Separate samples of $\text{Na}_2\text{CO}_3$ and $\text{NaHCO}_3$ react with $\text{HCl(aq)}$ to produce the same products, as shown in Table 2.1. Table 2.1 reaction equation $\Delta H / \text{kJ mol}^{-1}$ --- --- --- 1 $\text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2$ $\Delta H_1$ 2 $\text{NaHCO}_3 + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} + \text{CO}_2$ $\Delta H_2 = +27.2$ (a) Complete the reaction pathway…

(a) See diagram: endothermic profile with $E_\text{A}$ and $\Delta H$ arrows pointing upwards. · (b)(i) No more fizzing / effervescence stops. · (b)(ii) -34.5 · (b)(iii) +88.9 · (c) Ca(NO3)2; Ca(NO3)2 -> CaO + 2NO2 + 1/2 O2

Compound $\mathbf{W}$ has molecular formula $\text{C}_4\text{H}_{10}\text{O}$. It contains only one functional group. (a) Table 5.1 shows the two peaks with the greatest $m/e$ values in the mass spectrum of $\mathbf{W}$. Table 5.1 $m/e$ relative abundance --- --- 74 50 75 $x$ (i) Calculate the relative abundance, $x$, of the peak at $m/e = 75$ using the information from Table 5.1. $$ x = \text{..............................} $$ (ii) The mass spectrum of $\mathbf{W}$ also…

(a)(i) 2.2 · (a)(ii) m/e = 29: C2H5+; m/e = 59: C3H7O+ · (b)(i) A: O-H bond, hydroxy group; B: C=O bond, carbonyl group · (b)(ii) CH3COCH2CH3 (butanone) · (c)(i) Hydroxy (tertiary alcohol) · (c)(ii) 2C4H10O + 2Na -> 2C4H9ONa + H2 · (c)(iii) (CH3)3C-OH (2-methylpropan-2-ol)

9701/22 May/June 2024 — Chemistry Solved Past Paper

Q115MMediumAtomic StructureChemical BondingStates of MatterChemical PeriodicityAtoms, Molecules and Stoichiometry

(a)

Complete Table 1.1 using relevant information from the Periodic Table.

Table 1.1

	nucleon number	proton number	number of electrons
$\text{Mg}^{2+}$	24
$\text{Al}^{3+}$	27

2M Similar

DifficultyEasy

Worked solution

Answer

	nucleon number	proton number	number of electrons
$\text{Mg}^{2+}$	24	12	10
$\text{Al}^{3+}$	27	13	10

Final answer

Mg2+: 12, 10; Al3+: 13, 10

Detailed explanation

Background Concept

The atomic structure of an element is defined by its proton number (atomic number, $Z$ ), which is the number of protons in the nucleus and determines the element's identity. The nucleon number (mass number, $A$ ) is the total number of protons and neutrons. In a neutral atom, the number of electrons equals the proton number. When an atom forms an ion, it loses or gains electrons. A positive ion (cation) has lost electrons, so the number of electrons is $Z - \text{charge}$ . A negative ion (anion) has gained electrons, so the number of electrons is $Z + |\text{charge}|$ .

Understanding the Question

The question asks to complete a table for two ions, $\text{Mg}^{2+}$ and $\text{Al}^{3+}$ , by finding their proton numbers and the number of electrons they contain, given their nucleon numbers (24 and 27 respectively).

Approach

Look up the proton numbers for magnesium (Mg) and aluminium (Al) from the Periodic Table.
Calculate the number of electrons by subtracting the ionic charge from the proton number.

Step-by-Step Reasoning

For $\text{Mg}^{2+}$ : Magnesium is in Group 2, Period 3. Its proton number is 12. As a neutral atom, it has 12 electrons. The $2+$ charge means it has lost 2 electrons. Number of electrons = $12 - 2 = 10$ .
For $\text{Al}^{3+}$ : Aluminium is in Group 3, Period 3. Its proton number is 13. As a neutral atom, it has 13 electrons. The $3+$ charge means it has lost 3 electrons. Number of electrons = $13 - 3 = 10$ .

Key Takeaways

The proton number is fixed for a given element and is found directly from the Periodic Table. The number of electrons in an ion is adjusted by the charge: lose electrons for positive ions, gain for negative ions.

Common Mistakes

Confusing nucleon number (mass number) with proton number. The nucleon number is given (24 and 27) and is not needed to find the proton number or electron count for these specific questions, though it defines the isotope.
Forgetting to subtract the charge for cations (e.g., writing 12 for Mg2+ electrons).

Things to Be Careful About

Ensure you are reading the correct column in the Periodic Table. Proton number is the integer, not the relative atomic mass.

Techniques used

determine proton number from periodic tablecalculate number of electrons in an ion

(b)

State and explain the difference in the ionic radius of $\text{Al}^{3+}$ compared to $\text{Mg}^{2+}$ .

3M Similar

DifficultyMedium-Easy

Worked solution

Answer

$\text{Al}^{3+}$ is smaller than $\text{Mg}^{2+}$ .
Both ions have the same number of electrons (10) and the same shielding effect.
$\text{Al}^{3+}$ has a greater nuclear charge (13 protons) than $\text{Mg}^{2+}$ (12 protons).
The greater nuclear charge in $\text{Al}^{3+}$ exerts a stronger attraction on the electron cloud, pulling the electrons closer to the nucleus.

Final answer

Al3+ is smaller due to greater nuclear charge with same shielding.

Detailed explanation

Background Concept

Ionic radius is the measure of the size of an ion. When comparing isoelectronic species (ions with the same number of electrons), the size is determined by the nuclear charge (number of protons). A higher nuclear charge pulls the electron cloud more strongly towards the nucleus, resulting in a smaller ionic radius. Shielding is the effect where inner electrons repel outer electrons, reducing the effective nuclear charge felt by the outer electrons. For isoelectronic ions, the electron configuration is identical, so the shielding effect is the same.

Understanding the Question

The question asks to state and explain the difference in ionic radius between $\text{Al}^{3+}$ and $\text{Mg}^{2+}$ . We established in part (a) that both have 10 electrons (isoelectronic with neon).

Approach

State which ion is smaller.
Explain that they have the same number of electrons and thus the same shielding.
Compare the nuclear charges (proton numbers) and explain how this affects the attraction on the electron cloud.

Step-by-Step Reasoning

State the difference: $\text{Al}^{3+}$ is smaller than $\text{Mg}^{2+}$ .
Shielding: Both ions have 10 electrons with the configuration $1s^2 2s^2 2p^6$ . The inner shell (1s) shields the outer shells (2s, 2p) equally in both cases. Therefore, the shielding effect is the same.
Nuclear charge: Aluminium has a proton number of 13, while magnesium has a proton number of 12. $\text{Al}^{3+}$ has a greater nuclear charge ( $+13$ ) than $\text{Mg}^{2+}$ ( $+12$ ).
Explanation: The greater positive charge in the nucleus of $\text{Al}^{3+}$ attracts the 10 electrons more strongly than the nucleus of $\text{Mg}^{2+}$ . This stronger electrostatic attraction pulls the electron cloud closer to the nucleus, resulting in a smaller ionic radius.

Key Takeaways

For isoelectronic ions, ionic radius decreases as nuclear charge (proton number) increases. Always mention both the same shielding and the greater nuclear charge for full marks.

Common Mistakes

Saying "more protons" without specifying "greater nuclear charge" or "more positive charge in the nucleus".
Forgetting to mention that the shielding is the same. If shielding is different, the argument is invalid.
Stating that $\text{Al}^{3+}$ has more protons so it is larger (confusing the effect).

Things to Be Careful About

The mark scheme specifically looks for "same shielding" AND "greater nuclear charge". Omitting either point may cost a mark. Also, ensure you compare the correct ions (Al3+ is smaller, not larger).

Techniques used

compare ionic radii using nuclear charge and shieldingexplain effective nuclear attraction

(c)

Draw a labelled diagram to show the structure and bonding in sodium.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

A regular (lattice) arrangement of positive ions (cations) surrounded by a sea of delocalised electrons.

Final answer

Diagram showing positive ions in a regular lattice surrounded by delocalised electrons.

Detailed explanation

Background Concept

Metals, such as sodium, have metallic bonding. This is described as a regular arrangement of positive metal ions (cations) in a lattice, surrounded by a 'sea' of delocalised electrons. The delocalised electrons are the outer-shell (valence) electrons that are not associated with any particular atom but are free to move throughout the structure. The electrostatic attraction between the positive ions and the delocalised electrons holds the structure together.

Understanding the Question

The question asks for a labelled diagram showing the structure and bonding in sodium metal.

Approach

Draw a representation of the metallic lattice. Show at least a 2x2 arrangement of positive ions and some delocalised electrons around them. Label both components clearly.

Step-by-Step Reasoning

Structure: Sodium forms a giant metallic lattice. Draw circles to represent the positive sodium ions ( $\text{Na}^+$ ). Arrange them in a regular pattern (e.g., rows and columns). Put a '+' sign inside each circle to indicate the positive charge.
Bonding: Draw smaller circles or dashes with a '-' sign (or just '-' signs) scattered around and between the positive ions to represent the delocalised electrons. These are the valence electrons from the sodium atoms (3s1) that have been released into the 'sea'.
Labels: Clearly label the circles with '+' as 'positively charged ions' or 'cations' and the '-' symbols as 'delocalised electrons'.

Key Takeaways

The metallic bonding model must show both the regular arrangement of positive ions and the delocalised electrons. Simply drawing a lattice of atoms is incorrect; they must be shown as ions (+).

Common Mistakes

Drawing neutral atoms instead of positive ions. Sodium metal consists of $\text{Na}^+$ ions, not neutral Na atoms.
Forgetting to label the diagram. The question asks for a 'labelled diagram'.
Drawing the electrons as being attached to specific ions. They must be shown as free/delocalised throughout the structure.

Things to Be Careful About

Ensure the positive ions are clearly marked with a '+' and the electrons with a '-'. The arrangement should look regular, not random. At least a 2x2 grid of ions is typically expected to show the 'regular arrangement'.

Techniques used

draw metallic bonding modellabel positive ions and delocalised electrons

(d)

Fig. 1.1 shows the variation in melting point of some Period 3 elements in their standard states at room temperature and pressure.

Fig. 1.1

3M

(i)

Explain why Si has a high melting point.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

Silicon has a giant covalent (macromolecular) structure. A large amount of energy is required to break the strong covalent bonds throughout the lattice.

Final answer

Giant covalent structure; strong covalent bonds require large energy to break.

Detailed explanation

Background Concept

The melting point of a substance depends on the strength of the forces that must be overcome to separate the particles. For giant structures (giant ionic, giant metallic, giant covalent), melting involves breaking strong bonds (ionic, metallic, or covalent) throughout the entire lattice. Silicon is in Group 4 and forms a giant covalent structure similar to diamond, where each silicon atom is covalently bonded to four other silicon atoms in a tetrahedral arrangement. These strong covalent bonds extend throughout the entire crystal, requiring a large amount of thermal energy to break, hence the high melting point.

Understanding the Question

Part (d)(i) asks to explain why silicon (Si) has a high melting point, based on the provided graph showing a sharp peak at Si.

Approach

Identify the structure and bonding of silicon. State that it is a giant covalent structure. Explain that melting requires breaking the bonds within this structure.

Step-by-Step Reasoning

Structure: Silicon is a Group 4 element. In its standard state, it forms a giant covalent (macromolecular) lattice. Each Si atom is bonded to four others via strong covalent bonds.
Melting process: To melt silicon, the strong covalent bonds holding the atoms together in the giant lattice must be broken.
Energy: Because covalent bonds are strong and there are many of them in the giant structure, a large amount of energy (high temperature) is required to break them. This results in a high melting point (~1700 K).

Key Takeaways

High melting points in Period 3 elements (Na, Mg, Al, Si) are due to strong bonding (metallic or covalent) in giant structures. For Si, it is specifically the breaking of strong covalent bonds in a giant covalent lattice.

Common Mistakes

Saying 'strong forces' without specifying 'covalent bonds'. 'Forces' is too vague and could imply intermolecular forces, which are weak.
Saying 'breaking intermolecular forces'. Silicon does not have simple molecular structure; it has a giant covalent structure where bonds are broken, not just intermolecular forces.
Not mentioning 'giant covalent' or 'macromolecular'.

Things to Be Careful About

Be precise: 'strong covalent bonds' must be mentioned. 'Large amount of energy' or 'high temperature' is the consequence, but the reason is the bond breaking.

Techniques used

relate giant covalent structure to melting point

(ii)

Complete Fig. 1.1 to show the variation in the melting points of the elements P, S and Cl.

2M Similar

DifficultyMedium

Worked solution

Answer

The graph should show melting points for P, S, and Cl all lower than Mg (~920 K). The order of melting points must be $\text{Cl}_2 < \text{P}_4 < \text{S}_8$ (i.e., Cl is lowest, then P, then S). The line should drop sharply from Si to P, rise slightly to S, then fall to Cl.

Final answer

Graph showing melting points: P < S, and Cl < P. All lower than Mg.

Detailed explanation

Background Concept

The melting points of Period 3 elements after silicon depend on their molecular structures and the intermolecular forces (van der Waals forces) between the molecules. Simple molecular substances have low melting points because only weak intermolecular forces need to be overcome, not strong covalent bonds.

Phosphorus ( $\text{P}_4$ ): Exists as discrete $\text{P}_4$ molecules (tetrahedral). Moderate van der Waals forces.
Sulfur ( $\text{S}_8$ ): Exists as discrete $\text{S}_8$ rings (crown-shaped). Larger molecules have more electrons, leading to stronger van der Waals dispersion forces than $\text{P}_4$ . Thus, S has a higher melting point than P.
Chlorine ( $\text{Cl}_2$ ): Exists as small $\text{Cl}_2$ diatomic molecules. Very few electrons, so very weak van der Waals forces. Lowest melting point of the three.

Understanding the Question

Part (d)(ii) asks to complete the graph in Fig 1.1 for P, S, and Cl. The graph already shows Na, Mg, Al, Si. We need to plot the relative melting points for the remaining elements.

Approach

Recall that P, S, and Cl are simple molecular, so their melting points will be much lower than Si (and lower than Mg/Al).
Determine the relative order of melting points based on molecular size and van der Waals forces: $\text{Cl}_2 < \text{P}_4 < \text{S}_8$ .
Plot these points on the graph and connect them with a line.

Step-by-Step Reasoning

General trend: After the peak at Si (giant covalent), the melting points drop dramatically because P, S, and Cl form simple molecular structures with weak intermolecular forces. All three will be well below 1000 K (Mg is ~920 K).
Relative order:
- Chlorine ( $\text{Cl}_2$ ): Smallest molecule, fewest electrons, weakest van der Waals forces. Lowest melting point (~239 K). Plot a point low on the y-axis for Cl.
- Phosphorus ( $\text{P}_4$ ): Larger molecule than $\text{Cl}_2$ , more electrons, stronger van der Waals forces. Higher melting point than Cl (~317 K). Plot a point for P higher than Cl.
- Sulfur ( $\text{S}_8$ ): Largest molecule ( $\text{S}_8$ ring), most electrons, strongest van der Waals forces. Highest melting point of the three (~388 K). Plot a point for S higher than P.
Plotting: Draw a line from Si down to P, then up slightly to S, then down to Cl. Ensure all three points are clearly below the Mg/Al level (around 900-950 K) and below 500 K on the y-axis scale (0 to 500 is the first major interval). The exact values aren't critical, but the relative order and that they are low is essential.

Key Takeaways

For simple molecular substances, melting point increases with molecular size/number of electrons due to stronger van der Waals forces. $\text{S}_8 > \text{P}_4 > \text{Cl}_2$ .

Common Mistakes

Plotting P, S, Cl as having high melting points (confusing them with giant structures).
Getting the relative order wrong (e.g., thinking Cl > S because Cl is further right in the period). Remember it's about molecular size, not atomic number.
Drawing a straight line without the slight peak at S. The graph should show Cl < P < S.

Things to Be Careful About

The mark scheme requires 'all 3 melting points lower than Mg' and 'Cl < P < S'. Ensure the plotted points reflect this. Don't just draw a flat line; show the slight variation. The values should be roughly in the 200-400 K range, well below the 500 K mark on the axis.

Techniques used

predict melting points of molecular Period 3 elementsplot data on a graph

(e)

Two Period 3 elements react with an excess of oxygen at room pressure.

4M

(i)

Complete Table 1.2.

Table 1.2

1	2	3
Period 3 element	state of oxide at room temperature and pressure	approximate pH of solution made when oxide is added to water
Na
S

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Period 3 element	state of oxide at room temperature and pressure	approximate pH of solution made when oxide is added to water
Na	solid	10–14
S	gas	0–4

Final answer

Na2O: solid, pH 10-14; SO2: gas, pH 0-4

Detailed explanation

Background Concept

Period 3 elements react with oxygen to form oxides. The nature of these oxides changes across the period from basic (metallic) to acidic (non-metallic).

Sodium (Na): Reacts to form sodium oxide ( $\text{Na}_2\text{O}$ ), a giant ionic lattice. At room temperature, it is a solid. When added to water, it reacts to form sodium hydroxide ( $\text{NaOH}$ ), a strong alkali, giving a high pH (10-14).
Sulfur (S): Reacts to form sulfur dioxide ( $\text{SO}_2$ ), a simple molecular gas. At room temperature, it is a gas. When dissolved in water, it forms sulfurous acid ( $\text{H}_2\text{SO}_3$ ), a weak acid, giving a low pH (0-4, typically around 2-3).

Understanding the Question

Part (e)(i) asks to complete a table for Na and S, giving the state of their oxides at room temperature and the approximate pH when the oxide is added to water.

Approach

Identify the oxides formed by Na and S with excess oxygen.
State their physical state at room temperature.
Determine the pH of their aqueous solutions based on their acid-base nature.

Step-by-Step Reasoning

Sodium (Na): Oxide is $\text{Na}_2\text{O}$ (sodium oxide). It is an ionic solid, so state is solid. Reaction with water: $\text{Na}_2\text{O} + \text{H}_2\text{O} \rightarrow 2\text{NaOH}$ . NaOH is a strong base, so pH is 10–14 (or >7, but 10-14 is more precise for strong alkali).
Sulfur (S): Oxide is $\text{SO}_2$ (sulfur dioxide). It is a simple molecular substance, so state is gas. Reaction with water: $\text{SO}_2 + \text{H}_2\text{O} \rightarrow \text{H}_2\text{SO}_3$ . Sulfurous acid is acidic, so pH is 0–4 (or <7, but 0-4 is typical for acid rain/weak acid solutions in this context).

Key Takeaways

Metal oxides (like Na2O) are basic and form alkaline solutions (high pH). Non-metal oxides (like SO2) are acidic and form acidic solutions (low pH). State depends on structure: ionic = solid, simple molecular = gas/liquid.

Common Mistakes

Saying SO2 is a liquid or solid. It is a gas at room temperature.
Giving a pH of exactly 7 or saying 'neutral'. Na2O is basic, SO2 is acidic.
Confusing the oxide of S. Excess oxygen gives SO2 (and some SO3, but SO2 is the primary product asked for in this context, and it's a gas. SO3 is a solid/liquid but reacts to form H2SO4. The mark scheme accepts gas and pH 0-4, which fits SO2 perfectly).

Things to Be Careful About

The pH ranges given in mark schemes are often broad (e.g., 10-14, 0-4). Ensure you don't write a single value like '14' or '1' if a range is expected, though often a value within the range is accepted. 'Solid' and 'gas' must be exact words.

Techniques used

predict oxide properties from periodicityrelate oxide to pH in water

(ii)

The solutions made in column 3 of Table 1.2 are mixed together.
Name the type of reaction that occurs.

1M Similar

(iii)

Write an equation to describe the reaction between $\text{P}_4\text{O}_{10}$ and an excess of water.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

\text{P}_4\text{O}_{10} + 6\text{H}_2\text{O} \rightarrow 4\text{H}_3\text{PO}_4

Final answer

P4O10 + 6H2O -> 4H3PO4

Detailed explanation

Background Concept

Phosphorus burns in excess oxygen to form phosphorus(V) oxide, $\text{P}_4\text{O}_{10}$ (often written as $\text{P}_2\text{O}_5$ but the molecular formula is $\text{P}_4\text{O}_{10}$ ). When this acidic oxide reacts with water, it forms phosphoric(V) acid, $\text{H}_3\text{PO}_4$ .

Understanding the Question

Part (e)(iii) asks for an equation for the reaction between $\text{P}_4\text{O}_{10}$ and an excess of water.

Approach

Write the reactants: $\text{P}_4\text{O}_{10}$ and $\text{H}_2\text{O}$ .
Write the product: $\text{H}_3\text{PO}_4$ (phosphoric acid).
Balance the equation.

Step-by-Step Reasoning

Reactants: $\text{P}_4\text{O}_{10} + \text{H}_2\text{O}$
Product: $\text{H}_3\text{PO}_4$
Unbalanced: $\text{P}_4\text{O}_{10} + \text{H}_2\text{O} \rightarrow \text{H}_3\text{PO}_4$
Balance P: 4 P on left, so need 4 $\text{H}_3\text{PO}_4$ on right.
$\text{P}_4\text{O}_{10} + \text{H}_2\text{O} \rightarrow 4\text{H}_3\text{PO}_4$
Balance H: 4 * 3 = 12 H on right, so need 6 $\text{H}_2\text{O}$ on left.
$\text{P}_4\text{O}_{10} + 6\text{H}_2\text{O} \rightarrow 4\text{H}_3\text{PO}_4$
Check O: Left = 10 + 6 = 16. Right = 4 * 4 = 16. Balanced.

Key Takeaways

Acidic oxides react with water to form acids. $\text{P}_4\text{O}_{10}$ is the molecular formula for phosphorus(V) oxide. The product is $\text{H}_3\text{PO}_4$ .

Common Mistakes

Using $\text{P}_2\text{O}_5$ instead of $\text{P}_4\text{O}_{10}$ . While $\text{P}_2\text{O}_5$ is the empirical formula, the question specifically gives $\text{P}_4\text{O}_{10}$ , so use that.
Forgetting to balance the equation.
Writing the product as $\text{HPO}_3$ (metaphosphoric acid). With excess water, orthophosphoric acid ( $\text{H}_3\text{PO}_4$ ) is formed.

Things to Be Careful About

Ensure the equation is fully balanced. State symbols are not strictly required by the mark scheme for this specific entry (unlike some other parts), but it's good practice. The mark scheme gives $\text{P}_4\text{O}_{10} + 6\text{H}_2\text{O} \rightarrow 4\text{H}_3\text{PO}_4$ .

Techniques used

write and balance chemical equation for oxide hydration

(f)

Aluminium hydroxide is amphoteric.

2M

(i)

Explain what is meant by amphoteric.

1M Similar

(ii)

Write an equation to describe the reaction that occurs when aluminium hydroxide, $\text{Al(OH)}_3$ , reacts with $\text{NaOH(aq)}$ .

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

\text{Al(OH)}_3 + \text{NaOH} \rightarrow \text{NaAl(OH)}_4

(Alternatively: $\text{Al(OH)}_3 + \text{OH}^- \rightarrow \text{Al(OH)}_4^-$ )

Final answer

Al(OH)3 + NaOH -> NaAl(OH)4

Detailed explanation

Background Concept

Aluminium hydroxide is amphoteric. When it reacts with a strong base like sodium hydroxide ( $\text{NaOH}$ ), it acts as an acid and dissolves to form a soluble complex ion, tetrahydroxoaluminate(III), $\text{Al(OH)}_4^-$ . The salt formed is sodium aluminate, $\text{NaAl(OH)}_4$ (or $\text{Na}[\text{Al(OH)}_4]$ ).

Understanding the Question

Part (f)(ii) asks for an equation for the reaction between $\text{Al(OH)}_3$ and $\text{NaOH(aq)}$ .

Approach

Write the reactants: $\text{Al(OH)}_3$ and $\text{NaOH}$ . Write the product: $\text{NaAl(OH)}_4$ . Ensure it is balanced (it is 1:1:1).

Step-by-Step Reasoning

Reactants: $\text{Al(OH)}_3\text{(s)} + \text{NaOH(aq)}$
Product: $\text{NaAl(OH)}_4\text{(aq)}$ (sodium tetrahydroxoaluminate)
Equation: $\text{Al(OH)}_3 + \text{NaOH} \rightarrow \text{NaAl(OH)}_4$
Check balance: Al: 1=1, Na: 1=1, O: 3+1=4, H: 3+1=4. Balanced.
Ionic equation alternative: $\text{Al(OH)}_3 + \text{OH}^- \rightarrow \text{Al(OH)}_4^-$

Key Takeaways

Amphoteric hydroxides dissolve in excess strong base to form complex aluminate ions. The formula $\text{NaAl(OH)}_4$ is the standard molecular equation product.

Common Mistakes

Writing the product as $\text{NaAlO}_2$ (sodium meta-aluminate) or $\text{NaAlO}_3$ . While these exist in some contexts, the aqueous complex ion $\text{Al(OH)}_4^-$ is the correct product in aqueous NaOH.
Not balancing the equation or getting the formula of the product wrong.
Forgetting that $\text{Al(OH)}_3$ is a solid and dissolves (though state symbols aren't always strictly penalised if the chemistry is right, the mark scheme here just gives the formula equation).

Things to Be Careful About

The mark scheme specifically gives $\text{Al(OH)}_3 + \text{NaOH} \rightarrow \text{NaAl(OH)}_4$ . Use this exact formula. The ionic equation $\text{Al(OH)}_3 + \text{OH}^- \rightarrow \text{Al(OH)}_4^-$ is also acceptable and often preferred in advanced chemistry, but stick to the molecular one if unsure.

Techniques used

write equation for amphoteric hydroxide with base

Q210MMediumChemical EnergeticsGroup 2

Separate samples of $\text{Na}_2\text{CO}_3$ and $\text{NaHCO}_3$ react with $\text{HCl(aq)}$ to produce the same products, as shown in Table 2.1.

Table 2.1

reaction	equation	$\Delta H / \text{kJ mol}^{-1}$
1	$\text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2$	$\Delta H_1$
2	$\text{NaHCO}_3 + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} + \text{CO}_2$	$\Delta H_2 = +27.2$

(a)

Complete the reaction pathway diagram in Fig. 2.1 for reaction 2.

Label the diagram to show the enthalpy change, $\Delta H_2$ , and the activation energy, $E_\text{A}$ .

Fig. 2.1

2M Similar

DifficultyEasy

Worked solution

Answer

Draw a curve starting at the reactant level, rising to a peak (the hump), and finishing at a product level higher than the reactant level.
Draw an upward arrow from the reactant energy level to the peak of the curve, and label it $E_\text{A}$ .
Draw an upward arrow from the reactant energy level to the product energy level, and label it $\Delta H$ (or $+27.2$ ).

Final answer

See diagram: endothermic profile with $E_\text{A}$ and $\Delta H$ arrows pointing upwards.

Detailed explanation

Background Concept

A reaction pathway diagram (or energy profile) plots the energy of the system against the progress of the reaction. The vertical axis represents energy (often in kJ mol $^{-1}$ ), and the horizontal axis represents the progress of the reaction from reactants to products.

The peak of the curve represents the transition state (activated complex). The energy difference between the reactants and the peak is the activation energy ( $E_\text{A}$ ), which is the minimum energy required for the reaction to occur. The energy difference between the reactants and the products is the enthalpy change of reaction ( $\Delta H$ ).

If the products are at a higher energy level than the reactants, the reaction is endothermic ( $\Delta H > 0$ ). If the products are at a lower energy level, the reaction is exothermic ( $\Delta H < 0$ ).

Understanding the Question

The question asks to complete a reaction pathway diagram for reaction 2:
$\text{NaHCO}_3 + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} + \text{CO}_2 \quad \Delta H_2 = +27.2 \text{ kJ mol}^{-1}$

The given $\Delta H_2$ is positive, meaning the reaction is endothermic. The diagram must show the products at a higher energy level than the reactants. We also need to label the activation energy ( $E_\text{A}$ ) and the enthalpy change ( $\Delta H$ or $+27.2$ ).

Approach

Draw the reaction curve: start at the reactant line, curve up to a maximum (the hump), and then curve down to a product line that is higher than the reactant line.
Label the activation energy ( $E_\text{A}$ ) as an upward arrow from the reactant level to the peak.
Label the enthalpy change ( $\Delta H$ ) as an upward arrow from the reactant level to the product level.

Step-by-Step Reasoning

Profile shape: Since $\Delta H_2 = +27.2$ kJ mol $^{-1}$ , the reaction is endothermic. The product energy level ( $\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2$ ) must be drawn higher than the reactant energy level ( $\text{NaHCO}_3 + \text{HCl}$ ). A smooth curve with a peak connects them.
Activation energy ( $E_\text{A}$ ): This is the energy barrier from reactants to the transition state. Draw a vertical arrow from the reactant line up to the top of the peak. Label it $E_\text{A}$ .
Enthalpy change ( $\Delta H$ ): This is the net energy difference between products and reactants. Draw a vertical arrow from the reactant line up to the product line. Label it $\Delta H$ (or $+27.2$ ). The arrow must point upwards because energy is absorbed.

Key Takeaways

Endothermic reactions have products at a higher energy level than reactants.
$E_\text{A}$ is always measured from the reactant level to the peak.
$\Delta H$ is measured from the reactant level to the product level; its direction (up/down) indicates whether the reaction is endothermic or exothermic.

Common Mistakes

Drawing an exothermic profile (products lower than reactants) when $\Delta H$ is positive.
Labeling $E_\text{A}$ from the product level to the peak.
Forgetting to draw the arrow for $\Delta H$ or drawing it in the wrong direction (downwards for an endothermic reaction).

Things to Be Careful About

Ensure the $E_\text{A}$ arrow starts at the reactant energy level, not the x-axis.
The $\Delta H$ arrow must point upwards for an endothermic reaction.
The curve should have a clear peak (hump) representing the activation energy barrier.

Techniques used

draw reaction pathway diagramidentify activation energy and enthalpy change

(b)

The value for $\Delta H_1$ is determined by experiment using the following method.

$50.0\text{ cm}^3$ of $2.00\text{ mol dm}^{-3}\ \text{HCl(aq)}$ is added to a polystyrene cup.
The initial temperature of the acid is recorded as $19.6\ ^\circ\text{C}$ .
$0.0400\text{ mol}$ of $\text{Na}_2\text{CO}_3$ is added and the mixture is stirred.
All the solid $\text{Na}_2\text{CO}_3$ disappears and a colourless solution is produced.

The maximum temperature recorded during the reaction is $26.2\ ^\circ\text{C}$ .

6M

(i)

Describe one other observation that shows the reaction is complete.

1M Similar

DifficultyEasy

Worked solution

Answer

No more fizzing / effervescence stops.

Final answer

No more fizzing / effervescence stops.

Detailed explanation

Background Concept

When a carbonate reacts with an acid, carbon dioxide gas is produced. This is observed as fizzing or effervescence. The reaction continues as long as there is unreacted carbonate or acid present.

Understanding the Question

The experiment involves adding solid $\text{Na}_2\text{CO}_3$ to $\text{HCl}$ . The solid disappears and a colourless solution is produced. We need one other observation that indicates the reaction is complete.

Approach

Since $\text{CO}_2$ gas is produced during the reaction, the fizzing will stop when the limiting reagent is completely used up. Here, $\text{Na}_2\text{CO}_3$ is the limiting reagent ( $0.0400$ mol requires $0.0800$ mol $\text{HCl}$ , but $0.100$ mol $\text{HCl}$ is present). When all the solid carbonate has reacted, no more $\text{CO}_2$ will be produced.

Step-by-Step Reasoning

The reaction produces $\text{CO}_2$ gas: $\text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2$ .
Gas evolution is seen as fizzing or effervescence.
When the reaction is complete, no more gas is produced, so the fizzing stops.

Key Takeaways

Gas evolution is a common visual indicator of reaction progress.
The cessation of gas bubbles indicates the reaction has finished.

Common Mistakes

Saying 'the solution becomes clear' (this is already stated in the question as 'a colourless solution is produced').
Saying 'temperature stops rising' (this is harder to observe visually than fizzing and is not the primary visual indicator).

Things to Be Careful About

Use precise terminology: 'fizzing' or 'effervescence', not just 'bubbles' (though 'bubbles stop' is often accepted, 'fizzing stops' is better).
Do not say 'no more solid dissolves' as the question already states 'All the solid disappears'.

Techniques used

observe reaction progressidentify gas evolution

(ii)

Calculate the value of $\Delta H_1$ in $\text{kJ mol}^{-1}$ .

Assume the specific heat capacity of the reaction mixture is the same as for water and no heat is lost to the surroundings.

Show your working.

\Delta H_1 = \text{.............................. kJ mol}^{-1}

3M Similar

DifficultyMedium-Easy

Worked solution

Working

\begin{aligned} q &= mc\Delta T \\ &= 50.0 \times 4.18 \times (26.2 - 19.6) \\ &= 50.0 \times 4.18 \times 6.6 \\ &= 1379.4 \text{ J} \\ &= 1.3794 \text{ kJ} \end{aligned}

Moles of $\text{Na}_2\text{CO}_3$ reacted = $0.0400$ mol

\Delta H_1 = \frac{-q}{n} = \frac{-1.3794}{0.0400} = -34.485 \text{ kJ mol}^{-1}

Rounding to 3 significant figures:

\Delta H_1 = -34.5 \text{ kJ mol}^{-1}

Answer

$\Delta H_1 = -34.5 \text{ kJ mol}^{-1}$

Final answer

-34.5

Detailed explanation

Background Concept

Calorimetry is used to measure the heat energy change during a reaction. The formula is:
$q = mc\Delta T$
where:

$q$ is the heat energy change in Joules (J).
$m$ is the mass of the solution in grams (g). For dilute aqueous solutions, we assume the density is $1.00 \text{ g cm}^{-3}$ , so mass in g = volume in cm $^3$ .
$c$ is the specific heat capacity in J K $^{-1}$ g $^{-1}$ (or J K $^{-1}$ mol $^{-1}$ for water). We assume $c = 4.18 \text{ J K}^{-1} \text{g}^{-1}$ .
$\Delta T$ is the temperature change in K (or °C, as the magnitude is the same).

The enthalpy change $\Delta H$ is the heat energy change per mole of reaction. If the temperature increases, the reaction is exothermic, and $\Delta H$ is negative.

Understanding the Question

We are given:

Volume of $\text{HCl}$ = $50.0 \text{ cm}^3$ (assume mass = $50.0 \text{ g}$ )
$\Delta T = 26.2 - 19.6 = 6.6 \text{ °C}$
Moles of $\text{Na}_2\text{CO}_3$ = $0.0400 \text{ mol}$

We need to calculate $\Delta H_1$ in kJ mol $^{-1}$ .

Approach

Calculate $q$ using $q = mc\Delta T$ .
Convert $q$ to kJ.
Divide by the moles of $\text{Na}_2\text{CO}_3$ to find the energy change per mole.
Apply the correct sign: temperature increased, so the reaction is exothermic, $\Delta H$ is negative.

Step-by-Step Reasoning

Calculate $q$ :
$q = 50.0 \times 4.18 \times 6.6 = 1379.4 \text{ J}$
Convert to kJ:
$q = 1.3794 \text{ kJ}$
Calculate $\Delta H_1$ :
The reaction is exothermic (temperature rose), so $\Delta H$ is negative.
$\Delta H_1 = \frac{-1.3794 \text{ kJ}}{0.0400 \text{ mol}} = -34.485 \text{ kJ mol}^{-1}$
Significant figures: The temperature change ( $6.6$ ) has 2 significant figures, but typically in these calculations, we report to 3 significant figures or match the least precise data. The mark scheme accepts $-34.5$ .

Key Takeaways

Always convert J to kJ when asked for $\Delta H$ in kJ mol $^{-1}$ .
Remember the sign convention: temperature rise = exothermic = negative $\Delta H$ .
Assume mass of solution = volume of solution (in g) for dilute aqueous reactions.

Common Mistakes

Forgetting to convert J to kJ (resulting in $-34485$ ).
Forgetting the negative sign for an exothermic reaction.
Using the wrong mass (e.g., including the mass of the solid, which is not standard unless specified).

Things to Be Careful About

The mark scheme allows $50 \times 4.18 \times 6.6 / 1000 = 1.3794$ kJ directly.
Ensure $\Delta T$ is calculated correctly: $26.2 - 19.6 = 6.6$ , not $6.8$ .
The value $-34.5$ is correct to 3 significant figures.

Techniques used

apply calorimetry equation q = mc(delta T)calculate enthalpy change per moledetermine sign of enthalpy change

(iii)

Thermal decomposition occurs when $\text{NaHCO}_3$ is heated.

\text{reaction 3} \quad 2\text{NaHCO}_3 \rightarrow \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}_2

Calculate the enthalpy change for reaction 3, $\Delta H_\text{r}$ , using the data in Table 2.1 and the value of $\Delta H_1$ calculated in (b)(ii).

(If you were unable to calculate a value for $\Delta H_1$ in (b)(ii), assume the enthalpy change is $-38.4\text{ kJ mol}^{-1}$ . This is not the correct value.)

\Delta H_\text{r} = \text{.............................. kJ mol}^{-1}

2M Similar

DifficultyMedium

Worked solution

Working

Reaction 1: $\text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2 \quad \Delta H_1 = -34.5 \text{ kJ mol}^{-1}$
Reaction 2: $\text{NaHCO}_3 + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} + \text{CO}_2 \quad \Delta H_2 = +27.2 \text{ kJ mol}^{-1}$
Target: $2\text{NaHCO}_3 \rightarrow \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}_2$

Target = $2 \times$ (Reaction 2) $-$ (Reaction 1)

\begin{aligned} \Delta H_\text{r} &= 2\Delta H_2 - \Delta H_1 \\ &= 2(+27.2) - (-34.5) \\ &= 54.4 + 34.5 \\ &= +88.9 \text{ kJ mol}^{-1} \end{aligned}

Answer

$\Delta H_\text{r} = +88.9 \text{ kJ mol}^{-1}$

Final answer

+88.9

Detailed explanation

Background Concept

Hess's law states that the total enthalpy change for a reaction is independent of the pathway taken. This allows us to calculate $\Delta H$ for a reaction by combining other reactions whose $\Delta H$ values are known.

If we need to reverse a reaction, we change the sign of $\Delta H$ . If we multiply a reaction by a factor, we multiply $\Delta H$ by the same factor.

Understanding the Question

We need to find $\Delta H_\text{r}$ for:
$2\text{NaHCO}_3 \rightarrow \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}_2$

Given:

$\text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2 \quad \Delta H_1 = -34.5 \text{ kJ mol}^{-1}$
$\text{NaHCO}_3 + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} + \text{CO}_2 \quad \Delta H_2 = +27.2 \text{ kJ mol}^{-1}$

Approach

We need to manipulate equations 1 and 2 to sum to the target equation.

Target has $2\text{NaHCO}_3$ on the left. Equation 2 has $1\text{NaHCO}_3$ on the left. So multiply Equation 2 by 2.
Target has $\text{Na}_2\text{CO}_3$ on the right. Equation 1 has $\text{Na}_2\text{CO}_3$ on the left. So reverse Equation 1 (multiply by -1).

Target = $2 \times$ (Equation 2) $-$ (Equation 1)

Step-by-Step Reasoning

Multiply Equation 2 by 2:
$2\text{NaHCO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + 2\text{H}_2\text{O} + 2\text{CO}_2 \quad \Delta H = 2 \times (+27.2) = +54.4 \text{ kJ mol}^{-1}$
Reverse Equation 1:
$2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2 \rightarrow \text{Na}_2\text{CO}_3 + 2\text{HCl} \quad \Delta H = -(-34.5) = +34.5 \text{ kJ mol}^{-1}$
Add the two equations:
$2\text{NaHCO}_3 + 2\text{HCl} + 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2 \rightarrow 2\text{NaCl} + 2\text{H}_2\text{O} + 2\text{CO}_2 + \text{Na}_2\text{CO}_3 + 2\text{HCl}$
Cancel common terms:
$2\text{HCl}$ cancels. $2\text{NaCl}$ cancels. One $\text{H}_2\text{O}$ and one $\text{CO}_2$ cancel from both sides.
Result: $2\text{NaHCO}_3 \rightarrow \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}_2$
Calculate $\Delta H_\text{r}$ :
$\Delta H_\text{r} = 54.4 + 34.5 = +88.9 \text{ kJ mol}^{-1}$

Key Takeaways

Hess's law allows algebraic manipulation of chemical equations and their $\Delta H$ values.
Reversing a reaction changes the sign of $\Delta H$ .
Multiplying a reaction by a factor multiplies $\Delta H$ by the same factor.

Common Mistakes

Forgetting to reverse the sign of $\Delta H_1$ when subtracting the equation.
Incorrectly balancing the equations before combining.
Arithmetic errors in adding/subtracting the values.

Things to Be Careful About

The mark scheme accepts the alternative value $\Delta H_1 = -38.4$ , which gives $\Delta H_\text{r} = 2(27.2) - (-38.4) = +92.8 \text{ kJ mol}^{-1}$ . Always use your calculated value from part (b)(ii) if possible.
Ensure the final equation matches the target exactly.

Techniques used

construct Hess's law cyclecombine enthalpy changes algebraically

(c)

$\mathbf{Z}$ is a salt that contains a Period 4 element from Group 2. When $\mathbf{Z}$ is heated brown gas forms.

Identify the formula of $\mathbf{Z}$ and use it to write an equation for the reaction.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Group 2, Period 4 element is calcium ( $\text{Ca}$ ).
Brown gas is nitrogen dioxide ( $\text{NO}_2$ ), so the salt is calcium nitrate: $\text{Ca(NO}_3)_2$ .

Equation:
$\text{Ca(NO}_3)_2(\text{s}) \rightarrow \text{CaO}(\text{s}) + 2\text{NO}_2(\text{g}) + \frac{1}{2}\text{O}_2(\text{g})$

(or $2\text{Ca(NO}_3)_2(\text{s}) \rightarrow 2\text{CaO}(\text{s}) + 4\text{NO}_2(\text{g}) + \text{O}_2(\text{g})$ )

Final answer

Ca(NO3)2; Ca(NO3)2 -> CaO + 2NO2 + 1/2 O2

Detailed explanation

Background Concept

Group 2 elements are beryllium, magnesium, calcium, strontium, barium, and radium. They are in periods 2, 3, 4, 5, 6, and 7 respectively. The Period 4 Group 2 element is calcium ( $\text{Ca}$ ).

Nitrates of Group 2 metals decompose on heating to give the metal oxide, nitrogen dioxide gas, and oxygen gas. The general equation is:
$2\text{M(NO}_3)_2(\text{s}) \rightarrow 2\text{MO}(\text{s}) + 4\text{NO}_2(\text{g}) + \text{O}_2(\text{g})$

Nitrogen dioxide ( $\text{NO}_2$ ) is a brown gas. This is a characteristic test for nitrate ions or Group 2 nitrates.

Understanding the Question

Salt $\mathbf{Z}$ contains a Period 4 Group 2 element. This is calcium ( $\text{Ca}$ ).
When heated, brown gas forms. Brown gas from thermal decomposition is $\text{NO}_2$ , indicating the salt is a nitrate.
So $\mathbf{Z}$ is calcium nitrate, $\text{Ca(NO}_3)_2$ .
We need to write the balanced equation for its thermal decomposition.

Approach

Identify the Group 2 Period 4 element: Calcium ( $\text{Ca}$ ).
Identify the anion from the brown gas: Nitrate ( $\text{NO}_3^-$ ), producing $\text{NO}_2$ .
Write the formula: $\text{Ca(NO}_3)_2$ .
Write the balanced decomposition equation: $\text{Ca(NO}_3)_2 \rightarrow \text{CaO} + 2\text{NO}_2 + \frac{1}{2}\text{O}_2$ .

Step-by-Step Reasoning

Identify the metal: Group 2, Period 4 is calcium ( $\text{Ca}$ ).
Identify the anion: Brown gas on heating is $\text{NO}_2$ . This comes from the decomposition of a nitrate ( $\text{NO}_3^-$ ).
Formula of salt: Calcium nitrate is $\text{Ca(NO}_3)_2$ .
Balanced equation:
$\text{Ca(NO}_3)_2 \rightarrow \text{CaO} + 2\text{NO}_2 + \frac{1}{2}\text{O}_2$
To avoid fractions: $2\text{Ca(NO}_3)_2 \rightarrow 2\text{CaO} + 4\text{NO}_2 + \text{O}_2$ . Both are acceptable.

Key Takeaways

Period 4 Group 2 is calcium.
Brown gas in thermal decomposition of nitrates is $\text{NO}_2$ .
Group 2 nitrates decompose to oxide, $\text{NO}_2$ , and $\text{O}_2$ .

Common Mistakes

Identifying the wrong Group 2 element (e.g., magnesium is Period 3).
Writing the wrong products (e.g., $\text{N}_2$ instead of $\text{NO}_2$ ; Group 1 nitrates give nitrite and oxygen, but Group 2 give oxide, $\text{NO}_2$ , and $\text{O}_2$ ).
Forgetting to balance the equation.

Things to Be Careful About

The question asks for the formula of $\mathbf{Z}$ and the equation. Both are required for marks.
State symbols are often expected in CIE equations, though not always strictly penalized if omitted in some contexts, but best to include: $(\text{s})$ , $(\text{g})$ .

Techniques used

identify element from group and periodrecognize brown gas as NO2write balanced decomposition equation

Q312MMediumEquilibriaAtomic StructureChemical Bonding

(a)

Describe what is meant by dynamic equilibrium.

2M Similar

DifficultyEasy

Worked solution

Answer

The rates of the forward and reverse reactions are equal. The concentrations of reactants and products remain constant (no change in measurable macroscopic properties).

Final answer

Rates of forward and reverse reactions are equal; concentrations of reactants and products remain constant.

Detailed explanation

Background Concept

Dynamic equilibrium occurs in a closed system when a reversible reaction has proceeded to the point where the rate of the forward reaction equals the rate of the reverse reaction. Despite the continuous reaction at the molecular level (hence 'dynamic'), there is no net change in the amounts of reactants and products. This means that all macroscopic, measurable properties of the system—such as colour, concentration, pressure (if gases are involved), and pH—remain constant over time.

Understanding the Question

The question asks for a description of 'dynamic equilibrium'. This is a standard definition question requiring two key pieces of information: the kinetic aspect (rates of reaction) and the thermodynamic/macroscopic aspect (constant concentrations or properties).

Approach

Recall the two-part definition of dynamic equilibrium. First, state that the forward and reverse reaction rates are equal. Second, state that this results in no net change in the concentrations of reactants and products, meaning macroscopic properties are constant.

Step-by-Step Reasoning

Mark 1 (M1): State that the rate of the forward reaction is equal to the rate of the reverse (or backward) reaction. This captures the 'dynamic' nature of the equilibrium.
Mark 2 (M2): State that there is no change in measurable properties, or more specifically, that the concentrations of reactants and products remain constant. This captures the 'equilibrium' aspect.

Key Takeaways

Dynamic equilibrium is characterized by equal forward and reverse rates, leading to constant macroscopic properties. It is a dynamic process at the molecular level, not a static one.

Common Mistakes

Stating that the reaction has 'stopped'. The reaction continues; the rates are just equal.
Saying 'the concentrations are equal'. Concentrations are constant, not necessarily equal to each other.
Forgetting to mention that it must be a closed system (though often implied in simple definitions, it's good practice).

Things to Be Careful About

Use precise terminology: 'rates are equal', not 'speeds are the same' (though often accepted, 'rates' is better).
Specify 'concentrations of reactants and products' rather than just 'amounts', especially if the volume could change (though for solutions in a fixed volume, they are equivalent).
Ensure both parts of the definition are present for full marks.

Techniques used

define dynamic equilibriumdistinguish between macroscopic and microscopic properties

(b)

Reaction 4 describes the reversible reaction between yellow $\text{Fe}^{3+}\text{(aq)}$ and colourless $\text{SCN}^-\text{(aq)}$ to produce red $\text{FeSCN}^{2+}\text{(aq)}$ .

\begin{aligned} \text{reaction 4} \quad & \text{Fe}^{3+}\text{(aq)} + \text{SCN}^-\text{(aq)} \rightleftharpoons \text{FeSCN}^{2+}\text{(aq)} \\ & \text{yellow} \qquad \text{colourless} \qquad\quad \text{red} \end{aligned}

An equilibrium mixture contains $\text{Fe}^{3+}\text{(aq)}$ , $\text{SCN}^-\text{(aq)}$ and $\text{FeSCN}^{2+}\text{(aq)}$ . A few colourless crystals of soluble $\text{KSCN(s)}$ are added. The mixture is then left until it reaches equilibrium again. The temperature of both equilibrium mixtures is the same.

7M

(i)

Deduce the changes that occur, if any, in the equilibrium mixture after $\text{KSCN(s)}$ is added compared to the original equilibrium mixture.

change in appearance
change in relative concentration of $\text{Fe}^{3+}\text{(aq)}$
change in value of the equilibrium constant, $K_\text{c}$

3M Similar

DifficultyMedium-Easy

Worked solution

Answer

Change in appearance: The mixture becomes darker red (or more red).
Change in relative concentration of Fe³⁺(aq): Decreases.
Change in value of Kc: Remains constant (no change).

Final answer

Darker red; [Fe³⁺] decreases; Kc is constant.

Detailed explanation

Background Concept

Le Chatelier's principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change. Adding a reactant increases its concentration, so the system will try to reduce it by favoring the forward reaction. The equilibrium constant, Kc, is only dependent on temperature. If the temperature remains constant, Kc remains constant, even if concentrations change.

Understanding the Question

We have an equilibrium mixture of Fe³⁺(aq) (yellow), SCN⁻(aq) (colourless), and FeSCN²⁺(aq) (red). Soluble KSCN(s) is added, which dissociates to give more SCN⁻(aq) ions. We need to deduce the changes in appearance, [Fe³⁺], and Kc after the system re-equilibrates at the same temperature.

Approach

Identify the disturbance: addition of SCN⁻(aq) (a reactant).
Apply Le Chatelier's principle: equilibrium shifts to the right (forward direction) to consume the added SCN⁻.
Deduce the effect on species: Fe³⁺ is consumed (concentration decreases), FeSCN²⁺ is produced (concentration increases).
Deduce the effect on appearance: more red FeSCN²⁺ means a darker red colour.
Deduce the effect on Kc: temperature is unchanged, so Kc is constant.

Step-by-Step Reasoning

Appearance (M1): Adding KSCN increases [SCN⁻]. By Le Chatelier's principle, the equilibrium shifts to the right to reduce [SCN⁻]. This produces more FeSCN²⁺(aq), which is red. Therefore, the mixture becomes darker red.
Concentration of Fe³⁺ (M2): As the equilibrium shifts to the right, Fe³⁺(aq) reacts with the added SCN⁻(aq). Therefore, the relative concentration of Fe³⁺(aq) decreases.
Value of Kc (M3): The equilibrium constant Kc is a function of temperature only. Since the temperature of both mixtures is the same, the value of Kc remains constant.

Key Takeaways

Adding a reactant shifts the equilibrium to the right (products side).
The colour change reflects the change in concentration of the coloured species.
Kc only changes if the temperature changes.

Common Mistakes

Saying Kc increases or decreases. Kc is constant at constant temperature.
Saying the concentration of Fe³⁺ increases. It is consumed in the forward reaction.
Saying the colour becomes yellow. FeSCN²⁺ is red and its concentration increases.

Things to Be Careful About

The question asks for 'relative concentration' or just change. 'Decreases' is the correct term.
Ensure you distinguish between the position of equilibrium (which shifts) and the value of Kc (which stays constant).

Techniques used

apply Le Chatelier's principlepredict shift in equilibrium positionrecognize temperature dependence of Kc

(ii)

The expression for the equilibrium constant, $K_\text{c}$ , for reaction 4 is shown.

K_\text{c} = \frac{[\text{FeSCN}^{2+}\text{(aq)}]}{[\text{Fe}^{3+}\text{(aq)}] \times [\text{SCN}^-\text{(aq)}]}

$5.00 \times 10^{-5}\text{ mol}$ of $\text{Fe}^{3+}\text{(aq)}$ and $5.00 \times 10^{-5}\text{ mol}$ of $\text{SCN}^-\text{(aq)}$ are added together and allowed to reach equilibrium. The total volume of the mixture is $25.0\text{ cm}^3$ .

At equilibrium the concentration of $\text{FeSCN}^{2+}\text{(aq)}$ is $4.23 \times 10^{-4}\text{ mol dm}^{-3}$ .

Calculate the equilibrium constant, $K_\text{c}$ , for reaction 4.

Include the units in your answer.

K_\text{c} = \text{..............................}

\text{units } \text{..............................}

4M Similar

DifficultyMedium

Worked solution

Working

Step 1: Calculate initial concentrations

\text{Volume} = 25.0 \text{ cm}^3 = 0.0250 \text{ dm}^3

[\text{Fe}^{3+}]_{\text{initial}} = [\text{SCN}^-]_{\text{initial}} = \frac{5.00 \times 10^{-5} \text{ mol}}{0.0250 \text{ dm}^3} = 2.00 \times 10^{-3} \text{ mol dm}^{-3} = 0.00200 \text{ mol dm}^{-3}

Step 2: Determine equilibrium concentrations
Given $[\text{FeSCN}^{2+}]_{\text{eq}} = 4.23 \times 10^{-4} \text{ mol dm}^{-3}$ .
From the stoichiometry of reaction 4 (1:1:1 ratio), the concentration of Fe³⁺ and SCN⁻ that reacts is equal to $[\text{FeSCN}^{2+}]_{\text{eq}}$ .

[\text{Fe}^{3+}]_{\text{eq}} = [\text{SCN}^-]_{\text{eq}} = 0.00200 - 4.23 \times 10^{-4} = 0.001577 \text{ mol dm}^{-3}

Step 3: Calculate Kc

K_c = \frac{[\text{FeSCN}^{2+}]}{[\text{Fe}^{3+}] \times [\text{SCN}^-]} = \frac{4.23 \times 10^{-4}}{(0.001577) \times (0.001577)}

K_c = \frac{4.23 \times 10^{-4}}{2.4869 \times 10^{-6}} = 170.09

Rounding to 3 significant figures: $K_c = 170$ .

Step 4: Determine units

\text{Units} = \frac{\text{mol dm}^{-3}}{(\text{mol dm}^{-3}) \times (\text{mol dm}^{-3})} = \text{mol}^{-1} \text{ dm}^3

Answer

K_c = 170

\text{units } = \text{mol}^{-1} \text{ dm}^3

Final answer

170; mol⁻¹ dm³

Detailed explanation

Background Concept

The equilibrium constant Kc is calculated from the equilibrium concentrations of reactants and products. For a general reaction $aA + bB \rightleftharpoons cC + dD$ , $K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}$ . To find equilibrium concentrations, we often use an ICE (Initial, Change, Equilibrium) table or simple stoichiometric deduction. Units of Kc depend on the stoichiometry of the reaction and must be derived from the expression.

Understanding the Question

We are given initial moles of Fe³⁺ and SCN⁻ (5.00 × 10⁻⁵ mol each) and total volume (25.0 cm³). At equilibrium, [FeSCN²⁺] = 4.23 × 10⁻⁴ mol dm⁻³. We need to calculate Kc and its units.

Approach

Convert initial moles to initial concentrations (mol dm⁻³).
Use the stoichiometry (1:1:1) to find the equilibrium concentrations of Fe³⁺ and SCN⁻. Since 1 mol of FeSCN²⁺ is formed from 1 mol of Fe³⁺ and 1 mol of SCN⁻, the decrease in [Fe³⁺] and [SCN⁻] is equal to the increase in [FeSCN²⁺].
Substitute equilibrium concentrations into the Kc expression.
Derive the units for Kc.

Step-by-Step Reasoning

Initial Concentrations (M1):
Volume = 25.0 cm³ = 25.0 / 1000 = 0.0250 dm³.
$[\text{Fe}^{3+}]_{\text{initial}} = \frac{5.00 \times 10^{-5}}{0.0250} = 2.00 \times 10^{-3} \text{ mol dm}^{-3} = 0.00200 \text{ mol dm}^{-3}$ . Same for SCN⁻.
Equilibrium Concentrations (M2):
$[\text{FeSCN}^{2+}]_{\text{eq}} = 4.23 \times 10^{-4} \text{ mol dm}^{-3} = 0.000423 \text{ mol dm}^{-3}$ .
From the equation $\text{Fe}^{3+} + \text{SCN}^- \rightleftharpoons \text{FeSCN}^{2+}$ , the change in concentration for Fe³⁺ and SCN⁻ is $-0.000423 \text{ mol dm}^{-3}$ .
$[\text{Fe}^{3+}]_{\text{eq}} = [\text{SCN}^-]_{\text{eq}} = 0.00200 - 0.000423 = 0.001577 \text{ mol dm}^{-3}$ .
Calculate Kc (M3):
$K_c = \frac{0.000423}{(0.001577) \times (0.001577)} = \frac{0.000423}{0.000002486929} \approx 170.089$ .
To 3 significant figures (matching the data given: 5.00, 25.0, 4.23), $K_c = 170$ .
Units (M4):
$K_c = \frac{\text{mol dm}^{-3}}{(\text{mol dm}^{-3})(\text{mol dm}^{-3})} = \frac{1}{\text{mol dm}^{-3}} = \text{mol}^{-1} \text{ dm}^3$ (or $\text{dm}^3 \text{ mol}^{-1}$ ).

Key Takeaways

Always convert volumes to dm³ when calculating concentrations in mol dm⁻³.
Use stoichiometry to relate changes in concentration. For a 1:1:1 reaction, $\Delta[\text{reactants}] = -\Delta[\text{product}]$ .
Always include units for Kc unless they cancel out completely.

Common Mistakes

Forgetting to convert cm³ to dm³ (dividing by 1000).
Using moles directly in the Kc expression instead of concentrations.
Calculating the equilibrium concentration incorrectly (e.g., adding instead of subtracting).
Forgetting units or getting the units wrong (e.g., writing mol dm⁻³ instead of mol⁻¹ dm³).
Rounding too early in the calculation, leading to a final answer with incorrect significant figures.

Things to Be Careful About

Significant figures: The given data (5.00, 25.0, 4.23) has 3 significant figures, so the final answer should be to 3 s.f. (170, not 170.1).
The expression for Kc is given in the question; ensure you use it exactly as written (products over reactants, concentrations raised to stoichiometric coefficients).
Units: $\text{mol}^{-1} \text{ dm}^3$ is equivalent to $\text{dm}^3 \text{ mol}^{-1}$ . Both are acceptable.

Techniques used

calculate initial concentrations from moles and volumeuse ICE table or stoichiometry to find equilibrium concentrationssubstitute into Kc expressiondetermine units of Kc

(c)

Determine the full electronic configuration of $\text{Fe}^{3+}$ .

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

$1\text{s}^2 \ 2\text{s}^2 \ 2\text{p}^6 \ 3\text{s}^2 \ 3\text{p}^6 \ 3\text{d}^5$

Final answer

1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵

Detailed explanation

Background Concept

Electronic configuration describes the distribution of electrons in atomic or molecular orbitals. For neutral atoms, electrons fill orbitals in order of increasing energy: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, etc. (Aufbau principle). However, when transition metals form positive ions, electrons are removed from the highest principal quantum number shell first. For first-row transition metals, this means electrons are removed from the 4s orbital before the 3d orbital, even though 4s was filled before 3d in the neutral atom.

Understanding the Question

We need the full electronic configuration of the iron(III) ion, Fe³⁺. Iron (Fe) has atomic number 26.

Approach

Write the electronic configuration of the neutral Fe atom (26 electrons).
Remove 3 electrons to form Fe³⁺. Remove two from 4s and one from 3d.
Write the remaining configuration.

Step-by-Step Reasoning

Neutral Fe atom (Z=26): $1\text{s}^2 \ 2\text{s}^2 \ 2\text{p}^6 \ 3\text{s}^2 \ 3\text{p}^6 \ 4\text{s}^2 \ 3\text{d}^6$ (or $[\text{Ar}] \ 4\text{s}^2 \ 3\text{d}^6$ ).
Forming Fe³⁺: Remove 3 electrons. The 4s electrons are removed first (2 electrons), then one from 3d.
Fe³⁺ configuration: $1\text{s}^2 \ 2\text{s}^2 \ 2\text{p}^6 \ 3\text{s}^2 \ 3\text{p}^6 \ 3\text{d}^5$ (23 electrons remaining).

Key Takeaways

Transition metal ions lose 4s electrons before 3d electrons.
Fe³⁺ has a half-filled 3d subshell (3d⁵), which is particularly stable.

Common Mistakes

Writing the configuration as $[\text{Ar}] \ 3\text{d}^3 \ 4\text{s}^2$ (incorrect removal order).
Forgetting to remove the correct number of electrons (Fe³⁺ means 3 electrons removed, not 2).
Writing the configuration in order of filling (4s before 3d) for the ion. While sometimes accepted in simplified contexts, the standard convention for ions is to list 3d before 4s if 4s is empty, or simply omit 4s. The mark scheme gives $1\text{s}^2 \ 2\text{s}^2 \ 2\text{p}^6 \ 3\text{s}^2 \ 3\text{p}^6 \ 3\text{d}^5$ , which is the correct full configuration.

Things to Be Careful About

The question asks for the 'full' electronic configuration, so do not use the noble gas core abbreviation $[\text{Ar}]$ .
Ensure the total number of electrons is correct: 2 + 2 + 6 + 2 + 6 + 5 = 23. Fe has 26, Fe³⁺ has 23. Correct.

Techniques used

deduce electronic configuration of a transition metal ionremove electrons from the outermost shell first (4s before 3d)

(d)

$\text{SCN}^-\text{(aq)}$ is colourless.

Complete the dot-and-cross diagram in Fig. 3.1 to show the arrangement of outer electrons in an $\text{SCN}^-$ ion.

Fig. 3.1

2M Similar

DifficultyMedium

Worked solution

Answer

The diagram shows the thiocyanate ion, $[\text{SCN}]^-$ . Between S and C, there is a single bond (2 shared electrons, one dot and one cross). Between C and N, there is a triple bond (6 shared electrons, three pairs). S has 3 lone pairs (6 electrons). N has 1 lone pair (2 electrons). C has no lone pairs. The entire structure is enclosed in square brackets with a negative charge superscript.

Final answer

Dot-and-cross diagram: S-C single bond (2e), C≡N triple bond (6e), 3 lone pairs on S, 1 lone pair on N, overall charge -1.

Detailed explanation

Background Concept

A dot-and-cross diagram shows the arrangement of outer (valence) electrons in a molecule or ion. Dots and crosses are used to distinguish electrons from different atoms. Covalent bonds are formed by sharing electron pairs. Lone pairs are non-bonding pairs of electrons. For polyatomic ions, the structure is enclosed in brackets with the overall charge indicated.

Understanding the Question

We need to complete the dot-and-cross diagram for the thiocyanate ion, SCN⁻. The outline shows three overlapping circles for S, C, and N, with brackets and a negative charge. We need to place the correct number of bonding and non-bonding electrons.

Approach

Count total valence electrons: S (6) + C (4) + N (5) + 1 (for negative charge) = 16 electrons.
Determine the bonding arrangement. Carbon needs 4 bonds to complete its octet. Nitrogen typically forms 3 bonds and has 1 lone pair. Sulfur can form various bonds. The mark scheme indicates a single bond between S and C, and a triple bond between C and N.
Place bonding electrons: S-C (2e), C≡N (6e). Total bonding = 8e.
Place lone pairs to complete octets: S needs 6 more e⁻ (3 lone pairs), N needs 2 more e⁻ (1 lone pair), C has 8 e⁻ from bonds (no lone pairs).
Total electrons: 8 (bonding) + 6 (S lone pairs) + 2 (N lone pair) = 16. Correct.

Step-by-Step Reasoning

Valence electrons (M1 context): S has 6, C has 4, N has 5, plus 1 for the charge = 16 outer electrons.
Bonding (M1): To satisfy the octet rule for C (4 bonds) and N (3 bonds + lone pair), and considering the mark scheme guidance:
- Between C and N: 6 electrons (triple bond, $\text{C}\equiv\text{N}$ ).
- Between C and S: 2 electrons (single bond, $\text{C}-\text{S}$ ).
Lone pairs (M2):
- Sulfur (S): Has 2 bonding electrons, needs 6 more to complete octet. Place 3 lone pairs (6 electrons) around S.
- Carbon (C): Has 8 bonding electrons (2 from S-C, 6 from C≡N). Octet complete. No lone pairs.
- Nitrogen (N): Has 6 bonding electrons, needs 2 more. Place 1 lone pair (2 electrons) around N.
Overall: Enclose in brackets with a negative charge. Formal charges: S (-1), C (0), N (0). Sum = -1. Matches the ion charge.

Key Takeaways

Count total valence electrons carefully, including charge.
Carbon typically forms 4 bonds and has no lone pairs in stable molecules/ions.
Nitrogen in a nitrile-like group ( $\text{-C}\equiv\text{N}$ ) has 1 lone pair.
Sulfur can have expanded octets or lone pairs depending on bonding.

Common Mistakes

Drawing a double bond between S and C and a double bond between C and N ( $\text{S}=\text{C}=\text{N}$ ). While this is a valid resonance structure, the mark scheme specifically rewards the $\text{S}-\text{C}\equiv\text{N}$ structure with the given lone pairs. (Actually, $\text{S}=\text{C}=\text{N}$ with 2 lone pairs on S and 1 on N is also a valid Lewis structure, but CIE mark schemes often accept specific resonance forms or have a preferred one. The mark scheme explicitly says: '6 electrons between C and N AND 2 electrons between C and S'. This forces the single-triple bond structure).
Forgetting the overall charge or brackets.
Incorrect number of lone pairs (e.g., putting lone pairs on Carbon).
Not using dots and crosses to show the origin of electrons (though the mark scheme description focuses on the arrangement, a proper dot-and-cross diagram should use different symbols for electrons from different atoms).

Things to Be Careful About

The mark scheme is very specific: '6 electrons between C and N' (triple bond) and '2 electrons between C and S' (single bond). Do not draw double bonds.
'3 lone pairs around S AND none around C AND 1 lone pair around N'. Ensure these are exactly as specified.
The diagram must be enclosed in square brackets with a minus sign outside.

Techniques used

count valence electrons for a polyatomic iondraw Lewis structure with correct bonding and lone pairsuse dot-and-cross notation to show electron origin

Q415MMediumIntroduction to Organic ChemistryHalogen CompoundsHydroxy CompoundsHydrocarbons

$\text{CH}_3(\text{CH}_2)_5\text{CHBrCH}_3$ exists as a pair of stereoisomers.

(a)

Draw the three-dimensional structures of the two stereoisomers of $\text{CH}_3(\text{CH}_2)_5\text{CHBrCH}_3$ .
$\text{R}$ can be used to represent $\text{CH}_3(\text{CH}_2)_5$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

The carbon bearing Br is chiral (bonded to four different groups: H, Br, CH $_3$ , and (CH $_2$ ) $_5$ CH $_3$ ).

Final answer

Two enantiomers drawn as 3D structures with wedge and dashed bonds at the chiral carbon, showing non-superimposable mirror images

Detailed explanation

Background Concept

Optical isomerism (enantiomerism) occurs when a molecule contains a chiral centre — a carbon atom bonded to four different groups. The two enantiomers are non-superimposable mirror images of each other, related like left and right hands. They have identical physical properties except for the direction in which they rotate plane-polarised light.

Understanding the Question

The molecule CH $_3$ (CH $_2$ ) $_5$ CHBrCH $_3$ (2-bromooctane) has a carbon-2 that is bonded to: H, Br, CH $_3$ (carbon-1), and (CH $_2$ ) $_5$ CH $_3$ (the heptyl group, carbon-3 to carbon-8). These are four different groups, so carbon-2 is chiral. The question asks you to draw both enantiomers in three-dimensional form, using the R group abbreviation for CH $_3$ (CH $_2$ ) $_5$ .

Approach

Identify the chiral centre (C-2).
Draw one enantiomer using a solid wedge for a bond coming towards you and a dashed wedge for a bond going away.
Draw the mirror image by swapping two of the groups (or reflecting the entire structure).

Step-by-Step Reasoning

The chiral carbon has four different substituents: H, Br, CH $_3$ , and R (where R = CH $_3$ (CH $_2$ ) $_5$ ).
In the first enantiomer, place Br on a solid wedge (coming out), H on a dashed wedge (going in), with R and CH $_3$ in the plane.
The second enantiomer is the mirror image: swap the positions of Br and H (or equivalently, swap any two groups).
Both structures must clearly show the 3D arrangement using wedge and dash notation.

Key Takeaways

A chiral centre requires four different groups attached to a single carbon.
Enantiomers are drawn as non-superimposable mirror images using wedge (towards viewer) and dash (away from viewer) notation.
Swapping any two groups at a chiral centre inverts the stereochemistry.

Common Mistakes

Drawing the two structures as identical (superimposable) rather than as mirror images.
Using plain lines instead of wedge/dash bonds, failing to show the 3D arrangement.
Not recognising that the long alkyl chain R and the CH $_3$ are different groups.

Things to Be Careful About

Ensure the mirror image relationship is clear — the two structures must be non-superimposable.
The question allows R to represent CH $_3$ (CH $_2$ ) $_5$ , which simplifies the drawing.

Techniques used

identify the chiral centredraw enantiomers using wedge-dash notation

(b)

A sample of $\text{CH}_3(\text{CH}_2)_5\text{CHBrCH}_3$ reacts with $\text{NaOH}$ to make $\text{CH}_3(\text{CH}_2)_5\text{CH(OH)CH}_3$ in an $\text{S}_\text{N}1$ mechanism.

Complete Fig. 4.1 to show the mechanism for the reaction of $\text{CH}_3(\text{CH}_2)_5\text{CHBrCH}_3$ and $\text{NaOH}$ .

Include charges, dipoles, lone pairs of electrons and curly arrows, as appropriate.

Fig. 4.1

3M Similar

DifficultyMedium

Worked solution

Answer

Step 1: The C $-$ Br bond breaks heterolytically. A curly arrow goes from the C $-$ Br bond to Br, forming a carbocation intermediate and Br $^-$ . The C $-$ Br bond is polarised with $\delta^+$ on C and $\delta^-$ on Br.

Step 2: The lone pair on OH $^-$ attacks the positively charged carbon of the carbocation, forming the C $-$ OH bond.

Final answer

SN1 mechanism: (1) curly arrow from C-Br bond to Br showing heterolytic fission with dipole labels, forming carbocation CH3(CH2)5CH+CH3 + Br-; (2) curly arrow from lone pair on OH- to the C+ of the carbocation

Detailed explanation

Background Concept

The S $_N$ 1 (substitution nucleophilic unimolecular) mechanism proceeds in two steps. First, the C $-$ X bond undergoes heterolytic fission to form a carbocation intermediate and a halide ion. This is the rate-determining step and involves only the substrate molecule (hence 'unimolecular'). Second, the nucleophile attacks the planar carbocation to form the product. The carbocation is trigonal planar, which is why S $_N$ 1 reactions at a chiral centre give racemic mixtures.

Understanding the Question

The question provides the starting material (2-bromooctane) and the product (octan-2-ol) and asks you to fill in the mechanism between them. You must include: charges, dipoles, lone pairs of electrons, and curly arrows. The mechanism is specifically S $_N$ 1.

Approach

Draw the reactant with the C $-$ Br bond polarised ( $\delta^+$ on C, $\delta^-$ on Br).
Show a curly arrow from the C $-$ Br bond to Br, indicating heterolytic fission.
Draw the carbocation intermediate with a positive charge on C $^+$ .
Draw OH $^-$ with lone pairs on oxygen and a negative charge.
Show a curly arrow from a lone pair on O to the C $^+$ .

Step-by-Step Reasoning

M1: The C $-$ Br bond must show a dipole with $\delta^+$ on carbon and $\delta^-$ on bromine (because Br is more electronegative). A curly arrow starts from the bond (not from Br) and points to Br, showing the electron pair leaving with Br. This produces Br $^-$ (or NaBr).
M2: The intermediate is a carbocation: CH $_3$ (CH $_2$ ) $_5$ C $^+$ HCH $_3$ . The positive charge must be shown on carbon. The carbon is now trigonal planar (sp $^2$ hybridised) with an empty p orbital.
M3: OH $^-$ acts as the nucleophile. A curly arrow starts from a lone pair on the oxygen atom (which must be drawn) and points to the positively charged carbon, forming the new C $-$ OH bond.

Key Takeaways

In S $_N$ 1, the curly arrow for the first step starts from the bond (representing the electron pair in the bond), not from the atom.
The carbocation intermediate must carry a formal positive charge.
The nucleophile's curly arrow starts from a lone pair, which must be explicitly drawn.

Common Mistakes

Drawing a curly arrow from Br to C (wrong direction — the bond pair goes to Br).
Forgetting the positive charge on the carbocation intermediate.
Not drawing lone pairs on OH $^-$ before showing the curly arrow from them.
Drawing the S $_N$ 2 mechanism instead (single step with simultaneous attack and departure).
Starting the first curly arrow from Br rather than from the C $-$ Br bond.

Things to Be Careful About

The dipole labels ( $\delta^+$ and $\delta^-$ ) are required for M1.
The curly arrow must start from the bond, not from the atom.
The intermediate must be clearly a carbocation with the + charge on the correct carbon.
OH $^-$ must show at least one lone pair and the negative charge.

Techniques used

show heterolytic fission of C-Br bond with curly arrowdraw carbocation intermediateshow nucleophilic attack of OH- on carbocation

(c)

Separate samples of $\text{CH}_3(\text{CH}_2)_5\text{CHBrCH}_3$ , $\text{CH}_3(\text{CH}_2)_5\text{CH(OH)CH}_3$ and $\text{CH}_3(\text{CH}_2)_5\text{CHCH}_2$ are tested with different reagents.

Complete Table 4.1. If no reaction occurs, write $\times$ in the relevant box.

Table 4.1

3M Similar

DifficultyMedium

Worked solution

Answer

Reagent	CH $_3$ (CH $_2$ ) $_5$ CHBrCH $_3$	CH $_3$ (CH $_2$ ) $_5$ CH(OH)CH $_3$	CH $_3$ (CH $_2$ ) $_5$ CHCH $_2$
Br $_2$ (l) in the dark	$\times$	(not required)	orange/brown to colourless
PCl $_5$ (s)	(not required)	steamy fumes	$\times$
AgNO $_3$ (aq)	cream precipitate	$\times$	(not required)

Final answer

Br2 + 2-bromooctane: ✗; Br2 + oct-1-ene: orange/brown to colourless; PCl5 + octan-2-ol: steamy fumes; PCl5 + oct-1-ene: ✗; AgNO3 + 2-bromooctane: cream precipitate; AgNO3 + octan-2-ol: ✗

Detailed explanation

Background Concept

Three key chemical tests are being applied:

Bromine water / Br $_2$ (l): Tests for C=C double bonds (addition reaction decolourises bromine). Saturated compounds do not react in the dark.
PCl $_5$ : Tests for the $-$ OH group in alcohols (and carboxylic acids). The reaction produces HCl gas, observed as steamy fumes.
AgNO $_3$ (aq): Tests for halide ions. Halogenoalkanes react slowly via nucleophilic substitution to release halide ions, which precipitate as AgX. AgBr is cream/off-white.

Understanding the Question

Three compounds are tested with three reagents. Some cells are hatched (not to be filled). The six cells to complete test whether the student can predict reactions and non-reactions based on functional groups present.

CH $_3$ (CH $_2$ ) $_5$ CHBrCH $_3$ : a halogenoalkane (C $-$ Br bond, no C=C, no $-$ OH)
CH $_3$ (CH $_2$ ) $_5$ CH(OH)CH $_3$ : an alcohol ( $-$ OH group, no C $-$ Br, no C=C)
CH $_3$ (CH $_2$ ) $_5$ CHCH $_2$ : an alkene (C=C double bond, no C $-$ Br, no $-$ OH)

Approach

For each cell, identify the functional group in the compound and determine whether the reagent reacts with that group.

Step-by-Step Reasoning

Br $_2$ (l) in the dark:

With 2-bromooctane: This is a saturated halogenoalkane with no C=C. Bromine does not react with alkanes/halogenoalkanes in the dark. Answer: $\times$
With oct-1-ene: Contains a C=C double bond. Bromine adds across the double bond, decolourising the orange/brown solution. Answer: orange/brown to colourless

PCl $_5$ (s):

With octan-2-ol: Contains an $-$ OH group. PCl $_5$ reacts with alcohols to produce chloroalkane + POCl $_3$ + HCl. The HCl gas gives steamy fumes. Answer: steamy fumes
With oct-1-ene: No $-$ OH group. PCl $_5$ does not react with alkenes. Answer: $\times$

AgNO $_3$ (aq):

With 2-bromooctane: The C $-$ Br bond undergoes slow nucleophilic substitution with water (present in aqueous solution), releasing Br $^-$ ions which react with Ag $^+$ to form AgBr precipitate. AgBr is cream/off-white. Answer: cream precipitate
With octan-2-ol: No halogen present, so no halide ion can be released. No precipitate forms. Answer: $\times$

Key Takeaways

Bromine test is specific for C=C (addition reaction).
PCl $_5$ test is specific for $-$ OH (and $-$ COOH), producing HCl steamy fumes.
AgNO $_3$ test detects halide ions released from halogenoalkanes (cream = AgBr, white = AgCl, yellow = AgI).

Common Mistakes

Writing 'colourless' for the bromine test without mentioning the starting colour (must say orange/brown to colourless).
Confusing the PCl $_5$ observation with the sodium test (PCl $_5$ gives steamy fumes of HCl; sodium gives effervescence of H $_2$ ).
Thinking AgNO $_3$ reacts directly with the C $-$ Br bond without recognising that substitution must first release Br $^-$ .
Writing 'white' precipitate for AgBr (it is cream/off-white; white is AgCl).

Things to Be Careful About

The mark scheme accepts 'cream precipitate' OR 'off-white precipitate' for AgBr.
'No reaction' must be indicated with $\times$ in the appropriate boxes.
The bromine test must mention both the starting colour and the final colour.

Techniques used

predict observation of bromine addition to alkenepredict observation of PCl5 with alcoholpredict observation of AgNO3 with halogenoalkane

(d)

$\text{CH}_3(\text{CH}_2)_5\text{CHBrCH}_3$ is heated with $\mathbf{D}$ to produce three different molecules, $\mathbf{E}$ , $\mathbf{F}$ and $\mathbf{G}$ .

Fig. 4.2

2M

(i)

Name the type of reaction.

1M Similar

(ii)

Identify $\mathbf{D}$ and the conditions used.

1M Similar

(e)

5M

(i)

Both $\sigma$ and $\pi$ bonds are present in a molecule of $\mathbf{E}$ as a result of different types of hybridisation in the carbon atoms.

Complete Table 4.2 to show the number of carbon atoms with each type of hybridisation in a molecule of $\mathbf{E}$ .

Table 4.2

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Hybridisation	Number of carbon atoms
sp	0
sp $^2$	2
sp $^3$	6

Final answer

sp: 0, sp2: 2, sp3: 6

Detailed explanation

Background Concept

Carbon hybridisation depends on the number of electron domains (regions of electron density) around the carbon:

sp $^3$ : 4 electron domains (4 single bonds) — tetrahedral, 109.5°
sp $^2$ : 3 electron domains (2 single bonds + 1 double bond, or 1 single + 1 double + lone pair) — trigonal planar, 120°
sp: 2 electron domains (e.g. 2 double bonds, or 1 triple bond + 1 single bond) — linear, 180°

In an alkene, the two carbons of the C=C are sp $^2$ hybridised (each has 3 electron domains: the double bond counts as one domain, plus two single bonds). All other carbons with only single bonds are sp $^3$ .

Understanding the Question

Molecule E is trans-oct-2-ene: CH $_3$ CH=CH(CH $_2$ ) $_4$ CH $_3$ . It has 8 carbon atoms total. The question asks how many are sp, sp $^2$ , and sp $^3$ hybridised.

Approach

Draw or visualise the full structure of oct-2-ene.
Identify the C=C double bond carbons (sp $^2$ ).
Check for any triple bonds or allene-type carbons (sp) — there are none.
Count the remaining carbons (all sp $^3$ ).

Step-by-Step Reasoning

Oct-2-ene: CH $_3$ –CH=CH–CH $_2$ –CH $_2$ –CH $_2$ –CH $_2$ –CH $_3$
The C=C is between C-2 and C-3. Both are sp $^2$ hybridised (each has 3 electron domains: one double bond + two single bonds). That gives 2 sp $^2$ carbons.
No carbon has a triple bond or two double bonds, so there are 0 sp carbons.
The remaining 6 carbons (C-1, C-4, C-5, C-6, C-7, C-8) each have 4 single bonds, so they are all sp $^3$ . That gives 6 sp $^3$ carbons.
Check: 0 + 2 + 6 = 8 ✓

Key Takeaways

Each carbon in a C=C double bond is sp $^2$ hybridised.
All saturated carbons (only single bonds) are sp $^3$ .
sp hybridisation requires a triple bond or cumulated double bonds (allene).

Common Mistakes

Counting the sp $^2$ carbons as 1 instead of 2 (both carbons of the double bond are sp $^2$ ).
Assigning sp hybridisation to alkene carbons.
Miscounting the total number of carbons.

Things to Be Careful About

The double bond involves TWO sp $^2$ carbons, not one.
Verify that the total adds up to the number of carbon atoms in the molecule.

Techniques used

count carbon atoms by hybridisation type from structural formula

(ii)

Describe the essential feature of an unbranched hydrocarbon that causes its molecules to show stereoisomerism. Explain how this feature leads to stereoisomerism.

3M Similar

DifficultyMedium-Easy

Worked solution

Answer

The molecule contains a C=C double bond, which includes a $\pi$ bond.
The $\pi$ bond prevents/restricts rotation about the C $=$ C bond.
Each carbon of the double bond carries two different groups (e.g. one H and one alkyl group), so the groups can be arranged in two different spatial configurations (E and Z).

Final answer

Presence of a pi bond causing restricted rotation about C=C, with two different groups on each double-bonded carbon atom

Detailed explanation

Background Concept

E/Z (cis/trans) isomerism is a form of stereoisomerism that arises from restricted rotation about a C=C double bond. The double bond consists of one $\sigma$ bond and one $\pi$ bond. The $\pi$ bond is formed by sideways overlap of p orbitals above and below the internuclear axis. Rotation about the C=C would require breaking this $\pi$ bond, which requires significant energy (~270 kJ mol $^{-1}$ ), so rotation is effectively prevented at room temperature. If each carbon of the double bond carries two different groups, two distinct arrangements are possible: the groups can be on the same side (Z/cis) or opposite sides (E/trans).

Understanding the Question

The question asks for the essential structural feature in an unbranched hydrocarbon that causes stereoisomerism, and how this feature leads to the phenomenon. The answer requires three linked points: the $\pi$ bond, restricted rotation, and the requirement for different groups.

Approach

Build the explanation logically: (1) identify the structural feature ( $\pi$ bond in C=C), (2) explain its consequence (restricted rotation), (3) state the additional condition needed (different groups on each carbon).

Step-by-Step Reasoning

M1: The essential feature is the presence of a C=C double bond, which contains a $\pi$ bond. The $\pi$ bond is formed by overlap of p orbitals and locks the two carbon atoms in a fixed relative orientation.
M2: Because of the $\pi$ bond, there is restricted (or no) rotation about the C $=$ C bond. This means groups attached to the double-bonded carbons cannot freely exchange positions, so different spatial arrangements are 'frozen in'.
M3: For stereoisomerism to actually occur, each carbon of the double bond must carry two different groups. If either carbon has two identical groups (e.g. two H atoms), then swapping them gives the same molecule and no isomerism exists. In oct-2-ene, C-2 has H and CH $_3$ , and C-3 has H and (CH $_2$ ) $_4$ CH $_3$ — both have two different groups, so E and Z isomers exist.

Key Takeaways

The $\pi$ bond is the structural cause of restricted rotation.
Restricted rotation alone is not sufficient — the two groups on each carbon must also be different.
E/Z isomerism is a type of stereoisomerism (same connectivity, different spatial arrangement).

Common Mistakes

Saying 'the double bond prevents rotation' without mentioning the $\pi$ bond specifically.
Omitting the requirement that each carbon must have two different groups.
Confusing this with optical isomerism (which requires a chiral centre, not a double bond).
Saying 'different groups on the molecule' rather than 'different groups on each double-bonded carbon'.

Things to Be Careful About

All three points are needed for full marks: $\pi$ bond, restricted rotation, AND different groups on each carbon.
The mark scheme accepts various wordings for M3: 'two different groups on each double-bonded carbon', 'two different atoms on each', or 'one H and one alkyl on each'.

Techniques used

explain restricted rotation about C=Cidentify requirement for different groups on each double-bonded carbon

Q58MMediumAnalytical TechniquesHydroxy Compounds

Compound $\mathbf{W}$ has molecular formula $\text{C}_4\text{H}_{10}\text{O}$ . It contains only one functional group.

(a)

Table 5.1 shows the two peaks with the greatest $m/e$ values in the mass spectrum of $\mathbf{W}$ .

Table 5.1

$m/e$	relative abundance
74	50
75	$x$

3M

(i)

Calculate the relative abundance, $x$ , of the peak at $m/e = 75$ using the information from Table 5.1.

x = \text{..............................}

1M Similar

DifficultyMedium-Easy

Worked solution

Working

The M+1 peak at $m/e = 75$ arises from molecules containing one $^{13}\text{C}$ atom.

With 4 carbon atoms, the relative abundance of M+1 compared to M is:

\frac{4 \times 1.1}{100} = 0.044

Since the relative abundance of M ( $m/e = 74$ ) is 50:

x = 0.044 \times 50 = 2.2

Answer

$x = 2.2$

Final answer

2.2

Detailed explanation

Background Concept

In mass spectrometry, the molecular ion peak (M) corresponds to the molecule containing all the most abundant isotopes. A small peak appears at M+1 (one mass unit higher) due to the natural presence of $^{13}\text{C}$ , which has an abundance of 1.1% relative to $^{12}\text{C}$ . Each carbon atom in the molecule contributes independently to the probability of having one $^{13}\text{C}$ , so the relative height of the M+1 peak compared to M is approximately $n \times 1.1\%$ , where $n$ is the number of carbon atoms.

Understanding the Question

The question gives the relative abundance of the M peak ( $m/e = 74$ ) as 50 and asks for the relative abundance $x$ of the M+1 peak ( $m/e = 75$ ). Compound W has molecular formula $\text{C}_4\text{H}_{10}\text{O}$ , so it contains 4 carbon atoms. The command word is "calculate", so a numerical answer with working is required.

Approach

Use the formula: relative abundance of M+1 = (number of C atoms × 1.1/100) × relative abundance of M.

Step-by-Step Reasoning

Identify the number of carbon atoms in W: $\text{C}_4\text{H}_{10}\text{O}$ has 4 carbon atoms.
The probability that any one molecule contains exactly one $^{13}\text{C}$ (giving the M+1 peak) is approximately $4 \times 1.1\% = 4.4\%$ of the M peak intensity.
The M peak has a relative abundance of 50.
Therefore $x = \frac{4 \times 1.1}{100} \times 50 = \frac{4.4}{100} \times 50 = 2.2$ .

Key Takeaways

The M+1 peak in mass spectrometry is due to $^{13}\text{C}$ .
The ratio of M+1 to M peak heights gives the number of carbon atoms: $\frac{\text{height of M+1}}{\text{height of M}} \times \frac{100}{1.1} = n(\text{C})$ .
This is a straightforward application once the 1.1% rule is known.

Common Mistakes

Forgetting to multiply by the relative abundance of M (50) and just giving 4.4.
Using 1.1% as a decimal incorrectly (e.g., writing 1.1 instead of 0.011).
Confusing the M+1 peak with the M+2 peak (which relates to $^{37}\text{Cl}$ or $^{81}\text{Br}$ ).

Things to Be Careful About

The 1.1% figure must be used as given; do not confuse it with other isotopic abundances.
Ensure the final answer is dimensionless (it is a relative abundance, not a percentage).

Techniques used

apply the 1.1% natural abundance of carbon-13 per carbon atomcalculate relative abundance of the M+1 peak from the M peak

(ii)

The mass spectrum of $\mathbf{W}$ also shows peaks at $m/e = 29$ and $m/e = 59$ .

Suggest the molecular formulae of these fragments.

m/e = 29 \quad \text{..............................}

m/e = 59 \quad \text{..............................}

2M Similar

DifficultyMedium

Worked solution

Answer

$m/e = 29$ : $\text{C}_2\text{H}_5^+$

$m/e = 59$ : $\text{C}_3\text{H}_7\text{O}^+$ (or $[\text{CH}_3\text{CH}_2\text{CH}=\text{OH}]^+$ )

Final answer

m/e = 29: C2H5+; m/e = 59: C3H7O+

Detailed explanation

Background Concept

In mass spectrometry, the molecular ion ( $\text{M}^+$ ) fragments into smaller cations and neutral radicals. The $m/e$ value of each fragment peak tells us the mass of the cation. For alcohols, common fragmentation pathways include loss of an alkyl group (giving an oxonium ion) or loss of a methyl radical (giving a fragment containing the oxygen).

Understanding the Question

W has molecular formula $\text{C}_4\text{H}_{10}\text{O}$ with $M_r = 74$ . Peaks at $m/e = 29$ and $m/e = 59$ must be identified as fragment cations. The command word is "suggest", so a chemically reasonable formula is required.

Approach

Calculate what mass is lost from the molecular ion to give each fragment, then identify the neutral loss and the resulting cation.

Step-by-Step Reasoning

For $m/e = 29$ :

Mass lost from M: $74 - 29 = 45$ . A loss of 45 corresponds to $\text{CH}_3\text{CH}_2\text{O}\cdot$ or $\text{CH}_3\text{O}\cdot$ + $\text{CH}_2$ ... Actually, the fragment itself has mass 29. Common cations of mass 29 are $\text{C}_2\text{H}_5^+$ (ethyl cation) or $\text{CHO}^+$ (formyl cation). For an alcohol with 4 carbons, $\text{C}_2\text{H}_5^+$ is the most likely fragment from cleavage of the C–C bond.

For $m/e = 59$ :

Mass lost from M: $74 - 59 = 15$ , which corresponds to loss of a $\text{CH}_3\cdot$ radical.
The remaining cation has formula $\text{C}_3\text{H}_7\text{O}^+$ .
This is consistent with an oxonium ion $[\text{CH}_3\text{CH}_2\text{CH}=\text{OH}]^+$ formed by alpha-cleavage of the alcohol, where the $\text{CH}_3\cdot$ is lost and the positive charge resides on the oxygen-containing fragment.

Key Takeaways

Fragment peaks in mass spectra correspond to cations; the neutral loss accounts for the difference from M.
Loss of $\text{CH}_3\cdot$ (15) from a butanol gives $\text{C}_3\text{H}_7\text{O}^+$ at $m/e = 59$ .
$\text{C}_2\text{H}_5^+$ at $m/e = 29$ is a common alkyl fragment.

Common Mistakes

Writing $\text{C}_2\text{H}_5$ without the positive charge (fragments detected are cations).
For $m/e = 59$ , writing $\text{C}_2\text{H}_3\text{O}^+$ (mass 43) instead of $\text{C}_3\text{H}_7\text{O}^+$ .
Confusing the neutral loss with the fragment itself.

Things to Be Careful About

Always include the $^+$ charge on fragment formulae.
Check that the formula mass matches the stated $m/e$ value.

Techniques used

deduce fragment formulae from mass differencesidentify common fragmentation patterns of alcohols

(b)

A sample of $\mathbf{W}$ , $\text{C}_4\text{H}_{10}\text{O}$ , is heated under reflux with an excess of acidified $\text{K}_2\text{Cr}_2\text{O}_7$ until there is no further reaction. Only one organic product, $\mathbf{X}$ , is present in the mixture at the end of the reaction.

Fig. 5.1 shows the infrared spectrum of $\mathbf{W}$ .

Fig. 5.1

Fig. 5.2 shows the infrared spectrum of $\mathbf{X}$ .

Fig. 5.2

Table 5.2

bond	functional groups containing the bond	characteristic infrared absorption range (in wavenumbers) / $\text{cm}^{-1}$
$\text{C–O}$	hydroxy, ester	1040–1300
$\text{C=C}$	aromatic compound, alkene	1500–1680
$\text{C=O}$	amide carbonyl, carboxyl ester	1640–1690 1670–1740 1710–1750
$\text{C}\equiv\text{N}$	nitrile	2200–2250
$\text{C–H}$	alkane	2850–2950
$\text{N–H}$	amine, amide	3300–3500
$\text{O–H}$	carboxyl hydroxy	2500–3000 3200–3650

2M

(i)

Absorption A is shown in Fig. 5.1.
Absorption B is shown in Fig. 5.2.

Complete Table 5.3 using the information given in Fig. 5.1, Fig. 5.2 and Table 5.2.

Table 5.3

absorption	bond	functional group containing the bond
A
B

1M Similar

DifficultyEasy

Worked solution

Answer

absorption	bond	functional group containing the bond
A	$\text{O–H}$	hydroxy
B	$\text{C=O}$	carbonyl

Final answer

A: O-H bond, hydroxy group; B: C=O bond, carbonyl group

Detailed explanation

Background Concept

Infrared spectroscopy identifies functional groups by the characteristic wavenumber at which bonds absorb IR radiation. Each bond type vibrates at a specific frequency range. The O–H stretch in alcohols appears as a broad band at 3200–3650 $\text{cm}^{-1}$ (broad due to hydrogen bonding), while the C=O stretch of a carbonyl group appears as a strong, sharp band at 1670–1740 $\text{cm}^{-1}$ .

Understanding the Question

Absorption A in Fig. 5.1 is a broad band centred around 3350 $\text{cm}^{-1}$ . Absorption B in Fig. 5.2 is a strong sharp band around 1715 $\text{cm}^{-1}$ . Table 5.2 provides the reference ranges. The task is to match each absorption to the correct bond and functional group.

Approach

Compare the wavenumber of each labelled absorption to the ranges in Table 5.2.

Step-by-Step Reasoning

Absorption A (~3350 $\text{cm}^{-1}$ , broad):

Falls within the 3200–3650 $\text{cm}^{-1}$ range for O–H in hydroxy compounds.
The broadness confirms it is an O–H stretch involved in hydrogen bonding (alcohol).
Bond: $\text{O–H}$ ; Functional group: hydroxy.

Absorption B (~1715 $\text{cm}^{-1}$ , sharp and strong):

Falls within the 1670–1740 $\text{cm}^{-1}$ range for C=O in carbonyl compounds.
Bond: $\text{C=O}$ ; Functional group: carbonyl (which includes aldehydes and ketones).

Key Takeaways

Broad O–H absorption at 3200–3650 $\text{cm}^{-1}$ is diagnostic of alcohols (and carboxylic acids, but those also show C=O and a much broader O–H at 2500–3000 $\text{cm}^{-1}$ ).
Sharp C=O absorption at 1670–1740 $\text{cm}^{-1}$ indicates a carbonyl group (aldehyde, ketone, carboxylic acid, or ester).

Common Mistakes

Writing "alcohol" instead of "hydroxy" for the functional group (the table asks for the functional group containing the bond, not the compound class).
For absorption B, writing "ketone" instead of "carbonyl" — the table lists "carbonyl" as the functional group category.

Things to Be Careful About

Use the exact terminology from Table 5.2 for the functional group column.
Distinguish between the bond (e.g., O–H) and the functional group (e.g., hydroxy).

Techniques used

identify bond and functional group from characteristic IR absorption rangematch absorption wavenumber to data table

(ii)

Use the information in (a) and (b)(i) to draw the structure of $\mathbf{X}$ in the box in Fig. 5.3.

Fig. 5.3

1M Similar

DifficultyMedium

Worked solution

Answer

\text{CH}_3\text{COCH}_2\text{CH}_3

Final answer

CH3COCH2CH3 (butanone)

Detailed explanation

Background Concept

Secondary alcohols are oxidised by acidified potassium dichromate(VI) under reflux to ketones. The ketone retains the same number of carbon atoms as the parent alcohol. Tertiary alcohols are not oxidised by acidified dichromate. The IR spectrum of a ketone shows a strong C=O absorption but no O–H absorption.

Understanding the Question

W has formula $\text{C}_4\text{H}_{10}\text{O}$ and shows an O–H absorption (from part b(i)), so W is an alcohol. When W is heated under reflux with excess acidified $\text{K}_2\text{Cr}_2\text{O}_7$ , it gives a single organic product X. The IR spectrum of X shows a C=O absorption but no O–H absorption, so X is a ketone (not a carboxylic acid, which would show both C=O and a very broad O–H). A ketone from oxidation of a $\text{C}_4$ alcohol means W is a secondary alcohol, specifically butan-2-ol.

Approach

From the IR of W: it is an alcohol (O–H present, C=O absent).
From the IR of X: it has C=O but no O–H, so it is a ketone (not a carboxylic acid).
Oxidation of a $\text{C}_4$ alcohol to a ketone means the alcohol is butan-2-ol.
The ketone product is butanone: $\text{CH}_3\text{COCH}_2\text{CH}_3$ .

Step-by-Step Reasoning

W is $\text{C}_4\text{H}_{10}\text{O}$ with an O–H group → an alcohol.
X shows C=O at ~1715 $\text{cm}^{-1}$ but no broad O–H → X is a ketone (a carboxylic acid would show a very broad O–H at 2500–3000 $\text{cm}^{-1}$ ).
Oxidation of a primary alcohol gives an aldehyde then a carboxylic acid; oxidation of a secondary alcohol gives a ketone.
Since X is a ketone, W must be a secondary alcohol: butan-2-ol, $\text{CH}_3\text{CH(OH)CH}_2\text{CH}_3$ .
The oxidation product X is butanone: $\text{CH}_3\text{COCH}_2\text{CH}_3$ .
Cross-check with mass spectrum: loss of $\text{CH}_3\cdot$ from butan-2-ol gives $\text{C}_3\text{H}_7\text{O}^+$ at $m/e = 59$ ✓; $\text{C}_2\text{H}_5^+$ at $m/e = 29$ ✓.

Key Takeaways

The distinction between ketone and carboxylic acid in IR: carboxylic acids show both C=O AND a very broad O–H (2500–3000 $\text{cm}^{-1}$ ); ketones show only C=O.
Reflux with excess oxidising agent ensures complete oxidation; a primary alcohol would give a carboxylic acid under these conditions.
The number of carbon atoms is conserved in alcohol → ketone oxidation.

Common Mistakes

Drawing butanal or butanoic acid instead of butanone (these would come from a primary alcohol and would show different IR).
Drawing butan-2-ol as X instead of the oxidation product.
Forgetting to draw the full displayed or structural formula clearly.

Things to Be Careful About

The question asks for the structure of X (the product), not W.
Ensure the ketone has 4 carbons total: $\text{CH}_3\text{COCH}_2\text{CH}_3$ .

Techniques used

deduce structure from molecular formula and spectroscopic dataidentify secondary alcohol from oxidation product being a ketone

(c)

$\mathbf{Y}$ is a structural isomer of $\mathbf{W}$ .

Both $\mathbf{W}$ and $\mathbf{Y}$ produce colourless bubbles when sodium is added to them.

$\mathbf{Y}$ does not react when heated with acidified $\text{K}_2\text{Cr}_2\text{O}_7$ .

$\mathbf{Y}$ does not react when warmed with alkaline $\text{I}_2\text{(aq)}$ .

3M

(i)

Name the functional group present in $\mathbf{Y}$ .

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

Hydroxy (tertiary alcohol)

Final answer

Hydroxy (tertiary alcohol)

Detailed explanation

Background Concept

Alcohols react with sodium metal to produce hydrogen gas (colourless bubbles) and a sodium alkoxide. Primary and secondary alcohols are oxidised by acidified $\text{K}_2\text{Cr}_2\text{O}_7$ (orange to green), but tertiary alcohols are not, because the carbon bearing the OH group has no hydrogen atom available for removal. The iodoform test (alkaline $\text{I}_2$ ) is positive for methyl ketones and for alcohols containing the $\text{CH}_3\text{CH(OH)}-$ group (i.e., ethanol and secondary alcohols with a methyl on the carbinol carbon).

Understanding the Question

Y is a structural isomer of W ( $\text{C}_4\text{H}_{10}\text{O}$ ). Three tests are described:

Colourless bubbles with sodium → Y is an alcohol (has O–H).
No reaction with acidified $\text{K}_2\text{Cr}_2\text{O}_7$ → Y is a tertiary alcohol.
No reaction with alkaline $\text{I}_2$ → confirms no $\text{CH}_3\text{CH(OH)}-$ group (consistent with tertiary).

The question asks to name the functional group.

Approach

Since Y reacts with sodium, it must contain an O–H group → hydroxy functional group. The additional information confirms it is specifically a tertiary alcohol.

Step-by-Step Reasoning

Reaction with Na producing $\text{H}_2$ gas → Y is an alcohol → functional group is hydroxy.
No oxidation by acidified dichromate → tertiary alcohol (the C–OH carbon has no H).
No iodoform reaction → no $\text{CH}_3\text{CH(OH)}-$ unit, consistent with tertiary.
The functional group name is "hydroxy" (or equivalently, Y is a tertiary alcohol).

Key Takeaways

The hydroxy (–OH) group is the functional group in all alcohols.
Tertiary alcohols resist oxidation because there is no C–H bond on the carbinol carbon for the oxidising agent to remove.
The iodoform test distinguishes methyl carbinols from other alcohols.

Common Mistakes

Writing "alcohol" as the functional group name instead of "hydroxy".
Saying "hydroxyl" (which refers to the radical $\cdot\text{OH}$ ) instead of "hydroxy".

Things to Be Careful About

The mark scheme accepts "hydroxy (tertiary alcohol)" — naming the functional group as hydroxy and specifying it is a tertiary alcohol.

Techniques used

deduce tertiary alcohol from resistance to oxidation and negative iodoform testidentify functional group from chemical tests

(ii)

Complete the equation to describe the reaction of $\mathbf{W}$ or $\mathbf{Y}$ with sodium.

\text{..........C}_4\text{H}_{10}\text{O} + \text{..........Na} \rightarrow \text{........................................................................................}

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

2\text{C}_4\text{H}_{10}\text{O} + 2\text{Na} \rightarrow 2\text{C}_4\text{H}_9\text{ONa} + \text{H}_2

Final answer

2C4H10O + 2Na -> 2C4H9ONa + H2

Detailed explanation

Background Concept

Alcohols react with sodium metal in a similar way to water: the O–H bond is cleaved, the hydrogen is released as $\text{H}_2$ gas, and a sodium alkoxide is formed. The general equation is $2\text{ROH} + 2\text{Na} \rightarrow 2\text{RONa} + \text{H}_2$ . This is a redox reaction where sodium is oxidised and the hydrogen in the O–H bond is reduced.

Understanding the Question

The equation must be completed with coefficients and the organic product. Both W and Y react with sodium, so the equation applies to either (both are $\text{C}_4\text{H}_{10}\text{O}$ alcohols). The product is the sodium alkoxide $\text{C}_4\text{H}_9\text{ONa}$ .

Approach

Write the general alcohol + sodium reaction, substitute the molecular formula, and balance.

Step-by-Step Reasoning

The reaction is: alcohol + sodium → sodium alkoxide + hydrogen.
Unbalanced: $\text{C}_4\text{H}_{10}\text{O} + \text{Na} \rightarrow \text{C}_4\text{H}_9\text{ONa} + \text{H}_2$
Balancing: 2 H atoms are needed for $\text{H}_2$ , so 2 alcohol molecules provide them: $2\text{C}_4\text{H}_{10}\text{O} + 2\text{Na} \rightarrow 2\text{C}_4\text{H}_9\text{ONa} + \text{H}_2$
Check: Left: C = 8, H = 20, O = 2, Na = 2. Right: C = 8, H = 18 + 2 = 20, O = 2, Na = 2. ✓

Key Takeaways

The sodium alkoxide is formed by replacing the H of the O–H group with Na.
The stoichiometry is 2:2:2:1 (alcohol : Na : alkoxide : hydrogen).

Common Mistakes

Writing $\text{C}_4\text{H}_{10}\text{ONa}$ (forgetting that one H is lost from the alcohol).
Not balancing the equation (writing 1:1:1:1).
Writing the product as $\text{NaOH}$ instead of the alkoxide.

Things to Be Careful About

The organic product formula is $\text{C}_4\text{H}_9\text{ONa}$ (or equivalently $\text{C}_4\text{H}_9\text{O}^\ominus\text{Na}^+$ ).
State symbols are not required here but the equation must be balanced.

Techniques used

balance the equation for alcohol reacting with sodiumwrite the sodium alkoxide product

(iii)

Draw the structure of $\mathbf{Y}$ .

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

(\text{CH}_3)_3\text{C–OH}

Final answer

(CH3)3C-OH (2-methylpropan-2-ol)

Detailed explanation

Background Concept

For molecular formula $\text{C}_4\text{H}_{10}\text{O}$ , the possible alcohol isomers are:

Butan-1-ol (primary): $\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{OH}$
Butan-2-ol (secondary): $\text{CH}_3\text{CH(OH)CH}_2\text{CH}_3$
2-Methylpropan-1-ol (primary): $(\text{CH}_3)_2\text{CHCH}_2\text{OH}$
2-Methylpropan-2-ol (tertiary): $(\text{CH}_3)_3\text{COH}$

Only the tertiary alcohol resists oxidation by acidified dichromate and gives a negative iodoform test.

Understanding the Question

Y is a structural isomer of W. From part (c)(i), Y is a tertiary alcohol. The only tertiary alcohol with formula $\text{C}_4\text{H}_{10}\text{O}$ is 2-methylpropan-2-ol (tert-butanol).

Approach

Identify the unique tertiary alcohol isomer of $\text{C}_4\text{H}_{10}\text{O}$ and draw its structure.

Step-by-Step Reasoning

Y reacts with Na → alcohol (has O–H).
Y does not react with acidified $\text{K}_2\text{Cr}_2\text{O}_7$ → tertiary alcohol (no C–H on the carbinol carbon).
Y does not give iodoform → no $\text{CH}_3\text{CH(OH)}-$ group (consistent with tertiary).
The only $\text{C}_4$ tertiary alcohol is 2-methylpropan-2-ol: $(\text{CH}_3)_3\text{COH}$ .
Structure: a central carbon bonded to three methyl groups and one OH group.

Key Takeaways

There is only one tertiary alcohol with 4 carbon atoms.
Tertiary alcohols cannot be oxidised because the carbinol carbon has no hydrogen to lose.
The iodoform test is negative for all tertiary alcohols (no $\text{CH}_3\text{CH(OH)}-$ unit).

Common Mistakes

Drawing butan-2-ol (secondary) instead of the tertiary isomer.
Drawing 2-methylpropan-1-ol (primary) — this would react with dichromate.
Forgetting the OH group or placing it on the wrong carbon.

Things to Be Careful About

Draw the full structure showing all bonds (the central C bonded to three CH₃ groups and one OH).
The IUPAC name is 2-methylpropan-2-ol; the common name is tert-butanol.

Techniques used

deduce the only tertiary alcohol with formula C4H10Odraw the structure of 2-methylpropan-2-ol

Chemistry 9701/22 — May/June 2024

Answer

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Understanding the Question

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