Topic: 1.1
In this question Q is used to represent a halogen atom. Magnesium and calcium each form a compound with chlorine and a compound with bromine. One of these compounds contains:
- the element in Group 2 with the higher first ionisation energy and
- the element in Group 17 with the higher Q–Q bond energy.
What is the formula of this compound?
▶️ Answer/Explanation
Ans: A
In Group 2, magnesium (\(Mg\)) has a higher first ionisation energy than calcium (\(Ca\)) because ionisation energy decreases down the group. In Group 17, chlorine (\(Cl\)) has a higher \(Q-Q\) bond energy than bromine (\(Br\)) due to smaller atomic size and stronger bonding. Thus, the correct compound is \(MgCl_2\), as it satisfies both conditions.
Topic: 1.1
Compound X contains two elements, Y and Z. Element Y is in Period 2 of the Periodic Table. In one atom of element Y, the p sub-shell has all three orbitals occupied; only one of these three orbitals is fully occupied. Element Z is in Period 3 of the Periodic Table. In one atom of element Z, the p sub-shell has only two orbitals occupied. What is the formula of compound X?
▶️ Answer/Explanation
Ans: C
Element Y (Period 2) has three p-orbitals occupied, with only one fully occupied. This means its electron configuration is \(1s^2 2s^2 2p^4\), identifying Y as Oxygen (O). Element Z (Period 3) has two p-orbitals occupied, giving the configuration \(1s^2 2s^2 2p^6 3s^2 3p^2\), which is Silicon (Si). The stable compound formed between Si and O is \(SiO_2\), as Si shares its 4 valence electrons with two O atoms, each needing 2 electrons to complete their octet.
Topic: 4.1
Glauber’s salt consists of crystals of hydrated sodium sulfate, Na₂SO₄•xH₂O, which can be used for the manufacture of detergents. When a sample of Glauber’s salt was heated, 1.91 g of water was removed leaving 1.51 g of anhydrous Na₂SO₄. What is the value of x in Na₂SO₄•xH₂O?
▶️ Answer/Explanation
Ans: C
To find \( x \) in Na₂SO₄•xH₂O, we calculate the moles of anhydrous Na₂SO₄ and water. The molar mass of Na₂SO₄ is \( 142 \, \text{g/mol} \), so moles of Na₂SO₄ = \( \frac{1.51 \, \text{g}}{142 \, \text{g/mol}} = 0.0106 \, \text{mol} \). The moles of H₂O = \( \frac{1.91 \, \text{g}}{18 \, \text{g/mol}} = 0.106 \, \text{mol} \). The ratio \( \frac{0.106}{0.0106} = 10 \), confirming \( x = 10 \).
Topic: 2.1
What contains the greatest number of the named particles?
▶️ Answer/Explanation
Ans: D
To determine the greatest number of particles, we compare the number of moles for each option.
Option A: At room conditions, 6.0 dm³ of argon ≈ 0.25 mol.
Option B: 6.0 g of CO₂ (M = 44 g/mol) ≈ 0.136 mol.
Option C: 6.0 g of Mg (M = 24 g/mol) = 0.25 mol.
Option D: 6.0 g of H₂O (M = 18 g/mol) ≈ 0.333 mol.
Since water has the highest number of moles, it contains the greatest number of particles.
Topic: 3.2
Phosphorus forms a compound with hydrogen called phosphine, PH₃. This compound can react with a hydrogen ion, H⁺. Which type of interaction occurs between PH₃ and H⁺?
▶️ Answer/Explanation
Ans: A
Phosphine (\(PH_3\)) has a lone pair of electrons on the phosphorus atom. When \(H^+\) (which has an empty 1s orbital) interacts with \(PH_3\), the lone pair is donated to form a dative covalent bond (also called a coordinate bond). This results in the formation of \(PH_4^+\), where the bond is formed by the sharing of electrons provided entirely by \(PH_3\).
Topic: 9.2
The graphs show trends in four physical properties of elements in Period 3, excluding argon. Which graph has electronegativity on the y-axis?
▶️ Answer/Explanation
Ans: D
Electronegativity increases across Period 3 (Na to Cl) because the effective nuclear charge increases, attracting electrons more strongly. Graph D shows this upward trend, matching the expected electronegativity pattern. The other graphs depict different properties (e.g., atomic radius, melting point), making D the correct choice.
Topic: 10.1
The element tin exists in two forms, grey tin and white tin. Some properties of grey tin and white tin are shown.
Which structural change might take place when grey tin changes to white tin?
▶️ Answer/Explanation
Ans: B
Grey tin has a giant covalent structure (similar to diamond), while white tin has a giant metallic structure. The transition from grey tin to white tin involves breaking covalent bonds and forming a metallic lattice, which explains the change in conductivity and malleability. Thus, the correct structural change is giant covalent to giant metallic (Option B).
Topic: 10.1
Which solid has a simple molecular lattice?
A. calcium fluoride
B. nickel
C. silicon(IV) oxide
D. sulfur
▶️ Answer/Explanation
Ans: D
Sulfur (D) forms a simple molecular lattice, where \( \text{S}_8 \) molecules are held together by weak van der Waals forces. Calcium fluoride (A) has an ionic lattice, nickel (B) is a metallic solid, and silicon(IV) oxide (C) forms a giant covalent structure. Thus, sulfur is the only option with a simple molecular arrangement.
Topic: 5.1
The standard enthalpy change of combustion of carbon is –394 kJ mol⁻¹.
The standard enthalpy change of combustion of hydrogen is –286 kJ mol⁻¹.
The standard enthalpy change of formation of butane is –129 kJ mol⁻¹.
What is the standard enthalpy change of combustion of butane?
▶️ Answer/Explanation
Ans: B
The standard enthalpy change of combustion of butane can be calculated using Hess’s Law. The balanced combustion equation for butane (\( C_4H_{10} \)) is:
\[ C_4H_{10} + \frac{13}{2} O_2 \rightarrow 4 CO_2 + 5 H_2O \]
Using the given data:
- Combustion of carbon: \( C + O_2 \rightarrow CO_2 \), \( \Delta H = -394 \, \text{kJ mol}^{-1} \) (for 4 moles: \( 4 \times -394 = -1576 \, \text{kJ} \))
- Combustion of hydrogen: \( H_2 + \frac{1}{2} O_2 \rightarrow H_2O \), \( \Delta H = -286 \, \text{kJ mol}^{-1} \) (for 5 moles: \( 5 \times -286 = -1430 \, \text{kJ} \))
- Formation of butane: \( 4 C + 5 H_2 \rightarrow C_4H_{10} \), \( \Delta H = -129 \, \text{kJ mol}^{-1} \)
Applying Hess’s Law, the combustion enthalpy of butane is:
\[ \Delta H_{\text{combustion}} = (-1576 – 1430) – (-129) = -2877 \, \text{kJ mol}^{-1} \]
Thus, the correct answer is B.
Topic: 5.1
Three processes are described.
1 H⁺(aq) + OH⁻(aq) → H₂O(l)
2 CH₄(g) + 2 O₂(g) → CO₂(g) + 2H₂O(l)
3 NH₃(g) → NH₃(l)
Which statement is correct?
▶️ Answer/Explanation
Ans: A
Process 1: Neutralization (H⁺ + OH⁻ → H₂O) is exothermic (ΔH < 0).
Process 2: Combustion of methane (CH₄ + 2O₂ → CO₂ + 2H₂O) is exothermic (ΔH < 0).
Process 3: Condensation (NH₃(g) → NH₃(l)) is exothermic (ΔH < 0).
Since all three processes release heat, none have a positive ΔH. Thus, Option A is correct.
Topic: 6.1
In alkaline solution, MnO₄⁻ ions oxidise SO₃²⁻ ions to SO₄²⁻ ions. The MnO₄⁻ ions are reduced to MnO₂. What is the ratio of the two ions in the balanced chemical equation for this reaction?
▶️ Answer/Explanation
Ans: A
To balance the redox reaction, we first write the half-reactions:
Oxidation: \(\text{SO}_3^{2-} \rightarrow \text{SO}_4^{2-} + 2e^-\)
Reduction: \(\text{MnO}_4^- + 2\text{H}_2\text{O} + 3e^- \rightarrow \text{MnO}_2 + 4\text{OH}^-\)
Balancing the electrons, we multiply the oxidation half-reaction by 3 and the reduction half-reaction by 2. The ratio of \(\text{MnO}_4^-\) to \(\text{SO}_3^{2-}\) is 2:3, which corresponds to option A.
Topic: 7.1
Lithium reacts with nitrogen at room temperature to form solid Li₃N. Three vessels of equal volume are connected by taps 1 and 2 as shown.
At the start, taps 1 and 2 are closed, the left-hand vessel is evacuated, the middle vessel has the indicated reaction at equilibrium and the right-hand vessel contains lithium only. Which action would allow the equilibrium mixture to contain the most ammonia?
▶️ Answer/Explanation
Ans: A
To maximize ammonia (\(NH_3\)) in the equilibrium mixture, the system should remain undisturbed. Opening any tap would introduce changes (e.g., adding lithium or altering pressure), shifting the equilibrium away from ammonia production. Keeping both taps closed maintains the equilibrium conditions, ensuring the highest ammonia concentration.
Topic: 7.1
When 0.20 mol of hydrogen gas and 0.15 mol of iodine gas are heated at 723K until equilibrium is established, the equilibrium mixture contains 0.26 mol of hydrogen iodide. The equation for the reaction is as follows.
\(H_2(g) + I_2(g) \to 2HI(g)\)
What is the correct expression for the equilibrium constant \(K_c\)?
▶️ Answer/Explanation
Ans: C
1. **Write the equilibrium expression:** For the reaction \(H_2(g) + I_2(g) \rightleftharpoons 2HI(g)\), \[ K_c = \frac{[HI]^2}{[H_2][I_2]} \]
2. **Calculate equilibrium concentrations:** – Since 0.26 mol of HI is formed, the change in \(H_2\) and \(I_2\) is \( \frac{0.26}{2} = 0.13 \) mol (due to stoichiometry). – Equilibrium moles: \[ [H_2] = 0.20 – 0.13 = 0.07 \, \text{mol} \] \[ [I_2] = 0.15 – 0.13 = 0.02 \, \text{mol} \] \[ [HI] = 0.26 \, \text{mol} \]
3. **Substitute into \(K_c\) expression:** \[ K_c = \frac{(0.26)^2}{0.07 \times 0.02} \]
Topic: 8.1
In acidic conditions, iodine reacts with propanone in a substitution reaction.
CH₃COCH₃(aq) + I₂(aq) → CH₃COCH₂I(aq) + HI(aq)
The kinetics of the reaction are investigated using a colorimeter. As the I₂ reacts, the yellow/brown colour of the I₂(aq) fades to colourless, changing the absorbance of the solution. Known concentrations of I₂(aq) are used to prepare a calibration curve graph and the absorbance is then measured as the reaction proceeds.
What is the rate of reaction at 20 s?
▶️ Answer/Explanation
Ans: A
The rate of reaction is determined by the slope of the concentration-time graph at \( t = 20 \, \text{s} \). From the graph, the concentration of I₂ decreases linearly, and the slope (rate) is calculated as:
\[ \text{Rate} = -\frac{\Delta [\text{I}_2]}{\Delta t} = \frac{(0.00010 – 0.00020) \, \text{mol dm}^{-3}}{(30 – 10) \, \text{s}} = 5 \times 10^{-6} \, \text{mol dm}^{-3} \, \text{s}^{-1} \]
Since the rate is the magnitude of the slope, the correct answer is A.
Topic: 8.2
The diagram shows a Boltzmann distribution curve. The axes are not labelled.
Points X and Y are points on the vertical axis. What is represented by both points X and Y?
▶️ Answer/Explanation
Ans: A
In a Boltzmann distribution curve, the vertical axis represents the number of molecules with a given energy. Points X and Y lie on this axis, indicating they represent the number of molecules at specific energy levels. Since the question asks what is represented by both points, the correct interpretation is that they denote the number of molecules with particular energies, corresponding to option A.
Topic: 11.4
What are the acid–base nature and structure of SO₂?
▶️ Answer/Explanation
Ans: B
Sulfur dioxide (\(SO_2\)) is an acidic gas with a bent (V-shaped) molecular structure. The sulfur atom is \(sp^2\) hybridized, forming two sigma bonds with oxygen atoms and retaining one lone pair, giving a bond angle of approximately \(119^\circ\). In aqueous solution, \(SO_2\) dissolves to form sulfurous acid (\(H_2SO_3\)), confirming its acidic nature.
Topic: 9.3
Elements X and Y are in Period 3 of the Periodic Table. Element X is either phosphorus or sulfur. Element Y is either sodium or magnesium. Element X forms an oxide that reacts with water to give a solution containing the aqueous anion \(XO_4^{2–}\). One mole of element Y reacts with one mole of chlorine molecules. At the end of the reaction, all of the element Y and all of the chlorine molecules have been used up. What are elements X and Y?
▶️ Answer/Explanation
Ans: D
Element X: The oxide of sulfur (\(SO_3\)) reacts with water to form \(SO_4^{2–}\) (sulfate ion), while phosphorus oxide (\(P_4O_{10}\)) forms \(PO_4^{3–}\). Since the given anion is \(XO_4^{2–}\), X must be sulfur (S).
Element Y: Sodium (Na) reacts with chlorine in a 2:1 mole ratio (\(2Na + Cl_2 \rightarrow 2NaCl\)), while magnesium (Mg) reacts in a 1:1 mole ratio (\(Mg + Cl_2 \rightarrow MgCl_2\)). The condition matches magnesium, so Y is magnesium (Mg).
Thus, the correct combination is S and Mg (Option D).
Topic: 9.3
Q is a semi-conductor. The chloride of Q reacts with water to form white fumes and an acidic solution. Which Period 3 element is Q?
▶️ Answer/Explanation
Ans: C
1. **Semiconductor Property**: Silicon (Si) is the only Period 3 element that is a semiconductor (used in electronics). 2. **Chloride Reaction**: Silicon tetrachloride (SiCl4) reacts vigorously with water, producing white fumes of HCl and an acidic solution (H2SiO3). 3. **Elimination of Other Options**: – Magnesium (A) forms an ionic chloride (MgCl2), which dissolves without fumes. – Aluminium (B) forms AlCl3, which hydrolyzes but is not a semiconductor. – Phosphorus (D) forms PCl3/PCl5, but it is not a semiconductor. Thus, **Q must be silicon (C)**.
Topic: 10.1
V and W are two compounds. Each one contains a different Group 2 element. A sample of each solid is added to water, shaken, and the pH of the resulting solution is measured.
Which row could identify V and W? 
▶️ Answer/Explanation
Ans: C
Group 2 oxides (e.g., MgO, CaO) react with water to form hydroxides, increasing pH (strongly alkaline). The solubility of hydroxides increases down the group, so Ba(OH)₂ (from BaO) gives a higher pH than Mg(OH)₂ (from MgO). Thus, V = MgO (pH 10) and W = BaO (pH 13), matching option C.
Topic: 10.1
Compound L decomposes on heating. One of the products is gas M. M reacts with unburned hydrocarbons to form peroxyacetyl nitrate, PAN. What could be the formula of L?
▶️ Answer/Explanation
Ans: B
To determine the correct formula of compound L, we analyze the decomposition and the properties of gas M:
- Decomposition Reaction:
- Option B, \( \text{Ca(NO}_3\text{)}_2 \), decomposes on heating to produce \( \text{CaO} \), \( \text{NO}_2 \), and \( \text{O}_2 \). Here, \( \text{NO}_2 \) (gas M) is a key reactant in forming PAN.
- PAN (peroxyacetyl nitrate) is a secondary pollutant formed when \( \text{NO}_2 \) reacts with hydrocarbons and sunlight.
- Eliminating Incorrect Options:
- A (CaNO₃): Incorrect formula (does not exist as written).
- C (MgCO₃): Decomposes to \( \text{MgO} \) and \( \text{CO}_2 \), but \( \text{CO}_2 \) does not form PAN.
- D (Mg(CO₃)₂): Incorrect formula (magnesium carbonate is \( \text{MgCO}_3 \)).
Thus, the correct answer is B, \( \text{Ca(NO}_3\text{)}_2 \), as it produces \( \text{NO}_2 \), which reacts to form PAN.
Topic: 11.2
In reaction 1, concentrated sulfuric acid is added to potassium chloride and the fumes produced are bubbled into aqueous potassium iodide solution. In reaction 2, potassium chloride is dissolved in aqueous ammonia and this is then added to aqueous silver nitrate. What are the observations for reactions 1 and 2?
▶️ Answer/Explanation
Ans: C
Reaction 1: – Concentrated \( H_2SO_4 \) reacts with \( KCl \) to produce \( HCl \) gas (colorless fumes). – When \( HCl \) is bubbled into \( KI \) solution, no visible reaction occurs because \( HCl \) cannot oxidize \( I^- \). Reaction 2: – \( KCl \) in \( NH_3(aq) \) reacts with \( AgNO_3 \) to form a white precipitate of \( AgCl \), which dissolves in excess \( NH_3 \). Thus, the correct observations match Option C (fumes in Reaction 1, white precipitate soluble in excess \( NH_3 \) in Reaction 2).
Topic: 11.3
The table refers to the hydrogen halides. Which row is correct?
▶️ Answer/Explanation
Ans: C
Hydrogen halides (HX) exhibit trends in thermal stability and acidity:
- Thermal stability decreases down the group (HF > HCl > HBr > HI) due to weaker H-X bonds.
- Acidity increases down the group (HF < HCl < HBr < HI) because bond strength decreases, making proton release easier.
Row C correctly states that HI is the least thermally stable and most acidic, matching the observed trends.
Topic: 4.1
7.5 g of nitrogen monoxide reacts with 7.0 g of carbon monoxide on the surface of the catalytic converter in the exhaust system of a car. What is the total volume of the product gases measured at room conditions?
▶️ Answer/Explanation
Ans: C
The reaction is \(2NO + 2CO \rightarrow N_2 + 2CO_2\). Moles of NO = \(\frac{7.5}{30} = 0.25\) mol, and moles of CO = \(\frac{7.0}{28} = 0.25\) mol. Since the reaction consumes equal moles of NO and CO, all reactants are fully converted. The products are \(0.125\) mol \(N_2\) and \(0.25\) mol \(CO_2\), totaling \(0.375\) mol. At room conditions, 1 mol of gas occupies 24 dm³, so total volume = \(0.375 \times 24 = 9.0\) dm³.
Topic: 13.4
Three statements about ammonia molecules and ammonium ions are given.
1. In aqueous solution, ammonia molecules form coordinate bonds with hydroxide ions.
2. Ammonium ions are Brønsted–Lowry acids.
3. The H–N–H bond angle is larger in the ammonium ion than in the ammonia molecule.
Which statements are correct?
▶️ Answer/Explanation
Ans: C
Analysis of each statement:
1. False: Ammonia (NH3) forms coordinate bonds with H+ ions to form NH4+, not with hydroxide ions (OH–).
2. True: NH4+ can donate a proton (H+), making it a Brønsted-Lowry acid.
3. True: NH3 has a trigonal pyramidal shape (bond angle ≈ 107°), while NH4+ is tetrahedral (bond angle ≈ 109.5°), making the angle larger in the ion.
Therefore, only statements 2 and 3 are correct.
Topic: 14.2
Ethene reacts with steam in the presence of sulfuric acid.
\(C_2H_4 + H_2O \to CH_3CH_2OH\)
Which type of reaction is this?
▶️ Answer/Explanation
Ans: B
This is an addition reaction because water (\(H_2O\)) adds across the double bond of ethene (\(C_2H_4\)) to form ethanol (\(CH_3CH_2OH\)). The sulfuric acid acts as a catalyst. Addition reactions are characteristic of alkenes, where the π bond breaks and new atoms are added without any byproducts.
Topic: 16.1
Compound Z has the molecular formula \(C_4H_8O_2\). Compound Z reacts with propan-1-ol in the presence of concentrated H₂SO₄. The diagram shows the skeletal formulae of three compounds, S, T and U.
What are the possible skeletal formulae of the products of the reaction between compound Z and propan-1-ol?
▶️ Answer/Explanation
Ans: D
Given that compound Z (\(C_4H_8O_2\)) reacts with propan-1-ol under acidic conditions, the reaction is likely an esterification (forming an ester). The molecular formula \(C_4H_8O_2\) suggests Z is a carboxylic acid (e.g., butanoic acid or its isomers). The product of this reaction with propan-1-ol would be an ester (propyl butanoate or its isomers). Comparing the skeletal structures:
- S represents a different functional group (not an ester).
- T matches the structure of propyl butanoate (ester).
- U is not an ester and does not fit the expected product.
Thus, the only correct product is T, making option D correct.
Topic: 13.2
Geraniol and nerol are isomers of each other
Which type of isomerism is shown here?
▶️ Answer/Explanation
Ans: B
Geraniol and nerol are geometrical (cis/trans) isomers due to the different spatial arrangements around the double bond. In geraniol, the substituents are on opposite sides (trans configuration), while in nerol, they are on the same side (cis configuration). This difference in arrangement around the rigid \(C=C\) bond classifies them as stereoisomers under geometrical isomerism.
Topic: 13.2
Which compound has the greatest number of stereoisomers?
A. 2-methylhex-2-ene
B. 3-methylhex-2-ene
C. 4-methylhex-2-ene
D. 5-methylhex-2-ene
▶️ Answer/Explanation
Ans: C
Stereoisomers arise due to the presence of chiral centers or restricted rotation (E/Z isomerism). Here, all compounds exhibit E/Z isomerism (due to the C=C double bond at position 2), but 4-methylhex-2-ene (Option C) has an additional chiral center at carbon-4 (attached to four different groups: –H, –CH3, –C2H5, and –CH=CHCH3). This creates two stereoisomers (R/S) for each E/Z form, totaling 4 stereoisomers (22). The other options lack chiral centers, limiting them to only 2 stereoisomers (E/Z).
Topic: 14.1
Vitamin A contains retinol.
Under appropriate conditions, acidified KMnO₄ (aq) can be used to break C=C bonds. After these bonds have been broken, further oxidation of the fragments may occur. Under which conditions is the acidified KMnO₄ (aq) used and what do the final oxidation products include?
▶️ Answer/Explanation
Ans: D
1. Conditions for KMnO4 Oxidation: Acidified KMnO4 under hot and concentrated conditions cleaves C=C bonds and oxidizes the fragments further. 2. Final Products: – The terminal =CH2 groups oxidize to CO2 (a gas). – The non-terminal =CH– groups oxidize to carboxylic acids (e.g., RCOOH). – The –C(=O)– groups remain as ketones. 3. Application to Retinol: The structure of retinol (Vitamin A) contains multiple C=C bonds. Under these conditions, it would yield ketones, carboxylic acids, and CO2, matching option D.
Topic: 14.1
The structure of limonene is shown.
What are the number of moles of carbon dioxide and water produced when a sample of limonene is completely combusted in oxygen?
▶️ Answer/Explanation
Ans: B
The skeletal formula of limonene is \( \text{C}_{10}\text{H}_{16} \). The balanced combustion equation is:
\[ \text{C}_{10}\text{H}_{16} + 14\text{O}_2 \rightarrow 10\text{CO}_2 + 8\text{H}_2\text{O} \]
For 1 mole of limonene, complete combustion produces 10 moles of CO₂ and 8 moles of H₂O, corresponding to option B.
Topic: 15.1
The reaction of chlorine with methane is carried out in the presence of light. What is the function of the light?
▶️ Answer/Explanation
Ans: B
The reaction between chlorine (\( \text{Cl}_2 \)) and methane (\( \text{CH}_4 \)) is a free-radical substitution reaction. The role of light is crucial:
- Initiation Step:
- Light provides energy to break the \( \text{Cl–Cl} \) bond homolytically, forming two highly reactive chlorine radicals (\( \text{Cl}^• \)).
- This step is represented as: \( \text{Cl}_2 \xrightarrow{\text{light}} 2 \text{Cl}^• \).
- Why Not Other Options?:
- A: Light does not directly break \( \text{C–H} \) bonds; chlorine radicals do this in the propagation step.
- C: Light causes homolytic fission (forming radicals), not heterolytic fission (ions).
- D: While light may slightly heat the mixture, its primary role is to initiate the reaction by generating radicals.
Thus, the correct answer is B, as light initiates the reaction by breaking \( \text{Cl}_2 \) into chlorine atoms.
Topic: 15.1
When X is added to NaOH(aq) and heated under reflux, pentan-2-ol is made. Which organic product is made when X is heated with a solution of KCN dissolved in ethanol?
▶️ Answer/Explanation
Ans: D
Step 1: – X reacts with \( NaOH(aq) \) under reflux to form pentan-2-ol, indicating X is 2-chloropentane (\( CH_3CH_2CH_2CH(Cl)CH_3 \)). Step 2: – When 2-chloropentane reacts with \( KCN \) in ethanol, a nucleophilic substitution occurs, replacing \( Cl \) with \( CN \). – The product is 2-cyanopentane (\( CH_3CH_2CH_2CH(CN)CH_3 \)), which corresponds to Option D.
Topic: 15.1
1-chlorobutane and 1-iodobutane both react with aqueous sodium hydroxide by a nucleophilic substitution mechanism. Which reaction has the greatest rate under the same conditions and which mechanism is followed by this reaction?
▶️ Answer/Explanation
Ans: D
In nucleophilic substitution reactions:
- Reactivity: 1-iodobutane reacts faster than 1-chlorobutane because the C-I bond is weaker than the C-Cl bond (lower bond enthalpy), making it easier to break.
- Mechanism: The reaction follows an \( S_N2 \) mechanism, where the nucleophile (\( \text{OH}^- \)) attacks the carbon from the opposite side of the leaving group (I⁻ or Cl⁻), resulting in inversion of configuration.
Thus, 1-iodobutane has the greatest rate and follows the \( S_N2 \) mechanism, corresponding to option D.
Topic: 18.2
Compound Y reacts with alkaline \(I_2(aq)\). When the products of this reaction are acidified, a dicarboxylic acid is produced. The formula of the dicarboxylic acid is HOOC–R–COOH where R consists of one or more \(CH_2\) groups. Which compound is Y?
▶️ Answer/Explanation
Ans: D
The reaction with alkaline \(I_2\) oxidizes secondary alcohol groups (–CHOH–) to carboxyl groups (–COOH), forming a dicarboxylic acid. Pentan-2,4-diol (D) has two secondary alcohol groups at positions 2 and 4, which, upon oxidation, would produce HOOC–CH2–CH2–COOH (adipic acid). The other options either have primary alcohols (A, B) or would not form a linear dicarboxylic acid (C).
Topic: 18.2
Which alcohol gives only one possible oxidation product when warmed with dilute acidified potassium dichromate(VI)?
▶️ Answer/Explanation
Ans: B
Analysis of each option:
A. Butan-1-ol: Oxidizes first to butanal (aldehyde) and then to butanoic acid (carboxylic acid) → Two possible products
B. Butan-2-ol: Oxidizes only to butanone (ketone) which cannot be further oxidized → One product
C. 2-methylpropan-1-ol: Oxidizes to 2-methylpropanal and then to 2-methylpropanoic acid → Two products
D. 2-methylpropan-2-ol: Tertiary alcohol – Cannot be oxidized by K2Cr2O7
Only butan-2-ol (B) gives a single oxidation product (butanone).
Topic: 17.1
Which compound, on reaction with hydrogen cyanide, produces a compound with a chiral centre?
▶️ Answer/Explanation
Ans: A
The reaction of a carbonyl compound with HCN forms a cyanohydrin. A chiral centre is created when the product has four different groups attached to a carbon atom. Only \(CH_3CHO\) (ethanal) forms a cyanohydrin (\(CH_3CH(OH)CN\)) with a chiral centre (the carbon bonded to -OH, -CN, -CH3, and -H). The other options either do not react with HCN or form products without chiral centres.
Topic: 18.2
The diagram shows three reactions of ethanal. In each case, an excess of ethanal is used.
Observations are made after each of the three reactions. What are the colours of solution 1 and solids 2 and 3?
▶️ Answer/Explanation
Ans: B
The reactions shown are characteristic tests for aldehydes (ethanal in this case):
- Solution 1 (Fehling’s/Benedict’s test): Ethanal reduces Cu²⁺ (blue) to Cu⁺ (red precipitate of Cu₂O), leaving a colorless solution.
- Solid 2 (Tollens’ test): Ethanal reduces Ag⁺ to metallic silver, forming a grey/silver mirror.
Solid 3 (2,4-DNPH test):
- Ethanal forms a
yellow/orange precipitate
- of 2,4-dinitrophenylhydrazone.
Thus, the correct combination is B (colorless, grey, yellow).
Topic: 17.1
\((CH₃)₃CCN\) reacts to form alcohol Y via the reaction sequence shown.
Which row names the molecule X and the class of alcohol Y?
▶️ Answer/Explanation
Ans: C
The reaction sequence involves:
- Hydrolysis of nitrile \((CH₃)₃CCN\) to form carboxylic acid \((CH₃)₃CCOOH\) (X = 2,2-dimethylpropanoic acid).
- Reduction of the carboxylic acid using \(LiAlH₄\) to form alcohol Y = \((CH₃)₃CCH₂OH\) (2,2-dimethylpropan-1-ol).
Since the \(-OH\) group is attached to a primary carbon (bonded to only one alkyl group), alcohol Y is primary.
Topic: 20.1
The diagram shows a section of an addition polymer. The polymer is made using two different monomers.
What are the names of the two monomers needed to make this polymer?
A. 1,2-dichloropropene and 2-chlorobut-2-ene
B. 2,3-dichlorobut-2-ene and chloropropene
C. 1,2-dichloropropene and chloroethene
D. chloropropene and 2-chlorobut-2-ene
▶️ Answer/Explanation
Ans: A
To determine the monomers:
- Analyze the repeating unit: The polymer shows alternating units with:
- One unit: –CH2–CCl=CH– (from 1,2-dichloropropene)
- Other unit: –CH2–C(Cl)=CH–CH3 (from 2-chlorobut-2-ene)
- Verify options:
- Option A: 1,2-dichloropropene (CH2=CCl–CH2Cl) and 2-chlorobut-2-ene (CH3–C(Cl)=CH–CH3) match the polymer structure.
- Other options either mismatch the repeating units or don’t account for the alternating pattern.
Thus, the correct monomers are 1,2-dichloropropene and 2-chlorobut-2-ene (Option A).
Topic: 1.3
The diagram shows the mass spectrum of a sample of chlorine. Peaks V, W, X, Y and Z are labelled.
Which statements about this spectrum are correct?
1 The relative atomic mass of chlorine can be calculated from the abundances and m/ e values of 2 of the 5 peaks.
2 37.0 g of the species responsible for peak Z contains \(3.011 \times 10^{23}\) molecules.
3 The relative molecular mass of chlorine can be calculated from the abundances and m/ e values of peaks X, Y and Z.
▶️ Answer/Explanation
Ans: A
Statement 1: Correct. The relative atomic mass of chlorine is calculated using the isotopic peaks (V and W for \(^{35}\text{Cl}\) and \(^{37}\text{Cl}\)) and their abundances. Only these two peaks are needed.
Statement 2: Correct. Peak Z corresponds to \(\text{Cl}_2\) with \(m/e = 74\) (\(^{37}\text{Cl}-^{37}\text{Cl}\)). 37.0 g is 0.5 moles, containing \(0.5 \times 6.022 \times 10^{23} = 3.011 \times 10^{23}\) molecules.
Statement 3: Correct. The relative molecular mass of \(\text{Cl}_2\) is derived from peaks X (\(^{35}\text{Cl}-^{35}\text{Cl}\), \(m/e = 70\)), Y (\(^{35}\text{Cl}-^{37}\text{Cl}\), \(m/e = 72\)), and Z (\(^{37}\text{Cl}-^{37}\text{Cl}\), \(m/e = 74\)), weighted by their abundances.
Thus, all three statements (1, 2, and 3) are correct, making option A the answer.