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IB Chemistry – Reactivity 3: Mechanisms of Change

Acids and bases, redox, organic mechanisms.

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12 questions  ·  43 marks  ·  answers included  ·  IB Chemistry SL & HL (2023 syllabus)
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This worksheet covers:

  • This worksheet covers Reactivity 3
  • Proton transfer reactions (acids and bases)
  • Electron transfer reactions (redox)
  • Electron sharing reactions (organic mechanisms)

Questions marked HL extend beyond the SL syllabus. A calculator is allowed; use $K_w = 1.0\times10^{-14}$ at 298 K.

Questions

Section A – Proton Transfer Reactions

1

For the reaction $\text{HNO}_2 + \text{H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^+ + \text{NO}_2^-$:(a) Identify the Brønsted–Lowry acid on the left-hand side.(b) Identify the two conjugate acid–base pairs.

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(a) $\text{HNO}_2$ is the acid (proton donor).(b) $\text{HNO}_2$/$\text{NO}_2^-$ and $\text{H}_3\text{O}^+$/$\text{H}_2\text{O}$.
[3]
2

(a) Calculate the pH of $0.010\,\text{mol dm}^{-3}$ hydrochloric acid, a strong acid.(b) Calculate the pH of $0.0025\,\text{mol dm}^{-3}$ sodium hydroxide, a strong base.

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(a) $[\text{H}^+] = 0.010$, so $\text{pH} = -\log(0.010) = 2.00$(b) $[\text{OH}^-] = 0.0025$, $\text{pOH} = -\log(0.0025) = 2.60$, so $\text{pH} = 14.00 – 2.60 = 11.40$
[4]
3

A weak acid HA has $K_a = 1.8\times10^{-5}$ and an initial concentration of $0.10\,\text{mol dm}^{-3}$. Using the approximation $[\text{H}^+] \approx \sqrt{K_a c}$, calculate the pH of the solution.

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$[\text{H}^+] = \sqrt{(1.8\times10^{-5})(0.10)} = \sqrt{1.8\times10^{-6}} \approx 1.34\times10^{-3}\,\text{mol dm}^{-3}$$\text{pH} = -\log(1.34\times10^{-3}) \approx 2.87$
[3]
4

(HL) A buffer solution is prepared containing $0.20\,\text{mol dm}^{-3}$ ethanoic acid ($K_a = 1.8\times10^{-5}$) and $0.10\,\text{mol dm}^{-3}$ sodium ethanoate.(a) State the two components needed to make an acidic buffer.(b) Use the Henderson–Hasselbalch equation, $\text{pH} = \text{p}K_a + \log\dfrac{[\text{base}]}{[\text{acid}]}$, to calculate the pH of this buffer.

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(a) A weak acid and its conjugate base (usually as a soluble salt), present in comparable concentrations.(b) $\text{p}K_a = -\log(1.8\times10^{-5}) \approx 4.74$; $\text{pH} = 4.74 + \log\left(\dfrac{0.10}{0.20}\right) = 4.74 – 0.30 \approx 4.44$
[4]

Section B – Electron Transfer Reactions

5

For the reaction $\text{MnO}_4^- + 5\text{Fe}^{2+} + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}$:(a) State the change in oxidation number of manganese and of iron.(b) Identify the oxidising agent and the reducing agent.

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(a) Manganese: $+7 \rightarrow +2$ (reduced by 5). Iron: $+2 \rightarrow +3$ (oxidised by 1, per Fe atom).(b) Oxidising agent: $\text{MnO}_4^-$ (it is reduced). Reducing agent: $\text{Fe}^{2+}$ (it is oxidised).
[4]
6

Write the two half-equations for the reaction between acidified potassium manganate(VII) and iron(II) ions, and combine them to give the overall balanced ionic equation.

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Reduction: $\text{MnO}_4^- + 8\text{H}^+ + 5e^- \rightarrow \text{Mn}^{2+} + 4\text{H}_2\text{O}$Oxidation: $\text{Fe}^{2+} \rightarrow \text{Fe}^{3+} + e^-$ (× 5 to balance electrons)Overall: $\text{MnO}_4^- + 5\text{Fe}^{2+} + 8\text{H}^+ \rightarrow \text{Mn}^{2+} + 5\text{Fe}^{3+} + 4\text{H}_2\text{O}$
[4]
7

(HL) An electrochemical cell is set up using a zinc half-cell ($E^\ominus = -0.76\,\text{V}$) connected to a copper half-cell ($E^\ominus = +0.34\,\text{V}$) via a salt bridge.(a) State which electrode is the cathode (positive electrode).(b) Calculate the standard cell potential, $E^\ominus_{cell}$.(c) State whether the cell reaction is spontaneous, giving a reason.

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(a) Copper is the cathode (higher, less negative $E^\ominus$, so $\text{Cu}^{2+}$ is reduced).(b) $E^\ominus_{cell} = E^\ominus_{cathode} – E^\ominus_{anode} = 0.34 – (-0.76) = 1.10\,\text{V}$(c) Spontaneous, because $E^\ominus_{cell}$ is positive.
[4]
8

Explain, in terms of electron transfer, what happens to a species during oxidation and during reduction. Use the mnemonic OIL RIG in your answer.

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Oxidation Is Loss of electrons (the species loses electrons and its oxidation number increases); Reduction Is Gain of electrons (the species gains electrons and its oxidation number decreases).
[2]

Section C – Electron Sharing Reactions: Organic Mechanisms

HL emphasis
9

(HL) Bromoethane reacts with hydroxide ions in an $\text{S}_N2$ mechanism to form ethanol.Describe the mechanism, including the role of the nucleophile, the direction of attack, and the shape of the transition state.

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The hydroxide ion (nucleophile, with a lone pair) attacks the carbon bonded to bromine from the side opposite to the leaving group (backside attack), as the $\text{C}$–$\text{Br}$ bond breaks heterolytically and bromide ($\text{Br}^-$) leaves. The reaction proceeds in a single step through a five-coordinate transition state, and the carbon centre undergoes inversion of configuration.
[4]
10

(HL) 2-bromo-2-methylpropane (a tertiary haloalkane) reacts with water predominantly by an $\text{S}_N1$ mechanism, whereas bromoethane (a primary haloalkane) reacts by $\text{S}_N2$.Explain why the tertiary haloalkane favours $\text{S}_N1$ while the primary haloalkane favours $\text{S}_N2$.

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In $\text{S}_N1$, the rate-determining step is formation of a carbocation; tertiary carbocations are stabilised by the electron-donating (hyperconjugation/inductive) effect of three alkyl groups, so tertiary haloalkanes readily form a stable intermediate. Steric hindrance from the bulky alkyl groups also makes backside attack (needed for $\text{S}_N2$) very difficult. Primary haloalkanes have an unhindered carbon and a very unstable primary carbocation, so they instead react in a single concerted step ($\text{S}_N2$).
[3]
11

Propene reacts with hydrogen bromide by electrophilic addition to form predominantly 2-bromopropane rather than 1-bromopropane.(a) Outline the mechanism, stating which carbon the electrophile ($\text{H}^+$) adds to first.(b) Explain, in terms of carbocation stability, why 2-bromopropane is the major product (Markovnikov’s rule).

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(a) The $\pi$ electrons of the $\text{C}=\text{C}$ double bond act as the nucleophile and attack $\text{H}^+$ from $\text{HBr}$; $\text{H}^+$ adds to the terminal carbon (C1), forming a carbocation on the central carbon (C2), and $\text{Br}^-$ then bonds to this carbocation.(b) The secondary carbocation formed at C2 is more stable than the primary carbocation that would form at C1, because alkyl groups on either side donate electron density (inductive/hyperconjugation effect) and stabilise the positive charge; the more stable intermediate is formed faster, so 2-bromopropane is the major product.
[4]
12

Methane reacts with chlorine in the presence of ultraviolet light by a free-radical substitution mechanism.(a) Write an equation for the initiation step.(b) Write equations for the two propagation steps.(c) State the type of bond breaking that occurs in the initiation step.

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(a) $\text{Cl}_2 \rightarrow 2\text{Cl}^{\bullet}$ (initiated by UV light)(b) $\text{Cl}^{\bullet} + \text{CH}_4 \rightarrow \text{CH}_3^{\bullet} + \text{HCl}$ and $\text{CH}_3^{\bullet} + \text{Cl}_2 \rightarrow \text{CH}_3\text{Cl} + \text{Cl}^{\bullet}$(c) Homolytic fission (the $\text{Cl}$–$\text{Cl}$ bond breaks so that each chlorine atom keeps one electron, forming two radicals).
[4]

Related worksheets

More worksheets in IB Chemistry (SL & HL):

IB Chemistry – Structure 1: Particulate Nature of MatterIB Chemistry – Structure 2: Bonding and StructureIB Chemistry – Structure 3: Classification of MatterIB Chemistry – Reactivity 1: What Drives ReactionsIB Chemistry – Reactivity 2: How Much, How Fast, How FarIB Chemistry – Data-Based Questions (Paper 1B)

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