Magnesium nitrate, $\text{Mg(NO}_3)_2$, and strontium nitrate, $\text{Sr(NO}_3)_2$, both decompose when heated to form the metal oxide and a mixture of gases. (a) Write an equation for the thermal decomposition of $\text{Mg(NO}_3)_2$. (b) State which of $\text{Mg(NO}_3)_2$ or $\text{Sr(NO}_3)_2$ decomposes at a lower temperature. Explain your answer. compound that decomposes at a lower temperature ............................................................... explanation…

(a) Mg(NO3)2 → MgO + 2NO2 + ½O2 · (b) Mg(NO3)2; Mg2+ is smaller / higher charge density, polarises nitrate more · (c)(i) MgSO4 and H2O · (c)(ii) A (MgSO4) is more soluble because the difference in hydration enthalpy outweighs the difference in lattice enthalpy, giving a more exothermic ΔH_sol.

Asparagine and aspartic acid are two naturally occurring amino acids. Their structures and isoelectric points are shown in Table 8.1. Table 8.1 amino acid structure isoelectric point --- --- --- asparagine $\text{HOOCCH(NH}_2)\text{CH}_2\text{CONH}_2$ 5.41 aspartic acid $\text{HOOCCH(NH}_2)\text{CH}_2\text{COOH}$ 2.77 (a) Define isoelectric point. (b) Draw the structures of asparagine and aspartic acid at pH 2. (c) Asparagine and aspartic acid are treated separately with an…

(a) The pH at which an amino acid exists as a zwitterion · (b) Asparagine: HOOCCH(NH3+)CH2CONH2; Aspartic acid: HOOCCH(NH3+)CH2COOH · (c) Asparagine: HOCH2CH(NH2)CH2CH2NH2; Aspartic acid: HOCH2CH(NH2)CH2CH2OH · (d)(i) HOOCCH2COOH + 2SOCl2 -> ClOCCH2COCl + 2SO2 + 2HCl · (d)(ii) HOOCCH(CH2CONH2)NHCOCH2CONHCH(CH2CONH2)COOH · (e) -OOCCH(NH2)CH2COO- · (f) Polymer chain: -[NH-CH(CH2CONH2)-CO]-[NH-CH(CH2CONH2)-CO]-[NH-CH(CH2CONH2)-CO]- · (g)(i) Similarities: Rotate light, same…

Compound X is made from benzene by the route shown in Fig. 9.1. (a) Describe the bonding in benzene, $\text{C}_6\text{H}_6$. Your answer should include: the hybridisation of the six carbon atoms the types of bond between the carbon atoms the orbitals that overlap to produce the bonds between the carbon atoms the type of bond between the carbon atoms and the hydrogen atoms the orbitals that overlap to produce the bonds between the carbon atoms and the hydrogen atoms. (b)…

(a) All C atoms are sp² hybridised; C-C bonds are σ (sp²-sp² overlap) and π (p-p overlap); C-H bonds are σ (sp²-s overlap). · (b) CH3Cl and AlCl3 · (c) See diagram · (d) Concentrated HNO3 and concentrated H2SO4, 25-60 C · (e) Alkaline KMnO4 · (f) 4-nitrobenzoic acid · (g) Hot concentrated HCl and Sn · (h) Perform step 3 before step 2; COOH is meta-directing (3-directing).

9701/42 Oct/Nov 2025 — Chemistry Solved Past Paper

Q1MediumGroup 2

Magnesium nitrate, $\text{Mg(NO}_3)_2$ , and strontium nitrate, $\text{Sr(NO}_3)_2$ , both decompose when heated to form the metal oxide and a mixture of gases.

(a)

Write an equation for the thermal decomposition of $\text{Mg(NO}_3)_2$ .

1M Similar

DifficultyEasy

Worked solution

Answer

$\text{Mg(NO}_3)_2 \rightarrow \text{MgO} + 2\text{NO}_2 + \frac{1}{2}\text{O}_2$

(or the doubled version: $2\text{Mg(NO}_3)_2 \rightarrow 2\text{MgO} + 4\text{NO}_2 + \text{O}_2$ )

Final answer

Mg(NO3)2 → MgO + 2NO2 + ½O2

Detailed explanation

Background Concept

Group 2 nitrates decompose on heating to give the metal oxide, nitrogen dioxide, and oxygen. The general equation is:

$\text{M(NO}_3)_2 \rightarrow \text{MO} + 2\text{NO}_2 + \frac{1}{2}\text{O}_2$

The nitrate ion is decomposed by the polarising effect of the metal cation, which distorts the electron cloud of the nitrate ion, weakening the N–O bonds.

Understanding the Question

The question asks for a balanced equation for the thermal decomposition of magnesium nitrate. This is a recall task—you need to know the general form of the decomposition of a Group 2 nitrate and apply it to magnesium.

Approach

Write the general decomposition equation for a Group 2 nitrate and substitute Mg for M. Balance the equation by counting atoms on both sides.

Step-by-Step Reasoning

The general equation for the thermal decomposition of a Group 2 nitrate is:

$\text{M(NO}_3)_2 \rightarrow \text{MO} + 2\text{NO}_2 + \frac{1}{2}\text{O}_2$

Substituting M = Mg:

$\text{Mg(NO}_3)_2 \rightarrow \text{MgO} + 2\text{NO}_2 + \frac{1}{2}\text{O}_2$

Check the balance:

Mg: 1 on each side ✓
N: 2 on each side (2 in the nitrate, 2 in 2NO₂) ✓
O: 6 on the left (2 × 3), 1 + 4 + 1 = 6 on the right ✓

Alternatively, multiply the whole equation by 2 to avoid the fraction:

$2\text{Mg(NO}_3)_2 \rightarrow 2\text{MgO} + 4\text{NO}_2 + \text{O}_2$

Key Takeaways

Group 2 nitrates decompose to oxide + NO₂ + O₂.
Both the fractional and doubled equations are acceptable.
State symbols are not required here but would be (s), (g), (g) respectively.

Common Mistakes

Writing NO instead of NO₂ (NO₂ is brown, NO is colourless).
Forgetting to balance oxygen.
Writing Mg(NO₃)₂ → MgO + N₂ + O₂ (wrong—nitrogen is released as NO₂).

Things to Be Careful About

The equation must be balanced.
Either the fractional or whole-number version is accepted.
If state symbols are included, they must be correct: Mg(NO₃)₂(s) → MgO(s) + 2NO₂(g) + ½O₂(g).

Techniques used

write a balanced decomposition equationbalance the equation for a Group 2 nitrate decomposition

(b)

State which of $\text{Mg(NO}_3)_2$ or $\text{Sr(NO}_3)_2$ decomposes at a lower temperature.

Explain your answer.

compound that decomposes at a lower temperature ...............................................................

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

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Compound that decomposes at a lower temperature: $\text{Mg(NO}_3)_2$

Explanation:

$\text{Mg}^{2+}$ is smaller than $\text{Sr}^{2+}$ and has a higher charge density.
The $\text{Mg}^{2+}$ ion distorts/polarises the nitrate ion more strongly, weakening the N–O bonds, so $\text{Mg(NO}_3)_2$ decomposes at a lower temperature.

Final answer

Mg(NO3)2; Mg2+ is smaller / higher charge density, polarises nitrate more

Detailed explanation

Background Concept

Thermal stability of Group 2 nitrates decreases as you go down the group. This is due to the polarising power of the metal cation. A small, highly charged cation (high charge density) distorts the electron cloud of the nitrate ion more strongly, weakening the N–O bonds and making the nitrate easier to decompose. As you go down Group 2, the cation becomes larger with a lower charge density, so it polarises the nitrate less, making the nitrate more stable and requiring a higher temperature to decompose.

Understanding the Question

The question asks which nitrate—magnesium nitrate or strontium nitrate—decomposes at a lower temperature, and to explain why. This is a "state and explain" question: you must name the compound AND give the reason based on polarisation.

Approach

Recall the trend in thermal stability of Group 2 nitrates: it increases down the group. Therefore, the nitrate with the smaller cation (Mg) decomposes at a lower temperature. Explain using charge density and polarisation of the nitrate ion.

Step-by-Step Reasoning

Mg²⁺ is smaller than Sr²⁺ (Mg is above Sr in Group 2).
Mg²⁺ therefore has a higher charge density (charge per unit volume).
The higher charge density of Mg²⁺ means it polarises (distorts) the electron cloud of the nitrate ion more strongly.
This weakens the N–O bonds in the nitrate ion, making it less thermally stable.
Therefore, Mg(NO₃)₂ decomposes at a lower temperature than Sr(NO₃)₂.

Key Takeaways

Thermal stability of Group 2 nitrates increases down the group.
The explanation is based on charge density and polarisation of the anion.
The same argument applies to Group 2 carbonates.

Common Mistakes

Saying Sr(NO₃)₂ decomposes at a lower temperature (wrong—it's the opposite).
Giving only "Mg is smaller" without mentioning charge density or polarisation.
Confusing this with the trend for hydroxides (which become MORE soluble down the group).

Things to Be Careful About

The mark scheme requires both M1 (smaller/higher charge density) and M2 (nitrate ion more distorted/polarised).
"Charge density" is the key term—don't just say "smaller".
The polarisation weakens the N–O bond—this is the mechanism for the lower decomposition temperature.

Techniques used

compare ionic radii and charge densityapply polarisation of the nitrate ion

(c)

Magnesium oxide, $\text{MgO}$ , and strontium oxide, $\text{SrO}$ , both react with dilute sulfuric acid.

$\text{MgO}$ forms a soluble salt, A.

$\text{SrO}$ forms an insoluble salt, B.

4M

(i)

Identify the products formed when $\text{MgO}$ reacts with dilute sulfuric acid.

1M Similar

(ii)

Explain why A is more soluble than B.

3M Similar

DifficultyMedium

Worked solution

Answer

$\text{MgSO}_4$ (A) has a more exothermic $\Delta H_{\text{latt}}$ and a more exothermic $\Delta H_{\text{hyd}}$ than $\text{SrSO}_4$ (B).
The difference in $\Delta H_{\text{hyd}}$ is greater than the difference in $\Delta H_{\text{latt}}$ .
Therefore $\text{MgSO}_4$ (A) has a more exothermic $\Delta H_{\text{sol}}$ , so it is more soluble.

Final answer

A (MgSO4) is more soluble because the difference in hydration enthalpy outweighs the difference in lattice enthalpy, giving a more exothermic ΔH_sol.

Detailed explanation

Background Concept

The solubility of an ionic compound depends on the balance between lattice enthalpy (energy required to separate the ions in the solid) and hydration enthalpy (energy released when the ions are hydrated by water). The enthalpy of solution, ΔH_sol, is given by:

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

For a salt to dissolve, ΔH_sol should be exothermic (or at least not too endothermic). The more exothermic the ΔH_sol, the more soluble the salt tends to be.

As you go down Group 2, the sulfates become LESS soluble. This is because the hydration enthalpy decreases (becomes less exothermic) more rapidly than the lattice enthalpy decreases. For MgSO₄, the small Mg²⁺ ion has a very high hydration enthalpy, which makes ΔH_sol exothermic and the salt soluble. For SrSO₄, the larger Sr²⁺ ion has a lower hydration enthalpy, so the difference between ΔH_latt and ΔH_hyd is smaller, making ΔH_sol less exothermic (or even endothermic), so the salt is insoluble.

Understanding the Question

The question states that MgO reacts with dilute sulfuric acid to form a soluble salt A (MgSO₄), while SrO forms an insoluble salt B (SrSO₄). The question asks why A is more soluble than B. This requires comparing the lattice and hydration enthalpies of the two sulfates.

Approach

Recognise that solubility depends on ΔH_sol = ΔH_latt + ΔH_hyd.
For MgSO₄ (A), both ΔH_latt and ΔH_hyd are more exothermic than for SrSO₄ (B).
The key point is that the difference in ΔH_hyd is GREATER than the difference in ΔH_latt.
Therefore, ΔH_sol for MgSO₄ is more exothermic, making it more soluble.

Step-by-Step Reasoning

Mg²⁺ is smaller than Sr²⁺, so it has a higher charge density.
This means MgSO₄ has a more exothermic ΔH_latt than SrSO₄ (more energy released when the lattice forms).
Similarly, Mg²⁺ has a more exothermic ΔH_hyd than Sr²⁺ (more energy released when the ions are hydrated).
The difference in ΔH_hyd is greater than the difference in ΔH_latt.
Therefore, ΔH_sol = ΔH_latt + ΔH_hyd is more exothermic for MgSO₄ than for SrSO₄.
A more exothermic ΔH_sol means MgSO₄ is more soluble.

Key Takeaways

Solubility of Group 2 sulfates decreases down the group.
This is explained by the balance between lattice and hydration enthalpies.
The hydration enthalpy difference dominates, making the smaller cation's salt more soluble.

Common Mistakes

Saying "MgSO₄ has a more exothermic ΔH_latt" alone is not enough—you must mention the ΔH_hyd difference being greater.
Confusing the trend for sulfates (decreasing solubility down the group) with the trend for hydroxides (increasing solubility down the group).
Forgetting that ΔH_sol = ΔH_latt + ΔH_hyd.

Things to Be Careful About

The mark scheme requires all three points: more exothermic ΔH_latt and ΔH_hyd, the difference in ΔH_hyd is greater, and therefore more exothermic ΔH_sol.
"Exothermic" is the key term—use it precisely.
The explanation must link the enthalpy terms to solubility.

Techniques used

compare lattice and hydration enthalpiesdeduce which enthalpy difference dominates solubility

Q2Medium-EasyHydrocarbonsNitrogen CompoundsReaction Kinetics

Ethanal, $\text{CH}_3\text{CHO}$ , reacts with nitrogen dioxide, $\text{NO}_2$ . The products of the first step of this reaction are a $\text{CH}_3\dot{\text{C}}=\text{O}$ radical and a molecule of nitrous acid, $\text{HNO}_2$ .

\text{CH}_3\text{CHO} + \text{NO}_2 \rightarrow \text{CH}_3\dot{\text{C}}=\text{O} + \text{HNO}_2

(a)

3M

(i)

Use two words to complete the sentence.

This reaction involves ........................................ ........................................ of the single covalent bond between a hydrogen atom and a carbon atom in $\text{CH}_3\text{CHO}$ .

1M Similar

DifficultyEasy

Worked solution

Answer

homolytic fission

Final answer

homolytic fission

Detailed explanation

Background Concept

In organic chemistry, covalent bonds can break in two ways: homolytically or heterolytically. Homolytic fission occurs when the shared pair of electrons in a covalent bond is split equally between the two atoms, with each atom taking one electron. This process produces radicals (species with an unpaired electron). Homolytic fission is typically initiated by UV light or high temperatures and is the defining bond-breaking step in radical substitution mechanisms, such as the side-chain halogenation of alkanes or arenes.

Heterolytic fission, by contrast, involves the bond breaking unequally, with both electrons going to one atom to form ions (a cation and an anion). This is characteristic of polar reactions like nucleophilic substitution or electrophilic addition.

Understanding the Question

The question describes the first step of a reaction between ethanal ( $\text{CH}_3\text{CHO}$ ) and nitrogen dioxide ( $\text{NO}_2$ ). The products are a $\text{CH}_3\dot{\text{C}}=\text{O}$ radical and $\text{HNO}_2$ . We are asked to identify the specific term for the breaking of the C–H bond in ethanal that leads to these products.

Approach

Look at the products: one is a radical (indicated by the single dot $\dot{\text{C}}$ ). Since a radical is formed, the bond must have broken by sharing one electron with each fragment. This is the definition of homolytic fission.

Step-by-Step Reasoning

The reactant is $\text{CH}_3\text{CHO}$ . The bond broken is the single covalent bond between the hydrogen atom and the carbon atom (the C–H bond on the carbonyl carbon).
The products are $\text{CH}_3\dot{\text{C}}=\text{O}$ (an acetyl radical) and $\text{HNO}_2$ .
In $\text{CH}_3\dot{\text{C}}=\text{O}$ , the carbon atom has an unpaired electron. This means the hydrogen atom took one electron from the C–H bond, and the carbon atom took the other.
Equal splitting of a bonding pair to form radicals is called homolytic fission.

Key Takeaways

Radicals are formed by homolytic fission.
Ions are formed by heterolytic fission.
Always look at the products to determine the type of bond breaking: if radicals are produced, it is homolytic.

Common Mistakes

Writing "homolytic cleavage" instead of "homolytic fission" (both are acceptable, but "fission" is the standard mark scheme term).
Confusing it with "heterolytic fission" or "ionic bond breaking".
Writing "breakage" or "splitting" without the specific chemical terminology "homolytic fission".

Things to Be Careful About

The question asks for two words. "Homolytic fission" is exactly two words. "Homolysis" is one word and would not score.
Ensure spelling is correct: homolytic (not homolytical or homolyticly).

Techniques used

identify homolytic fission in radical mechanisms

(ii)

The hydrogen atom mentioned in (a)(i) forms a covalent bond with one of the oxygen atoms of an $\text{NO}_2$ molecule. An $\text{NO}_2$ molecule has a single, unpaired electron on the nitrogen atom. All electrons are paired in an $\text{HNO}_2$ molecule.

Draw dot-and-cross diagrams of $\text{NO}_2$ and $\text{HNO}_2$ in the boxes. Show outer shell electrons only.

Q2a_dotcross

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

NO₂:

Nitrogen atom (N) in the center.
One oxygen atom (O) double-bonded to N (sharing 4 electrons: 2 dots and 2 crosses).
The other oxygen atom (O) single-bonded to N (sharing 2 electrons: 1 dot and 1 cross, representing a coordinate bond from O to N).
N has one unpaired electron (a single dot or cross).
Each O has 3 lone pairs (6 non-bonding electrons).

HNO₂:

H atom single-bonded to an O atom (sharing 2 electrons).
This O atom is single-bonded to the N atom (sharing 2 electrons).
The N atom is double-bonded to the other O atom (sharing 4 electrons).
All electrons are paired; no unpaired electrons.
Each O has appropriate lone pairs to complete its octet (2 lone pairs on the O bonded to H, 2 lone pairs on the O double-bonded to N, 1 lone pair on N if considering formal charges, but typically N has 1 lone pair in HNO₂ structure: H–O–N=O with a lone pair on N).

(See diagram below for the exact dot-and-cross representation)

Final answer

See diagram description: NO₂ has N with one unpaired electron, double bond to one O, single coordinate bond to other O. HNO₂ has H-O-N=O structure with all electrons paired.

Detailed explanation

Background Concept

Dot-and-cross diagrams (Lewis structures) show the arrangement of valence (outer shell) electrons in a molecule. Dots (•) and crosses (×) are used to distinguish which atom contributed each electron, though in modern conventions, either symbol can be used as long as bonding pairs are clearly shown.

Covalent bond: A shared pair of electrons (one from each atom, or two from one atom in a coordinate/dative bond).
Radical: A species with an unpaired electron. In dot-and-cross diagrams, this is shown as a single dot or cross not paired with another.
Octet rule: Main group elements (like N and O) tend to form bonds to achieve 8 electrons in their outer shell (except H, which needs 2).

Understanding the Question

We need to draw dot-and-cross diagrams for $\text{NO}_2$ (nitrogen dioxide) and $\text{HNO}_2$ (nitrous acid), showing outer shell electrons only.

Key information given:

$\text{NO}_2$ : Single unpaired electron on the nitrogen atom.
$\text{HNO}_2$ : All electrons are paired. The H atom forms a covalent bond with one of the oxygen atoms.

Approach

Count the total outer shell electrons for each molecule.
Determine the connectivity (which atoms are bonded to which).
Distribute electrons to satisfy the octet rule (or duet for H), placing any remaining electrons as lone pairs or unpaired electrons.
Use dots for one atom's electrons and crosses for another's to show the bonding clearly.

Step-by-Step Reasoning

For $\text{NO}_2$ :

N has 5 outer shell electrons. Each O has 6. Total = $5 + 2(6) = 17$ electrons.
Since 17 is odd, there must be one unpaired electron (a radical).
Connectivity: O–N–O.
Draw a double bond between N and one O (4 electrons shared). Draw a single coordinate bond (or single covalent bond) between N and the other O (2 electrons shared).
N has the unpaired electron.
Distribute remaining electrons as lone pairs: each O gets 3 lone pairs (6 electrons). N has no lone pairs (5 valence - 4 in bonds - 1 unpaired = 0).
Diagram description: N in center. Left O: 3 lone pairs, double bond to N (2 dots, 2 crosses). Right O: 3 lone pairs, single bond to N (1 dot, 1 cross, coordinate from O). N: 1 unpaired electron.

For $\text{HNO}_2$ :

H has 1, N has 5, two O's have 12. Total = $1 + 5 + 12 = 18$ electrons.
All electrons paired (even number).
Connectivity given: H is bonded to an O. So we have H–O–N=O.
H–O bond: 2 electrons (1 from H, 1 from O).
O–N bond: 2 electrons (both from O, coordinate bond, or 1 from each).
N=O bond: 4 electrons (2 from N, 2 from O).
Lone pairs: O (bonded to H) has 2 lone pairs. N has 1 lone pair. O (double bonded to N) has 2 lone pairs.
Total electrons: 2 (H-O) + 4 (O lone pairs) + 2 (O-N) + 2 (N lone pair) + 4 (N=O) + 4 (O lone pairs) = 18. Correct.
Diagram description: H sharing a pair with O. That O sharing a pair with N. N sharing a double bond (two pairs) with the other O. Lone pairs placed to complete octets.

Key Takeaways

Odd number of outer shell electrons = radical (unpaired electron).
Coordinate bonds can be shown as both electrons coming from one atom (e.g., O to N in $\text{NO}_2$ ).
Always check the total electron count to ensure the diagram is correct.

Common Mistakes

Forgetting the unpaired electron on N in $\text{NO}_2$ .
Drawing $\text{NO}_2$ with a double bond to both oxygens and no unpaired electron (this would be 18 electrons, which is incorrect for $\text{NO}_2$ ; that's $\text{NO}_2^-$ or $\text{N}_2\text{O}_4$ ).
Drawing H bonded to N in $\text{HNO}_2$ instead of O. The question states H bonds to an oxygen atom.
Including inner shell electrons (the question specifies "outer shell electrons only").

Things to Be Careful About

The mark scheme allows either dots or crosses for bonding pairs, but consistency is key.
For $\text{NO}_2$ , the single bond to the second oxygen is often represented as a coordinate bond (both electrons from O), but a standard covalent bond (one from each) is also acceptable as long as the electron count and unpaired electron are correct.
In $\text{HNO}_2$ , ensure N has a lone pair to complete its octet (5 valence - 3 bonds - 0 unpaired = 1 lone pair).

Techniques used

draw dot-and-cross diagrams for radicalsshow outer shell electrons only

(iii)

Use VSEPR theory to predict the bond angle at the nitrogen atom in an $\text{HNO}_2$ molecule.

bond angle = ..............................

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

$115\text{--}120^\circ$

Final answer

115-120 degrees

Detailed explanation

Background Concept

VSEPR (Valence Shell Electron Pair Repulsion) theory predicts the geometry of molecules based on the repulsion between electron domains (bonding pairs, lone pairs, and sometimes single electrons in radicals) around a central atom. Electron domains arrange themselves as far apart as possible to minimize repulsion.

2 domains: Linear, $180^\circ$ .
3 domains: Trigonal planar, $120^\circ$ .
4 domains: Tetrahedral, $109.5^\circ$ .

Effect of lone pairs and multiple bonds: Lone pairs occupy more space than bonding pairs, so they compress bond angles. For example, in $\text{NH}_3$ (4 domains, 1 lone pair), the angle is $107^\circ$ instead of $109.5^\circ$ . In $\text{H}_2\text{O}$ (4 domains, 2 lone pairs), the angle is $104.5^\circ$ . Double bonds also exert slightly more repulsion than single bonds.

Understanding the Question

We need to predict the bond angle at the nitrogen atom in $\text{HNO}_2$ (nitrous acid) using VSEPR theory.

Approach

Determine the Lewis structure of $\text{HNO}_2$ to identify the electron domains around the central nitrogen atom.
Count the number of bonding domains (single, double, or triple bonds each count as one domain) and lone pairs.
Determine the electron geometry and then the molecular geometry.
Predict the bond angle, accounting for the compression caused by the lone pair.

Step-by-Step Reasoning

Lewis structure of $\text{HNO}_2$ : As established in part (a)(ii), the structure is H–O–N=O, with a lone pair on nitrogen.
Electron domains around N:
- One single bond to O (H–O–N): 1 domain.
- One double bond to O (N=O): 1 domain.
- One lone pair on N: 1 domain.
- Total domains = 3.
Electron geometry: 3 domains arrange in a trigonal planar geometry with an ideal angle of $120^\circ$ .
Molecular geometry: With 2 bonding domains and 1 lone pair, the shape is bent (or angular).
Bond angle prediction: The lone pair repels the bonding pairs more strongly than they repel each other, compressing the O–N–O bond angle to less than $120^\circ$ . Typically, for a trigonal planar electron geometry with one lone pair (like in $\text{SO}_2$ or ozone $\text{O}_3$ ), the angle is around $115\text{--}120^\circ$ . (In $\text{SO}_2$ , it is $119^\circ$ ; in $\text{O}_3$ , it is $116.8^\circ$ ). For $\text{HNO}_2$ , the accepted range is $115\text{--}120^\circ$ .

Key Takeaways

Count electron domains (bonds + lone pairs) around the central atom.
A double bond counts as ONE domain in VSEPR theory.
Lone pairs compress bond angles below the ideal value for that electron geometry.
For 3 domains (trigonal planar electron geometry) with 1 lone pair, the angle is slightly less than $120^\circ$ (typically $115\text{--}120^\circ$ ).

Common Mistakes

Counting the double bond as two domains (this would give 4 domains and a tetrahedral angle, which is wrong).
Forgetting the lone pair on nitrogen in $\text{HNO}_2$ .
Stating the angle is exactly $120^\circ$ without acknowledging the compression from the lone pair.
Confusing the electron geometry (trigonal planar) with the molecular shape (bent) and giving the wrong angle.

Things to Be Careful About

The question asks for the bond angle at the nitrogen atom, not the oxygen atom. The angle at the oxygen (H–O–N) would be different (around $104\text{--}109^\circ$ due to 2 lone pairs on O).
Give a range or a value within the accepted range ( $115\text{--}120^\circ$ ). Giving exactly $120^\circ$ may not score if the mark scheme requires recognition of lone pair compression.
Use the degree symbol ( $^\circ$ ) in the final answer.

Techniques used

apply VSEPR theory to predict bond angles

(b)

The rate equation for the reaction between $\text{CH}_3\text{CHO}$ and $\text{NO}_2$ is shown.

\text{rate} = k[\text{CH}_3\text{CHO}][\text{NO}_2]

Under certain conditions, when the concentrations of both $\text{CH}_3\text{CHO}$ and $\text{NO}_2$ are $0.200 \text{ mol dm}^{-3}$ , the rate of the reaction is $1.53 \times 10^{-4} \text{ mol dm}^{-3} \text{ s}^{-1}$ .

Calculate the value of the rate constant, $k$ , under these conditions. Give the units of $k$ .

$k$ = .............................. units ..............................

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Rate equation: $\text{rate} = k[\text{CH}_3\text{CHO}][\text{NO}_2]$

Rearrange for $k$ :

k = \frac{\text{rate}}{[\text{CH}_3\text{CHO}][\text{NO}_2]}

Substitute the given values:

$\text{rate} = 1.53 \times 10^{-4} \text{ mol dm}^{-3} \text{ s}^{-1}$
$[\text{CH}_3\text{CHO}] = 0.200 \text{ mol dm}^{-3}$
$[\text{NO}_2] = 0.200 \text{ mol dm}^{-3}$

k = \frac{1.53 \times 10^{-4}}{0.200 \times 0.200}

k = \frac{1.53 \times 10^{-4}}{0.0400}

k = 3.825 \times 10^{-3}

Units of $k$ :

\text{units} = \frac{\text{mol dm}^{-3} \text{ s}^{-1}}{(\text{mol dm}^{-3})(\text{mol dm}^{-3})} = \text{mol}^{-1} \text{ dm}^3 \text{ s}^{-1}

Answer

$k = 3.825 \times 10^{-3}$ (or $0.003825$ )
units = $\text{mol}^{-1} \text{ dm}^3 \text{ s}^{-1}$

Final answer

3.825e-3 mol^-1 dm^3 s^-1

Detailed explanation

Background Concept

The rate equation (or rate law) expresses the relationship between the rate of a reaction and the concentrations of the reactants. For a general reaction $a\text{A} + b\text{B} \rightarrow \text{products}$ , the rate equation is:

\text{rate} = k[\text{A}]^m[\text{B}]^n

where:

$k$ is the rate constant (specific to a given temperature).
$m$ and $n$ are the orders of reaction with respect to A and B, respectively.
The overall order is $m + n$ .

The units of $k$ depend on the overall order of the reaction. They can be derived by rearranging the rate equation and substituting the units of rate ( $\text{mol dm}^{-3} \text{ s}^{-1}$ ) and concentration ( $\text{mol dm}^{-3}$ ).

Understanding the Question

We are given:

Rate equation: $\text{rate} = k[\text{CH}_3\text{CHO}][\text{NO}_2]$
$[\text{CH}_3\text{CHO}] = 0.200 \text{ mol dm}^{-3}$
$[\text{NO}_2] = 0.200 \text{ mol dm}^{-3}$
$\text{rate} = 1.53 \times 10^{-4} \text{ mol dm}^{-3} \text{ s}^{-1}$

We need to calculate the value of $k$ and its units.

Approach

Rearrange the rate equation to solve for $k$ .
Substitute the numerical values and calculate $k$ .
Derive the units of $k$ by substituting the units into the rearranged equation.

Step-by-Step Reasoning

Calculating the value of $k$ :

Rearrange: $k = \frac{\text{rate}}{[\text{CH}_3\text{CHO}][\text{NO}_2]}$
Substitute: $k = \frac{1.53 \times 10^{-4}}{0.200 \times 0.200}$
Calculate the denominator: $0.200 \times 0.200 = 0.0400$
Calculate $k$ : $k = \frac{1.53 \times 10^{-4}}{0.0400} = 3.825 \times 10^{-3}$

Deriving the units of $k$ :

Units of rate = $\text{mol dm}^{-3} \text{ s}^{-1}$
Units of concentration = $\text{mol dm}^{-3}$
Substitute into the rearranged equation: $\text{units of } k = \frac{\text{mol dm}^{-3} \text{ s}^{-1}}{(\text{mol dm}^{-3})(\text{mol dm}^{-3})}$
Simplify: $\text{units of } k = \frac{\text{mol dm}^{-3} \text{ s}^{-1}}{\text{mol}^2 \text{ dm}^{-6}} = \text{mol}^{-1} \text{ dm}^3 \text{ s}^{-1}$ Alternatively, this can be written as $\text{dm}^3 \text{ mol}^{-1} \text{ s}^{-1}$ .

Key Takeaways

Always rearrange the rate equation before substituting values.
The units of $k$ can be found by dimensional analysis: $\text{units of } k = \frac{\text{units of rate}}{(\text{units of concentration})^{\text{overall order}}}$ .
For a second-order reaction (overall order = 2), the units of $k$ are always $\text{concentration}^{-1} \text{ time}^{-1}$ (e.g., $\text{mol}^{-1} \text{ dm}^3 \text{ s}^{-1}$ or $\text{M}^{-1} \text{ s}^{-1}$ ).

Common Mistakes

Forgetting to square the concentration in the denominator (i.e., calculating $0.200 \times 2$ instead of $0.200 \times 0.200$ ).
Writing the wrong units for $k$ . A common error is writing $\text{mol}^{-1} \text{ s}^{-1}$ (missing the $\text{dm}^3$ ) or $\text{dm}^3 \text{ mol} \text{ s}^{-1}$ (wrong sign on the mol exponent).
Using the wrong order of operations in the calculation.
Not giving the answer to an appropriate number of significant figures (3 s.f. is appropriate here, so $3.83 \times 10^{-3}$ is acceptable, but $3.825 \times 10^{-3}$ is also fine).

Things to Be Careful About

The mark scheme accepts $3.825 \times 10^{-3}$ , $0.003825$ , $0.00383$ , $0.0038$ , or $3.8 \times 10^{-3}$ . Any of these is correct.
Units must be exactly $\text{mol}^{-1} \text{ dm}^3 \text{ s}^{-1}$ . Writing $\text{M}^{-1} \text{ s}^{-1}$ is not accepted in Cambridge exams unless specified; always use $\text{mol dm}^{-3}$ based units.
Ensure the rate constant value is positive.

Techniques used

calculate rate constant from rate equationdetermine units of rate constant

(c)

The reaction mixture described in (b) is monitored over a period of time.

Predict whether the graph of $[\text{NO}_2]$ against time shows a constant half-life.

Explain your answer.

prediction ..................................................................................................................................

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

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

prediction: not constant half-life
explanation: because the overall order of the reaction is second order (or because the rate depends on the concentration of two reactants, so as concentration decreases, the rate decreases more rapidly than in a first-order reaction).

Final answer

not constant half-life because overall second order

Detailed explanation

Background Concept

The half-life ( $t_{1/2}$ ) of a reaction is the time taken for the concentration of a reactant to fall to half of its initial value.

For a first-order reaction, the half-life is constant. It is independent of the initial concentration: $t_{1/2} = \frac{\ln 2}{k} \approx \frac{0.693}{k}$ .
For a second-order reaction (or any reaction with overall order > 1), the half-life is not constant. As the reaction proceeds and the concentration of reactants decreases, the rate of reaction decreases, so it takes longer and longer for the concentration to halve. For a second-order reaction with rate equation $\text{rate} = k[\text{A}]^2$ , $t_{1/2} = \frac{1}{k[\text{A}]_0}$ , which is inversely proportional to the initial concentration.

Understanding the Question

We are asked to predict whether the graph of $[\text{NO}_2]$ against time for the reaction in part (b) shows a constant half-life, and to explain the answer.

From part (b), the rate equation is $\text{rate} = k[\text{CH}_3\text{CHO}][\text{NO}_2]$ . This means the reaction is first order with respect to $\text{CH}_3\text{CHO}$ and first order with respect to $\text{NO}_2$ , giving an overall order of 2 (second order).

Approach

Identify the overall order of the reaction from the rate equation.
Recall the relationship between reaction order and half-life constancy.
State the prediction and provide the explanation.

Step-by-Step Reasoning

The rate equation is $\text{rate} = k[\text{CH}_3\text{CHO}][\text{NO}_2]$ .
The overall order is $1 + 1 = 2$ (second order).
Only first-order reactions have a constant half-life. For second-order reactions, the half-life increases as the concentration decreases.
Therefore, the graph of $[\text{NO}_2]$ against time will not show a constant half-life.
Explanation: The reaction is overall second order (or the rate depends on the concentration of reactants, which decrease over time, causing the rate to decrease and the half-life to increase).

Key Takeaways

Constant half-life is a unique characteristic of first-order reactions.
For any reaction with overall order > 1, the half-life is not constant (it increases as concentration decreases).
Always determine the overall order from the rate equation before making predictions about half-life.

Common Mistakes

Assuming all reactions have a constant half-life.
Saying "the concentration decreases, so the rate decreases" without explicitly linking this to the overall order or the definition of half-life.
Confusing the half-life of a reactant with the half-life of a product.

Things to Be Careful About

The prediction must be clear: "not constant half-life" or "the half-life is not constant".
The explanation must mention the overall order (second order) or the fact that the rate depends on concentration (which is implied by the order). Simply saying "because the concentration decreases" is not sufficient; you must link it to the order of reaction.
Do not say "the half-life increases" without being asked; just predict whether it is constant or not.

Techniques used

relate reaction order to half-life constancy

(d)

$\text{NO}_2$ also reacts with ozone, $\text{O}_3$ .

The rate equation is shown.

\text{rate} = k_1[\text{NO}_2]

Under certain conditions, the value of $k_1$ is $0.0848 \text{ s}^{-1}$ .

The reaction has a constant half-life under these conditions.

Calculate the half-life in seconds.

half-life = .............................. s

1M Similar

DifficultyEasy

Worked solution

Working

The rate equation is $\text{rate} = k_1[\text{NO}_2]$ , which is first order with respect to $\text{NO}_2$ .

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

t_{1/2} = \frac{\ln 2}{k_1} = \frac{0.693}{k_1}

Given $k_1 = 0.0848 \text{ s}^{-1}$ :

t_{1/2} = \frac{0.693}{0.0848}

t_{1/2} = 8.172... \text{ s}

Rounding to 3 significant figures:

t_{1/2} = 8.17 \text{ s}

Answer

half-life = 8.17 s

Final answer

8.17

Detailed explanation

Background Concept

For a first-order reaction, the rate is directly proportional to the concentration of one reactant: $\text{rate} = k[\text{A}]$ .

The half-life ( $t_{1/2}$ ) of a first-order reaction is constant and is related to the rate constant $k$ by the equation:

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

This formula is derived from the integrated rate law for first-order reactions: $\ln[\text{A}] = \ln[\text{A}]_0 - kt$ . Setting $[\text{A}] = \frac{1}{2}[\text{A}]_0$ and solving for $t$ gives $t_{1/2} = \frac{\ln 2}{k}$ .

Understanding the Question

We are given:

Rate equation: $\text{rate} = k_1[\text{NO}_2]$
$k_1 = 0.0848 \text{ s}^{-1}$
The reaction has a constant half-life (confirming it is first order).

We need to calculate the half-life in seconds.

Approach

Recognize that the rate equation indicates a first-order reaction.
Use the first-order half-life formula: $t_{1/2} = \frac{\ln 2}{k}$ .
Substitute the given value of $k_1$ and calculate.

Step-by-Step Reasoning

The rate equation $\text{rate} = k_1[\text{NO}_2]$ shows that the reaction is first order with respect to $\text{NO}_2$ (overall order = 1).
For a first-order reaction, $t_{1/2} = \frac{\ln 2}{k_1}$ .
$\ln 2 \approx 0.693147$ .
Substitute $k_1 = 0.0848 \text{ s}^{-1}$ : $t_{1/2} = \frac{0.693147}{0.0848}$
Calculate: $t_{1/2} = 8.1739... \text{ s}$
Round to 3 significant figures (matching the precision of $k_1$ ): $t_{1/2} = 8.17 \text{ s}$ .

Key Takeaways

The formula $t_{1/2} = \frac{\ln 2}{k}$ applies only to first-order reactions.
Always check the units of $k$ : for a first-order reaction, $k$ has units of $\text{time}^{-1}$ (e.g., $\text{s}^{-1}$ ), which is consistent with the half-life having units of time.
Use $\ln 2$ (not $\log_{10} 2$ ) in the calculation.

Common Mistakes

Using the wrong formula for half-life (e.g., $t_{1/2} = \frac{1}{k}$ or $t_{1/2} = \frac{0.693}{k^2}$ ).
Forgetting to use the natural logarithm ( $\ln$ ) instead of the common logarithm ( $\log$ ).
Not rounding to the correct number of significant figures.
Confusing $k_1$ with the rate constant from part (b) (which had different units and value).

Things to Be Careful About

The mark scheme accepts $8.17$ or $8.2$ . Giving $8.17$ is more precise and preferred.
Ensure the unit is seconds (s), as $k_1$ is given in $\text{s}^{-1}$ .
Do not include units in the final answer box if the unit is already provided outside the box (as in this case, "half-life = .............................. s"). Just write the number.

Techniques used

calculate half-life from rate constant for first-order reaction

(e)

$\text{NO}_2$ is present in the exhaust gases of cars. It can react with carbon monoxide, $\text{CO}$ , on the surface of a heterogeneous catalyst in the car’s catalytic converter.

Describe the mode of action of this heterogeneous catalyst.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

The reactant molecules ( $\text{NO}_2$ and $\text{CO}$ ) are adsorbed onto the surface of the catalyst.
This adsorption weakens the bonds within the reactant molecules, lowering the activation energy.
The reaction occurs on the surface, and the product molecules are then desorbed from the catalyst surface.

(Any two of these points earn full marks)

Final answer

reactants adsorbed onto catalyst surface; bonds weakened; products desorbed

Detailed explanation

Background Concept

Heterogeneous catalysis occurs when the catalyst is in a different phase from the reactants (e.g., a solid catalyst with gaseous reactants). The most common example is the use of metal catalysts (like platinum, palladium, or rhodium) in a car's catalytic converter to convert toxic gases ( $\text{CO}$ , $\text{NO}_x$ ) into less harmful ones ( $\text{CO}_2$ , $\text{N}_2$ ).

The mechanism of heterogeneous catalysis involves three main steps:

Adsorption: Reactant molecules bind to the surface of the catalyst. This can be physical (weak van der Waals forces) or chemical (formation of new bonds between the reactant and the catalyst surface atoms). Chemical adsorption is the key step.
Reaction: The adsorption process weakens or breaks the bonds within the reactant molecules, lowering the activation energy for the reaction. The reactants are held in the correct orientation for the reaction to occur.
Desorption: The product molecules, which have a weaker attraction to the catalyst surface than the reactants, detach (desorb) from the surface, freeing up active sites for new reactant molecules.

Understanding the Question

We are asked to describe the mode of action of a heterogeneous catalyst in a car's catalytic converter, specifically for the reaction between $\text{NO}_2$ and $\text{CO}$ .

Approach

Recall the standard three-step mechanism for heterogeneous catalysis and apply it to the given context. The mark scheme rewards three key points: adsorption, bond weakening, and desorption.

Step-by-Step Reasoning

Adsorption: The reactant gases ( $\text{NO}_2$ and $\text{CO}$ ) come into contact with the solid catalyst surface (e.g., platinum or rhodium). They are adsorbed onto the surface. This is a crucial first step; without adsorption, the reaction cannot occur on the surface.
Bond weakening: When the molecules are adsorbed, they interact with the metal atoms on the catalyst surface. This interaction weakens the bonds within the reactant molecules (e.g., the N–O bonds in $\text{NO}_2$ or the C–O bond in $\text{CO}$ ). Weakening the bonds lowers the activation energy required for the reaction to proceed, allowing the reaction to occur at a lower temperature than it would without the catalyst.
Desorption: After the reaction occurs on the surface (forming $\text{CO}_2$ and $\text{N}_2$ or $\text{N}_2\text{O}$ ), the product molecules are desorbed (detached) from the catalyst surface. This frees up the active sites on the catalyst for new reactant molecules to adsorb, allowing the catalytic cycle to continue.

Any two of these three points are sufficient to earn the 2 marks.

Key Takeaways

Heterogeneous catalysis involves adsorption, reaction (with bond weakening), and desorption.
The catalyst provides an alternative reaction pathway with a lower activation energy.
The catalyst surface must have active sites where reactants can adsorb.

Common Mistakes

Saying the reactants are "absorbed" instead of "adsorbed". Absorption means taking in into the bulk of a material (like a sponge), while adsorption means sticking to the surface. This is a common and costly error.
Saying the catalyst "provides a surface for the reaction" without mentioning adsorption or bond weakening. This is too vague and may not score.
Forgetting to mention desorption. If products don't desorb, the catalyst becomes poisoned and the reaction stops.
Confusing heterogeneous catalysis with homogeneous catalysis (where the catalyst is in the same phase as the reactants).

Things to Be Careful About

Use precise terminology: adsorption (not absorption), desorption (not desorbing or leaving), weaken bonds (not break bonds completely, though bond breaking can occur, weakening is the key effect that lowers $E_a$ ).
The question asks for the "mode of action", so describe the process, not just list keywords. However, concise bullet points covering the three steps are acceptable.
Ensure you mention the reactants being adsorbed and the products being desorbed. Just saying "molecules are adsorbed" is slightly vague.

Techniques used

describe the mechanism of heterogeneous catalysis

Q3MediumEquilibriaCarboxylic Acids and Derivatives

(a)

2M

(i)

Define conjugate acid–base pair.

1M Similar

(ii)

Give the formulas of the conjugate acid and the conjugate base of the hydrogen phosphate ion, $\text{HPO}_4^{2-}$ .

conjugate acid of $\text{HPO}_4^{2-}$ ..............................

conjugate base of $\text{HPO}_4^{2-}$ ..............................

1M Similar

(b)

The $K_a$ of propanoic acid, $\text{CH}_3\text{CH}_2\text{COOH}$ , is $1.35 \times 10^{-5} \text{ mol dm}^{-3}$ at $298 \text{ K}$ .

Solution C is a solution of $\text{CH}_3\text{CH}_2\text{COOH}$ with a pH of $3.60$ at $298 \text{ K}$ .

7M

(i)

Calculate the concentration of $\text{CH}_3\text{CH}_2\text{COOH}$ in solution C.

$[\text{CH}_3\text{CH}_2\text{COOH}]$ = .............................. $\text{mol dm}^{-3}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

[\text{H}^+] = 10^{-3.60} = 2.51 \times 10^{-4} \text{ mol dm}^{-3}

For $\text{CH}_3\text{CH}_2\text{COOH} \rightleftharpoons \text{CH}_3\text{CH}_2\text{COO}^- + \text{H}^+$ :

K_a = \frac{[\text{CH}_3\text{CH}_2\text{COO}^-][\text{H}^+]}{[\text{CH}_3\text{CH}_2\text{COOH}]}

[\text{CH}_3\text{CH}_2\text{COOH}] = \frac{(2.51 \times 10^{-4})^2}{1.35 \times 10^{-5}} = 4.67 \times 10^{-3} \text{ mol dm}^{-3}

Answer

$[\text{CH}_3\text{CH}_2\text{COOH}] = 4.67 \times 10^{-3} \text{ mol dm}^{-3}$

Final answer

4.67 × 10^-3 mol dm^-3

Detailed explanation

Background Concept

A weak acid only partially dissociates, so at equilibrium $[\text{H}^+] \approx [\text{A}^-]$ and the undissociated acid concentration is approximately its initial concentration. $K_a = [\text{H}^+][\text{A}^-]/[\text{HA}]$ .

Understanding the Question

Given pH = 3.60 and $K_a = 1.35 \times 10^{-5}$ , find the acid concentration in solution C.

Approach

Convert pH to $[\text{H}^+]$ (M1), assume $[\text{A}^-] = [\text{H}^+]$ , then rearrange $K_a$ for $[\text{HA}]$ (M2).

Step-by-Step Reasoning

$[\text{H}^+] = 10^{-3.60} = 2.51 \times 10^{-4}$ . Since dissociation is small, $[\text{CH}_3\text{CH}_2\text{COO}^-] = [\text{H}^+]$ . Rearranging: $[\text{HA}] = (2.51 \times 10^{-4})^2 / (1.35 \times 10^{-5}) = 6.30 \times 10^{-8} / 1.35 \times 10^{-5} = 4.67 \times 10^{-3} \text{ mol dm}^{-3}$ .

Key Takeaways

The 'weak acid approximation' $[\text{A}^-] = [\text{H}^+]$ and $[\text{HA}] \approx$ initial concentration underpins all such calculations.

Common Mistakes

Using pH directly instead of $10^{-\text{pH}}$ ; forgetting to square $[\text{H}^+]$ ; subtracting $[\text{H}^+]$ from $[\text{HA}]$ unnecessarily (the approximation is expected at this level).

Things to Be Careful About

Give the answer to 3 significant figures with units; an arithmetic slip in the square still earns M1 (ecf).

Techniques used

convert pH to hydrogen ion concentrationrearrange the Ka expression to find acid concentration

(ii)

Calculate the concentration of hydroxide ions in solution C.

$[\text{OH}^-]$ = .............................. $\text{mol dm}^{-3}$

1M Similar

(iii)

Calculate the concentration of a solution of hydrochloric acid with the same pH as solution C.

concentration = .............................. $\text{mol dm}^{-3}$

1M Similar

(iv)

Table 3.1 shows three possible values of the $K_a$ of dimethylpropanoic acid, $(\text{CH}_3)_3\text{CCOOH}$ .

Place a tick in Table 3.1 to show the correct value. Explain your answer.

Table 3.1

value of $K_a$ / $\text{mol dm}^{-3}$	place one tick (✓) in this column
$9.33 \times 10^{-6}$
$1.35 \times 10^{-5}$
$3.35 \times 10^{-5}$

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

3M Similar

DifficultyMedium

Worked solution

Answer

Tick: $9.33 \times 10^{-6} \text{ mol dm}^{-3}$ (top box).

Explanation: $(\text{CH}_3)_3\text{CCOOH}$ is a weaker acid than $\text{CH}_3\text{CH}_2\text{COOH}$ , so its $K_a$ is smaller. The extra alkyl groups are electron donating (positive inductive effect), which strengthens the O–H bond and destabilises the carboxylate anion, so dissociation is less favourable.

Final answer

9.33 × 10^-6 mol dm^-3; weaker acid due to electron-donating alkyl groups

Detailed explanation

Background Concept

$K_a$ measures acid strength: larger $K_a$ = stronger acid = more dissociation. Carboxylic acid strength depends on how stable the carboxylate anion $\text{RCOO}^-$ is — the negative charge is shared between the two oxygens, and electron-withdrawing substituents stabilise it (raising $K_a$ ), while electron-donating alkyl groups destabilise it (lowering $K_a$ ).

Understanding the Question

Dimethylpropanoic acid, $(\text{CH}_3)_3\text{CCOOH}$ , has more alkyl groups attached near the COOH than propanoic acid. Choose the correct $K_a$ from three options and explain.

Approach

Decide whether the extra alkyl groups strengthen or weaken the acid, then pick the $K_a$ accordingly and justify with the inductive effect.

Step-by-Step Reasoning

M1: The three methyl groups push electron density towards the COOH group, so the acid is weaker than propanoic acid → $K_a$ must be smaller than $1.35 \times 10^{-5}$ → tick $9.33 \times 10^{-6}$ .
M2: The alkyl groups are electron donating (+I effect).
M3: This either strengthens the O–H bond (harder to break) or destabilises the carboxylate anion formed — either wording earns the mark. Less dissociation → smaller $K_a$ .

Key Takeaways

Electron-withdrawing groups increase acidity of carboxylic acids; electron-donating groups decrease it. The same logic explains why chloroethanoic acid is stronger than ethanoic acid.

Common Mistakes

Assuming more alkyl groups = stronger acid; ticking $3.35 \times 10^{-5}$ ; saying 'more alkyl groups make the anion more stable' — the opposite is true for electron donation.

Things to Be Careful About

All three marks are needed: correct tick, 'weaker acid' statement, and the electronic explanation including its consequence (stronger O–H bond or destabilised anion).

Techniques used

compare acid strengths via Kaexplain inductive electron donation of alkyl groups

(c)

Solution D is made by mixing $100 \text{ cm}^3$ of $0.100 \text{ mol dm}^{-3}$ $\text{CH}_3\text{CH}_2\text{COOH}$ and $100 \text{ cm}^3$ of $0.100 \text{ mol dm}^{-3}$ $\text{NaCl}$ .

The pH of solution D is measured as small amounts of $\text{H}_2\text{SO}_4\text{(aq)}$ are added to it, and when small amounts of $\text{NaOH(aq)}$ are added to it.

Solution D only acts as a buffer solution when one of these solutions is added to it.

2M

(i)

Complete the sentence and write an equation for the reaction that occurs.

Solution D acts as a buffer when ................................................................ is added to it.

equation ............................................................................................................................

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

Solution D acts as a buffer when NaOH is added to it.

\text{CH}_3\text{CH}_2\text{COOH} + \text{OH}^- \rightarrow \text{CH}_3\text{CH}_2\text{COO}^- + \text{H}_2\text{O}

Final answer

NaOH; CH3CH2COOH + OH− -> CH3CH2COO− + H2O

Detailed explanation

Background Concept

A buffer needs a weak acid and its conjugate base together. Solution D initially contains only the weak acid (NaCl is a spectator). Adding limited NaOH neutralises some acid, producing its conjugate base $\text{CH}_3\text{CH}_2\text{COO}^-$ ; the mixture then contains both HA and A⁻ — a buffer.

Understanding the Question

Small amounts of $\text{H}_2\text{SO}_4$ or NaOH are added to a propanoic acid/NaCl mixture; only one addition produces a buffer.

Approach

Adding base converts HA → A⁻, creating the pair. Adding acid adds no conjugate base, so no buffer.

Step-by-Step Reasoning

NaOH reacts with the acid: $\text{CH}_3\text{CH}_2\text{COOH} + \text{OH}^- \rightarrow \text{CH}_3\text{CH}_2\text{COO}^- + \text{H}_2\text{O}$ . Now both $\text{CH}_3\text{CH}_2\text{COOH}$ (excess) and $\text{CH}_3\text{CH}_2\text{COO}^-$ are present — a buffer. The equation must show $\text{OH}^-$ and the carboxylate ion (ionic form accepted).

Key Takeaways

Partial neutralisation of a weak acid with a strong base is a standard way to make a buffer.

Common Mistakes

Writing the molecular equation with NaOH/NaCH₃CH₂COO when the ionic equation is expected; choosing $\text{H}_2\text{SO}_4$ .

Things to Be Careful About

Both the word 'NaOH' and a correctly balanced equation are required for the single mark.

Techniques used

identify buffer componentswrite the neutralisation equation with added base

(ii)

Complete the sentence and explain why solution D does not act as a buffer when the other solution is added.

Solution D does not act as a buffer when .................................................. is added to it.

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

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

Solution D does not act as a buffer when $\text{H}_2\text{SO}_4$ is added to it.

Explanation: a buffer requires both the weak acid and its conjugate base; when $\text{H}_2\text{SO}_4$ is added, the conjugate base of $\text{CH}_3\text{CH}_2\text{COOH}$ , $\text{CH}_3\text{CH}_2\text{COO}^-$ , is not present in the mixture, so there is nothing to remove the added $\text{H}^+$ and resist the pH change.

Final answer

H2SO4; the conjugate base CH3CH2COO− is not present

Detailed explanation

Background Concept

A buffer resists pH change because it contains both a weak acid (to neutralise added base) and its conjugate base (to neutralise added acid). Remove either component and the mixture is no longer a buffer.

Understanding the Question

Explain why adding sulfuric acid to the propanoic acid/NaCl solution fails to create a buffer.

Approach

Check which buffer component would be present after each addition; adding acid supplies only $\text{H}^+$ , no conjugate base.

Step-by-Step Reasoning

Adding $\text{H}_2\text{SO}_4$ adds $\text{H}^+$ to a solution containing only $\text{CH}_3\text{CH}_2\text{COOH}$ and spectator ions. No $\text{CH}_3\text{CH}_2\text{COO}^-$ is generated, so the added $\text{H}^+$ is not removed by any conjugate base and the pH falls sharply — not buffer behaviour. The mark scheme requires the key point: the conjugate base of the acid is not present.

Key Takeaways

Buffer action depends on the presence of both members of the conjugate pair.

Common Mistakes

Saying 'strong acid' alone without explaining that the conjugate base is absent; naming NaOH instead of $\text{H}_2\text{SO}_4$ .

Things to Be Careful About

The explanation must mention the missing conjugate base, not merely that the pH changes.

Techniques used

explain absence of conjugate base in a mixture

(d)

Manganese(II) hydroxide, $\text{Mn(OH)}_2$ , is only slightly soluble in water.

The solubility of $\text{Mn(OH)}_2$ in water is $3.28 \times 10^{-3} \text{ g dm}^{-3}$ at $298 \text{ K}$ .

4M

(i)

Calculate the concentration of a saturated solution of $\text{Mn(OH)}_2$ at $298 \text{ K}$ .

$[\text{Mn(OH)}_2]$ = .............................. $\text{mol dm}^{-3}$

1M Similar

(ii)

Write an expression for the $K_{sp}$ of $\text{Mn(OH)}_2$ . Give the units of $K_{sp}$ .

$K_{sp}$ =

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

2M Similar

(iii)

Use your answers to (d)(i) and (d)(ii) to calculate the value of $K_{sp}$ of $\text{Mn(OH)}_2$ at $298 \text{ K}$ .

$K_{sp}$ = ..............................

1M Similar

DifficultyMedium-Easy

Worked solution

Working

$[\text{Mn}^{2+}] = 3.69 \times 10^{-5}$ and $[\text{OH}^-] = 2 \times 3.69 \times 10^{-5} = 7.38 \times 10^{-5} \text{ mol dm}^{-3}$

K_{\text{sp}} = (3.69 \times 10^{-5})(7.38 \times 10^{-5})^2 = 2.0 \times 10^{-13}

Answer

$K_{\text{sp}} = 2.0 \times 10^{-13} \text{ mol}^3 \text{ dm}^{-9}$

Final answer

2.0 × 10^-13

Detailed explanation

Background Concept

Each mole of dissolved $\text{Mn(OH)}_2$ produces 1 mol $\text{Mn}^{2+}$ and 2 mol $\text{OH}^-$ , so $[\text{Mn}^{2+}] = s$ and $[\text{OH}^-] = 2s$ where $s$ is the molar solubility.

Understanding the Question

Use $s = 3.69 \times 10^{-5}$ from (d)(i) and the expression from (d)(ii) to compute $K_{sp}$ .

Approach

Substitute $[\text{Mn}^{2+}] = s$ , $[\text{OH}^-] = 2s$ into $K_{sp} = [\text{Mn}^{2+}][\text{OH}^-]^2$ .

Step-by-Step Reasoning

$K_{sp} = (3.69 \times 10^{-5})(7.38 \times 10^{-5})^2 = (3.69 \times 10^{-5})(5.45 \times 10^{-9}) = 2.0 \times 10^{-13}$ .

Key Takeaways

For MX₂ salts, $K_{sp} = 4s^3$ ; the factor of 2 on the hydroxide concentration is essential.

Common Mistakes

Using $[\text{OH}^-] = s$ instead of $2s$ ; forgetting to square; arithmetic errors in cubing small numbers.

Things to Be Careful About

The answer $2.0 \times 10^{-13}$ is cao; a wrong value from (d)(i) can still earn the method mark via ecf.

Techniques used

substitute ion concentrations into the Ksp expression

Q4MediumChemical Energetics

(a)

Define enthalpy change of atomisation, $\Delta H_{at}$ .

1M Similar

DifficultyEasy

Worked solution

Answer

The enthalpy change of atomisation is the energy required to form one mole of gaseous atoms from the element in its standard state.

Final answer

Energy required to form one mole of gaseous atoms from the element in its standard state.

Detailed explanation

Background Concept

Enthalpy change of atomisation, $\Delta H_{at}$ , is a standard enthalpy change describing the conversion of an element in its standard state into its gaseous atoms. For a diatomic element such as chlorine, the standard state is $\text{Cl}_2(\text{g})$ , so atomisation corresponds to $\frac{1}{2}\text{Cl}_2(\text{g}) \rightarrow \text{Cl}(\text{g})$ . Breaking a covalent bond requires energy, so $\Delta H_{at}$ is always positive (endothermic). The definition is per mole of gaseous atoms formed, not per mole of bonds broken.

Understanding the Question

The command word "Define" requires the precise textbook definition. The mark is awarded for the exact conditions: one mole of gaseous atoms, formed from the element (in its standard state).

Approach

State the definition exactly, ensuring all three conditions are present: "one mole", "gaseous atoms", "from the element".

Step-by-Step Reasoning

The definition: the energy required when one mole of gaseous atoms is formed from the element in its standard state. For chlorine, this is the energy for $\frac{1}{2}\text{Cl}_2(\text{g}) \rightarrow \text{Cl}(\text{g})$ . Each condition matters: "one mole of gaseous atoms" fixes the quantity; "gaseous" fixes the state of the product; "from the element" fixes the starting material. The value is positive because energy must be supplied to break the Cl–Cl bond.

Key Takeaways

$\Delta H_{at}$ is endothermic, defined per mole of gaseous atoms, and is the first step in Born-Haber cycles for ionic compounds.

Common Mistakes

Writing "per mole of element" instead of "per mole of gaseous atoms".
Omitting "gaseous" for the atoms.
Confusing $\Delta H_{at}$ with bond dissociation enthalpy (which is per mole of bonds, e.g. $\text{Cl}_2 \rightarrow 2\text{Cl}$ , double the value per mole of Cl atoms).

Things to Be Careful About

The mark scheme rewards the phrase "one mole of gaseous atoms" and "from the element". State symbols matter: the product must be gaseous atoms.

Techniques used

state the definition of enthalpy change of atomisation

(b)

Define first electron affinity, EA.

1M Similar

DifficultyEasy

Worked solution

Answer

The first electron affinity is the energy released when one mole of gaseous atoms each gains one electron to form one mole of gaseous singly-charged negative ions.

Final answer

Energy released when one mole of gaseous atoms gains one mole of electrons to form one mole of gaseous 1- ions.

Detailed explanation

Background Concept

First electron affinity is the energy change when one mole of gaseous atoms each gains one electron to form one mole of gaseous singly-charged negative ions. For chlorine: $\text{Cl}(\text{g}) + \text{e}^- \rightarrow \text{Cl}^-(\text{g})$ . The incoming electron is attracted to the nucleus, so energy is released — the process is exothermic and $\Delta H$ is negative. "First" means one electron per atom.

Understanding the Question

"Define first electron affinity" — give the precise definition. The mark requires all the conditions: gaseous atoms, one mole of electrons, gaseous 1⁻ ions.

Approach

Recall the standard definition and state it precisely, including the gaseous states of both the atom and the ion.

Step-by-Step Reasoning

The definition: the energy released when one mole of gaseous atoms gains one mole of electrons to form one mole of gaseous singly-charged negative ions. For chlorine this is $\text{Cl}(\text{g}) + \text{e}^- \rightarrow \text{Cl}^-(\text{g})$ . Each part is essential: "one mole of gaseous atoms" (quantity and state), "one mole of electrons" (one electron per atom), "one mole of gaseous 1⁻ ions" (product state and charge).

Key Takeaways

First EA is exothermic for most atoms (negative $\Delta H$ ). It is a key quantity in Born-Haber cycles.

Common Mistakes

Forgetting "gaseous" for the atom or the ion.
Saying "an atom gains an electron" without specifying one mole.
Confusing first EA with second EA (which is endothermic because energy is needed to add an electron to a negative ion).

Things to Be Careful About

The mark scheme wording: "one mole of gaseous atoms gains one mole of electrons and becomes one mole of gaseous 1⁻ ions". Use "released" (exothermic) for the first EA.

Techniques used

state the definition of first electron affinity

(c)

Explain why the first electron affinity of chlorine is more exothermic than the first electron affinity of iodine.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

The iodine atom has a greater atomic radius than the chlorine atom.
The incoming electron experiences less attraction to the iodine nucleus (the outer electrons are further from the nucleus and more shielded), so less energy is released when the electron is added.

Final answer

Iodine has a greater atomic radius, so there is less attraction between the nucleus and the incoming electron.

Detailed explanation

Background Concept

The first electron affinity is exothermic because the incoming electron is attracted to the positively charged nucleus. The magnitude of the energy released depends on how strongly the nucleus attracts the incoming electron. Down Group 7 (Cl → I), each successive element has an additional electron shell, so atomic radius increases and the outer electrons are further from the nucleus and more strongly shielded by inner electrons. The effective nuclear charge felt by an incoming electron therefore decreases down the group.

Understanding the Question

Explain why the first EA of chlorine (−364 kJ mol⁻¹) is more exothermic than the first EA of iodine. Two marks: M1 greater atomic radius of iodine; M2 less attraction between the nucleus and the incoming electron in iodine.

Approach

Compare the atomic radii of Cl and I, then link the larger radius to weaker attraction for the incoming electron.

Step-by-Step Reasoning

M1: Iodine is below chlorine in Group 7, so it has more electron shells and a greater atomic radius. M2: Because the incoming electron is added at a greater distance from the nucleus and experiences more shielding from inner electrons, the attraction between the iodine nucleus and the incoming electron is weaker. Less energy is therefore released when the electron is added, so the first EA of iodine is less exothermic (less negative) than that of chlorine.

Key Takeaways

Down Group 7, first EA becomes less exothermic due to increasing atomic radius and shielding. The trend is the opposite of what a naive "more protons = more attraction" argument would predict.

Common Mistakes

Arguing "iodine has more protons, so it attracts the electron more strongly" — wrong, because the extra shielding outweighs the extra nuclear charge.
Not explicitly stating that the electron is added at a greater distance from the nucleus.
Confusing "more exothermic" with "more positive".

Things to Be Careful About

Both marking points are needed: greater radius of iodine AND less attraction/less energy released. Use the phrase "less attraction between the nucleus and the incoming electron".

Techniques used

compare atomic radii down a grouprelate atomic radius to attraction for an incoming electron

(d)

The enthalpy change for the reaction $\text{Cl}_2\text{(g)} + 2\text{e}^- \rightarrow 2\text{Cl}^-\text{(g)}$ is $-486 \text{ kJ mol}^{-1}$ .

The first electron affinity of chlorine is $-364 \text{ kJ mol}^{-1}$ .

Calculate the enthalpy change of atomisation of chlorine.

$\Delta H_{at}$ of chlorine = .............................. $\text{kJ mol}^{-1}$

2M Similar

DifficultyMedium

Worked solution

Working

The reaction can be split into two steps.

Step 1 — atomisation:

\text{Cl}_2(\text{g}) \rightarrow 2\text{Cl}(\text{g}), \quad \Delta H = 2\Delta H_{at}

Step 2 — electron gain:

2\text{Cl}(\text{g}) + 2\text{e}^- \rightarrow 2\text{Cl}^-(\text{g}), \quad \Delta H = 2 \times (-364) = -728 \text{ kJ mol}^{-1}

By Hess's law:

-486 = 2\Delta H_{at} + (-728)

2\Delta H_{at} = -486 + 728 = 242

\Delta H_{at} = +121 \text{ kJ mol}^{-1}

Answer

+121 kJ mol⁻¹

Final answer

+121 kJ mol^-1

Detailed explanation

Background Concept

This is a Hess's law calculation in the style of a Born-Haber cycle. The overall reaction $\text{Cl}_2(\text{g}) + 2\text{e}^- \rightarrow 2\text{Cl}^-(\text{g})$ can be achieved in two steps: first atomise the chlorine ( $\text{Cl}_2 \rightarrow 2\text{Cl}$ ), then add electrons ( $2\text{Cl} + 2\text{e}^- \rightarrow 2\text{Cl}^-$ ). Hess's law states that the enthalpy change of the overall reaction equals the sum of the enthalpy changes of the individual steps, regardless of the route.

Understanding the Question

Given the overall enthalpy change for $\text{Cl}_2(\text{g}) + 2\text{e}^- \rightarrow 2\text{Cl}^-(\text{g})$ is −486 kJ mol⁻¹, and the first EA of chlorine is −364 kJ mol⁻¹, find $\Delta H_{at}$ of chlorine. The key subtlety is the factor of 2: two moles of Cl atoms are formed and two moles of Cl⁻ ions are formed.

Approach

Write the two-step pathway, apply Hess's law, and solve for $\Delta H_{at}$ . Be careful with the factor of 2 on both the atomisation and the electron affinity terms.

Step-by-Step Reasoning

Step 1 (atomisation): $\text{Cl}_2(\text{g}) \rightarrow 2\text{Cl}(\text{g})$ , $\Delta H = 2 \times \Delta H_{at}$ . Two moles of Cl atoms are formed, so the atomisation enthalpy is doubled.

Step 2 (electron gain): $2\text{Cl}(\text{g}) + 2\text{e}^- \rightarrow 2\text{Cl}^-(\text{g})$ , $\Delta H = 2 \times (-364) = -728$ kJ mol⁻¹. Two moles of Cl⁻ ions are formed, each with first EA −364 kJ mol⁻¹.

By Hess's law:

-486 = 2\Delta H_{at} + (-728)

2\Delta H_{at} = -486 + 728 = 242

\Delta H_{at} = +121 \text{ kJ mol}^{-1}

The mark scheme expresses this as $\frac{1}{2}(-486 + (2 \times 364)) = \frac{1}{2}(242) = 121$ .

Key Takeaways

Hess's law allows enthalpy changes to be combined along a chosen pathway. Atomisation is endothermic (positive), electron affinity is exothermic (negative). Always account for the number of moles in each step.

Common Mistakes

Forgetting the factor of 2 on $\Delta H_{at}$ (two Cl atoms).
Forgetting the factor of 2 on the EA (two Cl⁻ ions).
Sign errors: EA is −364, so $2 \times (-364) = -728$ ; subtracting a negative is adding.
Reporting −121 instead of +121.

Things to Be Careful About

Units: kJ mol⁻¹. The answer is +121 kJ mol⁻¹. The mark scheme: M1 for the correct expression $\frac{1}{2}(-486 + (2 \times 364))$ , M2 for +121.

Techniques used

apply Hess's lawcombine atomisation and electron affinity enthalpiesaccount for mole ratios

Q5MediumTransition ElementsElectrochemistry

Cobalt is a transition element which forms compounds containing $\text{Co}^{2+}$ and $\text{Co}^{3+}$ ions. Cobalt(II) sulfate dissolves in water to form a solution containing the $[\text{Co(H}_2\text{O)}_6]^{2+}$ complex ion.

(a)

7M

(i)

Complete the electronic configurations of a $\text{Co}^{2+}$ ion and a $\text{Co}^{3+}$ ion.

$\text{Co}^{2+} = [\text{Ar}]$ ..............................

$\text{Co}^{3+} = [\text{Ar}]$ ..............................

1M Similar

(ii)

Explain why transition elements can form complex ions.

1M Similar

(iii)

An excess of concentrated $\text{HCl}$ is added to a solution containing $[\text{Co(H}_2\text{O)}_6]^{2+}$ .

Describe the colour change observed and the state of the cobalt-containing product.

The colour changes from .............................. to .............................. .

The state of the cobalt-containing product is .............................. .

2M Similar

(iv)

Write an equation for the reaction occurring in (a)(iii).

1M Similar

(v)

Name the type of reaction occurring in (a)(iii).

1M Similar

(vi)

Write an equation for the reaction that occurs when an excess of $\text{NaOH(aq)}$ is added to a solution containing $[\text{Co(H}_2\text{O)}_6]^{2+}$ .

1M Similar

(b)

Cobalt metal can be oxidised by acidified $\text{K}_2\text{Cr}_2\text{O}_7$ . The relevant half-equations, and their $E^\ominus$ values, are shown.

\begin{aligned} \text{Co}^{2+} + 2\text{e}^- &\rightleftharpoons \text{Co} & E^\ominus &= -0.28 \text{ V} \\ \text{Cr}_2\text{O}_7^{2-} + 14\text{H}^+ + 6\text{e}^- &\rightleftharpoons 2\text{Cr}^{3+} + 7\text{H}_2\text{O} & E^\ominus &= +1.33 \text{ V} \end{aligned}

5M

(i)

A $\text{Co}^{2+}/\text{Co}$ electrode is constructed in which $[\text{Co}^{2+}]$ is $0.020 \text{ mol dm}^{-3}$ at $298 \text{ K}$ .

Use the Nernst equation to show that the $E$ value for this $\text{Co}^{2+}/\text{Co}$ electrode is $-0.33 \text{ V}$ .

2M Similar

DifficultyMedium

Worked solution

Working

E = E^{\ominus} + \frac{0.059}{z} \log\left(\frac{[\text{oxidised}]}{[\text{reduced}]}\right)

For $\text{Co}^{2+} + 2\text{e}^- \rightleftharpoons \text{Co}$ , $z = 2$ , $[\text{oxidised}] = [\text{Co}^{2+}] = 0.020$ , $[\text{reduced}] = 1$ (pure solid):

E = -0.28 + \frac{0.059}{2} \log(0.020) = -0.28 + 0.0295 \times (-1.70) = -0.33 \text{ V}

Answer

$E = -0.33 \text{ V}$ (as required)

Final answer

E = -0.33 V

Detailed explanation

Background Concept

The Nernst equation adjusts a standard electrode potential for non-standard concentrations:

E = E^{\ominus} + \frac{0.059}{z} \log\left(\frac{[\text{oxidised}]}{[\text{reduced}]}\right)

where $z$ is the number of electrons in the half-equation. Pure solids and pure liquids have unit 'concentration' (they don't appear in the ratio).

Understanding the Question

Show that with $[\text{Co}^{2+}] = 0.020$ , the electrode potential is −0.33 V. The command 'show that' means full substitution must be visible.

Approach

Identify $z = 2$ from the half-equation, set the ratio to $[\text{Co}^{2+}]/1$ , and evaluate.

Step-by-Step Reasoning

$\log(0.020) = -1.70$ .
$\frac{0.059}{2} \times (-1.70) = -0.050$ V.
$E = -0.28 - 0.050 = -0.33$ V.
Lower $[\text{Co}^{2+}]$ than 1 mol dm⁻³ shifts the equilibrium left (less reduction), making $E$ more negative — consistent with the result.

Key Takeaways

The reduced form is a solid, so it contributes nothing to the log ratio; only $[\text{Co}^{2+}]$ appears.

Common Mistakes

Using $z = 1$ instead of 2; using $-\log$ or inverting the ratio; forgetting the log of 0.020 is negative.

Things to Be Careful About

The 0.059 form already includes the factor $2.303RT/F$ at 298 K; don't double-count.

Techniques used

apply the Nernst equationidentify z as the number of electrons transferredsubstitute concentrations into the log term

(ii)

An electrochemical cell is constructed using the $\text{Co}^{2+}/\text{Co}$ electrode described in (b)(i) and a $\text{Cr}_2\text{O}_7^{2-}/\text{Cr}^{3+}$ electrode in which all conditions are standard.

Calculate the value of $E_{cell}$ .

$E_{cell}$ = ..............................

1M Similar

(iii)

A current is drawn from the electrochemical cell described in (b)(ii).

Write an equation for the reaction taking place in the cell.

1M Similar

(iv)

Complete the sentences to identify the negative electrode and the direction of electron flow when a current is drawn from the cell described in (b)(ii).

The .............................. electrode is the negative electrode.

Electrons flow from the .............................. electrode to the .............................. electrode.

1M Similar

(c)

A molten $\text{Co}^{2+}$ salt is electrolysed using a current of $0.500 \text{ A}$ .

$0.547 \text{ g}$ of cobalt metal forms at the cathode. Under the conditions used no other reduction reaction occurs at the cathode.

Calculate the time in minutes for which the current flows to produce this mass of cobalt.

Give your answer to three significant figures.

time = .............................. min

3M Similar

DifficultyMedium

Worked solution

Working

n(\text{Co}) = \frac{0.547}{58.9} = 0.00929 \text{ mol}

Each $\text{Co}^{2+}$ needs 2 electrons:

Q = 0.00929 \times 96500 \times 2 = 1792 \text{ C}

t = \frac{Q}{I} = \frac{1792}{0.500} = 3584.8 \text{ s} = 59.7 \text{ min}

Answer

time = 59.7 min (3 s.f.)

Final answer

59.7 min

Detailed explanation

Background Concept

Faraday's laws: the charge $Q$ (in coulombs) passed is $Q = It$ . One mole of electrons carries 96 500 C (the Faraday constant). To deposit 1 mol of Co from $\text{Co}^{2+}$ , 2 mol of electrons (2 × 96 500 C) are required.

Understanding the Question

Find the electrolysis time for 0.547 g of cobalt at 0.500 A, to 3 significant figures, in minutes.

Approach

mass → moles → moles of electrons → charge → time (s) → time (min).

Step-by-Step Reasoning

$n(\text{Co}) = 0.547 / 58.9 = 0.00929$ mol.
$\text{Co}^{2+} + 2\text{e}^- \rightarrow \text{Co}$ , so electrons needed = $2 \times 0.00929 = 0.01858$ mol; $Q = 0.01858 \times 96500 = 1792$ C.
$t = Q/I = 1792 / 0.500 = 3584.8$ s = $3584.8 / 60 = 59.7$ min (3 s.f.).

Key Takeaways

Always multiply by the number of electrons in the cathode half-equation — here 2, not 1.

Common Mistakes

Forgetting the factor of 2 (giving ~29.9 min); giving the answer in seconds; rounding to 2 s.f. or 4 s.f.

Things to Be Careful About

The question asks for minutes and 3 significant figures — convert and round correctly.

Techniques used

calculate moles from mass and Arapply Faraday's constant with the number of electrons transferredconvert charge to time using Q = Itconvert seconds to minutes and round to 3 significant figures

Q6Medium-EasyTransition Elements

(a)

Nickel forms complexes.

4M

(i)

Give the formula and charge of the tetrahedral complex formed by $\text{Ni}$ atoms with carbon monoxide molecules. Carbon monoxide is a monodentate ligand. This is complex E.

E = ...............................................................................................................................

1M Similar

DifficultyEasy

Worked solution

Answer

E is $\text{Ni(CO)}_4$ , charge 0.

Final answer

Ni(CO)4, charge 0

Detailed explanation

Background Concept

A complex consists of a central metal atom or ion surrounded by ligands, each ligand donating at least one lone pair of electrons to form a coordinate (dative covalent) bond. A monodentate ligand has one donor atom, so it occupies one coordination site. A tetrahedral complex has four coordination sites arranged around the metal at approximately 109.5°. The overall charge of a complex is the sum of the charge on the metal and the charges on all ligands.

Understanding the Question

The stem specifies that neutral nickel atoms, not $\text{Ni}^{2+}$ ions, combine with carbon monoxide. Complex E is tetrahedral, so the coordination number is 4, and CO is described as monodentate. We need to state the formula and charge of E.

Approach

Because the geometry is tetrahedral, four monodentate ligands must bind. CO is a neutral ligand and Ni is present as the atom, so both contribute no charge. Therefore the formula is Ni with four CO groups and the charge is zero.

Step-by-Step Reasoning

Each CO ligand donates one lone pair. Tetrahedral geometry requires four coordinate bonds, so four CO molecules attach to one Ni atom. The oxidation state of Ni in an elemental atom is 0, and each CO ligand is neutral. Total charge = $0 + 4(0) = 0$ . The formula is $\text{Ni(CO)}_4$ .

Key Takeaways

The coordination number equals the number of donor atoms, not the number of atoms in the ligand.
A neutral complex has total charge 0.
Tetrahedral geometry fixes the number of monodentate ligands at four.

Common Mistakes

Writing the formula as $\text{NiC}_4\text{O}_4$ instead of $\text{Ni(CO)}_4$ .
Assuming nickel contributes a positive charge because transition metals often form ions; here nickel is present as the neutral atom.
Treating CO as a charged ligand.

Things to Be Careful About

The parentheses in $\text{Ni(CO)}_4$ show that CO is a ligand unit. The charge is zero, not 4+, because both Ni and CO are neutral.

Techniques used

determine coordination number from tetrahedral geometry and monodentate ligandwrite formula and charge of a neutral metal carbonyl complex

(ii)

Give the formula and charge of the octahedral complex formed by $\text{Ni}^{2+}$ ions with ethanedioate ions. This is complex F.

F = ...............................................................................................................................

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

F is $[\text{Ni}(\text{C}_2\text{O}_4)_3]^{4-}$ , charge 4−.

Final answer

[Ni(C2O4)3]^4-, charge 4-

Detailed explanation

Background Concept

The ethanedioate ion, $\text{C}_2\text{O}_4^{2-}$ (oxalate), is a bidentate ligand: each ion contains two oxygen atoms that each have a lone pair and can form a coordinate bond to the metal. A bidentate ligand therefore occupies two coordination sites. An octahedral complex has six coordination sites arranged around the central metal, so six donor atoms are required. The charge on a complex ion is the algebraic sum of the metal ion charge and the charges on all the ligands.

Understanding the Question

Nickel is present as $\text{Ni}^{2+}$ , and the ligand is the ethanedioate ion, $\text{C}_2\text{O}_4^{2-}$ . The complex F is octahedral, so we must use enough ethanedioate ions to provide six donor atoms, then work out the overall charge.

Approach

Since each ethanedioate is bidentate, only three ligands are needed to give a coordination number of 6. Combine one $\text{Ni}^{2+}$ with three $\text{C}_2\text{O}_4^{2-}$ ions and add the charges: $+2 + 3(-2) = -4$ . The formula is therefore $[\text{Ni}(\text{C}_2\text{O}_4)_3]^{4-}$ .

Step-by-Step Reasoning

The ion is octahedral, so there are six ligand donor atoms around nickel. Each ethanedioate ion donates two lone pairs (from two O atoms), so three ions supply the six coordination sites. Ligand charge: three $\text{C}_2\text{O}_4^{2-}$ ions contribute a total charge of $-6$ . Add the $\text{Ni}^{2+}$ charge: $-6 + (+2) = -4$ . Put the three ligands in square brackets: $[\text{Ni}(\text{C}_2\text{O}_4)_3]^{4-}$ .

Key Takeaways

Bidentate ligands count twice when filling the coordination sphere. Always calculate the net charge from both the metal and the ligands.

Common Mistakes

Writing six ethanedioate ions because octahedral means six; it means six donor atoms.
Forgetting the negative charge on ethanedioate and writing a neutral or positive complex.
Writing the charge as $2-$ instead of $4-$ .

Things to Be Careful About

The final answer must show one Ni and three ethanedioate ligands, and a charge of $4-$ : $[\text{Ni}(\text{C}_2\text{O}_4)_3]^{4-}$ .

Techniques used

recognise ethanedioate as bidentatecalculate overall charge from metal and ligand chargesconstruct formula of an octahedral complex

(iii)

Identify which complex, E or F, exists as a mixture of two stereoisomers and the type of stereoisomerism involved.

The complex which exists as a mixture of two stereoisomers is .............................. .

The type of stereoisomerism involved is .............................. .

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

The complex that exists as a mixture of two stereoisomers is F. The type of stereoisomerism involved is optical.

Final answer

F; optical

Detailed explanation

Background Concept

Stereoisomers have the same formula and same connectivity but differ in the spatial arrangement of atoms. In coordination chemistry the common types are geometric (cis/trans) and optical. An optical isomer is a molecule whose mirror image cannot be superimposed on itself; the molecule is chiral. Optical isomers are non-superimposable mirror images.

Understanding the Question

We must decide which of E and F can exist as two stereoisomers and name the type. E is tetrahedral and contains four identical CO ligands. F is octahedral and contains three identical bidentate ethanedioate ligands.

Approach

Check whether the molecular symmetry of each complex allows non-superimposable mirror images. A tetrahedron with four identical substituents is highly symmetric and is identical to its mirror image. An octahedron with three identical bidentate chelate rings can twist into a left-handed and a right-handed propeller, making the two mirror images non-identical.

Step-by-Step Reasoning

E: $\text{Ni(CO)}_4$ has four identical CO ligands at the corners of a tetrahedron. Any rotation or mirror operation maps the structure onto itself; no pair of stereoisomers exists.
F: $[\text{Ni}(\text{C}_2\text{O}_4)_3]^{4-}$ is octahedral. The three bidentate oxalate ligands wrap around the metal in a helical arrangement. The complex and its mirror image are not superimposable, so F exists as optical isomers.
Therefore the complex with two stereoisomers is F, and the stereoisomerism is optical.

Key Takeaways

All identical tetrahedral complexes usually show no stereoisomerism, while tris(bidentate) octahedral complexes are classic examples of optical isomerism.

Common Mistakes

Choosing E because it is tetrahedral.
Saying 'geometric' instead of 'optical'. There is no cis/trans possibility in the structures described.

Things to Be Careful About

Write 'optical' exactly, not just 'stereoisomerism'.

Techniques used

identify whether a complex has non-superimposable mirror imagesdistinguish optical isomerism from geometric isomerism

(b)

Cadmium forms complexes with methylamine, $\text{CH}_3\text{NH}_2$ , and 1,2-diaminoethane, en. The values of the stability constants, $K_{stab}$ , of these complex ions are given in Table 6.1.

Table 6.1

complex	$K_{stab}$
$[\text{Cd(CH}_3\text{NH}_2)_4]^{2+}$	$3.5 \times 10^6$
$[\text{Cd(en)}_2]^{2+}$	$4.0 \times 10^{10}$

3M

(i)

Explain, by reference to its structure, why $\text{CH}_3\text{NH}_2$ acts as a monodentate ligand.

1M Similar

(ii)

Some $\text{Cd}^{2+}\text{(aq)}$ is added to a solution containing equal concentrations of $\text{CH}_3\text{NH}_2$ and en.

Predict which of the two complexes in Table 6.1 forms at the higher concentration.

Explain your answer.

complex that forms at the higher concentration ................................................................

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

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

The complex that forms at the higher concentration is $[\text{Cd(en)}_2]^{2+}$ because it has the larger $K_{\text{stab}}$ and is therefore more stable.

Final answer

[Cd(en)2]^2+; larger Kstab / more stable

Detailed explanation

Background Concept

$K_{\text{stab}}$ is an equilibrium constant for formation of a complex ion. For $[\text{Cd(CH}_3\text{NH}_2)_4]^{2+}$ , $K_{\text{stab}} = \frac{[[\text{Cd(CH}_3\text{NH}_2)_4]^{2+}]}{[\text{Cd}^{2+}][\text{CH}_3\text{NH}_2]^4}$ . The larger the value, the more the equilibrium lies to the right (towards the complex). A larger $K_{\text{stab}}$ means a more stable complex and a higher concentration at equilibrium under otherwise comparable conditions.

Understanding the Question

Equal concentrations of methylamine and en are present. We are asked to predict which complex forms at higher concentration and to explain. The key data are the two stability constants in Table 6.1.

Approach

Compare the numerical values of $K_{\text{stab}}$ . The complex with the larger stability constant is the more stable one and will be present in higher concentration at equilibrium.

Step-by-Step Reasoning

From Table 6.1:

$[\text{Cd(CH}_3\text{NH}_2)_4]^{2+}$ : $K_{\text{stab}} = 3.5 \times 10^6$
$[\text{Cd(en)}_2]^{2+}$ : $K_{\text{stab}} = 4.0 \times 10^{10}$

The value for the en complex is much larger. A larger stability constant means that the formation equilibrium lies further to the right, so the concentration of that complex is higher. Hence $[\text{Cd(en)}_2]^{2+}$ forms at the higher concentration.

Key Takeaways

Higher $K_{\text{stab}}$ → more stable complex → higher concentration at equilibrium.

Common Mistakes

Choosing the methylamine complex because it is listed first.
Saying 'smaller Kstab means more complex forms'.

Things to Be Careful About

Do not compare the two complexes solely on formula size; the stability constant is the deciding factor. The bidentate en ligand gives the chelate effect, but the mark is awarded for the larger $K_{\text{stab}}$ .

Techniques used

compare stability constantspredict which complex predominates at equilibrium

(iii)

Complete the expression for the $K_{stab}$ of $[\text{Cd(CH}_3\text{NH}_2)_4]^{2+}$ .

$K_{stab}$ =

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

K_{\text{stab}} = \frac{[[\text{Cd}(\text{CH}_3\text{NH}_2)_4]^{2+}]}{[\text{Cd}^{2+}][\text{CH}_3\text{NH}_2]^4}

Final answer

Kstab = [[Cd(CH3NH2)4]^2+] / ([Cd2+][CH3NH2]^4)

Detailed explanation

Background Concept

The stability constant, $K_{\text{stab}}$ , is the equilibrium constant for the formation of a complex ion from the metal ion and its ligands. For a general reaction

\text{Cd}^{2+} + 4\text{CH}_3\text{NH}_2 \rightleftharpoons [\text{Cd}(\text{CH}_3\text{NH}_2)_4]^{2+}

the expression is written as concentration of the complex divided by the product of the concentrations of the reactants, each raised to the power of its stoichiometric coefficient.

Understanding the Question

The question gives the complex ion and asks us to complete the $K_{\text{stab}}$ expression. We need to include the complex ion as the numerator and $\text{Cd}^{2+}$ and four methylamine molecules as the denominator terms.

Approach

Write the formation equilibrium in words, then convert it into an equilibrium expression. The coefficient 4 in front of $\text{CH}_3\text{NH}_2$ becomes the power 4 in the denominator.

Step-by-Step Reasoning

The formation equilibrium is:

\text{Cd}^{2+}(\text{aq}) + 4\text{CH}_3\text{NH}_2(\text{aq}) \rightleftharpoons [\text{Cd}(\text{CH}_3\text{NH}_2)_4]^{2+}(\text{aq})

So:

K_{\text{stab}} = \frac{[[\text{Cd}(\text{CH}_3\text{NH}_2)_4]^{2+}]}{[\text{Cd}^{2+}][\text{CH}_3\text{NH}_2]^4}

Concentration terms are shown in square brackets. The ligand appears to the fourth power because four methylamine molecules are involved in the formation of the complex.

Key Takeaways

$K_{\text{stab}}$ always has the complex ion on top.
Each ligand concentration is raised to the power equal to the number of ligands in the formula.
Neutral ligands such as methylamine are included in the expression because they are present in aqueous solution.

Common Mistakes

Putting the ligand in the numerator.
Forgetting the power 4 on $[\text{CH}_3\text{NH}_2]$ .
Omitting square-bracket concentration notation.

Things to Be Careful About

The expression must include the four methylamine molecules to the power 4. There are no state symbols needed, but square brackets around every concentration term are essential.

Techniques used

write an equilibrium constant expression for complex formationuse ligand stoichiometric coefficients as powers

Q7MediumIntroduction to A Level Organic ChemistryAnalytical Techniques

P, Q, R, S, T, U, and V are the seven structural isomers with molecular formula $\text{C}_5\text{H}_{10}\text{O}$ that have a carbonyl group.

P $\text{CH}_3(\text{CH}_2)_3\text{CHO}$
Q $\text{CH}_3\text{CH}_2\text{CH(CH}_3)\text{CHO}$
R $(\text{CH}_3)_2\text{CHCH}_2\text{CHO}$
S $(\text{CH}_3)_3\text{CCHO}$
T $\text{CH}_3\text{CH}_2\text{CH}_2\text{COCH}_3$
U $\text{CH}_3\text{CH}_2\text{COCH}_2\text{CH}_3$
V $(\text{CH}_3)_2\text{CHCOCH}_3$

(a)

Only one of these seven compounds has stereoisomers.

Draw three-dimensional diagrams of the two stereoisomers of this compound.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

The compound with stereoisomers is Q: $\text{CH}_3\text{CH}_2\text{CH(CH}_3\text{)CHO}$ (2-methylbutanal).

The carbon atom bonded to $-\text{CHO}$ , $-\text{CH}_3$ , $-\text{C}_2\text{H}_5$ and $-\text{H}$ is a chiral centre (four different groups).

The two stereoisomers are non-superimposable mirror images (enantiomers).

Final answer

Q (2-methylbutanal); two enantiomers drawn with wedge/dash bonds showing the chiral carbon bonded to -H, -CH3, -C2H5 and -CHO

Detailed explanation

Background Concept

A carbon atom is a chiral centre (stereocentre) when it is bonded to four different groups. Such a carbon is $sp^3$ -hybridised and tetrahedral. A molecule with one chiral centre exists as a pair of enantiomers — non-superimposable mirror images. Enantiomers have identical physical properties except for the direction in which they rotate plane-polarised light.

In organic chemistry, the only carbonyl compounds among the $\text{C}_5\text{H}_{10}\text{O}$ isomers that can have a chiral centre are the aldehydes (since the carbonyl carbon in an aldehyde is bonded to H and cannot be chiral). The chiral centre must be elsewhere in the chain.

Understanding the Question

We are given seven structural isomers of $\text{C}_5\text{H}_{10}\text{O}$ containing a carbonyl group: four aldehydes (P, Q, R, S) and three ketones (T, U, V). We must identify which one has stereoisomers and draw both stereoisomers in three dimensions.

Approach

Examine each compound for a carbon atom bonded to four different substituents. For aldehydes, the $-\text{CHO}$ carbon is bonded to H, =O and the rest of the chain — only three groups, so it cannot be chiral. Check the other carbons.

Step-by-Step Reasoning

Checking each compound:

P $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{CHO}$ : no carbon has four different groups.
Q $\text{CH}_3\text{CH}_2\text{CH(CH}_3\text{)CHO}$ : the carbon bearing the methyl group is bonded to $-\text{H}$ , $-\text{CH}_3$ , $-\text{C}_2\text{H}_5$ and $-\text{CHO}$ — four different groups. This is a chiral centre. ✓
R $(\text{CH}_3)_2\text{CHCH}_2\text{CHO}$ : the CH carbon is bonded to two identical $\text{CH}_3$ groups — not chiral.
S $(\text{CH}_3)_3\text{CCHO}$ : the quaternary carbon has three identical $\text{CH}_3$ groups — not chiral.
T, U, V: ketones; none have a carbon with four different groups.

Drawing the enantiomers of Q:

The chiral carbon is drawn in the centre with four bonds in a tetrahedral arrangement. One enantiomer has $-\text{CH}_3$ on a wedge (coming out) and $-\text{H}$ on a dash (going in); the other has them swapped. The $-\text{CHO}$ and $-\text{C}_2\text{H}_5$ groups occupy the remaining two positions.

Key Takeaways

A chiral centre requires four different groups on one $sp^3$ carbon.
Enantiomers are drawn using wedge-and-dash notation to show 3D geometry.
Aldehydes with a branched alkyl chain can be chiral; ketones with a methyl group on each side of the carbonyl often have symmetry.

Common Mistakes

Drawing the carbonyl carbon as chiral — it only has three groups (H, =O, R), not four.
Forgetting that two identical groups (e.g. two $\text{CH}_3$ on R or S) destroy chirality.
Drawing flat 2D structures without wedge/dash bonds to show 3D arrangement.

Things to Be Careful About

The drawing must show four different groups on the central carbon with correct 3D representation (wedge = towards viewer, dash = away).
Both enantiomers must be drawn; they are mirror images, not the same molecule.
Do not confuse structural isomerism with stereoisomerism — only Q has a chiral centre among these seven.

Techniques used

identify the chiral centredraw 3D enantiomer representations using wedge-and-dash bonds

(b)

P, Q, R, S, T, U, and V are treated separately with alkaline $\text{I}_2\text{(aq)}$ and the product mixture is acidified.

3M

(i)

Identify the two compounds that give a positive result with alkaline $\text{I}_2\text{(aq)}$ .

............................................................... and ...............................................................

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

T and V

$\text{CH}_3\text{CH}_2\text{CH}_2\text{COCH}_3$ and $(\text{CH}_3)_2\text{CHCOCH}_3$

Both contain the $\text{CH}_3\text{CO}-$ (methyl ketone) group required for a positive iodoform test.

Final answer

T and V

Detailed explanation

Background Concept

The iodoform test uses alkaline iodine ( $\text{I}_2$ in $\text{NaOH}$ / $\text{KI}$ ) to detect compounds containing a methyl ketone group ( $\text{CH}_3\text{CO}-$ ) or a methyl carbinol group ( $\text{CH}_3\text{CH(OH)}-$ ). The reaction replaces all three hydrogens of the $\text{CH}_3$ group with iodine, forming $\text{CI}_3\text{CO}-$ , which then undergoes cleavage in base to give a carboxylate ion and iodoform ( $\text{CHI}_3$ ), a pale yellow precipitate.

After acidification, the carboxylate is converted to the carboxylic acid.

Understanding the Question

Seven carbonyl isomers of $\text{C}_5\text{H}_{10}\text{O}$ are treated with alkaline $\text{I}_2\text{(aq)}$ . We must identify which two give a positive result.

Approach

A positive iodoform test requires a $\text{CH}_3\text{CO}-$ group. Scan the seven compounds for this structural feature.

Step-by-Step Reasoning

P $\text{CH}_3(\text{CH}_2)_3\text{CHO}$ : aldehyde, no $\text{CH}_3\text{CO}-$ → negative.
Q $\text{CH}_3\text{CH}_2\text{CH(CH}_3\text{)CHO}$ : aldehyde, no $\text{CH}_3\text{CO}-$ → negative.
R $(\text{CH}_3)_2\text{CHCH}_2\text{CHO}$ : aldehyde, no $\text{CH}_3\text{CO}-$ → negative.
S $(\text{CH}_3)_3\text{CCHO}$ : aldehyde, no $\text{CH}_3\text{CO}-$ → negative.
T $\text{CH}_3\text{CH}_2\text{CH}_2\text{COCH}_3$ : methyl ketone ✓ → positive.
U $\text{CH}_3\text{CH}_2\text{COCH}_2\text{CH}_3$ : ethyl ketone, no $\text{CH}_3\text{CO}-$ → negative.
V $(\text{CH}_3)_2\text{CHCOCH}_3$ : methyl ketone ✓ → positive.

The two compounds are T (pentan-2-one) and V (3-methylbutan-2-one).

Key Takeaways

The iodoform test is positive for $\text{CH}_3\text{CO}-$ (methyl ketones) and $\text{CH}_3\text{CH(OH)}-$ (secondary alcohols with a methyl group).
Aldehydes (except ethanal) and ketones without a methyl group on the carbonyl side give negative results.

Common Mistakes

Selecting U ( $\text{CH}_3\text{CH}_2\text{COCH}_2\text{CH}_3$ ) — it has methyl groups but neither is directly attached to the carbonyl as $\text{CH}_3\text{CO}-$ .
Selecting any aldehyde — aldehydes other than ethanal do not give the iodoform test.

Things to Be Careful About

The $\text{CH}_3$ must be directly bonded to the carbonyl carbon ( $\text{CH}_3\text{CO}-$ ), not just present somewhere in the molecule.
After acidification, the product is the carboxylic acid, not the carboxylate salt.

Techniques used

recognise the methyl ketone structural requirement for the iodoform test

(ii)

Describe the observations when one of the compounds you have identified in (b)(i) is treated with alkaline $\text{I}_2\text{(aq)}$ and give the structural formulae of the two carbon-containing products of this reaction.

observations ......................................................................................................................

two carbon-containing products ........................................................................................

and ........................................................................................

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Observations: A pale yellow precipitate (solid) forms.

Two carbon-containing products (using T as the example):

$\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}$ (butanoic acid) and $\text{CHI}_3$ (iodoform)

(Using V: $(\text{CH}_3)_2\text{CHCOOH}$ (2-methylpropanoic acid) and $\text{CHI}_3$ )

Final answer

Yellow precipitate; CH3CH2CH2COOH and CHI3 (or (CH3)2CHCOOH and CHI3)

Detailed explanation

Background Concept

In the iodoform reaction, a methyl ketone $\text{R-CO-CH}_3$ is treated with $\text{I}_2$ / $\text{NaOH}$ . The three hydrogens of the methyl group are successively replaced by iodine to give $\text{R-CO-CI}_3$ . The $\text{CI}_3^-$ group is then a good leaving group and is expelled by hydroxide attack on the carbonyl carbon, producing $\text{R-COO}^-$ (carboxylate) and $\text{CHI}_3$ (iodoform).

Upon acidification, the carboxylate $\text{R-COO}^-$ is protonated to give the carboxylic acid $\text{R-COOH}$ .

Iodoform ( $\text{CHI}_3$ ) is a pale yellow solid — this is the characteristic positive observation.

Understanding the Question

We need to describe the visual observation and write the structural formulae of the two carbon-containing products after acidification, using one of the compounds identified in (b)(i).

Approach

State the observation (yellow precipitate). For the products, identify R in $\text{R-CO-CH}_3$ and write R-COOH and CHI3.

Step-by-Step Reasoning

Observation: The formation of a yellow precipitate of iodoform ( $\text{CHI}_3$ ) is the positive result.

Products from T ( $\text{CH}_3\text{CH}_2\text{CH}_2\text{COCH}_3$ ):

R = $\text{CH}_3\text{CH}_2\text{CH}_2-$
Carboxylic acid: $\text{CH}_3\text{CH}_2\text{CH}_2\text{COOH}$ (butanoic acid)
Iodoform: $\text{CHI}_3$

Products from V ( $(\text{CH}_3)_2\text{CHCOCH}_3$ ):

R = $(\text{CH}_3)_2\text{CH}-$
Carboxylic acid: $(\text{CH}_3)_2\text{CHCOOH}$ (2-methylpropanoic acid)
Iodoform: $\text{CHI}_3$

Either pair is acceptable.

Key Takeaways

The iodoform test gives a yellow precipitate of $\text{CHI}_3$ .
The organic product is a carboxylic acid with one fewer carbon than the original methyl ketone (the $\text{COCH}_3$ becomes $\text{COOH}$ ).
Acidification converts the carboxylate to the carboxylic acid.

Common Mistakes

Writing the carboxylate salt ( $\text{R-COO}^-\text{Na}^+$ ) instead of the carboxylic acid — the question says the mixture is acidified.
Forgetting $\text{CHI}_3$ as one of the products.
Writing incorrect structural formulae for the carboxylic acid.

Things to Be Careful About

The question asks for structural formulae, not names.
Both carbon-containing products must be given; $\text{CHI}_3$ counts as a carbon-containing product.

Techniques used

describe iodoform test observationswrite structural formulae of iodoform reaction products

(c)

The proton ( $^1\text{H}$ ) NMR spectra of P, Q, R, S, T, U, and V are compared.

4M

(i)

Identify the only compound that gives a spectrum with two singlets and no other peaks.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

S

$(\text{CH}_3)_3\text{CCHO}$

The nine $\text{CH}_3$ protons are all equivalent (one singlet) and the $\text{CHO}$ proton has no neighbouring protons (another singlet). Two singlets, no other peaks.

Final answer

S

Detailed explanation

Background Concept

In $^1\text{H}$ NMR spectroscopy, each set of chemically equivalent protons gives one peak. The number of peaks equals the number of different proton environments. The splitting pattern follows the $n+1$ rule: a proton signal is split into $n+1$ peaks by $n$ equivalent protons on adjacent carbons. A singlet means no neighbouring protons ( $n=0$ ).

Understanding the Question

Among the seven isomers, find the one whose $^1\text{H}$ NMR spectrum shows exactly two singlets and nothing else.

Approach

For each compound, count the number of proton environments and check whether each environment has neighbouring protons. We need exactly two environments, both with zero neighbouring protons.

Step-by-Step Reasoning

Checking S $(\text{CH}_3)_3\text{CCHO}$ (2,2-dimethylpropanal / pivaldehyde):

The three $\text{CH}_3$ groups are all equivalent (attached to the same quaternary carbon) → one environment, 9H. The quaternary carbon has no H, and the $\text{CHO}$ H is too far to couple effectively → singlet.
The $\text{CHO}$ proton → one environment, 1H. No adjacent H (the quaternary C has no H) → singlet.
Total: two singlets, no other peaks. ✓

Other compounds have more than two environments or have neighbouring protons causing splitting.

Key Takeaways

Symmetrical molecules can have very few NMR peaks.
A quaternary carbon between proton groups eliminates coupling, giving singlets.
The aldehyde proton ( $\text{CHO}$ ) typically appears at $\delta$ 9–10 ppm as a singlet if there are no adjacent protons.

Common Mistakes

Choosing P ( $\text{CH}_3(\text{CH}_2)_3\text{CHO}$ ) — it has multiple environments with splitting.
Forgetting that the $\text{CHO}$ proton is a separate peak.

Things to Be Careful About

"Two singlets and no other peaks" means exactly two peaks total, both singlets.

Techniques used

count chemically equivalent proton environmentsapply the n+1 splitting rule

(ii)

Fig. 7.1 shows the spectrum obtained from one of the compounds.

Fig. 7.1

Identify the compound that gives this spectrum.

1M Similar

DifficultyMedium

Worked solution

Answer

U

$\text{CH}_3\text{CH}_2\text{COCH}_2\text{CH}_3$ (pentan-3-one)

The spectrum shows two peaks: a triplet at $\delta \approx 1.0$ ppm (6H, two equivalent $\text{CH}_3$ groups) and a quartet at $\delta \approx 2.4$ ppm (4H, two equivalent $\text{CH}_2$ groups adjacent to the carbonyl). This matches the symmetric structure of U.

Final answer

U

Detailed explanation

Background Concept

In $^1\text{H}$ NMR:

Chemical shift ( $\delta$ ) indicates the electronic environment: $\text{CH}_3$ groups appear at $\delta$ 0.8–1.2 ppm; $\text{CH}_2$ adjacent to a carbonyl appears at $\delta$ 2.0–2.5 ppm.
Splitting follows $n+1$ : a $\text{CH}_3$ next to a $\text{CH}_2$ ( $n=2$ ) gives a triplet; a $\text{CH}_2$ next to a $\text{CH}_3$ ( $n=3$ ) gives a quartet.
Integration (peak area) is proportional to the number of protons.

Understanding the Question

Fig. 7.1 shows an NMR spectrum with two main peaks (plus the TMS reference at 0 ppm): a tall peak at $\delta \approx 1.0$ ppm and a shorter peak at $\delta \approx 2.4$ ppm. Identify which compound produces this spectrum.

Approach

Two peaks means only two proton environments. The chemical shifts suggest $\text{CH}_3$ (1.0 ppm) and $\text{CH}_2$ next to C=O (2.4 ppm). The splitting (triplet/quartet) indicates $\text{CH}_3$ - $\text{CH}_2$ coupling. Look for a symmetric molecule with this pattern.

Step-by-Step Reasoning

U $\text{CH}_3\text{CH}_2\text{COCH}_2\text{CH}_3$ (pentan-3-one):

The molecule is symmetric about the carbonyl.
Two equivalent $\text{CH}_3$ groups → 6H, coupled to adjacent $\text{CH}_2$ ( $n=2$ ) → triplet at $\delta \approx 1.0$ ppm.
Two equivalent $\text{CH}_2$ groups → 4H, coupled to adjacent $\text{CH}_3$ ( $n=3$ ) → quartet at $\delta \approx 2.4$ ppm (deshielded by carbonyl).
Total: two peaks (triplet + quartet). ✓

Other candidates:

T $\text{CH}_3\text{CH}_2\text{CH}_2\text{COCH}_3$ : four different environments → more than two peaks.
V $(\text{CH}_3)_2\text{CHCOCH}_3$ : three environments (isopropyl CH, isopropyl $\text{CH}_3$ , acetyl $\text{CH}_3$ ) → three peaks.
R, Q, P, S: aldehydes with more complex patterns or the aldehyde peak at $\delta$ 9–10.

Key Takeaways

Symmetric molecules give fewer NMR peaks.
A triplet at ~1 ppm and quartet at ~2.4 ppm is characteristic of an ethyl group next to a carbonyl.
Two equivalent ethyl groups on either side of a carbonyl give exactly two peaks.

Common Mistakes

Choosing T — it has four proton environments, not two.
Misreading the splitting pattern (triplet vs doublet, quartet vs septet).

Things to Be Careful About

The integration ratio (6H:4H = 3:2) confirms two equivalent $\text{CH}_3$ and two equivalent $\text{CH}_2$ groups.
The carbonyl carbon has no protons, so it doesn't appear in $^1\text{H}$ NMR.

Techniques used

interpret splitting patterns to identify proton environmentsmatch NMR spectrum to molecular structure

(iii)

Name the splitting pattern of the peak at $\delta = 1.1$ in Fig. 7.1.

Give the reason for this splitting.

name ..............................

reason ...............................................................................................................................

1M Similar

DifficultyEasy

Worked solution

Answer

Name: triplet

Reason: The protons are coupled to two hydrogen atoms on the neighbouring carbon atom ( $n=2$ , so $n+1=3$ peaks).

Final answer

triplet; two hydrogen atoms on neighbouring carbon atoms

Detailed explanation

Background Concept

The $n+1$ rule in $^1\text{H}$ NMR: a proton signal is split into $n+1$ peaks by $n$ equivalent protons on adjacent (neighbouring) carbon atoms. Protons on the same carbon do not split each other (in simple spectra). Protons separated by more than three bonds generally do not couple.

$n=0$ : singlet
$n=1$ : doublet
$n=2$ : triplet
$n=3$ : quartet

Understanding the Question

In Fig. 7.1, the peak at $\delta = 1.1$ ppm is the $\text{CH}_3$ signal of compound U ( $\text{CH}_3\text{CH}_2\text{COCH}_2\text{CH}_3$ ). Name the splitting and explain why.

Approach

The $\text{CH}_3$ group is adjacent to a $\text{CH}_2$ group (2 H atoms). Apply $n+1 = 2+1 = 3$ → triplet.

Step-by-Step Reasoning

The peak at $\delta \approx 1.1$ ppm corresponds to the $\text{CH}_3$ protons in U. These protons are on a carbon adjacent to a $\text{CH}_2$ group. The $\text{CH}_2$ has 2 hydrogen atoms ( $n=2$ ). By the $n+1$ rule, the signal is split into $2+1 = 3$ peaks → a triplet.

Key Takeaways

Triplet = coupled to 2 neighbouring H atoms.
The $n+1$ rule is the fundamental principle behind splitting patterns.

Common Mistakes

Saying "because there are 3 peaks" without explaining the cause (neighbouring protons).
Confusing which protons are causing the splitting (it's the neighbours, not the protons giving the signal).

Things to Be Careful About

The reason must mention the number of neighbouring hydrogen atoms (two), not just "coupling".
Use precise language: "two hydrogen atoms on the neighbouring carbon atom".

Techniques used

name splitting patternsapply the n+1 rule to explain coupling

(iv)

Identify the substance that gives the small peak at $\delta = 0$ in Fig. 7.1.

1M Similar

(d)

The carbon-13 NMR spectra of R, S, T and U are compared.

Complete Table 7.1 to state the number of peaks in the spectrum of each compound.

Table 7.1

compound	number of peaks
R $(\text{CH}_3)_2\text{CHCH}_2\text{CHO}$
S $(\text{CH}_3)_3\text{CCHO}$
T $\text{CH}_3\text{CH}_2\text{CH}_2\text{COCH}_3$
U $\text{CH}_3\text{CH}_2\text{COCH}_2\text{CH}_3$

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

compound	number of peaks
R $(\text{CH}_3)_2\text{CHCH}_2\text{CHO}$	4
S $(\text{CH}_3)_3\text{CCHO}$	3
T $\text{CH}_3\text{CH}_2\text{CH}_2\text{COCH}_3$	5
U $\text{CH}_3\text{CH}_2\text{COCH}_2\text{CH}_3$	3

Reasoning:

R: $(\text{CH}_3)_2$ (1), CH (1), $\text{CH}_2$ (1), CHO (1) = 4 unique carbons.
S: $(\text{CH}_3)_3$ (1), C (quaternary, 1), CHO (1) = 3 unique carbons.
T: $\text{CH}_3$ (1), $\text{CH}_2$ (1), $\text{CH}_2$ (1), CO (1), $\text{CH}_3$ (1) = 5 unique carbons (no symmetry).
U: $\text{CH}_3$ (1, symmetric), $\text{CH}_2$ (1, symmetric), CO (1) = 3 unique carbons.

Final answer

R: 4, S: 3, T: 5, U: 3

Detailed explanation

Background Concept

In $^{13}\text{C}$ NMR spectroscopy, each chemically unique carbon atom gives one peak. The number of peaks equals the number of different carbon environments. Molecular symmetry can make carbons equivalent, reducing the number of peaks. For example, two $\text{CH}_3$ groups on the same carbon are equivalent and give one peak.

Understanding the Question

Complete a table giving the number of peaks in the $^{13}\text{C}$ NMR spectrum of four compounds: R, S, T, U.

Approach

For each compound, identify all carbon atoms and determine which are chemically equivalent (related by symmetry or identical bonding environments). Count the unique environments.

Step-by-Step Reasoning

R $(\text{CH}_3)_2\text{CHCH}_2\text{CHO}$ (3-methylbutanal):

$(\text{CH}_3)_2$ : two equivalent methyl carbons → 1 peak
CH (methine): → 1 peak
$\text{CH}_2$ : → 1 peak
CHO (carbonyl): → 1 peak
Total: 4 peaks

S $(\text{CH}_3)_3\text{CCHO}$ (2,2-dimethylpropanal):

$(\text{CH}_3)_3$ : three equivalent methyl carbons → 1 peak
C (quaternary, bonded to 3 $\text{CH}_3$ and CHO): → 1 peak
CHO (carbonyl): → 1 peak
Total: 3 peaks

T $\text{CH}_3\text{CH}_2\text{CH}_2\text{COCH}_3$ (pentan-2-one):

$\text{CH}_3$ (C1, attached to CO): → 1 peak
$\text{CH}_2$ (C3, attached to CO): → 1 peak
$\text{CH}_2$ (C4): → 1 peak
$\text{CH}_2$ $CH_{2}$ (C3... wait, let me re-number):
- $\text{CH}_3$ -C1, $\text{CH}_2$ -C2, $\text{CH}_2$ -C3, CO-C4, $\text{CH}_3$ -C5
- C1 ( $\text{CH}_3$ next to $\text{CH}_2$ ): 1 peak
- C2 ( $\text{CH}_2$ next to C1 and C3): 1 peak
- C3 ( $\text{CH}_2$ next to C2 and C=O): 1 peak
- C4 (C=O): 1 peak
- C5 ( $\text{CH}_3$ next to C=O): 1 peak
No symmetry → 5 peaks

U $\text{CH}_3\text{CH}_2\text{COCH}_2\text{CH}_3$ (pentan-3-one):

Symmetric about the carbonyl.
Two equivalent $\text{CH}_3$ groups → 1 peak
Two equivalent $\text{CH}_2$ groups → 1 peak
C=O → 1 peak
Total: 3 peaks

Key Takeaways

$^{13}\text{C}$ NMR peak count = number of unique carbon environments.
Symmetry reduces the number of peaks (U has 3 peaks despite having 5 carbons).
Quaternary carbons (like the C in S) still give peaks.

Common Mistakes

Counting total carbons instead of unique environments (e.g. saying U has 5 peaks).
Forgetting that equivalent methyl groups give only one peak.
Missing the carbonyl carbon as a peak.

Things to Be Careful About

In T, the two $\text{CH}_3$ groups are in different environments (one next to C=O, one at the end of a propyl chain), so they give separate peaks.
In U, the two ethyl groups are mirror images, making the corresponding carbons equivalent.

Techniques used

count chemically equivalent carbon environmentsapply molecular symmetry to 13C NMR

Q8Medium-HardNitrogen CompoundsCarboxylic Acids and DerivativesPolymerisationIntroduction to A Level Organic Chemistry

Asparagine and aspartic acid are two naturally occurring amino acids. Their structures and isoelectric points are shown in Table 8.1.

Table 8.1

amino acid	structure	isoelectric point
asparagine	$\text{HOOCCH(NH}_2)\text{CH}_2\text{CONH}_2$	5.41
aspartic acid	$\text{HOOCCH(NH}_2)\text{CH}_2\text{COOH}$	2.77

(a)

Define isoelectric point.

1M Similar

DifficultyEasy

Worked solution

Answer

The pH at which an amino acid exists as a zwitterion (has no net charge).

Final answer

The pH at which an amino acid exists as a zwitterion

Detailed explanation

Background Concept

Amino acids contain both a basic amine group ( $-\text{NH}_2$ ) and an acidic carboxylic acid group ( $-\text{COOH}$ ). In aqueous solution, the amine group can accept a proton to form $-\text{NH}_3^+$ , and the carboxylic acid group can donate a proton to form $-\text{COO}^-$ . When both of these processes occur simultaneously within the same molecule, the result is a zwitterion (from the German for "hybrid ion"), which carries both a positive and a negative charge but has a net charge of zero.

Understanding the Question

The question asks for the definition of the isoelectric point (pI). This is a fundamental concept in the chemistry of amino acids and proteins, relating directly to the pH at which the molecule's ionization state changes.

Approach

Recall the definition of the isoelectric point in terms of the net charge or the specific structural form (zwitterion) of the amino acid.

Step-by-Step Reasoning

At very low pH (acidic conditions), the amino acid exists as a cation (e.g., $\text{H}_3\text{N}^+\text{CHR}\text{COOH}$ ). As the pH increases, the carboxylic acid group loses a proton to form the zwitterion ( $\text{H}_3\text{N}^+\text{CHR}\text{COO}^-$ ). As the pH increases further, the ammonium group loses a proton to form the anion ( $\text{H}_2\text{N}\text{CHR}\text{COO}^-$ ). The isoelectric point is the specific pH value where the concentration of the neutral zwitterion is at its maximum, and the average net charge of the population of molecules is zero.

Key Takeaways

The isoelectric point (pI) is the pH at which an amino acid is a zwitterion.
At the pI, the amino acid has no net electrical charge and will not migrate in an electric field.

Common Mistakes

Defining it as the pH where the solution is neutral (pH 7). The pI depends on the side chain and is rarely exactly 7.
Confusing the isoelectric point with the $pK_a$ values. The pI is often the average of the two $pK_a$ values flanking the zwitterion form.

Things to Be Careful About

Ensure the term "zwitterion" is used or clearly described (net charge of zero).

Techniques used

Define isoelectric pointDescribe zwitterion formation

(b)

Draw the structures of asparagine and aspartic acid at pH 2.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

At pH 2 (which is below the pI of both amino acids), both the amine and carboxyl groups are fully protonated, forming cations.

Asparagine:
$\text{HOOCCH(NH}_3^+)\text{CH}_2\text{CONH}_2$

Aspartic acid:
$\text{HOOCCH(NH}_3^+)\text{CH}_2\text{COOH}$

Final answer

Asparagine: HOOCCH(NH3+)CH2CONH2; Aspartic acid: HOOCCH(NH3+)CH2COOH

Detailed explanation

Background Concept

The ionization state of an amino acid depends on the pH of the solution relative to its $pK_a$ values.

If $\text{pH} < pK_a$ , the group is protonated.
If $\text{pH} > pK_a$ , the group is deprotonated.

For amino acids, the $pK_a$ of the $-\text{COOH}$ group is typically around 2-4, and the $pK_a$ of the $-\text{NH}_3^+$ group is around 9-10. At pH 2, the solution is very acidic. The amine group will definitely be protonated ( $-\text{NH}_3^+$ ). The carboxylic acid group will also be protonated ( $-\text{COOH}$ ) because the pH is at or below its $pK_a$ . Thus, the molecule exists as a positive ion (cation).

Understanding the Question

We need to draw the structures of asparagine and aspartic acid at pH 2.

Asparagine has a side chain with an amide group ( $-\text{CONH}_2$ ). Amides are not significantly basic or acidic in this pH range, so it remains unchanged.
Aspartic acid has a side chain with a second carboxylic acid group ( $-\text{COOH}$ ). Like the alpha-carboxyl group, this will be protonated at pH 2.

Approach

Start with the neutral structure.
Add a proton to the amine group to make $-\text{NH}_3^+$ .
Ensure all carboxylic acid groups are in the $-\text{COOH}$ form (not $-\text{COO}^-$ ).

Step-by-Step Reasoning

Asparagine:

Structure: $\text{HOOCCH(NH}_2)\text{CH}_2\text{CONH}_2$
At pH 2, $-\text{NH}_2$ becomes $-\text{NH}_3^+$ .
$-\text{COOH}$ remains $-\text{COOH}$ .
$-\text{CONH}_2$ remains $-\text{CONH}_2$ .
Result: $[\text{HOOCCH(NH}_3^+)\text{CH}_2\text{CONH}_2]$

Aspartic Acid:

Structure: $\text{HOOCCH(NH}_2)\text{CH}_2\text{COOH}$
At pH 2, $-\text{NH}_2$ becomes $-\text{NH}_3^+$ .
Both $-\text{COOH}$ groups remain $-\text{COOH}$ .
Result: $[\text{HOOCCH(NH}_3^+)\text{CH}_2\text{COOH}]$

Key Takeaways

At low pH, amino acids exist as cations with protonated amines and protonated carboxyls.
Side chains containing amides do not ionize in the same way as carboxylic acids or amines.

Common Mistakes

Leaving the carboxyl group as $-\text{COO}^-$ (zwitterion form) instead of $-\text{COOH}$ (cation form).
Forgetting to protonate the amine group.
Incorrectly ionizing the amide side chain of asparagine.

Things to Be Careful About

Include the positive charge on the nitrogen atom.
Ensure the correct number of hydrogens on the nitrogen ( $\text{NH}_3^+$ ).

Techniques used

Protonate amine group at low pHProtonate carboxylate group at low pHDraw structural formulas of cations

(c)

Asparagine and aspartic acid are treated separately with an excess of $\text{LiAlH}_4$ .

Draw the structures of the organic products of these reactions.

2M Similar

DifficultyMedium

Worked solution

Answer

$\text{LiAlH}_4$ reduces carboxylic acids to primary alcohols and amides to amines.

Asparagine product:
$\text{HOCH}_2\text{CH(NH}_2)\text{CH}_2\text{CH}_2\text{NH}_2$

Aspartic acid product:
$\text{HOCH}_2\text{CH(NH}_2)\text{CH}_2\text{CH}_2\text{OH}$

Final answer

Asparagine: HOCH2CH(NH2)CH2CH2NH2; Aspartic acid: HOCH2CH(NH2)CH2CH2OH

Detailed explanation

Background Concept

Lithium aluminium hydride ( $\text{LiAlH}_4$ ) is a powerful reducing agent. It acts as a source of hydride ions ( $\text{H}^-$ ).

Carboxylic acids ( $-\text{COOH}$ ) are reduced to primary alcohols ( $-\text{CH}_2\text{OH}$ ).
Amides ( $-\text{CONH}_2$ ) are reduced to primary amines ( $-\text{CH}_2\text{NH}_2$ ). The carbonyl oxygen is removed completely.
Amines ( $-\text{NH}_2$ ) are generally not reduced by $\text{LiAlH}_4$ .

Understanding the Question

We are treating asparagine and aspartic acid with excess $\text{LiAlH}_4$ . We must identify the functional groups in each and apply the reduction rules.

Asparagine: $\text{HOOC-CH(NH}_2)\text{-CH}_2\text{-CONH}_2$ . Contains a carboxylic acid, an amine, and an amide.
Aspartic acid: $\text{HOOC-CH(NH}_2)\text{-CH}_2\text{-COOH}$ . Contains two carboxylic acids and an amine.

Approach

Apply the reduction transformation to each reducible group independently.

Step-by-Step Reasoning

Asparagine:

The alpha-carboxyl group ( $-\text{COOH}$ ) reduces to a primary alcohol ( $-\text{CH}_2\text{OH}$ ).
The alpha-amino group ( $-\text{NH}_2$ ) is unaffected.
The side-chain amide group ( $-\text{CONH}_2$ ) reduces to a primary amine ( $-\text{CH}_2\text{NH}_2$ ).
Result: $\text{HOCH}_2\text{CH(NH}_2)\text{CH}_2\text{CH}_2\text{NH}_2$ .

Aspartic acid:

The alpha-carboxyl group ( $-\text{COOH}$ ) reduces to a primary alcohol ( $-\text{CH}_2\text{OH}$ ).
The alpha-amino group ( $-\text{NH}_2$ ) is unaffected.
The side-chain carboxyl group ( $-\text{COOH}$ ) reduces to a primary alcohol ( $-\text{CH}_2\text{OH}$ ).
Result: $\text{HOCH}_2\text{CH(NH}_2)\text{CH}_2\text{CH}_2\text{OH}$ .

Key Takeaways

$\text{LiAlH}_4$ converts $-\text{COOH}$ to $-\text{CH}_2\text{OH}$ .
$\text{LiAlH}_4$ converts $-\text{CONH}_2$ to $-\text{CH}_2\text{NH}_2$ .

Common Mistakes

Confusing amide reduction (to amine) with carboxylic acid reduction (to alcohol).
Thinking the existing amine group is reduced or removed.
Forgetting to reduce the side chain functional group.

Things to Be Careful About

Ensure the carbon count remains the same (the carbonyl carbon becomes a methylene carbon).
Draw the full displayed structure or correct semi-structure.

Techniques used

Reduce carboxylic acid to primary alcoholReduce amide to primary amineDraw organic product structures

(d)

Propanedioic acid, $\text{HOOCCH}_2\text{COOH}$ , is treated with an excess of thionyl chloride, $\text{SOCl}_2$ . Propanedioyl chloride, $\text{ClOCCH}_2\text{COCl}$ , is formed.

3M

(i)

Write an equation for this reaction.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

\text{HOOCCH}_2\text{COOH} + 2\text{SOCl}_2 \rightarrow \text{ClOCCH}_2\text{COCl} + 2\text{SO}_2 + 2\text{HCl}

Final answer

HOOCCH2COOH + 2SOCl2 -> ClOCCH2COCl + 2SO2 + 2HCl

Detailed explanation

Background Concept

Thionyl chloride ( $\text{SOCl}_2$ ) is a reagent used to convert carboxylic acids into acyl chlorides (acid chlorides). The reaction typically produces sulfur dioxide ( $\text{SO}_2$ ) and hydrogen chloride ( $\text{HCl}$ ) as gaseous by-products, which helps drive the reaction to completion.

Understanding the Question

Propanedioic acid (malonic acid) has two carboxylic acid groups. It reacts with an excess of thionyl chloride to form propanedioyl chloride (malonyl chloride). We need to write the balanced equation.

Approach

Write the reactants: $\text{HOOCCH}_2\text{COOH}$ and $\text{SOCl}_2$ .
Write the main organic product: $\text{ClOCCH}_2\text{COCl}$ .
Determine the stoichiometry: Since there are two $-\text{COOH}$ groups, two $\text{SOCl}_2$ molecules are needed.
Balance the by-products: Each substitution of $-\text{OH}$ by $-\text{Cl}$ releases one $\text{HCl}$ and the sulfur part forms $\text{SO}_2$ . So, 2 $\text{SO}_2$ and 2 $\text{HCl}$ .

Step-by-Step Reasoning

Reactants: $\text{C}_3\text{H}_4\text{O}_4 + 2\text{SOCl}_2$
Products: $\text{C}_3\text{H}_2\text{Cl}_2\text{O}_2 + 2\text{SO}_2 + 2\text{HCl}$
Check atoms:

C: 3 on both sides.
H: 4 on left, 2 (in organic) + 2 (in HCl) = 4 on right.
O: 4 on left, 2 (in organic) + 4 (in SO2) = 6? Wait. $\text{SOCl}_2$ has 1 O. Left: 4 (acid) + 2 (thionyl) = 6. Right: 2 (acid chloride) + 4 (SO2) = 6. Correct.
S: 2 on left, 2 on right.
Cl: 4 on left, 2 (in organic) + 2 (in HCl) = 4 on right.

Key Takeaways

Dicarboxylic acids react with 2 equivalents of $\text{SOCl}_2$ .
The by-products are $\text{SO}_2$ and $\text{HCl}$ .

Common Mistakes

Forgetting to double the coefficients for the dicarboxylic acid reaction.
Writing incorrect by-products (e.g., $\text{SO}_3$ ).

Things to Be Careful About

Ensure the equation is balanced.

Techniques used

Write balanced equation for acyl chloride formationIdentify by-products SO2 and HCl

(ii)

Propanedioyl chloride reacts with an excess of asparagine to form compound G with molecular formula $\text{C}_{11}\text{H}_{16}\text{N}_4\text{O}_8$ .

Each molecule of compound G has four amide groups.

Draw the structure of compound G.

2M Similar

DifficultyMedium-Hard

Worked solution

Answer

Asparagine has two nucleophilic groups: the amine ( $-\text{NH}_2$ ) and the amide nitrogen ( $-\text{CONH}_2$ ). Propanedioyl chloride has two electrophilic acyl chloride groups. The reaction forms amide linkages.

Structure of Compound G:

\text{H}_2\text{NCOCH}_2\text{CH}_2\text{CONHCH}(\text{CH}_2\text{CONH}_2)\text{COOH} \text{ (intermediate)}

Actually, the formula $\text{C}_{11}\text{H}_{16}\text{N}_4\text{O}_8$ suggests a 1:1 adduct where both the amine and the amide side chain of asparagine react with the diacid chloride? No, let's check the formula.
Asparagine: $\text{C}_4\text{H}_8\text{N}_2\text{O}_3$ . Propanedioyl chloride: $\text{C}_3\text{H}_2\text{Cl}_2\text{O}_2$ .
If 1 asparagine reacts with 1 propanedioyl chloride:
$\text{C}_7\text{H}_{10}\text{N}_2\text{O}_5$ . This is not $\text{C}_{11}$ .
If 2 asparagine molecules react with 1 propanedioyl chloride molecule:
$2(\text{C}_4\text{H}_8\text{N}_2\text{O}_3) + \text{C}_3\text{H}_2\text{Cl}_2\text{O}_2 - 2\text{HCl}$
$= \text{C}_8\text{H}_{16}\text{N}_4\text{O}_6 + \text{C}_3\text{H}_2\text{Cl}_2\text{O}_2 - 2\text{HCl}$
$= \text{C}_{11}\text{H}_{18}\text{N}_4\text{O}_8 - 2\text{HCl} = \text{C}_{11}\text{H}_{16}\text{N}_4\text{O}_8$ .
This matches the molecular formula exactly.

So, one molecule of propanedioyl chloride bridges two molecules of asparagine. The reaction occurs at the amine group of asparagine (primary amine is a better nucleophile than amide nitrogen).

Structure:

\text{HOOCCH(CH}_2\text{CONH}_2)\text{NHCOCH}_2\text{CONHCH(CH}_2\text{CONH}_2)\text{COOH}

Final answer

HOOCCH(CH2CONH2)NHCOCH2CONHCH(CH2CONH2)COOH

Detailed explanation

Background Concept

Amino acids react with acyl chlorides via nucleophilic addition-elimination. The primary amine group ( $-\text{NH}_2$ ) is a strong nucleophile and reacts readily with acyl chlorides to form amide (peptide) bonds. The amide group in the side chain of asparagine is much less nucleophilic and typically does not react under these conditions compared to the amine.

Understanding the Question

We are given the molecular formula of product G: $\text{C}_{11}\text{H}_{16}\text{N}_4\text{O}_8$ . We know the reactants are asparagine ( $\text{C}_4\text{H}_8\text{N}_2\text{O}_3$ ) and propanedioyl chloride ( $\text{C}_3\text{H}_2\text{Cl}_2\text{O}_2$ ). We need to determine the stoichiometry and connectivity.

Approach

Calculate the formula of a potential adduct.
Compare with the given formula to find the ratio of reactants.
Draw the structure based on the reactivity of the amine group with the acyl chloride groups.

Step-by-Step Reasoning

Stoichiometry Check:
- 1 Asparagine + 1 Propanedioyl chloride $\rightarrow$ Loss of 1 HCl.
  Formula: $(\text{C}_4\text{H}_8\text{N}_2\text{O}_3 + \text{C}_3\text{H}_2\text{Cl}_2\text{O}_2) - \text{HCl} = \text{C}_7\text{H}_9\text{N}_2\text{O}_5\text{Cl}$ . (Does not match given formula, and contains Cl).
- 2 Asparagine + 1 Propanedioyl chloride $\rightarrow$ Loss of 2 HCl.
  Formula: $2(\text{C}_4\text{H}_8\text{N}_2\text{O}_3) + \text{C}_3\text{H}_2\text{Cl}_2\text{O}_2 - 2\text{HCl}$
  $= \text{C}_8\text{H}_{16}\text{N}_4\text{O}_6 + \text{C}_3\text{H}_2\text{Cl}_2\text{O}_2 - \text{H}_2\text{Cl}_2$
  $= \text{C}_{11}\text{H}_{16}\text{N}_4\text{O}_8$ . Matches.
Connectivity:
- Propanedioyl chloride is a di-acid chloride: $\text{ClOC-CH}_2\text{-COCl}$ .
- It reacts with two molecules of asparagine.
- The nucleophilic site on asparagine is the $\alpha$ -amino group ( $-\text{NH}_2$ ).
- Two amide bonds are formed: $\text{Asn-NH-CO-CH}_2\text{-CO-NH-Asn}$ .
- The rest of the asparagine molecule (the $\text{CH(COOH)(CH}_2\text{CONH}_2)$ part) remains unchanged.
Structure Assembly:
- Left Asparagine residue: $\text{HOOC-CH(CH}_2\text{CONH}_2)\text{-NH-}$
- Central linker: $\text{-CO-CH}_2\text{-CO-}$
- Right Asparagine residue: $\text{-NH-CH(CH}_2\text{CONH}_2)\text{-COOH}$

Key Takeaways

Molecular formulas can be used to deduce the number of reactant molecules involved.
Acyl chlorides react with amines to form amides.
In amino acids, the amine group is the primary site of reaction with acylating agents.

Common Mistakes

Assuming the side-chain amide reacts.
Incorrect stoichiometry (1:1 instead of 2:1).
Forgetting that the carboxylic acid groups of asparagine remain as $-\text{COOH}$ (they do not react with the amine of another asparagine in this specific reaction context with the diacid chloride).

Things to Be Careful About

Ensure the final structure accounts for all atoms in the formula.
Draw the amide linkages correctly ( $\text{-CONH-}$ ).

Techniques used

Identify nucleophilic sites in amino acidForm amide bonds with acyl chlorideDetermine structure from molecular formula

(e)

Asparagine is hydrolysed with an excess of hot $\text{NaOH(aq)}$ .

Draw the structure of the organic product of this reaction.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Hot excess $\text{NaOH}$ hydrolyses the amide side chain to a carboxylate ion and neutralises the carboxylic acid group to a carboxylate ion.

Structure:

^-\text{OOCCH(NH}_2)\text{CH}_2\text{COO}^-

(Or drawn as the dianion with $-\text{NH}_2$ group intact)

Final answer

-OOCCH(NH2)CH2COO-

Detailed explanation

Background Concept

Amide Hydrolysis: Amides are hydrolysed by hot aqueous alkali ( $\text{NaOH}$ ) to form a carboxylate salt and ammonia (or an amine). The $-\text{CONH}_2$ group becomes $-\text{COO}^-$ .
Acid-Base Neutralization: Carboxylic acids react with bases like $\text{NaOH}$ to form carboxylate salts ( $-\text{COO}^-$ ).
Stability of Amines: The amine group ( $-\text{NH}_2$ ) is stable in basic conditions and does not react further (it remains neutral, not protonated).

Understanding the Question

Asparagine ( $\text{HOOCCH(NH}_2)\text{CH}_2\text{CONH}_2$ ) is treated with excess hot $\text{NaOH}$ . We need to determine the final organic product.

Approach

Identify the functional groups and their fate in hot excess alkali.

$-\text{COOH}$ (alpha): Neutralized to $-\text{COO}^-$ .
$-\text{NH}_2$ (alpha): Unchanged.
$-\text{CH}_2\text{-CONH}_2$ (side chain): Hydrolyzed to $-\text{CH}_2\text{-COO}^-$ .

Step-by-Step Reasoning

The alpha-carboxylic acid group loses a proton to the base: $\text{HOOC-} \rightarrow ^-\text{OOC-}$ .
The side-chain amide group undergoes hydrolysis. The $\text{C-N}$ bond breaks. The carbonyl carbon becomes a carboxylate ion ( $-\text{COO}^-$ ) in the alkaline medium, and the nitrogen leaves as ammonia ( $\text{NH}_3$ ).
Reaction: $\text{-CH}_2\text{CONH}_2 + \text{OH}^- \rightarrow \text{-CH}_2\text{COO}^- + \text{NH}_3$ .
The alpha-amine group remains as $-\text{NH}_2$ .
Combine these parts: The central $\text{CH}$ is bonded to $\text{NH}_2$ , $\text{COO}^-$ , and $\text{CH}_2\text{COO}^-$ .

Key Takeaways

Amides hydrolyse to carboxylates in base.
Carboxylic acids form carboxylates in base.
Amines are stable in base.

Common Mistakes

Leaving the carboxylic acid as $-\text{COOH}$ (forgetting neutralization).
Leaving the amide as $-\text{CONH}_2$ (forgetting hydrolysis).
Protonating the amine to $-\text{NH}_3^+$ (this happens in acid, not base).

Things to Be Careful About

The product is a dianion (charge of -2). Ensure both negative charges are shown.

Techniques used

Hydrolyse amide to carboxylateNeutralise carboxylic acid to carboxylateDraw anionic structure

(f)

A polymer can form from asparagine, $\text{HOOCCH(NH}_2)\text{CH}_2\text{CONH}_2$ , as the only monomer.

Draw a length of the polymer chain containing three monomer residues.

Clearly label the repeat unit of the polymer on your diagram.

3M Similar

DifficultyMedium

Worked solution

Answer

Asparagine has both amine ( $-\text{NH}_2$ ) and carboxylic acid ( $-\text{COOH}$ ) groups, allowing it to form a polyamide (polypeptide) via condensation polymerisation.

Structure of a length containing three residues:

\text{-[NH-CH(CH}_2\text{CONH}_2)\text{-CO]-[NH-CH(CH}_2\text{CONH}_2)\text{-CO]-[NH-CH(CH}_2\text{CONH}_2)\text{-CO]-}

Final answer

Polymer chain: -[NH-CH(CH2CONH2)-CO]-[NH-CH(CH2CONH2)-CO]-[NH-CH(CH2CONH2)-CO]-

Detailed explanation

Background Concept

Amino acids can undergo condensation polymerisation to form polyamides (proteins/polypeptides). This occurs between the amine group of one molecule and the carboxylic acid group of another, eliminating a water molecule and forming an amide (peptide) linkage ( $-\text{CONH}-$ ). The side chain (R group) remains unchanged.

Understanding the Question

We need to draw a polymer chain formed from asparagine ( $\text{R} = \text{-CH}_2\text{CONH}_2$ ). The chain must contain three monomer residues, and the repeat unit must be clearly labelled.

Approach

Draw the backbone: $\text{-N-C-C-}$ .
Add the peptide linkages between residues.
Add the side chain of asparagine to each alpha-carbon.
Bracket the repeat unit.

Step-by-Step Reasoning

Monomer: Asparagine is $\text{H}_2\text{N-CH(R)-COOH}$ .
Linkage: $\text{-CO-NH-}$ .
Side Chain (R): $\text{-CH}_2\text{CONH}_2$ .
Drawing the Chain:
- Start with an open bond (or H) at the N-terminus.
- Unit 1: $\text{-NH-CH(CH}_2\text{CONH}_2)\text{-CO-}$
- Unit 2: $\text{-NH-CH(CH}_2\text{CONH}_2)\text{-CO-}$
- Unit 3: $\text{-NH-CH(CH}_2\text{CONH}_2)\text{-CO-}$
- End with an open bond (or OH) at the C-terminus.
Repeat Unit: The segment $\text{-[NH-CH(CH}_2\text{CONH}_2)\text{-CO]-}$ is the repeat unit.

Key Takeaways

Polyamides from amino acids have the backbone $\text{-[NH-CHR-CO]-}$ .
The side chain must be correctly attached to the alpha-carbon.
Condensation polymerisation eliminates small molecules (water).

Common Mistakes

Drawing the side chain incorrectly (e.g., forgetting the amide in the side chain).
Incorrect linkage orientation (e.g., $\text{-NH-NH-}$ or $\text{-CO-CO-}$ ).
Failing to label the repeat unit with brackets.
Drawing only two residues instead of three.

Things to Be Careful About

Ensure the bonds extending from the ends of the chain are shown (dangling bonds) to indicate the polymer continues.
The side chain amide group ( $-\text{CONH}_2$ ) is distinct from the backbone amide linkages.

Techniques used

Identify functional groups for polymerisationDraw peptide linkagesConstruct polymer chain with repeat units

(g)

Aspartic acid exists in two optically active forms.

3M

(i)

Plane polarised light is passed through pure samples of these two optically active forms in solutions of the same concentration.

Describe two similarities and one difference in their effect on the plane polarised light.

similarities .........................................................................................................................
...........................................................................................................................................

difference ...........................................................................................................................
...........................................................................................................................................

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Similarities:

Both rotate the plane of plane polarised light.
The magnitude (angle) of rotation is the same for both.

Difference:

They rotate the plane in opposite directions (one clockwise/dextrorotatory, the other anticlockwise/levorotatory).

Final answer

Similarities: Rotate light, same angle. Difference: Opposite directions.

Detailed explanation

Background Concept

Optical isomers (enantiomers) are non-superimposable mirror images. They have identical physical properties (boiling point, density) except for their interaction with plane polarised light and other chiral environments. A chiral molecule rotates the plane of polarisation of light. One enantiomer rotates it to the right (+/dextro), and the other rotates it to the left (-/levo) by exactly the same amount.

Understanding the Question

We have two optically active forms of aspartic acid (enantiomers). We need to describe how they affect plane polarised light, giving two similarities and one difference.

Approach

Recall the definition of optical activity for a pair of enantiomers.

Step-by-Step Reasoning

Effect: Both isomers are optically active, meaning they will both cause rotation of the plane of polarised light. (Similarity 1)
Magnitude: The specific rotation $[\alpha]$ is a physical constant for a given compound. For enantiomers, the absolute value is identical. (Similarity 2)
Direction: The sign of the rotation is opposite. One is $(+)$ , the other is $(-)$ . (Difference)

Key Takeaways

Enantiomers rotate light by the same angle but in opposite directions.
Optical activity is a key distinguishing property of chiral molecules.

Common Mistakes

Saying one rotates light and the other doesn't (that would be a meso compound or achiral molecule, but the question states they are optically active).
Confusing optical activity with the effect on other physical properties.

Things to Be Careful About

Be precise with terms: "rotate the plane of polarised light", "same angle", "opposite directions".

Techniques used

Describe optical activity of enantiomersCompare magnitude and direction of rotation

(ii)

Give the term used to describe a mixture of equal amounts of the two optically active forms.

1M Similar

Q9MediumIntroduction to A Level Organic ChemistryHydrocarbonsNitrogen CompoundsOrganic Synthesis

Compound X is made from benzene by the route shown in Fig. 9.1.

Fig. 9.1

(a)

Describe the bonding in benzene, $\text{C}_6\text{H}_6$ .

Your answer should include:

the hybridisation of the six carbon atoms
the types of bond between the carbon atoms
the orbitals that overlap to produce the bonds between the carbon atoms
the type of bond between the carbon atoms and the hydrogen atoms
the orbitals that overlap to produce the bonds between the carbon atoms and the hydrogen atoms.

3M Similar

DifficultyEasy

Worked solution

Answer

All carbon atoms are sp² hybridised. Bonds between carbon atoms are σ and π bonds. C–C σ bonds are formed by overlap of sp² hybrid orbitals. C–C π bonds are formed by overlap of p orbitals. Bonds between C and H atoms are σ bonds formed by overlap of sp² hybrid orbitals and s orbitals.

Final answer

All C atoms are sp² hybridised; C-C bonds are σ (sp²-sp² overlap) and π (p-p overlap); C-H bonds are σ (sp²-s overlap).

Detailed explanation

Background Concept

Benzene ( $\text{C}_6\text{H}_6$ ) is a planar, hexagonal molecule. Each carbon atom is bonded to two other carbon atoms and one hydrogen atom, using three electron pairs for σ bonding. This requires three hybrid orbitals, so the carbon atoms are sp² hybridised. The remaining unhybridised p orbital on each carbon is perpendicular to the plane of the ring. These six p orbitals overlap sideways to form a delocalised π system above and below the ring, which is often represented by a circle inside the hexagon. This delocalisation gives benzene exceptional stability (resonance energy).

Understanding the Question

The question asks for a description of the bonding in benzene, specifically requiring details on hybridisation, bond types (σ/π), and the orbitals involved in overlap for both C–C and C–H bonds. This is a standard recall question testing the structural model of benzene.

Approach

List the six key facts about benzene's bonding as outlined in the mark scheme: hybridisation, bond types, and orbital overlaps for both C–C and C–H connections.

Step-by-Step Reasoning

Hybridisation: Each carbon forms 3 σ bonds (2 to C, 1 to H), so it uses sp² hybrid orbitals. All six carbon atoms are sp² hybridised.
C–C bonds: Between carbons, there is a σ bond and a π bond. The σ bond comes from head-on overlap of sp² orbitals. The π bond comes from sideways overlap of the unhybridised p orbitals.
C–H bonds: These are σ bonds formed by the overlap of a carbon sp² hybrid orbital and a hydrogen 1s orbital.

Key Takeaways

Benzene bonding involves sp² hybridised carbons forming a σ framework (sp²-sp² for C–C, sp²-s for C–H) and a delocalised π system from p-orbital overlap.

Common Mistakes

Stating that benzene has alternating single and double bonds (Kekulé structure) without mentioning delocalisation.
Saying p orbitals form σ bonds.
Forgetting that C–H bonds are σ bonds formed by sp² and s orbitals.

Things to Be Careful About

Ensure you distinguish between the σ framework and the π system. Specify which orbitals overlap for each bond type. The mark scheme awards marks for any combination of the key points, but providing all ensures full credit.

Techniques used

describe delocalised pi system in benzeneidentify hybridisation and orbital overlap

(b)

Describe the reagents and conditions required for step 1 in Fig. 9.1.

1M Similar

(c)

In step 1 of Fig. 9.1 benzene reacts with $^+\text{CH}_3$ .

Complete Fig. 9.2 to show the mechanism for this reaction, including:

the movement of electron pairs using curly arrows
the structure of the intermediate involved.

Fig. 9.2

3M Similar

DifficultyMedium-Easy

Worked solution

Answer

Final answer

See diagram

Detailed explanation

Background Concept

Electrophilic substitution in benzene involves two main stages: attack by the electrophile to form a carbocation intermediate (arenium ion), and loss of a proton to restore aromaticity. The curly arrows show the movement of electron pairs.

Understanding the Question

Complete the mechanism for the reaction of benzene with $^+\text{CH}_3$ . This involves drawing the initial attack, the intermediate structure, and the restoration of the ring.

Approach

Draw a curly arrow from the delocalised π system (or a double bond) of benzene to the positive carbon of $^+\text{CH}_3$ .
Draw the intermediate: a cyclohexadienyl cation with the methyl group and a hydrogen atom on the same carbon (tetrahedral, sp³). The positive charge is delocalised over the ring (horseshoe arrow).
Draw a curly arrow from the C–H bond (on the sp³ carbon) back into the ring to restore the π system, producing $\text{H}^+$ .

Step-by-Step Reasoning

M1: Curly arrow from the benzene ring (π electrons) to the $^+\text{C}$ of $^+\text{CH}_3$ . This forms a new C–C bond.
M2: The intermediate is a carbocation. Draw a hexagon with a circle (or alternating double bonds, but circle is better for benzene derivatives) and a + charge. The carbon bonded to $\text{CH}_3$ is also bonded to H. Use a horseshoe arrow to show delocalisation of the positive charge around the ring (positions 2, 4, 6 relative to the sp³ carbon).
M3: Curly arrow from the C–H bond on the sp³ carbon into the ring (to reform the π bond). The product is methylbenzene + $\text{H}^+$ .

Key Takeaways

In electrophilic aromatic substitution, the intermediate is a sigma complex (arenium ion) where aromaticity is temporarily lost. The base (or the conjugate base of the catalyst) removes a proton to restore aromaticity.

Common Mistakes

Drawing the arrow from a specific double bond instead of the delocalised system (though drawing from a double bond is often accepted, the circle is preferred).
Forgetting the positive charge on the intermediate.
Drawing the wrong intermediate (e.g., addition product without loss of H⁺).
Forgetting the $\text{H}^+$ product in the final step.

Things to Be Careful About

The intermediate must show the tetrahedral carbon with both H and $\text{CH}_3$ . The charge must be delocalised (horseshoe arrow). The final arrow must go from the C–H bond to the ring, not from the ring to H.

Techniques used

draw curly arrow mechanism for electrophilic aromatic substitutiondraw arenium ion intermediate

(d)

Describe the reagents and conditions required for step 2 of Fig. 9.1.

2M Similar

DifficultyEasy

Worked solution

Answer

Concentrated $\text{HNO}_3$ and concentrated $\text{H}_2\text{SO}_4$ at $25 \ ^\circ\text{C}$ to $60 \ ^\circ\text{C}$ .

Final answer

Concentrated HNO3 and concentrated H2SO4, 25-60 C

Detailed explanation

Background Concept

Step 2 converts methylbenzene to 1-methyl-4-nitrobenzene. This is an electrophilic substitution (nitration) where a nitro group ( $-\text{NO}_2$ ) replaces a hydrogen on the ring. The methyl group is an ortho/para director, so the major product is para (1-methyl-4-nitrobenzene).

Understanding the Question

Identify the reagents and conditions for nitration of an arene.

Approach

Recall the standard nitrating mixture: concentrated nitric acid and concentrated sulphuric acid. The temperature must be controlled (usually 25-60 °C) to prevent poly-nitration or oxidation.

Step-by-Step Reasoning

Reagents: Concentrated $\text{HNO}_3$ and concentrated $\text{H}_2\text{SO}_4$ . The sulphuric acid acts as a catalyst to generate the nitronium ion ( $\text{NO}_2^+$ ).
Conditions: Temperature between $25 \ ^\circ\text{C}$ and $60 \ ^\circ\text{C}$ . Higher temperatures can lead to multiple substitutions or side reactions.

Key Takeaways

Nitration of benzene derivatives requires concentrated $\text{HNO}_3$ /concentrated $\text{H}_2\text{SO}_4$ at moderate temperatures (25-60 °C).

Common Mistakes

Using dilute acids.
Forgetting the temperature range or stating room temperature only (60 °C is needed for less reactive arenes, though methylbenzene nitrates easily, the mark scheme accepts 25-60 °C).
Writing $\text{H}_2\text{SO}_4$ as a reactant instead of a catalyst (though it's part of the mixture).

Things to Be Careful About

Both acids must be concentrated. The temperature range is important; too high leads to di-nitration.

Techniques used

identify nitration reagents and conditions

(e)

Identify the reagents required for step 3 of Fig. 9.1. Compound W is the product of this step.

1M Similar

(f)

Name compound W.

1M Similar

(g)

The reagents commonly used for step 4 will not reduce the $-\text{COOH}$ group.

Identify the reagents and conditions required for step 4 of Fig. 9.1.

1M Similar

(h)

Benzene can also be used as a starting material to make compound Y.

compound Y

Describe how the route described in Fig. 9.1 (repeated below) can be changed to give compound Y instead of compound X.

Explain your answer.

Fig. 9.1 repeated

2M Similar

DifficultyMedium

Worked solution

Answer

Perform step 3 before step 2 (or perform step 2 before step 1).
Explanation: The $-\text{COOH}$ group (or $-\text{NO}_2$ group) is meta-directing (3-directing), so nitration occurs at the 3-position relative to the carboxylic acid/nitro group, giving 3-aminobenzoic acid.

Working

Original route: Benzene $\xrightarrow{\text{step 1}}$ Methylbenzene $\xrightarrow{\text{step 2}}$ 1-methyl-4-nitrobenzene ( $\text{CH}_3$ is ortho/para directing).
New route: Benzene $\xrightarrow{\text{step 1}}$ Methylbenzene $\xrightarrow{\text{step 3}}$ Benzoic acid $\xrightarrow{\text{step 2}}$ 3-nitrobenzoic acid ( $\text{COOH}$ is meta directing) $\xrightarrow{\text{step 4}}$ 3-aminobenzoic acid (Compound Y).

Answer

Rearrange steps: perform step 3 before step 2. Explanation: $-\text{COOH}$ is a meta-directing group (3-directing), so nitration of benzoic acid gives 3-nitrobenzoic acid, which reduces to 3-aminobenzoic acid (compound Y).

Final answer

Perform step 3 before step 2; COOH is meta-directing (3-directing).

Detailed explanation

Background Concept

Directing effects in electrophilic aromatic substitution determine the position of the second substituent. Alkyl groups (like $-\text{CH}_3$ ) are ortho/para directors (2,4-directing). Electron-withdrawing groups (like $-\text{NO}_2$ and $-\text{COOH}$ ) are meta directors (3-directing).

Understanding the Question

Compound Y is 3-aminobenzoic acid (meta-aminobenzoic acid). The original route produces 4-aminobenzoic acid (para). We need to change the sequence to get the meta isomer.

Approach

To get a meta-substituted product, we need a meta-directing group on the ring when the nitration (step 2) occurs. $-\text{CH}_3$ is ortho/para directing, so we cannot nitrate methylbenzene to get the meta product. We must convert $-\text{CH}_3$ to $-\text{COOH}$ (step 3) first, as $-\text{COOH}$ is meta-directing. Alternatively, nitrate benzene first (step 2) to get nitrobenzene ( $-\text{NO}_2$ is meta-directing), then try to add the carbon group (but Friedel-Crafts doesn't work well on nitrobenzene, so the mark scheme accepts "perform step 2 before step 1" theoretically, though practically step 3 before step 2 is the viable route). The mark scheme says: "perform step 3 before step 2 OR perform step 2 before step 1".

Step-by-Step Reasoning

Option 1 (Viable): Benzene $\xrightarrow{\text{step 1}}$ Methylbenzene $\xrightarrow{\text{step 3}}$ Benzoic acid. Now, perform step 2 (nitration). Since $-\text{COOH}$ is meta-directing, nitration gives 3-nitrobenzoic acid. Then step 4 (reduction) gives 3-aminobenzoic acid (Compound Y).
Explanation: The $-\text{COOH}$ group directs the incoming nitro group to the 3-position (meta).
Option 2 (Theoretical): Benzene $\xrightarrow{\text{step 2}}$ Nitrobenzene. $-\text{NO}_2$ is meta-directing. Then perform step 1 (Friedel-Crafts). However, nitrobenzene is strongly deactivated, so Friedel-Crafts alkylation fails or is very slow. Despite this, the mark scheme accepts the logic: if you nitrate first, the $-\text{NO}_2$ group is meta-directing. (Note: In practice, you can't do FC alkylation on nitrobenzene, but for exam purposes, the directing effect logic is what's tested). The mark scheme explicitly allows "perform step 2 before step 1" with the explanation that $\text{NO}_2$ is 3-directing.

Key Takeaways

The order of steps in aromatic synthesis is crucial. To get meta-substitution, ensure a meta-directing group is present before the electrophilic substitution step.

Common Mistakes

Trying to nitrate methylbenzene and expecting meta product (methyl is ortho/para directing).
Forgetting to explain why the order changes (directing effects).
Suggesting impossible reactions (like FC alkylation on nitrobenzene) without acknowledging the directing effect logic.

Things to Be Careful About

The mark scheme gives two acceptable answers for the rearrangement. Ensure the explanation links the reordering to the directing effect of the relevant group ( $-\text{COOH}$ or $-\text{NO}_2$ ).

Techniques used

plan synthesis using directing effectscompare reaction sequences

Chemistry 9701/42 — October/November 2025

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Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

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Things to Be Careful About

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Background Concept

Understanding the Question

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Understanding the Question

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Understanding the Question

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Background Concept

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Background Concept

Understanding the Question

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Step-by-Step Reasoning