9701/42 Feb/Mar 2025 — Chemistry Solved Past Paper

Q1MediumTransition ElementsChemical EnergeticsElectrochemistry

Silver, Ag, is a metal in the d-block of the Periodic Table.

(a)

Silver can form compounds containing either $\text{Ag}^+$ or $\text{Ag}^{2+}$ ions.

Explain why silver is a transition element.

1M Similar

DifficultyEasy

Worked solution

Answer

Silver is a transition element because it forms a stable ion, $\text{Ag}^{2+}$ , which has an incomplete d subshell / partially filled d orbitals.

Final answer

Silver forms a stable Ag2+ ion with an incomplete d subshell.

Detailed explanation

Background Concept

A transition element is defined as an element that forms at least one stable ion with a partially filled d subshell. The d-block elements are those whose highest-energy electrons are in d orbitals, but not all d-block elements are transition elements. For example, zinc (Zn) forms only Zn2+ with the configuration [Ar]3d10 — a full d subshell — so zinc is not a transition element. Silver (Ag) has the electron configuration [Kr]4d10 5s1. It commonly forms Ag+ ([Kr]4d10, full d subshell) but can also form Ag2+ ([Kr]4d9, incomplete d subshell). Because Ag2+ is a stable ion with a partially filled d subshell, silver satisfies the definition and is classified as a transition element.

Understanding the Question

The question asks us to explain why silver, a d-block metal, is classified as a transition element. We need to recall the definition of a transition element and apply it to silver.

Approach

Recall the definition: an element that forms at least one stable ion with a partially filled d subshell. Check whether silver forms such an ion. Ag+ has a full 4d10 subshell, but Ag2+ has 4d9 — partially filled. Since Ag2+ is stable, silver qualifies.

Step-by-Step Reasoning

The definition of a transition element: an element that forms at least one stable ion with an incomplete d subshell.
Silver's electron configuration is [Kr]4d10 5s1.
Ag+ has [Kr]4d10 — the d subshell is full, so Ag+ alone would not qualify.
However, Ag2+ has [Kr]4d9 — the d subshell is incomplete (partially filled).
Ag2+ is a stable ion of silver. Therefore, silver forms a stable ion with an incomplete d subshell and is a transition element.

Key Takeaways

The definition of a transition element is based on the existence of at least one stable ion with a partially filled d subshell.
Being in the d-block is not sufficient; the ion must have an incomplete d subshell.
Silver qualifies via its Ag2+ ion even though its most common ion Ag+ has a full d subshell.

Common Mistakes

Saying "silver is in the d-block so it is a transition element" — this is insufficient; the d-block alone does not define a transition element (zinc and scandium are d-block but not transition elements by this definition).
Stating that Ag+ has an incomplete d subshell — Ag+ has [Kr]4d10, a full d subshell.
Confusing "partially filled" with "unfilled" — the ion must have at least one d electron and at least one vacant d orbital.

Things to Be Careful About

The mark scheme specifically credits the Ag2+ ion with an incomplete d subshell.
Use precise terminology: "incomplete d subshell" or "partially filled d orbitals".

Techniques used

recall the definition of a transition elementidentify the d-electron configuration of Ag and Ag2+

(b)

Table 1.1 gives data relevant to the Born–Haber cycle for silver(I) fluoride, AgF.

Table 1.1

standard energy change	value / $\text{kJ mol}^{-1}$
first ionisation energy of silver	+732
enthalpy change of atomisation of silver	+289
enthalpy change of atomisation of fluorine	+79
enthalpy change of formation of silver(I) fluoride	-203
lattice energy of silver(I) fluoride	-955

6M

(i)

Write equations for the standard enthalpy changes described. Include state symbols.

standard enthalpy change of atomisation of silver
standard enthalpy change of formation of silver(I) fluoride

2M Similar

DifficultyEasy

Worked solution

Answer

Standard enthalpy change of atomisation of silver:

\text{Ag(s)} \rightarrow \text{Ag(g)}

Standard enthalpy change of formation of silver(I) fluoride:

\text{Ag(s)} + \frac{1}{2}\text{F}_2(\text{g}) \rightarrow \text{AgF(s)}

Final answer

Ag(s) → Ag(g); Ag(s) + ½F2(g) → AgF(s)

Detailed explanation

Background Concept

The standard enthalpy change of atomisation of an element is the enthalpy change when 1 mole of gaseous atoms is formed from the element in its standard state. For a solid metal like silver, this is simply the conversion of solid silver to gaseous silver atoms: Ag(s) → Ag(g).

The standard enthalpy change of formation of a compound is the enthalpy change when 1 mole of the compound is formed from its elements in their standard states under standard conditions (298 K, 1 atm). For AgF, the elements are silver (solid) and fluorine (gas, F2). Since one mole of AgF contains one mole of Ag and one mole of F, we need ½ mole of F2 to provide one mole of F atoms.

Understanding the Question

We are asked to write balanced equations, with state symbols, for two standard enthalpy changes: atomisation of silver and formation of silver(I) fluoride.

Approach

Apply the definitions directly:

Atomisation of silver: solid silver → gaseous silver atoms, 1 mole each side.
Formation of AgF: silver (solid) + fluorine (gas) → silver(I) fluoride (solid), balanced so that 1 mole of AgF is produced.

Step-by-Step Reasoning

Atomisation of Ag: one mole of Ag(s) becomes one mole of Ag(g). Equation: Ag(s) → Ag(g). State symbols: (s) for solid, (g) for gas.
Formation of AgF: the elements in their standard states are Ag(s) and F2(g). To make 1 mole of AgF(s), we need 1 mole of Ag atoms and 1 mole of F atoms. F2 provides 2 F atoms, so we use ½ F2. Equation: Ag(s) + ½F2(g) → AgF(s).

Key Takeaways

Atomisation always produces gaseous atoms from the element in its standard state.
Formation always produces exactly 1 mole of the compound from elements in their standard states.
State symbols are essential in these equations.

Common Mistakes

Writing Ag(g) → Ag(s) (reversing the atomisation).
Writing F(g) instead of ½F2(g) for the formation — fluorine's standard state is F2(g).
Forgetting state symbols entirely — the question explicitly asks for them.

Things to Be Careful About

The standard state of fluorine is F2(g), not F(g).
The coefficient ½ is required in front of F2 so that exactly 1 mole of AgF is formed.

Techniques used

write the equation for atomisationwrite the equation for formation

(ii)

Define lattice energy.

2M Similar

(iii)

Calculate the first electron affinity, $\text{EA}_1$ , of fluorine, using data from Table 1.1.

It may be helpful to draw a labelled energy cycle as part of the working for your answer.

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Using the Born-Haber cycle:

\Delta H_f = \Delta H_{\text{at}}(\text{Ag}) + \text{IE}_1(\text{Ag}) + \Delta H_{\text{at}}(\text{F}) + \text{EA}_1(\text{F}) + \text{LE}(\text{AgF})

Substituting:

-203 = (+289) + (+732) + (+79) + \text{EA}_1 + (-955)

\text{EA}_1 = -203 - 289 - 732 - 79 + 955 = -348 \text{ kJ mol}^{-1}

Answer

$\text{EA}_1(\text{F}) = -348 \text{ kJ mol}^{-1}$

Final answer

-348 kJ mol^-1

Detailed explanation

Background Concept

A Born-Haber cycle is an application of Hess's law to the formation of an ionic compound. The enthalpy change of formation of an ionic compound is the sum of several steps:

Atomisation of the metal (Ag(s) → Ag(g))
Ionisation of the metal (Ag(g) → Ag+(g) + e-)
Atomisation of the non-metal (½F2(g) → F(g))
Electron affinity of the non-metal (F(g) + e- → F-(g))
Lattice energy (Ag+(g) + F-(g) → AgF(s))

By Hess's law, the sum of these steps equals the enthalpy change of formation.

Understanding the Question

We are given all the data except the first electron affinity of fluorine and asked to calculate it using the Born-Haber cycle.

Approach

Set up the Born-Haber equation:
ΔHf = ΔHatom(Ag) + IE1(Ag) + ΔHatom(F) + EA1(F) + LE(AgF)
Substitute the known values and solve for EA1.

Step-by-Step Reasoning

Write the Born-Haber equation:
ΔHf = ΔHatom(Ag) + IE1(Ag) + ΔHatom(F) + EA1(F) + LE(AgF)
Substitute values:
-203 = +289 + 732 + 79 + EA1 + (-955)
Simplify the right side:
+289 + 732 + 79 - 955 = 145
So -203 = 145 + EA1
Solve for EA1:
EA1 = -203 - 145 = -348 kJ/mol

Key Takeaways

The Born-Haber cycle sums all the enthalpy changes in forming an ionic compound.
The electron affinity of a halogen is exothermic (negative).
Always check the sign of each term in the cycle.

Common Mistakes

Forgetting to include the atomisation of fluorine (79 kJ/mol).
Sign errors on the lattice energy (it is negative, -955).
Getting the final sign wrong — electron affinity of fluorine is negative.

Things to Be Careful About

The mark scheme requires the result to 3 significant figures minimum.
The electron affinity is negative because energy is released when an electron is added to a fluorine atom.

Techniques used

construct a Born-Haber cycleapply Hess's lawcalculate electron affinity

(c)

Table 1.2 shows some thermodynamic data at 298 K.

Table 1.2

energy change at 298 K	value / $\text{kJ mol}^{-1}$
lattice energy of AgF(s)	-955
enthalpy change of hydration of $\text{Ag}^+(\text{g})$	-464
enthalpy change of hydration of $\text{F}^-(\text{g})$	-506

2M

(i)

Use the data in Table 1.2 to calculate the enthalpy change of solution, $\Delta H_{\text{sol}}$ , of AgF(s).

1M Similar

DifficultyEasy

Worked solution

Working

\Delta H_{\text{sol}} = -\text{LE} + \Delta H_{\text{hyd}}(\text{Ag}^+) + \Delta H_{\text{hyd}}(\text{F}^-)

\Delta H_{\text{sol}} = -(-955) + (-464) + (-506) = 955 - 464 - 506 = -15 \text{ kJ mol}^{-1}

Answer

$\Delta H_{\text{sol}}(\text{AgF}) = -15 \text{ kJ mol}^{-1}$

Final answer

-15 kJ mol^-1

Detailed explanation

Background Concept

The enthalpy change of solution of an ionic compound is the energy change when 1 mole of the solid dissolves in water. It can be calculated from the lattice energy and the hydration enthalpies of the ions:
ΔHsol = -LE + ΔHhyd(cation) + ΔHhyd(anion)
The negative of the lattice energy represents the energy needed to break the lattice (lattice dissociation), and the hydration enthalpies represent the energy released when the ions are surrounded by water molecules.

Understanding the Question

We are given the lattice energy of AgF and the hydration enthalpies of Ag+ and F-, and asked to calculate the enthalpy change of solution.

Approach

Apply the formula ΔHsol = -LE + ΔHhyd(Ag+) + ΔHhyd(F-).

Step-by-Step Reasoning

Write the formula: ΔHsol = -LE + ΔHhyd(Ag+) + ΔHhyd(F-)
Substitute: ΔHsol = -(-955) + (-464) + (-506)
Simplify: ΔHsol = 955 - 464 - 506 = -15 kJ/mol

Key Takeaways

The enthalpy of solution combines lattice breaking (endothermic) and hydration (exothermic).
A negative ΔHsol indicates the dissolution is exothermic and thermodynamically favourable.

Common Mistakes

Forgetting the negative sign on the lattice energy term.
Adding the lattice energy instead of subtracting it.

Things to Be Careful About

The lattice energy is negative (-955), so -LE = +955.
The result is -15 kJ/mol, a small negative value.

Techniques used

apply the enthalpy of solution relationshipcombine lattice and hydration enthalpies

(ii)

Use your answer to (c)(i) to suggest whether AgF is soluble in water at 298 K. Explain your answer.

1M Similar

(d)

Table 1.3 shows some data relevant to the silver(I) halides, AgCl to AgI.

Table 1.3

silver(I) halide	first electron affinity of halogen / $\text{kJ mol}^{-1}$	lattice energy / $\text{kJ mol}^{-1}$
AgCl	-349	-905
AgBr	-325	-890
AgI	-295	-876

3M

(i)

Explain the trend in the first electron affinities of the halogens, Cl to I.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

As we go from Cl to I, the atomic radius increases / there are more electron shells / more shielding by inner shells. The nucleus has less attraction for the incoming electron, so the first electron affinity becomes less exothermic (less negative).

Final answer

Atomic radius increases down the group, more shielding, less attraction for the incoming electron, so electron affinity becomes less exothermic (less negative).

Detailed explanation

Background Concept

The first electron affinity is the energy change when one mole of gaseous atoms each gains one electron to form one mole of gaseous 1- ions:
X(g) + e- → X-(g)
For halogens, this is exothermic because the added electron enters a partially filled p subshell and is attracted to the nucleus. Down the group, the atomic radius increases and there is more shielding, so the attraction between the nucleus and the incoming electron decreases, making the electron affinity less exothermic (less negative).

Understanding the Question

We are asked to explain why the first electron affinities of the halogens become less negative from Cl to I.

Approach

Identify the trend: EA1 becomes less exothermic down the group. Explain using atomic size and shielding arguments.

Step-by-Step Reasoning

From Cl to I, the atomic radius increases because each successive element has an additional electron shell.
The additional inner shells provide more shielding of the nuclear charge.
The effective nuclear charge experienced by the incoming electron decreases.
Therefore, less energy is released when the electron is added, making EA1 less exothermic (less negative).

Key Takeaways

Electron affinity becomes less exothermic down a group.
The trend is explained by increasing atomic radius and shielding.

Common Mistakes

Saying electron affinity becomes more exothermic down the group — this is incorrect.
Confusing electron affinity with ionisation energy.

Things to Be Careful About

The mark scheme requires both the size/shielding reason and the consequence (less attraction, less exothermic).
Use precise terminology: "less exothermic" or "less negative".

Techniques used

explain the trend in electron affinity down a group

(ii)

Explain the trend in the lattice energies of the silver(I) halides, AgCl to AgI.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

From AgCl to AgI, the halide ions (X-) get larger. The attraction between the ions decreases, so the ionic bond is weaker and the lattice energy becomes less exothermic (less negative).

Final answer

Halide ions get larger down the group, decreasing attraction between ions, so lattice energy becomes less exothermic (less negative).

Detailed explanation

Background Concept

Lattice energy depends on the charges and sizes of the ions. For ions of the same charge, the lattice energy becomes less exothermic as the ionic radius increases, because the electrostatic attraction between the ions decreases with increasing distance. In the silver halides, the cation (Ag+) is the same, but the halide anion increases in size from Cl- to I-.

Understanding the Question

We are asked to explain why the lattice energies of AgCl, AgBr, and AgI become less negative.

Approach

Identify that the halide ions increase in size down the group, which reduces the electrostatic attraction and hence the lattice energy.

Step-by-Step Reasoning

All silver halides have the same cation Ag+.
The halide ions Cl-, Br-, I- increase in ionic radius down the group.
Larger ions mean greater distance between the centres of the positive and negative charges.
The electrostatic attraction between the ions decreases.
The ionic bond is weaker, so less energy is released when the lattice forms — lattice energy becomes less exothermic (less negative).

Key Takeaways

Lattice energy becomes less exothermic as ionic radius increases (for constant charge).
The trend in lattice energy follows the trend in ionic size.

Common Mistakes

Attributing the trend to the cation — the cation is the same in all three.
Saying lattice energy becomes more exothermic — the opposite is true.

Things to Be Careful About

The mark scheme requires both the size increase and the consequence (decreased attraction/weaker bond).
Use "less exothermic" or "less negative" rather than "smaller".

Techniques used

explain the trend in lattice energy with ionic radius

(e)

An electrochemical cell is constructed using the electrodes shown in Table 1.4.

Table 1.4

electrode	half-equation	$E^\ominus / \text{V}$
1	$\text{AgCl}(s) + e^- \rightleftharpoons \text{Ag}(s) + \text{Cl}^-(aq)$	+0.222
2	$\text{Cu}^{2+}(aq) + 2e^- \rightleftharpoons \text{Cu}(s)$	+0.342

5M

(i)

Calculate the standard cell potential, $E^\ominus_{\text{cell}}$ .

Construct an equation for the overall cell reaction.

2M Similar

DifficultyMedium-Easy

Worked solution

Working

E^\ominus_{\text{cell}} = E^\ominus(\text{Cu}^{2+}/\text{Cu}) - E^\ominus(\text{AgCl}/\text{Ag}) = +0.342 - (+0.222) = +0.120 \text{ V}

The more positive potential is the reduction (cathode): Cu2+ + 2e- → Cu.
The less positive potential is the oxidation (anode): Ag + Cl- → AgCl + e-.

Answer

$E^\ominus_{\text{cell}} = +0.120 \text{ V}$

Overall cell reaction:

\text{Cu}^{2+}(\text{aq}) + 2\text{Ag}(\text{s}) + 2\text{Cl}^-(\text{aq}) \rightarrow \text{Cu}(\text{s}) + 2\text{AgCl}(\text{s})

Final answer

E°cell = +0.120 V; Cu2+(aq) + 2Ag(s) + 2Cl-(aq) → Cu(s) + 2AgCl(s)

Detailed explanation

Background Concept

The standard cell potential is the difference between the electrode potentials of the two half-cells. The more positive electrode potential is the reduction (cathode), and the less positive is the oxidation (anode):
E°cell = E°(cathode) - E°(anode)
The overall cell reaction is obtained by combining the two half-equations so that the electrons cancel.

Understanding the Question

We are given two half-equations with their standard electrode potentials and asked to calculate E°cell and write the overall cell reaction.

Approach

Identify the cathode (more positive E°) and anode (less positive E°).
Calculate E°cell = E°cathode - E°anode.
Reverse the anode half-equation and combine with the cathode half-equation, balancing electrons.

Step-by-Step Reasoning

E°(Cu2+/Cu) = +0.342 V (more positive, cathode)
E°(AgCl/Ag) = +0.222 V (less positive, anode)
E°cell = +0.342 - (+0.222) = +0.120 V
Cathode: Cu2+ + 2e- → Cu
Anode (reversed): Ag + Cl- → AgCl + e-
To cancel electrons, multiply the anode equation by 2: 2Ag + 2Cl- → 2AgCl + 2e-
Combine: Cu2+ + 2Ag + 2Cl- → Cu + 2AgCl

Key Takeaways

The more positive electrode potential is the cathode.
E°cell = E°(cathode) - E°(anode).
The overall reaction combines the half-equations with balanced electrons.

Common Mistakes

Subtracting in the wrong order.
Not reversing the anode half-equation.
Forgetting to balance the electrons when combining.

Things to Be Careful About

The mark scheme requires the cell potential to 2 significant figures minimum.
The overall equation must be balanced — 2Ag and 2Cl- on the left, 2AgCl on the right.

Techniques used

calculate standard cell potentialconstruct the overall cell reaction

(ii)

In a different experiment, electrode 1 is set up using a saturated solution of KCl.

Saturated KCl(aq) contains 36.0 g of KCl per $100\text{ cm}^3$ of solution at 298 K.

The Nernst equation for electrode 1 is:

E = E^\ominus + \frac{0.059}{z} \log \frac{1}{[\text{Cl}^-(aq)]}

Calculate the electrode potential, $E$ , of electrode 1 under these conditions.

3M Similar

DifficultyMedium

Worked solution

Working

Molar mass of KCl = 39.1 + 35.5 = 74.6 g mol^-1

[\text{Cl}^-] = \frac{36.0 \text{ g}}{74.6 \text{ g mol}^{-1}} \times \frac{1000 \text{ cm}^3}{100 \text{ cm}^3} = 4.83 \text{ mol dm}^{-3}

Applying the Nernst equation:

E = 0.222 + \frac{0.059}{1} \log \frac{1}{4.83}

E = 0.222 + 0.059 \log(0.207) = 0.222 + 0.059 \times (-0.684)

E = 0.222 - 0.0404 = 0.182 \text{ V}

Answer

$E = +0.182 \text{ V}$

Final answer

+0.182 V

Detailed explanation

Background Concept

The Nernst equation relates the electrode potential to the concentration of the species involved. For the AgCl/Ag electrode:

\text{AgCl(s)} + e^- \rightleftharpoons \text{Ag(s)} + \text{Cl}^-(aq)

the Nernst equation is:

E = E^\ominus + \frac{0.059}{z} \log \frac{1}{[\text{Cl}^-]}

where z is the number of electrons transferred (z = 1 here). The concentration of Cl- is determined by the solubility of KCl in the saturated solution.

Understanding the Question

We are given a saturated KCl solution (36.0 g per 100 cm3) and asked to calculate the electrode potential using the Nernst equation.

Approach

Calculate the molar concentration of Cl- from the mass and volume of KCl.
Substitute into the Nernst equation.
Evaluate the logarithm and compute E.

Step-by-Step Reasoning

Molar mass of KCl: K = 39.1, Cl = 35.5, so M = 74.6 g/mol.
Moles of KCl in 100 cm3: 36.0 / 74.6 = 0.4826 mol.
Concentration: 0.4826 mol / 0.100 dm3 = 4.826 mol/dm3 ≈ 4.83 mol/dm3.
(Alternatively: 36.0/74.6 × 10 = 4.83 mol/dm3)
Since KCl dissociates completely, [Cl-] = 4.83 mol/dm3.
Nernst equation: E = 0.222 + (0.059/1) log(1/4.83)
log(1/4.83) = log(0.207) = -0.684
E = 0.222 + 0.059 × (-0.684) = 0.222 - 0.0404 = 0.1816 ≈ 0.182 V

Key Takeaways

The Nernst equation accounts for concentration effects on electrode potential.
Saturated KCl provides a known, high concentration of Cl-.
The electrode potential decreases when [Cl-] is high (the log term is negative).

Common Mistakes

Forgetting to convert the concentration to mol/dm3.
Using the wrong value of z (it is 1, since one electron is transferred).
Sign errors in the log term.

Things to Be Careful About

The mark scheme requires the result to 2 significant figures minimum.
The concentration calculation must be correct — 4.83 mol/dm3.
The final answer is +0.182 V, slightly less than the standard potential +0.222 V.

Techniques used

calculate concentration from mass and volumeapply the Nernst equationuse logarithms

Q2MediumReaction KineticsCarboxylic Acids and DerivativesHydroxy CompoundsEquilibria

Propanone, $\text{CH}_3\text{COCH}_3$ , is a common organic solvent and reagent.

(a)

Propanone reacts with methanol, $\text{CH}_3\text{OH}$ , under acidic conditions to form compound A, as shown by reaction 1.

Fig. 2.1

The overall order of reaction 1 can be found by studying experimental data.

Table 2.1 shows how the initial rate of reaction changes as $[\text{CH}_3\text{OH}]$ and $[\text{H}^+]$ are varied. In each experiment, a large excess of $\text{CH}_3\text{COCH}_3$ is used.

Table 2.1

experiment	$[\text{CH}_3\text{OH}] / \text{mol dm}^{-3}$	$[\text{H}^+] / \text{mol dm}^{-3}$	relative initial rate of reaction
1	0.010	0.010	1.00
2	0.015	0.015	2.25
3	0.015	0.020	3.00

4M

(i)

Explain why a large excess of $\text{CH}_3\text{COCH}_3$ is used in each experiment.

1M Similar

DifficultyEasy

Worked solution

Answer

The concentration of $\text{CH}_3\text{COCH}_3$ remains (almost) constant throughout the reaction, so any observed change in rate can be attributed solely to changes in $[\text{CH}_3\text{OH}]$ and $[\text{H}^+]$ .

Final answer

The concentration of CH3COCH3 stays (almost) constant, so it does not affect the observed rate changes.

Detailed explanation

Background Concept

In kinetics experiments, when we want to determine the order with respect to one reactant, we need to isolate its effect. If multiple reactants are changing simultaneously, we cannot distinguish their individual contributions to the rate change. One common strategy is to use a large excess of one reactant so that its concentration remains effectively constant throughout the reaction — this is called a pseudo-order or isolation method.

Understanding the Question

The question asks why propanone is used in large excess in each experiment. The table shows that only $[\text{CH}_3\text{OH}]$ and $[\text{H}^+]$ are varied between experiments, and we are asked to determine the order with respect to these two species. The role of propanone must be understood in this context.

Approach

Consider what would happen if propanone were NOT in excess: its concentration would decrease significantly as the reaction proceeds, and this change would be confounded with the changes in $[\text{CH}_3\text{OH}]$ and $[\text{H}^+]$ . By using a large excess, the fractional change in $[\text{CH}_3\text{COCH}_3]$ is negligible.

Step-by-Step Reasoning

The initial rate method requires measuring the rate at the very start of the reaction, when concentrations are known.
Even with initial rates, if propanone is not in large excess, its concentration varies between experiments (since it is a reactant), making it impossible to isolate the effect of $[\text{CH}_3\text{OH}]$ and $[\text{H}^+]$ .
By using a large excess of propanone, $[\text{CH}_3\text{COCH}_3]$ is essentially the same in all experiments and changes negligibly during each run.
This means the observed rate differences between experiments can be attributed entirely to the deliberate changes in $[\text{CH}_3\text{OH}]$ and $[\text{H}^+]$ .

Key Takeaways

The isolation method (pseudo-order conditions) allows determination of individual reaction orders.
A reactant in large excess has a concentration that is effectively constant, so its contribution to the rate is folded into the rate constant.
This is a standard experimental design principle in kinetics.

Common Mistakes

Saying 'to make the reaction go faster' — this is not the reason.
Saying 'propanone is a catalyst' — it is a reactant, not a catalyst.
Failing to connect the excess to the ability to isolate the effect of the other reactants.

Things to Be Careful About

The answer should mention that the concentration stays constant (or does not change significantly), not just that it is 'in excess'. The mark scheme requires the consequence of the excess to be stated.

Techniques used

apply pseudo-first-order conditionsexplain constant concentration approximation

(ii)

Use the data in Table 2.1 to determine the order of reaction 1 with respect to $\text{CH}_3\text{OH}$ and to $\text{H}^+$ ions. Explain your answers.

2M Similar

DifficultyMedium

Worked solution

Answer

Order with respect to $\text{H}^+$ :
Comparing experiments 2 and 3: $[\text{CH}_3\text{OH}]$ is constant at $0.015\text{ mol dm}^{-3}$ .
$[\text{H}^+]$ increases from $0.015$ to $0.020$ (factor of $\frac{4}{3}$ ).
Rate increases from $2.25$ to $3.00$ (factor of $\frac{4}{3}$ ).
Since rate changes by the same factor as $[\text{H}^+]$ , the reaction is first order with respect to $\text{H}^+$ .

Order with respect to $\text{CH}_3\text{OH}$ :
Comparing experiments 1 and 2: both $[\text{CH}_3\text{OH}]$ and $[\text{H}^+]$ increase by a factor of $1.5$ .
Rate increases from $1.00$ to $2.25$ (factor of $2.25 = 1.5^2$ ).
Since $[\text{H}^+]$ is first order, it contributes a factor of $1.5$ to the rate change.
The remaining factor of $\frac{2.25}{1.5} = 1.5$ is due to $[\text{CH}_3\text{OH}]$ changing by $1.5$ .
Therefore the reaction is first order with respect to $\text{CH}_3\text{OH}$ .

Final answer

First order with respect to CH3OH and first order with respect to H+

Detailed explanation

Background Concept

The method of initial rates determines reaction order by observing how the initial rate changes when one reactant concentration is varied while others are held constant. If doubling a concentration doubles the rate, the reaction is first order in that species. If doubling the concentration quadruples the rate, it is second order. When two concentrations change simultaneously, we must account for the known order of one to deduce the order of the other.

Understanding the Question

Three experiments are given with different combinations of $[\text{CH}_3\text{OH}]$ and $[\text{H}^+]$ . The task is to determine the order with respect to each reactant. The key challenge is that no two experiments differ in only one variable — we must use the information from one comparison to interpret another.

Approach

Find two experiments where one concentration is held constant — this directly gives the order with respect to the other.
Use that known order to interpret a comparison where both concentrations change.

Step-by-Step Reasoning

Step 1: Order with respect to H+

Experiments 2 and 3 have the same $[\text{CH}_3\text{OH}] = 0.015\text{ mol dm}^{-3}$ .
$[\text{H}^+]$ changes: $0.020 / 0.015 = 4/3$ .
Rate changes: $3.00 / 2.25 = 4/3$ .
Rate ratio equals concentration ratio, so order with respect to $\text{H}^+$ is 1.

Step 2: Order with respect to CH3OH

Experiments 1 and 2: $[\text{CH}_3\text{OH}]$ changes by $0.015/0.010 = 1.5$ , and $[\text{H}^+]$ also changes by $0.015/0.010 = 1.5$ .
Rate changes by $2.25/1.00 = 2.25$ .
If the rate equation is rate = $k[\text{CH}_3\text{OH}]^m[\text{H}^+]^n$ , then $2.25 = (1.5)^m \times (1.5)^n$ .
We know $n = 1$ , so $2.25 = (1.5)^m \times 1.5$ , giving $(1.5)^m = 1.5$ , so $m = 1$ .
Alternatively, comparing experiments 1 and 3: $[\text{CH}_3\text{OH}]$ changes by $1.5$ , $[\text{H}^+]$ changes by $2$ , rate changes by $3$ . Then $3 = (1.5)^m \times 2^1$ , so $(1.5)^m = 1.5$ , confirming $m = 1$ .

Key Takeaways

Always look for a pair of experiments where one variable is controlled.
When both variables change, use the known order to factor out its contribution.
The overall order here (excluding propanone) is 2, but including propanone (first order from part a(iii)) gives overall order 3.

Common Mistakes

Comparing experiments 1 and 2 and concluding second order overall without separating the two contributions.
Forgetting that $[\text{H}^+]$ also changes between experiments 1 and 2.
Writing 'rate × 2.25 and concentration × 1.5, so order = 2' without accounting for the simultaneous change in $[\text{H}^+]$ .

Things to Be Careful About

The mark scheme requires both the correct order AND the supporting numerical reasoning for each.
State which experiments you are comparing.
Show the factor by which rate and concentration change explicitly.

Techniques used

compare experiments to isolate one variablecalculate rate ratiosdeduce order from concentration-rate relationship

(iii)

In a separate experiment, a large excess of $\text{CH}_3\text{OH}$ and $\text{H}^+$ ions are added to a solution containing a known concentration of $\text{CH}_3\text{COCH}_3$ .

Fig. 2.2 shows how $[\text{CH}_3\text{COCH}_3]$ varies over time.

Fig. 2.2

Use Fig. 2.2 to show how, under these conditions, reaction 1 is first order with respect to $\text{CH}_3\text{COCH}_3$ .

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

From Fig. 2.2:

First half-life: $[\text{CH}_3\text{COCH}_3]$ falls from $100\%$ to $50\%$ in approximately $170\text{ s}$ .
Second half-life: $[\text{CH}_3\text{COCH}_3]$ falls from $50\%$ to $25\%$ in approximately $170\text{ s}$ (from $170\text{ s}$ to $340\text{ s}$ ).

The half-life is constant (approximately $170\text{ s}$ ), which is characteristic of a first-order reaction.

Final answer

Two half-lives are both approximately 170 s (constant), confirming first order with respect to CH3COCH3.

Detailed explanation

Background Concept

A defining feature of first-order reactions is that the half-life is independent of the initial concentration. For a first-order reaction, $t_{1/2} = \frac{\ln 2}{k}$ , which contains no concentration term. This means that whether the concentration starts at 100%, 50%, or 25%, the time taken to halve it is always the same. This is a powerful diagnostic tool: if successive half-lives read from a concentration-time graph are constant, the reaction is first order with respect to that species.

Understanding the Question

In this experiment, $[\text{CH}_3\text{OH}]$ and $[\text{H}^+]$ are in large excess, so they remain effectively constant. The graph shows $[\text{CH}_3\text{COCH}_3]$ decreasing over time. We must use the graph to demonstrate that the reaction is first order with respect to propanone.

Approach

Read off the time for the concentration to fall from 100% to 50% (first half-life), then from 50% to 25% (second half-life). If these are equal, the reaction is first order.

Step-by-Step Reasoning

From the graph, at $t = 0$ , the concentration is 100%.
The concentration reaches 50% at approximately $t = 170\text{ s}$ . First half-life = 170 s.
The concentration reaches 25% at approximately $t = 340\text{ s}$ . Second half-life = $340 - 170 = 170\text{ s}$ .
Since the half-life is constant regardless of the starting concentration, the reaction is first order with respect to $\text{CH}_3\text{COCH}_3$ .

Note: The mark scheme accepts values in the range 155–170 s for the half-lives, as reading from a graph involves some uncertainty.

Key Takeaways

Constant half-life is the hallmark of first-order kinetics.
When other reactants are in excess, the observed kinetics reflect only the species whose concentration is changing — this is the pseudo-first-order method.
Reading graphs requires identifying the time for each successive halving of concentration.

Common Mistakes

Only quoting one half-life — the question requires showing constancy, which needs at least two.
Reading the time for the concentration to reach zero rather than halving.
Stating 'the graph is exponential' without explicitly quoting and comparing half-lives.

Things to Be Careful About

Quote actual values from the graph (the mark scheme requires 'two quoted half-lives').
State that they are the same/constant.
The range 155–170 s is accepted; do not be overly precise about readings from a graph.

Techniques used

read half-lives from a concentration-time graphapply constant half-life criterion for first order

(b)

Propanone also reacts with acidified cyanide ions to form the hydroxynitrile compound B, as shown by reaction 2.

Fig. 2.3

The following rate equation is determined for reaction 2.

\text{rate} = k [\text{CH}_3\text{COCH}_3] [\text{H}^+]

Four possible mechanisms for reaction 2 are shown in Table 2.2.

Table 2.2

proposed reaction mechanism	steps
1	fast $\text{CH}_3\text{COCH}_3 + \text{H}^+ \rightarrow [\text{CH}_3\text{C(OH)CH}_3]^+$ \slow $[\text{CH}_3\text{C(OH)CH}_3]^+ + \text{CN}^- \rightarrow \text{CH}_3\text{C(OH)(CN)CH}_3$
2	fast $\text{H}^+ + \text{CN}^- \rightarrow \text{HCN}$ \slow $\text{CH}_3\text{COCH}_3 + \text{HCN} \rightarrow \text{CH}_3\text{C(OH)(CN)CH}_3$
3	slow $\text{CH}_3\text{COCH}_3 + \text{CN}^- \rightarrow \text{CH}_3\text{C(O}^-\text{)(CN)CH}_3$ \fast $\text{CH}_3\text{C(O}^-\text{)(CN)CH}_3 + \text{H}^+ \rightarrow \text{CH}_3\text{C(OH)(CN)CH}_3$
4	slow $\text{CH}_3\text{COCH}_3 + \text{H}^+ \rightarrow [\text{CH}_3\text{C(OH)CH}_3]^+$ \fast $[\text{CH}_3\text{C(OH)CH}_3]^+ + \text{CN}^- \rightarrow \text{CH}_3\text{C(OH)(CN)CH}_3$

Suggest which of these mechanisms is consistent with the rate equation for reaction 2. Explain your answer.

3M Similar

DifficultyMedium

Worked solution

Answer

Mechanism 4 is consistent with the rate equation.

The rate equation is $\text{rate} = k[\text{CH}_3\text{COCH}_3][\text{H}^+]$ , which is first order in $\text{CH}_3\text{COCH}_3$ and first order in $\text{H}^+$ .

In mechanism 4, the slow (rate-determining) step is:

\text{CH}_3\text{COCH}_3 + \text{H}^+ \rightarrow [\text{CH}_3\text{C(OH)CH}_3]^+

This step contains only $\text{CH}_3\text{COCH}_3$ and $\text{H}^+$ , in a 1:1 ratio, which matches the orders in the rate equation. The species in the rate-determining step and their stoichiometric coefficients correspond directly to the orders in the rate law.

Final answer

Mechanism 4

Detailed explanation

Background Concept

The rate-determining step (RDS) is the slowest step in a reaction mechanism, and it controls the overall rate. The rate equation reflects the species involved in the RDS (and any species in fast equilibrium steps preceding it). For a simple case where the RDS is the first step, the rate equation directly mirrors the stoichiometry of that step: each species in the RDS appears in the rate equation raised to the power of its coefficient in that step.

If a species appears in a fast step before the RDS, it may also appear in the rate equation (through the equilibrium approximation). But species appearing only AFTER the RDS (in fast steps) do not appear in the rate equation at all.

Understanding the Question

Four mechanisms are proposed for the reaction of propanone with cyanide ions under acidic conditions. The experimentally determined rate equation is $\text{rate} = k[\text{CH}_3\text{COCH}_3][\text{H}^+]$ . We must identify which mechanism is consistent with this rate equation and explain why.

Approach

Examine the slow step of each mechanism and check whether the species in that step match those in the rate equation, with the correct stoichiometric ratio.

Step-by-Step Reasoning

Mechanism 1: Slow step is $[\text{CH}_3\text{C(OH)CH}_3]^+ + \text{CN}^- \rightarrow \text{product}$ . This contains an intermediate and $\text{CN}^-$ , neither of which appears in the rate equation. Inconsistent.

Mechanism 2: Slow step is $\text{CH}_3\text{COCH}_3 + \text{HCN} \rightarrow \text{product}$ . This contains HCN, not $\text{H}^+$ and $\text{CN}^-$ separately. The rate equation would be $\text{rate} = k[\text{CH}_3\text{COCH}_3][\text{HCN}]$ , which does not match. Inconsistent.

Mechanism 3: Slow step is $\text{CH}_3\text{COCH}_3 + \text{CN}^- \rightarrow \text{intermediate}$ . This contains $\text{CN}^-$ but not $\text{H}^+$ . The rate equation would be $\text{rate} = k[\text{CH}_3\text{COCH}_3][\text{CN}^-]$ , which does not match. Inconsistent.

Mechanism 4: Slow step is $\text{CH}_3\text{COCH}_3 + \text{H}^+ \rightarrow [\text{CH}_3\text{C(OH)CH}_3]^+$ . This contains exactly $\text{CH}_3\text{COCH}_3$ and $\text{H}^+$ in a 1:1 ratio, giving $\text{rate} = k[\text{CH}_3\text{COCH}_3][\text{H}^+]$ . Consistent with the observed rate equation.

Key Takeaways

The rate equation tells us what is in (or before) the rate-determining step.
Species in fast steps after the RDS do not appear in the rate equation.
The stoichiometric coefficients in the RDS give the orders in the rate equation (for a first-step RDS).

Common Mistakes

Choosing mechanism 1 because it has a fast step involving $\text{CH}_3\text{COCH}_3$ and $\text{H}^+$ — but the fast step before the RDS does not determine the rate equation directly unless it is a pre-equilibrium (which is more complex and not indicated here).
Confusing which species appear in the rate equation with which appear in the overall reaction.
Not explicitly stating the connection between the slow step and the rate equation.

Things to Be Careful About

The mark scheme requires three points: identifying mechanism 4, stating the slow step contains only species from the rate equation, and that the stoichiometric ratio matches the orders. All three must be present for full marks.

Techniques used

compare rate-determining step composition to rate equationidentify species in the slow stepmatch stoichiometric coefficients to reaction orders

(c)

Carboxylic acid C, $\text{C}_4\text{H}_8\text{O}_3$ , forms when B is hydrolysed under hot acidic conditions.

3M

(i)

Draw the structure of C.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

Hydrolysis of the nitrile group ( $-\text{CN}$ ) in B under hot acidic conditions converts it to a carboxylic acid group ( $-\text{COOH}$ ).

Compound C is 2-hydroxy-2-methylpropanoic acid:

Final answer

2-hydroxy-2-methylpropanoic acid: (CH3)2C(OH)(COOH)

Detailed explanation

Background Concept

Nitriles ( $\text{R-C}\equiv\text{N}$ ) undergo hydrolysis under hot acidic conditions to form carboxylic acids ( $\text{R-COOH}$ ). The $\text{C}\equiv\text{N}$ triple bond is converted to $\text{C(=O)OH}$ through addition of water. The rest of the molecule is unchanged. This is a standard reaction in organic synthesis for extending a carbon chain by one carbon and then converting the nitrile to an acid.

Understanding the Question

Compound B is 2-hydroxy-2-methylpropanenitrile: $(\text{CH}_3)_2\text{C(OH)(CN)}$ . When hydrolysed under hot acidic conditions, the $-\text{CN}$ group becomes $-\text{COOH}$ , giving compound C with molecular formula $\text{C}_4\text{H}_8\text{O}_3$ .

Approach

Replace the $-\text{CN}$ group with $-\text{COOH}$ while keeping the rest of the structure (the central carbon with two methyl groups and one hydroxyl group) unchanged.

Step-by-Step Reasoning

B has the structure: central C bonded to $\text{CH}_3$ , $\text{CH}_3$ , $\text{OH}$ , and $\text{CN}$ .
Hot acidic hydrolysis converts $\text{CN} \rightarrow \text{COOH}$ .
C has the structure: central C bonded to $\text{CH}_3$ , $\text{CH}_3$ , $\text{OH}$ , and $\text{COOH}$ .
Molecular formula check: $\text{C}_4\text{H}_8\text{O}_3$ ✓ (4 carbons, 8 hydrogens, 3 oxygens).
Name: 2-hydroxy-2-methylpropanoic acid.

Key Takeaways

Nitrile hydrolysis gives carboxylic acids.
The carbon of the CN becomes the carbon of the COOH.
Other functional groups in the molecule are unaffected by this hydrolysis.

Common Mistakes

Drawing the wrong connectivity (e.g., placing OH and COOH on different carbons).
Forgetting to show all atoms and bonds in the displayed formula.
Drawing the structure of B instead of C (forgetting the hydrolysis).

Things to Be Careful About

The question says 'draw the structure', which means a displayed formula showing all atoms and bonds is expected.
Ensure the correct number of carbons (4 total) and the correct placement of functional groups on the same central carbon.

Techniques used

hydrolyse nitrile to carboxylic aciddraw displayed structural formula

(ii)

The $\text{p}K_a$ of C is 3.95. Calculate the pH of a $0.500\text{ mol dm}^{-3}$ solution of C.

Show your working.

2M Similar

DifficultyMedium-Easy

Worked solution

Working

K_a = 10^{-\text{p}K_a} = 10^{-3.95} = 1.122 \times 10^{-4}

For a weak acid: $K_a = \frac{[\text{H}^+]^2}{[\text{HA}]}$

[\text{H}^+] = \sqrt{K_a \times [\text{HA}]} = \sqrt{1.122 \times 10^{-4} \times 0.500} = \sqrt{5.61 \times 10^{-5}} = 7.49 \times 10^{-3}\text{ mol dm}^{-3}

\text{pH} = -\log(7.49 \times 10^{-3}) = 2.13

Answer

$\text{pH} = 2.13$

Final answer

2.13

Detailed explanation

Background Concept

For a weak acid HA that partially dissociates: $\text{HA} \rightleftharpoons \text{H}^+ + \text{A}^-$ . The acid dissociation constant is $K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}$ . When the acid is weak (small $K_a$ relative to $[\text{HA}]$ ), we can approximate $[\text{HA}]_{\text{equilibrium}} \approx [\text{HA}]_{\text{initial}}$ , and since $[\text{H}^+] = [\text{A}^-]$ , this gives $[\text{H}^+] = \sqrt{K_a \times [\text{HA}]}$ . The relationship between $\text{p}K_a$ and $K_a$ is $K_a = 10^{-\text{p}K_a}$ .

Understanding the Question

Compound C is a weak carboxylic acid with $\text{p}K_a = 3.95$ . We are given its concentration ( $0.500\text{ mol dm}^{-3}$ ) and asked to calculate the pH of this solution.

Approach

Convert $\text{p}K_a$ to $K_a$ .
Apply the weak acid approximation to find $[\text{H}^+]$ .
Calculate pH from $[\text{H}^+]$ .

Step-by-Step Reasoning

$K_a = 10^{-3.95} = 1.122 \times 10^{-4}$ .
The approximation $[\text{H}^+] = \sqrt{K_a \times [\text{HA}]}$ is valid here because $K_a/[\text{HA}] = 1.122 \times 10^{-4}/0.500 = 2.24 \times 10^{-4}$ , which is much less than 1, confirming the acid is weak enough for the approximation.
$[\text{H}^+] = \sqrt{1.122 \times 10^{-4} \times 0.500} = \sqrt{5.61 \times 10^{-5}} = 7.49 \times 10^{-3}\text{ mol dm}^{-3}$ .
$\text{pH} = -\log(7.49 \times 10^{-3}) = 2.13$ .

Key Takeaways

The weak acid approximation $[\text{H}^+] = \sqrt{K_a \times c}$ is the standard method for calculating pH of a weak acid solution.
Always convert $\text{p}K_a$ to $K_a$ before using it in calculations.
Check that the approximation is valid (though at A-level this is usually assumed).

Common Mistakes

Using $\text{p}K_a$ directly instead of converting to $K_a$ .
Forgetting to take the square root.
Reporting pH to only 1 decimal place when 2 significant figures (after the decimal) are required.
Using the strong acid formula $\text{pH} = -\log[\text{HA}]$ which would give $\text{pH} = 0.30$ — clearly wrong for a weak acid.

Things to Be Careful About

The mark scheme requires minimum 2 significant figures in the pH (2.13 is acceptable).
Units are not required for pH.
Ensure $K_a$ is calculated correctly from $\text{p}K_a$ — a common error is writing $K_a = 3.95 \times 10^{-1}$ instead of $10^{-3.95}$ .

Techniques used

convert pKa to Kaapply weak acid approximation [H+] = sqrt(Ka × [HA])calculate pH from hydrogen ion concentration

(d)

C can be used to form buffer solution D.

5M

(i)

Define a buffer solution.

2M Similar

(ii)

Buffer solution D is made when $20.0\text{ cm}^3$ of $1.00\text{ mol dm}^{-3}$ NaOH(aq) is added to $100\text{ cm}^3$ of a $0.500\text{ mol dm}^{-3}$ solution of C.

The $\text{p}K_a$ of C is 3.95.

Calculate the pH of buffer solution D.

Show your working.

3M Similar

DifficultyMedium

Worked solution

Working

Moles of NaOH added = $\frac{20.0}{1000} \times 1.00 = 0.0200\text{ mol}$

Moles of C initially = $\frac{100}{1000} \times 0.500 = 0.0500\text{ mol}$

NaOH neutralises C to form the conjugate base $\text{C}^-$ :

\text{HA} + \text{OH}^- \rightarrow \text{A}^- + \text{H}_2\text{O}

Moles of conjugate base $\text{A}^-$ formed = $0.0200\text{ mol}$

Moles of acid HA remaining = $0.0500 - 0.0200 = 0.0300\text{ mol}$

Using the $K_a$ expression:

K_a = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]} = [\text{H}^+] \times \frac{n(\text{A}^-)}{n(\text{HA})}

[\text{H}^+] = K_a \times \frac{n(\text{HA})}{n(\text{A}^-)} = 10^{-3.95} \times \frac{0.0300}{0.0200} = 1.122 \times 10^{-4} \times 1.5 = 1.68 \times 10^{-4}\text{ mol dm}^{-3}

\text{pH} = -\log(1.68 \times 10^{-4}) = 3.77

Answer

$\text{pH} = 3.77$

Final answer

3.77

Detailed explanation

Background Concept

When a strong base is added to a weak acid, it partially neutralises the acid, converting some of it to its conjugate base. The resulting solution contains both the weak acid and its conjugate base — a buffer. The pH of a buffer can be calculated using the $K_a$ expression rearranged to solve for $[\text{H}^+]$ . Since both species are in the same solution volume, the ratio of concentrations equals the ratio of moles, simplifying the calculation.

Understanding the Question

$20.0\text{ cm}^3$ of $1.00\text{ mol dm}^{-3}$ NaOH is added to $100\text{ cm}^3$ of $0.500\text{ mol dm}^{-3}$ of acid C ( $\text{p}K_a = 3.95$ ). We need to find the pH of the resulting buffer solution.

Approach

Calculate moles of NaOH and moles of acid C.
Determine which is in excess and calculate moles of acid remaining and conjugate base formed.
Use the $K_a$ expression with the mole ratio to find $[\text{H}^+]$ .
Calculate pH.

Step-by-Step Reasoning

Step 1: Moles

Moles NaOH = $0.0200 \times 1.00 = 0.0200\text{ mol}$
Moles C = $0.100 \times 0.500 = 0.0500\text{ mol}$

Step 2: Reaction
NaOH is the limiting reagent. It converts $0.0200\text{ mol}$ of HA to $\text{A}^-$ :

Moles $\text{A}^-$ = $0.0200\text{ mol}$
Moles HA remaining = $0.0500 - 0.0200 = 0.0300\text{ mol}$

Step 3: Buffer equation

[\text{H}^+] = K_a \times \frac{[\text{HA}]}{[\text{A}^-]} = K_a \times \frac{n(\text{HA})}{n(\text{A}^-)}

(The volume cancels in the ratio since both are in the same solution.)

[\text{H}^+] = 10^{-3.95} \times \frac{0.0300}{0.0200} = 1.122 \times 10^{-4} \times 1.5 = 1.68 \times 10^{-4}\text{ mol dm}^{-3}

Step 4: pH

\text{pH} = -\log(1.68 \times 10^{-4}) = 3.77

Note: Since $\text{pH} > \text{p}K_a$ is not the case here ( $3.77 < 3.95$ ), this makes sense because there is more acid than conjugate base ( $0.0300 > 0.0200$ ), so the pH should be below the $\text{p}K_a$ .

Key Takeaways

In a buffer calculation, the volume cancels when taking the ratio of acid to conjugate base moles.
The Henderson-Hasselbalch equation $\text{pH} = \text{p}K_a + \log\frac{[\text{A}^-]}{[\text{HA}]}$ gives the same result: $3.95 + \log(0.0200/0.0300) = 3.95 + \log(0.667) = 3.95 - 0.176 = 3.77$ .
Always check that the answer is reasonable relative to $\text{p}K_a$ .

Common Mistakes

Forgetting that NaOH only partially neutralises the acid (not all of it reacts since acid is in excess).
Using concentrations instead of moles without accounting for the total volume (though the ratio works either way since volume cancels).
Inverting the ratio: writing $[\text{H}^+] = K_a \times [\text{A}^-]/[\text{HA}]$ instead of $K_a \times [\text{HA}]/[\text{A}^-]$ .
Calculating moles of NaOH as $20 \times 1.00 = 20$ (forgetting to divide by 1000).

Things to Be Careful About

The mark scheme requires 2 significant figures minimum for the pH.
ECF is allowed from M1 (the mole calculation) into M2 and M3.
Ensure the correct species is identified as the conjugate base (formed from neutralisation) and the remaining acid.

Techniques used

calculate moles of acid and basedetermine moles of conjugate base formed and acid remainingapply Henderson-Hasselbalch or Ka expression for buffer pH

Q3Medium-HardTransition ElementsEquilibria

$\text{Fe}^{2+}$ and $\text{Fe}^{3+}$ ions are able to form a variety of complexes with different species.

(a)

5M

(i)

Define complex.

1M Similar

DifficultyEasy

Worked solution

Answer

A central metal atom or ion surrounded by ligands bonded by coordinate (dative covalent) bonds.

Final answer

A central metal atom or ion surrounded by ligands bonded by coordinate bonds.

Detailed explanation

Background Concept

In transition metal chemistry, a complex (or coordination compound) consists of a central metal atom or ion bonded to surrounding molecules or ions called ligands. The bond formed between a ligand and the metal is a coordinate (dative covalent) bond, meaning both electrons in the shared pair come from the ligand (which must have a lone pair) and are accepted by the metal (which must have an empty orbital, typically a d-orbital).

Understanding the Question

The question simply asks for the definition of a 'complex'. This is a foundational term in the study of transition metals and coordination chemistry.

Approach

Recall the standard IUPAC-style definition used in A-Level Chemistry: identify the central species (metal atom/ion) and what is around it (ligands), and specify the type of bonding (coordinate bonds).

Step-by-Step Reasoning

Identify the central component: a metal atom or, more commonly for these questions, a metal ion (e.g., $\text{Fe}^{2+}$ , $\text{Cu}^{2+}$ ).
Identify the surrounding species: ligands (e.g., $\text{H}_2\text{O}$ , $\text{NH}_3$ , $\text{Cl}^-$ ).
Specify the bonding: the ligands are attached via coordinate (dative covalent) bonds.
Combine these into a concise definition.

Key Takeaways

A complex is defined by three elements: a central metal atom/ion, surrounding ligands, and coordinate bonds. Remembering 'coordinate' is crucial, as simple 'covalent' is often not accepted.

Common Mistakes

Saying 'covalent bonds' instead of 'coordinate' or 'dative covalent' bonds.
Forgetting to mention the central metal atom or ion.
Calling the surrounding species 'molecules' instead of 'ligands' (though 'molecules or ions' is acceptable, 'ligands' is the precise term).

Things to Be Careful About

Ensure you include both 'central metal atom/ion' AND 'surrounded by ligands'. Mark schemes typically require both parts of the definition to award the mark.

Techniques used

define a complex ion

(ii)

Table 3.1 gives some details of different complexes of $\text{Fe}^{2+}$ and of $\text{Fe}^{3+}$ .

Complete Table 3.1.

Table 3.1

complex	ion	ligand	coordination number	formula and charge of complex
E	$\text{Fe}^{2+}$	$\text{NH}_3$	6
F				$[\text{FeCl}_4]^{2-}$
G		en		$[\text{Fe(en)}_3]^{3+}$

3M Similar

DifficultyMedium-Easy

Worked solution

Answer

complex	ion	ligand	coordination number	formula and charge of complex
E	$\text{Fe}^{2+}$	$\text{NH}_3$	6	$[\text{Fe(NH}_3)_6]^{2+}$
F	$\text{Fe}^{2+}$	$\text{Cl}^-$	4	$[\text{FeCl}_4]^{2-}$
G	$\text{Fe}^{3+}$	en	6	$[\text{Fe(en)}_3]^{3+}$

Final answer

E: [Fe(NH3)6]2+ ; F: Fe2+, Cl-, 4 ; G: Fe3+, 6

Detailed explanation

Background Concept

The formula and charge of a complex are determined by the charge of the central metal ion and the charges of the ligands. The overall charge of the complex is the algebraic sum of the metal ion charge and the ligand charges.
The coordination number is the total number of coordinate bonds formed between the ligands and the central metal ion. Monodentate ligands (like $\text{NH}_3$ , $\text{Cl}^-$ , $\text{H}_2\text{O}$ ) form one bond each. Bidentate ligands (like ethane-1,2-diamine, 'en') form two bonds each.

Understanding the Question

We are given a table with missing values for three complexes (E, F, G). We must deduce the missing ion, ligand, coordination number, or formula/charge based on the given information.

Approach

For E: Metal is $\text{Fe}^{2+}$ , ligand is neutral $\text{NH}_3$ , CN is 6. Formula is $[\text{Fe(NH}_3)_6]^{2+}$ .
For F: Formula is $[\text{FeCl}_4]^{2-}$ . Ligand is $\text{Cl}^-$ (charge -1). CN is 4 (four $\text{Cl}^-$ ). Charge of complex is -2. Let metal charge be $x$ . $x + 4(-1) = -2 \Rightarrow x = +2$ . So ion is $\text{Fe}^{2+}$ .
For G: Formula is $[\text{Fe(en)}_3]^{3+}$ . Ligand is 'en' (neutral). CN is $3 \times 2 = 6$ (since 'en' is bidentate). Charge is +3. Since ligands are neutral, metal charge is +3. So ion is $\text{Fe}^{3+}$ .

Step-by-Step Reasoning

Complex E:

Ion: Given as $\text{Fe}^{2+}$ .
Ligand: Given as $\text{NH}_3$ (neutral).
Coordination number: Given as 6.
Formula and charge: $[\text{Fe(NH}_3)_6]^{2+}$ .

Complex F:

Formula given: $[\text{FeCl}_4]^{2-}$ .
Ligand: $\text{Cl}^-$ (chloride ion).
Coordination number: 4 (four chloride ligands).
Ion charge: Let it be $x$ . $x + 4(-1) = -2 \Rightarrow x = +2$ . Ion is $\text{Fe}^{2+}$ .

Complex G:

Formula given: $[\text{Fe(en)}_3]^{3+}$ .
Ligand: 'en' (ethane-1,2-diamine, neutral).
Coordination number: 'en' is bidentate (forms 2 bonds). $3 \times 2 = 6$ .
Ion charge: Let it be $y$ . $y + 3(0) = +3 \Rightarrow y = +3$ . Ion is $\text{Fe}^{3+}$ .

Key Takeaways

Always check if a ligand is monodentate or bidentate to calculate the correct coordination number. Remember that the overall charge is the sum of the metal ion charge and the ligand charges.

Common Mistakes

Forgetting that 'en' is bidentate and writing a coordination number of 3 for complex G.
Miscalculating the metal ion charge in complex F (e.g., thinking $\text{Fe}^{3+}$ because of the context, but the math gives $\text{Fe}^{2+}$ ).

Things to Be Careful About

Ensure all missing cells in the table are filled. The mark scheme awards marks for any two correct rows, any four correct cells, or all six correct cells. Double-check your charge calculations.

Techniques used

deduce complex formula and chargedetermine coordination number and ligand

(iii)

Complete Fig. 3.1 to show the splitting of the d-orbitals in a tetrahedral complex.

Fig. 3.1

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

Final answer

See diagram: 3 higher energy levels, 2 lower energy levels.

Detailed explanation

Background Concept

When ligands approach a central metal ion, the five degenerate (equal energy) d-orbitals split into two energy levels due to electrostatic repulsion between the ligand electrons and the d-orbital electrons.
In an octahedral complex, the splitting results in two lower-energy $t_{2g}$ orbitals ( $d_{xy}, d_{xz}, d_{yz}$ ) and three higher-energy $e_g$ orbitals ( $d_{z^2}, d_{x^2-y^2}$ ).
In a tetrahedral complex, the geometry is inverted relative to the octahedral case (ligands approach between the axes rather than along them). This reverses the splitting pattern: the $e$ orbitals ( $d_{z^2}, d_{x^2-y^2}$ ) are lower in energy, and the $t_2$ orbitals ( $d_{xy}, d_{xz}, d_{yz}$ ) are higher in energy. The energy gap ( $\Delta_t$ ) is smaller than in octahedral complexes ( $\Delta_t \approx \frac{4}{9} \Delta_o$ ).

Understanding the Question

The question asks to complete a diagram showing the d-orbital splitting for a tetrahedral complex. The provided template has a vertical 'energy' axis and blank space to draw the orbitals.

Approach

Draw two groups of boxes (representing orbitals). For a tetrahedral complex, there must be two boxes at a lower energy level and three boxes at a higher energy level.

Step-by-Step Reasoning

Draw two horizontal boxes grouped together at a lower energy level (representing the $e$ set: $d_{z^2}, d_{x^2-y^2}$ ).
Draw three horizontal boxes grouped together at a higher energy level (representing the $t_2$ set: $d_{xy}, d_{xz}, d_{yz}$ ).
Ensure there is a clear energy gap between the two groups.

Key Takeaways

Tetrahedral splitting is the inverse of octahedral splitting: 2 lower, 3 higher. Memorise this to avoid the common mistake of drawing the octahedral pattern (3 lower, 2 higher).

Common Mistakes

Drawing 3 lower and 2 higher boxes (this is octahedral, not tetrahedral).
Forgetting to group the boxes (2 together, 3 together).

Things to Be Careful About

The question specifically asks for a tetrahedral complex. Read carefully! The mark scheme requires 'three higher AND two lower boxes'.

Techniques used

draw d-orbital splitting diagram for tetrahedral complex

(b)

Table 3.2 gives details of some complexes of $\text{Fe}^{3+}$ .

Table 3.2

complex	colour	value of $K_{\text{stab}}$
$[\text{Fe(H}_2\text{O)}_6]^{3+}$	violet	1
$[\text{Fe(H}_2\text{O)}_5\text{SCN}]^{2+}$	red	$1.40 \times 10^2$
$[\text{Fe(H}_2\text{O)}_5\text{F}]^{2+}$	colourless	$2.40 \times 10^5$

6M

(i)

Explain the reason for the difference in colour of the two complexes $[\text{Fe(H}_2\text{O)}_6]^{3+}$ and $[\text{Fe(H}_2\text{O)}_5\text{SCN}]^{2+}$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

The ligands are different ( $\text{H}_2\text{O}$ vs $\text{SCN}^-$ ), so the d–d energy gap ( $\Delta E$ ) is different.
This means a different frequency (or wavelength) of light is absorbed (and thus different light is transmitted/reflected, giving a different colour).

Final answer

Different ligands cause a different d-d energy gap, leading to absorption of different frequencies/wavelengths of light.

Detailed explanation

Background Concept

The colour of transition metal complexes arises from d-d electron transitions. When light shines on the complex, electrons in the lower-energy d-orbitals can absorb photons and jump to the higher-energy d-orbitals. The energy of the absorbed photon matches the energy gap ( $\Delta E$ ) between the split d-orbitals: $\Delta E = h\nu = \frac{hc}{\lambda}$ .
The colour observed is the complementary colour to the one absorbed. The value of $\Delta E$ depends on the nature of the ligands (spectrochemical series) and the oxidation state/identity of the metal ion. Different ligands cause different amounts of d-orbital splitting.

Understanding the Question

We are comparing $[\text{Fe(H}_2\text{O)}_6]^{3+}$ (violet) and $[\text{Fe(H}_2\text{O)}_5\text{SCN}]^{2+}$ (red). Both have $\text{Fe}^{3+}$ as the central ion, but one has $\text{H}_2\text{O}$ ligands and the other has one $\text{SCN}^-$ ligand replacing an $\text{H}_2\text{O}$ . We must explain why their colours differ.

Approach

Identify the difference between the two complexes: the ligands are different (mixed ligand complex vs pure aqua complex).
State that different ligands cause a different d-d energy gap ( $\Delta E$ ).
Link the different $\Delta E$ to the absorption of different frequencies/wavelengths of light.

Step-by-Step Reasoning

The two complexes have different ligands ( $\text{H}_2\text{O}$ only vs $\text{H}_2\text{O}$ and $\text{SCN}^-$ ).
Different ligands produce a different crystal field splitting, meaning the d–d energy gap ( $\Delta E$ ) is different.
Since $\Delta E = h\nu$ , a different energy gap means a different frequency (or wavelength) of light is absorbed.
Different light absorbed means different light is transmitted/reflected, resulting in a different observed colour.

Key Takeaways

Colour differences in complexes with the same metal ion are always due to different ligands causing a different d-d energy gap, which changes the absorbed wavelength.

Common Mistakes

Saying 'different energy levels' without specifying 'd-d energy gap' or 'd-orbital splitting'.
Saying 'different amount of light absorbed' without mentioning frequency/wavelength/energy.
Forgetting to link the energy gap to the frequency/wavelength of light.

Things to Be Careful About

The mark scheme specifically looks for 'd–d energy gap / $\Delta E$ is different' and 'different frequency / wavelength'. Ensure you mention both the gap and the light property.

Techniques used

relate ligand to d-d transition energy and light absorption

(ii)

Write an expression for $K_{\text{stab}}$ of $[\text{Fe(H}_2\text{O)}_5\text{SCN}]^{2+}$ .

1M Similar

DifficultyEasy

Worked solution

Answer

$K_{\text{stab}} = \frac{[[\text{Fe(H}_2\text{O)}_5\text{SCN}]^{2+}]}{[[\text{Fe(H}_2\text{O)}_6]^{3+}][\text{SCN}^-]}$

Final answer

K_stab = [[Fe(H2O)5SCN]2+] / ([[Fe(H2O)6]3+][SCN-])

Detailed explanation

Background Concept

The stability constant ( $K_{\text{stab}}$ ) describes the equilibrium position of a complex ion formation reaction. For the reaction where a ligand replaces water in an aqua complex:
$[\text{M(H}_2\text{O)}_n]^{m+} + \text{L} \rightleftharpoons [\text{M(H}_2\text{O)}_{n-1}\text{L}]^{m+} + \text{H}_2\text{O}$
Water is the solvent and its concentration is essentially constant, so it is omitted from the equilibrium expression.

Understanding the Question

Write the $K_{\text{stab}}$ expression for the formation of $[\text{Fe(H}_2\text{O)}_5\text{SCN}]^{2+}$ from $[\text{Fe(H}_2\text{O)}_6]^{3+}$ and $\text{SCN}^-$ .

Approach

Write the products over reactants, omitting water. Use square brackets for concentrations.

Step-by-Step Reasoning

Product: $[\text{Fe(H}_2\text{O)}_5\text{SCN}]^{2+}$
Reactants: $[\text{Fe(H}_2\text{O)}_6]^{3+}$ and $\text{SCN}^-$
Expression: $K_{\text{stab}} = \frac{[[\text{Fe(H}_2\text{O)}_5\text{SCN}]^{2+}]}{[[\text{Fe(H}_2\text{O)}_6]^{3+}][\text{SCN}^-]}$

Key Takeaways

Always omit $\text{H}_2\text{O}$ from $K_{\text{stab}}$ expressions as it is the solvent. Ensure charges on complex ions are correctly included in the formula.

Common Mistakes

Including $[\text{H}_2\text{O}]$ in the denominator.
Forgetting the charges on the complex ions in the formula (though strictly not required in the K expression itself, the formula must be correct).

Things to Be Careful About

The mark scheme shows the expression with the complex ion in square brackets inside square brackets: $[[\text{Fe(H}_2\text{O)}_5\text{SCN}]^{2+}]$ . This is correct notation: outer brackets for concentration, inner brackets for the complex ion formula.

Techniques used

write stability constant expression

(iii)

Use information in Table 3.2 to calculate the value of the equilibrium constant, $K_c$ , for the following reaction.

[\text{Fe(H}_2\text{O)}_5\text{SCN}]^{2+} + \text{F}^- \rightleftharpoons [\text{Fe(H}_2\text{O)}_5\text{F}]^{2+} + \text{SCN}^-

1M Similar

DifficultyMedium

Worked solution

Working

The target reaction is:
$[\text{Fe(H}_2\text{O)}_5\text{SCN}]^{2+} + \text{F}^- \rightleftharpoons [\text{Fe(H}_2\text{O)}_5\text{F}]^{2+} + \text{SCN}^-$

From Table 3.2, we have two formation equilibria:

$[\text{Fe(H}_2\text{O)}_6]^{3+} + \text{SCN}^- \rightleftharpoons [\text{Fe(H}_2\text{O)}_5\text{SCN}]^{2+} + \text{H}_2\text{O} \quad K_1 = 1.40 \times 10^2$
$[\text{Fe(H}_2\text{O)}_6]^{3+} + \text{F}^- \rightleftharpoons [\text{Fe(H}_2\text{O)}_5\text{F}]^{2+} + \text{H}_2\text{O} \quad K_2 = 2.40 \times 10^5$

The target reaction is equation (2) minus equation (1):
$K_c = \frac{K_2}{K_1} = \frac{2.40 \times 10^5}{1.40 \times 10^2} = 1714$

Answer

$K_c = 1714$ (or 1700 to 2 s.f.)

Final answer

1714 (or 1700 to 2 s.f.)

Detailed explanation

Background Concept

When a reaction is the sum or difference of other reactions, its equilibrium constant is the product or quotient of the equilibrium constants of those reactions. Specifically, if Reaction C = Reaction A - Reaction B, then $K_c = K_A / K_B$ .

Understanding the Question

Calculate $K_c$ for a ligand exchange reaction using the $K_{\text{stab}}$ values from Table 3.2.

Approach

Write the equilibrium equations corresponding to the two $K_{\text{stab}}$ values.
Manipulate these equations to obtain the target equation.
Calculate $K_c$ using the corresponding mathematical operation on the $K$ values.

Step-by-Step Reasoning

Target: $[\text{Fe(H}_2\text{O)}_5\text{SCN}]^{2+} + \text{F}^- \rightleftharpoons [\text{Fe(H}_2\text{O)}_5\text{F}]^{2+} + \text{SCN}^-$
Eq 1 (from $K_{\text{stab}}$ of SCN complex): $[\text{Fe(H}_2\text{O)}_6]^{3+} + \text{SCN}^- \rightleftharpoons [\text{Fe(H}_2\text{O)}_5\text{SCN}]^{2+} + \text{H}_2\text{O} \quad (K = 1.40 \times 10^2)$
Eq 2 (from $K_{\text{stab}}$ of F complex): $[\text{Fe(H}_2\text{O)}_6]^{3+} + \text{F}^- \rightleftharpoons [\text{Fe(H}_2\text{O)}_5\text{F}]^{2+} + \text{H}_2\text{O} \quad (K = 2.40 \times 10^5)$
To get the target, reverse Eq 1 and add to Eq 2:
Reverse Eq 1: $[\text{Fe(H}_2\text{O)}_5\text{SCN}]^{2+} + \text{H}_2\text{O} \rightleftharpoons [\text{Fe(H}_2\text{O)}_6]^{3+} + \text{SCN}^- \quad (K = 1 / 1.40 \times 10^2)$
Add Eq 2: $[\text{Fe(H}_2\text{O)}_6]^{3+} + \text{F}^- \rightleftharpoons [\text{Fe(H}_2\text{O)}_5\text{F}]^{2+} + \text{H}_2\text{O} \quad (K = 2.40 \times 10^5)$
Result: $[\text{Fe(H}_2\text{O)}_5\text{SCN}]^{2+} + \text{F}^- \rightleftharpoons [\text{Fe(H}_2\text{O)}_5\text{F}]^{2+} + \text{SCN}^-$
$K_c = \frac{2.40 \times 10^5}{1.40 \times 10^2} = 1714.28...$
Round to appropriate significant figures (minimum 2 s.f. from the data): 1700 or 1714.

Key Takeaways

To find $K_c$ for a new reaction from known $K$ values, write the equations, manipulate them (reverse = invert K, add = multiply K), and apply the operations to the K values.

Common Mistakes

Adding the K values instead of dividing.
Forgetting to invert the K value when reversing a reaction.
Incorrect significant figures (data has 3 s.f., so 1710 or 1714 is acceptable; mark scheme says min 2 s.f.).

Things to Be Careful About

The mark scheme accepts $1714$ min 2 s.f. Ensure you show the division clearly.

Techniques used

combine equilibrium constants to find Kc for ligand exchange

(iv)

A few drops of KF(aq) are added to a solution of $[\text{Fe(H}_2\text{O)}_6]^{3+}(aq)$ , followed by a few drops of KSCN(aq).

Use information in Table 3.2 to describe any observations after each addition. Explain your answer.

2M Similar

DifficultyMedium

Worked solution

Answer

After adding KF(aq): The solution turns from violet/purple to colourless.
After adding KSCN(aq): The solution remains colourless (no change / stays colourless).
Explanation: The $K_{\text{stab}}$ of $[\text{Fe(H}_2\text{O)}_5\text{F}]^{2+}$ ( $2.40 \times 10^5$ ) is greater than the $K_{\text{stab}}$ of $[\text{Fe(H}_2\text{O)}_5\text{SCN}]^{2+}$ ( $1.40 \times 10^2$ ). Therefore, $[\text{Fe(H}_2\text{O)}_5\text{F}]^{2+}$ is more stable, and the equilibrium lies further to the right for the fluoride complex. The red $[\text{Fe(H}_2\text{O)}_5\text{SCN}]^{2+}$ cannot form (or is immediately displaced) because the fluoride complex is more stable.

Final answer

Turns colourless with KF, stays colourless with KSCN. Kstab of Fe-F complex is larger, so it is more stable.

Detailed explanation

Background Concept

A higher $K_{\text{stab}}$ value indicates a more stable complex. In a mixture of ligands, the ligand that forms the more stable complex (higher $K_{\text{stab}}$ ) will predominantly coordinate to the metal ion. This is a competitive ligand exchange equilibrium.

Understanding the Question

We start with $[\text{Fe(H}_2\text{O)}_6]^{3+}$ (violet). We add a few drops of KF, then a few drops of KSCN. We must describe the colour changes and explain them using the $K_{\text{stab}}$ values.

Approach

Determine the colour of $[\text{Fe(H}_2\text{O)}_6]^{3+}$ (violet).
Add $\text{F}^-$ : compare $K_{\text{stab}}$ of $[\text{Fe(H}_2\text{O)}_5\text{F}]^{2+}$ with $[\text{Fe(H}_2\text{O)}_6]^{3+}$ . Since $K_{\text{stab}} > 1$ , the F complex forms. Its colour is colourless. So violet -> colourless.
Add $\text{SCN}^-$ : now we have $[\text{Fe(H}_2\text{O)}_5\text{F}]^{2+}$ in solution. Compare $K_{\text{stab}}$ values: $K_{\text{stab}}(\text{F}) = 2.40 \times 10^5$ and $K_{\text{stab}}(\text{SCN}) = 1.40 \times 10^2$ . Since $K_{\text{stab}}(\text{F}) \gg K_{\text{stab}}(\text{SCN})$ , the F complex is much more stable. Adding $\text{SCN}^-$ will not displace $\text{F}^-$ . The solution remains colourless.

Step-by-Step Reasoning

Addition of KF:

Initial solution: $[\text{Fe(H}_2\text{O)}_6]^{3+}$ is violet.
Reaction: $[\text{Fe(H}_2\text{O)}_6]^{3+} + \text{F}^- \rightleftharpoons [\text{Fe(H}_2\text{O)}_5\text{F}]^{2+} + \text{H}_2\text{O}$
$K_{\text{stab}} = 2.40 \times 10^5$ (large, equilibrium lies to the right).
Product $[\text{Fe(H}_2\text{O)}_5\text{F}]^{2+}$ is colourless.
Observation: Solution turns from violet/purple to colourless.

Addition of KSCN:

Solution now contains $[\text{Fe(H}_2\text{O)}_5\text{F}]^{2+}$ (colourless).
We add $\text{SCN}^-$ . Could it form $[\text{Fe(H}_2\text{O)}_5\text{SCN}]^{2+}$ (red)?
Compare stabilities: $K_{\text{stab}}([\text{Fe(H}_2\text{O)}_5\text{F}]^{2+}) = 2.40 \times 10^5$ is much greater than $K_{\text{stab}}([\text{Fe(H}_2\text{O)}_5\text{SCN}]^{2+}) = 1.40 \times 10^2$ .
The fluoride complex is more stable. The equilibrium for SCN complex formation does not proceed to the right (or is negligible).
Observation: Solution remains colourless (no change).
Explanation: $K_{\text{stab}}$ of $[\text{Fe(H}_2\text{O)}_5\text{F}]^{2+}$ is greater than that of $[\text{Fe(H}_2\text{O)}_5\text{SCN}]^{2+}$ , so the fluoride complex is more stable and SCN- cannot displace F-.

Key Takeaways

When multiple ligands are present, the one with the higher $K_{\text{stab}}$ will dominate. Sequential addition tests the relative stabilities. A more stable complex will not be displaced by a less stable one.

Common Mistakes

Saying the solution turns red after adding KSCN (ignoring the relative Kstab values).
Not describing the first colour change (violet to colourless).
Explaining the second observation incorrectly (e.g., saying 'no reaction occurs' without referencing the Kstab values or stability).

Things to Be Careful About

The mark scheme requires two marks: one for the observations (violet to colourless, then stays colourless) and one for the explanation (Kstab comparison or relative stability). Ensure both are present.

Techniques used

predict ligand exchange observations based on Kstab values

(c)

Hydrated compound J, $\text{K}_3\text{Fe(C}_2\text{O}_4)_3 \cdot x\text{H}_2\text{O}$ , contains the green complex ion $[\text{Fe(C}_2\text{O}_4)_3]^{3-}$ .

The value of $x$ can be determined by titration of a sample of J with acidified $\text{MnO}_4^-$ ions.

$\text{MnO}_4^-$ ions oxidise $\text{C}_2\text{O}_4^{2-}$ ions in acidic conditions.

2\text{MnO}_4^- + 5\text{C}_2\text{O}_4^{2-} + 16\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 10\text{CO}_2 + 8\text{H}_2\text{O}

6M

(i)

Write half equations for the oxidation of $\text{C}_2\text{O}_4^{2-}$ ions and for the reduction of $\text{MnO}_4^-$ ions.

oxidation of $\text{C}_2\text{O}_4^{2-}$
reduction of $\text{MnO}_4^-$

2M Similar

DifficultyEasy

Worked solution

Answer

oxidation of $\text{C}_2\text{O}_4^{2-}$ :
$\text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2 + 2\text{e}^-$

reduction of $\text{MnO}_4^-$ :
$\text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$

Final answer

Oxidation: C2O4^2- -> 2CO2 + 2e^- ; Reduction: MnO4^- + 8H^+ + 5e^- -> Mn^2+ + 4H2O

Detailed explanation

Background Concept

In redox titrations involving $\text{MnO}_4^-$ and ethanedioate ( $\text{C}_2\text{O}_4^{2-}$ ), we need the half-equations for the oxidation and reduction processes.

Oxidation of ethanedioate: Carbon in $\text{C}_2\text{O}_4^{2-}$ has oxidation state +3. In $\text{CO}_2$ , it is +4. Each carbon loses 1 electron, so the ion loses 2 electrons to form 2 $\text{CO}_2$ molecules.
Reduction of manganate(VII): Manganese in $\text{MnO}_4^-$ is +7. In acidic conditions, it is reduced to $\text{Mn}^{2+}$ (+2), gaining 5 electrons. Oxygen is balanced with water, hydrogen with $\text{H}^+$ .

Understanding the Question

Write the two half-equations for the reaction given in the question stem.

Approach

Recall or derive the standard half-equations for these common species in acidic conditions.

Step-by-Step Reasoning

Oxidation:

Reactant: $\text{C}_2\text{O}_4^{2-}$
Product: $\text{CO}_2$ (carbon is oxidised from +3 to +4)
Balance C: $\text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2$
Balance charge: Left is -2, right is 0. Add $2\text{e}^-$ to right.
Equation: $\text{C}_2\text{O}_4^{2-} \rightarrow 2\text{CO}_2 + 2\text{e}^-$

Reduction:

Reactant: $\text{MnO}_4^-$
Product: $\text{Mn}^{2+}$
Balance Mn: Already balanced.
Balance O: Add $4\text{H}_2\text{O}$ to right.
Balance H: Add $8\text{H}^+$ to left.
Balance charge: Left is $-1 + 8 = +7$ . Right is $+2$ . Add $5\text{e}^-$ to left.
Equation: $\text{MnO}_4^- + 8\text{H}^+ + 5\text{e}^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$

Key Takeaways

These are standard half-equations that should be memorised for A-Level Chemistry. Note the acidic conditions for permanganate reduction.

Common Mistakes

Forgetting to balance the oxygen and hydrogen in the permanganate half-equation.
Getting the electrons wrong (e.g., $4\text{e}^-$ instead of $5\text{e}^-$ for Mn).
Forgetting the charge on the ethanedioate ion.

Things to Be Careful About

Ensure state symbols are not required unless specified (mark scheme doesn't show them for half-equations here, but it's good practice. The mark scheme only awards marks for the correct species and electrons). The mark scheme gives 1 mark for each correct half-equation.

Techniques used

write half-equations for redox titration

(ii)

A student prepares a solution containing 0.100 g of J.

The student titrates this solution with $0.0200\text{ mol dm}^{-3}$ acidified $\text{KMnO}_4(aq)$ . The titre obtained is $12.20\text{ cm}^3$ .

Assume all of the $\text{C}_2\text{O}_4^{2-}$ ions are oxidised.

Calculate the value of $x$ in $\text{K}_3\text{Fe(C}_2\text{O}_4)_3 \cdot x\text{H}_2\text{O}$ .

Give your answer to the nearest whole number. Show your working.

[ $M_r$ : $\text{K}_3\text{Fe(C}_2\text{O}_4)_3$ , 437.1]

4M Similar

DifficultyMedium-Hard

Worked solution

Working

Step 1: Moles of $\text{MnO}_4^-$
$\text{Moles of } \text{MnO}_4^- = \frac{12.20}{1000} \times 0.0200 = 2.44 \times 10^{-4} \text{ mol}$

Step 2: Moles of $\text{C}_2\text{O}_4^{2-}$
From the balanced equation, $2\text{MnO}_4^-$ react with $5\text{C}_2\text{O}_4^{2-}$ .
$\text{Moles of } \text{C}_2\text{O}_4^{2-} = \frac{5}{2} \times 2.44 \times 10^{-4} = 6.10 \times 10^{-4} \text{ mol}$

Step 3: Moles of $\text{K}_3\text{Fe(C}_2\text{O}_4)_3$
Each formula unit contains 3 $\text{C}_2\text{O}_4^{2-}$ ions.
$\text{Moles of } \text{K}_3\text{Fe(C}_2\text{O}_4)_3 = \frac{6.10 \times 10^{-4}}{3} = 2.033 \times 10^{-4} \text{ mol}$

Step 4: Mass of anhydrous $\text{K}_3\text{Fe(C}_2\text{O}_4)_3$
$\text{Mass} = 2.033 \times 10^{-4} \times 437.1 = 0.08887 \text{ g}$

Step 5: Mass and moles of water
$\text{Mass of water} = 0.100 - 0.08887 = 0.01113 \text{ g}$
$\text{Moles of water} = \frac{0.01113}{18.0} = 6.183 \times 10^{-4} \text{ mol}$

Step 6: Value of $x$
$x = \frac{\text{Moles of water}}{\text{Moles of } \text{K}_3\text{Fe(C}_2\text{O}_4)_3} = \frac{6.183 \times 10^{-4}}{2.033 \times 10^{-4}} = 3.04 \approx 3$

Answer

$x = 3$

Final answer

3

Detailed explanation

Background Concept

This is a classic water of crystallisation calculation using a redox titration. The steps are:

Find moles of titrant ( $\text{MnO}_4^-$ ) used.
Use the stoichiometric ratio from the balanced equation to find moles of analyte ( $\text{C}_2\text{O}_4^{2-}$ ).
Use the formula of the complex to find moles of the complex.
Calculate the mass of the anhydrous complex.
Subtract from the total mass to find the mass of water.
Find moles of water and the ratio to determine $x$ .

Understanding the Question

We have $0.100 \text{ g}$ of hydrated $\text{K}_3\text{Fe(C}_2\text{O}_4)_3 \cdot x\text{H}_2\text{O}$ . It is titrated with $12.20 \text{ cm}^3$ of $0.0200 \text{ mol dm}^{-3}$ $\text{KMnO}_4$ . We need to find $x$ .

Approach

Follow the step-by-step calculation outlined above. Be careful with the stoichiometric ratio (5:2) and the fact that there are 3 oxalate ions per complex formula unit.

Step-by-Step Reasoning

Moles of $\text{MnO}_4^-$ :
$n = c \times V = 0.0200 \times \frac{12.20}{1000} = 2.44 \times 10^{-4} \text{ mol}$

Moles of $\text{C}_2\text{O}_4^{2-}$ :
Ratio is $5 \text{ C}_2\text{O}_4^{2-} : 2 \text{ MnO}_4^-$ .
$n(\text{C}_2\text{O}_4^{2-}) = \frac{5}{2} \times 2.44 \times 10^{-4} = 6.10 \times 10^{-4} \text{ mol}$

Moles of complex:
Formula is $\text{K}_3\text{Fe(C}_2\text{O}_4)_3$ , so 1 mol complex contains 3 mol $\text{C}_2\text{O}_4^{2-}$ .
$n(\text{complex}) = \frac{6.10 \times 10^{-4}}{3} = 2.0333 \times 10^{-4} \text{ mol}$

Mass of anhydrous complex:
$M_r(\text{K}_3\text{Fe(C}_2\text{O}_4)_3) = 437.1$
$\text{Mass} = 2.0333 \times 10^{-4} \times 437.1 = 0.088887 \text{ g}$

Mass of water:
$\text{Mass}(\text{H}_2\text{O}) = 0.100 - 0.088887 = 0.011113 \text{ g}$

Moles of water:
$n(\text{H}_2\text{O}) = \frac{0.011113}{18.0} = 6.174 \times 10^{-4} \text{ mol}$

Ratio $x$ :
$x = \frac{6.174 \times 10^{-4}}{2.0333 \times 10^{-4}} = 3.036 \approx 3$

Key Takeaways

Always track the stoichiometry carefully: the 5:2 ratio from the titration equation, and the 3:1 ratio from the complex formula. Round only at the final step to avoid rounding errors.

Common Mistakes

Forgetting to divide moles of oxalate by 3 to get moles of complex.
Using the wrong molar mass (e.g., including water in the $M_r$ before calculating the anhydrous mass).
Arithmetic errors in the final division.

Things to Be Careful About

Give the answer to the nearest whole number as requested: 3.
Keep intermediate values to at least 3-4 significant figures to avoid rounding errors.
The mark scheme awards method marks (M1, M2, M3, M4) for each stage of the calculation. Show your working clearly.

Techniques used

calculate moles from titreuse stoichiometric ratio to find molescalculate empirical formula or water of crystallisation

Q4MediumNitrogen CompoundsHalogen CompoundsHydrocarbonsOrganic Synthesis

(a)

State the difference in the basicities of ammonia, $\text{NH}_3$ , propanamide, $\text{CH}_3\text{CH}_2\text{CONH}_2$ , and propylamine, $\text{CH}_3\text{CH}_2\text{CH}_2\text{NH}_2$ .

Explain your answer.

\text{weakest base} < \text{ } < \text{ strongest base}

4M Similar

DifficultyMedium-Easy

Worked solution

Answer

Order: propanamide < ammonia < propylamine

Explanation:
Basicity depends on the availability of the lone pair on the nitrogen atom to accept a proton (H⁺).

Propanamide (weakest): The lone pair on the nitrogen atom is delocalised into the carbonyl (C=O) group. This makes it much less available to bond with a proton.
Ammonia (intermediate): The lone pair on nitrogen is fully available to accept a proton, but there are no electron-donating groups to increase electron density.
Propylamine (strongest): The propyl (alkyl) group is electron-donating (positive inductive effect). This increases the electron density on the nitrogen atom, making the lone pair more available to accept a proton.

Detailed explanation

Background Concept

The basicity of nitrogen-containing compounds depends on the availability of the lone pair of electrons on the nitrogen atom to act as a Lewis base (donate to a proton, H⁺).

Aliphatic amines (e.g., propylamine): Alkyl groups are electron-donating via the inductive effect. They push electron density towards the nitrogen, making the lone pair more negative and more attractive to protons. Thus, aliphatic amines are stronger bases than ammonia.
Ammonia (NH₃): Has a localized lone pair. It is the reference point. No electron-donating alkyl groups, but no electron-withdrawing or delocalising groups either.
Amides (e.g., propanamide): The nitrogen lone pair is adjacent to a carbonyl group (C=O). The lone pair delocalises (resonates) into the pi-system of the carbonyl group. This is represented by resonance structures where the N-C bond has partial double bond character and the oxygen bears a negative charge. Because the lone pair is involved in this delocalisation, it is not available to accept a proton. Consequently, amides are very weak bases (essentially neutral).

Understanding the Question

The question asks to order three nitrogen compounds—propanamide (an amide), ammonia (an inorganic amine), and propylamine (a primary aliphatic amine)—from weakest to strongest base, and to explain the reasons.

Approach

Identify the functional groups: Propanamide is an amide; ammonia is the reference base; propylamine is a primary aliphatic amine.
Determine the order: Amides are the least basic (lone pair delocalised). Amines are more basic than ammonia due to the inductive effect of alkyl groups. So: Amide < Ammonia < Amine.
Explain:
- For propanamide: Mention delocalisation of the nitrogen lone pair into the C=O group.
- For propylamine: Mention the electron-donating (positive inductive) effect of the alkyl/propyl group.
- For both: Mention that basicity requires the lone pair to accept/coordinate to a proton.

Step-by-Step Reasoning

Ordering: The correct order is propanamide < ammonia < propylamine.
General Principle: Basicity is the ability of the nitrogen lone pair to accept a proton (H⁺).
Propanamide: In amides, the lone pair on nitrogen is delocalised into the carbonyl group (C=O). The resonance involves the overlap of the nitrogen p-orbital with the pi-system of the carbonyl. This delocalisation significantly reduces the electron density on nitrogen, making the lone pair unavailable for protonation. Hence, it is the weakest base.
Ammonia: The lone pair is on an sp³ nitrogen and is localized. It can accept a proton, but there is no additional electron density being pushed onto the nitrogen.
Propylamine: The propyl group (CH₃CH₂CH₂–) is an alkyl group. Alkyl groups exert a positive inductive effect (+I), pushing electron density through the sigma bonds towards the nitrogen. This increases the electron density on the nitrogen lone pair, making it more attractive to protons (H⁺). Hence, it is the strongest base.

Key Takeaways

Amides are weak bases because the nitrogen lone pair is delocalised into the carbonyl group.
Aliphatic amines are stronger bases than ammonia due to the electron-donating inductive effect of alkyl groups.
Basicity is determined by the availability of the lone pair to accept a proton.

Common Mistakes

Stating that amides are basic because they have a nitrogen lone pair (ignoring delocalisation).
Confusing the inductive effect direction (alkyl groups are electron-donating, not withdrawing).
Forgetting to mention that basicity involves accepting a proton (coordination to H⁺).

Things to Be Careful About

Ensure the order is correct: weakest first.
Use precise terminology: "delocalised" for amides, "electron-donating" or "positive inductive effect" for alkyl groups.
Explicitly link the electronic effect to the availability of the lone pair for proton acceptance.

Techniques used

compare basicity of amines, amides, and ammoniaexplain delocalisation of lone pair in amidesexplain inductive effect of alkyl groups

(b)

Fig. 4.1 shows two different ways to synthesise propylamine.

Fig. 4.1

Identify compounds K and L and reagent M from Fig. 4.1.

3M Similar

DifficultyMedium-Easy

Worked solution

Answer

K: CH₃CH₂COCl (propanoyl chloride)
L: CH₃CH₂CN (propanenitrile)
M: LiAlH₄ (lithium tetrahydridoaluminate) or H₂ / Ni (hydrogen with nickel catalyst)

Detailed explanation

Background Concept

There are two common routes to synthesise primary amines from carboxylic acid derivatives or halogenoalkanes:

From Acyl Chlorides: An acyl chloride reacts with excess ammonia to form a primary amide. The amide can then be reduced to a primary amine using a strong reducing agent like LiAlH₄ or catalytic hydrogenation (H₂/Ni).
- Reaction: R-COCl + 2NH₃ → R-CONH₂ + NH₄Cl
- Reduction: R-CONH₂ + 4[H] → R-CH₂NH₂ + H₂O
From Halogenoalkanes: A halogenoalkane reacts with potassium cyanide (KCN) in ethanol to form a nitrile. The nitrile is then reduced to a primary amine.
- Reaction: R-Br + KCN → R-CN + KBr (Nucleophilic substitution)
- Reduction: R-CN + 4[H] → R-CH₂NH₂

Understanding the Question

Fig. 4.1 shows two pathways to synthesise propylamine (CH₃CH₂CH₂NH₂).

Pathway 1: Compound K reacts with NH₃ to form propanamide (CH₃CH₂CONH₂). This implies K is an acyl chloride with a 3-carbon chain.
Pathway 2: Bromoethane (CH₃CH₂Br) reacts with KCN to form compound L. This is a nucleophilic substitution extending the carbon chain by one. L is then reduced to propylamine.
Reagent M: Used in both pathways to reduce the intermediate (amide or nitrile) to the amine.

Approach

Identify K: Since K + NH₃ → propanamide (CH₃CH₂CONH₂), K must be propanoyl chloride (CH₃CH₂COCl). Reaction: CH₃CH₂COCl + 2NH₃ → CH₃CH₂CONH₂ + NH₄Cl.
Identify L: CH₃CH₂Br + KCN → L. The CN group replaces Br. So L is propanenitrile (CH₃CH₂CN).
Identify M: Both amides and nitriles are reduced to primary amines. Common reducing agents are LiAlH₄ (in dry ether, followed by acid workup) or H₂ with a Ni catalyst (high pressure/temp). LiAlH₄ is the standard A-level reagent for reducing amides and nitriles.

Step-by-Step Reasoning

Compound K: The product is propanamide (CH₃CH₂CONH₂). The reaction with ammonia to form an amide typically involves an acyl chloride. Therefore, K is propanoyl chloride, CH₃CH₂COCl.
Compound L: Bromoethane (CH₃CH₂Br) reacts with KCN. This is a nucleophilic substitution where CN⁻ replaces Br⁻. The product is propanenitrile, CH₃CH₂CN. Note the carbon chain increases from 2 to 3 carbons.
Reagent M: Both propanamide and propanenitrile are converted to propylamine (CH₃CH₂CH₂NH₂). This is a reduction reaction. LiAlH₄ is a strong reducing agent capable of reducing both amides and nitriles to amines. Alternatively, catalytic hydrogenation (H₂ / Ni) can be used.

Key Takeaways

Acyl chlorides react with ammonia to form amides.
Halogenoalkanes react with cyanide ions to form nitriles (chain extension).
Both amides and nitriles can be reduced to primary amines using LiAlH₄ or H₂/Ni.

Common Mistakes

Identifying K as propanoic acid (reacts slowly with ammonia to form salt, not amide directly without heat; acyl chloride is the standard reagent).
Identifying L as butanenitrile (wrong chain length; bromoethane has 2 carbons, KCN adds 1, total 3).
Forgetting that M is a reducing agent (some students might suggest NaBH₄, which is too weak to reduce amides/nitriles to amines).

Things to Be Careful About

Ensure the carbon count is correct: Propanamide has 3 carbons. Propanoyl chloride has 3 carbons.
Bromoethane (2C) + KCN (1C) = Propanenitrile (3C).
Reagent M must be a strong reducing agent (LiAlH₄) or catalytic hydrogenation.

Techniques used

identify reagents for amide synthesisidentify nitrile synthesis from halogenoalkaneidentify reducing agents for amide/nitrile to amine

(c)

Compound N is shown in Fig. 4.2.

Fig. 4.2

Compound N is treated with an excess of concentrated HCl(aq).

N undergoes complete hydrolysis to form three organic products.

The products are isolated from the reaction mixture at pH 4.

Draw the structures of the three organic products at pH 4.

Assume that the $\text{CH}_3\text{O}-$ group does not react.

4M Similar

DifficultyMedium

Worked solution

Answer

The three organic products at pH 4 are:

Benzylammonium ion:
$\text{C}_6\text{H}_5\text{CH}_2\text{NH}_3^+$
(Benzene ring attached to CH₂-NH₃⁺)
Protonated amino acid derivative:
$^+\text{H}_3\text{N}-\text{CH}(\text{CH}_2\text{OCH}_3)-\text{COOH}$
(HOOC-CH(CH₂OCH₃)-NH₃⁺)
Ethanoic acid:
$\text{CH}_3\text{COOH}$

(Note: At pH 4, amine groups are protonated to -NH₃⁺ and carboxylic acid groups remain as -COOH.)

Detailed explanation

Background Concept

Hydrolysis of Amides/Peptides:
Amide bonds (-CO-NH-) are hydrolysed by heating with aqueous acid (e.g., excess conc. HCl) or base. Acid hydrolysis produces carboxylic acids and ammonium salts (protonated amines).

R-CO-NH-R' + H₂O + HCl → R-COOH + R'-NH₃⁺Cl⁻

Protonation States and pH:

Carboxylic acids (-COOH): pKa ≈ 2-5. At pH < pKa, the group is protonated (-COOH). At pH > pKa, it is deprotonated (-COO⁻). At pH 4, a typical carboxylic acid (pKa ~2-4) will be partially protonated, but in the context of A-level questions with "excess acid" followed by isolation, we often consider the dominant species or follow the specific instruction. The mark scheme states "all acid groups as -COOH". This implies we treat pH 4 as acidic enough to keep COOH protonated, or simply that the hydrolysis produces COOH and at pH 4 it hasn't fully deprotonated (or we just draw the hydrolysis product). Actually, pKa of acetic acid is 4.76. At pH 4, [HA] > [A-]. So COOH is the major species. For the amino acid part, the carboxyl group has pKa ~2. At pH 4, it would be mostly COO-. However, the mark scheme explicitly says "all acid groups as -COOH". This is a specific instruction for the answer.
Amines (-NH₂ / -NH-): pKa of conjugate acid (R-NH₃⁺) ≈ 9-11. At pH < pKa, the amine is protonated (-NH₃⁺ or -NH₂R⁺). At pH 4, all amine groups will be fully protonated.

Understanding the Question

Compound N is a peptide-like molecule: Benzene-CH₂-NH-CO-CH(CH₂OCH₃)-NH-CO-CH₃.
It is treated with excess conc. HCl (complete hydrolysis of amide bonds).
Products are isolated at pH 4.
We need to draw the three organic products at this pH.
Assumption: The ether group (CH₃O-) does not react.

Approach

Hydrolyse the molecule: Break the amide bonds.
- Bond 1: Ph-CH₂-NH | CO-CH(...)
- Bond 2: ...CH(...) | NH-CO-CH₃
Identify fragments:
- Fragment 1: Ph-CH₂-NH₂ (benzylamine). From acid hydrolysis, this becomes Ph-CH₂-NH₃⁺.
- Fragment 2: HOOC-CH(CH₂OCH₃)-NH₂ (an amino acid derivative). From acid hydrolysis, the amine becomes -NH₃⁺. So +H₃N-CH(CH₂OCH₃)-COOH.
- Fragment 3: CH₃COOH (ethanoic acid).
Apply pH 4 conditions:
- Benzylamine (pKa ~9.4): At pH 4, it is protonated → Ph-CH₂-NH₃⁺.
- Amino acid part: Amine (pKa ~9) is protonated → -NH₃⁺. Carboxylic acid: Mark scheme says keep as -COOH.
- Ethanoic acid: Keep as -COOH.

Step-by-Step Reasoning

Hydrolysis: The molecule has two amide linkages.
- Structure: Ph-CH₂-NH-CO-CH(CH₂OCH₃)-NH-CO-CH₃
- Hydrolysis breaks C-N bonds in amides.
- Left part: Ph-CH₂-NH₂ (becomes Ph-CH₂-NH₃⁺ in acid).
- Middle part: HOOC-CH(CH₂OCH₃)-NH₂ (becomes HOOC-CH(CH₂OCH₃)-NH₃⁺ in acid).
- Right part: HOOC-CH₃ (ethanoic acid, CH₃COOH).
Protonation at pH 4:
- Benzylamine derivative: The amine group (pKa of conjugate acid ~9-10) will be protonated at pH 4. Structure: C₆H₅-CH₂-NH₃⁺.
- Amino acid derivative: The amine group on the central carbon will be protonated (-NH₃⁺). The carboxylic acid group: The mark scheme specifies "all acid groups as -COOH". So we draw HOOC-CH(CH₂OCH₃)-NH₃⁺. (Note: The ether group -OCH₃ remains unchanged).
- Ethanoic acid: CH₃COOH.

Key Takeaways

Acid hydrolysis of amides/peptides yields carboxylic acids and ammonium ions (protonated amines).
At acidic pH (like pH 4), amines are protonated (-NH₃⁺).
Carboxylic acids may be protonated (-COOH) depending on pKa and specific question instructions (here, explicitly instructed to keep as -COOH).

Common Mistakes

Drawing the products as neutral molecules (forgetting protonation of amines in acid hydrolysis).
Drawing the carboxylic acid as carboxylate (-COO⁻) at pH 4 (mark scheme says -COOH).
Breaking the wrong bonds (e.g., ether cleavage, which is not asked for / assumed not to happen).
Missing the ether group in the middle product.

Things to Be Careful About

Check the pH. pH 4 is acidic. Amines are definitely protonated.
Follow the mark scheme instruction for acid groups: "all acid groups as -COOH".
Ensure the ether linkage (CH₃O-) is preserved in the middle product.

Techniques used

hydrolyse peptide bonds with aciddetermine protonation state at specific pHdraw structures of hydrolysis products

(d)

Compound P can be synthesised from 1-methyl-4-nitrobenzene by the route shown in Fig. 4.3.

Fig. 4.3

7M

(i)

Step 1 is a reduction reaction.

Complete the equation for this reaction. Use [H] to represent an atom of hydrogen from the reducing agent.

\text{C}_7\text{H}_7\text{NO}_2 + \dots

1M Similar

DifficultyEasy

Worked solution

Answer

\text{C}_7\text{H}_7\text{NO}_2 + 6[\text{H}] \rightarrow \text{C}_7\text{H}_7\text{NH}_2 + 2\text{H}_2\text{O}

Detailed explanation

Background Concept

Reduction of nitroarenes (nitrobenzene derivatives) to amines (phenylamines) is a standard reaction.

Reagents: Tin (Sn) and concentrated hydrochloric acid (HCl), or Iron (Fe) and HCl, followed by neutralisation. Or catalytic hydrogenation (H₂/Ni).
The nitro group (-NO₂) is reduced to an amino group (-NH₂).
Using [H] to represent reducing agent (atoms of hydrogen):
- -NO₂ → -NH₂ requires addition of 6 hydrogen atoms and removal of 2 oxygen atoms (as water).
- Equation: R-NO₂ + 6[H] → R-NH₂ + 2H₂O.

Understanding the Question

Part (d)(i) asks to complete the equation for Step 1 in Fig 4.3.

Reactant: 1-methyl-4-nitrobenzene (C₇H₇NO₂). Formula check: Benzene ring (C6) + CH3 + NO2. C6H4(CH3)(NO2) = C7H7NO2. Correct.
Product: 1-methyl-4-aminobenzene (p-toluidine). Formula: C6H4(CH3)(NH2) = C7H9N. Wait. C7H7NH2 = C7H9N. The question writes product as C7H7NH2. Let's check atoms. Reactant C7H7NO2. Product C7H7NH2 (which is C7H9N). Difference: -O2 + 4H? No.
Let's balance: C7H7NO2 + x[H] -> C7H7NH2 + yH2O.
- N: 1 -> 1.
- O: 2 -> 2y. So y=2.
- H: 7 + x = 7 + 2 + 2(2) = 11. x = 4? Wait.
- Reactant: C7H7NO2. (p-nitrotoluene). H count: 4 (ring) + 3 (methyl) = 7. Correct.
- Product: C7H7NH2. (p-toluidine). H count: 4 (ring) + 3 (methyl) + 2 (amine) = 9. Written as C7H7NH2 implies C7H9N.
- Equation: C7H7NO2 + 6[H] -> C7H7NH2 + 2H2O.
- Check H: 7 + 6 = 13. Right side: 7+2 (in NH2) + 4 (in 2H2O) = 13. Correct.
- Check O: 2 -> 2. Correct.

Approach

Identify the change: -NO₂ becomes -NH₂.
Balance oxygen: 2 O atoms become 2 H₂O.
Balance hydrogen: Product has 2 extra H on N (NH2 vs NO2... wait. NO2 has no H. NH2 has 2 H. Water has 4 H. Total H needed on right side extra = 2+4=6. So 6[H].

Step-by-Step Reasoning

Reactant: C₇H₇NO₂ (4-nitrotoluene).
Product: C₇H₇NH₂ (4-methylaniline / p-toluidine).
Reduction of nitro group: -NO₂ + 6[H] → -NH₂ + 2H₂O.
Full equation: C₇H₇NO₂ + 6[H] → C₇H₇NH₂ + 2H₂O.

Key Takeaways

Nitro group reduction: R-NO₂ + 6[H] → R-NH₂ + 2H₂O.
Always check atom balance, especially H and O when using [H].

Common Mistakes

Wrong number of [H] (e.g., 4[H] or 8[H]).
Forgetting water as a product.
Wrong formula for product.

Things to Be Careful About

The question gives the product formula as C₇H₇NH₂. Be careful counting hydrogens.
Ensure the equation is balanced.

Techniques used

balance reduction equation using [H]reduce nitro group to amine

(ii)

Complete Table 4.1 to give details of each step of the synthesis shown in Fig. 4.3.

Table 4.1

step	reagents and conditions	type of reaction
1		reduction
2
3
4		condensation

6M Similar

DifficultyMedium

Worked solution

Answer

step	reagents and conditions	type of reaction
1	Sn / conc. HCl (+ heat)	reduction
2	CH₃Br (or CH₃Cl / CH₃I) in ethanol	nucleophilic substitution
3	hot KMnO₄ (or hot MnO₄⁻ / H⁺)	oxidation
4	3-methylbutan-1-ol (+ conc. H₂SO₄)	condensation

(Note for Step 4: The alcohol is 3-methylbutan-1-ol, structure HOCH₂CH₂CH(CH₃)₂)

Detailed explanation

Background Concept

Step 1: Reduction of Nitroarene to Phenylamine

Reagents: Tin (Sn) and concentrated HCl, or Iron (Fe) and HCl. Heat is usually required.
Product: Phenylamine derivative (p-toluidine here).

Step 2: Alkylation of Amine

Goal: Convert -NH₂ to -N(CH₃)₂.
Reagents: Excess alkyl halide (e.g., CH₃Br or CH₃Cl) in ethanol, or sequential methylation. Since we want a tertiary amine (dimethylamino), we use methyl halide. Often excess CH₃Br is used to drive it to the tertiary amine, though mixtures can occur. In synthesis questions, CH₃Br / ethanol is the standard answer for N-methylation.
Type: Nucleophilic substitution (amine acts as nucleophile, attacking alkyl halide).

Step 3: Side-chain Oxidation

Goal: Convert -CH₃ on benzene ring to -COOH.
Reagents: Hot alkaline or acidic potassium manganate(VII) (KMnO₄ / MnO₄⁻ / H⁺). Usually heated under reflux.
Note: The amine group -N(CH₃)₂ is sensitive to oxidation, but in this context (A-level), we assume the side-chain oxidation proceeds. Actually, N,N-dimethyl-p-toluidine oxidation... the amine might get oxidised, but for the purpose of the exam question, we focus on the methyl to carboxyl conversion. Reagents: hot KMnO₄.
Type: Oxidation.

Step 4: Esterification

Goal: Convert -COOH to ester -COO-R.
Reactant: 3-methylbutan-1-ol (from Fig 4.d.ii.4: HO-CH₂-CH₂-CH(CH₃)₂... wait, let's trace the structure in Fig 4.3 for P.
- P has ester group -O-CH₂-CH₂-CH(CH₃)-... wait. The image shows -O-CH₂-CH₂-CH(CH₃)₂? Let's look at Fig 4.d.ii.4. HO-CH₂-CH₂-CH(CH₃)-... it's a skeletal structure. HO-CH₂-CH₂-CH(CH₃)₂ is 3-methylbutan-1-ol. Yes.
Reagents: Carboxylic acid + alcohol + catalyst (conc. H₂SO₄). Heat.
Type: Condensation (or esterification).

Understanding the Question

Complete Table 4.1 for the synthesis of Compound P from 1-methyl-4-nitrobenzene.

Step 1: Nitro to Amine.
Step 2: Amine to Dimethylamine.
Step 3: Methyl to Carboxylic Acid.
Step 4: Carboxylic Acid to Ester.

Approach

Step 1: Standard nitro reduction. Sn/HCl.
Step 2: N-alkylation. CH₃Br / ethanol. Type: Substitution.
Step 3: Side chain oxidation. KMnO₄ / heat. Type: Oxidation.
Step 4: Esterification. Alcohol (3-methylbutan-1-ol) + acid catalyst. Type: Condensation.

Step-by-Step Reasoning

Step 1: Reactant is 1-methyl-4-nitrobenzene. Product is 1-methyl-4-aminobenzene. Reagents: Sn and conc. HCl (with heat). Type is given as reduction.
Step 2: Reactant is 1-methyl-4-aminobenzene (-NH₂). Product is 1-methyl-4-(dimethylamino)benzene (-N(CH₃)₂). This requires methylation. Reagents: CH₃Br (or CH₃Cl, CH₃I) in ethanol. Type: Nucleophilic substitution (the lone pair on N attacks the methyl halide).
Step 3: Reactant has a methyl group on the ring and a dimethylamino group. Product has a carboxylic acid group (-COOH) where the methyl was. The dimethylamino group remains. Reagents: Hot KMnO₄ (or hot MnO₄⁻ with acid). Type: Oxidation.
Step 4: Reactant is 4-(dimethylamino)benzoic acid. Product P is an ester. The alcohol part comes from Fig 4.d.ii.4, which is 3-methylbutan-1-ol (skeletal: HO-CH₂-CH₂-CH(CH₃)₂... wait, let's look at the structure in Fig 4.3 for P. The ester alkyl chain is -O-CH₂-CH₂-CH(CH₃)-... actually it looks like -O-CH₂-CH₂-CH(CH₃)₂. Yes, 3-methylbutan-1-ol. Reagents: 3-methylbutan-1-ol + conc. H₂SO₄ (catalyst). Type: Condensation (esterification is a type of condensation).

Key Takeaways

Nitroarenes are reduced to amines with Sn/HCl.
Amines can be alkylated with haloalkanes (nucleophilic substitution).
Alkyl side chains on benzene rings (with at least one H on the benzylic carbon) are oxidised to carboxylic acids by hot KMnO₄.
Carboxylic acids react with alcohols in the presence of acid catalyst to form esters (condensation).

Common Mistakes

Step 2: Using CH₃OH (methanol) instead of CH₃Br (methyl bromide). Alcohols don't alkylate amines easily without special conditions; haloalkanes are standard.
Step 3: Forgetting "hot" or "reflux" for KMnO₄ oxidation.
Step 4: Forgetting the acid catalyst (conc. H₂SO₄) or the specific alcohol name/structure.
Reaction types: Calling esterification "addition" (it's condensation/elimination of water).

Things to Be Careful About

Step 2 reagents: CH₃Br / ethanol is correct. Just "CH₃Br" might not be enough if conditions are required, but "in ethanol" covers it.
Step 3: KMnO₄ must be hot/concentrated to oxidise the side chain completely to COOH.
Step 4: The alcohol is 3-methylbutan-1-ol. Ensure the structure matches the ester in P.

Techniques used

identify reagents for nitro reductionidentify reagents for amine alkylationidentify reagents for side-chain oxidationidentify reagents for esterification

Q5MediumIntroduction to A Level Organic ChemistryHydrocarbonsChemical EnergeticsHydroxy CompoundsNitrogen Compounds

Cumene is an aromatic hydrocarbon used in the synthesis of other useful chemicals.

Fig. 5.1

(a)

Complete Table 5.1 to show the number of $\text{sp}^2$ and $\text{sp}^3$ hybridised carbon atoms that are present in a molecule of cumene.

Table 5.1

type of hybridisation	$\text{sp}^2$	$\text{sp}^3$
number of carbon atoms

1M Similar

DifficultyEasy

Worked solution

Answer

type of hybridisation	$\text{sp}^2$	$\text{sp}^3$
number of carbon atoms	6	3

Final answer

sp2 = 6, sp3 = 3

Detailed explanation

Background Concept

Hybridisation describes how atomic orbitals mix to form new hybrid orbitals used in bonding. Carbon with three regions of electron density (trigonal planar geometry, as in a benzene ring where each carbon forms two $\sigma$ bonds and contributes one electron to the $\pi$ system) is $\text{sp}^2$ hybridised. Carbon with four regions of electron density (tetrahedral geometry, as in saturated $\text{C}-\text{C}$ and $\text{C}-\text{H}$ single bonds) is $\text{sp}^3$ hybridised.

Understanding the Question

The question asks you to count the number of carbon atoms of each hybridisation type in a single molecule of cumene (isopropylbenzene), whose structure is given in Fig. 5.1. The molecule consists of a benzene ring bonded to an isopropyl group $-\text{CH}(\text{CH}_3)_2$ .

Approach

Examine each carbon atom and determine how many $\sigma$ -bonding regions it has. Ring carbons each have three regions (two C–C $\sigma$ bonds and one C–H or C–C bond to the substituent), so they are $\text{sp}^2$ . The three carbons in the isopropyl group each have four single bonds, so they are $\text{sp}^3$ .

Step-by-Step Reasoning

Benzene ring: The six carbon atoms in the aromatic ring each participate in the delocalised $\pi$ system. Each ring carbon forms three $\sigma$ bonds (to two neighbouring ring carbons and to either a hydrogen or the substituent), giving trigonal planar geometry. All six are $\text{sp}^2$ hybridised.

Isopropyl group: The central carbon of $-\text{CH}(\text{CH}_3)_2$ forms four $\sigma$ bonds (one to the ring, one to H, two to methyl carbons) — $\text{sp}^3$ . Each of the two $\text{CH}_3$ carbons also forms four $\sigma$ bonds (one to the central carbon, three to H) — $\text{sp}^3$ . Total $\text{sp}^3$ carbons = 3.

Key Takeaways

Aromatic ring carbons are always $\text{sp}^2$ hybridised.
Saturated alkyl carbons (only single bonds) are $\text{sp}^3$ hybridised.
Count carefully: the carbon bonded to the ring from the side chain is $\text{sp}^3$ , not $\text{sp}^2$ .

Common Mistakes

Counting the carbon attached to the ring as $\text{sp}^2$ — it is $\text{sp}^3$ because it forms four single bonds.
Forgetting that the ring carbon bonded to the substituent is still $\text{sp}^2$ (it retains its $\pi$ -system participation).
Miscounting the total number of carbons (cumene has 9 carbons total: 6 + 3).

Things to Be Careful About

The question asks for the number of carbon atoms of each type, not the total number of atoms or bonds.
Ensure you have identified the correct structure from Fig. 5.1 — it is isopropylbenzene, not propylbenzene.

Techniques used

identify hybridisation from bonding environmentcount carbon atoms by type

(b)

Cumene can be synthesised via a Friedel–Crafts alkylation reaction, as shown in Fig. 5.2.

Fig. 5.2

6M

(i)

Name the mechanism involved in the Friedel–Crafts alkylation shown in Fig. 5.2.

1M Similar

DifficultyEasy

Worked solution

Answer

Electrophilic substitution (electrophilic aromatic substitution)

Final answer

Electrophilic substitution

Detailed explanation

Background Concept

Friedel–Crafts alkylation is a method of attaching an alkyl group to an aromatic ring. The reaction proceeds via electrophilic aromatic substitution: a carbocation electrophile is generated, attacks the electron-rich $\pi$ system of benzene, forms a sigma complex (arenium ion) intermediate, and then loses a proton to restore aromaticity.

Understanding the Question

The question simply asks you to name the mechanism type. Fig. 5.2 shows benzene reacting with 2-bromopropane in the presence of a catalyst to form cumene and HBr — this is the classic Friedel–Crafts alkylation.

Approach

Recognise that Friedel–Crafts alkylation is a specific example of electrophilic aromatic substitution, where an electrophile (here, a carbocation) replaces a hydrogen on the aromatic ring.

Step-by-Step Reasoning

The mechanism involves:

Generation of an electrophile (the $\text{(CH}_3)_2\text{CH}^+$ carbocation) from the haloalkane using a Lewis acid catalyst.
Attack of the electrophile on the $\pi$ electrons of benzene.
Loss of $\text{H}^+$ to restore aromaticity.

This sequence is the definition of electrophilic substitution on an aromatic ring.

Key Takeaways

All Friedel–Crafts reactions (both alkylation and acylation) proceed via electrophilic aromatic substitution.
The key feature is that the aromatic ring acts as a nucleophile attacking an electrophile, and a hydrogen is substituted (not added across the ring).

Common Mistakes

Writing "nucleophilic substitution" — benzene is electron-rich and acts as a nucleophile; the reaction is electrophilic substitution.
Writing "addition" — the ring is not permanently broken; aromaticity is restored.

Things to Be Careful About

The full accepted name is "electrophilic substitution" or "electrophilic aromatic substitution." Either is creditable.

Techniques used

name the mechanism type for Friedel-Crafts alkylation

(ii)

The first step of the reaction forms the $(\text{CH}_3)_2\text{CH}^+$ cation.

Identify a suitable reagent for the formation of this cation from 2-bromopropane, $(\text{CH}_3)_2\text{CHBr}$ .

1M Similar

DifficultyEasy

Worked solution

Answer

$\text{AlBr}_3$ (aluminium bromide)

Final answer

AlBr3

Detailed explanation

Background Concept

In Friedel–Crafts alkylation, a Lewis acid catalyst is needed to polarise and ultimately cleave the C–Br bond of the haloalkane, generating a carbocation electrophile. Aluminium halides (typically $\text{AlCl}_3$ or $\text{AlBr}_3$ ) serve this role by accepting a lone pair from the halogen, weakening the C–X bond.

Understanding the Question

The question states that the first step forms the $(\text{CH}_3)_2\text{CH}^+$ cation from 2-bromopropane. You need to identify the reagent that facilitates this ionisation.

Approach

Since the haloalkane is a bromoalkane (contains Br), the appropriate aluminium halide catalyst is $\text{AlBr}_3$ . (If it were a chloroalkane, $\text{AlCl}_3$ would be used.)

Step-by-Step Reasoning

The mechanism of electrophile generation:

(\text{CH}_3)_2\text{CHBr} + \text{AlBr}_3 \rightarrow (\text{CH}_3)_2\text{CH}^+ + \text{AlBr}_4^-

$\text{AlBr}_3$ is a Lewis acid (electron-pair acceptor). The lone pair on bromine coordinates to the aluminium, polarising and breaking the C–Br bond, releasing the secondary carbocation.

Key Takeaways

The catalyst in Friedel–Crafts alkylation must match the halogen in the haloalkane: $\text{AlCl}_3$ for chloroalkanes, $\text{AlBr}_3$ for bromoalkanes.
The role of the catalyst is to generate the electrophile (carbocation).

Common Mistakes

Writing $\text{AlCl}_3$ when the substrate is a bromoalkane — while $\text{AlCl}_3$ might still function, $\text{AlBr}_3$ is the correct choice for a bromo substrate.
Writing "concentrated sulfuric acid" or " $\text{FeBr}_3$ " — these are catalysts for other reactions.

Things to Be Careful About

The question asks for a "suitable reagent" — the name "aluminium bromide" or the formula $\text{AlBr}_3$ are both acceptable.

Techniques used

identify the Lewis acid catalyst for generating a carbocation electrophile

(iii)

Complete Fig. 5.3 to show the mechanism for the reaction of benzene with the $(\text{CH}_3)_2\text{CH}^+$ cation.

Include all relevant curly arrows and charges.

Fig. 5.3

3M Similar

DifficultyMedium

Worked solution

Answer

Final answer

Curly arrow from ring to C+ of electrophile; intermediate with H and CH(CH3)2 on same carbon with delocalised positive charge in ring; curly arrow from C-H bond into ring with loss of H+

Detailed explanation

Background Concept

Electrophilic aromatic substitution proceeds in two stages. In the first stage, the $\pi$ electrons of the aromatic ring attack the electrophile, forming a new C–C bond. This destroys the aromaticity and produces a positively charged intermediate called the sigma complex (or arenium ion), in which the positive charge is delocalised over three carbon atoms (ortho and para positions relative to the site of attack). In the second stage, a base (often the $\text{AlBr}_4^-$ or a solvent molecule) removes the proton from the carbon bearing the electrophile, and the C–H bond electrons return to the ring, restoring aromaticity.

Understanding the Question

Fig. 5.3 provides a template: benzene + $(\text{CH}_3)_2\text{CH}^+$ → [intermediate box] → cumene + ___. You must fill in the intermediate structure, draw the curly arrows for both steps, and identify the by-product ( $\text{H}^+$ ).

Approach

Draw a curly arrow from the delocalised $\pi$ system (inside the hexagon) to the positively charged carbon of $(\text{CH}_3)_2\text{CH}^+$ .
Draw the intermediate: the ring now has one $\text{sp}^3$ carbon bearing both H and $\text{CH}(\text{CH}_3)_2$ , with a positive charge delocalised over the remaining five carbons (shown as a horseshoe with a + sign, or by drawing the three resonance structures).
Draw a curly arrow from the C–H bond (on the $\text{sp}^3$ carbon) back into the ring, and write $\text{H}^+$ as the product.

Step-by-Step Reasoning

M1 — Curly arrow from ring to electrophile: The arrow starts from inside the hexagon (representing the $\pi$ electrons) and points towards the carbon atom of $(\text{CH}_3)_2\text{CH}^+$ that bears the positive charge. This shows the ring acting as a nucleophile.

M2 — Intermediate structure: The carbon that was attacked is now $\text{sp}^3$ hybridised, bonded to H and $\text{CH}(\text{CH}_3)_2$ . The positive charge is delocalised over the other five ring carbons — typically drawn as a horseshoe (arc) inside the ring with a + sign, or by showing the three contributing resonance forms. The ring is no longer fully aromatic.

M3 — Curly arrow from C–H into ring and loss of H⁺: A curly arrow starts from the C–H bond on the $\text{sp}^3$ carbon and points into the ring (towards the adjacent C–C bond), showing the electrons returning to the $\pi$ system. $\text{H}^+$ is written as the by-product, restoring aromaticity.

Key Takeaways

The intermediate is a sigma complex (arenium ion) with delocalised positive charge.
The curly arrow convention: from electron source to electron destination.
Aromaticity is temporarily lost and then restored.

Common Mistakes

Drawing the curly arrow from the electrophile to the ring (wrong direction — arrows go from electrons to where they are going).
Forgetting to show the positive charge on the intermediate, or placing it on the wrong carbon.
Drawing the intermediate with a full benzene ring (aromatic) rather than showing the disrupted $\pi$ system.
Not showing the C–H bond curly arrow for the second step.

Things to Be Careful About

The intermediate must show that one carbon is $\text{sp}^3$ (bonded to both H and the electrophile group).
The positive charge must be delocalised (horseshoe + or resonance structures), not localised on one carbon.
The by-product is $\text{H}^+$ , not $\text{H}_2$ or anything else.

Techniques used

draw curly arrows from pi system to electrophiledraw the sigma complex intermediate with delocalised positive chargedraw curly arrow from C-H bond into ring to restore aromaticity

(iv)

The Friedel–Crafts alkylation of benzene by 1-bromopropane, $\text{CH}_3\text{CH}_2\text{CH}_2\text{Br}$ , also produces cumene as the major product.

The $\text{CH}_3\text{CH}_2\text{CH}_2^+$ cation formed in the first step quickly rearranges to form the $(\text{CH}_3)_2\text{CH}^+$ cation.

Suggest why this is the case. Explain your answer.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

The $(\text{CH}_3)_2\text{CH}^+$ cation is a secondary ( $2^\circ$ ) carbocation, which is more stable than the primary ( $1^\circ$ ) $\text{CH}_3\text{CH}_2\text{CH}_2^+$ cation because it has a greater positive inductive effect from two alkyl groups (which donate electron density towards the electron-deficient carbon).

\text{CH}_3\text{CH}_2\text{CH}_2^+ \rightarrow (\text{CH}_3)_2\text{CH}^+

Final answer

The secondary carbocation (CH3)2CH+ is more stable than the primary carbocation due to the greater positive inductive effect of two alkyl groups

Detailed explanation

Background Concept

Carbocations are electron-deficient species (the positively charged carbon has only six electrons in its valence shell). Their stability increases with the number of alkyl groups attached to the $\text{C}^+$ centre. This is because alkyl groups exhibit a positive inductive effect ( $+I$ ): they push electron density towards the electron-deficient carbon, partially dispersing the positive charge. The stability order is: tertiary $>$ secondary $>$ primary $>$ methyl.

Understanding the Question

When 1-bromopropane undergoes Friedel–Crafts alkylation, the initially formed primary carbocation $\text{CH}_3\text{CH}_2\text{CH}_2^+$ rearranges to the secondary carbocation $(\text{CH}_3)_2\text{CH}^+$ . You need to explain why this rearrangement occurs.

Approach

Compare the stability of the two carbocations. The secondary carbocation has two alkyl groups donating electron density to the positive centre, compared to only one alkyl group in the primary carbocation. The more stable carbocation is favoured.

Step-by-Step Reasoning

$\text{CH}_3\text{CH}_2\text{CH}_2^+$ is a primary ( $1^\circ$ ) carbocation with one alkyl group providing a positive inductive effect.
$(\text{CH}_3)_2\text{CH}^+$ is a secondary ( $2^\circ$ ) carbocation with two alkyl groups providing a greater positive inductive effect.
The secondary carbocation is more stable because the positive charge is better dispersed by the electron-donating effect of two methyl groups.
Therefore, a hydride shift (migration of $\text{H}^-$ from the adjacent carbon) converts the less stable primary cation to the more stable secondary cation.

Key Takeaways

Carbocation stability: $3^\circ > 2^\circ > 1^\circ > \text{methyl}$ .
The positive inductive effect of alkyl groups stabilises the electron-deficient carbon.
Rearrangements occur to form more stable intermediates.

Common Mistakes

Saying "more alkyl groups means more electrons" without specifying the inductive effect.
Confusing the inductive effect with hyperconjugation (though both contribute, the mark scheme specifically asks for the positive inductive effect).
Saying "tertiary" when the product is secondary.

Things to Be Careful About

The question asks you to both suggest WHY and EXPLAIN, so you must state that the secondary cation is more stable AND give the reason (greater positive inductive effect).

Techniques used

compare carbocation stability by degreeexplain stability using positive inductive effect

(c)

Cumene oxidises in air to form phenol, $\text{C}_6\text{H}_5\text{OH}$ , and propanone, $\text{CH}_3\text{COCH}_3$ .

\text{C}_6\text{H}_5\text{CH(CH}_3)_2 + \text{O}_2 \rightarrow \text{C}_6\text{H}_5\text{OH} + \text{CH}_3\text{COCH}_3 \quad \Delta H^\ominus = -371 \text{ kJ mol}^{-1}

Table 5.2 gives some relevant standard entropies for reaction 1.

Table 5.2

compound	$\text{C}_6\text{H}_5\text{CH(CH}_3)_2$	$\text{O}_2$	$\text{C}_6\text{H}_5\text{OH}$	$\text{CH}_3\text{COCH}_3$
standard entropy, $S^\ominus / \text{J K}^{-1} \text{mol}^{-1}$	278	205	146	200

3M

(i)

Calculate the standard entropy change, $\Delta S^\ominus$ , of reaction 1.

1M Similar

DifficultyMedium-Easy

Worked solution

Working

\Delta S^\ominus = \sum S^\ominus(\text{products}) - \sum S^\ominus(\text{reactants})

\Delta S^\ominus = (146 + 200) - (278 + 205)

\Delta S^\ominus = 346 - 483 = -137 \text{ J K}^{-1} \text{mol}^{-1}

Answer

$\Delta S^\ominus = -137 \text{ J K}^{-1} \text{mol}^{-1}$

Final answer

-137 J K^-1 mol^-1

Detailed explanation

Background Concept

The standard entropy change of a reaction is calculated using the formula:

\Delta S^\ominus_{\text{reaction}} = \sum S^\ominus(\text{products}) - \sum S^\ominus(\text{reactants})

Entropy is a measure of the number of ways energy can be distributed among particles. A negative $\Delta S$ indicates a decrease in disorder, which can occur when fewer moles of gas are produced than consumed, or when more complex molecules break into fewer products.

Understanding the Question

You are given the standard entropies of all four species in the reaction:

\text{C}_6\text{H}_5\text{CH(CH}_3)_2 + \text{O}_2 \rightarrow \text{C}_6\text{H}_5\text{OH} + \text{CH}_3\text{COCH}_3

You must calculate $\Delta S^\ominus$ for this reaction.

Approach

Identify products and reactants, sum their standard entropies, and subtract reactants from products.

Step-by-Step Reasoning

Products: $\text{C}_6\text{H}_5\text{OH}$ ( $S^\ominus = 146$ ) and $\text{CH}_3\text{COCH}_3$ ( $S^\ominus = 200$ )

\sum S^\ominus(\text{products}) = 146 + 200 = 346 \text{ J K}^{-1} \text{mol}^{-1}

Reactants: $\text{C}_6\text{H}_5\text{CH(CH}_3)_2$ ( $S^\ominus = 278$ ) and $\text{O}_2$ ( $S^\ominus = 205$ )

\sum S^\ominus(\text{reactants}) = 278 + 205 = 483 \text{ J K}^{-1} \text{mol}^{-1}

Entropy change:

\Delta S^\ominus = 346 - 483 = -137 \text{ J K}^{-1} \text{mol}^{-1}

The negative sign is chemically sensible: one mole of gaseous $\text{O}_2$ is consumed (gases have high entropy), and the products are both liquids/solids with lower entropy than the reactant cumene.

Key Takeaways

Always use products minus reactants.
The unit is $\text{J K}^{-1} \text{mol}^{-1}$ (not $\text{kJ}$ ).
A negative $\Delta S$ is expected when a gas is consumed.

Common Mistakes

Reversing the subtraction (reactants minus products), giving $+137$ .
Forgetting the negative sign.
Using kJ instead of J (the values in the table are in $\text{J K}^{-1} \text{mol}^{-1}$ ).

Things to Be Careful About

The answer must include units: $\text{J K}^{-1} \text{mol}^{-1}$ .
Ensure you include all species from the balanced equation (each has coefficient 1 here).

Techniques used

apply the formula delta S = sum S(products) - sum S(reactants)substitute standard entropy values from a table

(ii)

Show that reaction 1 is feasible at $25\text{ }^\circ\text{C}$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Working

\Delta G^\ominus = \Delta H^\ominus - T\Delta S^\ominus

T = 25 + 273 = 298 \text{ K}

\Delta S^\ominus = -137 \text{ J K}^{-1} \text{mol}^{-1} = -0.137 \text{ kJ K}^{-1} \text{mol}^{-1}

\Delta G^\ominus = -371 - 298 \times (-0.137)

\Delta G^\ominus = -371 + 40.8 = -330 \text{ kJ mol}^{-1}

Answer

Since $\Delta G^\ominus < 0$ (negative), reaction 1 is feasible at $25^\circ\text{C}$ .

\Delta G^\ominus = -330 \text{ kJ mol}^{-1}

Final answer

ΔG = -330 kJ mol^-1, which is negative, so the reaction is feasible

Detailed explanation

Background Concept

The Gibbs free energy change determines whether a reaction is thermodynamically feasible (spontaneous) at a given temperature:

\Delta G = \Delta H - T\Delta S

If $\Delta G < 0$ , the reaction is feasible. If $\Delta G > 0$ , it is not feasible under those conditions. Temperature must be in Kelvin, and $\Delta H$ and $\Delta S$ must be in consistent units (both kJ or both J).

Understanding the Question

You must show that reaction 1 is feasible at $25^\circ\text{C}$ , given $\Delta H^\ominus = -371 \text{ kJ mol}^{-1}$ and $\Delta S^\ominus = -137 \text{ J K}^{-1} \text{mol}^{-1}$ (from part c(i)).

Approach

Convert temperature to Kelvin: $T = 25 + 273 = 298 \text{ K}$ .
Convert $\Delta S$ from $\text{J}$ to $\text{kJ}$ : $-137 \text{ J} = -0.137 \text{ kJ}$ .
Substitute into $\Delta G = \Delta H - T\Delta S$ .
State that $\Delta G < 0$ , so the reaction is feasible.

Step-by-Step Reasoning

Step 1 — Temperature conversion:

T = 25 + 273 = 298 \text{ K}

Step 2 — Unit conversion:

\Delta S^\ominus = -137 \text{ J K}^{-1} \text{mol}^{-1} = -0.137 \text{ kJ K}^{-1} \text{mol}^{-1}

Step 3 — Substitution:

\Delta G^\ominus = -371 - (298 \times -0.137)

\Delta G^\ominus = -371 - (-40.8)

\Delta G^\ominus = -371 + 40.8 = -330.2 \approx -330 \text{ kJ mol}^{-1}

Step 4 — Conclusion:
Since $\Delta G^\ominus$ is negative ( $-330 \text{ kJ mol}^{-1}$ ), the reaction is thermodynamically feasible at $25^\circ\text{C}$ .

Note: The large negative $\Delta H$ dominates over the unfavourable (negative) $\Delta S$ term, making $\Delta G$ negative even at this temperature.

Key Takeaways

Always check units: $\Delta H$ in kJ, $\Delta S$ in J must be converted to match.
Temperature must be in Kelvin.
A reaction can be feasible even with a negative $\Delta S$ if $\Delta H$ is sufficiently negative.

Common Mistakes

Using $T = 25$ instead of $T = 298$ .
Forgetting to convert $\Delta S$ from J to kJ, giving $\Delta G = -371 - 298 \times (-137) = -371 + 40826 = +40455$ (absurd positive value).
Stating only the numerical answer without the conclusion about feasibility.
Mark scheme note: ecf from part c(i) is allowed, so if your $\Delta S$ was wrong but used correctly here, you can still earn M2.

Things to Be Careful About

The sign when substituting: $-T\Delta S$ where $\Delta S$ is negative gives $-T \times (-0.137) = +40.8$ , so $\Delta G = -371 + 40.8$ .
You must state the conclusion explicitly: "feasible because $\Delta G < 0$ ."
The question says "show that" — you must present the calculation clearly, not just state the answer.

Techniques used

apply the Gibbs free energy equation delta G = delta H - T delta Sconvert temperature to Kelvinconvert entropy from J to kJ for consistencyinterpret the sign of delta G for feasibility

(d)

Fig. 5.4 shows two reactions of phenol.

Fig. 5.4

9M

(i)

State the conditions for the bromination of phenol in reaction 2.

Explain why these conditions are different from those for the bromination of benzene.

4M Similar

DifficultyMedium

Worked solution

Answer

Conditions for bromination of phenol:

Use aqueous bromine (bromine water)
No catalyst ( $\text{AlBr}_3$ ) is needed
Room temperature

Why these conditions differ from benzene:

The lone pair on the oxygen atom of the $-\text{OH}$ group overlaps with (is delocalised into) the $\pi$ system of the ring
This increases the electron density in the ring, making it more electron-rich
The ring can therefore polarise $\text{Br}_2$ (the electrophile) more effectively, so no catalyst is required

Final answer

Aqueous bromine (bromine water), no AlBr3 catalyst needed. The lone pair on oxygen delocalises into the pi system, increasing electron density in the ring, making it more reactive towards electrophiles and able to polarise Br2 without a catalyst.

Detailed explanation

Background Concept

Phenol is significantly more reactive towards electrophilic substitution than benzene. This is because the $-\text{OH}$ group is an activating group: the lone pair on the oxygen atom can overlap with the $\pi$ system of the aromatic ring (p- $\pi$ conjugation), donating electron density into the ring. This makes the ring more nucleophilic and able to attract and polarise electrophiles more effectively, reducing the need for a strong catalyst.

In contrast, benzene has no activating substituent, so it requires a Lewis acid catalyst ( $\text{FeBr}_3$ or $\text{AlBr}_3$ ) to generate a sufficiently strong electrophile ( $\text{Br}^+$ ) from $\text{Br}_2$ .

Understanding the Question

The question has two parts: (1) state the conditions for bromination of phenol, and (2) explain why these are milder than those for benzene. You need four marking points: conditions, orbital overlap, increased electron density, and enhanced polarisation.

Approach

First, state the conditions clearly (bromine water, no catalyst). Then build the explanation logically: lone pair on O → overlap with ring $\pi$ system → increased electron density → ring polarises $\text{Br}_2$ more easily → no catalyst needed.

Step-by-Step Reasoning

M1 — Conditions: Phenol reacts with aqueous bromine (bromine water) at room temperature without any catalyst. This contrasts with benzene, which requires anhydrous $\text{Br}_2$ with an $\text{FeBr}_3$ or $\text{AlBr}_3$ catalyst.

M2 — Orbital overlap: The oxygen atom in the $-\text{OH}$ group has lone pairs of electrons in p-orbitals. One of these p-orbitals overlaps with the $\pi$ system of the aromatic ring, allowing delocalisation of electron density from oxygen into the ring.

M3 — Increased electron density: This delocalisation increases the electron density in the ring (particularly at the ortho and para positions), making the ring more nucleophilic and more attractive to electrophiles.

M4 — Polarisation of electrophile: Because the ring is more electron-rich, it can polarise the $\text{Br}_2$ molecule more effectively. The induced dipole in $\text{Br}_2$ is sufficient for the ring to attack, so no catalyst is needed to generate a full $\text{Br}^+$ electrophile.

Key Takeaways

The $-\text{OH}$ group activates the ring towards electrophilic substitution.
The mechanism of activation is p- $\pi$ overlap (lone pair delocalisation).
Phenol is so reactive that it undergoes tribromination (all ortho and para positions substituted) with bromine water.
The logical chain: lone pair → overlap → electron density → polarisation → no catalyst needed.

Common Mistakes

Saying "phenol has a lone pair" without specifying it is on the oxygen and that it overlaps with the ring $\pi$ system.
Saying "phenol is more reactive" without explaining WHY (the orbital overlap mechanism).
Confusing this with the inductive effect — the key effect here is resonance/delocalisation, not inductive.
Stating conditions as "Br2 with FeBr3" (that is for benzene, not phenol).

Things to Be Careful About

You must mention BOTH the orbital overlap AND the consequence (increased electron density) — they are separate marks.
"No catalyst needed" is part of the conditions mark, not a separate point.
The explanation must link back to why bromination of phenol is easier than bromination of benzene.

Techniques used

state conditions for bromination of phenolexplain the activating effect of the OH group on the aromatic ringrelate increased electron density to polarisation of electrophile

(ii)

Draw the structure of organic compound Q in Fig. 5.4.

1M Similar

DifficultyEasy

Worked solution

Answer

2,4,6-tribromophenol

Final answer

2,4,6-tribromophenol

Detailed explanation

Background Concept

Phenol is so activated towards electrophilic substitution that it undergoes multiple substitutions with bromine water. The $-\text{OH}$ group is strongly activating and ortho/para-directing, so all three available ortho and para positions (2, 4, and 6) are substituted by bromine in a single reaction with excess bromine water. The product is 2,4,6-tribromophenol, a white precipitate.

Understanding the Question

You are asked to draw the structure of compound Q, the product of reaction 2 (phenol + $\text{Br}_2$ ).

Approach

Since phenol is highly activated, bromination occurs at all ortho and para positions simultaneously. Draw a benzene ring with $-\text{OH}$ at position 1 and $-\text{Br}$ at positions 2, 4, and 6.

Step-by-Step Reasoning

The $-\text{OH}$ group directs incoming electrophiles to the ortho (2, 6) and para (4) positions.
With bromine water (excess $\text{Br}_2$ ), all three positions react.
The product is 2,4,6-tribromophenol: a benzene ring with $-\text{OH}$ at C1, and $-\text{Br}$ at C2, C4, and C6.

Key Takeaways

Phenol + bromine water gives 2,4,6-tribromophenol (white precipitate) — this is a test for phenol.
The $-\text{OH}$ group is ortho/para-directing and strongly activating.
Unlike benzene, phenol does not require a catalyst and gives multiple substitution.

Common Mistakes

Drawing only mono-bromination (4-bromophenol) — phenol with bromine water gives the tri-substituted product.
Placing bromines at meta positions.
Forgetting to include the $-\text{OH}$ group.

Things to Be Careful About

Draw all atoms and bonds clearly if the question asks for a displayed formula; a structural formula showing the ring with Br and OH substituents is acceptable.
Ensure the bromines are at the correct positions (2, 4, 6 relative to OH at position 1).

Techniques used

predict the product of bromination of phenol

(iii)

Identify organic reagent R.

1M Similar

DifficultyEasy

Worked solution

Answer

Benzenediazonium chloride ( $\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^-$ )

Final answer

Benzenediazonium chloride

Detailed explanation

Background Concept

Azo coupling is a reaction between a diazonium salt and an activated aromatic compound (such as phenol or an aromatic amine) to form an azo compound containing the $-\text{N}=\text{N}-$ linkage. The diazonium ion acts as a weak electrophile and attacks the electron-rich ring of phenol at the para position (or ortho if para is blocked). The reaction is carried out in alkaline solution (NaOH) to generate the phenoxide ion, which is even more reactive than phenol itself.

Understanding the Question

Fig. 5.4 shows phenol reacting with reagent R and NaOH(aq) to form compound S, which is 4-hydroxyphenylazobenzene. You must identify R.

Approach

The product S contains an azo group ( $-\text{N}=\text{N}-$ ) linking a phenol ring to a benzene ring. This is a classic azo coupling product. The reagent R must be a diazonium salt — specifically benzenediazonium chloride, which provides the $\text{C}_6\text{H}_5\text{N}_2^+$ electrophile.

Step-by-Step Reasoning

Compound S is 4-hydroxyphenylazobenzene: $\text{C}_6\text{H}_5-\text{N}=\text{N}-\text{C}_6\text{H}_4-\text{OH}$ .
The $\text{C}_6\text{H}_5\text{N}_2^+$ portion comes from the diazonium salt.
The $\text{C}_6\text{H}_4\text{OH}$ portion comes from phenol.
Therefore R = benzenediazonium chloride ( $\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^-$ ).
NaOH(aq) converts phenol to the phenoxide ion, making it more reactive towards the weak electrophile.

Key Takeaways

Azo coupling requires a diazonium salt and an activated aromatic compound.
The reaction is done in alkaline conditions to generate the more reactive phenoxide ion.
Benzenediazonium chloride is the standard reagent for coupling with phenol.

Common Mistakes

Writing "aniline" (the precursor to the diazonium salt, not the coupling reagent itself).
Writing "nitrosobenzene" or other nitrogen-containing compounds.
Not specifying that it is the chloride salt (though the cation structure alone may be accepted).

Things to Be Careful About

The full name "benzenediazonium chloride" or the structure of the benzenediazonium cation are both acceptable.
The diazonium ion is $\text{C}_6\text{H}_5-\text{N}\equiv\text{N}^+$ (positive charge on the nitrogen attached to the ring).

Techniques used

identify the reagent for azo coupling with phenol

(iv)

Name the functional group that is formed in reaction 3.

1M Similar

(v)

The reaction of phenol with $\text{HNO}_3$ produces a mixture of isomers with molecular formula $\text{C}_6\text{H}_5\text{NO}_3$ .

Identify the two isomers that are produced in the largest quantities.

Explain your answer.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

The two isomers produced in the largest quantities are 2-nitrophenol (ortho) and 4-nitrophenol (para).

The $-\text{OH}$ group is an electron-donating group (lone pair on oxygen delocalises into the ring $\pi$ system), so it directs incoming electrophiles to the ortho and para positions.

Final answer

2-nitrophenol and 4-nitrophenol; the -OH group is electron-donating and directs to ortho and para positions

Detailed explanation

Background Concept

When phenol undergoes electrophilic substitution, the $-\text{OH}$ group acts as an activating, ortho/para-directing substituent. This is because the lone pair on oxygen is delocalised into the ring $\pi$ system, increasing electron density specifically at the ortho and para positions. When phenol reacts with dilute nitric acid, a mixture of 2-nitrophenol (ortho) and 4-nitrophenol (para) is formed as the major products.

Understanding the Question

You are told that phenol + $\text{HNO}_3$ gives a mixture of isomers with molecular formula $\text{C}_6\text{H}_5\text{NO}_3$ (mononitrophenols). You must identify the two isomers formed in the largest quantities and explain why.

Approach

The $-\text{OH}$ group directs electrophilic substitution to the ortho (position 2) and para (position 4) positions. Therefore the two major products are 2-nitrophenol and 4-nitrophenol.

Step-by-Step Reasoning

The molecular formula $\text{C}_6\text{H}_5\text{NO}_3$ corresponds to a mononitrophenol (one $-\text{NO}_2$ group replacing one $-\text{H}$ on the ring).
Possible isomers: 2-nitrophenol (ortho), 3-nitrophenol (meta), 4-nitrophenol (para).
The $-\text{OH}$ group donates electron density into the ring via its lone pair (resonance effect), increasing electron density at ortho and para positions.
Therefore the electrophile ( $\text{NO}_2^+$ ) preferentially attacks at positions 2 and 4.
The two major products are 2-nitrophenol and 4-nitrophenol.

Key Takeaways

$-\text{OH}$ is an ortho/para director because it is electron-donating by resonance.
Meta products are minor because the meta position does not benefit from the increased electron density.
This directing effect applies to all electrophilic substitutions of phenol (nitration, bromination, etc.).

Common Mistakes

Including 3-nitrophenol as a major product (it is the minor product).
Saying "the $-\text{OH}$ group is electron-withdrawing" — it is electron-donating by resonance (though the oxygen is electronegative, the lone pair donation dominates).
Confusing the directing effect with the inductive effect.

Things to Be Careful About

You must give BOTH the structures/names AND the explanation (the $-\text{OH}$ group is electron-donating and ortho/para-directing).
The question asks for the two isomers "produced in the largest quantities" — these are the ortho and para products.

Techniques used

apply ortho/para directing effect of the hydroxyl groupidentify the major products of nitration of phenol

Q6Medium-HardPolymerisationIntroduction to A Level Organic ChemistryTransition ElementsAnalytical TechniquesCarboxylic Acids and Derivatives

(a)

Maleic anhydride is an unsaturated cyclic compound used in the formation of several polymers.

Maleic anhydride can be used to form maleic acid and tartaric acid.

Fig. 6.1

3M

(i)

Maleic acid reacts with ethane-1,2-diol to form a condensation polymer.

Draw a section of this polymer, showing only one repeat unit.

The new functional group formed should be shown fully displayed.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

The repeat unit is:

\left(\text{-CO-CH=CH-CO-}\overset{\displaystyle\|\|}{\text{O}}\text{-CH}_2\text{-CH}_2\text{-O-}\right)_n

with the ester group $\text{-C(=O)-O-}$ shown fully displayed (all bonds to oxygen drawn explicitly).

Final answer

See diagram — repeat unit of polyester from maleic acid and ethane-1,2-diol with fully displayed ester linkage

Detailed explanation

Background Concept

Condensation polymerisation occurs between monomers bearing two complementary functional groups (e.g. a dicarboxylic acid and a diol), with the elimination of a small molecule (usually water) at each linkage. When a diacid reacts with a diol, the product is a polyester, with ester linkages ( $\text{-COO-}$ ) joining the monomer units.

Understanding the Question

Maleic acid is $\text{HOOC-CH=CH-COOH}$ (a dicarboxylic acid with a $\text{C=C}$ double bond in the chain). Ethane-1,2-diol is $\text{HO-CH}_2\text{-CH}_2\text{-OH}$ . The question asks for one repeat unit of the condensation polymer, with the new functional group (the ester) fully displayed.

Approach

Identify that the reaction is between $\text{-COOH}$ groups and $\text{-OH}$ groups to form ester linkages. Each repeat unit contains one maleic acid residue and one diol residue joined by two ester bonds (one at each end). Draw the repeat unit in brackets with bonds extending through the brackets.

Step-by-Step Reasoning

Maleic acid has two $\text{-COOH}$ groups; ethane-1,2-diol has two $\text{-OH}$ groups.
Each $\text{-COOH}$ reacts with an $\text{-OH}$ to form an ester linkage $\text{-C(=O)-O-}$ , losing $\text{H}_2\text{O}$ .
The repeat unit is: $\text{-[CO-CH=CH-CO-O-CH}_2\text{-CH}_2\text{-O]-}$
The ester group must be fully displayed: show the $\text{C=O}$ double bond and the $\text{C-O}$ single bond explicitly.
The $\text{C=C}$ double bond from maleic acid remains in the polymer backbone.

Key Takeaways

A diacid + diol gives a polyester.
The repeat unit must show both ester linkages (one at each end of the maleic acid residue).
"Fully displayed" means showing all bonds to the ester oxygen atoms.

Common Mistakes

Drawing only one ester linkage per repeat unit.
Not showing the $\text{C=C}$ double bond in the maleic acid portion.
Forgetting brackets around the repeat unit with bonds extending through them.
Not fully displaying the ester group (e.g. writing $\text{-COO-}$ without showing the $\text{C=O}$ bond separately).

Things to Be Careful About

The question specifies ONE repeat unit only.
The ester group must be fully displayed (all bonds shown), while the rest can be in skeletal form.
State symbols are not required for polymer structures.

Techniques used

identify the functional groups on each monomerform ester linkages by condensationdraw the repeat unit with brackets

(ii)

Identify a suitable reagent and the conditions for reaction 2.

1M Similar

DifficultyEasy

Worked solution

Answer

Cold, dilute, acidified $\text{KMnO}_4$ (or cold, dilute $\text{MnO}_4^-$ ).

Final answer

Cold, dilute, acidified KMnO4

Detailed explanation

Background Concept

The conversion of maleic acid (which contains a $\text{C=C}$ double bond) to tartaric acid (which has two $\text{-OH}$ groups on adjacent carbons) is a syn-dihydroxylation of an alkene. The classic reagent for this in A-level chemistry is cold, dilute, acidified potassium manganate(VII), which adds two hydroxyl groups across the double bond.

Understanding the Question

Reaction 2 converts the $\text{C=C}$ in maleic acid into two $\text{C-OH}$ groups in tartaric acid. The question asks for a suitable reagent and conditions.

Approach

Recognise the transformation: alkene → vicinal diol. This is oxidation by cold, dilute $\text{MnO}_4^-$ . The key conditions are cold and dilute (if hot and concentrated, cleavage of the double bond would occur).

Step-by-Step Reasoning

Maleic acid: $\text{HOOC-CH=CH-COOH}$
Tartaric acid: $\text{HOOC-CH(OH)-CH(OH)-COOH}$
The $\text{C=C}$ has been converted to $\text{C-C}$ with an $\text{-OH}$ on each carbon.
This is syn-dihydroxylation, achieved by cold, dilute, acidified $\text{KMnO}_4$ .
If the $\text{KMnO}_4$ were hot and concentrated, oxidative cleavage would break the $\text{C=C}$ entirely, giving different products.

Key Takeaways

Cold, dilute $\text{KMnO}_4$ adds two $\text{-OH}$ groups across a $\text{C=C}$ (syn-dihydroxylation).
Hot, concentrated $\text{KMnO}_4$ cleaves the $\text{C=C}$ bond.
The purple colour of $\text{MnO}_4^-$ is discharged (decolourised) as $\text{Mn}^{2+}$ is formed.

Common Mistakes

Writing "hot, concentrated $\text{KMnO}_4$ " (this would cleave the double bond).
Writing only " $\text{KMnO}_4$ " without specifying cold and dilute.
Confusing this with bromine water (which adds $\text{Br}$ , not $\text{OH}$ ).

Things to Be Careful About

The conditions must include "cold" and "dilute" to earn the mark.
"Acidified" is often included but the mark scheme accepts the reagent name alone.

Techniques used

identify the functional group changerecall the reagent for syn-dihydroxylation of alkenes

(b)

Compound U can be formed from tartaric acid in two steps, as shown in Fig. 6.2.

Fig. 6.2

5M

(i)

Suggest the type of reaction that occurs in step 1.

1M Similar

DifficultyEasy

Worked solution

Answer

Condensation (or dehydration or elimination).

Final answer

Condensation

Detailed explanation

Background Concept

When a ketone (propanone) reacts with two alcohol groups on adjacent carbons (as in tartaric acid), a cyclic acetal (ketal) is formed with the elimination of a molecule of water. This is a condensation reaction — two molecules join with the loss of a small molecule ( $\text{H}_2\text{O}$ ). It can also be described as dehydration (loss of water) or elimination.

Understanding the Question

Tartaric acid has two $\text{-OH}$ groups on adjacent carbons. Propanone ( $\text{CH}_3\text{COCH}_3$ ) reacts with these to form compound T, which has a five-membered ring containing two oxygen atoms and a carbon bearing two methyl groups (an isopropylidene ketal/acetal). Water is lost in the process.

Approach

Compare the structures: tartaric acid + propanone → T + $\text{H}_2\text{O}$ . Two molecules combine with loss of a small molecule = condensation.

Step-by-Step Reasoning

Tartaric acid has two adjacent $\text{-OH}$ groups.
Propanone has a $\text{C=O}$ group.
The two $\text{-OH}$ groups react with the $\text{C=O}$ , forming two $\text{C-O-C}$ bonds and a five-membered ring.
One molecule of $\text{H}_2\text{O}$ is eliminated.
This is a condensation reaction (also acceptable: dehydration or elimination).

Key Takeaways

Acetal/ketal formation from a diol + ketone is a condensation reaction.
The $\text{H}^+$ acts as an acid catalyst.
"Condensation" in this context does not necessarily mean polymerisation — it simply means two molecules join with loss of a small molecule.

Common Mistakes

Writing "addition" (the $\text{C=O}$ is not simply added across; water is lost).
Writing "substitution" (no atom is replaced by another).

Things to Be Careful About

The mark scheme accepts condensation, dehydration, or elimination — any one of these.

Techniques used

compare structures of reactant and productidentify the type of reaction from the change

(ii)

U can act as a bidentate ligand.

Explain what is meant by a bidentate ligand.

2M Similar

DifficultyEasy

Worked solution

Answer

A bidentate ligand is a species with two lone pairs of electrons that forms two dative (coordinate) bonds to a central transition metal atom/ion.

Final answer

A species with two lone pairs that forms two dative (coordinate) bonds to a central transition metal atom/ion

Detailed explanation

Background Concept

In coordination chemistry, ligands are species that donate lone pairs of electrons to a central metal ion to form coordinate (dative) bonds. The number of coordinate bonds a ligand can form to the metal defines its denticity: monodentate (one bond), bidentate (two bonds), tridentate (three bonds), etc. Bidentate ligands such as ethane-1,2-diamine ( $\text{H}_2\text{NCH}_2\text{CH}_2\text{NH}_2$ ), ethanedioate ( $\text{C}_2\text{O}_4^{2-}$ ), and water in certain contexts form two bonds to the same metal centre.

Understanding the Question

The question asks for the meaning of "bidentate ligand" in the context of compound U acting as a ligand to $\text{Cu}^{2+}$ . Two marks are available, so two distinct points must be made.

Approach

Define both aspects: (1) the ligand has two lone pairs available for donation, and (2) it forms two dative/coordinate bonds to the central metal ion.

Step-by-Step Reasoning

A ligand is a species that donates electron pairs to a metal ion.
"Bi-dentate" means "two-toothed" — the ligand attaches at two points.
It has two lone pairs of electrons available for donation.
It forms two dative (coordinate) bonds to a central transition metal atom/ion.
Both points are needed for full marks.

Key Takeaways

A bidentate ligand forms exactly two coordinate bonds to one metal centre.
The word "dative" or "coordinate" is essential — it specifies the type of bond.
The metal must be identified as a transition element (or metal atom/ion).

Common Mistakes

Omitting "dative" or "coordinate" (writing just "bonds" loses the mark).
Saying "two bonds to different metal ions" (both bonds go to the SAME metal centre).
Not mentioning lone pairs.
Writing "it forms two bonds" without specifying they are dative/coordinate.

Things to Be Careful About

Both the lone pair point AND the dative bond point are needed (2 marks).
The metal must be specified as a transition element or central metal ion.

Techniques used

define key terms in coordination chemistryspecify the number of bonds and their type

(iii)

Complete the three-dimensional diagrams in Fig. 6.3 to show both stereoisomers of $[\text{Cu(U)}_3]^{4-}$ .

Use

ligand U symbol

to represent ligand U.

Fig. 6.3

2M Similar

DifficultyMedium

Worked solution

Answer

Both stereoisomers of $[\text{Cu(U)}_3]^{4-}$ are drawn as octahedral complexes with three bidentate ligands arranged in a propeller fashion around the central $\text{Cu}$ , using wedges (toward viewer) and dashes (away from viewer) to show 3D geometry. The two structures are non-superimposable mirror images (Δ and Λ optical isomers).

Final answer

Two octahedral diagrams showing Δ and Λ optical isomers of [Cu(U)3]4- with three bidentate ligands in propeller arrangement

Detailed explanation

Background Concept

Octahedral complexes with three bidentate ligands (such as $[\text{Co(en)}_3]^{3+}$ or $[\text{Cu(U)}_3]^{4-}$ ) exhibit optical isomerism. The three bidentate ligands can wrap around the metal centre in two non-superimposable ways, giving rise to a pair of enantiomers: the Δ (delta, right-handed propeller) and Λ (lambda, left-handed propeller) forms. These are mirror images that cannot be superimposed, analogous to left and right hands.

Understanding the Question

The question provides two empty boxes with a vertical line through "Cu" (representing the axial positions of the octahedron) and asks the candidate to complete them with the two stereoisomers. The ligand is represented by a curved line joining two O atoms (the bidentate symbol).

Approach

Recognise that three bidentate ligands give octahedral coordination (6 coordinate bonds to Cu).
The vertical line through Cu represents two axial bonds (top and bottom).
The remaining four coordination sites are filled by two ligands in the equatorial plane, shown using wedges (coming toward viewer) and dashes (going away).
One isomer has the ligands arranged in a right-handed (Δ) propeller; the other in a left-handed (Λ) propeller.

Step-by-Step Reasoning

Draw the first isomer: place one bidentate ligand vertically (using the given axial line), and two others in the remaining positions using wedge-dash notation to show 3D geometry. The curved lines connecting the O pairs show the chelate rings.
Draw the second isomer as the mirror image: reverse the wedge/dash assignments so the propeller direction is opposite.
Both must be enclosed in square brackets with the 4- charge shown outside.
The key feature is that the two diagrams are non-superimposable mirror images.

Key Takeaways

Three bidentate ligands around an octahedral centre always give a pair of optical isomers (Δ and Λ).
Wedge-dash notation is essential for showing 3D geometry.
The two isomers must be clearly mirror images (not rotations of the same structure).

Common Mistakes

Drawing the same isomer twice (just rotated) rather than a true mirror image.
Not using wedge/dash notation (drawing flat 2D structures).
Forgetting the square brackets and 4- charge.
Placing the ligand symbols incorrectly relative to the given axial line.

Things to Be Careful About

Use the given ligand symbol (curved line joining two O's) — do not draw the full structure.
The vertical line already drawn through Cu represents the axial bonds; build the equatorial ligands around it.
Both isomers must be clearly distinguishable as mirror images.

Techniques used

draw octahedral geometry with bidentate ligandsshow Δ and Λ optical isomers using wedge-dash notation

(c)

A student analyses an aromatic compound, X, $\text{C}_8\text{H}_8\text{O}_3$ , using NMR spectroscopy.

Fig. 6.4 shows the carbon-13 NMR spectrum of a sample of X dissolved in $\text{D}_2\text{O}$ .

Fig. 6.4

Separate samples of X were analysed using proton ( $^1\text{H}$ ) NMR spectroscopy.

Table 6.1 gives information obtained from this analysis.

Table 6.1

solvent	number of signals in proton ( $^1\text{H}$ ) NMR spectrum
$\text{CDCl}_3$	6
$\text{D}_2\text{O}$	4

6M

(i)

Identify the number of different carbon environments present in X.

1M Similar

(ii)

Explain why X is dissolved in $\text{D}_2\text{O}$ before obtaining its proton ( $^1\text{H}$ ) NMR spectrum.

1M Similar

DifficultyEasy

Worked solution

Answer

To remove peaks/signals from acidic (labile) protons such as $\text{-OH}$ (or $\text{-COOH}$ ) groups, because $\text{D}_2\text{O}$ exchanges with these protons and replaces them with deuterium, which does not appear in the $^1\text{H}$ NMR spectrum.

Final answer

To remove peaks from acidic/labile protons (OH/COOH) by exchange with deuterium

Detailed explanation

Background Concept

Protons attached to electronegative atoms (such as $\text{-OH}$ , $\text{-NH}$ , $\text{-COOH}$ ) are labile — they can exchange with deuterium from $\text{D}_2\text{O}$ . When the sample is dissolved in $\text{D}_2\text{O}$ , these exchangeable protons are replaced by $\text{D}$ (deuterium), which is not detected in $^1\text{H}$ NMR (deuterium resonates at a completely different frequency). This simplifies the spectrum by removing the often broad and variable signals from $\text{-OH}$ and $\text{-NH}$ protons.

Understanding the Question

Table 6.1 shows that X gives 6 signals in $\text{CDCl}_3$ but only 4 signals in $\text{D}_2\text{O}$ . The question asks why $\text{D}_2\text{O}$ is used as the solvent.

Approach

Recognise that the difference in signal count (6 vs 4) is due to two exchangeable protons being replaced by deuterium. The purpose of using $\text{D}_2\text{O}$ is to identify/remove these labile proton signals.

Step-by-Step Reasoning

$\text{D}_2\text{O}$ contains deuterium, which can exchange with labile protons ( $\text{-OH}$ , $\text{-NH}$ , $\text{-COOH}$ ).
After exchange, these positions contain $\text{D}$ instead of $\text{H}$ .
$\text{D}$ does not give a signal in $^1\text{H}$ NMR.
Therefore, the $\text{-OH}$ proton signals disappear, simplifying the spectrum.
The reduction from 6 to 4 signals confirms two exchangeable $\text{-OH}$ groups.

Key Takeaways

$\text{D}_2\text{O}$ exchange removes signals from labile protons ( $\text{-OH}$ , $\text{-NH}$ , $\text{-COOH}$ ).
The number of signals lost tells you how many exchangeable protons are present.
This is a useful diagnostic tool in structure determination.

Common Mistakes

Saying "to dissolve the compound" (solubility is not the primary reason for choosing $\text{D}_2\text{O}$ specifically).
Saying "to avoid interference from water peaks" (the reason is proton exchange, not avoiding a water peak per se).
Not mentioning that deuterium does not appear in $^1\text{H}$ NMR.

Things to Be Careful About

The key phrase is "remove peaks/signals from acidic/labile/OH protons" — simply saying "exchange" without stating the effect on the spectrum may not earn the mark.

Techniques used

explain the purpose of D2O exchange in 1H NMR

(iii)

Aromatic compound X gives a yellow precipitate when it reacts with alkaline $\text{I}_2(aq)$ .

Use the information in (c) to suggest a structure for X.

Explain your reasoning.

Answer box X

Table 6.2

4M Similar

DifficultyMedium-Hard

Worked solution

Answer

X is a dihydroxyphenyl methyl ketone (an acetyl dihydroxybenzene), e.g. 2,4-dihydroxyacetophenone.

Reasoning:

Yellow precipitate with alkaline $\text{I}_2\text{(aq)}$ → positive iodoform test → $\text{CH}_3\text{CO-}$ group present.
$^{13}\text{C}$ NMR peak at ~203 ppm (range 190–220) → ketone carbonyl ( $\text{C=O}$ ).
$^{13}\text{C}$ NMR peak at ~61 ppm (range 50–70) → carbon next to oxygen ( $\text{C-O}$ ), consistent with $\text{CH}_3$ adjacent to $\text{C=O}$ or aryl $\text{C-O}$ .
8 carbon environments with 8 carbon atoms → molecule is unsymmetrical.
6 signals in $\text{CDCl}_3$ vs 4 in $\text{D}_2\text{O}$ → 2 exchangeable protons → two $\text{-OH}$ groups on the ring.
Molecular formula $\text{C}_8\text{H}_8\text{O}_3$ : benzene ring ( $\text{C}_6$ ) + $\text{COCH}_3$ ( $\text{C}_2$ ) + 2 $\text{-OH}$ groups = $\text{C}_8\text{H}_8\text{O}_3$ ✓

Final answer

A dihydroxyphenyl methyl ketone (acetyl dihydroxybenzene), e.g. 2,4-dihydroxyacetophenone

Detailed explanation

Background Concept

Structure determination from spectroscopic data requires integrating multiple pieces of evidence: the molecular formula constrains the number of atoms; $^{13}\text{C}$ NMR tells us the number and types of carbon environments; $^1\text{H}$ NMR (especially the $\text{D}_2\text{O}$ exchange experiment) tells us about hydrogen environments and exchangeable protons; and chemical tests (like the iodoform test) confirm specific functional groups.

The iodoform test ( $\text{I}_2/\text{NaOH}$ ) gives a yellow precipitate of $\text{CHI}_3$ with methyl ketones ( $\text{CH}_3\text{CO-R}$ ) and with compounds containing $\text{CH}_3\text{CH(OH)-R}$ (which are oxidised to methyl ketones under the alkaline conditions).

Understanding the Question

Compound X has molecular formula $\text{C}_8\text{H}_8\text{O}_3$ and is aromatic. We are given:

$^{13}\text{C}$ NMR: 8 peaks at ~203, 158, 154, 137, 128, 120, 115, 61 ppm
$^1\text{H}$ NMR: 6 signals in $\text{CDCl}_3$ , 4 signals in $\text{D}_2\text{O}$
Positive iodoform test (yellow precipitate with alkaline $\text{I}_2$ )
Table 6.2 gives chemical shift ranges for different carbon environments

Approach

Use the iodoform test to identify a $\text{CH}_3\text{CO-}$ group.
Use the $^{13}\text{C}$ NMR peak at 203 ppm to confirm a ketone carbonyl.
Use the $\text{D}_2\text{O}$ exchange (6 → 4 signals) to identify two $\text{-OH}$ groups.
Use the 8 carbon environments (with 8 total carbons) to confirm the molecule is unsymmetrical.
Combine with the molecular formula and aromaticity to propose a structure.

Step-by-Step Reasoning

Iodoform test: Yellow precipitate with alkaline $\text{I}_2\text{(aq)}$ means $\text{CHI}_3$ is formed. This requires a $\text{CH}_3\text{CO-}$ group (methyl ketone). So X contains $\text{-COCH}_3$ .
$^{13}\text{C}$ NMR at ~203 ppm: This falls in the carbonyl range (190–220 ppm) from Table 6.2, confirming a ketone $\text{C=O}$ . This is consistent with the methyl ketone identified above.
$^{13}\text{C}$ NMR at ~61 ppm: This is in the "next to oxygen" range (50–70 ppm). In the context of a methyl ketone on a benzene ring, this could be the $\text{CH}_3$ carbon (which is next to the carbonyl, range 30–65 ppm) or an aromatic carbon bonded to oxygen. The mark scheme accepts "methyl/carbon next to C=O" for this peak.
$^{13}\text{C}$ NMR peaks at 110–160 ppm (five peaks): These are aromatic/alkene carbons. With 5 peaks in this region, we have 5 aromatic carbons visible as separate environments (the 6th aromatic carbon is the one bearing the $\text{COCH}_3$ group, which may overlap or be one of the 154/158 peaks).
$^1\text{H}$ NMR: 6 signals in $\text{CDCl}_3$ , 4 in $\text{D}_2\text{O}$ : The loss of 2 signals upon $\text{D}_2\text{O}$ exchange means 2 exchangeable protons → two $\text{-OH}$ groups.
Molecular formula check: $\text{C}_8\text{H}_8\text{O}_3$ . A benzene ring ( $\text{C}_6\text{H}_3$ , trisubstituted) + $\text{COCH}_3$ ( $\text{C}_2\text{H}_3\text{O}$ ) + 2 $\text{-OH}$ ( $\text{H}_2\text{O}_2$ ) = $\text{C}_8\text{H}_8\text{O}_3$ ✓
Symmetry: 8 carbon environments with 8 total carbons means no symmetry. The substituents ( $\text{-COCH}_3$ , $\text{-OH}$ , $\text{-OH}$ ) must be arranged so that no plane of symmetry exists. For example, 2,4-dihydroxyacetophenone has no symmetry (all 6 ring carbons are different, plus the carbonyl C and methyl C = 8 environments). However, 2,6-dihydroxyacetophenone has a mirror plane and would give only 6 environments, so it is excluded. Similarly, 3,5-dihydroxyacetophenone is excluded.
Viable structures: 2,3-; 2,4-; 2,5-; and 3,4-dihydroxyacetophenone all satisfy the criteria (8 environments, unsymmetrical). Any one of these is acceptable.

Key Takeaways

The iodoform test specifically identifies $\text{CH}_3\text{CO-}$ (methyl ketone) or $\text{CH}_3\text{CH(OH)-}$ groups.
$\text{D}_2\text{O}$ exchange identifies the number of labile protons ( $\text{-OH}$ , $\text{-NH}$ , $\text{-COOH}$ ).
The number of $^{13}\text{C}$ peaks vs total carbon atoms reveals molecular symmetry.
Structure determination requires combining ALL evidence — no single piece is sufficient.

Common Mistakes

Proposing a symmetrical structure (e.g. 3,5-dihydroxyacetophenone) which would give fewer than 8 carbon environments.
Proposing a structure with a $\text{-COOH}$ group (would give a peak at 160–185 ppm, not 190–220 ppm).
Forgetting to check the molecular formula against the proposed structure.
Not explaining the reasoning (the question requires explanation, not just a structure).
Proposing an aldehyde instead of a ketone (the iodoform test and the 203 ppm peak confirm a methyl ketone, not an aldehyde — though aldehydes also appear at 190–220 ppm, the iodoform test specifically requires $\text{CH}_3\text{CO-}$ ).

Things to Be Careful About

The molecule must be UNSYMMETRICAL (8 different carbon environments = 8 carbons total).
Two $\text{-OH}$ groups must be present (from the $\text{D}_2\text{O}$ exchange).
The $\text{CH}_3\text{CO-}$ must be present (from the iodoform test AND the 203 ppm peak).
The structure must account for all atoms in $\text{C}_8\text{H}_8\text{O}_3$ .
At least two linked pieces of reasoning are needed for full marks (the mark scheme requires the methyl ketone point plus any two linked statements from the list).

Techniques used

interpret iodoform test resultcorrelate 13C NMR chemical shifts with functional groupsuse 1H NMR signal loss in D2O to count OH groupsdeduce structure from molecular formula and spectroscopic datacheck for molecular symmetry

Chemistry 9701/42 — February/March 2025

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

Table 1.1

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

Table 1.2

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

Table 1.3

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

Table 1.4

Working

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About