The Group 17 elements are oxidising agents. (i) Explain how the Group 17 elements act as oxidising agents. (ii) Write an equation to show the reaction in which $\text{Cl}_2$ oxidises aluminium metal. (iii) A student heats equal amounts of $\text{I}_2\text{(g)}$ and $\text{H}_2\text{(g)}$ in a sealed flask. The student leaves the contents to cool. State what you would observe during the reaction. (b) $\text{Cl}_2$ and $\text{Br}_2$ can each react with $\text{NH}_3$ to give…

(a)(i) Halogens oxidise by removing electrons from the reactant (they are themselves reduced). · (a)(ii) 2Al + 3Cl2 -> 2AlCl3 · (a)(iii) The purple colour (of iodine vapour) disappears. · (b)(i) Br–Br is weaker than Cl–Cl because Br atoms are larger, giving less effective orbital overlap. · (b)(ii) HCl is more thermally stable than HBr. · (b)(iii) Enthalpy change when one mole of a compound is formed from its elements in their standard states. · (b)(iv) -460 kJ mol^-1 ·…

Propan-2-ol, $(\text{CH}_3)_2\text{CHOH}$, is sometimes added to fuel to help it burn. Fig. 3.1 shows some reactions of propan-2-ol. (i) Draw the structure of organic compound $\text{K}$. (ii) State an observation you would make in reaction 2. (iii) State the type of reaction that is shown in reaction 3. (iv) Complete Fig. 3.2 to show the pi ($\pi$) bond in $\text{L}$ that is formed from orbital overlap. (b) Propan-2-ol reacts with sodium to produce…

(a)(i) Propanone (CH3COCH3); skeletal structure: central C=O with two methyl groups · (a)(ii) Yellow precipitate · (a)(iii) Elimination · (a)(iv) See diagram for pi bond lobes above and below C-C axis · (b)(i) (CH3)2CHOH + Na -> (CH3)2CHONa + 1/2 H2 · (b)(ii) See mechanism diagram: C-Br dipole + arrow to Br -> carbocation + arrow from O lone pair to C+ · (b)(iii) Rate decreases; C-Cl bond is stronger/higher bond enthalpy than C-Br · (b)(iv) C6H14O + 9O2 -> 6CO2 + 7H2O

Compounds $\text{P}$ and $\text{Q}$ are structural isomers. (i) Define structural isomerism. (ii) $\text{P}$ shows geometrical isomerism. Draw the geometrical isomer of $\text{P}$. Explain why the two isomers are not identical. explanation (iii) Both $\text{P}$ and $\text{Q}$ react with aqueous bromine. Name the mechanism of this reaction. (iv) $\text{Q}$ is oxidised by hot concentrated acidified $\text{KMnO}_4\text{(aq)}$, forming two different organic products. Construct…

(a)(i) Same molecular formula but different structural formulae (different arrangement of atoms). · (a)(ii) Trans-pent-2-ene (E-pent-2-ene); restricted rotation about the C=C double bond. · (a)(iii) Electrophilic addition · (a)(iv) (CH3)2C=CHCH3 + 3[O] → CH3COCH3 + CH3COOH · (a)(v) S forms via the more stable tertiary carbocation; greater positive inductive effect of more alkyl groups stabilises the C+ intermediate. · (a)(vi) Molecular ions of R: [CH3CH(CH3)CH(79Br)CH3]+ at…

Chemistry 9701/24 — October/November 2025

Cambridge AS Level · AS Level Structured Questions · worked solutions for every part, with the mark scheme

questions

marks

minutes

Topics Chemical Bonding · States of Matter · Chemical Periodicity · Hydroxy Compounds · Halogen Compounds · Group 17 · +9 more

Q1Medium-EasyGroup 17Chemical BondingChemical EnergeticsStates of MatterChemical Periodicity

The Group 17 elements are oxidising agents.

(a)

(i)

Explain how the Group 17 elements act as oxidising agents.

1M Similar

(ii)

Write an equation to show the reaction in which $\text{Cl}_2$ oxidises aluminium metal.

1M Similar

(iii)

A student heats equal amounts of $\text{I}_2\text{(g)}$ and $\text{H}_2\text{(g)}$ in a sealed flask. The student leaves the contents to cool.

State what you would observe during the reaction.

1M Similar

(b)

$\text{Cl}_2$ and $\text{Br}_2$ can each react with $\text{NH}_3$ to give $\text{N}_2$ and a hydrogen halide, $\text{HX}$ .

3\text{X}_2 + 2\text{NH}_3 \rightarrow \text{N}_2 + 6\text{HX} \quad \text{X} = \text{Cl or Br}

The relative bond strengths of $\text{X–X}$ and $\text{H–X}$ determine the difference in enthalpy change of the two reactions.

(i)

Describe and explain the difference in the $\text{X–X}$ bond strengths of $\text{Cl}_2$ and $\text{Br}_2$ .

2M Similar

(ii)

Describe the relative thermal stabilities of $\text{HCl}$ and $\text{HBr}$ .

1M Similar

(iii)

Define enthalpy change of formation, $\Delta H_f$ .

2M Similar

(iv)

Table 1.1 gives data relevant to the reaction of $\text{Cl}_2\text{(g)}$ with $\text{NH}_3\text{(g)}$ .

Table 1.1

compound	enthalpy change of formation, $\Delta H_f / \text{kJ mol}^{-1}$
$\text{NH}_3\text{(g)}$	$-46$
$\text{HCl(g)}$	$-92$

Use the data in Table 1.1 to calculate the enthalpy change of the reaction of $\text{Cl}_2\text{(g)}$ with $\text{NH}_3\text{(g)}$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Working

For $3\text{Cl}_2 + 2\text{NH}_3 \rightarrow \text{N}_2 + 6\text{HCl}$ :

\Delta H = \sum \Delta H_f(\text{products}) - \sum \Delta H_f(\text{reactants})

\Delta H = (6 \times -92) - (2 \times -46)

\Delta H = -552 + 92 = -460 \text{ kJ mol}^{-1}

Answer

$\Delta H = -460 \text{ kJ mol}^{-1}$ (exothermic)

Final answer

-460 kJ mol^-1

Detailed explanation

Background Concept

ΔH_f of an element in its standard state is zero, so Cl₂(g) and N₂(g) contribute nothing. The reaction enthalpy from formation enthalpies is ΔH = ΣΔH_f(products) − ΣΔH_f(reactants), with each value multiplied by the stoichiometric coefficient.

Understanding the Question

Use the tabulated ΔH_f values for NH₃ (−46) and HCl (−92) with the balanced equation 3Cl₂ + 2NH₃ → N₂ + 6HCl to find ΔH for the reaction as written.

Approach

Multiply each ΔH_f by the number of moles in the equation, sum products, subtract reactants.

Step-by-Step Reasoning

Products: 6 mol HCl ⇒ 6 × (−92) = −552 kJ. Reactants: 2 mol NH₃ ⇒ 2 × (−46) = −92 kJ; Cl₂ and N₂ are elements, ΔH_f = 0. ΔH = −552 − (−92) = −552 + 92 = −460 kJ mol⁻¹. The negative sign shows the reaction is strongly exothermic.

Key Takeaways

Always multiply by coefficients and remember elements have ΔH_f = 0.

Common Mistakes

Forgetting to multiply HCl by 6 (giving −46 kJ). Subtracting wrongly: (−552) − (+92) = −644 is a sign error. Omitting the negative sign on the final answer.

Things to Be Careful About

The unit is kJ mol⁻¹ for the equation as written; keep the minus sign — the reaction is exothermic.

Techniques used

apply Hess's law using enthalpies of formationcalculate reaction enthalpy from ΔHf data

(v)

$\text{I}_2$ reacts with $\text{NH}_3$ to form $\text{NI}_3$ .

Predict the shape of a molecule of $\text{NI}_3$ . Explain your answer.

shape

explanation

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Shape: trigonal pyramidal

Explanation: nitrogen has three bonding pairs (N–I bonds) and one lone pair of electrons; the four electron pairs repel, and the lone pair repels more strongly, pushing the three bonding pairs into a pyramidal arrangement (bond angle about 107°).

Final answer

Trigonal pyramidal, due to three bonding pairs and one lone pair on N.

Detailed explanation

Background Concept

VSEPR theory: electron pairs around a central atom repel and arrange to minimise repulsion. Lone pair–lone pair > lone pair–bonding pair > bonding pair–bonding pair repulsion. Four electron pairs with one lone pair give a trigonal pyramidal shape (like NH₃).

Understanding the Question

NI₃ has N central with three I atoms attached. Predict the shape and explain it via electron-pair counting.

Approach

Count electron pairs on N: 3 bonding pairs + 1 lone pair = 4 pairs ⇒ tetrahedral electron arrangement, but with one position occupied by the lone pair the molecular shape is trigonal pyramidal.

Step-by-Step Reasoning

N has 5 outer electrons: three used in N–I bonds, one pair left as a lone pair. Four electron pairs arrange tetrahedrally; the lone pair occupies one position and repels the bonding pairs slightly more, so the three N–I bonds form a trigonal pyramid with bond angle slightly less than 109.5° (≈107°). M1: trigonal pyramidal. M2: three bonding pairs AND one lone pair.

Key Takeaways

Shape depends on the number of bonding and lone pairs; always count both.

Common Mistakes

Saying 'tetrahedral' (that is the electron-pair arrangement, not the molecular shape). Forgetting the lone pair in the explanation — the explanation mark requires both the three bonding pairs and the lone pair.

Things to Be Careful About

Name the shape precisely: 'trigonal pyramidal', not 'pyramidal' alone ideally, and justify with both pair types.

Techniques used

apply VSEPR to count bonding and lone pairsdeduce molecular shape

(c)

Table 1.2 shows some information about reactions of $\text{NaCl}$ , $\text{NaBr}$ and $\text{NaI}$ .

Table 1.2

	$\text{NaCl}$	$\text{NaBr}$	$\text{NaI}$
observation with $\text{Ag}^\text{+}\text{(aq)}$	white precipitate
type of reaction with concentrated $\text{H}_2\text{SO}_4$	acid–base	acid–base, then redox	acid–base, then redox
observations with concentrated $\text{H}_2\text{SO}_4$			• black solid • yellow solid • effervescence

(i)

Complete Table 1.2.

4M Similar

DifficultyMedium-Easy

Worked solution

Answer

Completed Table 1.2:

	NaCl	NaBr	NaI
with Ag⁺(aq)	white precipitate	cream precipitate	yellow precipitate
type of reaction with conc. H₂SO₄	acid–base	acid–base, then redox	acid–base, then redox
observations with conc. H₂SO₄	steamy fumes	brown liquid (or fizzing)	• black solid • yellow solid • effervescence

Final answer

NaBr: cream ppt with Ag+; NaI: yellow ppt with Ag+; NaCl with conc. H2SO4: steamy fumes; NaBr with conc. H2SO4: brown liquid (or fizzing).

Detailed explanation

Background Concept

Halide tests: Ag⁺(aq) + X⁻(aq) → AgX(s). AgCl is white, AgBr is cream, AgI is yellow. With concentrated H₂SO₄, halide ions are first protonated (acid–base: HX + steamy fumes), but Br⁻ and I⁻ are strong enough reducing agents to further reduce H₂SO₄ (redox), giving characteristic products: Br⁻ gives brown Br₂ vapour/liquid; I⁻ gives black I₂ solid, yellow sulfur, and (with deeper reduction) H₂S gas causing effervescence.

Understanding the Question

Complete the four empty cells: two precipitate colours and two sets of observations with concentrated sulfuric acid.

Approach

Fill in the standard halide-test results: AgBr cream, AgI yellow; NaCl + conc. H₂SO₄ gives only steamy fumes of HCl (chloride too weak a reducer for redox); NaBr gives brown bromine (or fizzing) after the initial steamy fumes.

Step-by-Step Reasoning

M1: NaBr + Ag⁺ → cream/off-white precipitate of AgBr. M2: NaI + Ag⁺ → yellow precipitate of AgI. M3: NaCl with conc. H₂SO₄: steamy fumes of HCl (acid–base only, no redox). M4: NaBr with conc. H₂SO₄: brown liquid/vapour of Br₂ (or fizzing), showing acid–base then redox.

Key Takeaways

Learn the AgX colour sequence white → cream → yellow down the group, and that reducing power of halide ions increases down the group, so only Br⁻ and I⁻ show redox with H₂SO₄.

Common Mistakes

Calling AgBr 'yellow' and AgI 'white' — order matters. Writing 'brown fumes' for NaCl (chloride does not produce bromine). Missing that NaCl shows no redox stage.

Things to Be Careful About

For NaBr the accepted observations are 'brown liquid' OR 'fizzing'; for NaCl 'steamy fumes' OR 'fizzing'. Use the precise colour words: cream, yellow, brown, black.

Techniques used

recall silver halide precipitate coloursrecall observations of halides with concentrated sulfuric acid

(ii)

Suggest an identity for the species that produces each observation in the reaction of $\text{NaI}$ with concentrated $\text{H}_2\text{SO}_4$ .

black solid

yellow solid

effervescence

2M Similar

(d)

Table 1.3 gives some information about $\text{MgCl}_2$ and $\text{SiCl}_4$ .

Table 1.3

	$\text{MgCl}_2$	$\text{SiCl}_4$
electrical conductivity when liquid	conducts	does not conduct
observation when added to water	dissolves	vigorous reaction

(i)

Explain the difference between the electrical conductivity of liquid $\text{MgCl}_2$ and of liquid $\text{SiCl}_4$ . Refer to bonding and relevant particles in your answer.

2M Similar

(ii)

Suggest the pH of the solutions that form when each chloride is added to water.

$\text{MgCl}_2$

$\text{SiCl}_4$

2M Similar

Q2MediumStates of MatterEquilibriaReaction KineticsHydroxy CompoundsChemical Periodicity

Aluminium oxide, $\text{Al}_2\text{O}_3$ , and phosphorus(V) oxide, $\text{P}_4\text{O}_{10}$ , are both used as reagents and catalysts.

(a)

The melting point of $\text{Al}_2\text{O}_3$ is 2072°C. The melting point of $\text{P}_4\text{O}_{10}$ is 340°C.

Explain the difference in the melting points of these two compounds.

3M Similar

DifficultyMedium-Easy

Worked solution

Answer

$\text{Al}_2\text{O}_3$ has a giant ionic lattice; melting it requires breaking strong ionic bonds.
$\text{P}_4\text{O}_{10}$ is a simple molecular substance; melting it only requires overcoming weak intermolecular (van der Waals') forces between molecules.
Ionic bonds are much stronger than intermolecular forces, so $\text{Al}_2\text{O}_3$ has the much higher melting point.

Final answer

Al2O3 is ionic with strong ionic bonds; P4O10 is simple molecular with weak intermolecular forces, so Al2O3 has a much higher melting point.

Detailed explanation

Background Concept

Melting point is the temperature at which the forces holding a substance together are overcome. In a giant ionic lattice, oppositely charged ions are held by strong electrostatic attractions; these ionic bonds extend throughout the whole crystal. In a simple molecular solid, atoms within each molecule are joined by strong covalent bonds, but the molecules attract each other only through weak intermolecular forces, usually van der Waals' forces.

Understanding the Question

This part asks you to explain why $\text{Al}_2\text{O}_3$ melts at 2072°C but $\text{P}_4\text{O}_{10}$ melts at only 340°C. The command word 'explain' means you must link the observed property to the type of bonding and structure of each compound.

Approach

First identify the structure of each compound. $\text{Al}_2\text{O}_3$ is a metal oxide with a giant ionic lattice. $\text{P}_4\text{O}_{10}$ is a covalent oxide that exists as simple molecules. Then compare the strength of the forces that must be broken when the solid melts.

Step-by-Step Reasoning

$\text{Al}_2\text{O}_3$ is ionic: it consists of $\text{Al}^{3+}$ and $\text{O}^{2-}$ ions in a giant lattice. Melting requires breaking the strong ionic bonds between these ions, which needs a very high temperature.
$\text{P}_4\text{O}_{10}$ is a simple molecular substance: the atoms are joined by covalent bonds within each molecule, but molecules are held together only by weak intermolecular forces. Melting only needs to overcome these weak forces, not break covalent bonds.
Because ionic bonds are much stronger than intermolecular forces, far more energy is needed to melt $\text{Al}_2\text{O}_3$ , so its melting point is much higher.

Key Takeaways

Physical properties such as melting point are determined by the type of structure and the strength of the forces between particles. A giant ionic lattice has a high melting point; a simple molecular solid has a low melting point.

Common Mistakes

Saying that $\text{P}_4\text{O}_{10}$ has covalent bonds broken when it melts. Covalent bonds stay intact; only intermolecular forces are overcome.
Writing 'strong bonds' without saying which bonds, or confusing ionic bonds with intermolecular forces.
Describing $\text{Al}_2\text{O}_3$ as molecular.

Things to Be Careful About

Use the precise terms: 'giant ionic lattice', 'simple molecular', 'ionic bonds', 'intermolecular forces'. The mark scheme rewards the comparison that ionic bonds are stronger than intermolecular forces.

Techniques used

identify the type of bonding and structurecompare strength of ionic bonds with intermolecular forcesrelate structure to melting point

(b)

A $5.00 \text{ dm}^3$ sealed flask contains $0.400 \text{ mol}$ of $\text{CO(g)}$ and $0.800 \text{ mol}$ of $\text{H}_2\text{(g)}$ and an $\text{Al}_2\text{O}_3$ catalyst. The flask is heated to a temperature of 290°C and allowed to reach equilibrium. Equation 1 shows the reaction.

equation 1: $\text{CO(g)} + 2\text{H}_2\text{(g)} \rightleftharpoons \text{CH}_3\text{OH(g)}$

The equilibrium constant, $K_c$ , of equation 1 is given.

K_c = \frac{[\text{CH}_3\text{OH}]}{[\text{CO}][\text{H}_2]^2}

(i)

State the units of $K_c$ .

1M Similar

DifficultyEasy

Worked solution

Answer

$\text{mol}^{-2}\,\text{dm}^6$

Final answer

mol^-2 dm^6

Detailed explanation

Background Concept

For a reaction $a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}$ , the equilibrium constant is $K_c = \frac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b}$ . Concentrations are measured in $\text{mol dm}^{-3}$ , so the units of $K_c$ depend on the powers in the expression.

Understanding the Question

You are given the expression for $K_c$ for $\text{CO} + 2\text{H}_2 \rightleftharpoons \text{CH}_3\text{OH}$ and asked only for the units. No calculation is needed.

Approach

Replace each concentration term in the expression by its unit $\text{mol dm}^{-3}$ , then simplify the powers of mol and dm.

Step-by-Step Reasoning

$K_c = \frac{[\text{CH}_3\text{OH}]}{[\text{CO}][\text{H}_2]^2}$
Substituting units:
$\frac{\text{mol dm}^{-3}}{(\text{mol dm}^{-3})(\text{mol dm}^{-3})^2} = \frac{\text{mol dm}^{-3}}{\text{mol}^3\text{dm}^{-9}} = \text{mol}^{-2}\text{dm}^6$
So the units are $\text{mol}^{-2}\,\text{dm}^6$ .

Key Takeaways

The units of $K_c$ are not fixed; they are found from the equilibrium expression. The denominator here has three concentration factors, giving $\text{mol}^{-2}\text{dm}^6$ .

Common Mistakes

Forgetting the square on $[\text{H}_2]$ , which would give $\text{mol}^{-1}\text{dm}^3$ .
Inverting the units and writing $\text{mol}^2\text{dm}^{-6}$ .
Writing only $\text{dm}^6$ without the mol power.

Things to Be Careful About

Treat the units algebraically: subtract powers in the denominator from those in the numerator. Include both mol and dm in the final unit.

Techniques used

derive units from equilibrium expressioncombine concentration units

(ii)

The equilibrium mixture contains $0.280 \text{ mol}$ of $\text{CH}_3\text{OH(g)}$ .

Calculate the value of $K_c$ .

Give your answer to three significant figures.

3M Similar

DifficultyMedium

Worked solution

Working

Moles at equilibrium:
$\text{CO} = 0.400 - 0.280 = 0.120 \text{ mol}$
$\text{H}_2 = 0.800 - 2(0.280) = 0.240 \text{ mol}$
Concentrations in $5.00 \text{ dm}^3$ :
$[\text{CH}_3\text{OH}] = \frac{0.280}{5.00} = 0.0560 \text{ mol dm}^{-3}$
$[\text{CO}] = \frac{0.120}{5.00} = 0.0240 \text{ mol dm}^{-3}$
$[\text{H}_2] = \frac{0.240}{5.00} = 0.0480 \text{ mol dm}^{-3}$
$K_c = \frac{0.0560}{0.0240 \times (0.0480)^2} = 1.01 \times 10^3 \text{ mol}^{-2}\,\text{dm}^6$

Answer

$K_c = 1.01 \times 10^3 \text{ mol}^{-2}\,\text{dm}^6$

Final answer

1.01 x 10^3 mol^-2 dm^6

Detailed explanation

Background Concept

At equilibrium, the concentrations in the $K_c$ expression are equilibrium concentrations, not initial amounts. The stoichiometry of the reaction tells you how much of each reactant is consumed when product forms.

Understanding the Question

A $5.00\text{ dm}^3$ flask initially contains $0.400\text{ mol CO}$ and $0.800\text{ mol H}_2$ . At equilibrium $0.280\text{ mol CH}_3\text{OH}$ is present. You must find equilibrium amounts of CO and H2, convert all amounts to concentrations, and substitute into $K_c$ .

Approach

Use the balanced equation: 1 mol CO + 2 mol H2 -> 1 mol CH3OH. Since 0.280 mol CH3OH formed, 0.280 mol CO and 0.560 mol H2 were consumed. Subtract from initial amounts, divide by volume to get concentrations, then substitute.

Step-by-Step Reasoning

Equilibrium moles:
$n(\text{CO}) = 0.400 - 0.280 = 0.120\text{ mol}$
$n(\text{H}_2) = 0.800 - 2(0.280) = 0.240\text{ mol}$
Concentrations:
$[\text{CH}_3\text{OH}] = \frac{0.280}{5.00} = 0.0560\text{ mol dm}^{-3}$
$[\text{CO}] = \frac{0.120}{5.00} = 0.0240\text{ mol dm}^{-3}$
$[\text{H}_2] = \frac{0.240}{5.00} = 0.0480\text{ mol dm}^{-3}$
Substitute:
$K_c = \frac{0.0560}{0.0240 \times (0.0480)^2} = \frac{0.0560}{0.000055296} = 1.01 \times 10^3\text{ mol}^{-2}\text{dm}^6$
The answer is given to three significant figures.

Key Takeaways

Equilibrium calculations require a stoichiometric table: initial amounts, change, equilibrium amounts. Always divide equilibrium amounts by the volume to get concentrations before using $K_c$ .

Common Mistakes

Using initial amounts instead of equilibrium amounts in $K_c$ .
Forgetting that 2 mol H2 are consumed per mol CH3OH.
Dividing by volume incorrectly or forgetting to convert moles to concentrations.
Giving too many or too few significant figures.

Things to Be Careful About

Keep track of units. The final value is large because the denominator is small. Check the arithmetic: $0.0240 \times 0.0480^2 = 5.5296 \times 10^{-5}$ .

Techniques used

calculate equilibrium moles from stoichiometryconvert moles to concentrations using volumesubstitute into Kc expressionexpress answer to three significant figures

(iii)

State and explain the effect, if any, on the value of $K_c$ when the overall pressure in the sealed flask is increased.

1M Similar

(c)

$\text{P}_4\text{O}_{10}$ catalyses the reversible reaction of $\text{CO}$ with $\text{H}_2$ to form $\text{CH}_3\text{OH}$ .

equation 1: $\text{CO(g)} + 2\text{H}_2\text{(g)} \rightleftharpoons \text{CH}_3\text{OH(g)}$

$\text{P}_4\text{O}_{10}$ then acts as a dehydrating agent, causing $\text{CH}_3\text{OH}$ to form $\text{CH}_3\text{OCH}_3$ .

(i)

Explain how the presence of a catalyst affects a chemical reaction.

1M Similar

(ii)

Construct an equation for the dehydration reaction of $\text{CH}_3\text{OH}$ to form $\text{CH}_3\text{OCH}_3$ .

1M Similar

(iii)

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

1M Similar

DifficultyEasy

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

$\text{P}_4\text{O}_{10}$ is the acidic oxide of phosphorus. Acidic oxides react with water to form acids. Phosphorus(V) oxide reacts with excess water to give phosphoric(V) acid, $\text{H}_3\text{PO}_4$ .

Understanding the Question

Write the equation for the reaction of $\text{P}_4\text{O}_{10}$ with an excess of water. 'Excess' ensures the product is the fully hydrated acid, not a partial product.

Approach

Identify the product as $\text{H}_3\text{PO}_4$ , then balance the equation.

Step-by-Step Reasoning

Each P atom ends up in $\text{H}_3\text{PO}_4$ , so 4 P atoms give 4 $\text{H}_3\text{PO}_4$ . Balance O: left has 10 O from $\text{P}_4\text{O}_{10}$ plus O from water; right has $4 \times 4 = 16$ O. So water must provide 6 O, requiring 6 $\text{H}_2\text{O}$ . Balance H: left $6 \times 2 = 12$ H; right $4 \times 3 = 12$ H. Equation:
$\text{P}_4\text{O}_{10} + 6\text{H}_2\text{O} \rightarrow 4\text{H}_3\text{PO}_4$

Key Takeaways

Acidic oxides react with water to form acids. Balancing an oxide-water reaction requires matching atoms of the non-metal, then oxygen and hydrogen.

Common Mistakes

Writing $\text{P}_2\text{O}_5 + 3\text{H}_2\text{O} \rightarrow 2\text{H}_3\text{PO}_4$ instead of using $\text{P}_4\text{O}_{10}$ as given.
Forgetting the coefficient 6 on water or 4 on $\text{H}_3\text{PO}_4$ .
Writing $\text{HPO}_3$ or another partial acid.

Things to Be Careful About

Use the formula given in the question, $\text{P}_4\text{O}_{10}$ . Check that atoms of P, O, and H balance.

Techniques used

write reaction of acidic oxide with waterbalance equation with state symbols

Q3Medium-HardHydroxy CompoundsChemical BondingHalogen CompoundsAtoms, Molecules and Stoichiometry

Propan-2-ol, $(\text{CH}_3)_2\text{CHOH}$ , is sometimes added to fuel to help it burn.

Fig. 3.1 shows some reactions of propan-2-ol.

Fig. 3.1

(a)

(i)

Draw the structure of organic compound $\text{K}$ .

1M Similar

DifficultyEasy

Worked solution

Answer

The structure of compound K (propanone) is:

\text{CH}_3-\text{C}(=\text{O})-\text{CH}_3

(Skeletal formula: a central carbon double-bonded to oxygen, with two methyl groups attached.)

Final answer

Propanone (CH3COCH3); skeletal structure: central C=O with two methyl groups

Detailed explanation

Background Concept

Secondary alcohols, such as propan-2-ol, contain the -OH group attached to a carbon atom bonded to two other carbon atoms. When treated with an oxidising agent like acidified potassium dichromate(VI) ( $\text{K}_2\text{Cr}_2\text{O}_7$ ), secondary alcohols are oxidised to ketones. The reaction involves the removal of two hydrogen atoms: one from the -OH group and one from the carbon atom bearing the -OH group, forming a carbon-oxygen double bond (C=O).

Understanding the Question

The question asks for the structure of organic compound K, which is formed when propan-2-ol reacts with acidified $\text{K}_2\text{Cr}_2\text{O}_7$ (Reaction 1). Since propan-2-ol is a secondary alcohol, the oxidation product is a ketone, specifically propanone (also known as acetone).

Approach

Identify the functional group transformation: secondary alcohol $\rightarrow$ ketone. Draw the structure of propanone, ensuring the carbonyl group (C=O) is correctly placed on the central carbon.

Step-by-Step Reasoning

Identify reactant: Propan-2-ol is $\text{CH}_3\text{CH(OH)CH}_3$ . The -OH is on carbon-2, a secondary carbon.
Identify reagent: Acidified $\text{K}_2\text{Cr}_2\text{O}_7$ is a strong oxidising agent.
Determine product: Oxidation of a secondary alcohol yields a ketone. For propan-2-ol, the product is propanone ( $\text{CH}_3\text{COCH}_3$ ).
Draw structure: The mark scheme accepts a skeletal formula. Draw a central carbon atom double-bonded to an oxygen atom, with two single bonds extending to represent the methyl groups ( $\text{CH}_3$ ). In skeletal form, this is a 'Y' shape with a double bond to O at the junction.

Key Takeaways

Secondary alcohols oxidise to ketones.
Primary alcohols oxidise to aldehydes (and then carboxylic acids).
Tertiary alcohols do not oxidise under these conditions.

Common Mistakes

Drawing propanal (an aldehyde) instead of propanone. This happens if the student confuses primary and secondary alcohol oxidation.
Forgetting the double bond to oxygen in the ketone structure.

Things to Be Careful About

Ensure the carbon chain length is maintained (3 carbons).
The carbonyl carbon in propanone is bonded to two other carbons, not a hydrogen.

Techniques used

identify oxidation product of secondary alcoholdraw skeletal formula of ketone

(ii)

State an observation you would make in reaction 2.

1M Similar

DifficultyEasy

Worked solution

Answer

Yellow precipitate.

Final answer

Yellow precipitate

Detailed explanation

Background Concept

The reaction of propan-2-ol with alkaline iodine ( $\text{I}_2$ /NaOH) is the triiodomethane (or iodoform) test. This test is positive for:

Ethanol.
Secondary alcohols containing a $\text{CH}_3\text{CH(OH)-}$ group (like propan-2-ol).
Aldehydes and ketones containing a $\text{CH}_3\text{CO-}$ group (methyl carbonyls).

The reaction produces triiodomethane ( $\text{CHI}_3$ ), which is a pale yellow, crystalline solid with a characteristic antiseptic smell. In the context of the reaction scheme, propan-2-ol is first oxidised to propanone ( $\text{CH}_3\text{COCH}_3$ ), which then undergoes halogenation and cleavage to form the ethanoate ion ( $\text{CH}_3\text{CO}_2^-$ ) and triiodomethane ( $\text{CHI}_3$ ).

Understanding the Question

The question asks for an observation in Reaction 2, where propan-2-ol reacts with alkaline $\text{I}_2$ (aq) to form $\text{CH}_3\text{CO}_2^-$ and $\text{CHI}_3$ . The formation of $\text{CHI}_3$ is the key observation.

Approach

Identify the product $\text{CHI}_3$ as triiodomethane (iodoform) and state its physical appearance.

Step-by-Step Reasoning

Identify reaction: Alkaline iodine test (iodoform test) on a methyl carbinol ( $\text{CH}_3\text{CH(OH)-}$ group).
Identify products: The scheme shows $\text{CH}_3\text{CO}_2^-$ and $\text{CHI}_3$ .
Observation: $\text{CHI}_3$ is a yellow precipitate. The solution may also turn from brown/orange (iodine) to colourless as iodine is consumed, but the formation of a yellow precipitate is the definitive positive test observation.

Key Takeaways

Positive iodoform test: yellow precipitate of $\text{CHI}_3$ .
Tests for: ethanol, methyl ketones, secondary alcohols with a methyl group on the carbinol carbon.

Common Mistakes

Saying 'white precipitate' (that's for halide tests with $\text{Ag}^+$ ).
Forgetting to mention 'precipitate' or 'solid'.

Things to Be Careful About

The precipitate is specifically yellow. Do not just say 'precipitate'.

Techniques used

recognise iodoform test conditionsstate observation for positive iodoform test

(iii)

State the type of reaction that is shown in reaction 3.

1M Similar

DifficultyEasy

Worked solution

Answer

Elimination (or dehydration).

Final answer

Elimination

Detailed explanation

Background Concept

Alcohols can react with concentrated sulfuric acid ( $\text{H}_2\text{SO}_4$ ) or phosphoric acid ( $\text{H}_3\text{PO}_4$ ) at high temperatures to form alkenes. This reaction involves the removal of a water molecule ( $\text{H}_2\text{O}$ ) from the alcohol molecule. Because a small molecule is removed to form a double bond, it is classified as an elimination reaction. Specifically, it is a dehydration reaction.

Understanding the Question

Reaction 3 shows propan-2-ol reacting with concentrated $\text{H}_2\text{SO}_4$ to form compound L. The image for L shows propene (an alkene). The question asks for the type of reaction.

Approach

Identify the transformation: alcohol $\rightarrow$ alkene + water. Classify this transformation.

Step-by-Step Reasoning

Reactant: Propan-2-ol ( $\text{C}_3\text{H}_8\text{O}$ ).
Reagent: Concentrated $\text{H}_2\text{SO}_4$ (catalyst and dehydrating agent).
Product L: The image shows a 3-carbon chain with a double bond, which is propene ( $\text{C}_3\text{H}_6$ ).
Process: $\text{C}_3\text{H}_8\text{O} \rightarrow \text{C}_3\text{H}_6 + \text{H}_2\text{O}$ . Water is lost.
Classification: Removal of atoms/groups to form a double bond is elimination.

Key Takeaways

Alcohol $\rightarrow$ Alkene is elimination (dehydration).
Reagents: conc. $\text{H}_2\text{SO}_4$ or conc. $\text{H}_3\text{PO}_4$ , heat ( $\sim 170^\circ\text{C}$ for $\text{H}_2\text{SO}_4$ ).

Common Mistakes

Calling it 'substitution' or 'oxidation'.
Writing 'dehydration' is correct, but 'elimination' is the broader mechanistic class often required.

Things to Be Careful About

Both 'elimination' and 'dehydration' are acceptable. 'Elimination' is the fundamental reaction type.

Techniques used

identify reaction type from reagents and productrecognise dehydration as elimination

(iv)

Complete Fig. 3.2 to show the pi ( $\pi$ ) bond in $\text{L}$ that is formed from orbital overlap.

Fig. 3.2

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

The $\pi$ bond is formed by the sideways overlap of p-orbitals above and below the plane of the sigma bond framework. In the diagram, add two shaded oval lobes (electron clouds) above and below the C-C single bond line.

Final answer

See diagram for pi bond lobes above and below C-C axis

Detailed explanation

Background Concept

A carbon-carbon double bond (C=C) consists of one sigma ( $\sigma$ ) bond and one pi ( $\pi$ ) bond.

The sigma bond is formed by head-on overlap of $sp^2$ hybridised orbitals along the internuclear axis. It is the strong bond shown in the provided skeletal structure.
The pi bond is formed by the sideways (lateral) overlap of unhybridised p-orbitals on each carbon atom. These p-orbitals are perpendicular to the plane of the sigma bond framework (the plane containing the C-C bond and the attached atoms). The electron density of the pi bond is concentrated in two lobes, one above and one below the plane of the molecule.

Understanding the Question

Compound L is propene ( $\text{CH}_3\text{CH=CH}_2$ ). Fig 3.2 shows the sigma bond framework with wedges and dashes indicating 3D geometry (trigonal planar around the double-bonded carbons). The question asks to complete the diagram to show the pi bond formed from orbital overlap.

Approach

Draw the electron density lobes representing the pi bond. These must be above and below the plane defined by the sigma bonds.

Step-by-Step Reasoning

Identify the sigma framework: The C-C bond is horizontal. The attached atoms (H, H, H3C) are in the plane (represented by wedges/dashes for 3D perspective, but effectively in a plane perpendicular to the p-orbitals).
Locate p-orbitals: Each carbon in the C=C bond has an unhybridised p-orbital perpendicular to the molecular plane.
Draw overlap: The sideways overlap creates two regions of electron density. Draw an oval/lobe above the C-C bond and an oval/lobe below the C-C bond. These should be aligned vertically with the C-C bond axis.
Shading: Typically, the lobes are shaded or filled to represent electron density. One lobe might be shaded darker or both shaded to show the continuous cloud, but standard exam convention is two lobes (top and bottom) indicating the pi system.

Key Takeaways

Double bond = 1 sigma + 1 pi.
Pi bond electron density is above and below the internuclear axis.
Pi bond prevents rotation around the C=C bond.

Common Mistakes

Drawing the pi bond in the same plane as the atoms (this would be a second sigma bond or incorrect geometry).
Forgetting that the p-orbitals are perpendicular to the plane of the paper/sigma bonds.

Things to Be Careful About

The lobes must be clearly above and below the C-C line. Do not draw them to the sides.

Techniques used

draw pi bond orbital overlap diagramidentify electron density regions in alkene

(b)

Propan-2-ol reacts with sodium to produce $(\text{CH}_3)_2\text{CH–O}^-$ anions.

These anions react with 2-bromopropane to form compound $\text{N}$ , as shown in Fig. 3.3.

Fig. 3.3

(i)

Write an equation for the reaction of propan-2-ol with sodium.

1M Similar

DifficultyEasy

Worked solution

Answer

(\text{CH}_3)_2\text{CHOH} + \text{Na} \rightarrow (\text{CH}_3)_2\text{CHO}^-\text{Na}^+ + \frac{1}{2}\text{H}_2

(Alternatively: $2(\text{CH}_3)_2\text{CHOH} + 2\text{Na} \rightarrow 2(\text{CH}_3)_2\text{CHO}^-\text{Na}^+ + \text{H}_2$ )

Final answer

(CH3)2CHOH + Na -> (CH3)2CHONa + 1/2 H2

Detailed explanation

Background Concept

Alcohols are weakly acidic and react with reactive metals (like sodium, potassium, magnesium) to produce metal alkoxides (or alkoxides) and hydrogen gas. The reaction is similar to the reaction of water with sodium, but slower. The O-H bond is broken, releasing $\text{H}^+$ which forms $\text{H}_2$ .

2\text{ROH} + 2\text{Na} \rightarrow 2\text{RO}^-\text{Na}^+ + \text{H}_2

Understanding the Question

The question asks for the equation for the reaction of propan-2-ol with sodium. The product anion is given as $(\text{CH}_3)_2\text{CH-O}^-$ .

Approach

Write the balanced chemical equation. Reactant: propan-2-ol + Na. Products: sodium propan-2-oxide + hydrogen.

Step-by-Step Reasoning

Reactants: $(\text{CH}_3)_2\text{CHOH}$ and $\text{Na}$ .
Products: The anion is $(\text{CH}_3)_2\text{CHO}^-$ . The cation is $\text{Na}^+$ . So the salt is $(\text{CH}_3)_2\text{CHONa}$ . Hydrogen gas ( $\text{H}_2$ ) is evolved.
Balancing: One mole of alcohol reacts with one mole of sodium to give half a mole of hydrogen. Or 2 moles alcohol + 2 moles Na -> 2 moles salt + 1 mole H2.
Equation: $(\text{CH}_3)_2\text{CHOH} + \text{Na} \rightarrow (\text{CH}_3)_2\text{CHONa} + \frac{1}{2}\text{H}_2$ .

Key Takeaways

Alcohols + Na -> alkoxide + H2.
Similar to water + Na -> NaOH + H2, but slower.

Common Mistakes

Forgetting the hydrogen gas product.
Writing $\text{H}$ instead of $\text{H}_2$ .
Incorrectly balancing the sodium (needing 2 Na for 1 H2 if writing integer coefficients, or 1/2 H2 for 1 Na).

Things to Be Careful About

The mark scheme accepts $(\text{CH}_3)_2\text{CHO}^-\text{Na}^+$ . Ensure charges are balanced if written as ions.

Techniques used

write equation for alcohol with metalbalance redox equation for alcohol and sodium

(ii)

The reaction of 2-bromopropane with $(\text{CH}_3)_2\text{CH–O}^-$ anions follows an $\text{S}_\text{N}1$ mechanism.

Complete Fig. 3.4 to show this mechanism. Include charges, dipoles, lone pairs of electrons and curly arrows, as appropriate.

Fig. 3.4

3M Similar

DifficultyMedium-Hard

Worked solution

Answer

Step 1: Heterolytic fission of C-Br bond.

Show $\delta^+$ on C and $\delta^-$ on Br.
Curly arrow from the center of the C-Br bond to the Br atom.
Intermediate formed: secondary carbocation $(\text{CH}_3)_2\text{CH}^+$ .

Step 2: Nucleophilic attack.

Curly arrow from a lone pair on the oxygen of $(\text{CH}_3)_2\text{CH-O}^-$ to the positively charged carbon of the carbocation.

Final answer

See mechanism diagram: C-Br dipole + arrow to Br -> carbocation + arrow from O lone pair to C+

Detailed explanation

Background Concept

The $\text{S}_\text{N}1$ (Substitution Nucleophilic Unimolecular) mechanism occurs in two steps, primarily with secondary and tertiary halogenoalkanes.

Step 1 (Slow, rate-determining): The C-X bond breaks heterolytically. The halogen takes both electrons, forming a halide ion and a carbocation intermediate. A dipole must be shown on the C-X bond ( $\delta^+$ on C, $\delta^-$ on X) before the arrow is drawn.
Step 2 (Fast): The nucleophile attacks the carbocation. A curly arrow is drawn from a lone pair on the nucleophile to the positive carbon.

Understanding the Question

The reaction is between 2-bromopropane and the propan-2-oxide anion ( $(\text{CH}_3)_2\text{CH-O}^-$ ). The mechanism is specified as $\text{S}_\text{N}1$ . We need to complete Fig 3.4 showing the mechanism.

Approach

Draw the two steps: bond breaking to form carbocation, then nucleophilic attack. Include dipoles, lone pairs, and curly arrows.

Step-by-Step Reasoning

Reactants: 2-bromopropane ( $\text{CH}_3\text{CHBrCH}_3$ ) and $(\text{CH}_3)_2\text{CH-O}^-$ (with 3 lone pairs on O, though 1 is needed for the arrow).
Step 1 - Bond breaking:
- The C-Br bond is polar. Draw $\delta^+$ on the central C and $\delta^-$ on Br.
- Draw a curly arrow starting from the middle of the C-Br bond and pointing to the Br atom (representing the pair of electrons moving to Br).
- This forms the intermediate: a secondary carbocation, $(\text{CH}_3)_2\text{CH}^+$ (isopropyl cation). Draw this with a positive charge on the central carbon.
- Also form $\text{Br}^-$ (though the mark scheme focuses on the organic intermediate and the arrow to Br).
Step 2 - Nucleophilic attack:
- The nucleophile is $(\text{CH}_3)_2\text{CH-O}^-$ . The oxygen has a negative charge and lone pairs.
- Draw a curly arrow from a lone pair on the oxygen of $(\text{CH}_3)_2\text{CH-O}^-$ to the positively charged carbon of the carbocation $(\text{CH}_3)_2\text{CH}^+$ .
- This forms the product: diisopropyl ether, $(\text{CH}_3)_2\text{CH-O-CH(CH}_3)_2$ .

Key Takeaways

$\text{S}_\text{N}1$ is two-step: carbocation formation then nucleophilic attack.
Must show dipole on bond breaking.
Curly arrow from bond to atom for heterolysis.
Curly arrow from lone pair to positive center for attack.

Common Mistakes

Drawing a curly arrow from the negative oxygen to the carbon in step 1 (wrong, step 1 is just bond breaking).
Forgetting the positive charge on the carbocation intermediate.
Not showing the dipole ( $\delta^+$ / $\delta^-$ ) on the C-Br bond.
Drawing the arrow from the C-Br bond to the C (wrong direction).

Things to Be Careful About

The mark scheme requires M1 (dipole + arrow to Br), M2 (intermediate carbocation), M3 (arrow from O lone pair to C+). All three are needed for full marks.
Ensure the intermediate is clearly a carbocation (positive charge on carbon).

Techniques used

draw SN1 mechanism curly arrowsidentify carbocation intermediateshow dipole on C-Br bond

(iii)

Suggest how the rate of the $\text{S}_\text{N}1$ reaction would change, if at all, if 2-chloropropane were used instead of 2-bromopropane.

Explain your answer.

2M Similar

DifficultyMedium

Worked solution

Answer

The rate would decrease.

Explanation: The C—Cl bond has a higher bond enthalpy (is stronger) than the C—Br bond, so it takes longer to break (or requires more energy to break).

Working

Bond enthalpy: C—Cl > C—Br.
Rate determining step in $\text{S}_\text{N}1$ is C-X bond fission.
Stronger bond -> slower fission -> slower rate.

Final answer

Rate decreases; C-Cl bond is stronger/higher bond enthalpy than C-Br

Detailed explanation

Background Concept

In $\text{S}_\text{N}1$ reactions, the rate-determining step (slow step) is the heterolytic fission of the carbon-halogen bond (C-X) to form a carbocation and a halide ion. Therefore, the rate of reaction depends on the strength of the C-X bond.

Bond enthalpy trend for carbon-halogen bonds: C—F > C—Cl > C—Br > C—I.

Bond length increases down the group (Cl < Br < I).
Bond enthalpy decreases down the group (Cl > Br > I).
Weaker bonds break more easily/faster.

Understanding the Question

The question asks how the rate changes if 2-chloropropane is used instead of 2-bromopropane, and to explain why.

Approach

Compare the C-Cl and C-Br bonds. Relate bond strength to the rate-determining step of the $\text{S}_\text{N}1$ mechanism.

Step-by-Step Reasoning

Identify rate-determining step: In $\text{S}_\text{N}1$ , the first step (C-X bond breaking) is slow and rate-determining.
Compare bonds: We are changing from C-Br to C-Cl. Chlorine is above bromine in Group 17. The C-Cl bond is shorter and stronger (higher bond enthalpy) than the C-Br bond.
Effect on rate: Since the C-Cl bond is stronger, it requires more energy to break and breaks more slowly.
Conclusion: The rate of reaction decreases.

Key Takeaways

$\text{S}_\text{N}1$ rate depends on C-X bond strength.
C-I > C-Br > C-Cl > C-F in terms of reactivity (inverse of bond strength).
Iodoalkanes react fastest in $\text{S}_\text{N}1$ ; chloroalkanes are slower.

Common Mistakes

Saying the rate increases (confusing bond strength trend).
Saying 'chlorine is more electronegative' (true, but doesn't directly explain rate; bond enthalpy is the key factor for the rate-determining step).
Forgetting to say 'decrease'.

Things to Be Careful About

Must mention bond strength or bond enthalpy. 'Bond is stronger' is a good keyword.
Must link bond strength to the breaking of the bond in the mechanism.

Techniques used

compare C-Cl and C-Br bond enthalpiesrelate bond strength to reaction rate

(iv)

$\text{N}$ is also added to petrol to make it burn more smoothly.

Construct an equation for the complete combustion of $\text{N}$ , $\text{C}_6\text{H}_{14}\text{O}$ .

\text{C}_6\text{H}_{14}\text{O} \dots

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

\text{C}_6\text{H}_{14}\text{O} + 9\text{O}_2 \rightarrow 6\text{CO}_2 + 7\text{H}_2\text{O}

(Or: $2\text{C}_6\text{H}_{14}\text{O} + 18\text{O}_2 \rightarrow 12\text{CO}_2 + 14\text{H}_2\text{O}$ )

Final answer

C6H14O + 9O2 -> 6CO2 + 7H2O

Detailed explanation

Background Concept

Complete combustion of an organic compound containing C, H, and O in excess oxygen produces carbon dioxide ( $\text{CO}_2$ ) and water ( $\text{H}_2\text{O}$ ).
General equation: $\text{C}_x\text{H}_y\text{O}_z + (x + y/4 - z/2)\text{O}_2 \rightarrow x\text{CO}_2 + (y/2)\text{H}_2\text{O}$ .

Understanding the Question

Compound N is diisopropyl ether with molecular formula $\text{C}_6\text{H}_{14}\text{O}$ . The question asks for the equation for its complete combustion.

Approach

Balance the combustion equation: $\text{C}_6\text{H}_{14}\text{O} + \text{O}_2 \rightarrow \text{CO}_2 + \text{H}_2\text{O}$ .

Step-by-Step Reasoning

Carbon balance: 6 C in reactant -> 6 $\text{CO}_2$ .
$\text{C}_6\text{H}_{14}\text{O} + \text{O}_2 \rightarrow 6\text{CO}_2 + \text{H}_2\text{O}$
Hydrogen balance: 14 H in reactant -> 7 $\text{H}_2\text{O}$ .
$\text{C}_6\text{H}_{14}\text{O} + \text{O}_2 \rightarrow 6\text{CO}_2 + 7\text{H}_2\text{O}$
Oxygen balance:
- Products: $(6 \times 2) + (7 \times 1) = 12 + 7 = 19$ oxygen atoms.
- Reactant N provides 1 oxygen atom.
- Need 18 oxygen atoms from $\text{O}_2$ .
- So, 9 $\text{O}_2$ molecules.
  $\text{C}_6\text{H}_{14}\text{O} + 9\text{O}_2 \rightarrow 6\text{CO}_2 + 7\text{H}_2\text{O}$

Key Takeaways

Combustion products: $\text{CO}_2$ and $\text{H}_2\text{O}$ .
Balance C, then H, then O last (accounting for O in the fuel).

Common Mistakes

Forgetting the oxygen in the fuel molecule ( $\text{C}_6\text{H}_{14}\text{O}$ has 1 O).
Wrong coefficient for water (7, not 14).
Wrong coefficient for oxygen (9, not 19 or 18).

Things to Be Careful About

State symbols are not explicitly asked for in the mark scheme snippet, but good practice is $(\text{g})$ for all. The mark scheme just shows the balanced equation.

Techniques used

balance combustion equationcalculate oxygen stoichiometry for ether

Q4MediumIntroduction to Organic ChemistryHydrocarbonsAnalytical TechniquesHalogen CompoundsNitrogen CompoundsCarboxylic Acids and Derivatives

Compounds $\text{P}$ and $\text{Q}$ are structural isomers.

Fig. 4.1

(a)

11M

(i)

Define structural isomerism.

2M Similar

DifficultyEasy

Worked solution

Answer

Structural isomers are molecules with the same molecular formula but different structural formulae (different arrangement of atoms / different connectivity).

Final answer

Same molecular formula but different structural formulae (different arrangement of atoms).

Detailed explanation

Background Concept

Isomerism in organic chemistry arises when two or more compounds share the same molecular formula (same number of each type of atom) but differ in how those atoms are connected or arranged in space. Structural (or constitutional) isomerism is the broadest category: the atoms are joined in a different order, giving different structural formulae. This contrasts with stereoisomerism, where the connectivity is identical but the spatial arrangement differs (e.g. cis/trans or optical isomers).

Understanding the Question

The command word is 'Define'. The question asks for the formal definition of structural isomerism. The two marks correspond to the two essential components of the definition: (M1) same molecular formula and (M2) different structural formulae. The stem provides compounds P (pent-2-ene) and Q (2-methylbut-2-ene) as examples, both with molecular formula $\text{C}_5\text{H}_{10}$ .

Approach

State the definition in two parts matching the two marks. No working is needed; this is pure recall.

Step-by-Step Reasoning

M1 (1 mark): Structural isomers must first share the same molecular formula — the same number of atoms of each element. For P and Q, both are $\text{C}_5\text{H}_{10}$ .
M2 (1 mark): They must differ in their structural formula — the atoms are connected in a different order. In P the carbon skeleton is a straight five-carbon chain with the double bond at position 2; in Q the skeleton is branched (a methyl substituent on carbon 2).

Key Takeaways

The definition of structural isomerism has two mandatory components: identical molecular formula AND different structural formulae. Omitting either half loses a mark.

Common Mistakes

Saying 'same formula' without specifying 'molecular formula' — this could be confused with empirical formula.
Saying 'different shapes' or 'different arrangements in space' — that describes stereoisomerism, not structural isomerism.
Writing 'different structures' without first stating the same molecular formula.

Things to Be Careful About

The mark scheme requires 'same molecular formula' specifically. 'Same formula' alone may not earn M1. Also, 'different structural formulae' or 'different structures' or 'different arrangement of atoms' all satisfy M2.

Techniques used

define structural isomerismstate molecular formula requirementstate structural formula requirement

(ii)

$\text{P}$ shows geometrical isomerism.

Draw the geometrical isomer of $\text{P}$ . Explain why the two isomers are not identical.

explanation

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

The geometrical isomer of P is trans-pent-2-ene (E-pent-2-ene):

The two isomers are not identical because there is restricted (no) rotation about the $\text{C}=\text{C}$ double bond.

Final answer

Trans-pent-2-ene (E-pent-2-ene); restricted rotation about the C=C double bond.

Detailed explanation

Background Concept

Geometrical (cis/trans or E/Z) isomerism occurs in alkenes when each carbon of the $\text{C}=\text{C}$ double bond carries two different substituents. The $\pi$ bond prevents free rotation about the $\text{C}=\text{C}$ axis, so the two substituents on each carbon are locked either on the same side (cis/Z) or opposite sides (trans/E). These are genuinely different molecules with different physical properties (different boiling points, dipoles, etc.).

Understanding the Question

Compound P is pent-2-ene, $\text{CH}_3\text{CH}_2\text{CH}=\text{CHCH}_3$ , drawn in the cis (Z) configuration in Fig. 4.1. The question asks you to draw the trans (E) geometrical isomer and explain why the two are not the same molecule.

Approach

Draw the trans isomer with the $\text{CH}_3$ and $\text{CH}_2\text{CH}_3$ groups on opposite sides of the $\text{C}=\text{C}$ bond. Then state the reason: the $\pi$ bond in the $\text{C}=\text{C}$ prevents rotation, so the two spatial arrangements cannot interconvert without breaking the $\pi$ bond.

Step-by-Step Reasoning

M1 (drawing): Draw trans-pent-2-ene. In the skeletal formula the double bond is drawn horizontally (or at an angle) with the ethyl group ( $\text{CH}_2\text{CH}_3$ ) extending downward on one carbon and the methyl group ( $\text{CH}_3$ ) extending upward on the other carbon — i.e. the two alkyl groups are on opposite sides of the double bond. The remaining substituents (H atoms, implicit in skeletal notation) are on the other side.
M2 (explanation): The $\text{C}=\text{C}$ double bond consists of one $\sigma$ and one $\pi$ bond. The $\pi$ bond locks the two carbons so that free rotation is not possible at ordinary temperatures. Therefore the cis and trans arrangements are distinct, non-interconverting molecules — they are not identical.

Key Takeaways

Geometrical isomerism in alkenes is only possible when each $\text{C}=\text{C}$ carbon bears two different groups. The trans isomer is the 'opposite-side' arrangement. The explanation always rests on restricted rotation about the $\text{C}=\text{C}$ .

Common Mistakes

Drawing the cis isomer again (i.e. not actually changing the geometry).
Drawing a structural isomer (e.g. pent-1-ene) instead of the geometrical isomer.
Explaining the difference as 'different shapes' without mentioning restricted rotation about the $\text{C}=\text{C}$ .
Forgetting that the explanation must reference the double bond specifically, not just say 'the molecule cannot rotate'.

Things to Be Careful About

The skeletal drawing must clearly show the trans configuration: the main carbon chain must zigzag across the double bond so the two alkyl groups are on opposite sides. A poorly drawn 'trans' that looks like a straight chain may be ambiguous.

Techniques used

draw the trans geometrical isomerexplain restricted rotation about the C=C bond

(iii)

Both $\text{P}$ and $\text{Q}$ react with aqueous bromine.

Name the mechanism of this reaction.

1M Similar

DifficultyEasy

Worked solution

Answer

Electrophilic addition.

Final answer

Electrophilic addition

Detailed explanation

Background Concept

Alkenes react with halogens (e.g. $\text{Br}_2$ ) by electrophilic addition. The electron-rich $\pi$ bond of the $\text{C}=\text{C}$ polarises the approaching $\text{Br}_2$ molecule, generating a temporary dipole ( $\text{Br}^{\delta+}\text{Br}^{\delta-}$ ). The $\delta+$ bromine acts as an electrophile, accepting electrons from the $\pi$ bond to form a cyclic bromonium ion intermediate, which is then attacked by $\text{Br}^-$ to give a 1,2-dibromoalkane. The overall process adds two atoms across the double bond — hence 'addition' — and the first step is initiated by an electrophile — hence 'electrophilic'.

Understanding the Question

The command word is 'Name'. The question asks only for the mechanism name. Both P (pent-2-ene) and Q (2-methylbut-2-ene) contain a $\text{C}=\text{C}$ bond, so both undergo electrophilic addition with aqueous bromine.

Approach

Recognise the reaction type: alkene + bromine → dibromoalkane. This is the classic electrophilic addition reaction.

Step-by-Step Reasoning

Aqueous bromine ( $\text{Br}_2$ ) is an electrophile in this context. The $\pi$ electrons of the $\text{C}=\text{C}$ bond attack $\text{Br}^{\delta+}$ , forming a bromonium ion, then $\text{Br}^-$ (or $\text{OH}^-$ in aqueous solution) opens the ring. The net result is addition across the double bond.
The mechanism name required is 'electrophilic addition' (B1).

Key Takeaways

The reaction of alkenes with halogens ( $\text{Br}_2$ , $\text{Cl}_2$ ) and with hydrogen halides ( $\text{HBr}$ , $\text{HCl}$ ) is electrophilic addition. This is one of the most frequently tested mechanism names at AS level.

Common Mistakes

Writing 'addition reaction' alone (too vague — must specify 'electrophilic').
Writing 'nucleophilic addition' (that is for carbonyl compounds, not alkenes).
Writing 'free radical substitution' (that is for alkanes with halogens under UV).

Things to Be Careful About

The full two-word term 'electrophilic addition' is required. 'Electrophilic' alone or 'addition' alone will not earn the mark.

Techniques used

identify the reaction mechanism type for alkene + bromine

(iv)

$\text{Q}$ is oxidised by hot concentrated acidified $\text{KMnO}_4\text{(aq)}$ , forming two different organic products.

Construct an equation for this reaction. Use $\text{[O]}$ to represent an atom of oxygen from the oxidising agent.

(\text{CH}_3)_2\text{C=CHCH}_3 \dots

2M Similar

DifficultyMedium

Worked solution

Working

Hot concentrated acidified $\text{KMnO}_4$ cleaves the $\text{C}=\text{C}$ bond. The carbon bearing two alkyl groups becomes a ketone; the carbon bearing one H and one alkyl group becomes a carboxylic acid.

(\text{CH}_3)_2\text{C}=\text{CHCH}_3 + 3[\text{O}] \rightarrow \text{CH}_3\text{COCH}_3 + \text{CH}_3\text{COOH}

Answer

$(\text{CH}_3)_2\text{C}=\text{CHCH}_3 + 3[\text{O}] \rightarrow \text{CH}_3\text{COCH}_3 + \text{CH}_3\text{COOH}$

Final answer

(CH3)2C=CHCH3 + 3[O] → CH3COCH3 + CH3COOH

Detailed explanation

Background Concept

Hot, concentrated, acidified potassium manganate(VII) ( $\text{KMnO}_4$ ) is a powerful oxidising agent that cleaves carbon-carbon double bonds completely. Each carbon of the $\text{C}=\text{C}$ is oxidised to its highest stable oxidation state given its substituents:

A $\text{C}=\text{C}$ carbon with two alkyl substituents (no H) → ketone ( $\text{C}=\text{O}$ ).
A $\text{C}=\text{C}$ carbon with one H and one alkyl substituent → carboxylic acid ( $\text{COOH}$ ).
A $\text{C}=\text{C}$ carbon with two H atoms → $\text{CO}_2$ (and $\text{H}_2\text{O}$ ).

This is in contrast to cold, dilute, alkaline $\text{KMnO}_4$ , which gives a diol (syn-hydroxylation) without cleavage.

Understanding the Question

Q is 2-methylbut-2-ene, $(\text{CH}_3)_2\text{C}=\text{CHCH}_3$ . The question asks for the equation for its oxidation by hot concentrated acidified $\text{KMnO}_4$ , using $[\text{O}]$ to represent an oxygen atom from the oxidising agent. Two marks are available: M1 for correct products, M2 for the balanced equation.

Approach

Identify the two carbons of the double bond and their substituents, determine the product from each, then balance the oxygen atoms using $[\text{O}]$ .

Step-by-Step Reasoning

Left-hand carbon of the $\text{C}=\text{C}$ : bears two $\text{CH}_3$ groups (no H). Oxidative cleavage gives a ketone: $\text{CH}_3\text{COCH}_3$ (propanone / acetone).
Right-hand carbon of the $\text{C}=\text{C}$ : bears one H and one $\text{CH}_3$ . Oxidative cleavage gives a carboxylic acid: $\text{CH}_3\text{COOH}$ (ethanoic acid / acetic acid).
Balancing oxygen: The ketone requires 1 oxygen atom ( $\text{C}\rightarrow\text{C}=\text{O}$ ). The carboxylic acid requires 2 oxygen atoms ( $\text{CH}\rightarrow\text{COOH}$ , adding one O for the C=O and one O for the OH, but one of the original C–H is replaced). Net oxygen atoms needed = 3, so $3[\text{O}]$ .

(\text{CH}_3)_2\text{C}=\text{CHCH}_3 + 3[\text{O}] \rightarrow \text{CH}_3\text{COCH}_3 + \text{CH}_3\text{COOH}

Atom check: C: 5 = 3 + 2 ✓; H: 10 = 6 + 4 ✓; O: 3 = 1 + 2 ✓.

Key Takeaways

Hot concentrated $\text{KMnO}_4$ cleaves $\text{C}=\text{C}$ completely. The product from each alkene carbon depends on how many H atoms that carbon originally carried: 0 H → ketone, 1 H → carboxylic acid, 2 H → $\text{CO}_2$ . Always use $[\text{O}]$ notation when the question specifies it.

Common Mistakes

Writing $\text{CO}_2$ as a product (only occurs when the alkene carbon has two H atoms, which is not the case here).
Forgetting to balance the equation (M2 requires the coefficient of 3 before $[\text{O}]$ ).
Writing the products as aldehydes instead of carboxylic acids (aldehydes are further oxidised to acids under these harsh conditions).
Using $\text{O}_2$ instead of $[\text{O}]$ when the question specifies $[\text{O}]$ .

Things to Be Careful About

The question explicitly says 'Use $[\text{O]}$ to represent an atom of oxygen'. Do not write $\text{O}_2$ or $\text{KMnO}_4$ in the equation. The coefficient 3 before $[\text{O}]$ is essential for the balance.

Techniques used

cleave the C=C bond by hot concentrated KMnO4identify ketone and carboxylic acid products from each alkene carbonbalance the oxidation equation using [O] notation

(v)

$\text{Q}$ reacts with $\text{HBr(g)}$ to produce two structural isomers, $\text{R}$ and $\text{S}$ , as shown in Fig. 4.2.

Fig. 4.2

State and explain why isomer $\text{S}$ is the major product of the reaction.

2M Similar

DifficultyMedium

Worked solution

Answer

Isomer S is the major product because the reaction proceeds via a more stable (tertiary) carbocation intermediate. The greater number of alkyl groups attached to the positively charged carbon exerts a stronger positive inductive effect ( $+I$ ), which stabilises the $\text{C}^+$ centre more effectively than in the secondary carbocation leading to R.

Final answer

S forms via the more stable tertiary carbocation; greater positive inductive effect of more alkyl groups stabilises the C+ intermediate.

Detailed explanation

Background Concept

When an unsymmetrical alkene reacts with a hydrogen halide ( $\text{HBr}$ , $\text{HCl}$ ), the electrophilic addition can proceed via two possible carbocation intermediates, depending on which carbon of the $\text{C}=\text{C}$ the $\text{H}^+$ adds to. The reaction preferentially follows the pathway through the more stable carbocation (Markovnikov's rule). Carbocation stability increases with the number of alkyl groups attached to the $\text{C}^+$ centre: tertiary ( $3^\circ$ ) > secondary ( $2^\circ$ ) > primary ( $1^\circ$ ). This is because alkyl groups are electron-donating via the positive inductive effect ( $+I$ effect): they push electron density toward the electron-deficient carbon, dispersing and stabilising the positive charge.

Understanding the Question

Q (2-methylbut-2-ene) reacts with $\text{HBr(g)}$ to give two products: R (2-bromo-3-methylbutane, a secondary bromide) and S (2-bromo-2-methylbutane, a tertiary bromide). The question asks you to state and explain why S is the major product.

Approach

Identify the two possible carbocation intermediates. S arises from a tertiary carbocation; R from a secondary one. Explain that the tertiary carbocation is more stable due to the greater $+I$ effect of three alkyl groups versus two.

Step-by-Step Reasoning

$\text{H}^+$ can add to either carbon of the $\text{C}=\text{C}$ :
- If $\text{H}^+$ adds to the $\text{CH}$ carbon (right-hand carbon), the positive charge falls on the carbon bearing two methyl groups → tertiary carbocation $(\text{CH}_3)_2\text{C}^+\text{CH}_2\text{CH}_3$ .
- If $\text{H}^+$ adds to the $\text{C(CH}_3)_2$ carbon (left-hand carbon), the positive charge falls on the carbon bearing one methyl and one H → secondary carbocation $(\text{CH}_3)_2\text{CHCH}^+\text{CH}_3$ .
The tertiary carbocation is more stable because it has three alkyl groups (two methyls and one ethyl) exerting the $+I$ effect, compared to only two alkyl groups in the secondary carbocation.
$\text{Br}^-$ then attacks the carbocation. Attack on the tertiary carbocation gives S (2-bromo-2-methylbutane); attack on the secondary gives R (2-bromo-3-methylbutane).
Since the tertiary carbocation is lower in energy, it forms faster (lower activation energy), so S is the major product.
M1: The reaction proceeds via a more stable intermediate / carbocation.
M2: Because of the greater positive inductive effect of more alkyl groups.

Key Takeaways

Markovnikov's rule is a consequence of carbocation stability. Always identify the intermediate carbocation and count the alkyl groups on the $\text{C}^+$ centre. The $+I$ effect of alkyl groups is the standard explanation for carbocation stability at AS level.

Common Mistakes

Saying 'S has more alkyl groups' without linking it to the carbocation intermediate.
Saying 'tertiary is more stable' without mentioning the inductive effect (the mark scheme requires the $+I$ explanation for M2).
Confusing the products: R is the secondary bromide, S is the tertiary bromide.
Saying 'more substituted alkene' — this is not the relevant argument here.

Things to Be Careful About

The explanation must explicitly reference the intermediate carbocation (M1) and the positive inductive effect of alkyl groups (M2). Simply saying 'tertiary carbocation is more stable' without the inductive effect reasoning may lose M2.

Techniques used

identify the more stable carbocation intermediateexplain carbocation stability via positive inductive effect of alkyl groups

(vi)

The mass spectrum of $\text{R}$ shows peaks at $m/e = 150$ and $m/e = 152$ .

Suggest structures for the ions responsible for these peaks.

2M Similar

DifficultyMedium

Worked solution

Answer

The peaks at $m/e = 150$ and $m/e = 152$ are the molecular ion peaks of R ( $\text{C}_5\text{H}_{11}\text{Br}$ ), corresponding to the two naturally occurring isotopes of bromine.

The ion at $m/e = 150$ contains $^{79}\text{Br}$ :

The ion at $m/e = 152$ contains $^{81}\text{Br}$ :

Final answer

Molecular ions of R: [CH3CH(CH3)CH(79Br)CH3]+ at m/e = 150 and [CH3CH(CH3)CH(81Br)CH3]+ at m/e = 152

Detailed explanation

Background Concept

Bromine has two naturally occurring isotopes: $^{79}\text{Br}$ (50.7%) and $^{81}\text{Br}$ (49.3%), in an approximately 1:1 ratio. In a mass spectrum, a molecule containing one bromine atom therefore shows two molecular ion peaks of nearly equal height, separated by 2 mass units: the M peak (containing $^{79}\text{Br}$ ) and the M+2 peak (containing $^{81}\text{Br}$ ). This 1:1 doublet is a diagnostic signature of a bromine-containing compound.

Understanding the Question

R is 2-bromo-3-methylbutane, $\text{CH}_3\text{CH(CH}_3)\text{CHBrCH}_3$ , with molecular formula $\text{C}_5\text{H}_{11}\text{Br}$ . The mass spectrum shows peaks at $m/e = 150$ and $m/e = 152$ . The question asks you to suggest the structures of the ions responsible. Since $5(12) + 11(1) + 79 = 150$ and $5(12) + 11(1) + 81 = 152$ , these are the molecular ions $\text{M}^+$ containing $^{79}\text{Br}$ and $^{81}\text{Br}$ respectively.

Approach

Calculate the molecular mass with each bromine isotope. Confirm the peaks match the molecular ion. Draw the full molecular ion structure (in brackets with a + charge) showing the correct isotope label.

Step-by-Step Reasoning

M1 (m/e = 150): The molecular ion containing $^{79}\text{Br}$ . Structure: $[(\text{CH}_3)\text{CH}(\text{CH}_3)\text{CH}(^{79}\text{Br})\text{CH}_3]^+$ . In skeletal form, this is the 2-bromo-3-methylbutane skeleton with $^{79}\text{Br}$ on C-2, enclosed in square brackets with a superscript +.
M2 (m/e = 152): The molecular ion containing $^{81}\text{Br}$ . Identical structure but with $^{81}\text{Br}$ : $[(\text{CH}_3)\text{CH}(\text{CH}_3)\text{CH}(^{81}\text{Br})\text{CH}_3]^+$ .
These are not fragment ions — they are the intact molecular ions. The 2-unit separation and roughly equal intensity confirm a single bromine atom.

Key Takeaways

A 1:1 doublet at M and M+2 in a mass spectrum is the fingerprint of one bromine atom. Always check whether the peaks correspond to the molecular ion or a fragment by calculating the mass.

Common Mistakes

Drawing a fragment ion (e.g. loss of $\text{CH}_3$ ) instead of the molecular ion. The masses 150 and 152 match the full molecule, not a fragment.
Forgetting the brackets and + charge around the ion structure.
Not labelling the bromine isotope ( $^{79}\text{Br}$ vs $^{81}\text{Br}$ ) — the mark scheme requires the isotope label.
Drawing the wrong carbon skeleton (e.g. the S skeleton instead of R).

Things to Be Careful About

The structures must be drawn in square brackets with a superscript + to indicate a positive ion. The bromine isotope must be explicitly labelled as $^{79}\text{Br}$ or $^{81}\text{Br}$ . The carbon skeleton must match R (2-bromo-3-methylbutane), not S.

Techniques used

identify molecular ion peaks from isotopic composition of brominedraw the molecular ion structures containing 79Br and 81Br

(b)

Fig. 4.3 shows a synthesis starting from $\text{T}$ , a different isomer of $\text{R}$ and $\text{S}$ .

Fig. 4.3

(i)

Identify the reagent and conditions for reaction 1.

1M Similar

DifficultyEasy

Worked solution

Answer

$\text{KCN}$ in ethanol, heated under reflux.

Final answer

KCN in ethanol, heat under reflux

Detailed explanation

Background Concept

Halogenoalkanes react with cyanide ions ( $\text{CN}^-$ ) by nucleophilic substitution. The $\text{CN}^-$ ion acts as a nucleophile, attacking the electron-deficient carbon bonded to the halogen and displacing the halide ion ( $\text{Br}^-$ ). This converts a halogenoalkane into a nitrile ( $\text{R-CN}$ ), extending the carbon chain by one carbon. The standard conditions are potassium cyanide ( $\text{KCN}$ ) dissolved in ethanol (to dissolve the organic halogenoalkane) and heating under reflux.

Understanding the Question

Reaction 1 converts T (3-bromopentane) to U (3-cyanopentane): the $\text{Br}$ is replaced by $\text{CN}$ . The question asks for the reagent and conditions.

Approach

Recognise the transformation: halogenoalkane → nitrile. This is nucleophilic substitution with $\text{CN}^-$ . State the reagent ( $\text{KCN}$ ), solvent (ethanol), and condition (heat/reflux).

Step-by-Step Reasoning

The $\text{Br}$ on C-3 of T is replaced by $\text{CN}$ on C-3 of U. This is a one-for-one substitution.
Reagent: $\text{KCN}$ (provides $\text{CN}^-$ nucleophile).
Solvent: ethanol (the organic solvent needed to dissolve the halogenoalkane; aqueous conditions would favour hydrolysis to an alcohol instead).
Condition: heat (under reflux) to provide sufficient activation energy.
B1: $\text{KCN}$ in ethanol (and heat).

Key Takeaways

The conversion halogenoalkane → nitrile uses $\text{KCN}$ /ethanol/heat. The choice of ethanol (not water) is critical: aqueous $\text{KCN}$ would give an alcohol via hydrolysis. This is a standard AS-level reagent-and-conditions question.

Common Mistakes

Writing 'NaCN' instead of 'KCN' — NaCN is also acceptable in practice, but the mark scheme specifies KCN.
Omitting the solvent (ethanol) — without it, the reaction does not proceed effectively.
Writing 'aqueous KCN' — this would give the alcohol, not the nitrile.
Forgetting 'heat' or 'reflux' as the condition.

Things to Be Careful About

The mark scheme credits 'KCN in ethanol (and heat)'. Both the reagent and the solvent must be stated. 'Heat' or 'reflux' should be mentioned as the condition.

Techniques used

identify reagent and conditions for nucleophilic substitution of halogenoalkane with cyanide

(ii)

Reaction 2 is a hydrolysis reaction.

Construct an equation for reaction 2.

(\text{C}_2\text{H}_5)_2\text{CHCN} \dots

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

$(\text{C}_2\text{H}_5)_2\text{CHCN} + 2\text{H}_2\text{O} + \text{HCl} \rightarrow (\text{C}_2\text{H}_5)_2\text{CHCOOH} + \text{NH}_4\text{Cl}$

Final answer

(C2H5)2CHCN + 2H2O + HCl → (C2H5)2CHCOOH + NH4Cl

Detailed explanation

Background Concept

Nitriles ( $\text{R-CN}$ ) undergo acidic hydrolysis when heated with aqueous acid (e.g. $\text{HCl(aq)}$ or $\text{H}_2\text{SO}_4\text{(aq)}$ ). The $\text{C}\equiv\text{N}$ triple bond is cleaved and converted into a $\text{COOH}$ group, while the nitrogen is released as ammonium ion ( $\text{NH}_4^+$ ). The overall reaction consumes 2 water molecules and 1 acid molecule, producing the carboxylic acid and the ammonium salt of the acid.

\text{R-CN} + 2\text{H}_2\text{O} + \text{HCl} \rightarrow \text{R-COOH} + \text{NH}_4\text{Cl}

Understanding the Question

Reaction 2 is the hydrolysis of U, $(\text{C}_2\text{H}_5)_2\text{CHCN}$ (3-cyanopentane), with $\text{HCl(aq)}$ to give V, $(\text{C}_2\text{H}_5)_2\text{CHCOOH}$ (2-ethylbutanoic acid). The question asks for the balanced equation.

Approach

Write the nitrile on the left, add $2\text{H}_2\text{O}$ and $\text{HCl}$ , and produce the carboxylic acid and $\text{NH}_4\text{Cl}$ on the right. Check atom balance.

Step-by-Step Reasoning

The $\text{C}\equiv\text{N}$ group is hydrolysed: the C becomes $\text{COOH}$ and the N becomes $\text{NH}_4^+$ .
Oxygen atoms needed: 2 (for the $\text{COOH}$ group) → supplied by $2\text{H}_2\text{O}$ .
Hydrogen: the $\text{NH}_4^+$ needs 4 H; 2 from each $\text{H}_2\text{O}$ gives 4 H to the nitrogen; the $\text{HCl}$ provides the $\text{Cl}^-$ to pair with $\text{NH}_4^+$ forming $\text{NH}_4\text{Cl}$ .
Balanced equation:

$(\text{C}_2\text{H}_5)_2\text{CHCN} + 2\text{H}_2\text{O} + \text{HCl} \rightarrow (\text{C}_2\text{H}_5)_2\text{CHCOOH} + \text{NH}_4\text{Cl}$

Atom check: C: 7 = 7 ✓; H: 13+4+1 = 18; right: 14+4 = 18 ✓; N: 1 = 1 ✓; O: 2 = 2 ✓; Cl: 1 = 1 ✓.

Key Takeaways

Acidic hydrolysis of a nitrile: $\text{R-CN} + 2\text{H}_2\text{O} + \text{HCl} \rightarrow \text{R-COOH} + \text{NH}_4\text{Cl}$ . The coefficient of 2 before water is essential. The nitrogen leaves as $\text{NH}_4^+$ , not $\text{NH}_3$ (in acidic solution).

Common Mistakes

Writing $\text{NH}_3$ instead of $\text{NH}_4\text{Cl}$ (in acidic solution, ammonia is protonated).
Writing $\text{H}_2\text{O}$ instead of $2\text{H}_2\text{O}$ (unbalanced).
Forgetting $\text{HCl}$ on the left or $\text{NH}_4\text{Cl}$ on the right.
Writing the organic product incorrectly (e.g. losing the ethyl branches).

Things to Be Careful About

The equation must be fully balanced. The mark scheme gives the exact form; any deviation in coefficients loses the mark. $\text{NH}_4\text{Cl}$ (not $\text{NH}_3 + \text{HCl}$ ) is the correct product in acidic solution.

Techniques used

construct the hydrolysis equation of a nitrile with aqueous HClbalance the equation to give carboxylic acid and ammonium chloride

(iii)

$\text{V}$ reacts with propan-2-ol in the presence of a catalytic amount of $\text{H}_2\text{SO}_4$ to form organic compound $\text{W}$ .

Complete Table 4.1 to give details of this reaction.

Table 4.1

	reaction of $\text{V}$ with propan-2-ol
type of reaction
functional group formed
molecular formula of organic product $\text{W}$

3M Similar

DifficultyMedium-Easy

Worked solution

Answer

	reaction of V with propan-2-ol
type of reaction	condensation
functional group formed	ester
molecular formula of organic product W	$\text{C}_9\text{H}_{18}\text{O}_2$

Final answer

Condensation; ester; C9H18O2

Detailed explanation

Background Concept

A carboxylic acid reacts with an alcohol in the presence of a catalytic amount of concentrated $\text{H}_2\text{SO}_4$ to form an ester and water. This is a condensation reaction (also called esterification or Fischer esterification): two molecules combine with the loss of a small molecule ( $\text{H}_2\text{O}$ ). The functional group formed is the ester group, $\text{R-COO-R'}$ .

Understanding the Question

V is 2-ethylbutanoic acid, $(\text{C}_2\text{H}_5)_2\text{CHCOOH}$ , with molecular formula $\text{C}_6\text{H}_{12}\text{O}_2$ . It reacts with propan-2-ol, $\text{CH}_3\text{CH(OH)CH}_3$ ( $\text{C}_3\text{H}_8\text{O}$ ), catalysed by $\text{H}_2\text{SO}_4$ . The table asks for: (1) the type of reaction, (2) the functional group formed, and (3) the molecular formula of the ester W.

Approach

Identify the reaction as esterification (condensation). The functional group is an ester. Calculate the molecular formula of W by adding the two reactants and subtracting $\text{H}_2\text{O}$ .

Step-by-Step Reasoning

M1 — Type of reaction: The reaction of a carboxylic acid with an alcohol to form an ester and water is a condensation reaction. (Esterification is also acceptable but the mark scheme credits 'condensation'.)
M2 — Functional group formed: The product contains the ester group ( $\text{-COO-}$ ).
M3 — Molecular formula of W:

\text{C}_6\text{H}_{12}\text{O}_2 + \text{C}_3\text{H}_8\text{O} \rightarrow \text{W} + \text{H}_2\text{O}

\text{W} = \text{C}_6\text{H}_{12}\text{O}_2 + \text{C}_3\text{H}_8\text{O} - \text{H}_2\text{O} = \text{C}_9\text{H}_{18}\text{O}_2

C: $6 + 3 = 9$ ✓
H: $12 + 8 - 2 = 18$ ✓
O: $2 + 1 - 1 = 2$ ✓

So W is $\text{C}_9\text{H}_{18}\text{O}_2$ , which is propan-2-yl 2-ethylbutanoate.

Key Takeaways

Carboxylic acid + alcohol → ester + water is a condensation reaction. The molecular formula of the ester is obtained by summing the reactant formulae and subtracting $\text{H}_2\text{O}$ . Always check the atom balance.

Common Mistakes

Writing 'esterification' for the reaction type when the mark scheme expects 'condensation' (though esterification may be accepted, 'condensation' is the credited answer).
Writing 'carboxylic acid' or 'acid' as the functional group instead of 'ester'.
Arithmetic error in the molecular formula: forgetting to subtract $\text{H}_2\text{O}$ , giving $\text{C}_9\text{H}_{20}\text{O}_3$ instead of $\text{C}_9\text{H}_{18}\text{O}_2$ .
Writing the structural formula instead of the molecular formula (the question asks for molecular formula).

Things to Be Careful About

The question asks for the molecular formula, not the structural formula. The answer must be in the form $\text{C}_9\text{H}_{18}\text{O}_2$ . The 'type of reaction' must be 'condensation' (or 'esterification' if accepted). The functional group is 'ester', not 'carboxylic acid' or 'alcohol'.

Techniques used

identify the reaction type as condensation/esterificationidentify the ester functional groupcalculate the molecular formula of the ester product