be Careful About

The question provides blanks: $\dots\text{VO}_3^- + \dots\text{Zn} + \dots\rightarrow \dots\text{V}^{2+} + \dots\text{Zn}^{2+} + \dots$
Ensure the coefficients are the simplest whole numbers (2, 3, 12, 2, 3, 6 cannot be simplified further).
State symbols are not explicitly asked for in the blanks, but if included, H $^+$ is (aq), H $_2$ O is (l), etc. The mark scheme focuses on the coefficients and species.

Techniques used

balance redox equations using oxidation number changesbalance oxygen and hydrogen using water and H+ ions

Q6Medium-EasyAnalytical Techniques

(a)

Thin-layer and gas/liquid chromatography can be used to separate mixtures into their individual components.

3M

(i)

Define the following terms used in chromatography.

$R_{\text{f}}$ value
retention time

2M Similar

DifficultyEasy

Worked solution

Answer

$R_{\text{f}}$ value: The ratio of the distance moved by the solute (or component/spot) to the distance moved by the solvent (or mobile phase) from the origin.

Retention time: The time between injection of the sample and the detection of the component at the detector.

Final answer

See answer

Detailed explanation

Background Concept

Chromatography is a separation technique that relies on the differential partitioning of mixture components between two phases: a stationary phase (which remains fixed) and a mobile phase (which moves through or over the stationary phase). Components that interact more strongly with the stationary phase move slowly, while those that prefer the mobile phase move quickly.

In thin-layer chromatography (TLC), the stationary phase is a solid adsorbent (like silica gel or alumina) coated on a glass or plastic plate, and the mobile phase is a liquid solvent that travels up the plate via capillary action. In gas-liquid chromatography (GLC), the stationary phase is a non-volatile liquid coating the inside of a column or a solid support, and the mobile phase is an inert carrier gas.

Understanding the Question

This part asks for the precise definitions of two fundamental terms used to quantify and record chromatographic results: the $R_{\text{f}}$ value (specific to planar chromatography like TLC) and retention time (specific to column chromatography like GLC).

Approach

Recall the standard IUPAC-style definitions for both terms. Ensure the $R_{\text{f}}$ definition is stated as a ratio of distances measured from the same origin, and that retention time is defined as a duration between two specific events (injection and detection).

Step-by-Step Reasoning

$R_{\text{f}}$ value: This is a unitless ratio used in TLC. It is calculated by measuring the distance from the origin (the line where the sample was spotted) to the centre of the spot, and dividing it by the distance from the origin to the solvent front. The mark scheme accepts either "distance moved by component divided by distance moved by solvent" or "ratio of distance moved by component to distance moved by solvent".
Retention time: In GLC or HPLC, the sample is injected into the mobile phase. As components travel through the column, they elute at different times. The retention time is the elapsed time from the moment of injection to the moment the component reaches the detector and produces a peak.

Key Takeaways

Always define $R_{\text{f}}$ as a ratio of two distances from the origin. Never define it as a percentage or absolute distance. Retention time is always a duration measured from injection to detection.

Common Mistakes

Stating $R_{\text{f}}$ is the distance moved by the solvent divided by the distance moved by the solute (inverted ratio).
Defining retention time as the time the component spends in the stationary phase (that is residence time, not retention time).
Forgetting to specify that the distances in the $R_{\text{f}}$ definition must be measured from the origin.

Things to Be Careful About

Use the word "ratio" or "divided by" clearly for $R_{\text{f}}$ .
Ensure "injection" and "detection" are explicitly named for retention time; vague terms like "start" and "end" may not score.

Techniques used

define Rf valuedefine retention time

(ii)

Each type of chromatography makes use of a stationary phase and a mobile phase.

Complete Table 6.1 with a description of each of these.

Table 6.1

1M Similar

DifficultyEasy

Worked solution

Answer

thin-layer chromatography (mobile phase): polar solvent (or non-polar solvent / named solvent such as hexane or ethyl ethanoate)
gas/liquid chromatography (stationary phase): non-volatile liquid (or high boiling point liquid)

Final answer

See table

Detailed explanation

Background Concept

Chromatography requires a stationary phase and a mobile phase. The choice of these phases determines the type of chromatography and what it can separate.

In thin-layer chromatography (TLC), the stationary phase is a solid (typically silica, $\text{SiO}_2$ , or alumina, $\text{Al}_2\text{O}_3$ ). Because the stationary phase is polar, the mobile phase is usually a liquid solvent. Depending on the polarity of the analytes, this can be a polar solvent (like water or ethanol) or a less polar solvent (like hexane or ethyl ethanoate) to adjust the solvent strength.

In gas-liquid chromatography (GLC), the mobile phase is an inert carrier gas (like helium or nitrogen). The stationary phase must be a liquid that does not vaporise at the operating temperature of the column; hence it is a non-volatile liquid or a high boiling point liquid (often a polymer like polyethylene glycol) coated onto an inert solid support inside the column.

Understanding the Question

Table 6.1 has two blanks to fill: the mobile phase for TLC, and the stationary phase for GLC. The other two cells (TLC stationary phase and GLC mobile phase) are already crossed out as they are implicitly known (solid adsorbent and carrier gas).

Approach

Recall the physical state and key properties of the missing phases in each technique.

Step-by-Step Reasoning

TLC mobile phase: The liquid that moves up the plate. It can be described generally as a "polar solvent" or "non-polar solvent", or a candidate can name a specific solvent like hexane or ethyl ethanoate. The key is that it is a liquid solvent.
GLC stationary phase: The liquid coating the column. It must not vaporise, so it is described as a "non-volatile liquid" or a "high boiling point liquid".

Key Takeaways

Remember the physical states: TLC uses a solid stationary phase and liquid mobile phase; GLC uses a liquid stationary phase and gaseous mobile phase.

Common Mistakes

Calling the GLC stationary phase a "solid" (that is gas-solid chromatography, GSC, which is rarely used).
Calling the TLC mobile phase a "gas" or "carrier gas".
Writing "water" as the TLC mobile phase without qualification, as pure water is rarely used alone on silica due to strong adsorption; a mixed solvent system is typical.

Things to Be Careful About

The question asks for a description, not necessarily a specific chemical name, though naming a valid solvent is acceptable. "Non-volatile liquid" is the precise phrase expected for the GLC stationary phase.

Techniques used

identify stationary and mobile phases in chromatography

(b)

A mixture of two substances A and B is analysed by thin-layer chromatography.

The $R_{\text{f}}$ value of substance A is larger than that of substance B.

Suggest why substance A has a larger $R_{\text{f}}$ value.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

Substance A is more soluble in the mobile phase (or has less affinity / less adsorption to the stationary phase) than substance B.

Final answer

See answer

Detailed explanation

Background Concept

In chromatography, a component's position on the plate (or time in the column) is determined by how it partitions between the stationary and mobile phases. A higher $R_{\text{f}}$ value means the component has travelled further with the solvent front. This happens when the component spends more time in the mobile phase and less time interacting with the stationary phase.

For TLC on a polar stationary phase (like silica):

Components that are more polar will hydrogen-bond or dipole-dipole interact strongly with the silica, leading to lower $R_{\text{f}}$ values.
Components that are more non-polar will be more soluble in the (typically less polar) mobile phase and will travel further, giving higher $R_{\text{f}}$ values.

Understanding the Question

Substance A has a larger $R_{\text{f}}$ value than substance B. We need to explain why in terms of the interactions with the two phases.

Approach

Relate the distance travelled to the partitioning behaviour: more time in mobile phase = higher $R_{\text{f}}$ .

Step-by-Step Reasoning

Since A has a larger $R_{\text{f}}$ , it has moved further up the plate.
This means A has a greater affinity for the mobile phase (it is more soluble in it) OR a weaker interaction with the stationary phase (less adsorption).
Either explanation is chemically valid and scores a mark.

Key Takeaways

High $R_{\text{f}}$ = high solubility in mobile phase / low adsorption to stationary phase. Low $R_{\text{f}}$ = low solubility in mobile phase / high adsorption to stationary phase.

Common Mistakes

Saying "A is less polar" without linking it to solubility or adsorption. While often true in TLC, the mark scheme specifically rewards the direct link to phase interactions.
Stating "A is lighter" or "A has a lower molecular mass". Molecular mass does not directly determine $R_{\text{f}}$ ; polarity and intermolecular forces do.

Things to Be Careful About

Use precise terminology: "soluble in the mobile phase" or "adsorption to the stationary phase". Vague phrases like "moves faster" or "likes the solvent more" may not score unless they clearly imply the correct physicochemical principle.

Techniques used

interpret Rf values in terms of phase partitioning

(c)

The two isomeric compounds Y and Z are analysed by proton ( $^1\text{H}$ ) NMR spectroscopy.

Fig. 6.1

3M

(i)

Complete Table 6.2 to predict the number of peaks observed in the proton ( $^1\text{H}$ ) NMR spectra for Y and Z.

compound	number of peaks observed
Y
Z

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

compound	number of peaks observed
Y	2
Z	2

Final answer

Y = 2, Z = 2

Detailed explanation

Background Concept

In $^1\text{H}$ NMR spectroscopy, each chemically distinct environment of hydrogen atoms produces a separate signal (peak). Hydrogens are in the same chemical environment if they are equivalent by symmetry or rapid rotation (like in a methyl group, $-\text{CH}_3$ ).

To predict the number of peaks:

Draw the full structural formula or clearly identify all hydrogen atoms.
Look for planes of symmetry or axes of symmetry that make groups of hydrogens equivalent.
Count the number of unique sets of hydrogens.

Understanding the Question

We are given two isomeric diketones:

Y: hexane-2,5-dione ( $\text{CH}_3-\text{CO}-\text{CH}_2-\text{CH}_2-\text{CO}-\text{CH}_3$ )
Z: hexane-3,4-dione ( $\text{CH}_3-\text{CH}_2-\text{CO}-\text{CO}-\text{CH}_2-\text{CH}_3$ )
We need to predict the number of $^1\text{H}$ NMR peaks for each.

Approach

Identify the symmetry in each molecule and count the sets of equivalent protons.

Step-by-Step Reasoning

Compound Y (hexane-2,5-dione):
- The molecule has a plane of symmetry between C3 and C4.
- The two $\text{CH}_3$ groups are equivalent (6H).
- The two $\text{CH}_2$ groups are equivalent (4H).
- Total unique proton environments = 2. So, 2 peaks.
Compound Z (hexane-3,4-dione):
- The molecule has a plane of symmetry between C3 and C4.
- The two $\text{CH}_3$ groups are equivalent (6H).
- The two $\text{CH}_2$ groups are equivalent (4H).
- Total unique proton environments = 2. So, 2 peaks.

Key Takeaways

Always check for molecular symmetry first. Symmetrical molecules often have fewer NMR peaks than their asymmetrical isomers, even if they have the same molecular formula.

Common Mistakes

Counting the number of carbon atoms instead of proton environments.
Forgetting that all three hydrogens in a $-\text{CH}_3$ group are equivalent due to rapid rotation around the C-C bond.
Assuming Y and Z must have different numbers of peaks just because they are structural isomers.

Things to Be Careful About

The question asks for the number of peaks, not the integration ratios or chemical shifts. Keep the answer concise.

Techniques used

predict number of 1H NMR peaks from symmetry and chemical environments

(ii)

Name all the different splitting patterns observed in the proton ( $^1\text{H}$ ) NMR spectra for Y and Z.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Y: singlet
Z: triplet and quartet

Final answer

Y: singlet; Z: triplet and quartet

Detailed explanation

Background Concept

The splitting pattern (multiplicity) of an NMR signal is determined by the number of equivalent protons on adjacent carbon atoms, following the $n+1$ rule.

If a set of protons has $n$ equivalent protons on the adjacent carbon(s), its signal will be split into $n+1$ peaks:

$n=0$ → singlet (1 peak)
$n=1$ → doublet (2 peaks)
$n=2$ → triplet (3 peaks)
$n=3$ → quartet (4 peaks)

Note: Protons separated by more than 3 bonds (or by heteroatoms like O or C=O in many cases) do not typically split each other in routine $^1\text{H}$ NMR.

Understanding the Question

We need to name the splitting patterns observed in the $^1\text{H}$ NMR spectra for Y and Z, based on their structures.

Approach

For each compound, identify the adjacent protons for each unique proton environment and apply the $n+1$ rule.

Step-by-Step Reasoning

Compound Y (hexane-2,5-dione: $\text{CH}_3-\text{CO}-\text{CH}_2-\text{CH}_2-\text{CO}-\text{CH}_3$ ):
- The $\text{CH}_3$ protons are adjacent to a carbonyl carbon ( $\text{C=O}$ ), which has 0 protons. $n=0 \rightarrow 0+1 = 1$ → singlet.
- The $\text{CH}_2$ protons are adjacent to two carbonyl carbons ( $\text{C=O}$ ), which have 0 protons. $n=0 \rightarrow 0+1 = 1$ → singlet.
- Both signals are singlets. The mark scheme accepts "Y (two) singlet(s)".
Compound Z (hexane-3,4-dione: $\text{CH}_3-\text{CH}_2-\text{CO}-\text{CO}-\text{CH}_2-\text{CH}_3$ ):
- The $\text{CH}_3$ protons are adjacent to a $\text{CH}_2$ group, which has 2 protons. $n=2 \rightarrow 2+1 = 3$ → triplet.
- The $\text{CH}_2$ protons are adjacent to a $\text{CH}_3$ group (3 protons) and a carbonyl carbon (0 protons). Total adjacent protons $n = 3+0 = 3 \rightarrow 3+1 = 4$ → quartet.
- The signals are a triplet and a quartet.

Key Takeaways

Carbonyl groups ( $\text{C=O}$ ) and other heteroatoms often break spin-spin coupling. Protons on either side of a $\text{C=O}$ do not split each other. Always count only the protons on directly adjacent saturated carbons.

Common Mistakes

Forgetting that the carbonyl carbon has no protons, leading to incorrect $n$ values for Y.
Counting protons across the carbonyl in Z (e.g., thinking $\text{CH}_3$ is split by the other $\text{CH}_3$ group). Coupling does not occur across the $\text{C-C(=O)-C-C}$ framework in this way; only adjacent $\text{C-H}$ bonds couple.
Naming the patterns incorrectly (e.g., calling a triplet a "doublet of doublets").

Things to Be Careful About

The question asks to "name all the different splitting patterns". For Y, since both environments give singlets, the only pattern observed is a singlet. Do not write "singlet and singlet"; write "singlet" (or "two singlets" is also acceptable per the mark scheme, but "singlet" is the pattern name).
Ensure you list the patterns for both compounds clearly.

Techniques used

predict splitting patterns using the n+1 rule

Q7MediumCarboxylic Acids and DerivativesHydrocarbonsPolymerisationNitrogen Compounds

(a)

State the relative acidities of bromoethanoic acid, $\text{BrCH}_2\text{COOH}$ , chloroethanoic acid, $\text{ClCH}_2\text{COOH}$ , ethanoic acid, $\text{CH}_3\text{COOH}$ and ethanol, $\text{CH}_3\text{CH}_2\text{OH}$ .

Explain your answer.

4M Similar

DifficultyMedium

Worked solution

Answer

Order of acidity: $\text{ClCH}_2\text{COOH} > \text{BrCH}_2\text{COOH} > \text{CH}_3\text{COOH} > \text{CH}_3\text{CH}_2\text{OH}$

Explanation:

Chlorine and bromine are electron-withdrawing (negative inductive effect), which weakens the O–H bond and stabilises the conjugate base anion. Chlorine is more electronegative than bromine, so $\text{ClCH}_2\text{COOH}$ is the most acidic.
The C=O group in ethanoic acid is electron-withdrawing, making it more acidic than ethanol.
The ethyl group in ethanol is electron-donating (positive inductive effect), which strengthens the O–H bond and destabilises the conjugate base, making it the least acidic.

Final answer

ClCH2COOH > BrCH2COOH > CH3COOH > CH3CH2OH

Detailed explanation

Background Concept

The acidity of an organic compound depends on the ease with which it donates a proton (H⁺) and the stability of the resulting conjugate base (anion). For compounds containing an O–H bond, electron-withdrawing groups (EWGs) pull electron density away from the O–H bond, weakening it and making the proton easier to release. They also spread the negative charge over a larger area in the conjugate base (delocalisation or inductive effect), stabilising it. Conversely, electron-donating groups (EDGs) push electron density towards the O–H bond, strengthening it and concentrating the negative charge on the anion, making it less stable.

Electronegativity plays a key role: more electronegative atoms exert a stronger inductive effect. The inductive effect diminishes with distance from the functional group.

Understanding the Question

The question asks to order four compounds—bromoethanoic acid, chloroethanoic acid, ethanoic acid, and ethanol—by acidity and to explain the order. The compounds contain different substituents attached to the carbon chain or functional group: halogens (Cl, Br), a carbonyl group (C=O in COOH), and an alkyl group (ethyl in ethanol).

Approach

Identify the acidic proton in each molecule (the O–H proton in the carboxylic acids and ethanol).
Compare the substituents attached to the acidic group: halogens are EWGs, the carbonyl in COOH is an EWG, and the alkyl group is an EDG.
Rank the strength of the electron-withdrawing/donating effects to determine the relative stability of the conjugate bases and the strength of the O–H bond.
State the final order and link each difference to the specific inductive effect.

Step-by-Step Reasoning

Most acidic: $\text{ClCH}_2\text{COOH}$ . Both Cl and the C=O group are electron-withdrawing. Chlorine is more electronegative than bromine, so it exerts a stronger negative inductive effect, weakening the O–H bond more and stabilising the anion more than Br does.
Second most acidic: $\text{BrCH}_2\text{COOH}$ . Bromine is electron-withdrawing but less so than chlorine.
Third most acidic: $\text{CH}_3\text{COOH}$ . The C=O group is electron-withdrawing, making the carboxylic acid more acidic than an alcohol. However, it lacks the halogen substituent.
Least acidic: $\text{CH}_3\text{CH}_2\text{OH}$ . The ethyl group is electron-donating (positive inductive effect), which strengthens the O–H bond and destabilises the ethoxide anion.

Key Takeaways

Electron-withdrawing groups increase acidity by weakening the O–H bond and stabilising the conjugate base.
Electronegativity determines the strength of the inductive effect (Cl > Br).
Carboxylic acids are more acidic than alcohols due to the electron-withdrawing C=O group and resonance stabilisation (though resonance isn't explicitly required here, inductive effects from C=O are).

Common Mistakes

Stating that halogens "attract electrons" without linking it to the O–H bond weakening or anion stabilisation.
Forgetting that the C=O group in ethanoic acid is electron-withdrawing, incorrectly placing it as less acidic than ethanol.
Confusing the inductive effect direction: electron-withdrawing is negative inductive effect (-I), electron-donating is positive inductive effect (+I).

Things to Be Careful About

Ensure the order is strictly from most acidic to least acidic.
Use precise terminology: "electron-withdrawing", "negative inductive effect", "weaken O–H bond", "stabilise the anion/conjugate base".
Do not just say "Cl is more electronegative"; explain the consequence (stronger inductive effect).

Techniques used

compare acidities using inductive effectsexplain bond weakening and conjugate base stabilisation

(b)

Fig. 7.1 shows the reaction of methylbenzene and ethanedioic acid with $\text{KMnO}_4$ .

Predict the major carbon-containing product for each of these reactions.

Fig. 7.1

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Methylbenzene:

Ethanedioic acid:
$\text{CO}_2$

Final answer

Benzoic acid and CO2

Detailed explanation

Background Concept

Hot potassium manganate(VII) ( $\text{KMnO}_4$ ) is a strong oxidising agent. Its behaviour depends on the conditions (acidified or alkaline) and the substrate.

Alkylbenzenes (Arenes): Any alkyl side chain on a benzene ring that has at least one hydrogen atom on the benzylic carbon (the carbon attached directly to the ring) is oxidised to a carboxylic acid group ( $-\text{COOH}$ ), regardless of the length of the side chain. The ring itself is not oxidised. This works for hot alkaline or hot acidified $\text{KMnO}_4$ .
Ethanedioic acid (Oxalic acid, $\text{HOOC-COOH}$ ): This is an alpha-dicarboxylic acid. Under hot, strong oxidising conditions (especially acidified $\text{KMnO}_4$ ), the C–C bond between the two carboxyl groups is cleaved. One carbon becomes $\text{CO}_2$ and the other becomes carbonic acid (which decomposes to $\text{CO}_2$ and water). Essentially, the entire molecule is oxidised to $\text{CO}_2$ .

Understanding the Question

The question provides two reaction schemes with hot $\text{KMnO}_4$ :

Methylbenzene + hot alkaline $\text{KMnO}_4$ .
Ethanedioic acid + hot acidified $\text{KMnO}_4$ .
We need to predict the major carbon-containing organic/inorganic product for each.

Approach

For methylbenzene, apply the rule for side-chain oxidation of arenes: the methyl group ( $-\text{CH}_3$ ) becomes a carboxyl group ( $-\text{COOH}$ ). The product is benzoic acid.
For ethanedioic acid, apply the rule for oxidation of oxalic acid: the C–C bond breaks, producing carbon dioxide ( $\text{CO}_2$ ).

Step-by-Step Reasoning

Methylbenzene reaction: The benzylic carbon has 3 hydrogens. Hot $\text{KMnO}_4$ oxidises the entire side chain down to the carboxyl group attached to the ring. The product is benzoic acid ( $\text{C}_6\text{H}_5\text{COOH}$ ). In alkaline conditions, it initially forms the benzoate ion, but upon acidification (often implied or asked for as the final product), or simply stating the organic product, benzoic acid is the standard answer for the carbon-containing product structure. The mark scheme accepts the structure of benzoic acid.
Ethanedioic acid reaction: $\text{HOOC-COOH} \xrightarrow{[\text{O}]} 2\text{CO}_2 + \text{H}_2\text{O}$ . The major carbon-containing product is carbon dioxide ( $\text{CO}_2$ ).

Key Takeaways

Alkylbenzenes with benzylic hydrogens oxidise to benzoic acid derivatives.
Ethanedioic acid is easily oxidised and decomposes to $\text{CO}_2$ .

Common Mistakes

Writing methanoic acid or carbon monoxide for ethanedioic acid oxidation (it goes all the way to $\text{CO}_2$ ).
Forgetting that the ring is intact in methylbenzene oxidation.
Writing the benzoate ion ( $\text{C}_6\text{H}_5\text{COO}^-$ ) instead of benzoic acid if the question asks for the product (usually the acid is expected, or the structure is accepted either way, but mark scheme shows $-\text{COOH}$ ).

Things to Be Careful About

Ensure the structure of benzoic acid is drawn correctly with the $-\text{COOH}$ group attached to the benzene ring.
$\text{CO}_2$ is a gas, but it is the correct carbon-containing product.

Techniques used

predict oxidation products of arenespredict oxidative cleavage of alpha-dicarboxylic acids

(c)

Polyamide X can be synthesised from ethanedioic acid and benzene-1,4-diamine.

Fig. 7.2

3M

(i)

Draw the repeat unit of polyamide X in the box.

The new functional group formed should be shown displayed.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Final answer

Repeat unit: -NH-C6H4-NH-CO-CO-

Detailed explanation

Background Concept

Polyamides are formed by condensation polymerisation between a diamine (containing two $-\text{NH}_2$ groups) and a dicarboxylic acid (containing two $-\text{COOH}$ groups). During the reaction, an amide bond ( $-\text{CO}-\text{NH}-$ ) is formed, and a water molecule is lost for each bond.

The repeat unit of a condensation polymer contains the structural remnants of both monomers, linked by the new functional group. Continuation bonds (dashed lines) at the ends indicate the polymer chain continues.

Understanding the Question

Polyamide X is made from ethanedioic acid ( $\text{HOOC-COOH}$ ) and benzene-1,4-diamine ( $\text{H}_2\text{N}-\text{C}_6\text{H}_4-\text{NH}_2$ ). We need to draw the repeat unit, showing the new amide functional group in displayed form (all atoms and bonds shown).

Approach

Identify the functional groups reacting: $-\text{NH}_2$ and $-\text{COOH}$ .
Form the amide linkage: $-\text{CO}-\text{NH}-$ .
Combine the remaining parts of the monomers: the benzene ring with para-substitution and the ethanedioyl group ( $-\text{CO}-\text{CO}-$ ).
Draw the repeat unit with continuation bonds at both ends.

Step-by-Step Reasoning

Monomer 1: Benzene-1,4-diamine. The active parts are the two $-\text{NH}_2$ groups. After losing H, we have $-\text{NH}-\text{C}_6\text{H}_4-\text{NH}-$ .
Monomer 2: Ethanedioic acid. The active parts are the two $-\text{COOH}$ groups. After losing OH, we have $-\text{CO}-\text{CO}-$ .
Link them: The nitrogen from the diamine bonds to the carbonyl carbon from the acid. The repeat unit is $-\text{NH}-\text{C}_6\text{H}_4-\text{NH}-\text{CO}-\text{CO}-$ .
The new functional group is the amide bond ( $-\text{CO}-\text{NH}-$ ). It must be shown displayed: C double bonded to O, single bonded to N, which has an H attached.
The benzene ring is para-substituted (positions 1 and 4).

Key Takeaways

Condensation polymers lose small molecules (water) to form linkages.
Polyamides have amide linkages ( $-\text{CO}-\text{NH}-$ ).
Display formula means showing all bonds in the new functional group.

Common Mistakes

Drawing the monomers instead of the repeat unit.
Forgetting continuation bonds at the ends of the repeat unit.
Not showing the amide bond in displayed form (e.g., just writing -CONH-).
Incorrect connectivity (e.g., O bonded to N instead of C to N).

Things to Be Careful About

The benzene ring must have the substituents at 1,4 positions (para).
The amide bond is $-\text{C}(=\text{O})-\text{NH}-$ , not $-\text{O}-\text{C}(=\text{O})-\text{NH}-$ (that would be a carbonate or similar).
Ensure the H on the nitrogen is shown.

Techniques used

draw repeat unit of condensation polymerdisplay amide linkage

(ii)

Benzene-1,4-diamine can be formed by reduction of 1,4-dinitrobenzene.

Complete the equation for this reduction.

[H] represents one atom of hydrogen from a reducing agent.

\dots\dots + \dots\dots\dots\dots [\text{H}] \rightarrow \text{H}_2\text{N} - \text{C}_6\text{H}_4 - \text{NH}_2 + \dots\dots\dots\dots

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

$\text{O}_2\text{N}-\text{C}_6\text{H}_4-\text{NO}_2 + 12[\text{H}] \rightarrow \text{H}_2\text{N}-\text{C}_6\text{H}_4-\text{NH}_2 + 4\text{H}_2\text{O}$

Final answer

O2N-C6H4-NO2 + 12[H] -> H2N-C6H4-NH2 + 4H2O

Detailed explanation

Background Concept

Nitroarenes (compounds with $-\text{NO}_2$ on a benzene ring) can be reduced to phenylamines (aromatic amines, $-\text{NH}_2$ ) using a reducing agent. In equation balancing using the $[\text{H}]$ notation, $[\text{H}]$ represents a hydrogen atom from the reducing agent (like $\text{Sn/HCl}$ or $\text{H}_2$ /Ni).

Reduction of a nitro group ( $-\text{NO}_2$ ) to an amino group ( $-\text{NH}_2$ ) requires 6 hydrogen atoms:
$-\text{NO}_2 + 6[\text{H}] \rightarrow -\text{NH}_2 + 2\text{H}_2\text{O}$

Understanding the Question

We are reducing 1,4-dinitrobenzene (two $-\text{NO}_2$ groups) to benzene-1,4-diamine (two $-\text{NH}_2$ groups). We need to complete the equation with the correct number of $[\text{H}]$ and the by-product.

Approach

Write the reduction half-equation for one $-\text{NO}_2$ group to $-\text{NH}_2$ .
Multiply by 2 for the two nitro groups.
Balance the oxygen atoms by forming water.

Step-by-Step Reasoning

One $-\text{NO}_2$ group needs 6[H] to become $-\text{NH}_2$ and $2\text{H}_2\text{O}$ .
There are two $-\text{NO}_2$ groups in 1,4-dinitrobenzene.
Total $[\text{H}]$ needed = $2 \times 6 = 12[\text{H}]$ .
Total water produced = $2 \times 2 = 4\text{H}_2\text{O}$ .
Equation: $\text{O}_2\text{N}-\text{C}_6\text{H}_4-\text{NO}_2 + 12[\text{H}] \rightarrow \text{H}_2\text{N}-\text{C}_6\text{H}_4-\text{NH}_2 + 4\text{H}_2\text{O}$ .

Key Takeaways

Reduction of $-\text{NO}_2$ to $-\text{NH}_2$ requires 6[H] per group.
Balance oxygen as water.

Common Mistakes

Using the wrong number of $[\text{H}]$ (e.g., 6[H] for the whole molecule instead of 12[H]).
Forgetting to produce water as a by-product.
Incorrectly balancing the hydrogen atoms on the right side.

Things to Be Careful About

The question states $[\text{H}]$ represents one atom of hydrogen. Do not write $\text{H}_2$ .
Ensure the organic structures are correct (para-substitution).

Techniques used

balance reduction equation with [H]

(d)

Fig. 7.3 shows the two-step synthesis of the azo compound W.

Fig. 7.3

3M

(i)

Suggest structures for compounds V and W and draw them in the boxes in Fig. 7.3.

2M Similar

DifficultyMedium

Worked solution

Answer

Compound V:

Compound W:

Final answer

V: 1,4-benzenediazonium dichloride; W: bis-azo dye (see diagram)

Detailed explanation

Background Concept

Azo coupling is a reaction between a diazonium salt (electrophile) and an activated aromatic ring like phenol or an amine (nucleophile). Diazonium salts are formed from primary aromatic amines (like phenylamine) by reaction with nitrous acid ( $\text{HNO}_2$ , generated in situ from $\text{NaNO}_2$ and $\text{HCl}$ ) at temperatures below 10°C.

Diazonium salt: Contains the $-\text{N}_2^+$ group (or $-\text{N}\equiv\text{N}^+$ ). With chloride ions present, it's often written as $\text{Ar}-\text{N}_2^+\text{Cl}^-$ .
Coupling with phenol: Phenol is activated by the $-\text{OH}$ group (electron-donating, +M effect). Coupling occurs at the ortho or para position relative to the $-\text{OH}$ group. In alkaline conditions (NaOH), phenol forms the phenoxide ion ( $\text{C}_6\text{H}_5\text{O}^-$ ), which is even more activated, and coupling predominantly occurs at the para position to give the azo dye.

Understanding the Question

Step 1: Benzene-1,4-diamine reacts to form V ( $\text{C}_6\text{H}_4\text{N}_4\text{Cl}_2$ ). Since the starting material has two $-\text{NH}_2$ groups, and the formula has 4 nitrogens and 2 chlorines, both amino groups must have been converted to diazonium chloride groups ( $-\text{N}_2^+\text{Cl}^-$ ). So V is 1,4-benzenediazonium dichloride (or p-benzenediazonium chloride, but with two groups).
Step 2: V reacts with phenol and NaOH to form W ( $\text{C}_{18}\text{H}_{14}\text{N}_4\text{O}_2$ ). This is a bis-azo coupling. Each diazonium group couples with a phenol molecule. Since phenol couples at the para position (in alkaline conditions), the product W will have the central benzene ring linked via two $-\text{N}=\text{N}-$ groups to two phenol rings at their para positions.

Let's check the formula for W: Central ring $\text{C}_6\text{H}_4$ . Two azo groups $2 \times \text{N}_2 = \text{N}_4$ . Two phenol rings (para-substituted): $2 \times \text{C}_6\text{H}_4\text{OH} = \text{C}_{12}\text{H}_8\text{O}_2$ . Total: $\text{C}_6\text{H}_4 + \text{N}_4 + \text{C}_{12}\text{H}_8\text{O}_2 = \text{C}_{18}\text{H}_{12}\text{N}_4\text{O}_2$ ... wait.
Let's recount:
Central ring: $\text{C}_6\text{H}_4$ (para substituted, so 4 H).
Two azo linkages: $2 \times (-\text{N}=\text{N}-)$ .
Two phenol rings attached at para position: Each is $-\text{C}_6\text{H}_4-\text{OH}$ .
Total C: $6 + 6 + 6 = 18$ .
Total H: $4 (\text{central}) + 4 (\text{ring 1}) + 4 (\text{ring 2}) + 2 (\text{OH}) = 14$ .
Total N: 4.
Total O: 2.
Formula: $\text{C}_{18}\text{H}_{14}\text{N}_4\text{O}_2$ . Matches W.

Approach

Deduce V from the molecular formula and starting material. Two $-\text{NH}_2$ become two $-\text{N}_2^+\text{Cl}^-$ .
Deduce W by coupling two molecules of phenol (at para position due to NaOH) with the bis-diazonium salt V.

Step-by-Step Reasoning

Structure V: Benzene-1,4-diamine has $-\text{NH}_2$ at 1 and 4. Diazotisation converts $-\text{NH}_2$ to $-\text{N}_2^+$ . With Cl⁻ counterions, the structure is $\text{Cl}^- \text{N}\equiv\text{N}^+-\text{C}_6\text{H}_4-\text{N}^+\equiv\text{N} \text{Cl}^-$ . (Or written as $\text{Ar}-\text{N}_2^+\text{Cl}^-$ twice).
Structure W: The diazonium groups are electrophiles. Phenol in NaOH is phenoxide, an activated nucleophile. Coupling occurs para to the $-\text{OH}$ group.
Product: $\text{HO}-\text{C}_6\text{H}_4-\text{N}=\text{N}-\text{C}_6\text{H}_4-\text{N}=\text{N}-\text{C}_6\text{H}_4-\text{OH}$ . The central ring is from V, the outer rings are from phenol. The $-\text{OH}$ groups are para to the azo linkages.

Key Takeaways

Primary aromatic amines form diazonium salts with $\text{HNO}_2$ /HCl at <10°C.
Diazonium salts couple with phenols in alkaline solution at the para position.
Bis-amines can form bis-diazonium salts and couple with two equivalents of phenol.

Common Mistakes

Drawing the coupling at the ortho position (possible, but para is major product, especially in alkali).
Forgetting the chloride ions in the diazonium salt structure (though sometimes accepted without, mark scheme shows them).
Incorrect molecular formula counting.
Drawing the azo linkage as $-\text{N}-\text{N}-$ instead of $-\text{N}=\text{N}-$ .

Things to Be Careful About

The question asks for structures in boxes. Ensure clarity.
For V, the formula is $\text{C}_6\text{H}_4\text{N}_4\text{Cl}_2$ . This confirms two diazonium groups with two chlorides.
For W, the coupling is para-para.

Techniques used

identify diazonium salt structurepredict azo coupling product

(ii)

Give the reagents and conditions for step 1.

1M Similar

DifficultyEasy

Worked solution

Answer

$\text{NaNO}_2$ (or $\text{HNO}_2$ ) and $\text{HCl}$ , temperature $\leqslant 10\text{ }^\circ\text{C}$

Final answer

NaNO2 and HCl at <= 10 C

Detailed explanation

Background Concept

Diazotisation is the reaction of a primary aromatic amine with nitrous acid ( $\text{HNO}_2$ ) to form a diazonium salt. Nitrous acid is unstable and is generated in situ by mixing sodium nitrite ( $\text{NaNO}_2$ ) with a strong acid like hydrochloric acid ( $\text{HCl}$ ) at low temperatures.

$\text{NaNO}_2 + \text{HCl} \rightarrow \text{HNO}_2 + \text{NaCl}$
$\text{Ar}-\text{NH}_2 + \text{HNO}_2 + \text{HCl} \rightarrow \text{Ar}-\text{N}_2^+\text{Cl}^- + 2\text{H}_2\text{O}$

The temperature must be kept below 10°C (usually 0-5°C) because diazonium salts are unstable and can decompose explosively or form phenols at higher temperatures.

Understanding the Question

Step 1 converts benzene-1,4-diamine to the bis-diazonium salt V. We need the reagents and conditions for this transformation (diazotisation of both amino groups).

Approach

State the reagents for generating nitrous acid and the acid required, plus the critical temperature condition.

Step-by-Step Reasoning

Reagents: Sodium nitrite ( $\text{NaNO}_2$ ) and hydrochloric acid ( $\text{HCl}$ ). Alternatively, nitrous acid ( $\text{HNO}_2$ ) and $\text{HCl}$ can be stated.
Conditions: Temperature must be $\leqslant 10\text{ }^\circ\text{C}$ (or 0-5°C). This is crucial to prevent decomposition of the diazonium salt.

Key Takeaways

Diazotisation requires $\text{NaNO}_2$ /HCl and low temperature (<10°C).
Applies to primary aromatic amines.

Common Mistakes

Forgetting the temperature condition.
Writing $\text{HNO}_3$ instead of $\text{HNO}_2$ .
Using $\text{NaNO}_3$ .

Things to Be Careful About

The question asks for reagents AND conditions. Both are needed for the mark.
$\text{NaNO}_2$ is the solid salt usually used, so stating $\text{NaNO}_2$ and $\text{HCl}$ is the most precise answer.

Techniques used

state diazotisation reagents and conditions

Q8Medium-HardHydrocarbonsOrganic SynthesisHalogen CompoundsCarboxylic Acids and Derivatives

(a)

In the electrophilic substitution of arenes, different substituents can direct to different ring positions.

2M

(i)

Describe the directing effect of the $-\text{NO}_2$ group. Explain your answer.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

The $-\text{NO}_2$ group directs electrophilic substitution to the 3 (and 5) / meta position. This is because the $-\text{NO}_2$ group is an electron-withdrawing (or electronegative) group.

Final answer

Directs to meta (3,5) position; electron-withdrawing group.

Detailed explanation

Background Concept

In electrophilic aromatic substitution, existing substituents on the benzene ring influence where the new group attaches. Substituents are classified as ortho/para-directing (activating or deactivating) or meta-directing (deactivating). The directing effect is determined by the stability of the intermediate carbocation (arenium ion) formed during the reaction.

Understanding the Question

The question asks for the directing effect of the nitro group ( $-\text{NO}_2$ ) and the reason for this effect. We need to state the position (meta) and the electronic nature of the group.

Approach

Recall that electron-withdrawing groups (EWG) deactivate the ring and direct meta. The $-\text{NO}_2$ group is a strong EWG due to the electronegativity of nitrogen and oxygen and the formal positive charge on nitrogen in resonance structures.

Step-by-Step Reasoning

Directing Effect: The nitro group is a meta-director. This means new substituents will attach primarily at positions 3 and 5 relative to the nitro group (position 1).
Explanation: The $-\text{NO}_2$ group is electron-withdrawing (via induction and resonance). It withdraws electron density from the ring, making it less reactive (deactivated). Specifically, if substitution were to occur at the ortho or para positions, the intermediate carbocation would place a positive charge adjacent to the electron-deficient nitrogen of the nitro group, which is highly destabilizing. Substitution at the meta position avoids this direct destabilization, making it the preferred pathway (though slower than benzene).

Key Takeaways

Electron-withdrawing groups like $-\text{NO}_2$ , $-\text{CN}$ , and $-\text{COOH}$ are meta-directors and deactivators. Electron-donating groups (except halogens) are ortho/para-directors and activators.

Common Mistakes

Stating that $-\text{NO}_2$ is electron-donating.
Confusing meta with ortho/para.
Not linking the directing effect to the electronic nature (electron-withdrawing) of the group.

Things to Be Careful About

Ensure you use the correct terminology: 'electron-withdrawing' or 'electronegative'. 'Meta' is the key position.

Techniques used

identify directing effects of substituentsrelate electronic effects to orientation

(ii)

The nitration of arenes uses a mixture of concentrated $\text{HNO}_3$ and concentrated $\text{H}_2\text{SO}_4$ to generate the $\text{NO}_2^+$ electrophile.

Write an equation for the formation of the $\text{NO}_2^+$ electrophile.

1M Similar

DifficultyEasy

Worked solution

Answer

$\text{HNO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + \text{HSO}_4^- + \text{H}_2\text{O}$
OR
$\text{HNO}_3 + 2\text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + 2\text{HSO}_4^- + \text{H}_3\text{O}^+$

Final answer

HNO3 + H2SO4 -> NO2+ + HSO4- + H2O

Detailed explanation

Background Concept

Nitration of arenes requires a nitrating mixture of concentrated nitric acid and concentrated sulfuric acid. Sulfuric acid is a stronger acid than nitric acid, so it protonates the nitric acid, leading to the loss of water and formation of the nitronium ion ( $\text{NO}_2^+$ ), which is the active electrophile.

Understanding the Question

Write the balanced equation for the formation of the nitronium ion ( $\text{NO}_2^+$ ) from concentrated $\text{HNO}_3$ and $\text{H}_2\text{SO}_4$ .

Approach

Nitric acid acts as a base here, accepting a proton from sulfuric acid. The protonated nitric acid ( $\text{H}_2\text{NO}_3^+$ ) loses a water molecule to form $\text{NO}_2^+$ .

Step-by-Step Reasoning

Protonation: $\text{HNO}_3 + \text{H}_2\text{SO}_4 \rightleftharpoons \text{H}_2\text{NO}_3^+ + \text{HSO}_4^-$
Loss of water: $\text{H}_2\text{NO}_3^+ \rightarrow \text{NO}_2^+ + \text{H}_2\text{O}$
Overall: $\text{HNO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + \text{HSO}_4^- + \text{H}_2\text{O}$

Alternatively, if the proton is transferred to another sulfuric acid molecule (or water formed): $\text{HNO}_3 + 2\text{H}_2\text{SO}_4 \rightarrow \text{NO}_2^+ + 2\text{HSO}_4^- + \text{H}_3\text{O}^+$ .

Key Takeaways

The nitrating mixture generates the nitronium ion. Sulfuric acid acts as the catalyst/acid.

Common Mistakes

Writing $\text{HNO}_2$ instead of $\text{HNO}_3$ .
Forgetting the sulfate ion ( $\text{HSO}_4^-$ ) or water.
Unbalanced equations.

Things to Be Careful About

State symbols are not strictly required here unless specified, but the species must be correct. The nitronium ion is $\text{NO}_2^+$ , not $\text{NO}_3^+$ .

Techniques used

write equation for electrophile generationbalance acid-base reaction

(b)

Carbon-carbon bond formation is an important reaction in organic synthesis.

Fig. 8.1 shows the synthesis of compound Q from benzene in two reaction steps.

Fig. 8.1

3M

(i)

Draw the structure of compound P in the box in Fig. 8.1.

1M Similar

DifficultyMedium

Worked solution

Answer

Compound P is phenylethanone (acetophenone).

Structure: A benzene ring attached to a $-\text{COCH}_3$ group.

Final answer

Phenylethanone (benzene ring with -COCH3 attached)

Detailed explanation

Background Concept

In multi-step synthesis of substituted benzenes, the order of reactions is crucial due to directing effects.

Alkyl groups (like isopropyl) are ortho/para-directing and activating.
Acyl groups (like acetyl, $-\text{COCH}_3$ ) are meta-directing and deactivating.

To get a meta-substituted product (1,3-disubstituted), the meta-directing group must be present on the ring before the second group is added.

Understanding the Question

We start with benzene and end with compound Q: 1-(3-isopropylphenyl)ethan-1-one. This has an acetyl group at position 1 and an isopropyl group at position 3 (meta). We need to find compound P (the intermediate).

Reaction 1: Benzene $\rightarrow$ P
Reaction 2: P $\rightarrow$ Q

Approach

If we add the isopropyl group first (Friedel-Crafts alkylation), we get isopropylbenzene. The isopropyl group is ortho/para-directing. Adding the acetyl group next would give primarily 1-(4-isopropylphenyl)ethan-1-one (para product). This is not Q.
If we add the acetyl group first (Friedel-Crafts acylation), we get phenylethanone (P). The acetyl group is meta-directing. Adding the isopropyl group next will go to the meta position. This gives Q.
Therefore, P is phenylethanone.

Step-by-Step Reasoning

Analyze product Q: Meta-substituted benzene with $-\text{COCH}_3$ and $-\text{CH(CH}_3)_2$ .
Directing effects: $-\text{COCH}_3$ is meta-directing. $-\text{CH(CH}_3)_2$ is ortho/para-directing.
To get meta substitution, the meta-director ( $-\text{COCH}_3$ ) must be on the ring first.
Reaction 1 must be the introduction of the acetyl group. Benzene + $\text{CH}_3\text{COCl}$ / $\text{AlCl}_3$ $\rightarrow$ Phenylethanone ( $\text{C}_6\text{H}_5\text{COCH}_3$ ).
So P is phenylethanone.

Key Takeaways

Always consider directing effects when planning aromatic synthesis. To get meta products, install the meta-directing group first.

Common Mistakes

Assuming alkylation comes before acylation.
Drawing the wrong structure for P (e.g., isopropylbenzene).

Things to Be Careful About

The structure must show the benzene ring and the acetyl group clearly. The bond from the ring to the carbonyl carbon.

Techniques used

deduce intermediate structure from productapply directing effects to synthesis planning

(ii)

Suggest reagents and conditions for reactions 1 and 2 in Fig. 8.1.

2M Similar

DifficultyMedium

Worked solution

Answer

Reaction 1:
Reagents: Ethanoyl chloride ( $\text{CH}_3\text{COCl}$ ) and anhydrous $\text{AlCl}_3$ .
Conditions: Room temperature / warm.

Reaction 2:
Reagents: 2-bromopropane ( $(\text{CH}_3)_2\text{CHBr}$ ) and $\text{FeBr}_3$ (or $\text{AlBr}_3$ ).
Conditions: Room temperature / warm.

Final answer

Rxn 1: CH3COCl, AlCl3. Rxn 2: (CH3)2CHBr, FeBr3.

Detailed explanation

Background Concept

Friedel-Crafts reactions are used to attach alkyl or acyl groups to benzene rings.

Acylation: Benzene + acyl chloride ( $\text{RCOCl}$ ) + Lewis acid catalyst ( $\text{AlCl}_3$ ) $\rightarrow$ phenyl ketone ( $\text{C}_6\text{H}_5\text{COR}$ ). This installs a meta-directing group.
Alkylation: Benzene + haloalkane ( $\text{R-X}$ ) + Lewis acid catalyst ( $\text{FeX}_3$ or $\text{AlX}_3$ ) $\rightarrow$ alkylbenzene ( $\text{C}_6\text{H}_5\text{R}$ ). This installs an ortho/para-directing group.

Understanding the Question

We established in (b)(i) that Reaction 1 is acylation (Benzene $\rightarrow$ Phenylethanone) and Reaction 2 is alkylation (Phenylethanone $\rightarrow$ meta-isopropyl product).

Approach

Identify the specific reagents for each transformation.

Reaction 1: Need to add $-\text{COCH}_3$ . Use ethanoyl chloride ( $\text{CH}_3\text{COCl}$ ) and $\text{AlCl}_3$ .
Reaction 2: Need to add isopropyl group ( $-\text{CH(CH}_3)_2$ ). Use 2-bromopropane ( $\text{CH}_3\text{CHBrCH}_3$ ) and a Lewis acid like $\text{FeBr}_3$ or $\text{AlBr}_3$ .

Step-by-Step Reasoning

Reaction 1 (Acylation): To convert benzene to phenylethanone ( $\text{C}_6\text{H}_5\text{COCH}_3$ ), use ethanoyl chloride ( $\text{CH}_3\text{COCl}$ ) and aluminium chloride ( $\text{AlCl}_3$ ) catalyst. Conditions are typically room temperature or slightly warm.
Reaction 2 (Alkylation): To convert phenylethanone to 1-(3-isopropylphenyl)ethan-1-one, we need to add an isopropyl group. The reagent is 2-bromopropane (isopropyl bromide, $(\text{CH}_3)_2\text{CHBr}$ ). The catalyst is iron(III) bromide ( $\text{FeBr}_3$ ) or aluminium bromide ( $\text{AlBr}_3$ ). Note: Since the ring is deactivated by the acetyl group, heating may be required, but 'warm' or 'room temp' is often accepted in marking schemes for simplicity, though strictly it needs more vigorous conditions than benzene. The mark scheme accepts $\text{FeBr}_3$ .

Key Takeaways

Friedel-Crafts acylation uses $\text{RCOCl}/\text{AlCl}_3$ . Friedel-Crafts alkylation uses $\text{RX}/\text{FeX}_3$ .

Common Mistakes

Using $\text{H}_2\text{SO}_4$ as catalyst for Friedel-Crafts (that's for nitration/sulfonation).
Forgetting the catalyst ( $\text{AlCl}_3$ , $\text{FeBr}_3$ ).
Wrong alkyl halide (e.g., 1-bromopropane would give a straight chain, though rearrangement can occur, 2-bromopropane is direct).

Things to Be Careful About

Anhydrous conditions are required for Friedel-Crafts reactions (water destroys the Lewis acid catalyst). The mark scheme specifically lists $\text{FeBr}_3$ for the second step.

Techniques used

identify Friedel-Crafts acylation reagentsidentify Friedel-Crafts alkylation reagents

(c)

Separate samples of $\text{C}_6\text{H}_5\text{Br}$ and $\text{C}_6\text{H}_5\text{CH}_2\text{Br}$ are added to warm $\text{AgNO}_3\text{(aq)}$ .

State the expected observations, if any. Explain your answer.

3M Similar

DifficultyMedium

Worked solution

Answer

Observations:

With $\text{C}_6\text{H}_5\text{Br}$ (bromobenzene): No change / no precipitate.
With $\text{C}_6\text{H}_5\text{CH}_2\text{Br}$ (benzyl bromide): Cream precipitate.

Explanation:
In $\text{C}_6\text{H}_5\text{Br}$ , the lone pair on the bromine atom is delocalised into the pi-system of the benzene ring. This gives the $\text{C-Br}$ bond partial double bond character, making it stronger and harder to break. In $\text{C}_6\text{H}_5\text{CH}_2\text{Br}$ , the bromine is attached to an $sp^3$ carbon, so the $\text{C-Br}$ bond is a standard polar bond and can be hydrolysed by water/ $\text{Ag}^+$ to form $\text{AgBr}$ precipitate.

Final answer

C6H5Br: no change. C6H5CH2Br: cream ppt. Explanation: delocalisation strengthens C-Br bond in C6H5Br.

Detailed explanation

Background Concept

Halogenoarenes (aryl halides) like bromobenzene are unreactive towards nucleophilic substitution under normal conditions (like warming with aqueous silver nitrate). Halogenoalkanes (alkyl halides) like benzyl bromide react readily.

The difference is due to the bonding. In bromobenzene, the lone pairs on the halogen are in p-orbitals that overlap with the delocalized pi-system of the benzene ring. This delocalization gives the $\text{C-Br}$ bond partial double bond character, increasing its bond enthalpy and making it difficult to break.

In benzyl bromide ( $\text{C}_6\text{H}_5\text{CH}_2\text{Br}$ ), the bromine is attached to an $sp^3$ hybridized carbon in the side chain. The lone pairs on bromine do not delocalize into the ring (they are orthogonal or separated by the $sp^3$ carbon). The $\text{C-Br}$ bond is a normal single bond, polar, and susceptible to nucleophilic attack (hydrolysis) or ionization to form $\text{Br}^-$ ions.

Understanding the Question

We are adding warm $\text{AgNO}_3$ (aq) to two compounds: bromobenzene ( $\text{C}_6\text{H}_5\text{Br}$ ) and benzyl bromide ( $\text{C}_6\text{H}_5\text{CH}_2\text{Br}$ ). We need to state observations and explain the difference.

Approach

Recall the test: $\text{Ag}^+$ ions react with free $\text{Br}^-$ ions to form a cream precipitate of $\text{AgBr}$ .
Determine which compound releases $\text{Br}^-$ ions. Benzyl bromide hydrolyzes to release $\text{Br}^-$ . Bromobenzene does not.
Explain why bromobenzene doesn't react (delocalization, bond strength).

Step-by-Step Reasoning

Observation for $\text{C}_6\text{H}_5\text{Br}$ : No reaction. The solution remains clear (or no precipitate forms). Mark scheme: 'no change / no precipitate'.
Observation for $\text{C}_6\text{H}_5\text{CH}_2\text{Br}$ : Hydrolysis occurs. $\text{C}_6\text{H}_5\text{CH}_2\text{Br} + \text{H}_2\text{O} \rightarrow \text{C}_6\text{H}_5\text{CH}_2\text{OH} + \text{H}^+ + \text{Br}^-$ . The $\text{Br}^-$ ions react with $\text{Ag}^+$ to form $\text{AgBr}$ . $\text{AgBr}$ is a cream precipitate. Mark scheme: 'cream precipitate'.
Explanation for $\text{C}_6\text{H}_5\text{Br}$ : The lone pair on the Br atom is delocalised into the benzene ring (pi system). This creates resonance structures where there is a $\text{C=Br}^+$ bond character. This makes the $\text{C-Br}$ bond stronger (higher bond enthalpy) and shorter, so it doesn't break easily to release $\text{Br}^-$ .

Key Takeaways

Aryl halides are unreactive due to delocalization. Benzylic halides are very reactive (even more than typical alkyl halides due to stabilization of the carbocation intermediate, though here we just focus on bond strength/delocalization argument for the unreactive one).

Common Mistakes

Saying 'no reaction' for both (wrong, benzyl bromide reacts).
Saying 'white precipitate' for bromine (AgBr is cream, AgCl is white, AgI is yellow).
Not mentioning delocalization or partial double bond character.

Things to Be Careful About

Color of precipitate: AgBr is cream (not white, not yellow).
The explanation must link delocalization to bond strength/character.

Techniques used

predict observations from nucleophilic substitutionexplain C-Br bond strength differences

(d)

Acyl bromides, $\text{RCOBr}$ , react readily with $\text{H}_2\text{O}$ .

The mechanism of this reaction is similar to that of the reaction of $\text{H}_2\text{O}$ with acyl chlorides, $\text{RCOCl}$ .

5M

(i)

Name the mechanism of this reaction.

1M Similar

(ii)

Complete the mechanism in Fig. 8.2 for the reaction of $\text{RCOBr}$ with $\text{H}_2\text{O}$ .

Include all relevant lone pairs of electrons, curly arrows, charges and dipoles.

Draw the structure of the intermediate.

Fig. 8.2

4M Similar

DifficultyMedium-Hard

Worked solution

Answer

Mechanism Steps:

Nucleophilic Attack: A lone pair on the oxygen of $\text{H}_2\text{O}$ attacks the $\delta^+$ carbon of the $\text{C=O}$ group in $\text{RCOBr}$ . Simultaneously, the $\text{C=O}$ pi bond breaks, with electrons moving to the oxygen atom. (Curly arrow from O lone pair to C; curly arrow from C=O bond to O). Dipoles: $\text{C=O}$ has $\delta^+$ on C and $\delta^-$ on O.
Intermediate: A tetrahedral intermediate is formed. The central carbon is bonded to R, Br, $\text{O}^-$ (with 3 lone pairs and negative charge), and $\text{OH}_2^+$ (oxygen with positive charge, bonded to 2 H and the central C, with 1 lone pair).
Elimination: A lone pair on the $\text{O}^-$ moves to reform the $\text{C=O}$ double bond. Simultaneously, the $\text{C-Br}$ bond breaks, with electrons moving to the bromine atom to form $\text{Br}^-$ . (Curly arrow from $\text{O}^-$ lone pair to C-O bond; curly arrow from C-Br bond to Br).

Final answer

See mechanism diagram: H2O attacks C=O, tetrahedral intermediate forms, then C=O reforms and Br leaves.

Detailed explanation

Background Concept

The mechanism of nucleophilic addition-elimination for acyl halides:

The carbonyl carbon is electrophilic ( $\delta^+$ ) due to the electronegativity of oxygen and the halogen.
The carbonyl oxygen is nucleophilic/basic ( $\delta^-$ ).
Step 1: Nucleophile ( $\text{H}_2\text{O}$ ) attacks $\text{C}$ . Pi electrons move to O. Tetrahedral intermediate forms.
Step 2: Lone pair on O- reforms pi bond. Leaving group (Br-) leaves.
Final step (often not drawn in detail or implied): Deprotonation of the $\text{OH}_2^+$ group to give the carboxylic acid and $\text{HBr}$ .

Understanding the Question

Complete the mechanism for $\text{RCOBr} + \text{H}_2\text{O}$ . Show lone pairs, curly arrows, charges, dipoles, and the intermediate structure.

Approach

Draw reactants with dipoles and lone pairs.
Draw arrows for the first step (addition).
Draw the intermediate (tetrahedral).
Draw arrows for the second step (elimination).

Step-by-Step Reasoning

Reactants:

$\text{R-C(=O)-Br}$ : Carbon is $\delta^+$ , Oxygen is $\delta^-$ . Bromine is electronegative but C-Br bond is less polarized than C=O in terms of reactivity here, though Br is the leaving group. Actually, C is $\delta^+$ due to both O and Br. O is $\delta^-$ .
$\text{H}_2\text{O}$ : Oxygen has 2 lone pairs.

Step 1 (Addition):

Curly arrow from a lone pair on $\text{H}_2\text{O}$ oxygen to the carbonyl carbon ( $\text{C}$ ).
Curly arrow from the $\text{C=O}$ double bond to the oxygen atom ( $\text{O}$ ).
Dipoles: Show $\delta^+$ on C and $\delta^-$ on O of the carbonyl group.

Intermediate:

Central Carbon bonded to: R, Br, $\text{O}^-$ , and $\text{O}^+\text{H}_2$ .
The $\text{O}^-$ has 3 lone pairs and a negative charge.
The $\text{O}^+$ (from water) has 1 lone pair and a positive charge (bonded to 2 H and C).
Note: The mark scheme image shows the intermediate with $\text{O}^-$ and $\text{O}^+\text{H}_2$ .

Step 2 (Elimination):

Curly arrow from a lone pair on the $\text{O}^-$ to the C-O single bond (to reform $\text{C=O}$ ).
Curly arrow from the C-Br bond to the Br atom (to form $\text{Br}^-$ ).

Products (implied):

$\text{RCOOH}$ (after loss of $\text{H}^+$ ) and $\text{HBr}$ (or $\text{Br}^-$ and $\text{H}^+$ ). The question asks for the mechanism up to products, usually showing the elimination step leading to the acid and HBr.

Key Takeaways

Arrow goes from lone pair to electron-deficient atom.
Arrow goes from bond to atom (heterolytic fission).
Intermediate is tetrahedral.
Charges must balance.

Common Mistakes

Arrow from O of water to H (that's acid-base, not the main mechanism step here, though proton transfer happens).
Forgetting lone pairs on O in water and intermediate.
Wrong charges in intermediate (must have $\text{O}^-$ and $\text{O}^+$ ).
Arrow from C-Br bond to C (wrong direction).

Things to Be Careful About

Dipoles on $\text{C=O}$ ( $\delta^+$ on C, $\delta^-$ on O) are often required for marks.
The intermediate must show the correct connectivity and charges.

Techniques used

draw curly arrow mechanismidentify tetrahedral intermediateshow dipoles and lone pairs

Q9MediumNitrogen CompoundsIntroduction to A Level Organic Chemistry

(a)

Explain why amides are much weaker bases than amines.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Basicity depends on the availability of the lone pair on nitrogen to accept a proton ( $\text{H}^+$ ). In amides, the lone pair on nitrogen is delocalised into the $\text{C=O}$ group by resonance, making it unavailable for protonation.

Final answer

The lone pair on N in amides is delocalised into the C=O group, reducing its availability to accept a proton.

Detailed explanation

Background Concept

A base is a proton acceptor (Brønsted–Lowry definition). In amines, the nitrogen atom has a lone pair of electrons in a $\text{sp}^3$ hybrid orbital that is available to form a dative covalent bond with $\text{H}^+$ . The strength of a base depends on how readily this lone pair can be donated — the more available and higher in energy the lone pair, the stronger the base.

Amides ( $\text{RCONH}_2$ ) also have a nitrogen with a lone pair, but they are essentially neutral in aqueous solution and do not act as bases under normal conditions. The key difference lies in the electronic environment of the nitrogen.

Understanding the Question

The question asks you to explain why amides are much weaker bases than amines. The command word is "explain", so you must give the reason, not just state the observation. You need to address two linked points: what makes a species basic (lone pair availability to accept $\text{H}^+$ ), and why that lone pair is unavailable in amides specifically.

Approach

Define basicity in terms of lone pair donation to a proton.
Identify the structural feature in amides that removes the lone pair from availability — resonance delocalisation into the carbonyl group.

Step-by-Step Reasoning

Point 1 (M1): Basicity is the ability of a species to donate a lone pair to accept a proton ( $\text{H}^+$ ). In amines, the nitrogen lone pair is localised on the nitrogen atom in an $\text{sp}^3$ orbital, making it available for protonation to form an ammonium ion.

Point 2 (M2): In amides, the nitrogen is directly bonded to a carbonyl carbon ( $\text{C=O}$ ). The lone pair on nitrogen can overlap with the $\pi$ -system of the $\text{C=O}$ group, giving resonance structures where the lone pair is delocalised onto the oxygen atom:

\text{R}-\overset{\cdot\cdot}{\text{N}}\text{H}_2-\text{C}(=\text{O})\text{R}' \leftrightarrow \text{R}-\text{NH}_2^+=\text{C}(-\text{O}^-)\text{R}'

This delocalisation means the lone pair is no longer fully available on nitrogen for protonation, so the amide is essentially neutral and a far weaker base than an amine.

Key Takeaways

Basicity requires an available lone pair to donate to $\text{H}^+$ .
Resonance delocalisation of a lone pair into an adjacent $\pi$ -system (as in amides) removes its availability for protonation.
Amides are neutral; amines are basic.

Common Mistakes

Saying only "the lone pair is delocalised" without explaining that this makes it unavailable for protonation (misses M1).
Confusing amides with carboxylic acids — amides are not acidic; they are neutral because the lone pair is tied up in resonance.
Writing that the $\text{C=O}$ group "withdraws electrons inductively" — while there is some inductive effect, the dominant explanation is resonance delocalisation.

Things to Be Careful About

The mark scheme requires BOTH points: the definition of basicity (lone pair accepts $\text{H}^+$ ) AND the reason for reduced basicity (delocalisation into $\text{C=O}$ ).
Use precise language: "delocalised into the $\text{C=O}$ group" or "into the carbonyl group" — not just "into the molecule".

Techniques used

explain delocalisation of lone pair into carbonylrelate lone pair availability to basicity

(b)

Fig. 9.1 shows the preparation of 2-phenylethylamine, $\text{C}_6\text{H}_5\text{CH}_2\text{CH}_2\text{NH}_2$ , by three different routes.

Fig. 9.1

3M

(i)

Suggest structures for compounds M and N and draw them in the boxes in Fig. 9.1.

2M Similar

DifficultyMedium

Worked solution

Answer

M (reduced by $\text{H}_2/\text{Ni}$ to give the amine): 2-phenylethanenitrile (benzyl cyanide)

\text{C}_6\text{H}_5\text{CH}_2\text{CN}

N (reacts with $\text{NH}_3$ to give the amine): (2-bromoethyl)benzene

\text{C}_6\text{H}_5\text{CH}_2\text{CH}_2\text{Br}

Final answer

M = C6H5CH2CN (2-phenylethanenitrile); N = C6H5CH2CH2Br (2-bromoethylbenzene)

Detailed explanation

Background Concept

Primary amines can be prepared by several routes:

Reduction of nitriles: $\text{RCN} + 2\text{H}_2 \xrightarrow{\text{Ni}} \text{RCH}_2\text{NH}_2$ . The $\text{C}\equiv\text{N}$ triple bond is reduced to $\text{CH}_2\text{NH}_2$ , adding one carbon to the chain relative to the nitrile carbon.
Nucleophilic substitution of halogenoalkanes with ammonia: $\text{RX} + \text{NH}_3 \rightarrow \text{RNH}_2 + \text{HX}$ . The halogen is replaced by $\text{NH}_2$ .
Reduction of amides: $\text{RCONH}_2 \xrightarrow{\text{LiAlH}_4} \text{RCH}_2\text{NH}_2$ . The $\text{C=O}$ is reduced to $\text{CH}_2$ .

Understanding the Question

The reaction scheme (Fig. 9.1) shows three routes to 2-phenylethylamine ( $\text{C}_6\text{H}_5\text{CH}_2\text{CH}_2\text{NH}_2$ ). You are given:

Route 1: 2-phenylethanamide $\xrightarrow{\text{reaction 1}}$ product
Route 2: M $\xrightarrow{\text{H}_2/\text{Ni}}$ product
Route 3: N $\xrightarrow{\text{NH}_3}$ product

You must deduce M and N by working backwards from the reagents.

Approach

For M: $\text{H}_2/\text{Ni}$ is a catalytic hydrogenation reagent. It reduces nitriles ( $\text{C}\equiv\text{N}$ ) to primary amines ( $\text{CH}_2\text{NH}_2$ ). The product is $\text{C}_6\text{H}_5\text{CH}_2\text{CH}_2\text{NH}_2$ , so removing the $\text{CH}_2\text{NH}_2$ and replacing with $\text{CN}$ gives $\text{C}_6\text{H}_5\text{CH}_2\text{CN}$ .

For N: $\text{NH}_3$ acts as a nucleophile in substitution reactions with halogenoalkanes. The product has $\text{CH}_2\text{NH}_2$ , so the starting material must have a halogen on that carbon: $\text{C}_6\text{H}_5\text{CH}_2\text{CH}_2\text{Br}$ (or Cl or I).

Step-by-Step Reasoning

M (benzyl cyanide):

$\text{H}_2/\text{Ni}$ reduces nitriles to primary amines.
Product: $\text{C}_6\text{H}_5\text{CH}_2\text{CH}_2\text{NH}_2$
The $\text{CH}_2\text{NH}_2$ group comes from reduction of $\text{CN}$ .
Therefore M = $\text{C}_6\text{H}_5\text{CH}_2\text{CN}$ (2-phenylethanenitrile)

N (2-phenylethyl bromide):

$\text{NH}_3$ substitutes a halogen on a halogenoalkane.
Product: $\text{C}_6\text{H}_5\text{CH}_2\text{CH}_2\text{NH}_2$
The $\text{NH}_2$ replaces a halogen on the terminal $\text{CH}_2$ .
Therefore N = $\text{C}_6\text{H}_5\text{CH}_2\text{CH}_2\text{Br}$ (or Cl, I)

Key Takeaways

$\text{H}_2/\text{Ni}$ reduces nitriles to primary amines (adds one $\text{CH}_2$ from the nitrile carbon).
$\text{NH}_3$ (excess, heat, pressure) converts halogenoalkanes to primary amines by nucleophilic substitution.
Working backwards from reagents to deduce starting materials is a key A2 skill.

Common Mistakes

For M, drawing $\text{C}_6\text{H}_5\text{CH}_2\text{CH}_2\text{CN}$ (which would give a 9-carbon amine, not 8).
For N, drawing $\text{C}_6\text{H}_5\text{CHBrCH}_3$ (wrong position of halogen — would give a different amine).
Confusing the nitrile route with the amide route — both give primary amines but the nitrile adds a carbon while the amide does not.

Things to Be Careful About

The nitrile carbon becomes the $\text{CH}_2$ next to $\text{NH}_2$ in the product, so the nitrile must have one fewer carbon in the chain than the product.
Any halogen (Cl, Br, I) is acceptable for N.

Techniques used

deduce starting material from reduction conditionsdeduce starting material from nucleophilic substitution reagentidentify nitrile reduction with H2/Niidentify halogenoalkane amination with NH3

(ii)

Give the reagents and conditions for reaction 1.

1M Similar

DifficultyEasy

Worked solution

Answer

$\text{LiAlH}_4$ (in dry ether, followed by hydrolysis)

Final answer

LiAlH4

Detailed explanation

Background Concept

Amides can be reduced to primary amines using a strong reducing agent. The $\text{C=O}$ group is reduced to $\text{CH}_2$ , converting $\text{RCONH}_2$ to $\text{RCH}_2\text{NH}_2$ . Lithium tetrahydridoaluminate(III) ( $\text{LiAlH}_4$ ) is the standard reagent for this reduction. It is used in dry ether (anhydrous conditions) because it reacts violently with water, followed by hydrolysis with dilute acid or water to release the amine.

Understanding the Question

Reaction 1 converts 2-phenylethanamide ( $\text{C}_6\text{H}_5\text{CH}_2\text{CONH}_2$ ) to 2-phenylethylamine ( $\text{C}_6\text{H}_5\text{CH}_2\text{CH}_2\text{NH}_2$ ). This is a reduction of an amide to an amine. You need to state the reagent.

Approach

Identify the transformation: amide $\rightarrow$ amine (loss of oxygen, gain of two hydrogens on the carbonyl carbon). This requires a strong hydride reducing agent.

Step-by-Step Reasoning

The $\text{C=O}$ in the amide is reduced to $\text{CH}_2$ .
$\text{NaBH}_4$ is not strong enough to reduce amides (it reduces aldehydes and ketones only).
$\text{LiAlH}_4$ is the correct reagent — it reduces amides, carboxylic acids, and their derivatives.
Conditions: dry ether (anhydrous solvent), then hydrolysis.

Key Takeaways

$\text{LiAlH}_4$ reduces amides to amines; $\text{NaBH}_4$ does not.
$\text{LiAlH}_4$ must be used in dry ether due to its violent reaction with water.

Common Mistakes

Writing $\text{NaBH}_4$ — this cannot reduce amides.
Writing "heat with $\text{H}_2/\text{Ni}$ " — this reduces nitriles but not amides under normal conditions.
Omitting the anhydrous/dry ether condition (though the mark scheme here only requires the reagent name).

Things to Be Careful About

The mark scheme accepts $\text{LiAlH}_4$ alone. Adding "dry ether" or "followed by hydrolysis" shows understanding but is not required for the mark.

Techniques used

identify reagent for reduction of amide to amine

(c)

Fig. 9.2 shows compound H which is a useful starting material in organic synthesis.

Fig. 9.2

H contains an alkene and an amine functional group.

Name the other functional group and give the classification of the amine group in H.

1M Similar

DifficultyEasy

Worked solution

Answer

Other functional group: ketone (carbonyl)

Classification of amine: secondary ( $2^\circ$ )

Final answer

ketone (carbonyl) and secondary (2°) amine

Detailed explanation

Background Concept

In organic chemistry, functional groups are identified by their characteristic bonding patterns:

Ketone: a $\text{C=O}$ group where the carbonyl carbon is bonded to two other carbon atoms (not to H, OH, or N).
Amine classification: Primary ( $1^\circ$ ) amines have one carbon group attached to N ( $\text{RNH}_2$ ); secondary ( $2^\circ$ ) have two ( $\text{R}_2\text{NH}$ ); tertiary ( $3^\circ$ ) have three ( $\text{R}_3\text{N}$ ).

Understanding the Question

Compound H (Fig. 9.2) is given. You are told it contains an alkene and an amine. You must name the other functional group and classify the amine. Looking at the structure: there is a $\text{C=O}$ bonded to two carbons (a methyl group and a $\text{CH}_2$ group) — this is a ketone. The nitrogen has two carbon groups attached (a $\text{CH}_2\text{COCH}_3$ group and a $\text{CH}_2$ -cyclohexenyl group) and one H — this is a secondary amine.

Approach

Scan the structure for all functional groups beyond the given alkene and amine.
Identify the $\text{C=O}$ with two carbon substituents as a ketone.
Count carbon groups on nitrogen to classify the amine.

Step-by-Step Reasoning

Ketone identification: The $\text{C=O}$ group has a methyl ( $\text{CH}_3$ ) on one side and a $\text{CH}_2$ on the other — both are carbon groups, so it is a ketone (not an aldehyde, which would need H on one side, nor a carboxylic acid/ester/amide which would need heteroatom on the other side).

Amine classification: The nitrogen is bonded to: (1) a $\text{CH}_2$ -cyclohexenyl group, (2) a $\text{CH}_2\text{COCH}_3$ group, and (3) a hydrogen. Two carbon groups + one H = secondary amine.

Key Takeaways

A ketone is a carbonyl flanked by two carbon groups.
Amine classification depends on the number of carbon groups directly bonded to nitrogen, not the number of carbons in those groups.

Common Mistakes

Calling the $\text{C=O}$ an "aldehyde" — it has no H directly on the carbonyl carbon.
Calling the amine "primary" because it has an N-H — the classification is based on carbon groups on N, not H atoms.
Saying "carbonyl" without specifying "ketone" — while carbonyl is technically correct, ketone is more specific and preferred.

Things to Be Careful About

The mark scheme requires BOTH answers (ketone AND secondary) for the single mark. Getting one right but not the other scores zero.
"Carbonyl" is accepted as an alternative to "ketone" per the mark scheme.

Techniques used

identify functional groups from skeletal structureclassify amine by number of carbon groups on nitrogen

(d)

Ozonolysis involves the oxidative cleavage of a C=C bond in alkenes using ozone, $\text{O}_3$ , as shown in Fig. 9.3.

Fig. 9.3

Fig. 9.4 shows the first step in this reaction which involves the formation of an ozonide intermediate.

Fig. 9.4

3M

(i)

On Fig. 9.4, draw three curly arrows to complete the mechanism of this step.

2M Similar

DifficultyMedium

Worked solution

Answer

Three curly arrows:

From the lone pair on $\text{O}^-$ to one carbon of the $\text{C=C}$ double bond.
From the $\text{C=C}$ bond to the terminal (double-bonded) oxygen of ozone.
From the $\text{O=O}$ $\pi$ -bond to the central $\text{O}^+$ .

Final answer

Curly arrow 1: O- lone pair to C=C carbon; Curly arrow 2: C=C bond to terminal O; Curly arrow 3: O=O bond to central O+

Detailed explanation

Background Concept

Ozonolysis begins with a 1,3-dipolar cycloaddition between ozone (a 1,3-dipole) and an alkene (a dipolarophile). Ozone has a resonance structure where the terminal oxygen carries a negative charge (with three lone pairs) and the central oxygen carries a positive charge: $^-\text{O}-\text{O}^+=\text{O}$ . The negatively charged terminal oxygen acts as a nucleophile, attacking one carbon of the electron-rich $\text{C=C}$ double bond. Simultaneously, the $\pi$ -electrons of the alkene attack the other terminal oxygen, and the $\pi$ -electrons of the $\text{O=O}$ bond shift onto the central oxygen to neutralise its positive charge. This concerted process forms the five-membered cyclic ozonide (molozonide).

Understanding the Question

You are shown Fig. 9.4 with the alkene and the polarised ozone molecule ( $^-\text{O}-\text{O}^+=\text{O}$ ). You must draw three curly arrows showing how electrons move to form the cyclic ozonide intermediate. The product (ozonide) is already shown, so you can verify your arrows by checking they produce the correct bonding changes.

Approach

Identify the nucleophilic site on ozone: the $\text{O}^-$ with its lone pair.
Identify the electrophilic site on the alkene: either carbon of the $\text{C=C}$ (both are electron-rich, but one will bond to $\text{O}^-$ ).
Trace the electron flow that forms the two new C-O bonds and breaks the O=O $\pi$ -bond.

Step-by-Step Reasoning

Arrow 1 (nucleophilic attack): The lone pair on $\text{O}^-$ attacks one carbon of the $\text{C=C}$ , forming a new C-O single bond. This is the nucleophile-to-electrophile arrow.

Arrow 2 (electrophilic capture): The $\pi$ -electrons of the $\text{C=C}$ bond shift to form a bond with the other terminal oxygen (the one double-bonded to the central O). This creates the second C-O bond of the ring.

Arrow 3 (charge neutralisation): The $\pi$ -electrons of the $\text{O=O}$ bond shift onto the central $\text{O}^+$ , neutralising its positive charge and converting the double bond to a single bond. This gives the final ozonide structure with all oxygens neutral.

The three arrows together show a concerted cycloaddition: two new sigma bonds form (C-O and C-O), the C=C $\pi$ -bond breaks, and the O=O $\pi$ -bond breaks.

Key Takeaways

Curly arrows always go from electron source (lone pair or bond) to electron destination (atom or space between atoms).
In concerted mechanisms, multiple arrows are drawn simultaneously to show all electron movements in one step.
The negative oxygen is the nucleophile; the positive oxygen needs its charge neutralised.

Common Mistakes

Drawing the arrow from the $\text{O}^-$ to the wrong carbon (either carbon works, but must be consistent with the product shown).
Drawing arrow 2 from the carbon to the oxygen (wrong direction — arrows go FROM electrons TO destination).
Forgetting arrow 3 — without it, the central oxygen retains its positive charge and the product doesn't match the ozonide shown.
Drawing the arrow from the $\text{O}^-$ to the $\text{O}^+$ directly (this doesn't form the ring).

Things to Be Careful About

Arrow tails must start at the electron source (lone pair dot or bond line), not at the atom symbol.
Arrow heads must point to where the electrons are going (between atoms for bond formation, or onto an atom for bond breaking).
The three arrows must be consistent with the product structure shown on the right of Fig. 9.4.

Techniques used

draw curly arrows for nucleophilic attack on electrophilic carbondraw curly arrows for electron movement in cycloadditionshow bond breaking with curly arrow to positive centre

(ii)

L is formed from alkene K, $\text{C}_8\text{H}_{14}$ , by a similar reaction to that shown in Fig. 9.3.

Fig. 9.5

Suggest the structure of K.

1M Similar

DifficultyMedium

Worked solution

Answer

Compound L is octane-2,6-dione: $\text{CH}_3\text{COCH}_2\text{CH}_2\text{CH}_2\text{COCH}_2\text{CH}_3$

The two carbonyl carbons (C2 and C6) were originally joined by a $\text{C=C}$ in a ring. Joining them gives a five-membered ring (C2-C3-C4-C5-C6) with methyl on C2 and ethyl on C6.

K is 1-methyl-2-ethylcyclopentene:

Molecular formula: $\text{C}_8\text{H}_{14}$ ✓

Final answer

1-methyl-2-ethylcyclopentene (C8H14)

Detailed explanation

Background Concept

Ozonolysis cleaves a $\text{C=C}$ double bond, converting each carbon of the double bond into a carbonyl carbon ( $\text{C=O}$ ). If the alkene is part of a ring, cleavage opens the ring to give a dicarbonyl compound (a dialdehyde, diketone, or keto-aldehyde depending on substitution). To work backwards from the product to the starting alkene, you identify the two carbonyl carbons and join them with a $\text{C=C}$ bond, reconnecting the chain that was broken.

Understanding the Question

L is a diketone with the structure $\text{CH}_3\text{COCH}_2\text{CH}_2\text{CH}_2\text{COCH}_2\text{CH}_3$ (octane-2,6-dione), formed by ozonolysis of cyclic alkene K ( $\text{C}_8\text{H}_{14}$ ). You must deduce K.

Approach

Identify the two carbonyl carbons in L — these were the two carbons of the $\text{C=C}$ in K.
Count the carbons between the two carbonyl carbons in the chain — these form the rest of the ring.
Join the two carbonyl carbons with a $\text{C=C}$ to close the ring.
Verify the molecular formula matches $\text{C}_8\text{H}_{14}$ .

Step-by-Step Reasoning

Step 1: The carbonyl carbons are at positions 2 and 6 in the chain.

Step 2: Between C2 and C6 in the chain are C3, C4, C5 — three $\text{CH}_2$ groups. These form part of the ring.

Step 3: On the other side of C2 (outside the ring) is $\text{CH}_3$ (C1). On the other side of C6 is $\text{CH}_2\text{CH}_3$ (C7-C8, an ethyl group).

Step 4: Joining C2 to C6 with a $\text{C=C}$ gives a ring of: C2-C3-C4-C5-C6 (five atoms = cyclopentene ring). C2 bears a methyl group, C6 bears an ethyl group.

Step 5: Structure = 1-methyl-2-ethylcyclopent-1-ene.

Verification: Cyclopentene ring = $\text{C}_5\text{H}_8$ . Adding methyl (replacing one H): $\text{C}_6\text{H}_{10}$ . Adding ethyl (replacing one H): $\text{C}_8\text{H}_{14}$ . ✓

Key Takeaways

Ozonolysis of a cyclic alkene gives a dicarbonyl compound with the chain intact between the two carbonyls.
To reverse ozonolysis, join the two carbonyl carbons with a double bond to reform the ring.
Always verify the molecular formula of your proposed structure.

Common Mistakes

Drawing a six-membered ring by miscounting the carbons between the carbonyls.
Placing the methyl and ethyl groups on the wrong carbons.
Drawing an open-chain alkene instead of a cyclic one (the product is a single molecule with two carbonyls, so the starting material must be cyclic).
Forgetting that the substituents on the carbonyl carbons in L become the substituents on the ring carbons in K.

Things to Be Careful About

The ring size is determined by counting ALL atoms in the ring (including the two carbonyl carbons that become the double-bond carbons): C2, C3, C4, C5, C6 = 5-membered ring.
The formula $\text{C}_8\text{H}_{14}$ has a degree of unsaturation of 2 (one ring + one double bond), consistent with a cycloalkene.

Techniques used

work backwards from ozonolysis product to deduce cyclic alkenejoin carbonyl carbons with C=C to reconstruct ringverify molecular formula

Chemistry 9701/41 — October/November 2025

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Working

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Working

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning