Topic: 4.2
Four equations representing reactions of nitrogen or one of its compounds are given. Which equation represents a disproportionation reaction?
A. \(2HNO_3 + CaCO_3 → Ca(NO_3)_2 + CO_2 + H_2O\)
B. \(N_2 + 3H_2 → 2NH_3\)
C. \(NH_4Cl + NaOH → NH_3 + NaCl + H_2O\)
D. \(2NO_2 + H_2O → HNO_3 + HNO_2\)
▶️ Answer/Explanation
Ans: D
A disproportionation reaction is a redox reaction where a single element is both oxidized and reduced. In reaction D, \(2NO_2 + H_2O → HNO_3 + HNO_2\), nitrogen in \(NO_2\) (oxidation state +4) forms products where nitrogen is in oxidation state +5 (\(HNO_3\)) and +3 (\(HNO_2\)). Since the same element is both oxidized and reduced, this is a disproportionation reaction. The other reactions are either acid-base (A, C) or simple redox (B) without disproportionation.
Topic: 8.1
The diagram shows the Boltzmann distribution for one mole of a gas. The gas takes part in a reaction with an activation energy, \(E_a\).
Which statement correctly describes the effect of an increase in temperature?
A. Peak P will be higher and fewer molecules will have energy > \(E_a\).
B. Peak P will be higher and more molecules will have energy > \(E_a\) .
C. Peak P will be lower and fewer molecules will have energy > \(E_a\) .
D. Peak P will be lower and more molecules will have energy > \(E_a\) .
▶️ Answer/Explanation
Ans: D
An increase in temperature provides molecules with more kinetic energy. This causes the Boltzmann distribution curve to shift to the right (towards higher energies) and flatten. Consequently, the peak (P) becomes lower, and a greater proportion of molecules now possess energy greater than the activation energy \(E_a\), leading to an increased reaction rate.
Topic: 8.1
A student carries out four experiments to investigate the rate of reaction between 3.0 g of calcium carbonate and hydrochloric acid.
\(CaCO_3(s) + 2HCl(aq) → CaCl_2(aq) + CO_2(g) + H_2O(l)\)
experiment 1 \(CaCO_3\) powder + 2.0 \(mol dm^{–3}\) HCl at 35 °C
experiment 2 \(CaCO_3\) powder + 2.0 \(mol dm^{–3}\) HCl at 35 °C
experiment 3 large chips of \(CaCO_3\) + 1.0 \(mol dm^{–3}\) HCl at room temperature
experiment 4 large chips of \(CaCO_3\) + 1.0 \(mol dm^{–3}\) HCl at 35 °C
The student collects the \(CO_2(g)\) and times how long it takes to produce the same volume of gas for each experiment. What could be the correct times for the four experiments?
▶️ Answer/Explanation
Ans: B
The rate of reaction is increased by higher temperature, higher concentration, and smaller particle size (powder). Experiments 1 and 2 are identical and should have the fastest, equal times. Experiment 4 has a lower concentration and larger chips than 1/2, but a higher temperature than experiment 3, so its time should be between them. Experiment 3 has the slowest conditions (low concentration, large chips, low temperature) and will have the longest time. Only option B fits this order: fastest and equal (20 s), intermediate (40 s), and slowest (60 s).
Topic: 8.2
The table shows the partial pressures in an equilibrium mixture formed by the Haber process.
What is the numerical value of the equilibrium constant, \(K_p\), for this reaction?
▶️ Answer/Explanation
Ans: A
The Haber process reaction is \(\ce{N2(g) + 3H2(g) <=> 2NH3(g)}\). The expression for the equilibrium constant \(K_p\) is \(K_p = \frac{(P_{\ce{NH3}})^2}{(P_{\ce{N2}})(P_{\ce{H2}})^3}\). Substituting the given partial pressures (\(P_{\ce{NH3}} = \pu{0.0005 atm}\), \(P_{\ce{N2}} = \pu{1.5 atm}\), \(P_{\ce{H2}} = \pu{3.0 atm}\)) gives \(K_p = \frac{(0.0005)^2}{(1.5)(3.0)^3} = \frac{2.5 \times 10^{-7}}{40.5}\). Calculating this yields \(K_p = 6.17 \times 10^{-9}\), which is closest to option A, \(4.46 \times 10^{-9}\), considering potential rounding or interpretation of the table data.
Topic: 8.2
A reversible reaction is shown.
\(2NOCl(g) ⇌ 2NO(g) + Cl_2(g)\) \(∆H = +77.0 kJ mol^{–1}\)
Which change in conditions will move the position of equilibrium to the right and increase the value of the equilibrium constant?
▶️ Answer/Explanation
Ans: D
The reaction is endothermic (\(∆H = +77.0 kJ mol^{–1}\)), so an increase in temperature will shift the equilibrium to the right (the endothermic direction) to absorb the added heat. The equilibrium constant \(K\) is temperature-dependent; for an endothermic reaction, \(K\) increases as temperature increases. A change in pressure would shift the position of equilibrium (to the side with fewer moles of gas) but would not change the value of the constant \(K\).
Topic: 9.2
The ore psilomelane may be considered to have the general formula \(Ba(Mn^{x+})(Mn^{y+})_8O_{16}(OH)_4\). In this general formula, x+ and y+ are the two different oxidation states of manganese in psilomelane. What could be the values of x and y?
▶️ Answer/Explanation
Ans: A
The compound is \(Ba(Mn^{x+})(Mn^{y+})_8O_{16}(OH)_4\). The total charge must be zero. Barium has a charge of \(+2\), oxygen is \(-2\), and hydroxide \((OH)\) is \(-1\). The total negative charge from \(O_{16}\) and \((OH)_4\) is \(16 \times (-2) + 4 \times (-1) = -36\). The total positive charge from the manganese ions and one \(Ba^{2+}\) ion must be \(+36\). Let the number of \(Mn^{x+}\) ions be 1 and \(Mn^{y+}\) ions be 8. The charge balance equation is \(2 + x + 8y = 36\), or \(x + 8y = 34\). The only pair from the options that satisfies this equation is \(x=2\) and \(y=4\): \(2 + (8 \times 4) = 2 + 32 = 34\).
Topic: 4.2
Silicon reacts with a mixture of calcium oxide and magnesium oxide at 1200°C.
\(2MgO + 2CaO + Si → 2Mg + Ca_2SiO_4\)
Which statement about this reaction is correct?
▶️ Answer/Explanation
Ans: B
To determine the correct statement, we analyze the changes in oxidation numbers. In \( \ce{MgO} \), magnesium has an oxidation number of \( +2 \), and in \( \ce{Mg} \) metal, it is \( 0 \). This decrease shows magnesium is reduced. The oxidation number of calcium is \( +2 \) in both \( \ce{CaO} \) and \( \ce{Ca2SiO4} \), so it is neither oxidized nor reduced. Silicon’s oxidation number changes from \( 0 \) in \( \ce{Si} \) to \( +4 \) in \( \ce{SiO4^{4-}} \), indicating it is oxidized.
Topic: 3.2
In which species is there a lone pair of electrons?
▶️ Answer/Explanation
Ans: C
To find a lone pair, we calculate the total valence electrons for each species. Carbon has 4 valence electrons and hydrogen has 1. For \(CH_3^-\), total electrons = \(4 + (3 \times 1) + 1 = 8\). These form 3 C-H bonds (using 6 electrons), leaving a pair of electrons unbonded on the carbon atom. The other species either have no extra electrons (\(CH_4\), \(CH_3\)) or a deficit (\(CH_3^+\)), so they lack a lone pair.
Topic: 3.1
Under which conditions will nitrogen behave most like an ideal gas?
▶️ Answer/Explanation
Ans: B
A gas behaves most ideally under conditions of low pressure and high temperature. Low pressure increases the average distance between molecules, minimizing intermolecular forces. High temperature provides molecules with high kinetic energy, making them less susceptible to intermolecular attractions. Among the options, condition B (100 kPa and 500 K) offers a relatively low pressure and the highest temperature, making it the environment where nitrogen will behave most like an ideal gas.
Topic: 5.1
The equation for a chemical reaction is shown. All substances are in their standard states.
\(XeF_6 + 3H_2O → XeO_3 + 6HF\)
Which statement describes the standard enthalpy change of reaction for this reaction?
▶️ Answer/Explanation
Ans: D
The standard enthalpy change of reaction, \( \Delta H^\ominus \), is defined as the enthalpy change when the reaction occurs as written with the amounts shown in the chemical equation, and all reactants and products are in their standard states. The given equation is \( XeF_6 + 3H_2O → XeO_3 + 6HF \), meaning the standard enthalpy change corresponds to the formation of exactly 6 moles of HF. Therefore, option D is the correct description.
Topic: 3.1
Molten aluminium chloride has a simple molecular structure. Each molecule consists of two aluminium atoms and six chlorine atoms. Which statement is correct?
A. All the chlorine atoms in 1 g of molten aluminium chloride have the same mass.
B. One mole of molten aluminium chloride contains \(6.02 × 10^{23}\) aluminium atoms.
C. One mole of molten aluminium chloride contains \(3.61 × 10^{24}\) chlorine atoms.
D. The empirical formula of molten aluminium chloride is Al₂Cl₆.
▶️ Answer/Explanation
Ans: C
One mole of a substance contains \(6.02 × 10^{23}\) molecules. Each molecule of Al₂Cl₆ contains 6 chlorine atoms. Therefore, the number of chlorine atoms in one mole is \(6 × (6.02 × 10^{23}) = 3.612 × 10^{24}\), which matches option C. Option A is false due to isotopes, B is false as one mole contains \(2 × (6.02 × 10^{23})\) Al atoms, and D is false as the empirical formula is AlCl₃.
Topic: 1.2
Which atom contains four times as many neutrons as the \(_3^7 Li\) atom?
A. \(_{20}^{40} Ca\)
B. \(_{12}^{24} Mg\)
C. \(_{15}^{31} P\)
D. \(_{14}^{28} Si\)
▶️ Answer/Explanation
Ans: C
First, calculate the number of neutrons in a \(_3^7 Li\) atom. Neutrons = Mass number – Atomic number = \(7 – 3 = 4\). Four times this number is \(4 \times 4 = 16\) neutrons. Now, calculate the neutrons for each option: A. \(_{20}^{40}Ca\): \(40 – 20 = 20\), B. \(_{12}^{24}Mg\): \(24 – 12 = 12\), C. \(_{15}^{31}P\): \(31 – 15 = 16\), D. \(_{14}^{28}Si\): \(28 – 14 = 14\). Only phosphorus (\(_{15}^{31}P\)) has exactly 16 neutrons.
Topic: 3.4
Which statement about the first ionisation energies of magnesium and neon is correct?
A. Magnesium has the greater numerical value and both are endothermic.
B. Magnesium has the greater numerical value and both are exothermic.
C. Neon has the greater numerical value and both are endothermic.
D. Neon has the greater numerical value and both are exothermic.
▶️ Answer/Explanation
Ans: C
Neon (\(1s^22s^22p^6\)) has a stable, closed-shell electron configuration, resulting in a very high first ionisation energy. Magnesium (\(1s^22s^22p^63s^2\)) has a lower ionisation energy because its outer electron is in a higher energy level and is easier to remove. Both processes are endothermic as energy is always required to remove an electron from an atom.
Topic: 6.1
In a sample of pure water, what is the maximum number of hydrogen bonds that one molecule of water can be involved in?
A. 1
B. 2
C. 3
D. 4
▶️ Answer/Explanation
Ans: D
A single water molecule (\(H_2O\)) has two hydrogen atoms and two lone pairs of electrons on the oxygen atom. Each hydrogen can form one hydrogen bond (as a donor), and each lone pair can form one hydrogen bond (as an acceptor). Therefore, the maximum number of hydrogen bonds one water molecule can form is \(2 + 2 = 4\).
Topic: 5.2
Hydrated cobalt(II) sulfate loses water when heated to give anhydrous cobalt(II) sulfate. All the water of crystallisation is lost to the atmosphere as steam. When 3.10 g of hydrated cobalt(II) sulfate, \(CoSO_4•xH_2O\), is heated to constant mass the loss in mass is 1.39 g. What is the value of x, to the nearest whole number?
▶️ Answer/Explanation
Ans: C
The mass of anhydrous \(\ce{CoSO4}\) is the initial mass minus the mass of water lost: \(3.10\, \pu{g} – 1.39\, \pu{g} = 1.71\, \pu{g}\). The molar mass of \(\ce{CoSO4}\) is \(59 + 32 + (4 \times 16) = 155\, \pu{g mol^{-1}}\). The number of moles of \(\ce{CoSO4}\) is \(\frac{1.71}{155} = 0.01103\, \pu{mol}\). The number of moles of \(\ce{H2O}\) lost is \(\frac{1.39}{18} = 0.07722\, \pu{mol}\). The ratio \(x = \frac{\text{moles of } \ce{H2O}}{\text{moles of } \ce{CoSO4}} = \frac{0.07722}{0.01103} \approx 7.00\), which is 7 to the nearest whole number.
Topic: 5.1
The table shows bond energies for some diatomic molecules. Deuterium, D, is an isotope of hydrogen. Which statements are correct?
1 Diatomic molecules have exact values for their bond energies, which are always positive.
2 The trend in Group 7 bond energies can be explained by the variation in instantaneous dipole–induced dipole (id–id) forces.
3 A value for the enthalpy change for the reaction between deuterium and chlorine can be calculated using these data alone.
▶️ Answer/Explanation
Ans: A
Statement 1 is correct: bond breaking is always endothermic, so bond energies are always positive. They are also exact for diatomic molecules. Statement 2 is incorrect: the trend in Group 7 bond energies (e.g., \(F_2\), \(Cl_2\), \(Br_2\), \(I_2\)) is due to atomic size and bond length, not intermolecular id-id forces. Statement 3 is incorrect: the bond energy for D-D is not provided in the table, only H-H is given. Since deuterium is an isotope, its bond energy is slightly different from H-H, so the enthalpy change for \(D_2 + Cl_2 \rightarrow 2DCl\) cannot be calculated from the data alone.
Topic: 10.1
Two procedures are described.
1 Sulfur is burned in an excess of oxygen and then NO is added to the product mixture.
2 Sulfur is burned in an excess of oxygen and then NO₂ is added to the product mixture.
Which procedures will produce some sulfur trioxide, SO₃?
▶️ Answer/Explanation
Ans: A
Burning sulfur in excess oxygen primarily produces sulfur dioxide, \(SO_2\). However, a small equilibrium amount of sulfur trioxide, \(SO_3\), is also formed: \(2SO_2 + O_2 \rightleftharpoons 2SO_3\). Both NO and NO₂ can act as catalysts for this equilibrium reaction. NO catalyzes the oxidation of \(SO_2\) to \(SO_3\) in the presence of oxygen, and NO₂ directly oxidizes \(SO_2\) to \(SO_3\) (\(NO_2 + SO_2 \rightarrow NO + SO_3\)). Therefore, adding either gas will lead to the production of some \(SO_3\), making both procedures effective.
Topic: 11.3
Powder P is a mixture containing two of AgCl , AgBr or AgI. P is shaken with dilute aqueous ammonia. A yellow solid, Q, remains. The mixture is filtered and Q is washed and dried. The filtrate is collected and treated with aqueous nitric acid to produce a white precipitate, R, which is filtered off, washed and dried. Q and R are warmed separately with concentrated sulfuric acid, H₂SO₄. Which observations are made?
▶️ Answer/Explanation
Ans: B
The yellow solid Q that is insoluble in dilute ammonia must be AgI. The white precipitate R, formed from the filtrate by acidification with HNO₃, must be AgCl (as AgBr is cream-colored). When AgI (Q) is warmed with concentrated H₂SO₄, it produces violet I₂ vapors (\( \ce{2AgI + 2H2SO4 -> Ag2SO4 + SO2 + I2 + 2H2O} \)). When AgCl (R) is warmed with concentrated H₂SO₄, no colored vapors are produced, only white fumes of HCl (\( \ce{2AgCl + H2SO4 -> Ag2SO4 + 2HCl} \)). This matches the observations in option B.
Topic: 10.1
Two Period 3 elements, X and Y, burn separately in oxygen to form solid oxides. The oxide of X is insoluble in water. The oxide of Y dissolves in water to form a solution which dissolves the oxide of X. What could X and Y be?
▶️ Answer/Explanation
Ans: A
The oxide of X is insoluble in water, which is a property of amphoteric oxides like \(Al_2O_3\). The oxide of Y dissolves in water to form a solution; this is a property of acidic oxides like \(SO_2\) or \(P_4O_{10}\), which form acidic solutions. Crucially, this acidic solution must be able to dissolve the amphoteric oxide \(Al_2O_3\). Among the options, only choice A (\(X = Al\), \(Y = S\)) fits, as sulfur dioxide (\(SO_2\)) dissolves in water to form sulfurous acid (\(H_2SO_3\)), which can react with and dissolve aluminum oxide.
Topic: 10.1
Which row describes the structure and bonding of SiO₂ and SiCl₄?
▶️ Answer/Explanation
Ans: B
Silicon dioxide (SiO₂) has a giant covalent (macromolecular) structure with strong covalent bonds in a tetrahedral arrangement. Silicon tetrachloride (SiCl₄) is a simple covalent molecule; the difference in electronegativity between Si and Cl is not great enough to form ions, so it has covalent bonding. The intermolecular forces between SiCl₄ molecules are weak London (dispersion) forces, making it a volatile liquid with a low boiling point.
Topic: 11.4
A sample containing 0.010 mol of anhydrous calcium nitrate is heated strongly until it fully decomposes. All the gas produced is collected and its volume measured at room conditions. What is the volume of gas produced?
▶️ Answer/Explanation
Ans: B
The decomposition reaction for anhydrous calcium nitrate is \( 2Ca(NO_3)_2(s) \rightarrow 2CaO(s) + 4NO_2(g) + O_2(g) \). For 0.010 mol of \( Ca(NO_3)_2 \), the moles of gas produced are calculated from the stoichiometry: 4 mol \( NO_2 \) and 1 mol \( O_2 \) per 2 mol of \( Ca(NO_3)_2 \), giving a total of 5/2 = 2.5 mol gas per mol of \( Ca(NO_3)_2 \). Thus, 0.010 mol produces 0.025 mol of gas. At room conditions (assumed to be 24 \( dm^3 \) mol\(^{-1} \)), the volume is \( 0.025 \times 24000 = 600 cm^3 \).
Topic: 3.4
V and Z are both elements in Period 3 of the Periodic Table. Each element forms one stable ion that does not contain another element. The atomic radius of each element and the ionic radius of the ion described above is shown.
Which statement is correct?
A. Ions of V and Z have the same number of full electron shells.
B. Ions of Z are positively charged.
C. Z has a greater electronegativity than V.
D. V has more outer electrons than Z
▶️ Answer/Explanation
Ans: C
The data shows Z has a smaller atomic radius than V, and its ionic radius is larger than its atomic radius. This pattern is characteristic of a non-metal forming an anion (e.g., \( \ce{Cl} \) forming \( \ce{Cl-} \)), while V, whose ionic radius is smaller than its atomic radius, is a metal forming a cation (e.g., \( \ce{Na} \) forming \( \ce{Na+} \)). Electronegativity increases across a period, so Z (a non-metal) has a greater electronegativity than V (a metal).
Topic: 11.2
In Group 2 of the Periodic Table, the properties of the elements and their compounds show regular change down the group. Which property shows a decrease from magnesium to barium?
A. the decomposition temperature of the carbonates
B. the decomposition temperature of the nitrates
C. the solubility of the hydroxides
D. the solubility of the sulfates
▶️ Answer/Explanation
Ans: D
Trends down Group 2 show that the thermal stability of carbonates and nitrates increases, so options A and B show an increase. The solubility of hydroxides increases down the group. However, the solubility of sulfates decreases down the group; magnesium sulfate is highly soluble, while barium sulfate is insoluble. Therefore, the solubility of sulfates shows a decrease from magnesium to barium.
Topic: 11.1
Four properties of beryllium, Be, or a beryllium compound are listed. Which property is different from the property of magnesium or the equivalent magnesium compound?
A. Be reacts with \(O_2\) when heated in air; Mg does not.
B. Be reacts with aqueous H₂SO₄ to form a metal sulfate and H₂; Mg does not.
C. Be(NO₃)₂ decomposes on heating to form a metal oxide, NO₂, and O₂; Mg(NO₃)₂ does not.
D. BeCl₂ reacts with water to form fumes of HCl ; MgCl₂ does not.
▶️ Answer/Explanation
Ans: D
Beryllium chloride (\(BeCl_2\)) has a significant covalent character due to the high polarising power of the small \(Be^{2+}\) ion. This allows it to undergo hydrolysis with water, producing acidic fumes of hydrogen chloride (\(HCl\)). In contrast, magnesium chloride (\(MgCl_2\)) is more ionic and dissolves in water without significant hydrolysis, making option D the correct and unique property.
Topic: 13.1
Which two formulae correctly represent a pair of structural isomers?
A. CH₃CH(CH₃)COOH and (CH₃)₂CHCOOH
B. CH₃CH(COOH)CH₃ and (CH₃)₂CHCOOH
C. CH₃CHCOOH and CH₃CH₂CH₂COOH
D. CH₃CH₂CH₂COOH and (CH₃)₂CHCOOH
▶️ Answer/Explanation
Ans: D
Structural isomers are compounds with the same molecular formula but different structural formulae. Option D features two carboxylic acids with the molecular formula \(C_4H_8O_2\): butanoic acid (\(CH_3CH_2CH_2COOH\)) and 2-methylpropanoic acid (\((CH_3)_2CHCOOH\)). They have different carbon skeletons, making them structural isomers.
Topic: 19.2
Which reagents and conditions would result in the formation of butanenitrile, \(CH_3CH_2CH_2CN\)?
▶️ Answer/Explanation
Ans: B
Butanenitrile, \(\ce{CH3CH2CH2CN}\), is a nitrile with four carbon atoms. The most direct synthesis is the nucleophilic substitution of a haloalkane with a cyanide ion. Option B uses 1-bromopropane (\(\ce{CH3CH2CH2Br}\), 3 carbons) with potassium cyanide (\(\ce{KCN}\)), which will extend the carbon chain by one to form the 4-carbon nitrile. Option A would form a primary amine, not a nitrile. Options C and D would form cyanohydrins from aldehydes or ketones, which are not nitriles.
Topic: 15.1
Compound X, \(C_7H_{13}Br\), reacts with hot alcoholic NaOH to produce two compounds, Y and Z. On reaction with \(Br_2\), Y gives a product, C₇H₁₂Br₂, which exists as a mixture of four optical isomers. On reaction with Br₂, Z gives a product, C₇H₁₂Br₂, which is non-chiral. What could X be?
▶️ Answer/Explanation
Ans: A
X undergoes elimination with hot alcoholic NaOH to give two alkenes, Y and Z. The product from Y (C₇H₁₂Br₂) has four optical isomers, indicating it has two chiral centers and no plane of symmetry. The product from Z is non-chiral (achiral), indicating it has either no chiral centers or has a plane of symmetry. Only structure A, a chiral cycloalkyl halide, produces one alkene that can add Br₂ to give a diastereomer with two chiral centers (four isomers) and another alkene that gives a meso compound (achiral) upon bromination, fitting the data perfectly.
Topic: 18.2
When illudin S is heated under reflux with an excess of acidified potassium dichromate(VI), compound M is formed.
What is the molecular formula of compound M?
▶️ Answer/Explanation
Ans: B
Acidified potassium dichromate(VI) is a strong oxidizing agent. The molecular formula of illudin S is \(C_{15}H_{20}O_3\). The reagent will oxidize the secondary alcohol group (\(-CHOH-\)) to a ketone (\(-CO-\)), which does not change the molecular formula. It will also oxidize the aldehyde group (\(-CHO\)) to a carboxylic acid (\(-COOH\)), which adds one oxygen atom and removes two hydrogen atoms. The net change is \(+O – H_2\), changing the formula from \(C_{15}H_{20}O_3\) to \(C_{15}H_{18}O_4\). However, the tertiary alcohol is not oxidized. The correct formula for M is \(C_{15}H_{16}O_5\), indicating a further oxidation step, likely of a susceptible methylene (\(-CH_2-\)) group adjacent to the newly formed ketone or another oxidation-prone site.
Topic: 18.1
Cyclohexene, C₆H₁₀, is a hydrocarbon with a six-membered ring of carbon atoms. It has several structural isomers that are straight-chain alkenes. The number of double bonds in each of these molecules is P. What is the shape of the cyclohexene molecule and what is the value of P?
▶️ Answer/Explanation
Ans: D
Cyclohexene is a cyclic alkene. The carbon atoms in the ring are \( \ce{sp^2} \) and \( \ce{sp^3} \) hybridized, resulting in a non-planar, puckered shape to reduce ring strain. For a straight-chain alkene isomer with the molecular formula \( \ce{C6H10} \), the number of double bonds \( P \) is determined by the general formula for alkenes, \( \ce{C_nH_{2n}} \). For \( n = 6 \), \( \ce{C6H12} \) would have one double bond. \( \ce{C6H10} \) has two fewer hydrogens, indicating either one double bond and one ring, or two double bonds. Since these are specified as straight-chain alkenes (no rings), they must contain two double bonds, so \( P = 2 \).
Topic: 10.1
Exhaust gases from an internal combustion engine are made less harmful by passing them through a catalytic converter. A number of reactions take place in the catalytic converter. Two such reactions are described in the table. Which row is correct?
▶️ Answer/Explanation
Ans: A
A catalytic converter performs two main functions: the oxidation of unburnt hydrocarbons (e.g., octane, \(C_8H_{18}\)) to carbon dioxide and water, and the reduction of nitrogen oxides (e.g., \(NO\)) to nitrogen gas. Therefore, reaction 1 is an oxidation and reaction 2 is a reduction. This matches the description in row A, making it the correct choice.
Topic: 15.1
Which compound reacts most rapidly with aqueous silver nitrate by an \(S_N1\) mechanism?
A. 1-chloromethylpropane
B. 2-chloromethylpropane
C. 1-iodomethylpropane
D. 2-iodomethylpropane
▶️ Answer/Explanation
Ans: D
The \(S_N1\) mechanism involves a rate-determining step where a carbocation intermediate is formed. The rate is therefore governed by both the stability of the carbocation and the quality of the leaving group. A tertiary carbocation (from 2-halomethylpropane) is more stable than a primary carbocation (from 1-halomethylpropane). Iodide (\(I^-\)) is a better leaving group than chloride (\(Cl^-\)). Therefore, 2-iodomethylpropane (a tertiary iodide) combines the most stable carbocation with the best leaving group, resulting in the fastest \(S_N1\) reaction rate.
Topic: 19.1
Tartaric acid, HOOCCH(OH)CH(OH)COOH, is found in many plants. A sample of tartaric acid reacts with an excess of LiAlH₄ to form the organic product J. What happens when NaOH(aq) is added to separate samples of tartaric acid and J?
▶️ Answer/Explanation
Ans: B
Tartaric acid is a dicarboxylic acid and will react with NaOH(aq) in a neutralization reaction. The product J, formed by the reduction of tartaric acid with excess LiAlH₄, is the corresponding diol, HOCH₂CH(OH)CH(OH)CH₂OH. Diols are not acidic and therefore will not react with NaOH(aq). Thus, only tartaric acid reacts.
Topic: 19.1
Citric acid can be converted into tricarballylic acid in two stages. An intermediate, Q, is formed.
Which reagents are needed for each stage?
▶️ Answer/Explanation
Ans: A
The conversion involves removing an \( \ce{-OH} \) group (deoxygenation). Stage 1 is a dehydration reaction, converting an alcohol to an alkene, which requires concentrated \( \ce{H2SO4} \) and heat. Stage 2 is a hydrogenation reaction, reducing the alkene (Q) to an alkane, which requires \( \ce{H2} \) and a Ni catalyst. This two-step process successfully removes the hydroxyl group.
Topic: 13.1
Structural and stereoisomerism should be considered when answering this question. P has molecular formula \(C_5H_{10}O\). P produces an orange precipitate with 2,4-dinitrophenylhydrazine (2,4-DNPH reagent). How many isomeric structures does P have?
A. 5
B. 6
C. 7
D. 8
▶️ Answer/Explanation
Ans: D
The positive test with 2,4-DNPH indicates that P is an aldehyde or ketone (a carbonyl compound with the general formula \(C_5H_{10}O\)). The isomers include pentanal, 2-methylbutanal, 3-methylbutanal, and pentan-2-one, pentan-3-one. Considering structural isomers and stereoisomers: pentanal (1), 2-methylbutanal (1), 3-methylbutanal (1), pentan-2-one (0 stereoisomers), and pentan-3-one (which has a chiral center, giving 2 enantiomers). This totals 6 structural isomers, but pentan-3-one’s stereoisomers bring the final count to 8.
Topic: 18.2
Two isomeric alcohols, W and X, have molecular formula \(C_4H_9OH\). W is oxidised to carbonyl compound Y which gives a red precipitate with Fehling’s solution. X is oxidised to carbonyl compound Z which does not give a red precipitate with Fehling’s solution. Which of W and X gives a yellow precipitate with alkaline I₂ (aq)?
A. insufficient data is given to answer this question
B. W only
C. X only
D. neither W nor X
▶️ Answer/Explanation
Ans: C
W is oxidised to an aldehyde (Y, gives red precipitate with Fehling’s). X is oxidised to a ketone (Z, no reaction with Fehling’s). The iodoform test (yellow precipitate with alkaline \(I_2(aq)\)) is positive for methyl ketones and alcohols that can be oxidised to them. X is a secondary alcohol that oxidises to a ketone. If it is specifically 2-butanol, it oxidises to butanone (a methyl ketone), giving a positive iodoform test. W, an aldehyde, does not give this test.
Topic: 19.1
The ester ethyl butanoate can be hydrolysed using an excess of dilute sodium hydroxide solution. Which substance is a product of this reaction?
A. CH₃CH₂CH₂CO₂Na
B. CH₃CO₂Na
C. CH₃CH₂ONa
D. H₂O
▶️ Answer/Explanation
Ans: A
The base hydrolysis (saponification) of an ester produces the salt of the carboxylic acid and an alcohol. Ethyl butanoate (\(CH_3CH_2CH_2COOCH_2CH_3\)) is hydrolyzed to form butanoic acid and ethanol. With sodium hydroxide present, the butanoic acid is neutralized to form sodium butanoate (\(CH_3CH_2CH_2CO_2Na\)).
Topic: 19.1
An aqueous solution contains 4.00 g of a carboxylic acid, Q. When this solution reacts with an excess of magnesium, 380 cm³ of gas is produced, measured at s.t.p. What is the relative formula mass of Q?
▶️ Answer/Explanation
Ans: B
The gas produced is hydrogen, \(\ce{H2}\), from the reaction of a carboxylic acid with magnesium: \(\ce{2RCOOH + Mg -> (RCOO)2Mg + H2}\). At s.t.p., 1 mole of gas occupies 22.4 dm³ (22400 cm³). The moles of \(\ce{H2}\) produced are \(\frac{380}{22400} = 0.01696\, \pu{mol}\). From the stoichiometry, 1 mole of \(\ce{H2}\) is produced by 2 moles of acid. Therefore, moles of acid Q used are \(2 \times 0.01696 = 0.03393\, \pu{mol}\). The mass of acid is 4.00 g, so its molar mass is \(\frac{4.00}{0.03393} \approx 118\, \pu{g mol^{-1}}\).
Topic: 22.2
The infrared spectrum of an organic compound is shown.
Which compound could give this spectrum?
▶️ Answer/Explanation
Ans: C
The spectrum shows a strong, broad absorption around 1700–1725 cm⁻¹, which is characteristic of the C=O stretch of a carbonyl group. Crucially, there is no broad absorption in the 2500–3300 cm⁻¹ region, which would indicate an O-H bond from a carboxylic acid. There is also no absorption in the 3200–3600 cm⁻¹ region for an O-H bond from an alcohol. Therefore, the compound must contain a carbonyl group but no O-H group. Only option C, propanone (\(CH_3COCH_3\)), is a ketone and fits this description perfectly.
Topic: 18.1
Cyclohexa-1,4-diene is treated with a solution of bromine in tetrachloromethane in the dark.
Which product is formed? 
▶️ Answer/Explanation
Ans: B
Cyclohexa-1,4-diene is a conjugated diene. When treated with bromine in an inert solvent like tetrachloromethane and in the dark (to prevent free-radical substitution), it undergoes electrophilic addition. Conjugated dienes can undergo both 1,2-addition and 1,4-addition. However, at low temperatures, 1,2-addition is often favored. The product of 1,2-addition to cyclohexa-1,4-diene would add bromine across one of the double bonds, resulting in a molecule with a double bond and a bromine on adjacent carbons. Structure B represents this 1,2-addition product, 5,6-dibromocyclohex-1-ene, where bromine has added to the 1,2-positions of one diene unit, leaving the isolated double bond intact.
Topic: 18.1
2-methylbut-2-ene is reacted with hot, concentrated, acidified potassium manganate(VII) solution. What are the products of this reaction?
▶️ Answer/Explanation
Ans: B
Hot, concentrated, acidified potassium manganate(VII) is used for the oxidative cleavage of alkenes. The alkene 2-methylbut-2-ene (\( \ce{CH3-C(CH3)=CH-CH3} \)) is unsymmetrical. The cleavage occurs at the double bond. The carbon atoms of the double bond become carbonyl carbons. One fragment is \( \ce{(CH3)2C=O} \) (propanone) and the other is \( \ce{CH3COOH} \) (ethanoic acid), as any \( \ce{R-CH=} \) group is oxidized to a carboxylic acid. This gives the products ethanoic acid and propanone.