2021 jupeb chemistry

CHM 001: GENERAL CHEMISTRY

Question 1

(a) Distinguish between ionization energy and electron affinity, giving an example in each case. [2 marks]

(b) Briefly describe and give reasons for the observed trends of the following atomic properties in the periodic table: (i) Atomic size (ii) Ionization energy (iii) Electron affinity. [3 marks]

© Balance the following oxidation-reduction reaction, which occurs in acidic solution, using the half-reaction method:

Cr₂O₇²⁻(aq) + Cl⁻(aq) → Cr³⁺(aq) + Cl₂(g) [3 marks]

(d) Assign oxidation numbers to each atom in the following reactions and identify the oxidizing agent and reducing agent:

(i) 2Al(s) + 3H₂SO₄(aq) → Al₂(SO₄)₃(aq) + 3H₂(g)

(ii) PbO(s) + CO(g) → Pb(s) + CO₂(g) [2 marks]

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Question 2

(a) A compound X contains carbon, hydrogen, and oxygen only. When burnt completely in excess oxygen, CO₂ and H₂O are the only products. It is found that 0.46 g of X gives 0.88 g of carbon dioxide and 0.54 g of water. Find the empirical formula of compound X. (R.A.M: H = 1.0, C = 12.0, O = 16.0) [4 marks]

(b)

(i) Differentiate between extensive and intensive properties.

(ii) Give one example each of extensive and intensive properties. [2 marks]

©

(i) State one piece of information that each of the three quantum numbers n, l, and m give about the atomic orbital.

(ii) Identify the orbitals described by the following quantum numbers and arrange them in order of increasing energy:

| n | l | m |

|—|---|—|

| 4 | 0 | 0 |

| 3 | 2 | −1 |

| 4 | 2 | 2 |

| 3 | 1 | 0 |

[4 marks]

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CHM 002: PHYSICAL CHEMISTRY

Question 3

(a)

(i) Write the van der Waals equation and state all the parameters. [2 marks]

(ii) If sulfur dioxide (SO₂) were an ideal gas, the pressure at 0.0 °C exerted by 1.0 mol occupying 22.4 L would be 1.0 atm. Use the van der Waals equation to estimate the actual pressure of 1.0 mol SO₂ at 0.0 °C. (a = 6.865 L²·atm/mol², b = 0.05679 L/mol) [2 marks]

(b) Account for the following observation: when solutions of sodium iodide and potassium nitrate are mixed, no visible reaction occurs; however, when lead nitrate is substituted for potassium nitrate, a precipitate forms. [2 marks]

©

(i) Distinguish between natural and induced radioactivity.

(ii) A 0.600 g sample of ⁹⁰₃₈Sr diminishes to 0.395 g in ten years. Calculate the half-life. [4 marks]

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Question 4

(a)

(i) State the law of mass action.

(ii) Using a relevant illustrative reaction, distinguish between the reaction quotient (Q) and the equilibrium constant (K). [3 marks]

(b) 70 g of ethanol and 30 g of methanol are mixed to form a solution. The vapour pressure of pure ethanol is 75 mmHg and that of pure methanol is 22 mmHg at 20 °C. (C = 12, O = 16, H = 1)

(i) Calculate the mole fraction of ethanol and methanol in the solution.

(ii) Calculate the partial pressure of each component and the total vapour pressure of the solution. [4 marks]

© Using the standard electrode potentials below, answer the following questions:

| Half-equation | E°/V |

|—|---|

| I₂(aq) + 2e⁻ ⇌ 2I⁻(aq) | +0.54 |

| Fe³⁺(aq) + e⁻ ⇌ Fe²⁺(aq) | +0.77 |

| VO₂⁺ + 2H⁺(aq) + e⁻ ⇌ VO²⁺(aq) + H₂O(l) | +1.00 |

| Cl₂(g) + 2e⁻ ⇌ 2Cl⁻(aq) | +1.36 |

A reaction occurs when gaseous chlorine is passed into iron(II) sulfate solution.

(i) Write the half-equations and the overall cell equation for the reaction.

(ii) Calculate the standard cell e.m.f. [3 marks]

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CHM 003: INORGANIC CHEMISTRY

Question 5

(a)

(i) Using chemical equations only, show the amphoteric behaviour of aluminium (Group 13). [2 marks]

(ii) Using chemical equations only, describe what happens when fluorine and bromine are separately added to water. Name the type of reaction that occurs. [3 marks]

(b)

(i) What is meant by diagonal relationship?

(ii) Give two properties of lithium that illustrate its diagonal relationship with magnesium. [3 marks]

© Mention two consequences of ozone layer depletion. [2 marks]

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Question 6

(a)

(i) Write balanced equations for the reactions of sodium and phosphorus with oxygen and chlorine respectively.

(ii) Write chemical equations to show what happens when each oxide in (i) is in contact with water.

(iii) Using chemical equations only, describe what happens when water is added to the chlorides of the elements in (i). [4 marks]

(b) Consider the coordination compound [Co(NH₃)₆]Cl₃.

(i) State the primary valency of the central metal ion.

(ii) State the secondary valency of the central metal ion.

(iii) How many moles of silver chloride will be precipitated when 1 mole of the compound reacts with silver nitrate?

(iv) Which ions are formed when the compound is dissolved in water? [4 marks]

©

(i) Arrange the hydrogen halides in order of increasing acidic strength.

(ii) Account for the order in (i). [2 marks]

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CHM 004: ORGANIC CHEMISTRY

Question 7

(a)

(i) Give three characteristics of a homologous series.

(ii) Name three classes of compounds that form a homologous series.

(iii) Define the term hybridization. [4 marks]

(b) A saturated organic compound A containing two carbon atoms reacted with ethanoic acid in the presence of a mineral acid to form compound B with a sweet smell.

(i) Name the functional group present in A.

(ii) Draw the structure of A.

(iii) Write a chemical equation to show the formation of B.

(iv) Name compound B. [4 marks]

©

(i) Mention two types of bond scission that organic compounds undergo.

(ii) What type of organic reagents result from each type of scission? [2 marks]

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Question 8

(a) Write the name and structure of the major organic product from each of the following reactions:

(i) CH₃CONH₂ heated with dilute HCl

(ii) CH₃CH₂CH₂CONH₂ heated with LiAlH₄

(iii) CH₃CH₂CO₂H refluxed with SOCl₂

(iv) CH₃COCH₂CH₃ shaken with H₂NNH₂ [4 marks]

(b)

(i) Name four types of addition reactions an alkene can undergo.

(ii) Write the structures of the following compounds: butoxypentane and N-propylpropanoate. [4 marks]

© Potassium manganate(VII) (potassium permanganate) reacts with but-2-ene in the cold. Describe the colour change in the potassium manganate(VII) solution under:

(i) acidic conditions

(ii) alkaline conditions. [2 marks]

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ANSWERS

CHM 001: GENERAL CHEMISTRY

Answer 1

(a)

Ionization energy is the minimum energy required to remove an electron from a gaseous atom in its ground state.

Example: Na(g) → Na⁺(g) + e⁻

Electron affinity is the energy released when a gaseous atom in its ground state gains an electron to form a negative ion.

Example: Cl(g) + e⁻ → Cl⁻(g)

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(b)

(i) Atomic size

• Across a period (left → right): Atomic size decreases. As nuclear charge (number of protons) increases across a period, the electrons are pulled closer to the nucleus while remaining in the same shell, reducing the atomic radius.

• Down a group: Atomic size increases. Each successive element has an additional electron shell, so the outermost electrons are farther from the nucleus.

(ii) Ionization energy

• Across a period: Ionization energy increases. Increasing nuclear charge with no additional shielding means electrons are held more tightly and are harder to remove.

• Down a group: Ionization energy decreases. Outer electrons are farther from the nucleus and experience greater shielding from inner shells, making them easier to remove.

(iii) Electron affinity

• Across a period: Electron affinity generally increases (becomes more negative). Smaller, more electronegative atoms attract an incoming electron more strongly.

• Down a group: Electron affinity generally decreases. Larger atomic size and increased shielding reduce the attraction for an additional electron.

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© Balancing by the half-reaction method in acidic solution:

Oxidation half-reaction:

2Cl⁻ → Cl₂ + 2e⁻

Reduction half-reaction:

Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O

Multiply oxidation half-reaction by 3:

6Cl⁻ → 3Cl₂ + 6e⁻

Overall balanced equation:

$$\text{Cr}_2\text{O}_7^{2-}(aq) + 14\text{H}^+(aq) + 6\text{Cl}^-(aq) \rightarrow 2\text{Cr}^{3+}(aq) + 7\text{H}_2\text{O}(l) + 3\text{Cl}_2(g)$$

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(d)

(i) 2Al(s) + 3H₂SO₄(aq) → Al₂(SO₄)₃(aq) + 3H₂(g)

| Atom | Reactant O.N. | Product O.N. | Change |

|—|---|—|---|

| Al | 0 | +3 | Oxidized |

| H (in H₂SO₄) | +1 | 0 | Reduced |

| S | +6 | +6 | No change |

| O | −2 | −2 | No change |

• Oxidizing agent: H₂SO₄ (hydrogen is reduced from +1 to 0)

• Reducing agent: Al (aluminium is oxidized from 0 to +3)

(ii) PbO(s) + CO(g) → Pb(s) + CO₂(g)

| Atom | Reactant O.N. | Product O.N. | Change |

|—|---|—|---|

| Pb | +2 | 0 | Reduced |

| C | +2 | +4 | Oxidized |

| O | −2 | −2 | No change |

• Oxidizing agent: PbO (lead is reduced from +2 to 0)

• Reducing agent: CO (carbon is oxidized from +2 to +4)

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Answer 2

(a) Empirical formula of compound X

Step 1 – Find mass of each element:

Mass of C in CO₂:

$$\text{C} = \frac{12}{44} \times 0.88 = 0.24 \text{ g}$$

Mass of H in H₂O:

$$\text{H} = \frac{2}{18} \times 0.54 = 0.06 \text{ g}$$

Mass of O = 0.46 − 0.24 − 0.06 = 0.16 g

Step 2 – Convert to moles:

| Element | Mass (g) | Moles | Ratio |

|—|---|—|---|

| C | 0.24 | 0.24/12 = 0.02 | 0.02/0.02 = 1 |

| H | 0.06 | 0.06/1 = 0.06 | 0.06/0.02 = 3 |

| O | 0.16 | 0.16/16 = 0.01 | 0.01/0.02 = 0.5 |

Multiply all by 2 to get whole numbers → C:H:O = 2:6:1

$$\boxed{\text{Empirical formula: } \text{C}_2\text{H}_6\text{O}}$$

(This corresponds to ethanol, CH₃CH₂OH)

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(b)

(i)

• An extensive property depends on the amount of matter present in a sample.

• An intensive property does not depend on the amount of matter present; it is characteristic of the substance.

(ii)

• Extensive property example: Mass (or volume)

• Intensive property example: Density (or boiling point, melting point)

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©

(i)

• n (principal quantum number): indicates the energy level (shell) of the electron and the size of the orbital.

• l (azimuthal/angular momentum quantum number): indicates the shape of the orbital.

• m (magnetic quantum number): indicates the spatial orientation of the orbital.

(ii)

| n | l | m | Orbital |

|—|---|—|---|

| 4 | 0 | 0 | 4s |

| 3 | 2 | −1 | 3d |

| 4 | 2 | 2 | 4d |

| 3 | 1 | 0 | 3p |

Order of increasing energy:

$$3p < 4s < 3d < 4d$$

(Using the Aufbau principle and the (n + l) rule: 3p: 3+1=4; 4s: 4+0=4 but lower n wins; 3d: 3+2=5; 4d: 4+2=6)

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CHM 002: PHYSICAL CHEMISTRY

Answer 3

(a)(i) Van der Waals Equation:

$$\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT$$

Where:

• P = pressure of the gas (atm)

• V = volume of the gas (L)

• n = number of moles of gas (mol)

• T = absolute temperature (K)

• R = universal gas constant (0.08206 L·atm/mol·K)

• a = constant accounting for intermolecular attractive forces (L²·atm/mol²)

• b = constant accounting for the finite volume of gas molecules (L/mol)

(a)(ii) Given: n = 1.0 mol, V = 22.4 L, T = 273.15 K, a = 6.865 L²·atm/mol², b = 0.05679 L/mol

$$P = \frac{nRT}{V - nb} - \frac{an^2}{V^2}$$

$$P = \frac{(1.0)(0.08206)(273.15)}{22.4 - (1.0)(0.05679)} - \frac{(6.865)(1.0)^2}{(22.4)^2}$$

$$P = \frac{22.40}{22.343} - \frac{6.865}{501.76}$$

$$P = 1.00255 - 0.01368$$

$$\boxed{P \approx 0.989 \text{ atm}}$$

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(b)

When NaI and KNO₃ are mixed, the potential products are NaNO₃ and KI — both of which are soluble in water. Since no insoluble product forms, no precipitation occurs and no visible reaction is observed.

When Pb(NO₃)₂ is mixed with NaI, the reaction produces lead(II) iodide (PbI₂), which is insoluble in water and precipitates as a bright yellow solid:

$$\text{Pb}^{2+}(aq) + 2\text{I}^-(aq) \rightarrow \text{PbI}_2(s)\downarrow$$

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©(i)

• Natural radioactivity: The spontaneous disintegration of unstable nuclei of certain naturally occurring isotopes without any external influence.

• Induced (artificial) radioactivity: Radioactivity produced by bombarding stable nuclei with high-energy particles (e.g., alpha particles, neutrons), converting them into unstable radioactive nuclei.

©(ii) Half-life of ⁹⁰Sr:

Using the radioactive decay law:

$$N = N_0 e^{-\lambda t}$$

$$\frac{N}{N_0} = \frac{0.395}{0.600} = 0.6583$$

$$\ln(0.6583) = -\lambda \times 10$$

$$-0.4193 = -10\lambda \Rightarrow \lambda = 0.04193 \text{ yr}^{-1}$$

$$t_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.6931}{0.04193}$$

$$\boxed{t_{1/2} \approx 16.5 \text{ years}}$$

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Answer 4

(a)(i) Law of Mass Action:

At constant temperature, the rate of a chemical reaction is directly proportional to the product of the molar concentrations of the reactants, each raised to the power of its stoichiometric coefficient.

(a)(ii)

For the reaction: aA + bB ⇌ cC + dD

$$Q = \frac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b} \quad \text{and} \quad K_c = \frac{[\text{C}]^c[\text{D}]^d}{[\text{A}]^a[\text{B}]^b} \text{ at equilibrium}$$

• Q (reaction quotient): Calculated using concentrations at any point during the reaction. It predicts the direction the reaction will proceed.

• If Q < K: reaction proceeds forward

• If Q > K: reaction proceeds in reverse

• If Q = K: system is at equilibrium

• K (equilibrium constant): Calculated using concentrations only at equilibrium at a given temperature. It is a fixed value for a given reaction at a given temperature.

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(b)

Molar masses: Ethanol (C₂H₅OH) = 46 g/mol; Methanol (CH₃OH) = 32 g/mol

Moles:

• Moles of ethanol = 70/46 = 1.522 mol

• Moles of methanol = 30/32 = 0.9375 mol

• Total moles = 1.522 + 0.9375 = 2.459 mol

(i) Mole fractions:

$$x_{\text{ethanol}} = \frac{1.522}{2.459} = \boxed{0.619}$$

$$x_{\text{methanol}} = \frac{0.9375}{2.459} = \boxed{0.381}$$

(ii) Partial and total pressures (Raoult’s Law):

$$P_{\text{ethanol}} = x_{\text{ethanol}} \times P°_{\text{ethanol}} = 0.619 \times 75 = \boxed{46.4 \text{ mmHg}}$$

$$P_{\text{methanol}} = x_{\text{methanol}} \times P°_{\text{methanol}} = 0.381 \times 22 = \boxed{8.4 \text{ mmHg}}$$

$$P_{\text{total}} = 46.4 + 8.4 = \boxed{54.8 \text{ mmHg}}$$

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©

(i) When Cl₂ is passed into FeSO₄ solution, Cl₂ (E° = +1.36 V) oxidises Fe²⁺ (E° = +0.77 V) to Fe³⁺ since Cl₂ has the higher reduction potential.

Reduction half-equation:

$$\text{Cl}_2(g) + 2e^- \rightarrow 2\text{Cl}^-(aq) \quad E° = +1.36\text{ V}$$

Oxidation half-equation:

$$2\text{Fe}^{2+}(aq) \rightarrow 2\text{Fe}^{3+}(aq) + 2e^- \quad E° = -0.77\text{ V}$$

Overall cell equation:

$$\text{Cl}_2(g) + 2\text{Fe}^{2+}(aq) \rightarrow 2\text{Cl}^-(aq) + 2\text{Fe}^{3+}(aq)$$

(ii) Standard cell e.m.f.:

$$E°_{\text{cell}} = E°_{\text{cathode}} - E°_{\text{anode}} = 1.36 - 0.77 = \boxed{+0.59 \text{ V}}$$

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CHM 003: INORGANIC CHEMISTRY

Answer 5

(a)(i) Amphoteric behaviour of aluminium:

Reaction with acid (acting as a base):

$$2\text{Al}(s) + 6\text{HCl}(aq) \rightarrow 2\text{AlCl}_3(aq) + 3\text{H}_2(g)$$

Reaction with alkali (acting as an acid):

$$2\text{Al}(s) + 2\text{NaOH}(aq) + 2\text{H}_2\text{O}(l) \rightarrow 2\text{NaAlO}_2(aq) + 3\text{H}_2(g)$$

(a)(ii)

Fluorine with water (disproportionation/redox — fluorine oxidises water):

$$2\text{F}_2(g) + 2\text{H}_2\text{O}(l) \rightarrow 4\text{HF}(aq) + \text{O}_2(g)$$

Type of reaction: Redox (oxidation-reduction)

Bromine with water (disproportionation):

$$\text{Br}_2(l) + \text{H}_2\text{O}(l) \rightleftharpoons \text{HBr}(aq) + \text{HOBr}(aq)$$

Type of reaction: Disproportionation (a type of redox)

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(b)

(i) The diagonal relationship refers to the similarity in properties between certain elements in Period 2 and the elements diagonally adjacent to them in Period 3, due to similar charge density (charge-to-size ratio).

(ii) Two properties of lithium illustrating its diagonal relationship with magnesium:

1. Both form normal oxides when burned in excess oxygen (Li₂O and MgO), unlike other Group 1 metals which form peroxides or superoxides.

$$4\text{Li} + \text{O}_2 \rightarrow 2\text{Li}_2\text{O}; \quad 2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$$

2. Both form insoluble carbonates that decompose on heating to give the metal oxide and CO₂:

$$\text{Li}_2\text{CO}_3 \rightarrow \text{Li}_2\text{O} + \text{CO}_2; \quad \text{MgCO}_3 \rightarrow \text{MgO} + \text{CO}_2$$

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© Two consequences of ozone layer depletion:

1. Increased UV-B radiation reaching the Earth’s surface, leading to higher rates of skin cancer, cataracts, and immune system suppression in humans.

2. Disruption of ecosystems — particularly marine ecosystems, where phytoplankton (base of the aquatic food chain) are harmed by increased UV radiation, affecting biodiversity and food security.

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Answer 6

(a)(i) Reactions with oxygen and chlorine:

Sodium with oxygen:

$$4\text{Na}(s) + \text{O}_2(g) \rightarrow 2\text{Na}_2\text{O}(s)$$

Sodium with chlorine:

$$2\text{Na}(s) + \text{Cl}_2(g) \rightarrow 2\text{NaCl}(s)$$

Phosphorus with oxygen:

$$4\text{P}(s) + 5\text{O}_2(g) \rightarrow \text{P}_4\text{O}_{10}(s)$$

Phosphorus with chlorine (excess):

$$2\text{P}(s) + 5\text{Cl}_2(g) \rightarrow 2\text{PCl}_5(s)$$

(a)(ii) Oxides in contact with water:

$$\text{Na}_2\text{O}(s) + \text{H}_2\text{O}(l) \rightarrow 2\text{NaOH}(aq)$$

$$\text{P}_4\text{O}_{10}(s) + 6\text{H}_2\text{O}(l) \rightarrow 4\text{H}_3\text{PO}_4(aq)$$

(a)(iii) Chlorides in contact with water:

NaCl simply dissolves (no hydrolysis):

$$\text{NaCl}(s) \xrightarrow{\text{H}_2\text{O}} \text{Na}^+(aq) + \text{Cl}^-(aq)$$

PCl₅ undergoes vigorous hydrolysis:

$$\text{PCl}_5(s) + 4\text{H}_2\text{O}(l) \rightarrow \text{H}_3\text{PO}_4(aq) + 5\text{HCl}(aq)$$

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(b) [Co(NH₃)₆]Cl₃

(i) Primary valency: 3 (the oxidation state of Co is +3, satisfied by the 3 Cl⁻ ions outside the coordination sphere)

(ii) Secondary valency: 6 (the coordination number of Co³⁺ is 6, satisfied by the 6 NH₃ ligands)

(iii) The 3 Cl⁻ ions are outside the coordination sphere (ionisable). Therefore, 3 moles of AgCl are precipitated per mole of compound:

$$[\text{Co(NH}_3)_6]\text{Cl}_3 + 3\text{AgNO}_3 \rightarrow 3\text{AgCl}(s) + [\text{Co(NH}_3)_6](\text{NO}_3)_3$$

(iv) When dissolved in water, the compound produces:

$$[\text{Co(NH}_3)_6]^{3+}(aq) \text{ and } 3\text{Cl}^-(aq)$$

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©

(i) Order of increasing acidic strength of hydrogen halides:

$$\text{HF} < \text{HCl} < \text{HBr} < \text{HI}$$

(ii) Account for the order:

Acidic strength depends on the ease of dissociation (H–X bond breaking). As we descend Group 17, the H–X bond length increases and bond dissociation enthalpy decreases (the bond becomes weaker), making it easier to release H⁺. Thus acid strength increases from HF to HI. HF is the weakest acid because the H–F bond is exceptionally strong and short due to fluorine’s high electronegativity, and it only partially dissociates in water.

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CHM 004: ORGANIC CHEMISTRY

Answer 7

(a)(i) Characteristics of a homologous series:

1. Members have the same general formula and the same functional group.

2. Successive members differ by a –CH₂– unit (14 mass units).

3. Members show a gradual change in physical properties (e.g., boiling point, melting point) with increasing molecular mass, but have similar chemical properties.

(a)(ii) Three classes of compounds forming homologous series:

1. Alkanes (e.g., methane, ethane, propane…)

2. Alcohols (e.g., methanol, ethanol, propanol…)

3. Carboxylic acids (e.g., methanoic, ethanoic, propanoic acid…)

(a)(iii) Hybridization:

Hybridization is the process of mixing atomic orbitals of similar energy within the same atom to form new hybrid orbitals of equal energy and shape, suitable for bond formation. The number of hybrid orbitals formed equals the number of atomic orbitals mixed.

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(b) Compound A is a two-carbon saturated compound that reacts with ethanoic acid (in the presence of a mineral acid catalyst) to form a sweet-smelling ester — this is an esterification reaction. The compound must be ethanol (C₂H₅OH).

(i) Functional group: Hydroxyl group (–OH)

(ii) Structure of A (ethanol):

$$\text{CH}_3\text{CH}_2\text{OH}$$

(iii) Chemical equation for formation of B:

$$\text{CH}_3\text{CH}_2\text{OH} + \text{CH}_3\text{COOH} \underset{\Delta}{\overset{\text{H}^+}{\rightleftharpoons}} \text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O}$$

(iv) Name of compound B: Ethyl ethanoate (ethyl acetate)

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©

(i) Two types of bond scission:

1. Homolytic fission — the bond breaks symmetrically, with each atom retaining one electron from the shared pair, producing free radicals.

2. Heterolytic fission — the bond breaks unsymmetrically, with both electrons going to one atom, producing ions (carbocations or carbanions).

(ii) Reagents resulting from each type:

1. Homolytic fission → Free radicals (reactive species with unpaired electrons); reactions proceed via radical mechanisms.

2. Heterolytic fission → Electrophiles (electron-seeking, e.g., carbocations) or nucleophiles (electron-donating, e.g., carbanions); reactions proceed via ionic mechanisms.

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Answer 8

(a)

(i) CH₃CONH₂ + HCl(dil) + H₂O → CH₃COOH + NH₄Cl

• Name: Ethanoic acid (and ammonium chloride)

• Product structure: CH₃COOH (amide hydrolysis in acid gives the carboxylic acid)

(ii) CH₃CH₂CH₂CONH₂ + LiAlH₄ → CH₃CH₂CH₂CH₂NH₂

• Name: Butan-1-amine (1-butylamine)

• Reduction of an amide with LiAlH₄ gives a primary amine

(iii) CH₃CH₂COOH + SOCl₂ → CH₃CH₂COCl + SO₂ + HCl

• Name: Propanoyl chloride (propionyl chloride)

• Structure: CH₃CH₂C(=O)Cl

(iv) CH₃COCH₂CH₃ + H₂NNH₂ → CH₃C(=NNH₂)CH₂CH₃ + H₂O

• Name: Butanone hydrazone (methyl ethyl ketone hydrazone)

• Reaction of a ketone with hydrazine gives a hydrazone

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(b)(i) Four types of addition reactions of alkenes:

1. Hydrogenation — addition of H₂ (e.g., in the presence of Ni catalyst)

2. Halogenation — addition of X₂ (e.g., Br₂, Cl₂)

3. Hydrohalogenation — addition of HX (e.g., HBr, HCl)

4. Hydration — addition of H₂O (in the presence of an acid catalyst)

(b)(ii) Structures:

Butoxypentane (an ether — butyl pentyl ether):

$$\text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2 - \text{O} - \text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_2\text{CH}_3$$

N-propylpropanoate (an ester — propyl propanoate):

$$\text{CH}_3\text{CH}_2\text{C}(=\text{O})\text{O} - \text{CH}_2\text{CH}_2\text{CH}_3$$

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© Reaction of KMnO₄ with but-2-ene in the cold:

(i) Acidic conditions:

The purple/violet colour of KMnO₄ is decolourised (turns colourless). Mn⁷⁺ is reduced to Mn²⁺ (colourless Mn²⁺ ions), and the alkene is oxidised to a diol (glycol), then further to carboxylic acids/ketones under acidic conditions.

(ii) Alkaline conditions:

The purple/violet colour of KMnO₄ changes to a brown precipitate (MnO₂). Mn⁷⁺ is reduced to Mn⁴⁺ (brown manganese(IV) oxide), and the alkene is oxidised to a diol (cis-diol) under mild alkaline conditions.