Chemistry 9701 / 44 — October/November 2025

DifficultyEasy

Worked solution

Answer

A transition element is an element that forms one or more stable ions with incomplete (partially filled) d orbitals.

Final answer

An element that forms one or more stable ions with incomplete (partially filled) d orbitals.

Detailed explanation

Background Concept

The IUPAC definition of a transition element is strictly based on the electronic configuration of its ions, not just the neutral atom. An element is a transition metal if it has an incomplete d sub-shell in at least one of its stable oxidation states (ions). This is why scandium (Sc, $3d^1 4s^2$ ) and zinc (Zn, $3d^{10} 4s^2$ ) are not considered transition elements: Sc forms only $\text{Sc}^{3+}$ ( $3d^0$ , empty d sub-shell) and Zn forms only $\text{Zn}^{2+}$ ( $3d^{10}$ , full d sub-shell).

Understanding the Question

The question asks for the formal definition of a transition element. This is a foundational concept in transition metal chemistry, setting the stage for understanding properties like variable oxidation states, catalytic activity, and coloured complexes.

Approach

Recall the precise IUPAC definition. It is crucial to mention that the incomplete d sub-shell must be present in a stable ion (or compound/oxidation state), not necessarily the neutral atom.

Step-by-Step Reasoning

The definition requires the formation of at least one stable ion.
This ion must have an incomplete or partially filled d sub-shell (d orbitals).
Combining these: "forms (one or more) stable ions with incomplete (partially filled) d orbitals."

Key Takeaways

Transition elements are defined by the electron configuration of their ions, not just their neutral atoms.
"Incomplete d sub-shell" is the key phrase.

Common Mistakes

Defining a transition element as simply "an element with incomplete d orbitals" (missing the requirement for stable ions).
Confusing the definition with properties like "forms coloured compounds" or "has variable oxidation states", which are consequences, not the definition.

Things to Be Careful About

Ensure you use the word "ions" or "oxidation states". Saying "an element with a partially filled d sub-shell" is technically incorrect for elements like Zinc where the neutral atom is full but the ion is also full, or Scandium where the ion is empty.

Techniques used

define transition element

(b)

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

2M

(i)

Define the term degenerate.

(ii)

Complete the electronic configuration of $\text{Cu}^{2+}$ .

$1\text{s}^2$ ...............................................................................................................................

DifficultyMedium-Easy

Worked solution

Answer

$2s^2 2p^6 3s^2 3p^6 3d^9$

Final answer

2s^2 2p^6 3s^2 3p^6 3d^9

Detailed explanation

Background Concept

Copper (Cu, atomic number 29) has an anomalous electron configuration in its neutral state: $[\text{Ar}] 3d^{10} 4s^1$ (or $1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^1$ ). This is due to the extra stability of a fully filled d sub-shell.

When forming ions, transition metals lose electrons from the 4s orbital before the 3d orbital. This is a common point of confusion; students often remove 3d electrons first.

Understanding the Question

We need to complete the electronic configuration for the $\text{Cu}^{2+}$ ion, starting from $1s^2$ .

Approach

Write the full configuration for neutral Cu.
Remove 2 electrons to form $\text{Cu}^{2+}$ , removing them from 4s first, then 3d.

Step-by-Step Reasoning

Neutral Cu: $1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^1$ .
To form $\text{Cu}^{2+}$ , remove 2 electrons.
First electron removed: from $4s^1$ -> $4s^0$ .
Second electron removed: from $3d^{10}$ -> $3d^9$ .
Resulting configuration: $1s^2 2s^2 2p^6 3s^2 3p^6 3d^9$ .
The question provides $1s^2$ , so we fill in the rest: $2s^2 2p^6 3s^2 3p^6 3d^9$ .

Key Takeaways

Always remove 4s electrons before 3d electrons when forming transition metal ions.
Remember the anomalous configuration of Cu ( $3d^{10} 4s^1$ ) and Cr ( $3d^5 4s^1$ ).

Common Mistakes

Writing $3d^8 4s^1$ (removing one 3d electron and keeping 4s).
Writing $3d^9 4s^0$ (technically correct but usually the 4s is omitted if empty, and the question implies filling the rest of the core).
Forgetting the anomalous configuration of neutral Cu and using $3d^9 4s^2$ .

Things to Be Careful About

Ensure the configuration is balanced and complete up to the d orbitals. The mark scheme accepts $2s^2 2p^6 3s^2 3p^6 3d^9$ .

Techniques used

write electron configuration of ion

(c)

4M

(i)

State the colours of the aqueous solutions for the two copper(II) complex ions shown.

$[\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}(\text{aq})$ .............................................................................................
$[\text{CuCl}_4]^{2-}(\text{aq})$ .............................................................................................................

DifficultyEasy

Worked solution

Answer

$[\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}(\text{aq})$ : deep / dark / royal blue
$[\text{CuCl}_4]^{2-}(\text{aq})$ : yellow

Final answer

[Cu(NH3)4(H2O)2]2+(aq): deep blue; [CuCl4]2-(aq): yellow

Detailed explanation

Background Concept

The colour of transition metal complexes depends on the ligands attached to the central metal ion. Different ligands cause different amounts of splitting of the d orbitals (d-d splitting). For copper(II):

With water ligands (or ammonia replacing water), the complex is typically blue.
With chloride ligands (high concentration of Cl-), the complex is yellow/green.

Understanding the Question

State the colours of two specific aqueous copper(II) complex ions: the tetraamminediaquacopper(II) ion and the tetrachloridocuprate(II) ion.

Approach

Recall the standard colours associated with these specific ligand environments for Cu2+.

Step-by-Step Reasoning

$[\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}$ : Ammonia is a stronger field ligand than water, but for Cu2+, the resulting complex is a deep, dark, or royal blue.
$[\text{CuCl}_4]^{2-}$ : Chloride is a weaker field ligand. The tetrachlorocuprate(II) ion is yellow (often observed as green in mixtures with the blue aqua complex, but the pure complex ion is yellow).

Key Takeaways

Ligand identity affects colour. $[\text{Cu}(\text{H}_2\text{O})_6]^{2+}$ is pale blue, $[\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}$ is deep blue, $[\text{CuCl}_4]^{2-}$ is yellow.

Common Mistakes

Saying "blue" for both (ignoring the ligand effect).
Saying "green" for $[\text{CuCl}_4]^{2-}$ (green is the colour of a mixture of blue and yellow, not the pure complex ion).

Things to Be Careful About

Use precise colour terms: "deep blue" or "royal blue" for the ammonia complex, "yellow" for the chloride complex.

Techniques used

recall colours of copper complexes

(ii)

Explain why aqueous complex ions of transition elements are usually coloured.

DifficultyMedium-Easy

Worked solution

Answer

The d orbitals split into two different energy levels (d-d splitting occurs).
Electrons are promoted (excited) from the lower energy d orbitals to the higher energy d orbitals.
Light (photons) of a specific wavelength/frequency is absorbed, and the complementary colour is seen.

Final answer

d orbitals split; electrons excited; light absorbed, complementary colour seen

Detailed explanation

Background Concept

In an isolated ion, d orbitals are degenerate. When ligands approach, the electrostatic repulsion between the ligand electrons and the d orbital electrons causes the d orbitals to split into two energy levels (e.g., $t_{2g}$ and $e_g$ in an octahedral field). The energy difference ( $\Delta E$ ) between these levels often corresponds to the energy of visible light photons.

Understanding the Question

Explain why aqueous complex ions of transition elements are usually coloured. This is a 3-mark explanation requiring a logical chain of events.

Approach

Mention d-orbital splitting.
Mention electron promotion (excitation).
Mention light absorption and the resulting colour (complementary colour).

Step-by-Step Reasoning

M1: In a complex ion, the ligands cause the d orbitals to split into two different energy levels (d-d splitting). This creates an energy gap.
M2: Electrons in the lower energy d orbitals can absorb energy to be promoted (excited) to the higher energy d orbitals.
M3: The energy absorbed corresponds to a specific wavelength/frequency of visible light (a photon is absorbed). The colour we see is the complementary colour to the absorbed light (i.e., the light that is transmitted or reflected).

Key Takeaways

Colour arises from d-d transitions.
Splitting -> excitation -> absorption of visible light -> transmission of complementary colour.

Common Mistakes

Saying "electrons jump between energy levels" without specifying d orbitals.
Saying "light is emitted" (it's absorbed, and the complementary colour is seen/transmitted).
Forgetting to mention the complementary colour.

Things to Be Careful About

Use precise terminology: "d orbitals split", "electrons promoted/excited", "light/photon absorbed", "complementary colour seen".

Techniques used

explain d-d transition colour

(d)

5M

(i)

When an excess of $\text{NH}_3(\text{aq})$ is added to a solution of $[\text{CuCl}_4]^{2-}(\text{aq})$ , $[\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}(\text{aq})$ is formed.

State the type of reaction.
Complete the equation for this reaction. State symbols are not required.

type of reaction ..................................................................................................................

equation

$[\text{CuCl}_4]^{2-} +$ ..................................... $\rightarrow [\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+} +$ .....................................

DifficultyMedium-Easy

Worked solution

Answer

type of reaction: ligand exchange (or ligand substitution / replacement / displacement)

equation:

[\text{CuCl}_4]^{2-} + 4\text{NH}_3 + 2\text{H}_2\text{O} \rightarrow [\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+} + 4\text{Cl}^-

Final answer

ligand exchange; [CuCl4]2- + 4NH3 + 2H2O -> [Cu(NH3)4(H2O)2]2+ + 4Cl-

Detailed explanation

Background Concept

Ligand exchange (or ligand substitution) is a reaction where one or more ligands in a complex ion are replaced by different ligands. This often occurs when a ligand that forms a more stable complex (or is present in large excess) is added to a solution of a different complex.

Understanding the Question

When excess ammonia is added to $[\text{CuCl}_4]^{2-}$ , it forms $[\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}$ . State the reaction type and complete the equation.

Approach

Identify that ligands are being swapped: Cl- and H2O are replaced by NH3.
Balance the equation by ensuring the number of each atom/ligand is conserved on both sides.

Step-by-Step Reasoning

Type of reaction: Ligands are being replaced, so it is a ligand exchange, substitution, replacement, or displacement reaction.
Equation balancing:
- Reactants: $[\text{CuCl}_4]^{2-}$ + ammonia + water (since the product has water ligands, water must be a reactant from the aqueous solution).
- Products: $[\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}$ + chloride ions (displaced from the original complex).
- To get 4 NH3 in the product, we need 4 NH3 on the reactant side.
- To get 2 H2O in the product, we need 2 H2O on the reactant side.
- The 4 Cl- ligands are displaced, so 4 Cl- are on the product side.
- Check charges: Left = -2. Right = +2 + 4(-1) = -2. Balanced.

Key Takeaways

Ligand exchange equations must balance both atoms and charges.
Water is often a ligand in aqueous complexes and may appear as a reactant or product.

Common Mistakes

Forgetting to include H2O as a reactant (the product has 2 water ligands, so water must come from somewhere).
Incorrectly balancing the chloride ions (must be 4 Cl-).
Writing "precipitation" or "redox" (no change in oxidation state, Cu is +2 throughout).

Things to Be Careful About

State symbols are not required, so don't add them.
Ensure the equation is fully balanced with correct stoichiometry.

Techniques used

identify ligand exchange reactionbalance ligand substitution equation

(ii)

The $[\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}$ complex ion shows stereoisomerism.

Complete the three-dimensional diagrams in Fig. 1.1 to show the two different stereoisomers of $[\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}$ .

Fig. 1.1

DifficultyMedium

Worked solution

Answer

Isomer 1 (trans): H₂O ligands at top and bottom (axial positions), NH₃ ligands in the equatorial plane.
Isomer 2 (cis): H₂O ligands adjacent to each other (one axial, one equatorial), NH₃ ligands in the remaining positions.

Final answer

See diagram: trans isomer (H2O axial) and cis isomer (H2O adjacent)

Detailed explanation

Background Concept

Octahedral complexes of the type $\text{MA}_4\text{B}_2$ (where M is the metal, A and B are different ligands) can exhibit geometric stereoisomerism (cis-trans isomerism).

Trans isomer: The two B ligands are opposite each other (180° apart), typically at the axial positions (top and bottom).
Cis isomer: The two B ligands are adjacent to each other (90° apart), typically one axial and one equatorial.

Understanding the Question

Complete the 3D diagrams for the two stereoisomers of $[\text{Cu}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}$ . The central atom is Cu, with 4 NH₃ and 2 H₂O ligands.

Approach

Draw two octahedral structures. Use wedges and dashes to show 3D geometry. Place the two H₂O ligands in the trans positions for one isomer, and in cis positions for the other.

Step-by-Step Reasoning

Isomer 1 (trans): Place H₂O at the top (axial) and bottom (axial) positions. Place the four NH₃ ligands in the equatorial plane (using wedges and dashes to show they are in the plane but projecting towards/away from the viewer).
Isomer 2 (cis): Place one H₂O at the top (axial) and one H₂O at an equatorial position (e.g., front-left). Place the four NH₃ ligands in the remaining positions (bottom axial, and three equatorial).
The mark scheme shows:
- Left box (isomer 1): H₂O top/bottom (axial), NH₃ equatorial (wedges/dashes).
- Right box (isomer 2): H₂O top (axial) and front-left (equatorial), NH₃ bottom (axial) and remaining equatorial.

Key Takeaways

Cis-trans isomerism in octahedral complexes requires two identical ligands.
Trans = opposite (180°), Cis = adjacent (90°).
Use wedges (coming out) and dashes (going in) to represent 3D structure clearly.

Common Mistakes

Drawing a square planar complex (this is octahedral, coordination number 6).
Placing all 4 NH₃ in axial positions (only 2 axial positions exist).
Forgetting to use wedges/dashes to show 3D geometry.

Things to Be Careful About

Ensure the coordination number is 6 (octahedral).
The diagrams must clearly show the relative positions of the H₂O ligands.

Techniques used

draw stereoisomers of octahedral complex

(iii)

Deduce which stereoisomer in (d)(ii) is polar.

Explain your answer.

polar isomer ......................................................................................................................

explanation ........................................................................................................................

(e)

The dianion $\mathbf{P}$ can act as a tridentate ligand.

2M

(i)

Suggest how $\mathbf{P}$ can form three dative covalent bonds.

(ii)

$2$ moles of dianion $\mathbf{P}$ , $\text{C}_4\text{H}_5\text{NO}_4^{2-}$ , react with $1$ mole of aqueous cobalt(III) ions, $[\text{Co}(\text{H}_2\text{O})_6]^{3+}$ to form $1$ mole of complex ion $\mathbf{Q}$ .

Deduce the formula and charge of $\mathbf{Q}$ .

DifficultyMedium-Easy

Worked solution

Answer

$[\text{Co}(\text{C}_4\text{H}_5\text{NO}_4)_2]^-$

Final answer

[Co(C4H5NO4)2]-

Detailed explanation

Background Concept

When ligands coordinate to a central metal ion, the resulting complex ion has a formula that combines the metal and ligands, with an overall charge equal to the sum of the metal ion charge and the ligand charges.

Understanding the Question

2 moles of dianion P ( $\text{C}_4\text{H}_5\text{NO}_4^{2-}$ ) react with 1 mole of $[\text{Co}(\text{H}_2\text{O})_6]^{3+}$ to form complex Q. Deduce the formula and charge of Q.

Approach

The ligand P replaces the water ligands. Since P is tridentate and Co(III) has a coordination number of 6, 2 moles of P (each providing 3 bonds) will replace all 6 water ligands.
Calculate the overall charge: Co is +3, each P is -2. Total charge = +3 + 2(-2) = -1.
Write the formula with the ligands in brackets.

Step-by-Step Reasoning

Metal ion: $\text{Co}^{3+}$ (from $[\text{Co}(\text{H}_2\text{O})_6]^{3+}$ , waters are displaced).
Ligands: 2 moles of $\text{C}_4\text{H}_5\text{NO}_4^{2-}$ .
Formula: $[\text{Co}( ext{C}_4\text{H}_5\text{NO}_4)_2]^{\text{charge}}$ .
Charge calculation: +3 (from Co) + 2 × (-2) (from two P dianions) = +3 - 4 = -1.
Final formula: $[\text{Co}(\text{C}_4\text{H}_5\text{NO}_4)_2]^-$ .

Key Takeaways

Charge of complex = charge of metal ion + sum of charges of ligands.
Coordination number determines how many ligands fit (Co(III) is typically octahedral, CN=6; tridentate ligand P needs 2 to fill CN=6).

Common Mistakes

Forgetting to calculate the charge correctly (e.g., +3 - 2 = +1, forgetting there are 2 ligands).
Including the water ligands in the final formula (they are displaced).

Things to Be Careful About

Ensure the charge is correct: -1, not -2 or +1.
Use the molecular formula of the ligand as given: $\text{C}_4\text{H}_5\text{NO}_4$ .

Techniques used

deduce formula and charge of complex ion

(f)

Table 1.1 shows values for the stability constants, $K_{\text{stab}}$ , of some silver(I) complexes.

Table 1.1

complex	value of $K_{\text{stab}}$
$[\text{Ag}(\text{CN})_2]^-(\text{aq})$	$1.1 \times 10^{18}$
$[\text{Ag}(\text{NH}_3)_2]^+(\text{aq})$	$1.2 \times 10^7$
$[\text{Ag}(\text{S}_2\text{O}_3)_2]^{3-}(\text{aq})$	$2.9 \times 10^{13}$

2M

(i)

Define the stability constant of a complex.

(ii)

Use the information in Table 1.1 to identify the most stable silver(I) complex.

Explain your answer.

most stable ........................................................................................................................

explanation ........................................................................................................................

DifficultyEasy

Worked solution

Answer

most stable: $[\text{Ag}(\text{CN})_2]^-(\text{aq})$
explanation: It has the largest value of $K_{\text{stab}}$ ( $1.1 \times 10^{18}$ ), indicating the greatest extent of complex formation.

Final answer

[Ag(CN)2]-(aq); largest Kstab value

Detailed explanation

Background Concept

The stability constant ( $K_{\text{stab}}$ ) is a measure of how strongly the ligands bind to the metal ion. A larger $K_{\text{stab}}$ value means the equilibrium lies further to the right (towards the complex ion), indicating a more stable complex.

Understanding the Question

Use Table 1.1 to identify the most stable silver(I) complex and explain why.

Approach

Look at the $K_{\text{stab}}$ values in the table.
Identify the largest value.
State that the complex with the largest $K_{\text{stab}}$ is the most stable.

Step-by-Step Reasoning

$[\text{Ag}(\text{CN})_2]^-$ : $K_{\text{stab}} = 1.1 \times 10^{18}$
$[\text{Ag}(\text{NH}_3)_2]^+$ : $K_{\text{stab}} = 1.2 \times 10^7$
$[\text{Ag}(\text{S}_2\text{O}_3)_2]^{3-}$ : $K_{\text{stab}} = 2.9 \times 10^{13}$
The largest value is $1.1 \times 10^{18}$ , which corresponds to $[\text{Ag}(\text{CN})_2]^-$ .
Explanation: It has the largest stability constant ( $K_{\text{stab}}$ ), meaning the equilibrium for its formation lies furthest to the right, so it is the most stable.

Key Takeaways

Higher $K_{\text{stab}}$ = more stable complex.
Compare the powers of 10 first, then the coefficients.

Common Mistakes

Choosing the complex with the most negative charge (irrelevant).
Not explaining why (just naming the complex).

Things to Be Careful About

Ensure the explanation explicitly mentions "largest $K_{\text{stab}}$ " or "largest equilibrium constant".

Techniques used

compare stability constants to identify most stable complex

(g)

Sodium sulfite, $\text{Na}_2\text{SO}_3$ , is used as a food preservative.

A $3.75\text{ g}$ sample of impure $\text{Na}_2\text{SO}_3$ is dissolved in distilled water and made up to $250\text{ cm}^3$ in a volumetric flask.

$10.0\text{ cm}^3$ of this solution requires $18.70\text{ cm}^3$ of acidified $0.0150\text{ mol dm}^{-3}$ $\text{MnO}_4^-(\text{aq})$ to reach the end-point.

The equation for the reaction is shown.

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

Calculate the percentage by mass of $\text{Na}_2\text{SO}_3$ in the sample.

DifficultyMedium

Worked solution

Working

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

\text{moles} = \frac{0.0150 \times 18.70}{1000} = 2.805 \times 10^{-4} \text{ mol}

Moles of $\text{SO}_3^{2-}$ in $10.0 \text{ cm}^3$ :
From the equation, $2 \text{ mol } \text{MnO}_4^- : 5 \text{ mol } \text{SO}_3^{2-}$

\text{moles } \text{SO}_3^{2-} = 2.805 \times 10^{-4} \times \frac{5}{2} = 7.0125 \times 10^{-4} \text{ mol}

Moles of $\text{SO}_3^{2-}$ in $250 \text{ cm}^3$ :

\text{moles} = 7.0125 \times 10^{-4} \times \frac{250}{10.0} = 1.753125 \times 10^{-2} \text{ mol}

Mass of $\text{Na}_2\text{SO}_3$ :
$M_r(\text{Na}_2\text{SO}_3) = (2 \times 23.0) + 32.1 + (3 \times 16.0) = 126.1$

\text{mass} = 1.753125 \times 10^{-2} \times 126.1 = 2.2107 \text{ g}

Percentage by mass:

\% \text{ purity} = \frac{2.2107}{3.75} \times 100 = 58.95\% \approx 59.0\%

Answer

$59.0\%$

Final answer

59.0%

Detailed explanation

Background Concept

This is a redox titration calculation. Permanganate(VII) ( $\text{MnO}_4^-$ ) is a strong oxidizing agent in acidic solution, and sulfite ( $\text{SO}_3^{2-}$ ) is a reducing agent. The balanced equation gives the stoichiometric ratio between the two reactants.

Understanding the Question

Calculate the percentage by mass of pure $\text{Na}_2\text{SO}_3$ in an impure sample, given titration data with acidified $\text{MnO}_4^-$ .

Approach

Calculate moles of $\text{MnO}_4^-$ used from concentration and volume.
Use the stoichiometric ratio (2:5) to find moles of $\text{SO}_3^{2-}$ in the $10.0 \text{ cm}^3$ aliquot.
Scale up to find moles of $\text{SO}_3^{2-}$ in the total $250 \text{ cm}^3$ solution.
Convert moles to mass using $M_r$ of $\text{Na}_2\text{SO}_3$ .
Calculate percentage purity: $(\text{mass of pure} / \text{mass of sample}) \times 100$ .

Step-by-Step Reasoning

Moles of $\text{MnO}_4^-$ : $n = C \times V = 0.0150 \text{ mol dm}^{-3} \times (18.70 / 1000) \text{ dm}^3 = 2.805 \times 10^{-4} \text{ mol}$ .
Stoichiometry: The equation is $2\text{MnO}_4^- + 5\text{SO}_3^{2-} + 6\text{H}^+ \rightarrow 2\text{Mn}^{2+} + 5\text{SO}_4^{2-} + 3\text{H}_2\text{O}$ . Ratio is 2:5.
Moles of $\text{SO}_3^{2-}$ in 10 cm³: $n = 2.805 \times 10^{-4} \times (5/2) = 7.0125 \times 10^{-4} \text{ mol}$ .
Moles in 250 cm³: The 10 cm³ is a fraction of the total 250 cm³. Scale factor = $250 / 10 = 25$ .
$n_{\text{total}} = 7.0125 \times 10^{-4} \times 25 = 0.01753125 \text{ mol}$ .
Mass of $\text{Na}_2\text{SO}_3$ : $M_r = 2(23) + 32.1 + 3(16) = 126.1 \text{ g mol}^{-1}$ .
$\text{mass} = 0.01753125 \times 126.1 = 2.2107 \text{ g}$ .
Percentage purity: $(2.2107 / 3.75) \times 100 = 58.952\%$ .
Rounding to 3 significant figures (consistent with data like 3.75, 10.0, 18.70, 0.0150): $59.0\%$ .

Key Takeaways

Always use the balanced equation for the mole ratio.
Remember to scale up from the aliquot volume to the total volume.
Calculate $M_r$ correctly.

Common Mistakes

Forgetting to divide volume by 1000 to convert cm³ to dm³.
Using the wrong mole ratio (e.g., 1:1 instead of 2:5).
Forgetting to scale up from 10 cm³ to 250 cm³.
Incorrect $M_r$ calculation (e.g., using atomic mass of S as 32 instead of 32.1, though 32.1 is standard).

Things to Be Careful About

Significant figures: The final answer should be to 3 sf (59.0%), as the data (3.75 g, 0.0150 M) is given to 3 sf.
Ensure the equation is balanced correctly before using the ratio.

Techniques used

calculate percentage purity from titration datause stoichiometric ratio in redox titration

Q2Medium-HardGroup 2Chemical EnergeticsEquilibria

(a)

The Group 2 sulfates and the Group 2 chromates show similar trends in solubility.

Suggest the trend in the solubility of the Group 2 chromates down the group.

Explain your answer.

DifficultyMedium

Worked solution

Answer

The solubility of the Group 2 chromates decreases down the group.

Down the group, the cation radius increases, so both the lattice enthalpy and the hydration enthalpy become less exothermic (less negative). The hydration enthalpy decreases by a larger amount than the lattice enthalpy, so the enthalpy change of solution becomes less exothermic / more endothermic. Therefore solubility decreases.

Final answer

Solubility decreases down the group.

Detailed explanation

Background Concept

Solubility of an ionic solid is governed by the enthalpy change of solution, $\Delta H_{\text{sol}}$ , which is the sum of the lattice enthalpy and the hydration enthalpy:

\Delta H_{\text{sol}} = \Delta H_{\text{latt}} + \Delta H_{\text{hyd}}

Lattice enthalpy is the enthalpy change when one mole of a solid ionic compound forms from its gaseous ions. Hydration enthalpy is the enthalpy change when one mole of gaseous ions becomes hydrated in water. Both terms are exothermic for small, highly charged ions and become less exothermic as ionic radius increases and charge density falls. A compound tends to be more soluble when $\Delta H_{\text{sol}}$ is more exothermic (or less endothermic).

Understanding the Question

The question tells us that Group 2 sulfates and Group 2 chromates show similar trends in solubility. We need to predict the trend in solubility of the Group 2 chromates down the group and explain it in terms of lattice and hydration enthalpies. The command words "Suggest" and "Explain" require a trend statement plus a reason.

Approach

Identify how ionic radius changes down Group 2. Then state the effect on lattice enthalpy and on hydration enthalpy. Finally, compare the magnitudes of these changes to decide which effect dominates and hence whether $\Delta H_{\text{sol}}$ becomes more or less favourable.

Step-by-Step Reasoning

Down Group 2, the cation radius increases.
Lattice enthalpy becomes less exothermic because larger ions are held less tightly in the lattice.
Hydration enthalpy also becomes less exothermic because larger ions attract water molecules less strongly.
The change in hydration enthalpy is greater than the change in lattice enthalpy, so the net $\Delta H_{\text{sol}}$ becomes less exothermic / more endothermic.
Therefore solubility decreases down the group.

Key Takeaways

Solubility trends are rationalised by comparing lattice and hydration enthalpies. The dominant change decides the trend.

Common Mistakes

Stating that solubility increases down the group.
Saying both lattice and hydration enthalpies become more exothermic.
Failing to identify which enthalpy change is greater.

Things to Be Careful About

Use "less exothermic / less negative" rather than "smaller" without a sign. The mark scheme accepts either "hydration enthalpy decreases more" or "lattice enthalpy decreases less".

Techniques used

compare lattice and hydration enthalpy changes down a groupapply solubility as a function of enthalpy of solutionrelate ionic radius to enthalpy changes

(b)

Silver(I) chromate, $\text{Ag}_2\text{CrO}_4$ , is sparingly soluble in water.

4M

(i)

Write an ionic equation to show the equilibrium between solid $\text{Ag}_2\text{CrO}_4$ and its aqueous solution.

Include state symbols.

(ii)

The value of the solubility product, $K_{\text{sp}}$ , of $\text{Ag}_2\text{CrO}_4$ is $1.12 \times 10^{-12}$ at $298\text{ K}$ .

Calculate the equilibrium concentration of $\text{Ag}^+$ , in $\text{mol dm}^{-3}$ , in a saturated solution of $\text{Ag}_2\text{CrO}_4$ at $298\text{ K}$ .

DifficultyMedium

Worked solution

Working

Let the solubility of $\text{Ag}_2\text{CrO}_4$ be $s\ \text{mol dm}^{-3}$ .

[\text{CrO}_4^{2-}] = s, \quad [\text{Ag}^+] = 2s

K_{\text{sp}} = [\text{Ag}^+]^2[\text{CrO}_4^{2-}] = (2s)^2(s) = 4s^3

4s^3 = 1.12 \times 10^{-12}

s = \sqrt[3]{\frac{1.12 \times 10^{-12}}{4}} = 6.54 \times 10^{-5}\ \text{mol dm}^{-3}

[\text{Ag}^+] = 2s = 2 \times 6.54 \times 10^{-5} = 1.31 \times 10^{-4}\ \text{mol dm}^{-3}

Answer

$1.31 \times 10^{-4}\ \text{mol dm}^{-3}$

Final answer

1.31 × 10^-4 mol dm^-3

Detailed explanation

Background Concept

For $\text{Ag}_2\text{CrO}_4$ , the solubility product expression is:

K_{\text{sp}} = [\text{Ag}^+]^2[\text{CrO}_4^{2-}]

If the molar solubility of the salt is $s$ , then each mole of dissolved salt gives $2$ moles of $\text{Ag}^+$ and $1$ mole of $\text{CrO}_4^{2-}$ .

Understanding the Question

We are given $K_{\text{sp}} = 1.12 \times 10^{-12}$ and asked for the equilibrium concentration of $\text{Ag}^+$ in a saturated solution. The stoichiometry of the salt must be used to connect $[\text{Ag}^+]$ and $[\text{CrO}_4^{2-}]$ .

Approach

Set up $K_{\text{sp}}$ in terms of the solubility $s$ , solve for $s$ , then multiply by 2 to obtain $[\text{Ag}^+]$ .

Step-by-Step Reasoning

Let $s = [\text{CrO}_4^{2-}]$ .
Then $[\text{Ag}^+] = 2s$ .
Substitute into $K_{\text{sp}}$ :

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

Solve for $s$ :

s = \sqrt[3]{\frac{1.12 \times 10^{-12}}{4}} = 6.54 \times 10^{-5}\ \text{mol dm}^{-3}

Therefore:

[\text{Ag}^+] = 2 \times 6.54 \times 10^{-5} = 1.31 \times 10^{-4}\ \text{mol dm}^{-3}

The answer is given to 3 significant figures, which is more than the minimum 2 significant figures required.

Key Takeaways

Stoichiometric coefficients appear both as powers in the $K_{\text{sp}}$ expression and as multipliers when converting solubility to ion concentration.

Common Mistakes

Using $[\text{Ag}^+] = s$ instead of $2s$ .
Forgetting to take the cube root.
Forgetting to multiply by 2 at the end.
Giving the solubility $s$ instead of $[\text{Ag}^+]$ .

Things to Be Careful About

The $K_{\text{sp}}$ value has units, but concentrations are quoted in $\text{mol dm}^{-3}$ . Give the final answer to at least 2 significant figures.

Techniques used

write the Ksp expressionrelate ion concentrations to solubility stoichiometrysolve a cubic equation

(c)

The hydrogenchromate ion, $\text{HCrO}_4^-$ , is a weak acid. The $\text{p}K_{\text{a}}$ of $\text{HCrO}_4^-$ is $6.49$ .

3M

(i)

Calculate the $\text{pH}$ of a $0.0250\text{ mol dm}^{-3}$ $\text{HCrO}_4^-$ solution.

DifficultyMedium-Easy

Worked solution

Working

K_{\text{a}} = 10^{-6.49} = 3.24 \times 10^{-7}

For a weak acid:

[\text{H}^+] \approx \sqrt{K_{\text{a}}c}

[\text{H}^+] = \sqrt{(3.24 \times 10^{-7})(0.0250)} = 8.99 \times 10^{-5}\ \text{mol dm}^{-3}

\text{pH} = -\log(8.99 \times 10^{-5}) = 4.05

Answer

pH = 4.05

Final answer

4.05

Detailed explanation

Background Concept

A weak acid only partially dissociates in water. For a weak acid HA:

K_{\text{a}} = \frac{[\text{H}^+][\text{A}^-]}{[\text{HA}]}

If dissociation is small, $[\text{HA}] \approx c$ , the initial concentration, and $[\text{H}^+] \approx [\text{A}^-]$ , so:

K_{\text{a}} \approx \frac{[\text{H}^+]^2}{c}

Hence $[\text{H}^+] \approx \sqrt{K_{\text{a}}c}$ . Also, $\text{p}K_{\text{a}} = -\log K_{\text{a}}$ .

Understanding the Question

We are given the concentration of $\text{HCrO}_4^-$ and its $\text{p}K_{\text{a}}$ . We need to calculate the pH of the solution.

Approach

First convert $\text{p}K_{\text{a}}$ to $K_{\text{a}}$ . Then use the weak acid approximation to find $[\text{H}^+]$ . Finally calculate pH.

Step-by-Step Reasoning

Convert $\text{p}K_{\text{a}}$ to $K_{\text{a}}$ :

K_{\text{a}} = 10^{-6.49} = 3.24 \times 10^{-7}

Use $[\text{H}^+] \approx \sqrt{K_{\text{a}}c}$ :

[\text{H}^+] = \sqrt{(3.24 \times 10^{-7})(0.0250)} = 8.99 \times 10^{-5}\ \text{mol dm}^{-3}

Calculate pH:

\text{pH} = -\log(8.99 \times 10^{-5}) = 4.05

Key Takeaways

The weak acid approximation is valid when the acid is weak and the concentration is not extremely dilute.

Common Mistakes

Using $[\text{H}^+] = K_{\text{a}}c$ instead of $\sqrt{K_{\text{a}}c}$ .
Forgetting to convert $\text{p}K_{\text{a}}$ to $K_{\text{a}}$ .
Calculating $\text{pH} = \text{p}K_{\text{a}}$ directly.

Things to Be Careful About

Give the pH to at least 2 significant figures. pH has no units.

Techniques used

convert pKa to Kaapply the weak acid approximationcalculate pH from [H+]

(ii)

$\text{HCrO}_4^-$ can show amphoteric behaviour.

State the formula of:

the conjugate acid of $\text{HCrO}_4^-$ ....................................................................................
the conjugate base of $\text{HCrO}_4^-$ ..................................................................................

(d)

Table 2.1 shows some energy changes.

Table 2.1

energy change	value / $\text{kJ mol}^{-1}$
first ionisation energy of silver	$+731$
second ionisation energy of silver	$+2074$
first ionisation energy of sulfur	$+1000$
second ionisation energy of sulfur	$+2251$
first electron affinity of sulfur	$-200$
second electron affinity of sulfur	$+532$
enthalpy change of atomisation of sulfur	$+279$
enthalpy change of formation of silver(I) sulfide, $\text{Ag}_2\text{S}(\text{s})$	$-33$
lattice energy of silver(I) sulfide, $\text{Ag}_2\text{S}(\text{s})$	$-2677$

6M

(i)

Define the term first electron affinity.

(ii)

Explain why the value for the second electron affinity of sulfur is positive.

(iii)

Construct an equation for the lattice energy of $\text{Ag}_2\text{S}$ .

Include state symbols.

(iv)

Calculate the enthalpy change of atomisation, $\Delta H_{\text{at}}$ , in $\text{kJ mol}^{-1}$ , of silver using relevant data from Table 2.1.

It may be helpful to draw a labelled Born–Haber cycle.

Show your working.

DifficultyMedium-Hard

Worked solution

Working

Born–Haber cycle for $\text{Ag}_2\text{S}$ :

\Delta H_{\text{f}} = 2\Delta H_{\text{at}}(\text{Ag}) + 2\Delta H_{\text{iE1}}(\text{Ag}) + \Delta H_{\text{at}}(\text{S}) + \Delta H_{\text{EA1}}(\text{S}) + \Delta H_{\text{EA2}}(\text{S}) + \Delta H_{\text{latt}}

-33 = 2\Delta H_{\text{at}}(\text{Ag}) + (2 \times 731) + 279 + (-200) + 532 + (-2677)

-33 = 2\Delta H_{\text{at}}(\text{Ag}) - 604

2\Delta H_{\text{at}}(\text{Ag}) = 571

\Delta H_{\text{at}}(\text{Ag}) = +285.5\ \text{kJ mol}^{-1}

Answer

$+285.5\ \text{kJ mol}^{-1}$

Final answer

+285.5 kJ mol^-1

Detailed explanation

Background Concept

A Born–Haber cycle applies Hess's law to ionic compound formation. For $\text{Ag}_2\text{S}$ , the formation enthalpy can be broken into:

atomisation of 2 mol Ag atoms: $2\Delta H_{\text{at}}(\text{Ag})$
first ionisation of 2 mol Ag atoms: $2 \times \Delta H_{\text{iE1}}(\text{Ag})$
atomisation of sulfur: $\Delta H_{\text{at}}(\text{S})$
first and second electron affinities of sulfur: $\Delta H_{\text{EA1}}(\text{S}) + \Delta H_{\text{EA2}}(\text{S})$
lattice energy: $\Delta H_{\text{latt}}$

Only the first ionisation energy of silver is used because silver forms $\text{Ag}^+$ , not $\text{Ag}^{2+}$ . Only the first and second electron affinities of sulfur are used because sulfur forms $\text{S}^{2-}$ .

Understanding the Question

We need to calculate $\Delta H_{\text{at}}$ of silver using the relevant data from Table 2.1. The question suggests drawing a labelled Born–Haber cycle, but the calculation can be done algebraically.

Approach

Write the overall energy cycle equation, substitute the correct values, and solve for the unknown $\Delta H_{\text{at}}(\text{Ag})$ .

Step-by-Step Reasoning

The formation of $\text{Ag}_2\text{S}(\text{s})$ from its elements is:

2\text{Ag}(\text{s}) + \text{S}(\text{s}) \rightarrow \text{Ag}_2\text{S}(\text{s}) \qquad \Delta H_{\text{f}} = -33\ \text{kJ mol}^{-1}

The cycle gives:

\Delta H_{\text{f}} = 2\Delta H_{\text{at}}(\text{Ag}) + 2\Delta H_{\text{iE1}}(\text{Ag}) + \Delta H_{\text{at}}(\text{S}) + \Delta H_{\text{EA1}}(\text{S}) + \Delta H_{\text{EA2}}(\text{S}) + \Delta H_{\text{latt}}

Substitute the values:

-33 = 2\Delta H_{\text{at}}(\text{Ag}) + (2 \times 731) + 279 + (-200) + 532 + (-2677)

Simplify:

-33 = 2\Delta H_{\text{at}}(\text{Ag}) - 604

Solve:

2\Delta H_{\text{at}}(\text{Ag}) = 571

\Delta H_{\text{at}}(\text{Ag}) = +285.5\ \text{kJ mol}^{-1}

Key Takeaways

Born–Haber cycles require careful selection of data and correct stoichiometry. The charge on the ion determines which ionisation energies and electron affinities are included.

Common Mistakes

Including the second ionisation energy of silver.
Including the first ionisation energy of sulfur.
Forgetting to multiply the silver terms by 2.
Sign errors with electron affinities or lattice energy.

Things to Be Careful About

Lattice energy is negative. The second electron affinity of sulfur is positive. The final answer must be in $\text{kJ mol}^{-1}$ .

Techniques used

construct a Born-Haber cyclesum enthalpy changes using Hess's lawsolve for an unknown atomisation enthalpy

(e)

Suggest how the magnitude for the lattice energy of $\text{Ag}_2\text{S}(\text{s})$ differs from the lattice energy of $\text{Cu}_2\text{S}(\text{s})$ .

Explain your answer.

DifficultyMedium-Easy

Worked solution

Answer

The lattice energy of $\text{Ag}_2\text{S}$ is smaller in magnitude (less exothermic / less negative) than that of $\text{Cu}_2\text{S}$ , because $\text{Ag}^+$ has a larger ionic radius than $\text{Cu}^+$ , so the electrostatic attraction between the ions is weaker.

Final answer

Ag2S lattice energy is less exothermic (smaller magnitude) than Cu2S; Ag+ has a larger ionic radius than Cu+.

Detailed explanation

Background Concept

The magnitude of lattice energy depends on ionic charge and ionic radius. For ions of the same charge, a smaller ionic radius gives a stronger electrostatic attraction and therefore a more exothermic (more negative) lattice energy.

Understanding the Question

We need to compare the lattice energy of $\text{Ag}_2\text{S}$ with that of $\text{Cu}_2\text{S}$ and explain the difference.

Approach

Compare the radii of $\text{Ag}^+$ and $\text{Cu}^+$ . Both cations have the same charge and the anion is the same, so the difference is due to ionic radius.

Step-by-Step Reasoning

$\text{Ag}^+$ is below $\text{Cu}^+$ in the periodic table, so it has a larger ionic radius.
A larger cation means the ions in the lattice are farther apart.
The electrostatic attraction between $\text{Ag}^+$ and $\text{S}^{2-}$ is weaker than between $\text{Cu}^+$ and $\text{S}^{2-}$ .
Therefore the lattice energy of $\text{Ag}_2\text{S}$ is less exothermic / less negative / smaller in magnitude than that of $\text{Cu}_2\text{S}$ .

Key Takeaways

For compounds with the same ionic charges, lattice energy becomes less exothermic as ionic radius increases.

Common Mistakes

Saying $\text{Ag}_2\text{S}$ has a more exothermic lattice energy.
Confusing the trend in ionic radius.
Not mentioning that the ionic charges are the same.

Things to Be Careful About

Use "less exothermic / less negative" or "smaller magnitude" rather than simply "smaller".

Techniques used

compare ionic radiirelate ionic radius to lattice energy

Q3Medium-HardChemical Energetics

(a)

Define the term entropy.

(b)

3M

(i)

Place one tick ( $\checkmark$ ) in each row of Table 3.1 to show the sign of the entropy change, $\Delta S$ , for each process.

Table 3.1

process	$\Delta S$ is negative	$\Delta S$ is positive
steam condensing into water
solid $\text{KCl}$ dissolving in water

DifficultyMedium-Easy

Worked solution

Answer

process	$\Delta S$ is negative	$\Delta S$ is positive
steam condensing into water	$\checkmark$
solid $\text{KCl}$ dissolving in water		$\checkmark$

Final answer

steam condensing: negative; KCl dissolving: positive

Detailed explanation

Background Concept

The sign of the entropy change ( $\Delta S$ ) depends on whether the disorder of the system increases or decreases. Gases have much higher entropy than liquids, which have higher entropy than solids. Dissolving a solid into an aqueous solution typically increases entropy because the ordered crystal lattice breaks down into freely moving hydrated ions.

Understanding the Question

We need to determine the sign of $\Delta S$ for two specific processes: condensation of steam and dissolution of solid KCl.

Approach

Compare the states of matter and the degree of disorder before and after each process.

Step-by-Step Reasoning

Steam condensing into water: $\text{H}_2\text{O}(\text{g}) \rightarrow \text{H}_2\text{O}(\text{l})$ . A gas is turning into a liquid. The particles become more ordered and have fewer possible arrangements. Therefore, entropy decreases, and $\Delta S$ is negative.
Solid KCl dissolving in water: $\text{KCl}(\text{s}) \rightarrow \text{K}^+(\text{aq}) + \text{Cl}^-(\text{aq})$ . An ordered solid lattice breaks apart into mobile, hydrated ions in solution. The number of possible arrangements increases significantly. Therefore, entropy increases, and $\Delta S$ is positive.

Key Takeaways

Gas $\rightarrow$ liquid/solid: $\Delta S$ is negative.
Solid $\rightarrow$ aqueous/gas: $\Delta S$ is positive.

Common Mistakes

Assuming all dissolving processes have positive entropy (some can be negative if the solvent becomes highly ordered around the ions, though this is rare for simple salts like KCl).
Confusing the sign: negative means entropy decreases.

Things to Be Careful About

Ensure the tick is placed in the correct column. The table has columns for "negative" and "positive".

Techniques used

predict sign of entropy change from physical state changes

(ii)

Chlorine trifluoride, $\text{ClF}_3$ , decomposes on heating into its elements, as shown.

\text{reaction 1} \quad 2\text{ClF}_3(\text{g}) \rightarrow \text{Cl}_2(\text{g}) + 3\text{F}_2(\text{g})

Standard entropies are shown in Table 3.2.

Table 3.2

substance	$\text{ClF}_3(\text{g})$	$\text{Cl}_2(\text{g})$	$\text{F}_2(\text{g})$
$S^\ominus / \text{J K}^{-1} \text{mol}^{-1}$	$+281.6$	$+223.1$	$+203.0$

Calculate the standard entropy change, $\Delta S^\ominus$ , in $\text{J K}^{-1} \text{mol}^{-1}$ , for reaction 1.

DifficultyMedium-Easy

Worked solution

Working

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

\Delta S^\ominus = [3 \times S^\ominus(\text{F}_2) + 1 \times S^\ominus(\text{Cl}_2)] - [2 \times S^\ominus(\text{ClF}_3)]

\Delta S^\ominus = [(3 \times 203.0) + 223.1] - [2 \times 281.6]

\Delta S^\ominus = [609.0 + 223.1] - [563.2]

\Delta S^\ominus = 832.1 - 563.2 = +268.9 \text{ J K}^{-1} \text{mol}^{-1}

Answer

$+268.9 \text{ J K}^{-1} \text{mol}^{-1}$

Final answer

+268.9 J K^-1 mol^-1

Detailed explanation

Background Concept

The standard entropy change of a reaction ( $\Delta S^\ominus$ ) is calculated by subtracting the sum of the standard entropies of the reactants from the sum of the standard entropies of the products. It is crucial to multiply each entropy value by the stoichiometric coefficient from the balanced chemical equation.

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

Understanding the Question

We are given the balanced equation $2\text{ClF}_3(\text{g}) \rightarrow \text{Cl}_2(\text{g}) + 3\text{F}_2(\text{g})$ and the standard entropy values for each substance. We need to calculate $\Delta S^\ominus$ .

Approach

Identify products and reactants and their coefficients.
Sum the entropies of the products: $3 \times S^\ominus(\text{F}_2) + 1 \times S^\ominus(\text{Cl}_2)$ .
Sum the entropies of the reactants: $2 \times S^\ominus(\text{ClF}_3)$ .
Subtract reactants from products.

Step-by-Step Reasoning

Products: $1 \text{ mol } \text{Cl}_2$ and $3 \text{ mol } \text{F}_2$ .
Sum $= (1 \times 223.1) + (3 \times 203.0) = 223.1 + 609.0 = 832.1 \text{ J K}^{-1} \text{mol}^{-1}$ .
Reactants: $2 \text{ mol } \text{ClF}_3$ .
Sum $= 2 \times 281.6 = 563.2 \text{ J K}^{-1} \text{mol}^{-1}$ .
$\Delta S^\ominus = 832.1 - 563.2 = +268.9 \text{ J K}^{-1} \text{mol}^{-1}$ .

The positive sign is consistent with the reaction producing more moles of gas (4 moles from 2 moles), which increases disorder.

Key Takeaways

Always multiply by the stoichiometric coefficient. The unit is $\text{J K}^{-1} \text{mol}^{-1}$ (per mole of reaction as written).

Common Mistakes

Forgetting to multiply $S^\ominus(\text{F}_2)$ by 3 or $S^\ominus(\text{ClF}_3)$ by 2.
Subtracting in the wrong order (reactants - products).
Forgetting the sign.

Things to Be Careful About

Check the balancing of the equation. Ensure units are correct ( $\text{J}$ , not $\text{kJ}$ ).

Techniques used

apply formula for standard entropy change of reaction

(c)

Group 2 carbonates decompose on heating. The decomposition for one of the Group 2 carbonates, $\text{MCO}_3$ , is shown in reaction 2.

\text{reaction 2} \quad \text{MCO}_3(\text{s}) \rightarrow \text{MO}(\text{s}) + \text{CO}_2(\text{g})

5M

(i)

Predict the sign of the entropy change, $\Delta S$ , for reaction 2.

Explain your answer.

(ii)

The Gibbs equation is shown.

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

Fig. 3.1 shows values of the Gibbs free energy change, $\Delta G^\ominus$ , in $\text{kJ mol}^{-1}$ , at different temperatures, $T$ , in $\text{K}$ , for reaction 2.

Assume $\Delta H^\ominus$ and $\Delta S^\ominus$ values for this reaction remain constant over this temperature range.

Fig. 3.1

Use the gradient and intercept on the $y$ -axis in Fig. 3.1 and the Gibbs equation to determine:

$\Delta S^\ominus$ , in $\text{J K}^{-1} \text{mol}^{-1}$ , for reaction 2
the minimum temperature, $T$ , in $\text{K}$ , at which the reaction is feasible
$\Delta H^\ominus$ , in $\text{kJ mol}^{-1}$ , for reaction 2.

DifficultyMedium-Hard

Worked solution

Working

The Gibbs equation is $\Delta G^\ominus = \Delta H^\ominus - T\Delta S^\ominus$ , which is in the form $y = c + mx$ , where:

$y = \Delta G^\ominus$ (in $\text{kJ mol}^{-1}$ )
$x = T$ (in $\text{K}$ )
gradient $m = -\Delta S^\ominus$ (in $\text{kJ K}^{-1} \text{mol}^{-1}$ )
$y$ -intercept $c = \Delta H^\ominus$ (in $\text{kJ mol}^{-1}$ )

1. Calculate $\Delta S^\ominus$ from the gradient:
Using points on the line, e.g., $(600, 80)$ and $(1500, -60)$ :

\text{gradient} = \frac{\Delta(\Delta G^\ominus)}{\Delta T} = \frac{-60 - 80}{1500 - 600} = \frac{-140}{900} = -0.1556 \text{ kJ K}^{-1} \text{mol}^{-1}

-\Delta S^\ominus = -0.1556 \text{ kJ K}^{-1} \text{mol}^{-1}

\Delta S^\ominus = 0.1556 \times 1000 = +155.6 \text{ J K}^{-1} \text{mol}^{-1}

2. Minimum temperature for feasibility:
The reaction is feasible when $\Delta G^\ominus \leq 0$ . From the graph, $\Delta G^\ominus = 0$ at the $x$ -intercept.

T \approx 1120 \text{ K}

3. Calculate $\Delta H^\ominus$ from the $y$ -intercept:
Using $\Delta G^\ominus = \Delta H^\ominus - T\Delta S^\ominus$ and the point $(1120, 0)$ :

0 = \Delta H^\ominus - (1120)(0.1556)

\Delta H^\ominus = 174.3 \text{ kJ mol}^{-1}

(Alternatively, extrapolating the line to $T=0$ gives $\Delta H^\ominus \approx +172 \text{ kJ mol}^{-1}$ )

Answer

$\Delta S^\ominus = +156 \text{ J K}^{-1} \text{mol}^{-1}$ (accept $150-161$ )
Minimum $T = 1120 \text{ K}$ (accept $1115-1125$ )
$\Delta H^\ominus = +172 \text{ kJ mol}^{-1}$ (accept $162-182$ )

Final answer

ΔS° = +156 J K⁻¹ mol⁻¹; T = 1120 K; ΔH° = +172 kJ mol⁻¹

Detailed explanation

Background Concept

The Gibbs free energy equation is $\Delta G^\ominus = \Delta H^\ominus - T\Delta S^\ominus$ . This is a linear equation of the form $y = c + mx$ if we plot $\Delta G^\ominus$ on the $y$ -axis and $T$ on the $x$ -axis.

$y = \Delta G^\ominus$
$x = T$
$m \text{ (gradient)} = -\Delta S^\ominus$
$c \text{ (y-intercept)} = \Delta H^\ominus$

A reaction is feasible (spontaneous) when $\Delta G^\ominus < 0$ . The minimum temperature for feasibility is where $\Delta G^\ominus = 0$ (the $x$ -intercept).

Understanding the Question

We are given a graph of $\Delta G^\ominus$ vs $T$ . We must use the gradient and intercepts to find $\Delta S^\ominus$ , the minimum temperature for feasibility, and $\Delta H^\ominus$ .

Approach

Identify the gradient of the line and relate it to $\Delta S^\ominus$ . Remember to convert from $\text{kJ}$ to $\text{J}$ .
Read the $x$ -intercept directly from the graph for the minimum temperature.
Use the $y$ -intercept or calculate using $\Delta H^\ominus = T\Delta S^\ominus$ at the $x$ -intercept to find $\Delta H^\ominus$ .

Step-by-Step Reasoning

1. Finding $\Delta S^\ominus$ :

Select two clear points on the line. From the graph description: $(600 \text{ K}, 80 \text{ kJ mol}^{-1})$ and $(1500 \text{ K}, -60 \text{ kJ mol}^{-1})$ .
$\text{Gradient} = \frac{-60 - 80}{1500 - 600} = \frac{-140}{900} = -0.1556 \text{ kJ K}^{-1} \text{mol}^{-1}$ .
Since $\text{gradient} = -\Delta S^\ominus$ , then $-\Delta S^\ominus = -0.1556 \text{ kJ K}^{-1} \text{mol}^{-1}$ .
$\Delta S^\ominus = +0.1556 \text{ kJ K}^{-1} \text{mol}^{-1} = +155.6 \text{ J K}^{-1} \text{mol}^{-1}$ .
(Allowable range from mark scheme: $150-161 \text{ J K}^{-1} \text{mol}^{-1}$ )

2. Minimum temperature ( $T$ ):

Feasibility requires $\Delta G^\ominus \leq 0$ . This occurs when the line crosses the $x$ -axis ( $\Delta G^\ominus = 0$ ).
Reading from the graph, the line crosses the $x$ -axis at approximately $T = 1120 \text{ K}$ .
(Allowable range: $1115-1125 \text{ K}$ )

3. Finding $\Delta H^\ominus$ :

The $y$ -intercept is $\Delta H^\ominus$ . Extrapolating the line to $T=0$ :
$\Delta H^\ominus = \Delta G^\ominus + T\Delta S^\ominus$ . Using the point $(1120, 0)$ :
$0 = \Delta H^\ominus - (1120)(0.1556) \Rightarrow \Delta H^\ominus = +174.3 \text{ kJ mol}^{-1}$ .
Alternatively, reading the $y$ -intercept directly from an extrapolated graph gives $\approx +172 \text{ kJ mol}^{-1}$ .
(Allowable range: $162-182 \text{ kJ mol}^{-1}$ )

Key Takeaways

Plotting $\Delta G$ vs $T$ gives a straight line with gradient $-\Delta S$ and intercept $\Delta H$ .
Always check units: gradient is in $\text{kJ}$ , so multiply by 1000 to get $\Delta S$ in $\text{J}$ .
Feasibility is determined by $\Delta G < 0$ .

Common Mistakes

Forgetting to multiply the gradient by 1000 to convert $\text{kJ}$ to $\text{J}$ for $\Delta S$ .
Reading the wrong intercept (e.g., reading $\Delta S$ from the $y$ -intercept).
Not using a large triangle for the gradient calculation, leading to inaccurate readings.

Things to Be Careful About

The gradient is $-\Delta S$ , not $+\Delta S$ . The negative sign is crucial.
Ensure the final answer for $\Delta S$ is in $\text{J K}^{-1} \text{mol}^{-1}$ , not $\text{kJ}$ .
When reading off the graph, use points that are far apart to minimize reading errors.

Techniques used

determine thermodynamic quantities from Gibbs free energy graphcalculate gradient from graph

Q4MediumReaction Kinetics

(a)

Nitrogen monoxide, $\text{NO}$ , reacts with hydrogen, as shown in reaction 3.

\text{reaction 3} \quad 2\text{NO} + 2\text{H}_2 \rightarrow \text{N}_2 + 2\text{H}_2\text{O}

7M

(i)

The rate equation for reaction 3 is shown.

\text{rate} = k[\text{H}_2][\text{NO}]^2

Complete Table 4.1.

Table 4.1


the order of reaction with respect to $[\text{H}_2]$
the order of reaction with respect to $[\text{NO}]$
the overall order of the reaction

DifficultyEasy

Worked solution

Answer

order with respect to $[\text{H}_2]$ : 1
order with respect to $[\text{NO}]$ : 2
overall order: 3

Final answer

Order wrt H2: 1; order wrt NO: 2; overall: 3

Detailed explanation

Background Concept

The rate equation expresses how the rate of a reaction depends on the concentrations of the reactants. Each concentration term is raised to a power — the order with respect to that reactant — which is simply the exponent of that concentration in the rate equation. The overall order is the sum of the individual orders. Orders must be determined experimentally (or given in the question); they cannot be predicted from the stoichiometric coefficients of the balanced equation.

Understanding the Question

The question gives the rate equation rate $= k[\text{H}_2][\text{NO}]^2$ and asks for the orders with respect to each reactant and the overall order. This is a direct read-off task — no calculation is needed.

Approach

Read the exponent of each concentration in the rate equation: $[\text{H}_2]$ has exponent 1, $[\text{NO}]$ has exponent 2. Add them to find the overall order.

Step-by-Step Reasoning

The rate equation is rate $= k[\text{H}_2]^1[\text{NO}]^2$ . The exponent on $[\text{H}_2]$ is 1, so the order with respect to $\text{H}_2$ is 1. The exponent on $[\text{NO}]$ is 2, so the order with respect to $\text{NO}$ is 2. The overall order is the sum: $1 + 2 = 3$ .

Key Takeaways

Orders come straight from the exponents in the rate equation. The overall order is the sum of the individual orders, never the sum of the stoichiometric coefficients.

Common Mistakes

Do not use the stoichiometric coefficients (2 for NO, 2 for H $_2$ ) — those are unrelated to the orders. The rate equation is given, so the orders are read from it directly.

Things to Be Careful About

Make sure you sum correctly: $1 + 2 = 3$ , not $2 + 2 = 4$ (which would be the stoichiometric sum, not the kinetic order).

Techniques used

read orders directly from the rate equationsum the individual orders

(ii)

Predict how the initial rate for reaction 3 changes when the concentration of $\text{NO}$ is halved.

(iii)

Predict how the initial rate for reaction 3 changes when the concentrations of $\text{NO}$ and $\text{H}_2$ are both increased three times.

(iv)

Suggest why reaction 3 is unlikely to proceed by a mechanism involving only a single step.

(v)

Suggest equations for the three steps of the reaction mechanism for reaction 3.

Each step involves a reaction between two molecules.

step 1 .................................................. $\rightarrow$ ..................................................

step 2 ...................... $+$ ...................... $\rightarrow \text{N}_2\text{O} +$ .....................................

step 3 $\text{N}_2\text{O} +$ ..................................... $\rightarrow$ ....................... $+$ .......................

DifficultyMedium

Worked solution

Answer

step 1: $2\text{NO} \rightarrow \text{N}_2\text{O}_2$

step 2: $\text{N}_2\text{O}_2 + \text{H}_2 \rightarrow \text{N}_2\text{O} + \text{H}_2\text{O}$

step 3: $\text{N}_2\text{O} + \text{H}_2 \rightarrow \text{N}_2 + \text{H}_2\text{O}$

Final answer

step 1: 2NO -> N2O2; step 2: N2O2 + H2 -> N2O + H2O; step 3: N2O + H2 -> N2 + H2O

Detailed explanation

Background Concept

A reaction mechanism is a series of elementary steps whose sum is the overall equation. Each step is typically bimolecular (two molecules collide). Intermediates are species formed in one step and consumed in a later step; they do not appear in the overall equation.

Understanding the Question

Provide a three-step mechanism for $2\text{NO} + 2\text{H}_2 \rightarrow \text{N}_2 + 2\text{H}_2\text{O}$ , with each step involving two molecules. Steps 2 and 3 are partially given, so the blanks must be filled consistently.

Approach

Combine the two NO molecules first to form a dimer $\text{N}_2\text{O}_2$ (a well-known step for NO). Then let each H $_2$ reduce one oxygen: $\text{N}_2\text{O}_2 + \text{H}_2 \rightarrow \text{N}_2\text{O} + \text{H}_2\text{O}$ , then $\text{N}_2\text{O} + \text{H}_2 \rightarrow \text{N}_2 + \text{H}_2\text{O}$ . Verify the sum equals the overall equation.

Step-by-Step Reasoning

Step 1: $2\text{NO} \rightarrow \text{N}_2\text{O}_2$ (dimerisation of NO).
Step 2: $\text{N}_2\text{O}_2 + \text{H}_2 \rightarrow \text{N}_2\text{O} + \text{H}_2\text{O}$ .
Step 3: $\text{N}_2\text{O} + \text{H}_2 \rightarrow \text{N}_2 + \text{H}_2\text{O}$ .

Summing: $2\text{NO} + \text{N}_2\text{O}_2 + \text{N}_2\text{O} + 2\text{H}_2 \rightarrow \text{N}_2\text{O}_2 + \text{N}_2\text{O} + \text{N}_2 + 2\text{H}_2\text{O}$ . Cancelling $\text{N}_2\text{O}_2$ and $\text{N}_2\text{O}$ gives $2\text{NO} + 2\text{H}_2 \rightarrow \text{N}_2 + 2\text{H}_2\text{O}$ , which matches reaction 3.

The mechanism is also consistent with the given rate equation if step 2 is rate-determining: rate $= k[\text{N}_2\text{O}_2][\text{H}_2]$ , and from the fast pre-equilibrium step 1, $[\text{N}_2\text{O}_2] \propto [\text{NO}]^2$ , giving rate $= k[\text{NO}]^2[\text{H}_2]$ .

Key Takeaways

Mechanisms must sum to the overall equation; intermediates cancel out. A proposed mechanism should also be consistent with the rate equation (the slow step determines the order).

Common Mistakes

Leaving out H $_2$ O, or writing a step in which three molecules collide (violates the "two molecules" instruction). Also, not checking the sum of the steps against the overall equation.

Things to Be Careful About

Each step must involve exactly two molecules. Check the overall sum balances atoms and charge.

Techniques used

propose a multi-step mechanism consistent with the rate equationcheck that the steps sum to the overall equation

(vi)

Suggest the role of $\text{N}_2\text{O}$ in this mechanism.

Explain your reasoning.

(b)

Iodine, $\text{I}_2$ , reacts with thiosulfate ions, $\text{S}_2\text{O}_3^{2-}$ , as shown in reaction 4.

\text{reaction 4} \quad 2\text{S}_2\text{O}_3^{2-} + \text{I}_2 \rightarrow \text{S}_4\text{O}_6^{2-} + 2\text{I}^-

Reaction 4 is carried out in the presence of a large excess of $\text{I}_2$ . Under these conditions, the reaction is first order with respect to $[\text{S}_2\text{O}_3^{2-}]$ and zero order with respect to $[\text{I}_2]$ .

The half-life, $t_{\frac{1}{2}}$ , for reaction 4 is $720\text{ s}$ under certain conditions.

Calculate the value of the rate constant, $k$ , for reaction 4. Include the units of $k$ .

DifficultyMedium-Easy

Worked solution

Working

For a first-order reaction:

t_{\frac{1}{2}} = \frac{\ln 2}{k} \quad \Rightarrow \quad k = \frac{0.693}{t_{\frac{1}{2}}}

k = \frac{0.693}{720\text{ s}} = 9.6 \times 10^{-4}\text{ s}^{-1} \text{ (2 s.f.)}

Answer

$k = 9.6 \times 10^{-4}\text{ s}^{-1}$

Final answer

9.6 x 10^-4 s^-1

Detailed explanation

Background Concept

For a first-order reaction, the half-life is constant and related to the rate constant by $t_{\frac{1}{2}} = \frac{\ln 2}{k}$ . This constancy of half-life is a hallmark of first-order kinetics.

Understanding the Question

Reaction 4 is first order with respect to $\text{S}_2\text{O}_3^{2-}$ and zero order with respect to $\text{I}_2$ , so overall it is first order. Given $t_{\frac{1}{2}} = 720\text{ s}$ , we must calculate $k$ and include its units.

Approach

Rearrange $t_{\frac{1}{2}} = \frac{\ln 2}{k}$ to $k = \frac{\ln 2}{t_{\frac{1}{2}}}$ , substitute the value, and derive the units from the first-order rate equation.

Step-by-Step Reasoning

$k = \frac{0.693}{720\text{ s}} = 9.625 \times 10^{-4}\text{ s}^{-1} \approx 9.6 \times 10^{-4}\text{ s}^{-1}$ (to 2 s.f.).

For the units: for a first-order reaction, rate $= k[\text{A}]$ , so $k = \frac{\text{rate}}{[\text{A}]}$ . Rate has units $\text{mol dm}^{-3}\text{ s}^{-1}$ and concentration has units $\text{mol dm}^{-3}$ , so $k$ has units $\text{s}^{-1}$ . (Since $\ln 2$ is dimensionless, the units come entirely from the reciprocal of the time.)

Key Takeaways

First-order half-life: $t_{\frac{1}{2}} = \frac{\ln 2}{k}$ . Units of a first-order rate constant are always $\text{s}^{-1}$ (or more generally time $^{-1}$ ).

Common Mistakes

Using $t_{\frac{1}{2}} = \frac{1}{k}$ (wrong) or giving the units of a second-order rate constant ( $\text{mol}^{-1}\text{ dm}^3\text{ s}^{-1}$ ). Also forgetting to divide by 720.

Things to Be Careful About

Report to an appropriate number of significant figures (2 s.f. is fine here). The units must be $\text{s}^{-1}$ — the mark scheme requires them.

Techniques used

use the first-order half-life equationderive the units of a first-order rate constant

(c)

The reaction between iodide ions, $\text{I}^-(\text{aq})$ , and peroxydisulfate ions, $\text{S}_2\text{O}_8^{2-}(\text{aq})$ , is catalysed by $\text{Co}^{3+}(\text{aq})$ . The mechanism is similar to the mechanism of this reaction when $\text{Fe}^{3+}(\text{aq})$ is used as the catalyst.

4M

(i)

State the type of catalysis that occurs in this reaction.

Explain your reasoning.

(ii)

Write two equations to show how $\text{Co}^{3+}(\text{aq})$ catalyses this reaction.

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

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

DifficultyMedium

Worked solution

Answer

equation 1: $2\text{Co}^{3+}(\text{aq}) + 2\text{I}^-(\text{aq}) \rightarrow 2\text{Co}^{2+}(\text{aq}) + \text{I}_2(\text{aq})$

equation 2: $2\text{Co}^{2+}(\text{aq}) + \text{S}_2\text{O}_8^{2-}(\text{aq}) \rightarrow 2\text{Co}^{3+}(\text{aq}) + 2\text{SO}_4^{2-}(\text{aq})$

Final answer

2Co3+ + 2I- -> 2Co2+ + I2; 2Co2+ + S2O8^2- -> 2Co3+ + 2SO4^2-

Detailed explanation

Background Concept

A homogeneous catalyst participates in the reaction by being oxidised or reduced, then regenerated. The catalyst is consumed in one step and reformed in another, so it does not appear in the overall equation.

Understanding the Question

Write two equations showing how $\text{Co}^{3+}(\text{aq})$ catalyses the reaction between $\text{I}^-$ and $\text{S}_2\text{O}_8^{2-}$ . The mechanism mirrors the $\text{Fe}^{3+}$ case, where the metal ion shuttles between two oxidation states.

Approach

$\text{Co}^{3+}$ oxidises $\text{I}^-$ to $\text{I}_2$ (itself being reduced to $\text{Co}^{2+}$ ), then $\text{Co}^{2+}$ is re-oxidised by $\text{S}_2\text{O}_8^{2-}$ back to $\text{Co}^{3+}$ . This regenerates the catalyst.

Step-by-Step Reasoning

Step 1: $\text{Co}^{3+}$ is a strong oxidising agent; it oxidises $\text{I}^-$ to $\text{I}_2$ , itself being reduced to $\text{Co}^{2+}$ . Balancing: $2\text{Co}^{3+} + 2\text{I}^- \rightarrow 2\text{Co}^{2+} + \text{I}_2$ (two electrons transferred, two I $^-$ oxidised).

Step 2: $\text{Co}^{2+}$ is re-oxidised by $\text{S}_2\text{O}_8^{2-}$ (which is reduced to $\text{SO}_4^{2-}$ ), regenerating $\text{Co}^{3+}$ . Balancing: $2\text{Co}^{2+} + \text{S}_2\text{O}_8^{2-} \rightarrow 2\text{Co}^{3+} + 2\text{SO}_4^{2-}$ .

Summing the two equations: $2\text{I}^- + \text{S}_2\text{O}_8^{2-} \rightarrow \text{I}_2 + 2\text{SO}_4^{2-}$ . This is the net catalysed reaction. $\text{Co}^{3+}$ is regenerated, confirming it acts as a catalyst.

Key Takeaways

Catalytic cycles: the catalyst is consumed in one step and regenerated in another. The two steps must balance atoms and charge, and their sum gives the overall (uncatalysed net) reaction.

Common Mistakes

Unbalanced equations, wrong charges on ions, or writing $\text{Co}^{3+} \rightarrow \text{Co}^{2+}$ without the partner species (I $^-$ and S $_2$ O $_8^{2-}$ ). Also forgetting the 2:1 stoichiometry.

Things to Be Careful About

Both equations must balance charge and atoms. The catalyst must be regenerated ( $\text{Co}^{3+}$ reformed in equation 2).

Techniques used

write redox half-equations for catalytic regenerationbalance charge and atoms in each step

(iii)

Suggest why this reaction is slow in the absence of $\text{Co}^{3+}(\text{aq})$ .

Q5MediumIntroduction to A Level Organic ChemistryHydrocarbons

(a)

Describe and explain the shape of benzene.

In your answer, include:

the shape and bond angle in the ring
the hybridisation of the carbon atoms
how orbital overlap forms $\sigma$ and $\pi$ bonds between the carbon atoms in the ring.

DifficultyMedium-Easy

Worked solution

Answer

The ring is planar (trigonal planar geometry around each carbon) with bond angles of 120°.
Each carbon atom is sp² hybridised.
The unhybridised p orbitals on each carbon atom overlap sideways (laterally) above and below the plane of the ring to form a delocalised pi (π) system.
The sp² hybridised orbitals overlap end-on-end (head-on) to form sigma (σ) bonds between carbon atoms (and between C and H).

Final answer

Planar ring, 120°, sp² hybridised, p orbitals overlap sideways for π bonds, sp² orbitals overlap end-on for σ bonds.

Detailed explanation

Background Concept

Benzene ( $C_6H_6$ ) is the simplest arene and its structure is a classic example of delocalisation. Each carbon atom is bonded to two other carbons and one hydrogen. To achieve this, carbon undergoes sp² hybridisation, leaving one unhybridised p orbital per carbon. The sp² orbitals form a planar hexagonal sigma framework, while the p orbitals overlap laterally to create a continuous ring of electron density above and below the plane. This delocalisation gives benzene exceptional thermodynamic stability (lower enthalpy of hydrogenation than expected for a cyclic triene).

Understanding the Question

The question asks for a description and explanation of benzene's shape. It specifically requires four elements: the overall shape and bond angle, the hybridisation of the carbons, and the orbital overlap that forms both σ and π bonds. This is a foundational organic chemistry question testing the student's understanding of benzene's molecular orbital theory.

Approach

To answer this, recall the VSEPR theory applied to sp² hybridised atoms (trigonal planar, 120°). Then, describe the hybridisation process: mixing one s and two p orbitals to form three sp² orbitals (forming σ bonds) and leaving one p orbital unhybridised (forming the π system). Finally, specify the type of overlap for each bond type: end-on for σ, sideways for π.

Step-by-Step Reasoning

Shape and bond angle: Each carbon has 3 regions of electron density (two C-C bonds, one C-H bond). According to VSEPR, these arrange themselves at 120° to minimise repulsion, resulting in a planar, hexagonal ring with trigonal planar geometry around each carbon.
Hybridisation: To form three σ bonds in a plane, each carbon uses sp² hybrid orbitals. This is formed from one 2s and two 2p orbitals.
π bond formation: The remaining unhybridised 2p orbital on each carbon is perpendicular to the plane of the ring. These six p orbitals overlap sideways (laterally) with their neighbours, creating a continuous delocalised π system above and below the ring.
σ bond formation: The sp² hybrid orbitals on adjacent carbon atoms (and between C and H) overlap end-on-end (head-on) along the internuclear axis to form strong σ bonds.

Key Takeaways

Benzene's planar structure, 120° bond angles, sp² hybridisation, and delocalised π system are fundamental concepts. Understanding the distinction between σ (end-on) and π (sideways) overlap is crucial for explaining arene reactivity.

Common Mistakes

Stating the bond angle is 109.5° (confusing with sp³ tetrahedral geometry).
Saying 'double bonds' instead of 'delocalised π system' or 'pi bonds'.
Describing p orbitals overlapping 'end-on' for π bonds.
Forgetting to specify that the ring is planar.

Things to Be Careful About

Use precise terminology: 'planar', 'trigonal planar', 'sp²', 'sideways/lateral overlap', 'end-on/head-on overlap'.
Ensure you address all four bullet points in the question prompt.

Techniques used

describe benzene molecular geometryexplain sp2 hybridisation and delocalisationdescribe sigma and pi bond formation in benzene

(b)

Fig. 5.1 shows two reactions of benzoic acid.

Fig. 5.1

3M

(i)

Suggest reagents and conditions for reaction 5 and for reaction 6 in Fig. 5.1.

reaction 5 ..........................................................................................................................

reaction 6 ..........................................................................................................................

DifficultyMedium

Worked solution

Answer

Reaction 5: $H_2$ with $Pt$ or $Ni$ catalyst and heat.

Reaction 6: $(CH_3)_2CHCOCl$ (isobutyryl chloride) with $AlCl_3$ catalyst and heat.

Final answer

Reaction 5: H2, Pt or Ni, heat. Reaction 6: (CH3)2CHCOCl, AlCl3, heat.

Detailed explanation

Background Concept

The benzene ring is unusually stable due to delocalisation, so it resists addition reactions that would break the π system. However, under forcing conditions, it can be hydrogenated to cyclohexane. Conversely, benzene readily undergoes electrophilic substitution reactions, such as Friedel-Crafts acylation, to attach alkyl or acyl groups to the ring without destroying the aromatic system.

Understanding the Question

Fig 5.1 shows two reactions of benzoic acid. Reaction 5 converts the benzene ring to a cyclohexane ring (ring hydrogenation). Reaction 6 adds an isobutyryl group ( $-COCH(CH_3)_2$ ) to the ring at the meta position (Friedel-Crafts acylation). We need to suggest the reagents and conditions for both.

Approach

For Reaction 5, recognise that converting an arene to an alicyclic ring requires catalytic hydrogenation. For Reaction 6, recognise that adding an acyl group ( $-COR$ ) to an arene is a Friedel-Crafts acylation, which requires an acyl chloride and a Lewis acid catalyst.

Step-by-Step Reasoning

Reaction 5: The product is cyclohexanecarboxylic acid. The benzene ring has been fully hydrogenated. Reagents for benzene hydrogenation are $H_2$ gas with a metal catalyst like $Pt$ or $Ni$ , and heat (since benzene is stable, higher temperatures/pressures are needed than for alkenes).
Reaction 6: The product has an isobutyryl group ( $-COCH(CH_3)_2$ ) added to the ring. This is an acylation. The reagent for Friedel-Crafts acylation is an acyl chloride. The acyl group is $-COCH(CH_3)_2$ , so the acyl chloride is $(CH_3)_2CHCOCl$ (isobutyryl chloride or 2-methylpropanoyl chloride). The catalyst is a Lewis acid, typically $AlCl_3$ (anhydrous), and heat is required.

Key Takeaways

Benzene ring hydrogenation requires $H_2$ / $Ni$ or $Pt$ / heat.
Friedel-Crafts acylation uses an acyl chloride ( $RCOCl$ ) and $AlCl_3$ catalyst.

Common Mistakes

Using $FeBr_3$ or $AlBr_3$ for acylation (these are for halogenation; $AlCl_3$ is standard for acylation/alkylation).
Forgetting the catalyst or heat.
Suggesting an alkyl halide for reaction 6 (that would give an alkyl group, not an acyl group).

Things to Be Careful About

Ensure the acyl chloride formula matches the group added: $-COCH(CH_3)_2$ comes from $(CH_3)_2CHCOCl$ .
State symbols are not strictly required here, but conditions (heat, catalyst) are essential marks.

Techniques used

identify reagents for benzene ring hydrogenationidentify reagents for Friedel-Crafts acylation

(ii)

State the type of reaction for reaction 5 in Fig. 5.1.

(c)

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

6M

(i)

Describe the directing effect of the $-\text{CH}_2\text{CH}_3$ group.

Explain your answer.

DifficultyMedium-Easy

Worked solution

Answer

The $-CH_2CH_3$ group directs incoming electrophiles to the 2-, 4- (and 6-) positions (ortho and para positions).
This is because the alkyl group is an electron-donating group (by induction/hyperconjugation), which increases the electron density at the ortho and para positions of the ring, making them more attractive to electrophiles.

Final answer

Directs to 2-, 4- (and 6-) positions; it is an electron-donating group.

Detailed explanation

Background Concept

Substituents already on a benzene ring influence where new substituents will attach in subsequent electrophilic substitution reactions. These are called directing groups. Electron-donating groups (EDGs) like alkyl groups activate the ring and direct ortho/para. Electron-withdrawing groups (EWGs) like $-NO_2$ or $-COOH$ deactivate the ring and direct meta.

Understanding the Question

The question asks about the directing effect of an ethyl group ( $-CH_2CH_3$ ) and requires an explanation based on electronic effects.

Approach

Recall that alkyl groups are activating and ortho/para-directing. Explain this by referencing their electron-donating nature via induction or hyperconjugation, which stabilises the intermediate carbocation at ortho/para positions.

Step-by-Step Reasoning

Directing effect: Alkyl groups like $-CH_2CH_3$ are ortho/para directors. This means substitution occurs at positions 2, 4, and 6 relative to the substituent.
Explanation: Alkyl groups push electron density into the ring through the sigma bonds (inductive effect) and through hyperconjugation. This increases the overall electron density of the ring (activating it) and specifically increases electron density at the ortho and para positions. Electrophiles are attracted to regions of high electron density, so they attack these positions.

Key Takeaways

Alkyl groups are ortho/para directing and activating.
The reason is electron donation (induction/hyperconjugation) increasing electron density at ortho/para positions.

Common Mistakes

Saying alkyl groups are meta-directing (confusing with EWGs).
Saying they are electron-withdrawing.
Not explaining the link between electron density and electrophile attraction.

Things to Be Careful About

The question asks to 'describe' (state the positions) and 'explain' (give the reason). Ensure both parts are answered.
'Electron-donating' is the key phrase required for the explanation.

Techniques used

determine directing effect of alkyl substituentexplain electron-donating nature of alkyl groups

(ii)

The alkylation of arenes uses a mixture of $\text{CH}_3\text{CH}_2\text{Br}$ and $\text{FeBr}_3$ to generate the $\text{CH}_3\text{CH}_2^+$ electrophile.

Write an equation for the formation of the $\text{CH}_3\text{CH}_2^+$ electrophile.

DifficultyMedium-Easy

Worked solution

Answer

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

Final answer

CH3CH2Br + FeBr3 -> CH3CH2+ + FeBr4-

Detailed explanation

Background Concept

In Friedel-Crafts alkylation, a strong electrophile (carbocation) is needed to attack the electron-rich benzene ring. Alkyl halides are not electrophilic enough on their own, so a Lewis acid catalyst like $FeBr_3$ or $AlCl_3$ is used to polarise the C-Br bond and generate the carbocation.

Understanding the Question

Write the equation showing how $CH_3CH_2Br$ reacts with $FeBr_3$ to form the $CH_3CH_2^+$ electrophile.

Approach

The Lewis acid ( $FeBr_3$ ) accepts a lone pair from the bromine in $CH_3CH_2Br$ , weakening the C-Br bond and causing it to heterolytically cleave, releasing $Br^-$ which complexes with $FeBr_3$ to form $FeBr_4^-$ , and leaving $CH_3CH_2^+$ .

Step-by-Step Reasoning

Reactants: $CH_3CH_2Br$ and $FeBr_3$ .
The Br atom has lone pairs. $FeBr_3$ is electron-deficient (Lewis acid).
The Br donates a lone pair to Fe, forming a complex.
The C-Br bond breaks heterolytically. Both electrons go to Br.
Products: $CH_3CH_2^+$ (ethyl carbocation electrophile) and $FeBr_4^-$ (tetrabromoferrate(III) ion).
Equation: $CH_3CH_2Br + FeBr_3 \rightarrow CH_3CH_2^+ + FeBr_4^-$

Key Takeaways

Lewis acid catalysts generate electrophiles from alkyl halides.
The byproduct is a complex halide ion like $FeBr_4^-$ or $AlCl_4^-$ .

Common Mistakes

Writing $FeBr_3$ on the right side of the equation (it is a catalyst, consumed in step 1 but regenerated later; in this specific equation showing electrophile formation, it is a reactant).
Forgetting the charge on the ions.
Writing $Br^-$ instead of $FeBr_4^-$ .

Things to Be Careful About

Ensure charges are balanced: left side 0, right side (+1) + (-1) = 0.
Use correct formulas for the complex ion: $FeBr_4^-$ .

Techniques used

write electrophile generation equation for Friedel-Crafts alkylation

(iii)

Complete the mechanism in Fig. 5.2.

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

Fig. 5.2

DifficultyMedium

Worked solution

Answer

Step 1 (Electrophilic attack):

Curly arrow from the centre of the benzene ring (or one of the double bonds) to the $C^+$ of the $^+CH_2CH_3$ electrophile.

Intermediate (Sigma complex / Wheland intermediate):

A cyclohexadienyl cation structure.
The benzene ring has two double bonds and one $sp^3$ hybridised carbon.
The $sp^3$ carbon is bonded to the original $H$ atom and the new $CH_2CH_3$ group.
The positive charge is delocalised within the ring (often shown as a '+' inside a broken circle or on one of the adjacent carbons).

Step 2 (Loss of proton / Rearomatisation):

Curly arrow from the C-H bond (on the $sp^3$ carbon) into the ring (to reform the double bond/delocalised system).
The other product is $H^+$ (and $FeBr_4^-$ is regenerated in the next step).

(See diagram for visual representation)

Final answer

Curly arrow from ring to C+; intermediate is cyclohexadienyl cation with CH2CH3 and H on same C; curly arrow from C-H bond into ring; product is H+.

Detailed explanation

Background Concept

Electrophilic substitution in arenes occurs in two main steps:

Electrophilic attack: The electrophile ( $E^+$ ) attacks the delocalised π system, forming a sigma bond with one carbon. This breaks the aromaticity, creating a positively charged intermediate called a sigma complex or Wheland intermediate. The carbon attacked becomes $sp^3$ hybridised.
Loss of proton: A base (often the catalyst complex like $FeBr_4^-$ ) removes the proton from the $sp^3$ carbon. The electrons from the C-H bond move back into the ring to restore the delocalised π system (rearomatisation), releasing $H^+$ (or $HBr$ if the base is $FeBr_4^-$ ).

Understanding the Question

Complete the mechanism for the electrophilic substitution of benzoic acid with the ethyl carbocation ( $^+CH_2CH_3$ ). We need to show the attack, draw the intermediate, and show the rearomatisation step.

Approach

Draw a curly arrow from the benzene π system to the electrophile.
Draw the intermediate: a ring with two double bonds, an $sp^3$ carbon bonded to H and $CH_2CH_3$ , and a positive charge delocalised in the ring.
Draw a curly arrow from the C-H bond to reform the π system, releasing $H^+$ .

Step-by-Step Reasoning

Attack: The benzene ring is electron-rich. Draw a curly arrow starting from inside the ring (representing the π electrons) pointing to the positive carbon of $^+CH_2CH_3$ . This forms a C-C bond.
Intermediate Structure: The carbon that was attacked is now $sp^3$ hybridised. It is bonded to the ring carbons, the original H atom, and the new $CH_2CH_3$ group. The ring now has only two double bonds (conjugated diene system) and a positive charge. In CIE mark schemes, drawing the '+' inside a broken circle within the ring is often accepted to show delocalisation, or placing it on an adjacent carbon. The key is showing the $sp^3$ carbon with both H and $CH_2CH_3$ .
Rearomatisation: To restore aromaticity, the H atom must leave as $H^+$ . Draw a curly arrow from the C-H bond (on the $sp^3$ carbon) pointing into the ring (towards the adjacent double bond or into the ring system). This reforms the π bond. The product shown is 3-ethylbenzoic acid (meta product, directed by the electron-withdrawing COOH group) and $H^+$ .

Key Takeaways

The intermediate in electrophilic substitution is a sigma complex (Wheland intermediate) with an $sp^3$ carbon.
Rearomatisation involves loss of $H^+$ and electrons moving back into the ring.
Curly arrows must show the movement of electron pairs.

Common Mistakes

Drawing the intermediate with three double bonds (wrong, aromaticity is lost).
Forgetting the H atom on the $sp^3$ carbon in the intermediate.
Drawing the curly arrow for rearomatisation from the C-C bond instead of the C-H bond.
Not showing the charge on the intermediate.

Things to Be Careful About

The question asks for 'all relevant curly arrows and charges'.
Ensure the intermediate structure matches the product: if the product is 3-ethylbenzoic acid (meta), the ethyl group and H must be at position 3 in the intermediate.
The blank space after the product is for $H^+$ .

Techniques used

draw curly arrows for electrophilic attack on benzene ringdraw Wheland intermediate structuredraw curly arrows for rearomatisation

(iv)

Write an equation to show how $\text{FeBr}_3$ is regenerated after the reaction in Fig. 5.2.

DifficultyMedium-Easy

Worked solution

Answer

\text{H}^+ + \text{FeBr}_4^- \rightarrow \text{HBr} + \text{FeBr}_3

Final answer

H+ + FeBr4- -> HBr + FeBr3

Detailed explanation

Background Concept

In electrophilic substitution using a Lewis acid catalyst (like $FeBr_3$ or $AlCl_3$ ), the catalyst is consumed in the first step to generate the electrophile, forming a complex ion (e.g., $FeBr_4^-$ ). In the final step, when the proton ( $H^+$ ) is lost from the intermediate to restore aromaticity, it reacts with the complex ion to release the catalyst and form the byproduct (e.g., $HBr$ or $HCl$ ).

Understanding the Question

Write the equation showing how $FeBr_3$ is regenerated after the reaction in Fig 5.2. The reaction produced $H^+$ and $FeBr_4^-$ (from the electrophile generation step).

Approach

The $H^+$ released in the rearomatisation step combines with the $FeBr_4^-$ complex ion. The $H^+$ takes a $Br^-$ from the complex, forming $HBr$ and leaving behind the regenerated $FeBr_3$ catalyst.

Step-by-Step Reasoning

Reactants: The proton $H^+$ (from the intermediate) and the tetrabromoferrate(III) ion $FeBr_4^-$ (from step 1).
Reaction: $H^+$ reacts with one of the $Br^-$ ligands on the iron complex.
Products: Hydrogen bromide ( $HBr$ ) and the regenerated catalyst, anhydrous iron(III) bromide ( $FeBr_3$ ).
Equation: $H^+ + FeBr_4^- \rightarrow HBr + FeBr_3$

Key Takeaways

Lewis acid catalysts are regenerated in the final step of electrophilic substitution.
The byproduct is a hydrogen halide (e.g., $HBr$ , $HCl$ ).

Common Mistakes

Writing $H^+ + Br^- \rightarrow HBr$ (forgetting that the $Br^-$ is part of the $FeBr_4^-$ complex in this context, though the net result is similar, the mark scheme specifically looks for the interaction with $FeBr_4^-$ ).
Forgetting to balance the equation (charges and atoms).
Writing $FeBr_3$ as a reactant instead of a product.

Things to Be Careful About

Ensure charges are balanced: left side (+1) + (-1) = 0; right side 0 + 0 = 0.
Use correct formulas: $FeBr_4^-$ , not $FeBr_3Br^-$ or similar.
State symbols are not required here.

Techniques used

write catalyst regeneration equation for Friedel-Crafts

Q6MediumIntroduction to A Level Organic ChemistryCarboxylic Acids and DerivativesNitrogen CompoundsAnalytical Techniques

(a)

Compound $\mathbf{Z}$ is used in organic synthesis.

Compound Z

Complete Table 6.1 to show the number of $\text{sp}$ , $\text{sp}^2$ and $\text{sp}^3$ hybridised carbon atoms present in one molecule of $\mathbf{Z}$ .

Table 6.1

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

DifficultyEasy

Worked solution

Answer

type of hybridisation	sp	sp²	sp³
number of carbon atoms	1	1	3

Final answer

sp = 1, sp² = 1, sp³ = 3

Detailed explanation

Background Concept

Carbon atoms can be hybridised as sp, sp², or sp³ depending on their bonding:

sp³: Carbon with four single bonds (tetrahedral geometry, ~109.5°).
sp²: Carbon with one double bond and two single bonds (trigonal planar, ~120°). This includes carbons in C=C, C=O, and aromatic rings.
sp: Carbon with one triple bond and one single bond, or two double bonds (linear, 180°). This includes nitriles (C≡N) and alkynes (C≡C).

Understanding the Question

We are given the structure of compound Z: H₂N–CO–CH₂–CH(NH₂)–CH₂–CN. We need to count how many carbons fall into each hybridisation category.

Approach

Identify the bonding environment for each carbon atom in the chain:

C in C=O (amide): Double bond to O, single to N, single to C. -> sp².
CH₂ (next to C=O): Four single bonds (2 to C, 2 to H). -> sp³.
CH (with NH₂): Four single bonds (1 to C, 1 to C, 1 to N, 1 to H). -> sp³.
CH₂ (next to CN): Four single bonds (1 to C, 1 to C, 2 to H). -> sp³.
C in CN (nitrile): Triple bond to N, single to C. -> sp.

Step-by-Step Reasoning

sp carbon: The carbon in the nitrile group (–C≡N). Count = 1.
sp² carbon: The carbonyl carbon in the amide group (H₂N–C(=O)–). Count = 1.
sp³ carbons: The three saturated carbons in the chain (–CH₂–, –CH–, –CH₂–). Count = 3.

Key Takeaways

Amide carbonyl carbons are sp² hybridised.
Nitrile carbons are sp hybridised.
Alkyl chain carbons with only single bonds are sp³ hybridised.

Common Mistakes

Forgetting that the nitrile carbon is sp hybridised.
Counting the nitrogen atoms or other atoms as carbons.
Misidentifying the amide carbon as sp³ (it has a double bond to oxygen).

Things to Be Careful About

Ensure you are only counting carbon atoms.
Remember that the carbon in C=O is sp², not sp³.

Techniques used

identify hybridisation from bonding environmentcount carbon atoms by type

(b)

$\mathbf{Z}$ can undergo different reactions, as shown in Fig. 6.1.

Fig. 6.1

5M

(i)

Name the two types of reaction occurring in reaction 7 in Fig. 6.1.

(ii)

Draw the structures of the organic products of reactions 7, 8 and 9 in Fig. 6.1.

DifficultyMedium

Worked solution

Answer

Reaction 7 (excess HCl, heat):
HOOC–CH₂–CH(NH₃⁺Cl⁻)–CH₂–COOH

Reaction 8 (CH₃COCl):
H₂N–CO–CH₂–CH(NHCOCH₃)–CH₂–CN

Reaction 9 (excess LiAlH₄):
H₂N–CH₂–CH₂–CH(NH₂)–CH₂–CH₂–NH₂

Final answer

See working for structures

Detailed explanation

Background Concept

Hydrolysis of amides/nitriles with acid: Produces carboxylic acid and ammonium salt.
Acylation of amines: Amines react with acyl chlorides (CH₃COCl) to form amides. The lone pair on nitrogen attacks the carbonyl carbon.
Reduction by LiAlH₄: A strong reducing agent.
- Reduces amides (R–CONH₂) to primary amines (R–CH₂NH₂).
- Reduces nitriles (R–CN) to primary amines (R–CH₂NH₂).
- Does not reduce isolated C=C or benzene rings, but reduces polar multiple bonds.

Understanding the Question

Draw the organic products for three reactions starting from Z (H₂N–CO–CH₂–CH(NH₂)–CH₂–CN).

Approach

Analyze each reagent's effect on the functional groups:

Reaction 7 (excess HCl, heat):
- Amide (H₂N–CO–) hydrolyses to carboxylic acid (HO–CO–) and NH₄⁺.
- Nitrile (–CN) hydrolyses to carboxylic acid (–COOH) and NH₄⁺.
- Amine (–NH₂) is protonated to –NH₃⁺Cl⁻.
- Product: HOOC–CH₂–CH(NH₃⁺Cl⁻)–CH₂–COOH.
Reaction 8 (CH₃COCl):
- Acyl chloride reacts with the most nucleophilic group: the aliphatic amine (–NH₂).
- Amide nitrogen is less nucleophilic due to resonance with C=O.
- Product: H₂N–CO–CH₂–CH(NHCOCH₃)–CH₂–CN.
Reaction 9 (excess LiAlH₄):
- Reduces amide (H₂N–CO–) to amine (H₂N–CH₂–).
- Reduces nitrile (–CN) to amine (–CH₂NH₂).
- Existing amine (–NH₂) is unchanged.
- Product: H₂N–CH₂–CH₂–CH(NH₂)–CH₂–CH₂–NH₂.

Step-by-Step Reasoning

Reaction 7: The acid hydrolyses both the amide and nitrile groups to carboxylic acids. The basic amine group picks up a proton from the excess acid to form a chloride salt. Structure: HOOC–CH₂–CH(NH₃⁺Cl⁻)–CH₂–COOH.
Reaction 8: The acyl chloride acylates the primary amine group (–NH₂) to form an amide (–NHCOCH₃). The amide and nitrile groups do not react under these conditions. Structure: H₂N–CO–CH₂–CH(NHCOCH₃)–CH₂–CN.
Reaction 9: LiAlH₄ reduces the carbonyl of the amide to a methylene group (–CH₂–) and the nitrile to a methylene group with an amine (–CH₂NH₂). The central amine is unaffected. Structure: H₂N–CH₂–CH₂–CH(NH₂)–CH₂–CH₂–NH₂.

Key Takeaways

Acid hydrolysis converts amides and nitriles to carboxylic acids and protonates amines.
Acyl chlorides acylate amines to amides.
LiAlH₄ reduces amides and nitriles to amines.

Common Mistakes

Forgetting to protonate the amine in reaction 7 (leaving it as –NH₂ instead of –NH₃⁺Cl⁻).
Reducing the amide incorrectly (e.g., to an alcohol instead of an amine).
Acylating the amide nitrogen in reaction 8 instead of the amine nitrogen.

Things to Be Careful About

Ensure structures are fully drawn with all atoms/bonds shown clearly.
In reaction 7, the product is a salt (chloride), so include Cl⁻ or write it as a zwitterion/salt form correctly.

Techniques used

predict products of acid hydrolysispredict acylation of aminespredict reduction by LiAlH4

(c)

Compound $\mathbf{Z}$ is dissolved in $\text{D}_2\text{O}$ and analysed by carbon-13 NMR and proton ( $^1\text{H}$ ) NMR spectroscopy.

3M

(i)

Predict the number of peaks in the carbon-13 NMR spectrum of $\mathbf{Z}$ .

(ii)

The proton ( $^1\text{H}$ ) NMR spectrum of $\mathbf{Z}$ in $\text{D}_2\text{O}$ gives three peaks for the proton environments, labelled $\mathbf{a}$ , $\mathbf{b}$ and $\mathbf{c}$ , as shown on Fig. 6.2.

Fig. 6.2

Complete Table 6.2 for the proton ( $^1\text{H}$ ) NMR spectrum of $\mathbf{Z}$ in $\text{D}_2\text{O}$ .

Table 6.2

proton environment	$\mathbf{a}$	$\mathbf{b}$	$\mathbf{c}$
name of splitting pattern
chemical shift range, $\delta / \text{ppm}$

Table 6.3

environment of proton	example	chemical shift range, $\delta / \text{ppm}$
alkane	$-\text{CH}_3$ , $-\text{CH}_2-$ , $>\text{CH}-$	$0.9-1.7$
alkyl next to $\text{C}=\text{O}$	$\text{CH}_3-\text{C}=\text{O}$ , $-\text{CH}_2-\text{C}=\text{O}$ , $>\text{CH}-\text{C}=\text{O}$	$2.2-3.0$
alkyl next to nitrile	$-\text{CH}_2-\text{CN}$	$2.0-3.0$
alkyl next to electronegative atom	$\text{CH}_3-\text{O}$ , $-\text{CH}_2-\text{O}$ , $-\text{CH}_2-\text{N}$	$3.2-4.0$
attached to alkene	$=\text{CHR}$	$4.5-6.0$
alkyl amine	$\text{R}-\text{NH}-$	$1.0-5.0$
amide	$\text{RCONHR}$	$5.0-12.0$

DifficultyMedium

Worked solution

Answer

proton environment	a	b	c
name of splitting pattern	doublet	multiplet	doublet
chemical shift range, $\delta$ / ppm	2.2–3.0	3.2–4.0	2.0–3.0

Final answer

See table

Detailed explanation

Background Concept

Splitting (n+1 rule): A proton environment is split by protons on adjacent carbons. If there are $n$ adjacent protons, the peak is split into $n+1$ peaks.
Chemical Shift ( $\delta$ ): Depends on the electronic environment. Electronegative atoms or pi systems deshield protons, shifting them downfield (higher ppm).
D₂O exchange: Protons attached to heteroatoms like O or N (e.g., –NH₂, –OH) exchange with deuterium in D₂O and disappear from the ¹H NMR spectrum (or appear as broad peaks that are often ignored in simple analysis). Amide protons can also exchange, but the question specifies only environments a, b, c are labelled and considered.

Understanding the Question

Complete the table for environments a, b, c in compound Z dissolved in D₂O. Use the provided data table for shifts.

Approach

Identify the splitting for each labelled CH group based on neighbours.
Identify the chemical shift range using the provided table and the specific group next to the CH.

Step-by-Step Reasoning

Environment a (–CH₂– next to C=O):
- ** neighbours**: The adjacent carbon is b (–CH–), which has 1 proton. So $n=1$ , splitting = $1+1 = 2$ (doublet).
- Shift: Table says "alkyl next to C=O" is 2.2–3.0 ppm.
Environment b (–CH– next to –NH₂):
- ** neighbours**: Adjacent carbons are a (–CH₂–, 2 protons) and c (–CH₂–, 2 protons). Total $n=4$ . Splitting = $4+1 = 5$ (quintet), but often described as a multiplet in mark schemes for complex coupling or when exact splitting isn't simple. The mark scheme accepts "multiplet".
- Shift: Table says "alkyl next to electronegative atom" (N is electronegative) is 3.2–4.0 ppm.
Environment c (–CH₂– next to –CN):
- ** neighbours**: Adjacent carbon is b (–CH–), which has 1 proton. So $n=1$ , splitting = $1+1 = 2$ (doublet).
- Shift: Table says "alkyl next to nitrile" is 2.0–3.0 ppm.
Note on D₂O: The –NH₂ protons (on the amine and amide) exchange with D₂O and do not appear as the labelled peaks a, b, c. The question asks for the table for a, b, c only.

Key Takeaways

Use the n+1 rule for splitting.
Use the provided table to match functional groups to shift ranges.
Remember D₂O removes exchangeable protons (NH, OH).

Common Mistakes

Incorrectly counting neighbours for splitting (e.g., counting protons on the same carbon).
Using the wrong shift range (e.g., using alkane range for a carbon next to C=O).
Forgetting that the amine protons exchange and don't show up (though the question guides us to a, b, c).

Things to Be Careful About

The shift for b is next to Nitrogen. The table gives "alkyl next to electronegative atom" as 3.2-4.0 ppm. Nitrogen is electronegative.
Splitting for b: neighbours are 2 (from a) + 2 (from c) = 4. 4+1=5 (quintet). However, mark schemes often accept "multiplet" for complex coupling patterns involving different coupling constants. Stick to the mark scheme's accepted answer: multiplet.

Techniques used

predict splitting from adjacent protonsuse chemical shift tablesaccount for D2O exchange

Q7MediumCarboxylic Acids and DerivativesHydroxy CompoundsPolymerisation

Phenylmethanol and 4-methylphenol are isomers.

phenylmethanol and 4-methylphenol

(a)

Complete Table 7.1 to show the relative acidities of benzoic acid ( $\text{C}_6\text{H}_5\text{COOH}$ ), phenylmethanol, 4-methylphenol and water.

Explain your answer.

Table 7.1

	name of compound
most acidic


least acidic

DifficultyMedium

Worked solution

Answer

Order of acidity (most to least acidic):

Position	Compound
Most acidic	Benzoic acid
2nd	4-methylphenol
3rd	Water
Least acidic	Phenylmethanol

Explanation:

Benzoic acid is most acidic because the negative charge on the carboxylate anion is delocalised (resonance) across two electronegative oxygen atoms, stabilising the conjugate base and making H⁺ more easily lost.
4-methylphenol is more acidic than water because the lone pair on oxygen in the phenoxide ion is delocalised into the benzene ring π system, stabilising the conjugate base.
Phenylmethanol is the least acidic because the $\text{CH}_2$ group (alkyl group) exerts a positive inductive effect (+I), destabilising the alkoxide conjugate base and making H⁺ less readily lost.

Final answer

Most acidic: benzoic acid > 4-methylphenol > water > phenylmethanol (least acidic). Benzoic acid: –COO⁻ charge delocalised over two O atoms. 4-methylphenol: lone pair on O delocalised into ring π system. Phenylmethanol: +I effect of alkyl group destabilises conjugate base.

Detailed explanation

Background Concept

Acidity in organic chemistry is determined by the stability of the conjugate base formed when a proton (H⁺) is lost. The more stable the conjugate base (anion), the stronger the acid. Two key factors stabilise anions:

Resonance (delocalisation): spreading the negative charge over multiple atoms, especially electronegative ones like oxygen.
Inductive effects: electron-withdrawing groups stabilise negative charge; electron-donating groups (such as alkyl groups) destabilise it.

For compounds with an O–H bond (carboxylic acids, phenols, alcohols, water), the acid strength depends on how well the resulting oxygen-centred anion is stabilised.

Understanding the Question

The question asks you to rank four compounds by acidity and explain the ranking. The compounds are:

Benzoic acid ( $\text{C}_6\text{H}_5\text{COOH}$ ) — a carboxylic acid
4-methylphenol — a phenol (–OH directly attached to the ring)
Water — the reference compound
Phenylmethanol — an alcohol (–OH on a $\text{CH}_2$ group attached to the ring, not directly on the ring)

The command word is "explain," so each ranking point must carry a structural reason.

Approach

Identify the conjugate base of each compound and assess its stability:

Carboxylate anion: charge shared over two oxygens by resonance → very stable → strongest acid.
Phenoxide anion: charge delocalised into the aromatic π system → moderately stable.
Hydroxide ion: no additional stabilisation beyond oxygen's electronegativity.
Alkoxide (from phenylmethanol): the $\text{CH}_2$ group has a +I effect that pushes electron density toward the already negatively charged oxygen → destabilises the anion → weakest acid.

Step-by-Step Reasoning

Benzoic acid (most acidic):
When benzoic acid loses H⁺, the benzoate anion forms. The negative charge is delocalised equally over two oxygen atoms through resonance. This is the most effective stabilisation among the four conjugate bases, so benzoic acid is the strongest acid.

4-methylphenol (second):
When 4-methylphenol loses H⁺, the 4-methylphenoxide anion forms. The lone pair on oxygen can overlap with the p orbitals of the aromatic ring, delocalising electron density into the π system. This provides significant (though less than carboxylate) stabilisation. Note: the methyl group at position 4 has a slight +I effect, but the dominant factor is the resonance stabilisation of the phenoxide.

Water (third):
Water loses H⁺ to form OH⁻. The negative charge is localised on one oxygen atom with no additional delocalisation or stabilisation. This is less stable than the phenoxide ion, so water is less acidic than 4-methylphenol.

Phenylmethanol (least acidic):
Phenylmethanol loses H⁺ to form $\text{C}_6\text{H}_5\text{CH}_2\text{O}^-$ . The negative charge is on a single oxygen with no resonance stabilisation into the ring (the $\text{CH}_2$ group acts as a spacer, preventing overlap of the oxygen lone pair with the ring π system). Additionally, the alkyl $\text{CH}_2$ group exerts a positive inductive effect (+I), pushing electron density toward the negatively charged oxygen, further destabilising the anion. This makes phenylmethanol the weakest acid — even weaker than water.

Key Takeaways

Carboxylic acids > phenols > water > alcohols is the general order of acidity for O–H compounds.
The key to explaining acidity is always the stability of the conjugate base.
Resonance delocalisation over two electronegative atoms (as in carboxylate) is the strongest stabilising factor.
Delocalisation into an aromatic ring (as in phenoxide) provides moderate stabilisation.
Alkyl groups destabilise anions via +I effect, making alcohols weaker acids than water.

Common Mistakes

Confusing phenylmethanol with a phenol: the –OH is NOT directly on the ring, so there is no resonance stabilisation of the conjugate base into the ring.
Saying the methyl group in 4-methylphenol makes it less acidic than phenol without acknowledging it is still more acidic than water — the question compares it to water, not to unsubstituted phenol.
Failing to link the explanation to conjugate base stability (the mark scheme requires this link explicitly).
Saying "the O–H bond is stronger" without explaining why in terms of anion stability.

Things to Be Careful About

The mark scheme requires a general statement about conjugate base stability (M2) AND two specific structural reasons (M3, M4). Missing the general link costs a mark even if the specific reasons are correct.
For phenylmethanol, the key point is the +I effect of the alkyl group destabilising the anion — do not say the ring stabilises it (it doesn't, because of the $\text{CH}_2$ spacer).
Use precise language: "delocalised across two oxygen atoms," "lone pair on oxygen overlaps with the ring π system," "positive inductive effect."

Techniques used

rank compounds by acidityexplain stabilisation of conjugate base by resonanceapply inductive effect arguments to conjugate base stability

(b)

4-methylphenol reacts readily with sodium.

Complete the equation for this reaction.

4-methylphenol reaction

DifficultyEasy

Worked solution

Answer

\text{4-methylphenol} + \text{Na} \rightarrow \text{sodium 4-methylphenoxide} + \tfrac{1}{2}\text{H}_2

In structural form: the –OH group is replaced by –ONa, and hydrogen gas is evolved.

Final answer

4-methylphenol + Na → sodium 4-methylphenoxide + 1/2 H₂

Detailed explanation

Background Concept

Phenols react with reactive metals such as sodium to produce the corresponding sodium phenoxide salt and hydrogen gas. This is analogous to the reaction of alcohols with sodium, but phenols react more readily because they are more acidic than alcohols (the phenoxide ion is resonance-stabilised).

The general reaction is:

2\text{ArOH} + 2\text{Na} \rightarrow 2\text{ArO}^-\text{Na}^+ + \text{H}_2

Understanding the Question

The question provides the skeletal structure of 4-methylphenol and asks you to complete the equation by adding the missing reactant (Na) and the products. The mark scheme shows the equation written with 1/2 H₂ (i.e. using one mole of phenol and one mole of sodium), which is an acceptable way to balance it.

Approach

Identify that sodium metal reacts with the acidic O–H proton of the phenol. The products are the sodium salt (sodium 4-methylphenoxide, where –OH becomes –ONa) and hydrogen gas.

Step-by-Step Reasoning

The O–H bond in phenol is sufficiently acidic to react with sodium metal.
Sodium displaces hydrogen from the hydroxyl group, forming the ionic compound sodium 4-methylphenoxide ( $\text{CH}_3\text{C}_6\text{H}_4\text{ONa}$ ).
The displaced hydrogen atoms combine to form $\text{H}_2$ gas.
The equation can be written with fractional coefficients (1/2 H₂) or doubled to give whole numbers.

Key Takeaways

Phenols react with sodium to give sodium phenoxide and hydrogen gas.
This reaction demonstrates the acidic nature of phenols (they react with metals like acids do).
The skeletal structure shows the –OH group converted to –ONa.

Common Mistakes

Forgetting to include hydrogen gas as a product.
Writing the product as a covalent structure rather than showing –ONa (the ionic salt).
Using incorrect balancing (though the mark scheme accepts the 1/2 H₂ form).

Things to Be Careful About

The mark scheme accepts the equation written with 1/2 H₂. If writing with whole numbers, use 2 phenol + 2 Na → 2 sodium phenoxide + H₂.
Ensure the methyl group position is maintained in the product structure.

Techniques used

write a balanced equation for phenol reacting with sodiumidentify the organic product as a sodium phenoxide salt

(c)

Under certain conditions, ethane-1,2-diol, $\text{HOCH}_2\text{CH}_2\text{OH}$ , reacts with propane-1,3-dioic acid, $\text{HOOCCH}_2\text{COOH}$ , to form different organic products, as shown in Fig. 7.1.

Fig. 7.1

4M

(i)

$\mathbf{X}$ does not react with $\text{Na}$ metal.

Draw the structure of the organic product $\mathbf{X}$ , $\text{C}_5\text{H}_6\text{O}_4$ , shown in Fig. 7.1.

DifficultyMedium

Worked solution

Answer

X is a cyclic diester (dilactone) formed by intramolecular double esterification of propane-1,3-dioic acid with ethane-1,2-diol. The structure is a nine-membered ring containing two ester linkages:

Molecular formula: $\text{C}_5\text{H}_6\text{O}_4$ ✓ (consistent with loss of 2 H₂O from $\text{C}_3\text{H}_4\text{O}_4 + \text{C}_2\text{H}_6\text{O}_2$ ).

Final answer

Cyclic diester (dilactone): a nine-membered ring with two ester groups, structure -C(=O)-CH₂-C(=O)-O-CH₂-CH₂-O- forming a closed ring. Molecular formula C₅H₆O₄.

Detailed explanation

Background Concept

When a dicarboxylic acid reacts with a diol, two types of product are possible:

Intermolecular reaction (polymerisation): many molecules join end-to-end to form a polyester.
Intramolecular reaction (cyclisation): one molecule of diacid reacts with one molecule of diol at both ends to form a cyclic diester (a dilactone).

The question states that X does not react with Na metal, confirming there are no free –OH or –COOH groups remaining — all four reactive groups have participated in ester formation.

Understanding the Question

Given: propane-1,3-dioic acid (HOOC–CH₂–COOH) + ethane-1,2-diol (HO–CH₂–CH₂–OH) → X, C₅H₆O₄.

The molecular formula C₅H₆O₄ tells us:

Starting materials: C₃H₄O₄ + C₂H₆O₂ = C₅H₁₀O₆
Product: C₅H₆O₄
Difference: H₄O₂ = 2 H₂O lost

This confirms two ester bonds have formed (each esterification loses one H₂O), and since only one molecule of each reactant is involved, the product must be cyclic.

Approach

Calculate the number of water molecules lost: (C₅H₁₀O₆) − (C₅H₆O₄) = 2 H₂O.
Two ester bonds formed from one diacid + one diol → cyclic diester.
Draw the ring: connect both –COOH groups to both –OH groups, forming a ring.

Step-by-Step Reasoning

The ring is formed as follows:

One –COOH of the diacid reacts with one –OH of the diol → ester bond + H₂O
The other –COOH reacts with the other –OH → second ester bond + H₂O
The result is a closed ring: –C(=O)–CH₂–C(=O)–O–CH₂–CH₂–O– (cyclic)

Counting atoms in the ring: 5 C, 6 H, 4 O = C₅H₆O₄ ✓

The ring size: C(=O) + CH₂ + C(=O) + O + CH₂ + CH₂ + O = 7 atoms in the ring backbone (9-membered ring if counting all atoms including the carbonyl carbons and oxygens).

Key Takeaways

A cyclic diester (dilactone) forms when a diacid and diol undergo intramolecular double esterification.
The molecular formula allows you to determine how many water molecules are lost and hence how many bonds form.
The absence of reaction with Na confirms no free –OH or –COOH groups remain.

Common Mistakes

Drawing an open-chain diester instead of a cyclic structure.
Incorrectly counting atoms and getting the wrong ring size.
Forgetting that the –CH₂– between the two carbonyls must be included in the ring.

Things to Be Careful About

The ring must contain BOTH ester linkages (–C(=O)–O–) and the –CH₂– spacer from the diacid.
Ensure the molecular formula matches: C₅H₆O₄.

Techniques used

deduce a cyclic diester structure from molecular formulaapply condensation reaction logic between a diacid and a diol

(ii)

Reactions 10 and 11 in Fig. 7.1 are different types of reaction.

Name the type of reaction for reaction 10 and for reaction 11.

reaction 10 ........................................................................................................................

reaction 11 ........................................................................................................................

DifficultyMedium-Easy

Worked solution

Answer

Reaction 10: Condensation (esterification) / addition–elimination

Reaction 11: Condensation polymerisation (polyesterification) / addition–elimination

Final answer

Reaction 10: condensation (esterification); Reaction 11: condensation polymerisation

Detailed explanation

Background Concept

Both reactions involve the formation of ester bonds between a carboxylic acid and an alcohol, which is a condensation reaction (small molecule, H₂O, is eliminated) proceeding via an addition–elimination mechanism.

The difference lies in the product:

Reaction 10 produces a single cyclic molecule (X) — this is a simple condensation/esterification.
Reaction 11 produces a polymer (Y) — this is a condensation polymerisation, where many monomer units join end-to-end, each linkage formed by loss of water.

Understanding the Question

The question asks you to name the type of reaction for each. Both involve ester bond formation, but one forms a discrete cyclic product and the other forms a long-chain polymer.

Approach

Identify the common mechanism (condensation/esterification/addition-elimination) and then distinguish by whether the product is a single molecule or a polymer.

Step-by-Step Reasoning

Reaction 10: A diacid reacts with a diol to form a cyclic diester. This is a condensation reaction (esterification). The mechanism is addition–elimination at the carbonyl carbon.

Reaction 11: The same monomers react intermolecularly, with each –COOH reacting with an –OH of a different molecule, building a long chain. This is condensation polymerisation (specifically, polyesterification). The mechanism at each linkage is still addition–elimination.

Key Takeaways

Condensation and condensation polymerisation share the same mechanism (addition–elimination) but differ in product size.
Both are types of esterification.
The mark scheme accepts multiple valid names for each reaction type.

Common Mistakes

Saying reaction 10 is 'addition' (it is addition–elimination, or condensation).
Not distinguishing between simple condensation and condensation polymerisation.

Things to Be Careful About

The mark scheme requires BOTH names to be correct for the single mark. Acceptable answers include: condensation, esterification, addition–elimination for reaction 10; condensation polymerisation, polyesterification, addition–elimination for reaction 11.

Techniques used

distinguish between intramolecular cyclisation and intermolecular polymerisationname esterification reaction types

(iii)

Draw a section of polymer $\mathbf{Y}$ showing only one repeat unit.

The new functional group formed should be displayed.

DifficultyMedium-Easy

Worked solution

Answer

The repeat unit is: $-\text{C}(=\text{O})-\text{CH}_2-\text{C}(=\text{O})-\text{O}-\text{CH}_2-\text{CH}_2-\text{O}-$ with continuation bonds (dashed lines) at both ends.

Final answer

Repeat unit: -C(=O)-CH₂-C(=O)-O-CH₂-CH₂-O- with continuation bonds at both ends, ester linkage displayed showing C=O and C-O-C

Detailed explanation

Background Concept

A condensation polymer forms when bifunctional monomers react repeatedly, each reaction eliminating a small molecule (water). The repeat unit of a polyester contains the ester linkage (–COO–) formed between the diacid and the diol.

For propane-1,3-dioic acid + ethane-1,2-diol:

The diacid contributes: –C(=O)–CH₂–C(=O)–
The diol contributes: –O–CH₂–CH₂–O–
Together they form the repeat unit with ester links at both ends.

Understanding the Question

Draw ONE repeat unit of polymer Y. The ester functional group must be displayed (show the C=O and C–O–C bonds explicitly, not abbreviated as –COO–). Continuation bonds (short dashed lines) must appear at both ends to show the chain continues.

Approach

Identify the ester linkage formed between –COOH and –OH.
Build the repeat unit by connecting the diacid residue to the diol residue.
Display the ester group fully and add continuation bonds.

Step-by-Step Reasoning

Starting from the monomers:

HOOC–CH₂–COOH loses two H (from –OH of each COOH)
HO–CH₂–CH₂–OH loses two OH (from each –OH)
The remaining fragments join: –C(=O)–CH₂–C(=O)–O–CH₂–CH₂–O–

This is the repeat unit. The ester linkage is –C(=O)–O– and must be shown explicitly with the double bond on carbon and the single bond to oxygen.

Key Takeaways

The repeat unit of a polyester shows the full ester linkage with C=O displayed.
Continuation bonds at both ends indicate the polymer chain extends in both directions.
The repeat unit contains contributions from BOTH monomers.

Common Mistakes

Not showing the ester linkage explicitly (writing –COO– instead of –C(=O)–O–).
Forgetting continuation bonds at both ends.
Including the eliminated water molecule in the structure.
Drawing the repeat unit with incorrect number of carbons (the diacid contributes 3 carbons, the diol contributes 2 carbons).

Things to Be Careful About

The mark scheme awards M1 for showing the ester linkage correctly and M2 for the rest of the structure being correct with continuation bonds.
The ester group must be 'displayed' — meaning show the C=O double bond and the C–O single bond explicitly.
Ensure the –CH₂– from the diacid is between the two carbonyl groups.

Techniques used

draw a polyester repeat unit with ester linkage displayedinclude continuation bonds at both ends of the repeat unit

Q8Medium-HardHalogen CompoundsIntroduction to A Level Organic ChemistryCarboxylic Acids and DerivativesNitrogen CompoundsAnalytical Techniques

(a)

Describe the difference in reactivity between ethanoyl chloride and chlorobenzene with water.

Explain your answer.

DifficultyMedium

Worked solution

Answer

Chlorobenzene is less reactive than ethanoyl chloride towards water.

In chlorobenzene, the lone pair on the chlorine atom overlaps with the $\pi$ system of the benzene ring (delocalisation into the ring), giving the C–Cl bond partial double-bond character and strengthening it, making it harder to break.

In ethanoyl chloride, the electron-withdrawing C=O group makes the carbonyl carbon highly electron-deficient and weakens the C–Cl bond, making it more susceptible to nucleophilic attack by water.

Final answer

Chlorobenzene is less reactive because delocalisation of the Cl lone pair into the ring gives partial double-bond character to C–Cl, strengthening it. Ethanoyl chloride has an electron-deficient carbonyl carbon and a weakened C–Cl bond.

Detailed explanation

Background Concept

The reactivity of a C–Cl bond towards nucleophilic substitution depends on two key factors: (1) the strength of the C–Cl bond itself, and (2) the electrophilicity of the carbon atom to which the chlorine is attached. In halogenoarenes such as chlorobenzene, the lone pair on the halogen can overlap with the $\pi$ -electron system of the aromatic ring through p-orbital conjugation. This delocalisation gives the C–Cl bond partial double-bond character, making it shorter and stronger than a normal C–Cl single bond. In acyl chlorides such as ethanoyl chloride, the adjacent carbonyl group is strongly electron-withdrawing due to the electronegative oxygen, which polarises the C–Cl bond and makes the carbonyl carbon highly electrophilic.

Understanding the Question

The question asks for both a description of the difference in reactivity (which compound reacts more readily with water) and an explanation of why. The command word "describe" requires the observation, while "explain" requires the underlying electronic reasoning. Two marks are available for any two of the three key points.

Approach

First, state which is more reactive. Then provide the electronic explanation for chlorobenzene's low reactivity (lone pair delocalisation into the ring → partial double bond → stronger C–Cl bond). Finally, contrast with ethanoyl chloride (electron-withdrawing C=O → electron-deficient carbon → weakened C–Cl bond → susceptible to nucleophilic attack).

Step-by-Step Reasoning

Observation: Chlorobenzene does not react with water under normal conditions, whereas ethanoyl chloride reacts vigorously (fuming, producing HCl and ethanoic acid). So chlorobenzene is less reactive.
Explanation for chlorobenzene: The chlorine atom has lone pairs in p orbitals that can overlap side-on with the $\pi$ system of the benzene ring. This conjugation delocalises electron density from Cl into the ring, giving the C–Cl bond partial double-bond character. A partial double bond is shorter and stronger than a pure single bond, so it is much harder to break, and nucleophilic substitution does not occur readily.
Explanation for ethanoyl chloride: The carbonyl group (C=O) contains a highly electronegative oxygen that withdraws electron density from the carbon. This makes the carbonyl carbon strongly $\delta+$ (electron-deficient), attracting nucleophiles like water. Additionally, the C–Cl bond is weakened because the carbon is already electron-poor. The combination of a strong electrophilic centre and a good leaving group (Cl⁻) makes hydrolysis rapid.

Key Takeaways

Delocalisation of a halogen lone pair into an aromatic ring strengthens the C–X bond and reduces reactivity towards nucleophilic substitution.
The electron-withdrawing effect of a carbonyl group activates the adjacent C–Cl bond towards nucleophilic attack.
Reactivity differences between acyl chlorides and halogenoarenes are fundamentally about bond strength and electrophilicity.

Common Mistakes

Saying chlorobenzene is unreactive because "the C–Cl bond is in a ring" without mentioning delocalisation.
Confusing the direction of electron flow: it is the lone pair on Cl that delocalises INTO the ring, not the ring electrons moving towards Cl.
Failing to mention that the partial double-bond character makes the C–Cl bond stronger/harder to break.
Not contrasting both compounds — the mark scheme requires at least two of the three points.

Things to Be Careful About

Use precise terminology: "delocalisation," "partial double-bond character," "electron-deficient," "electron-withdrawing."
The mark scheme accepts either "p orbital/lone pair on Cl overlaps/delocalises into the ring" OR "C of COCl is most electron deficient / has electronegative oxygen / electron-withdrawing C=O group" as the second point — you need any two of the three listed points for full marks.

Techniques used

compare reactivity of acyl chloride with halogenoareneexplain delocalisation of lone pair into aromatic ringrelate bond strength to reactivity

(b)

The structure of compound $\mathbf{V}$ is shown.

Compound V

4M

(i)

Name all the functional groups in $\mathbf{V}$ .

DifficultyMedium-Easy

Worked solution

Answer

The functional groups present in compound V are:

Nitrile ( $-\text{C}\equiv\text{N}$ )
Amide ( $-\text{CONH}-$ )
Ketone ( $-\text{CO}-$ )
Ester (cyclic ester / lactone, $-\text{COO}-$ )

Final answer

Nitrile, amide, ketone, ester

Detailed explanation

Background Concept

Functional group identification from skeletal structures is a fundamental skill in organic chemistry. Each functional group has a characteristic arrangement of atoms: a nitrile is a carbon triple-bonded to nitrogen ( $-\text{C}\equiv\text{N}$ ); an amide has a carbonyl adjacent to a nitrogen ( $-\text{CONH}-$ ); a ketone has a carbonyl bonded to two carbon groups ( $-\text{CO}-$ ); and an ester has a carbonyl bonded to an oxygen that is in turn bonded to another carbon ( $-\text{COO}-$ ). When an ester is part of a ring, it is called a lactone.

Understanding the Question

The question asks you to name ALL functional groups in compound V. Looking at the skeletal structure, you must systematically identify each distinct functional group. The mark scheme requires any two for one mark and all four for two marks.

Approach

Scan the molecule from left to right: the $-\text{CN}$ group is a nitrile; the $-\text{C}(=\text{O})\text{NH}-$ linkage is an amide; the $-\text{C}(=\text{O})-$ between two carbons is a ketone; the five-membered ring containing $-\text{O}-\text{C}(=\text{O})-$ is a cyclic ester (lactone).

Step-by-Step Reasoning

Nitrile: The $-\text{CN}$ group attached to the cyclohexyl ring. This is a carbon triple-bonded to nitrogen.
Amide: The $-\text{C}(=\text{O})\text{NH}-$ group connecting the cyclohexyl ring to the $\text{CH}_2$ chain. A carbonyl directly bonded to nitrogen defines an amide.
Ketone: The $\text{C}=\text{O}$ group in the middle of the chain, bonded to the $\text{CH}_2$ on one side and the ring carbon on the other. A carbonyl flanked by two carbon groups is a ketone.
Ester (lactone): The five-membered ring contains $-\text{O}-\text{C}(=\text{O})-$ , which is an ester linkage within a ring (a cyclic ester, or lactone).

Key Takeaways

A lactone is simply a cyclic ester — the functional group is still "ester."
Distinguishing amide from ketone: an amide has $\text{C}=\text{O}$ bonded to N; a ketone has $\text{C}=\text{O}$ bonded to two C atoms.
Systematic scanning of the molecule prevents missing functional groups.

Common Mistakes

Calling the cyclic ester a "cyclic ketone" — the ring oxygen makes it an ester, not a ketone.
Confusing the amide carbonyl with a ketone carbonyl — the adjacent nitrogen distinguishes them.
Writing "carbonyl" instead of "ketone" or "amide" — "carbonyl" is a general term, not a specific functional group name.

Things to Be Careful About

The mark scheme accepts "carbonyl" as an alternative to "ketone," but "ester" must be specified (not just "carbonyl" for the lactone).
Do not list "benzene ring" or "alkane" — these are not functional groups in the A-Level sense.

Techniques used

identify functional groups from a skeletal structurerecognise amide, nitrile, ketone, and ester linkages

(ii)

Deduce the number of possible optical isomers for $\mathbf{V}$ .

DifficultyMedium

Worked solution

Answer

Compound V has 3 chiral centres (the two carbons on the cyclohexyl ring bearing the $-\text{CN}$ and $-\text{CONH}-$ groups, and the ring carbon of the lactone bearing the ketone group).

Number of optical isomers $= 2^3 = 8$

Answer

8

Final answer

8

Detailed explanation

Background Concept

A chiral centre (asymmetric carbon) is a carbon atom bonded to four different groups. For a molecule with $n$ chiral centres and no internal symmetry (no meso compounds), the number of optical isomers (stereoisomers) is $2^n$ . Each chiral centre can be $R$ or $S$ , giving $2^n$ combinations.

Understanding the Question

The question asks for the number of possible optical isomers of compound V. This requires identifying all chiral centres in the molecule and applying the $2^n$ rule. One mark is available.

Approach

Examine each carbon in the molecule and determine whether it is bonded to four different groups. Focus on the ring carbons and the lactone ring carbon, as these are the likely chiral centres.

Step-by-Step Reasoning

Cyclohexyl ring carbon bearing $-\text{CN}$ : This carbon is bonded to: (i) $-\text{CN}$ , (ii) H, (iii) the ring path going towards the $-\text{CONH}-$ substituent, and (iv) the ring path going the other way around. Since the two ring paths are different (one leads to the amide-substituted carbon, the other does not), this carbon has four different groups → chiral centre.
Cyclohexyl ring carbon bearing $-\text{CONH}-$ : This carbon is bonded to: (i) $-\text{CONH}-$ , (ii) H, (iii) the ring path towards the $-\text{CN}$ substituent, and (iv) the ring path going the other way. The two ring paths are different → chiral centre.
Lactone ring carbon bearing the ketone group: This carbon is bonded to: (i) the ketone $\text{C}=\text{O}$ group, (ii) H, (iii) the $-\text{O}-$ of the ester linkage, and (iv) the $-\text{CH}_2-$ of the ring. All four groups are different → chiral centre.
Total: 3 chiral centres → $2^3 = 8$ optical isomers.

Key Takeaways

Ring carbons can be chiral if the two paths around the ring are different (i.e., the ring is asymmetrically substituted).
The $2^n$ rule applies when there is no internal plane of symmetry (no meso forms).
Always check each potential chiral centre by listing all four groups attached.

Common Mistakes

Missing the lactone ring carbon as a chiral centre because it is part of a ring.
Thinking the cyclohexyl ring carbons are not chiral because "they're in a ring" — they ARE chiral if the two paths around the ring differ.
Calculating $2^2 = 4$ by missing one chiral centre.
Confusing optical isomers with geometric isomers.

Things to Be Careful About

The question says "possible optical isomers" — this includes all stereoisomers arising from chiral centres, not just enantiomers.
No meso compound is possible here because the three chiral centres are all in different environments.

Techniques used

identify chiral centres in a complex moleculecalculate number of optical isomers using 2^n rule

(iii)

Suggest one reason, other than better biological activity and lower dosage required, why it is beneficial to synthesise a single optical isomer of $\mathbf{V}$ for use as a drug.

(c)

A sample of $\mathbf{V}$ is hydrolysed with an excess of hot aqueous alkali.

The products are isolated from the reaction mixture at $\text{pH } 12$ .

Draw the structures of the two organic products of the complete alkaline hydrolysis of $\mathbf{V}$ in Fig. 8.1.

Fig. 8.1

DifficultyMedium-Hard

Worked solution

Answer

At pH 12, all acidic groups are deprotonated. Complete alkaline hydrolysis cleaves three bonds:

The amide bond hydrolyses → $-\text{COO}^-$ (on the cyclohexyl ring) and $\text{H}_2\text{N}-$ (on the chain)
The nitrile hydrolyses → $-\text{COO}^-$ (on the cyclohexyl ring)
The ester (lactone) hydrolyses → ring opens to give $-\text{COO}^-$ and $-\text{OH}$

Product 1 (from the cyclohexyl portion):

A cyclohexane ring with two adjacent $-\text{COO}^-$ groups (one from nitrile hydrolysis, one from amide hydrolysis).

Product 2 (from the chain + lactone portion):

$\text{H}_2\text{N}-\text{CH}_2-\text{C}(=\text{O})-\text{CH}(\text{OH})-\text{CH}_2-\text{CH}_2-\text{COO}^-$

(The ketone remains unchanged; the amine is free at pH 12; the lactone ring has opened to give an alcohol and a carboxylate.)

Final answer

Product 1: cyclohexane-1,2-dicarboxylate (two COO⁻ groups on adjacent ring carbons). Product 2: H₂N-CH₂-C(=O)-CH(OH)-CH₂-CH₂-COO⁻

Detailed explanation

Background Concept

Complete alkaline hydrolysis (hot excess NaOH) cleaves multiple hydrolysable functional groups simultaneously. Amides hydrolyse to give a carboxylate and an amine. Nitriles hydrolyse (via an amide intermediate) to give a carboxylate. Esters hydrolyse to give a carboxylate and an alcohol. At pH 12, all carboxylic acid groups exist as carboxylate anions ( $-\text{COO}^-$ ), and amines remain as free bases ( $-\text{NH}_2$ ). Ketones are not hydrolysed under these conditions.

Understanding the Question

Compound V contains four functional groups: nitrile, amide, ketone, and ester (lactone). The question asks for the TWO organic products of COMPLETE alkaline hydrolysis at pH 12. Three marks are available: one for the amide bond cleavage, one for the nitrile hydrolysis, and one for the ester/lactone hydrolysis.

Approach

Identify each hydrolysable bond and determine where the molecule splits. The amide bond is the cleavage point that divides the molecule into two fragments. Then within each fragment, identify further hydrolysis products.

Step-by-Step Reasoning

Identify the amide bond cleavage: The $-\text{C}(=\text{O})\text{NH}-$ bond breaks. On the left side, the carbonyl becomes $-\text{COO}^-$ (still attached to the cyclohexyl ring). On the right side, the nitrogen becomes $\text{H}_2\text{N}-$ (now the terminus of the chain fragment).
Nitrile hydrolysis (left fragment): The $-\text{CN}$ on the cyclohexyl ring is hydrolysed by hot alkali to $-\text{COO}^-$ . This gives a cyclohexane ring with two adjacent $-\text{COO}^-$ groups.
Lactone hydrolysis (right fragment): The cyclic ester opens. The $-\text{C}(=\text{O})-\text{O}-$ bond of the lactone breaks, giving $-\text{COO}^-$ on one end and $-\text{OH}$ on the other. The ring carbon that bore the OH now has an $-\text{OH}$ group.
Ketone unchanged: The $\text{C}=\text{O}$ in the middle of the chain is a ketone and does not hydrolyse.
Final products:
- Product 1: Cyclohexane ring with $-\text{COO}^-$ at position 1 (from nitrile) and $-\text{COO}^-$ at position 2 (from amide).
- Product 2: $\text{H}_2\text{N}-\text{CH}_2-\text{C}(=\text{O})-\text{CH}(\text{OH})-\text{CH}_2-\text{CH}_2-\text{COO}^-$

Key Takeaways

Nitriles hydrolyse to carboxylates under hot alkaline conditions (via amide intermediate, but complete hydrolysis gives $-\text{COO}^-$ ).
Lactones are cyclic esters and hydrolyse by ring-opening to give a hydroxy-carboxylate.
At pH 12, always write $-\text{COO}^-$ not $-\text{COOH}$ , and $-\text{NH}_2$ not $-\text{NH}_3^+$ .
Ketones are stable to alkaline hydrolysis.

Common Mistakes

Writing $-\text{COOH}$ instead of $-\text{COO}^-$ at pH 12.
Forgetting to hydrolyse the nitrile (it does hydrolyse under hot alkaline conditions).
Drawing three products instead of two — the molecule splits into exactly two fragments at the amide bond.
Hydrolysing the ketone — ketones do not undergo hydrolysis.
Writing $\text{NH}_3^+$ instead of $\text{NH}_2$ at pH 12.

Things to Be Careful About

The question specifies pH 12, so all acidic protons are removed — carboxyl groups must be drawn as $\text{COO}^-$ .
The amine product must be drawn as $\text{H}_2\text{N}-$ (free base), not protonated.
The lactone ring must be shown as OPENED — draw the chain with $-\text{OH}$ and $-\text{COO}^-$ at the former ring positions.
Each of the three hydrolysis events (amide, nitrile, ester) earns one mark.

Techniques used

predict products of complete alkaline hydrolysisidentify bond cleavage sites in amide, nitrile, and esterdraw organic products at high pH as carboxylate salts

(d)

A polypeptide formed from four amino acids, $\mathbf{A}$ , $\mathbf{B}$ , $\mathbf{C}$ and $\mathbf{D}$ , is completely hydrolysed and then analysed by gas–liquid chromatography.

The chromatogram produced is shown in Fig. 8.2.

Fig. 8.2

The number above each peak represents the area under the peak.

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

2M

(i)

Calculate the percentage by mass of amino acid $\mathbf{A}$ in the original mixture.

(ii)

The retention time for amino acid $\mathbf{D}$ is the longest.

Explain why $\mathbf{D}$ has a longer retention time than the other amino acids $\mathbf{A}$ , $\mathbf{B}$ and $\mathbf{C}$ .