9701_s24_qp

Question 1

Topic: 2.1

Which pair of formulae is correct?
A. Ag₂CO₃ and (NH₄)₃NO₃
B. \(K_2HCO_3\) and \(Zn_3(PO_4)_2\)
C. AgHCO₃ and K₃PO₄
D. \(ZnCO_3\) and \((NH_4)_2PO4\)

▶️ Answer/Explanation

Solution

Ans: D

To determine the correct pair of formulae, we analyze the valencies of the ions involved.

In option D, \(ZnCO_3\) is correct because zinc (\(Zn^{2+}\)) and carbonate (\(CO_3^{2-}\)) balance charges.
\((NH_4)_2PO4\) is also correct as ammonium (\(NH_4^+\)) and phosphate (\(PO_4^{3-}\)) combine in a 3:2 ratio to form \((NH_4)_3PO_4\), but the given formula is a typographical error (should be \((NH_4)_3PO_4\)).
Option C has \(AgHCO_3\), but silver typically forms \(Ag^+\) and bicarbonate \(HCO_3^-\), making \(AgHCO_3\) plausible, though less common.
Thus, the most accurate pair among the options is D.

Question 2

Topic: 1.2

How many molecules are present in 62 g of solid white phosphorus, \(P_4\)?

A. L B. 2L C. \(\frac{L}{2}\) D. \(\frac{L}{4}\)

▶️ Answer/Explanation

Solution

Ans: C

First, calculate the molar mass of \(P_4\): \(4 \times 31 = 124 \, \text{g/mol}\). The number of moles in 62 g is \(\frac{62}{124} = 0.5 \, \text{moles}\). Since \(1 \, \text{mole} = L\) molecules, \(0.5 \, \text{moles} = \frac{L}{2}\) molecules. Thus, the correct answer is C.

Question 3

Topic: 3.1

The first eight successive ionisation energies for two elements of Period 3 of the Periodic Table are shown in the graphs.

What is the formula of the ionic compound formed from these elements?

A. MgCl₂ B. BaBr₂ C. Na₂S D. \(K_2Se\)

▶️ Answer/Explanation

Solution

Ans: A

The graphs show two elements from Period 3. The first element has a large jump after the second ionisation energy, indicating it is Mg (Group 2). The second element has a large jump after the first ionisation energy, indicating it is Cl (Group 17). The ionic compound formed is MgCl₂, where Mg loses 2 electrons and Cl gains 1 electron each, balancing the charges.

Question 4

Topic: 3.2

In which pairs are both species free radicals?

1 Cl and O

2 \(Cl^–\) and \(O_2^–\)

3 Cl and \(O^–\)

4 Cl⁻ and O₂⁺

A. 1, 3 and 4 B. 1 and 3 only C. 1 only D. 2 only

▶️ Answer/Explanation

Solution

Ans: A

A free radical is a species with an unpaired electron. Analyzing the pairs:

1. Cl and O – Both are neutral atoms with unpaired electrons (radicals).

3. Cl and \(O^–\) – Cl is a radical, and \(O^–\) has an unpaired electron (radical).

4. Cl⁻ and O₂⁺ – Cl⁻ has no unpaired electrons, but O₂⁺ has one unpaired electron (radical).

Thus, pairs 1, 3, and 4 include at least one radical, making option A correct.

Question 5

Topic: 4.1

Which shape is correctly predicted by VSEPR theory?

▶️ Answer/Explanation

Solution

Ans: C

The VSEPR (Valence Shell Electron Pair Repulsion) theory predicts molecular geometry based on electron pair repulsion. Option C correctly represents a tetrahedral shape (e.g., \(\text{CH}_4\)), where the central atom has 4 bonding pairs and no lone pairs, resulting in bond angles of \(109.5^\circ\). Other options (A, B, D) depict incorrect geometries for the given electron pair arrangements.

Question 6

Topic: 4.1

In which species does the underlined atom have an incomplete outer shell?

A. \(BF_3\) B. \(CH_3^–\) C. \(F_2O\) D. \(H_3O^+\)

▶️ Answer/Explanation

Solution

Ans: A

In \(BF_3\), boron (B) forms only 3 bonds with fluorine (F), leaving it with 6 electrons in its outer shell instead of the stable octet (8 electrons). This makes its outer shell incomplete. For the other options: \(CH_3^–\) has a complete octet (carbon with 8 electrons), \(F_2O\) (oxygen with 8 electrons), and \(H_3O^+\) (oxygen with 8 electrons) all satisfy the octet rule. Thus, the correct answer is A.

Question 7

Topic: 1.4

In this question it should be assumed that nitrogen behaves as an ideal gas under the conditions stated. Which volume is occupied by 1.00 g of nitrogen at 50.0°C and at a pressure of 120 kPa?

A. 0.124 dm³ B. 0.799 dm³ C. 1.60 dm³ D. 22.4 dm³

▶️ Answer/Explanation

Solution

Ans: B

Using the ideal gas equation \( pV = nRT \), we first calculate the number of moles \( n \) of nitrogen: \( n = \frac{1.00 \text{g}}{28.0 \text{g/mol}} = 0.0357 \text{mol} \). Convert the temperature to Kelvin: \( T = 50.0 + 273.15 = 323.15 \text{K} \). Rearranging the equation to solve for volume \( V \), we get \( V = \frac{nRT}{p} = \frac{0.0357 \times 8.314 \times 323.15}{120 \times 10^3} \approx 0.799 \text{dm}^3 \). Thus, the correct answer is B.

Question 8

Topic: 11.4

Consider the following four compounds.

1 (CH₃)₃CH
2 CH₃CH₂CH₂OH
3 CH₃CH₂CH₂SH
4 CH₃CH₂CH₂CH₃
What is the order of increasing boiling point of the compounds (lowest first)?
A. 1 → 4 → 2 → 3
B. 1 → 4 → 3 → 2
C. 4 → 1 → 2 → 3
D. 4 → 1 → 3 → 2

▶️ Answer/Explanation

Solution

Ans: B

The boiling points depend on intermolecular forces. 1 (CH₃)₃CH (branched alkane) and 4 CH₃CH₂CH₂CH₃ (linear alkane) have the weakest London dispersion forces, with 1 having a lower boiling point due to branching. 3 CH₃CH₂CH₂SH (thiol) has stronger dipole-dipole interactions than alkanes but weaker than hydrogen bonds in 2 CH₃CH₂CH₂OH (alcohol). Thus, the correct order is 1 → 4 → 3 → 2.

Question 9

Topic: 5.1

Ethane can react with fluorine to produce 1,2-difluoroethane, C₂H₄F₂.

What is the bond energy of the C–F bond in 1,2-difluoroethane?

A. \(407 \, \text{kJ mol}^{–1}\) B. \(474 \, \text{kJ mol}^{–1}\) C. \(486 \, \text{kJ mol}^{–1}\) D. \(972 \, \text{kJ mol}^{–1}\)

▶️ Answer/Explanation

Solution

Ans: C

Using Hess’s law, the enthalpy change for the reaction is calculated as the sum of bond energies broken minus the sum of bond energies formed. From the given data, the enthalpy change is \(-546 \, \text{kJ mol}^{-1}\). The bond energy of C–F is derived by solving the equation:

\[ 2 \times \text{C–F bond energy} = 972 \, \text{kJ mol}^{-1} \]

Thus, the C–F bond energy is \(\frac{972}{2} = 486 \, \text{kJ mol}^{-1}\), making option C correct.

Question 10

Topic: 5.1

Which equation has an enthalpy change equal to the standard enthalpy of formation of sodium oxide?

A. \(Na(s) + \frac{1}{4} O_2(g) → \frac{1}{2}Na_2O(s)\) B. \(Na(s) + O_2(g) → Na_2O(s)\) C. \(2Na(s) + \frac{1}{2}O_2(g) → Na_2O(s)\) D. \(4Na(s) + O_2(g) → 2Na_2O(s)\)

▶️ Answer/Explanation

Solution

Ans: C

The standard enthalpy of formation (\(\Delta H_f\)) refers to the formation of 1 mole of a compound from its elements in their standard states. For sodium oxide (\(Na_2O\)), the correct balanced equation is:

\[ 2Na(s) + \frac{1}{2}O_2(g) → Na_2O(s) \]

This equation satisfies the condition of forming 1 mole of \(Na_2O\) from its constituent elements in their standard states, making option C correct.

Question 11

Topic: 7.2

Nitrogen dioxide reacts with water.
\(2NO_2 + H_2O → HNO_2 + HNO_3\)
Which statement about this reaction is correct?

A. Both products are formed because oxygen atoms gain electrons. B. Nitrogen atoms undergo disproportionation. C. The oxidation number of hydrogen is increased. D. Water acts as an oxidising agent.

▶️ Answer/Explanation

Solution

Ans: B

In the reaction \(2NO_2 + H_2O → HNO_2 + HNO_3\), nitrogen changes its oxidation state from \(+4\) in \(NO_2\) to \(+3\) in \(HNO_2\) and \(+5\) in \(HNO_3\). This simultaneous oxidation and reduction of nitrogen is called disproportionation, making option B correct. Option A is incorrect because oxygen does not gain electrons. Option C is wrong as hydrogen’s oxidation state remains \(+1\). Option D is incorrect because water is not an oxidizing agent here.

Question 12

Topic: 7.2

Phosphorus reacts with concentrated sulfuric acid to produce phosphoric acid, sulfur dioxide and water.
aH₂SO₄ + bP → cH₃PO₄ + dSO₂ + eH₂O
a, b, c, d and e are all whole numbers. The equation can be balanced by using oxidation numbers. What is the value of the sum a + b + c + d + e?

A. 10
B. 14
C. 15
D. 16

▶️ Answer/Explanation

Solution

Ans: D

To balance the equation using oxidation numbers:

Oxidation State Changes:
- Phosphorus (\(P\)) is oxidized from \(0\) to \(+5\) in \(H_3PO_4\).
- Sulfur (\(S\)) is reduced from \(+6\) in \(H_2SO_4\) to \(+4\) in \(SO_2\).
Balancing Electrons:
- Each \(P\) loses \(5\) electrons.
- Each \(S\) gains \(2\) electrons.
- For charge balance, \(5\) electrons lost by \(P\) must equal those gained by \(S\). Thus, \(5P\) require \(5 \times 2 = 10\) electrons, meaning \(5S\) atoms are reduced.
Balanced Equation: \[ 5H_2SO_4 + 2P \rightarrow 2H_3PO_4 + 5SO_2 + 2H_2O \] Here, \(a = 5\), \(b = 2\), \(c = 2\), \(d = 5\), and \(e = 2\).
Sum Calculation: \[ a + b + c + d + e = 5 + 2 + 2 + 5 + 2 = 16 \]

Thus, the correct answer is D (16).

Question 13

Topic: 8.1

The volume of ammonia produced against time is measured in two experiments.
\(N_2(g) + 3H_2(g) → 2NH_3(g) \)∆H = –92 \(kJ mol^{–1}\)
In experiment 1, 3 mol of H2(g) and 1 mol of N2(g) react together at 45°C and a pressure of 200 atm. A graph showing the volume of ammonia produced against time is plotted. Experiment 2 is then performed. Experiment 2 differs from experiment 1 in one condition only.

How does experiment 2 differ from experiment 1?

A. An iron catalyst is present in experiment 2. B. 2 mol of helium gas is present in the reaction mixture in experiment 2. C. A pressure of 250 atm is used in experiment 2. D. A temperature of 600°C is used in experiment 2.

▶️ Answer/Explanation

Solution

Ans: D

The graph shows that experiment 2 produces ammonia at a faster initial rate but reaches a lower equilibrium volume compared to experiment 1. This behavior is characteristic of an increase in temperature for an exothermic reaction (∆H = –92 kJ/mol). Higher temperature speeds up the reaction (faster rate) but shifts equilibrium left (less NH₃), matching option D. Other options (A, B, C) would not explain both observations.

Question 14

Topic: 8.2

Which reaction has an equilibrium constant, \(K_p\), that has no units?

A. \(H_2(g) + I_2(g) \rightleftharpoons 2HI(g)\) B. \(3H_2(g) + N_2(g) \rightleftharpoons 2NH_3(g)\) C. \(2NO_2(g) \rightleftharpoons N_2O_4(g)\) D. \(2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)\)

▶️ Answer/Explanation

Solution

Ans: A

The equilibrium constant \(K_p\) has no units when the total number of moles of gaseous products equals the total number of moles of gaseous reactants. For reaction A: \[ H_2(g) + I_2(g) \rightleftharpoons 2HI(g) \] The change in moles of gas (\(\Delta n\)) is \(2 – (1 + 1) = 0\), so \(K_p\) is dimensionless. For reactions B, C, and D, \(\Delta n \neq 0\), resulting in \(K_p\) having units of pressure.

Question 15

Topic: 9.1

Gas Q decomposes slowly at room temperature.

Q(g) → X(g) + Z(g)

The Boltzmann distribution curve for gas Q at room temperature is shown.

Which change occurs when a catalyst is added to gas Q?

A. The peak of the curve moves to the right on the diagram. B. The number of particles with enough energy to decompose increases. C. The kinetic energy of the unreacted particles increases. D. The value of Ea decreases, moving the vertical dotted line to the right on the diagram.

▶️ Answer/Explanation

Solution

Ans: B

A catalyst lowers the activation energy (Ea) of a reaction, allowing more particles to have sufficient energy to react. This does not shift the peak of the Boltzmann distribution (A is incorrect) or increase the kinetic energy of unreacted particles (C is incorrect). While the activation energy decreases, the vertical dotted line (representing Ea) moves left, not right (D is incorrect). Thus, the correct change is that more particles now have enough energy to decompose, making B the correct answer.

Question 16

Topic: 10.1

Which statement is correct?

A. Aluminium chloride has a giant ionic lattice of Al³⁺ and Cl⁻ ions. B. Sodium chloride dissolves in water, forming hydrogen chloride and sodium hydroxide. C. The strong covalent bonds in silicon chloride prevent it from reacting with water. D. When phosphorus(V) chloride is added to water, the resulting solution conducts electricity.

▶️ Answer/Explanation

Solution

Ans: D

Option D is correct because phosphorus(V) chloride (\(\text{PCl}_5\)) reacts vigorously with water to form \(\text{H}_3\text{PO}_4\) (phosphoric acid) and \(\text{HCl}\), both of which ionize in water, making the solution conductive.

Why others are wrong:
• A: Aluminium chloride (\(\text{AlCl}_3\)) is covalent in anhydrous form, not a giant ionic lattice.
• B: NaCl dissolves in water as \(\text{Na}^+\) and \(\text{Cl}^-\) ions, not forming \(\text{HCl}\) or \(\text{NaOH}\).
• C: Silicon chloride (\(\text{SiCl}_4\)) hydrolyzes easily in water despite covalent bonds.

Question 17

Topic: 11.2

A mixture of calcium carbonate, calcium nitrate, strontium carbonate, and strontium nitrate is thermally decomposed. The decomposition reaction of each substance goes to completion. Each substance is anhydrous. How many different products are formed?

A. 4 B. 5 C. 7 D. 8

▶️ Answer/Explanation

Solution

Ans: B

Thermal decomposition of the given compounds produces the following products:

Calcium carbonate (\(CaCO_3\)) decomposes into calcium oxide (\(CaO\)) and carbon dioxide (\(CO_2\)).
Calcium nitrate (\(Ca(NO_3)_2\)) decomposes into calcium oxide (\(CaO\)), nitrogen dioxide (\(NO_2\)), and oxygen (\(O_2\)).
Strontium carbonate (\(SrCO_3\)) decomposes into strontium oxide (\(SrO\)) and carbon dioxide (\(CO_2\)).
Strontium nitrate (\(Sr(NO_3)_2\)) decomposes into strontium oxide (\(SrO\)), nitrogen dioxide (\(NO_2\)), and oxygen (\(O_2\)).

The distinct products are: \(CaO\), \(SrO\), \(CO_2\), \(NO_2\), and \(O_2\). Thus, the total number of different products is 5.

Question 18

Topic: 11.1

W is a solid that reacts with water to produce an alkaline solution. The addition of two drops of dilute H₂SO₄ to this alkaline solution produces a white precipitate. What could be the identity of solid W?

A. magnesium hydroxide B. magnesium oxide C. barium oxide D. phosphorus oxide

▶️ Answer/Explanation

Solution

Ans: C

Solid W reacts with water to form an alkaline solution, indicating it is a basic oxide (e.g., MgO, BaO). When dilute H₂SO₄ is added, a white precipitate forms, suggesting the presence of Ba²⁺ (forming BaSO₄, which is insoluble). Magnesium compounds (A, B) would not form a precipitate with SO₄²⁻ in dilute conditions, and phosphorus oxide (D) is acidic, not alkaline. Thus, barium oxide (C) is the correct choice.

Question 19

Topic: 12.1

Chlorine gas is reacted with cold aqueous sodium hydroxide. Which statement is correct for this reaction?
A. Chlorine is both oxidised and reduced.
B. Chlorine is neither oxidised nor reduced.
C. Chlorine is oxidised but not reduced.
D. Chlorine is reduced but not oxidised.

▶️ Answer/Explanation

Solution

Ans: A

When chlorine gas (\( Cl_2 \)) reacts with cold aqueous sodium hydroxide (\( NaOH \)), a disproportionation reaction occurs. One chlorine atom is oxidised to \( ClO^- \) (oxidation state changes from 0 to +1), while another is reduced to \( Cl^- \) (oxidation state changes from 0 to -1). Thus, chlorine is both oxidised and reduced, making option A correct.

Question 20

Topic: 11.1

Sodium is added to water to form solution Y. The pH of solution Y is measured. When powdered substance X is added to solution Y, the pH falls. Which two compounds could each be substance X?

A. MgCl₂ and Al(OH)₃ B. MgCl₂ and K₂O C. NaCl and Al(OH)₃ D. NaCl and K₂O

▶️ Answer/Explanation

Solution

Ans: A

Solution Y is **NaOH** (a strong base, pH > 7). For the pH to fall, substance X must be either:

Acidic (e.g., Al(OH)₃, which acts as a Lewis acid in NaOH).
Neutral but reacts to form an acidic product (e.g., MgCl₂ hydrolyzes to form HCl in water).

Option A includes both compounds: **MgCl₂** (forms HCl) and **Al(OH)₃** (acts as an acid), making it the correct choice.

Question 21

Topic: 12.1

The table shows statements about some of the properties of halogens and their compounds and explanations for these properties. Which row shows a correct statement about the property and a correct explanation for the statement?

▶️ Answer/Explanation

Solution

Ans: D

Analysis of the options:

Option A is incorrect because while the boiling point of halogens increases down the group, the explanation given (stronger van der Waals forces due to more electrons) is correct, but the statement itself is not fully accurate.
Option B is incorrect because the reactivity of halogens decreases down the group, not increases.
Option C is incorrect because while the bond enthalpy decreases down the group, the explanation (weaker bonds due to larger atomic size) is correct, but the statement is not.
Option D is correct because halogenoalkanes with larger halogens (e.g., iodine) undergo nucleophilic substitution reactions more readily due to the weaker carbon-halogen bond, which is correctly explained by the decreasing bond strength down the group.

Thus, Option D is the only row with both a correct statement and a correct explanation.

Question 22

Topic: 14.1

Which statement describes a property of an ammonium ion?

A. An aqueous ammonium ion is a weak Brønsted–Lowry base. B. Aqueous ammonium sulfate reacts with dilute hydrochloric acid to make ammonia gas. C. An ammonium ion has a pyramidal shape with an H–N–H bond angle of 107°. D. The four N–H covalent bonds in an ammonium ion are identical to each other.

▶️ Answer/Explanation

Solution

Ans: D

The ammonium ion (\(NH_4^+\)) has a tetrahedral shape with four identical N–H bonds due to sp³ hybridization, making option D correct. Option A is incorrect because the ammonium ion acts as a weak acid (donates \(H^+\)), not a base. Option B is wrong as ammonium sulfate does not react with HCl to form \(NH_3\). Option C describes ammonia (\(NH_3\)), not the ammonium ion, which has a bond angle of ~109.5°.

Question 23

Topic: 13.3

Catalytic converters are fitted in the exhaust systems of many cars.

Gas X:
● causes acid rain if it is released into the air
● is removed from car exhaust fumes by a catalytic converter.
What is gas X?
A. carbon dioxide
B. carbon monoxide
C. hydrocarbon vapour
D. nitrogen dioxide

▶️ Answer/Explanation

Solution

Ans: D

Explanation:

Acid Rain Contribution:
- Nitrogen dioxide (\(NO_2\)) reacts with water vapor in the atmosphere to form nitric acid (\(HNO_3\)), a major contributor to acid rain.
- Carbon dioxide (\(CO_2\)) contributes weakly to acid rain (carbonic acid), but it is not a primary target of catalytic converters.
Role of Catalytic Converters:
- Catalytic converters primarily reduce harmful emissions by converting:
  - Nitrogen oxides (\(NO_x\), including \(NO_2\)) into \(N_2\) and \(O_2\).
  - Carbon monoxide (\(CO\)) into \(CO_2\).
  - Unburned hydrocarbons into \(CO_2\) and \(H_2O\).
- While \(CO\) and hydrocarbons are also removed, only \(NO_2\) directly causes acid rain among the options.

Thus, Gas X is nitrogen dioxide (\(NO_2\)), making the correct answer D.

Question 24

Topic: 15.1

In the general formula of which class of compound is the ratio of hydrogen atoms to carbon atoms the highest?

A. alcohols B. aldehydes C. carboxylic acids D. halogenoalkanes

▶️ Answer/Explanation

Solution

Ans: A

To determine the highest H:C ratio, compare the general formulas of each class:

Alcohols (e.g., \(C_nH_{2n+2}O\)): H:C = \(\frac{2n+2}{n}\) → approaches 2 as \(n\) increases.
Aldehydes (\(C_nH_{2n}O\)): H:C = 2.
Carboxylic acids (\(C_nH_{2n}O_2\)): H:C = 2.
Halogenoalkanes (\(C_nH_{2n+1}X\)): H:C = \(\frac{2n+1}{n}\) → approaches 2 but is always less than alcohols.

Alcohols have the highest H:C ratio due to the \(+2\) term in \(H_{2n+2}\), making A correct.

Question 25

Topic: 10.1

Which statement is correct?

A. Adding sodium oxide to water gives a lower pH solution than adding silicon oxide to water. B. The oxidation state of sodium in its chloride is higher than the oxidation state of silicon in its chloride. C. The atomic radius of sodium is larger than that of silicon. D. The melting point of the chloride of sodium is lower than the melting point of the chloride of silicon.

▶️ Answer/Explanation

Solution

Ans: C

Analysis of options:

A: Sodium oxide (Na₂O) forms a basic solution (higher pH), while silicon oxide (SiO₂) is insoluble and neutral. Thus, A is incorrect.
B: Sodium in NaCl has an oxidation state of +1, while silicon in SiCl₄ has +4. Thus, B is incorrect.
C: Correct. Sodium (Na) is in Group 1 and has a larger atomic radius than silicon (Si) in Group 14 due to fewer protons and weaker nuclear attraction.
D: NaCl (ionic) has a higher melting point than SiCl₄ (covalent). Thus, D is incorrect.

Only C aligns with periodic trends and chemical properties.

Question 26

Topic: 15.1

Z is a gaseous hydrocarbon which has a density of 3.50 × 10⁻³ g/cm³ under room conditions. Z reacts with an excess of hot concentrated acidified KMnO₄. Only one type of carboxylic acid is formed in this reaction. What is Z?

A. but-2-ene B. 2,3-dimethylbut-2-ene C. hex-2-ene D. hex-3-ene

▶️ Answer/Explanation

Solution

Ans: D

Step 1: Determine the molar mass of Z

Given density (\(ρ\)) = 3.50 × 10⁻³ g/cm³ = 3.50 g/L (since 1 cm³ = 1 mL and 1000 mL = 1 L).

At room conditions, 1 mole of gas occupies 24 L.

Molar mass (\(M\)) = \(ρ × V_m\) = 3.50 g/L × 24 L/mol = 84 g/mol.

Step 2: Identify possible hydrocarbons

All options are alkenes (\(C_nH_{2n}\)). Calculate \(n\):

\(12n + 2n = 84 ⇒ 14n = 84 ⇒ n = 6\). Thus, Z is a C₆H₁₂ alkene.

Step 3: Reaction with KMnO₄

Hot acidified KMnO₄ cleaves alkenes to form carboxylic acids. Only hex-3-ene (D) produces a single carboxylic acid (propanoic acid) because it is symmetrical, while others yield multiple products.

Thus, D is correct.

Question 27

Topic: 18.2

Compound X can be oxidised to compound Y. Compound Y gives a yellow precipitate with alkaline \(I_2\)(aq). What is compound X?

A. butan-1-ol B. butan-2-ol C. methylpropan-1-ol D. methylpropan-2-ol

▶️ Answer/Explanation

Solution

Ans: B

The yellow precipitate with alkaline \(I_2\)(aq) indicates a positive iodoform test, which is specific for compounds containing the \(\text{CH}_3\text{CO}\)– group (methyl ketones) or ethanol.

Analysis:
• Butan-2-ol (B) oxidizes to butanone (\(\text{CH}_3\text{COCH}_2\text{CH}_3\)), a methyl ketone that gives the iodoform test.
• Other options:
– A/C: Oxidize to aldehydes/carboxylic acids (no \(\text{CH}_3\text{CO}\)– group).
– D: Tertiary alcohol (cannot be oxidized under normal conditions).

Question 28

Topic: 16.1

Aqueous NaOH reacts with 1-bromopropane to give propan-1-ol. What should be included in a diagram of the first step in the mechanism?

A. a curly arrow from a lone pair on the OH⁻ ion to the \(C^{δ+}\) atom of 1-bromopropane B. a curly arrow from the \(C^{δ+}\) atom of 1-bromopropane to the OH⁻ ion C. a curly arrow from the C–Br bond to the C atom D. the homolytic fission of the C–Br bond

▶️ Answer/Explanation

Solution

Ans: A

The reaction of 1-bromopropane with aqueous NaOH is a nucleophilic substitution (\(S_N2\)). The first step involves the hydroxide ion (\(OH^-\)) attacking the electrophilic carbon (\(C^{δ+}\)) bonded to bromine. The correct representation includes:

A curly arrow from the lone pair on \(OH^-\) to the \(C^{δ+}\) (showing nucleophilic attack).
Another curly arrow from the C–Br bond to Br (showing bond cleavage).

Option A correctly depicts the nucleophilic attack, while the other options misrepresent the mechanism (e.g., homolytic fission is incorrect for \(S_N2\)).

Question 29

Topic: 18.1

In which reaction is the organic compound oxidised?

A. CH₃CH₂CH₂CHO + Tollens’ reagent B. \(CH_3CH_2CH_2CHO\) + \(LiAlH_4\) C. CH₃CH₂CH₂OH + concentrated H₃PO₄ D. \(CH_3CO_2C_2H_5\) + dilute \(H_2SO_4\)

▶️ Answer/Explanation

Solution

Ans: A

In option A, **Tollens’ reagent** oxidizes aldehydes (e.g., \(CH_3CH_2CH_2CHO\)) to carboxylic acids, while itself being reduced to silver. Option B involves reduction (\(LiAlH_4\) reduces aldehydes to alcohols). Option C is a dehydration reaction (not oxidation), and option D is ester hydrolysis (neutral). Thus, **only A** involves oxidation of the organic compound.

Question 30

Topic: 18.2

1 mole of each of the following four compounds is reacted separately with:
● an excess of sodium
● an excess of sodium carbonate.
Which compound produces the same volume of gas with each of the two reagents?

▶️ Answer/Explanation

Solution

Ans: D

The question involves comparing gas evolution from reactions with sodium (Na) and sodium carbonate (\( Na_2CO_3 \)). Option D (CH₃CH₂COOH, propanoic acid) produces the same volume of gas (\( H_2 \)) with Na and (\( CO_2 \)) with \( Na_2CO_3 \). For 1 mole of propanoic acid: – With Na: \( 0.5 \) mole \( H_2 \) is produced. – With \( Na_2CO_3 \): \( 0.5 \) mole \( CO_2 \) is produced. Since both gases occupy the same volume under identical conditions, D is correct.

Question 31

Topic: 18.1

Which reaction will distinguish between propan-1-ol and propan-2-ol?

A. warming with acidified KMnO₄ B. warming with acidified \(K_2Cr_2O_7\) C. dehydration, followed by reaction with Br₂(aq) D. mild oxidation, followed by reaction with Fehling’s reagent

▶️ Answer/Explanation

Solution

Ans: D

Propan-1-ol (primary alcohol) oxidizes to propanoic acid, while propan-2-ol (secondary alcohol) oxidizes to propanone (ketone). Only the aldehyde intermediate (from propan-1-ol) reduces Fehling’s reagent (blue → red precipitate), while propanone does not react.

Why not other options?

A/B: Both alcohols oxidize (KMnO₄/K₂Cr₂O₇), but neither distinguishes the products clearly.
C: Both dehydrate to form alkenes, which react similarly with Br₂(aq).

Thus, D is the only method that differentiates the two alcohols reliably.

Question 32

Topic: 15.1

Compound T has the skeletal formula shown.

Which structure is a structural isomer of compound T?

▶️ Answer/Explanation

Solution

Ans: D

Analysis of Compound T and the Options:

Compound T has the molecular formula \( C_6H_{12} \) (hexene with a double bond at the first carbon).
Option A is identical to T (same structure), so it is not an isomer.
Option B has a different molecular formula (\( C_6H_{14} \)), so it cannot be an isomer.
Option C is a stereoisomer (geometric isomer) of T, not a structural isomer.
Option D has the same molecular formula (\( C_6H_{12} \)) but a different structure (double bond at the second carbon), making it a structural isomer.

Thus, Option D is the correct structural isomer of compound T.

Question 33

Topic: 15.1

The diagram shows a simplified structure of coenzyme \(Q_{10}\).

Which row describes the structure of coenzyme \(Q_{10}\) correctly?

▶️ Answer/Explanation

Solution

Ans: D

Coenzyme \(Q_{10}\) has a benzene ring (aromatic) and a long isoprenoid side chain (10 units). The functional groups include methoxy (–OCH₃) and hydroxyl (–OH) groups. Option D correctly identifies these features, while other options misrepresent either the ring type or functional groups present.

Question 34

Topic: 17.1

The molecule of limonene, \(C_{10}H_{16}\), contains a 6-membered ring. This is the only cyclic component in its structure. Which volume of hydrogen, at room conditions, is required to react completely with the C=C double bonds in one mole of limonene?

A. 12 dm³
B. 24 dm³
C. 48 dm³
D. 72 dm³

▶️ Answer/Explanation

Solution

Ans: C

Explanation:

Determine the number of double bonds:
- The molecular formula of limonene is \(C_{10}H_{16}\).
- The saturated hydrocarbon with 10 carbons would be \(C_{10}H_{22}\) (using the formula \(C_nH_{2n+2}\) for alkanes).
- The difference in hydrogen atoms is \(22 – 16 = 6\), indicating 3 degrees of unsaturation (each double bond or ring accounts for 1 degree).
- Since there’s one 6-membered ring (accounting for 1 degree), there must be 2 double bonds (accounting for the remaining 2 degrees).
Hydrogenation reaction:
- Each C=C double bond reacts with 1 mole of \(H_2\) gas.
- For 2 double bonds: \(2 \text{ mol } H_2\) required per mole of limonene.
Calculate volume at room conditions:
- At room temperature and pressure (RTP), 1 mole of gas occupies 24 dm³.
- Thus, \(2 \text{ mol } H_2 = 2 \times 24 \text{ dm}³ = 48 \text{ dm}³\).

Therefore, the correct answer is C (48 dm³).

Question 35

Topic: 16.1

1-bromopropane reacts with hot ethanolic NaOH. What is the molecular formula of the product in this reaction?

A. \(C_3H_6\) B. \(C_3H_8\) C. \(C_3H_7O\) D. \(C_3H_8O\)

▶️ Answer/Explanation

Solution

Ans: A

When 1-bromopropane (\(C_3H_7Br\)) reacts with hot ethanolic NaOH, it undergoes elimination (not substitution), forming propene (\(C_3H_6\)) and HBr. The reaction is:

\[ C_3H_7Br + NaOH \xrightarrow{\text{hot ethanol}} C_3H_6 + NaBr + H_2O \]

Key observations:

Elimination reduces the H count by 2 (from \(C_3H_8\) in propane to \(C_3H_6\) in propene).
Substitution (cold NaOH) would give propanol (\(C_3H_8O\)), but hot conditions favor elimination.

Thus, the product is propene (A), with molecular formula \(C_3H_6\).

Question 36

Topic: 15.1

A sample of pent-2-en-4-ol, \(C_5H_{10}O\), contains all the possible stereoisomers of this compound. How many stereoisomers are there in the sample?

A. 2 B. 3 C. 4 D. 5

▶️ Answer/Explanation

Solution

Ans: C

Step-by-Step Analysis:

Structure of pent-2-en-4-ol: \[ CH_2=CH-CH(OH)-CH_2-CH_3 \] This compound has:
- A C=C double bond at position 2 (geometric isomerism possible: E/Z).
- A chiral center at carbon 4 (the −OH-bearing carbon, with four different substituents).
Stereoisomerism:
- Geometric isomers: The C=C bond gives E and Z forms.
- Optical isomers: The chiral center at C-4 gives two enantiomers (R and S) for each geometric isomer.
Total stereoisomers: \[ 2 \text{ (geometric)} \times 2 \text{ (optical)} = 4 \text{ stereoisomers.} \]

Thus, the correct answer is C (4).

Question 37

Topic: 15.1

Which pair of reagents reacts to form a product with a chiral carbon atom?

A. CH₃CH=CH₂ + HBr B. (CH₃)₂C=O + NaBH₄ C. CH₃CH₂CHO + HCN D. \(CH_3COOH + CH_3CH_2OH\)

▶️ Answer/Explanation

Solution

Ans: C

Step 1: Understand chiral carbon

A chiral carbon is a carbon atom bonded to four different groups. We analyze each reaction:

Option A (CH₃CH=CH₂ + HBr): Forms CH₃CHBrCH₃ (no chiral carbon, as the central carbon has two identical CH₃ groups).

Option B ((CH₃)₂C=O + NaBH₄): Reduces to (CH₃)₂CHOH (no chiral carbon, same reason as A).

Option C (CH₃CH₂CHO + HCN): Forms CH₃CH₂CH(OH)CN. The second carbon (CH(OH)CN) is chiral (bonded to -H, -OH, -CN, and -CH₂CH₃).

Option D (CH₃COOH + CH₃CH₂OH): Forms CH₃COOCH₂CH₃ (ester, no chiral carbon).

Conclusion: Only C produces a product with a chiral carbon.

Question 38

Topic: 19.2

The diagrams show the structures of two esters, X and Y, that are formed in ripening apples.

Which carboxylic acids are formed when these esters are hydrolysed by H₂SO₄ (aq)?

▶️ Answer/Explanation

Solution

Ans: B

Hydrolysis of Esters X and Y:
• Ester X (methyl butanoate) hydrolyzes to butanoic acid and methanol.
• Ester Y (ethyl propanoate) hydrolyzes to propanoic acid and ethanol.

Why Option B is Correct:
The image in option B shows the correct carboxylic acids (butanoic acid and propanoic acid) formed from the hydrolysis of X and Y, respectively. The other options either show incorrect acids or mismatch the structures.

Question 39

Topic: 20.1

An addition polymer is made from monomer Z.

What is the structure of the polymer made from this monomer?

▶️ Answer/Explanation

Solution

Ans: D

The monomer Z is a substituted alkene that undergoes addition polymerization. The double bond opens up to form the polymer chain, with the substituents (methyl group and COOCH₃ group) alternating on every other carbon atom in the polymer backbone.

Key features of the correct polymer structure (Option D):

The polymer backbone consists of -CH₂-CH- repeating units
Each CH unit has two substituents: a methyl group (CH₃) and a methoxycarbonyl group (COOCH₃)
The substituents maintain the same relative positions as in the monomer

This arrangement is characteristic of addition polymers formed from asymmetrically substituted alkenes, where the substituents appear on every other carbon in the polymer chain.

Question 40

Topic: 22.2

Compound X reacts with acidified K₂Cr₂O₇ to form compound Y. The infrared spectrum of compound Y is shown.

What is the identity of compound X?

A. propan-1-ol B. propan-2-ol C. propanone D. propanoic acid

▶️ Answer/Explanation

Solution

Ans: A

Acidified K₂Cr₂O₇ oxidizes primary alcohols (propan-1-ol, A) to carboxylic acids (propanoic acid), secondary alcohols (propan-2-ol, B) to ketones (propanone), and has no effect on ketones (C) or carboxylic acids (D). The IR spectrum of Y shows a broad peak at ~3000 cm⁻¹ (O-H) and a sharp peak at ~1700 cm⁻¹ (C=O), confirming it as a carboxylic acid. Thus, X must be propan-1-ol, the only primary alcohol option.

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