IB Chemistry HL · Reactivity 3: Mechanisms of Chemical Change · Solutions

Q1EASYPaper 13.1.1 Conjugate Pairs

Conjugate base of $\mathrm{HSO_4^-}$?$\mathrm{HSO_4^-}$ 的共轭碱？

(A) $\mathrm{H_2SO_4}$
(B) $\mathrm{SO_4^{2-}}$
(C) $\mathrm{HSO_4^-}$
(D) $\mathrm{H_3O^+}$

Answer:答案： (B)

A conjugate base is formed by removing one $\mathrm{H^+}$ from an acid. Removing $\mathrm{H^+}$ from $\mathrm{HSO_4^-}$ leaves $\mathrm{SO_4^{2-}}$. Trap (A) is what you get by adding $\mathrm{H^+}$ to $\mathrm{HSO_4^-}$, so it is the conjugate acid, not the conjugate base.共轭碱是从酸中移去一个 $\mathrm{H^+}$ 得到的物种。从 $\mathrm{HSO_4^-}$ 中移去 $\mathrm{H^+}$ 得到 $\mathrm{SO_4^{2-}}$。陷阱 (A) 是给 $\mathrm{HSO_4^-}$ 加上 $\mathrm{H^+}$ 得到的物种，是共轭酸而非共轭碱。

Q2EASYPaper 13.1.2 pH / pOH / Kw

At 298 K, $\mathrm{pH} = 3.0$. Find $\mathrm{pOH}$.298 K 下 $\mathrm{pH} = 3.0$。求 $\mathrm{pOH}$。

(A) 3.0
(B) 7.0
(C) 11.0
(D) 14.0

Answer:答案： (C)

At 298 K, $K_w = [\mathrm{H^+}][\mathrm{OH^-}] = 1.0\times10^{-14}$, so $\mathrm{pH} + \mathrm{pOH} = 14.00$. Therefore $\mathrm{pOH} = 14.00 - 3.0 = 11.0$.298 K 下 $K_w = [\mathrm{H^+}][\mathrm{OH^-}] = 1.0\times10^{-14}$，故 $\mathrm{pH} + \mathrm{pOH} = 14.00$。因此 $\mathrm{pOH} = 14.00 - 3.0 = 11.0$。

Q3EASYPaper 13.2.1 Oxidation States

Oxidation state of Cr in $\mathrm{Cr_2O_7^{2-}}$?$\mathrm{Cr_2O_7^{2-}}$ 中 Cr 的氧化数？

(A) $+3$
(B) $+6$
(C) $+7$
(D) $+12$

Answer:答案： (B)

Let $x$ be the oxidation state of each Cr. Oxygen is $-2$, and the overall charge is $-2$: $2x + 7(-2) = -2$, so $2x = 12$ and $x = +6$.设每个 Cr 的氧化数为 $x$。氧为 $-2$，总电荷为 $-2$：$2x + 7(-2) = -2$，故 $2x = 12$，$x = +6$。

$$2x - 14 = -2 \;\Longrightarrow\; x = +6$$

Q4EASYPaper 13.2.3 Oxidizing / Reducing Agent

Oxidizing agent in $\mathrm{Zn(s) + Cu^{2+}(aq) \to Zn^{2+}(aq) + Cu(s)}$?$\mathrm{Zn(s) + Cu^{2+}(aq) \to Zn^{2+}(aq) + Cu(s)}$ 中的氧化剂？

(A) $\mathrm{Zn}$
(B) $\mathrm{Cu^{2+}}$
(C) $\mathrm{Zn^{2+}}$
(D) $\mathrm{Cu}$

Answer:答案： (B)

The oxidizing agent is the species that is itself reduced, since it causes another species to be oxidized. $\mathrm{Cu^{2+}}$ gains two electrons and its oxidation state falls from $+2$ to $0$, so $\mathrm{Cu^{2+}}$ is reduced and is therefore the oxidizing agent. Zn loses electrons ($0 \to +2$), so Zn is oxidized and is the reducing agent.氧化剂是本身被还原的物种，因为它使另一物种被氧化。$\mathrm{Cu^{2+}}$ 得到两个电子，氧化数从 $+2$ 降为 $0$，故 $\mathrm{Cu^{2+}}$ 被还原，是氧化剂。Zn 失去电子（$0 \to +2$），故 Zn 被氧化，是还原剂。

Q5MEDIUMPaper 13.1.3 Strong vs Weak Acids

Equal-concentration $\mathrm{HCl}$ vs $\mathrm{CH_3COOH}$: which observation distinguishes them?等浓度 $\mathrm{HCl}$ 与 $\mathrm{CH_3COOH}$：哪项观察能区分二者？

(A) $\mathrm{HCl}$ has a higher pH.$\mathrm{HCl}$ 的 pH 更高。
(B) $\mathrm{HCl}$ has a lower pH and conducts better.$\mathrm{HCl}$ 的 pH 更低，导电性更好。
(C) Identical pH.pH 相同。
(D) $\mathrm{CH_3COOH}$ reacts faster with Mg.$\mathrm{CH_3COOH}$ 与镁反应更快。

Answer:答案： (B)

$\mathrm{HCl}$ is a strong acid and dissociates completely, so $0.10~\mathrm{mol\,dm^{-3}}$ $\mathrm{HCl}$ gives $[\mathrm{H^+}] = 0.10~\mathrm{mol\,dm^{-3}}$, $\mathrm{pH} = 1.00$. $\mathrm{CH_3COOH}$ is a weak acid and only partially dissociates ($K_a = 1.8\times10^{-5}$), giving $[\mathrm{H^+}] \approx \sqrt{K_a C} \approx 1.3\times10^{-3}~\mathrm{mol\,dm^{-3}}$, $\mathrm{pH} \approx 2.9$. Because $\mathrm{HCl}$ produces far more ions in solution, it also conducts electricity better. Both concentration and identity are equal, so pH is not the same, ruling out (C); (D) reverses the logic, since a lower $[\mathrm{H^+}]$ means a slower reaction with Mg, not a faster one.$\mathrm{HCl}$ 是强酸，完全电离，故 $0.10~\mathrm{mol\,dm^{-3}}$ 的 $\mathrm{HCl}$ 给出 $[\mathrm{H^+}] = 0.10~\mathrm{mol\,dm^{-3}}$，$\mathrm{pH} = 1.00$。$\mathrm{CH_3COOH}$ 是弱酸，只部分电离（$K_a = 1.8\times10^{-5}$），给出 $[\mathrm{H^+}] \approx \sqrt{K_a C} \approx 1.3\times10^{-3}~\mathrm{mol\,dm^{-3}}$，$\mathrm{pH} \approx 2.9$。由于 $\mathrm{HCl}$ 在溶液中产生的离子多得多，其导电性也更好。二者浓度相同但 pH 不相同，排除 (C)；(D) 逻辑相反，$[\mathrm{H^+}]$ 更低意味着与镁反应更慢，而非更快。

Q6MEDIUMPaper 13.2.4 Voltaic Cells

Reduction occurs at which electrode in a $\mathrm{Zn/Cu}$ voltaic cell, and what is its polarity?$\mathrm{Zn/Cu}$ 原电池中还原反应发生在哪个电极，其极性是什么？

(A) anode, negative.阳极，负极。
(B) anode, positive.阳极，正极。
(C) cathode, positive.阴极，正极。
(D) cathode, negative.阴极，负极。

Answer:答案： (C)

By definition, reduction always occurs at the cathode (oxidation at the anode) in any cell. In a voltaic (galvanic) cell, Zn is more reactive and is oxidized at the anode, releasing electrons that flow through the external circuit to the Cu electrode, where $\mathrm{Cu^{2+}}$ is reduced. Because electrons flow toward the cathode from the external circuit, the cathode is the positive terminal in a voltaic cell (this polarity is reversed in an electrolytic cell).按定义，任何电池中还原反应总发生在阴极（氧化发生在阳极）。在原电池中，Zn 更活泼，在阳极被氧化，释放的电子经外电路流向 Cu 电极，在此 $\mathrm{Cu^{2+}}$ 被还原。由于电子经外电路流向阴极，原电池中阴极是正极（在电解池中该极性相反）。

Q7MEDIUMPaper 13.3.1 Radical Substitution

Initiation step of methane / chlorine radical substitution?甲烷/氯气自由基取代的引发步骤？

(A) Heterolytic fission of $\mathrm{Cl-Cl}$.$\mathrm{Cl-Cl}$ 的异裂。
(B) Homolytic fission of $\mathrm{Cl-Cl}$ under UV light, giving two $\mathrm{Cl}$ radicals.紫外光下 $\mathrm{Cl-Cl}$ 的均裂，生成两个 $\mathrm{Cl}$ 自由基。
(C) Homolytic fission of a $\mathrm{C-H}$ bond.$\mathrm{C-H}$ 键的均裂。
(D) Formation of $\mathrm{C-Cl}$ releasing a proton.形成 $\mathrm{C-Cl}$ 并释放质子。

Answer:答案： (B)

Initiation supplies the radicals that start the chain. UV light provides enough energy to break the weak $\mathrm{Cl-Cl}$ bond homolytically, meaning each chlorine atom keeps one electron from the shared pair, so two neutral $\mathrm{Cl}\bullet$ radicals form: $\mathrm{Cl_2 \xrightarrow{h\nu} 2Cl\bullet}$. Heterolytic fission (A) would give ions, not radicals, and is not how this reaction is initiated. Breaking a $\mathrm{C-H}$ bond (C) does not occur until the propagation step.引发步骤生成开启链式反应的自由基。紫外光提供足够能量使较弱的 $\mathrm{Cl-Cl}$ 键发生均裂，即每个氯原子各保留共用电子对中的一个电子，生成两个中性的 $\mathrm{Cl}\bullet$ 自由基：$\mathrm{Cl_2 \xrightarrow{h\nu} 2Cl\bullet}$。异裂（A）会生成离子而非自由基，并非该反应的引发方式。断裂 $\mathrm{C-H}$ 键（C）要到传播步骤才发生。

Q8MEDIUMPaper 13.3.2 Product Mixtures

Why does chlorination of methane give a mixture of $\mathrm{CH_3Cl}$, $\mathrm{CH_2Cl_2}$, $\mathrm{CHCl_3}$, $\mathrm{CCl_4}$?为何甲烷氯化会生成 $\mathrm{CH_3Cl}$、$\mathrm{CH_2Cl_2}$、$\mathrm{CHCl_3}$、$\mathrm{CCl_4}$ 的混合物？

(A) $\mathrm{Cl}$ radicals react with product molecules as well as $\mathrm{CH_4}$, giving further substitution.$\mathrm{Cl}$ 自由基不仅与 $\mathrm{CH_4}$ 反应，也与产物分子反应，造成进一步取代。
(B) Initiation randomly gives different radicals.引发步骤随机生成不同自由基。
(C) Methane acts as both oxidizing and reducing agent.甲烷既作氧化剂也作还原剂。
(D) The $\mathrm{HCl}$ formed inhibits further reaction.生成的 $\mathrm{HCl}$ 抑制进一步反应。

Answer:答案： (A)

Once $\mathrm{CH_3Cl}$ has formed, it is still present in the reaction mixture along with unreacted $\mathrm{CH_4}$ and $\mathrm{Cl_2}$. A $\mathrm{Cl}\bullet$ radical cannot distinguish between abstracting a hydrogen from $\mathrm{CH_4}$ or from $\mathrm{CH_3Cl}$, so propagation continues on product molecules too, giving $\mathrm{CH_2Cl_2}$, then $\mathrm{CHCl_3}$, then $\mathrm{CCl_4}$ as the chain reaction proceeds statistically. This is an inherent feature of the chain mechanism, not a matter of random initiation (B) or a change in the redox role of methane (C).$\mathrm{CH_3Cl}$ 一旦生成，仍与未反应的 $\mathrm{CH_4}$、$\mathrm{Cl_2}$ 共存于反应混合物中。$\mathrm{Cl}\bullet$ 自由基无法区分从 $\mathrm{CH_4}$ 还是从 $\mathrm{CH_3Cl}$ 中夺取氢原子，因此传播步骤也会持续作用于产物分子，随着链反应统计性地进行，依次生成 $\mathrm{CH_2Cl_2}$、$\mathrm{CHCl_3}$、$\mathrm{CCl_4}$。这是链式机理本身固有的特征，而非引发的随机性（B）或甲烷氧化还原角色的改变（C）所致。

Q9HARDPaper 1HL3.1.4 Amphiprotic Species

Which species is amphiprotic?哪种物种是两性质子物种？

(A) $\mathrm{CO_3^{2-}}$
(B) $\mathrm{HCO_3^-}$
(C) $\mathrm{H_2CO_3}$
(D) $\mathrm{Na^+}$

Answer:答案： (B)

An amphiprotic species can both donate and accept a proton. $\mathrm{HCO_3^-}$ can donate $\mathrm{H^+}$ to become $\mathrm{CO_3^{2-}}$ (acting as an acid), or accept $\mathrm{H^+}$ to become $\mathrm{H_2CO_3}$ (acting as a base). $\mathrm{CO_3^{2-}}$ has no proton left to donate, so it can only act as a base; $\mathrm{H_2CO_3}$ has no lone pair available to accept a proton without first losing one of its own, so it can only act as an acid; $\mathrm{Na^+}$ has no acid-base behaviour in this context.两性质子物种既能给出质子，又能接受质子。$\mathrm{HCO_3^-}$ 可给出 $\mathrm{H^+}$ 变为 $\mathrm{CO_3^{2-}}$（作酸），也可接受 $\mathrm{H^+}$ 变为 $\mathrm{H_2CO_3}$（作碱）。$\mathrm{CO_3^{2-}}$ 已无质子可给出，只能作碱；$\mathrm{H_2CO_3}$ 不先失去自身质子就无法再接受质子，只能作酸；$\mathrm{Na^+}$ 在此语境下无酸碱行为。

Q10HARDPaper 1HL3.4.1 SN1 vs SN2

Tertiary substrate: rate $= k[\mathrm{substrate}]$. Primary substrate: rate $= k[\mathrm{substrate}][\mathrm{OH^-}]$. Why?叔底物：rate $= k[\mathrm{substrate}]$。伯底物：rate $= k[\mathrm{substrate}][\mathrm{OH^-}]$。为什么？

(A) Tertiary is $S_N2$; primary is $S_N1$.叔底物按 $S_N2$；伯底物按 $S_N1$。
(B) Tertiary is $S_N1$ (carbocation-forming rate-determining step, no $\mathrm{OH^-}$ involved); primary is $S_N2$ (concerted, $\mathrm{OH^-}$ attacks directly).叔底物按 $S_N1$（决速步生成碳正离子，不涉及 $\mathrm{OH^-}$）；伯底物按 $S_N2$（协同过程，$\mathrm{OH^-}$ 直接进攻）。
(C) Same mechanism; tertiary is just hindered.机理相同；只是叔底物位阻更大。
(D) Leaving-group ability alone explains it.仅由离去基团能力解释。

Answer:答案： (B)

A tertiary carbon is heavily hindered, which blocks the backside attack that $S_N2$ requires, but it readily forms a stable tertiary carbocation (three alkyl groups donate electron density by hyperconjugation and induction). The rate-determining step is therefore the unimolecular, slow ionization of the substrate, $\mathrm{R-Br \to R^+ + Br^-}$, with $\mathrm{OH^-}$ attacking the carbocation only in a fast second step; rate depends on $[\mathrm{substrate}]$ alone. A primary carbon is unhindered and cannot form a stable primary carbocation, so instead $\mathrm{OH^-}$ attacks the substrate directly from the side opposite the leaving group in a single bimolecular step, giving rate $= k[\mathrm{substrate}][\mathrm{OH^-}]$.叔碳位阻很大，阻碍了 $S_N2$ 所需的背面进攻，但它容易生成稳定的叔碳正离子（三个烷基通过超共轭与诱导效应提供电子云密度）。因此决速步是底物的单分子慢速电离：$\mathrm{R-Br \to R^+ + Br^-}$，$\mathrm{OH^-}$ 仅在随后的快步中进攻碳正离子；速率只取决于 $[\mathrm{substrate}]$。伯碳位阻小，无法生成稳定的伯碳正离子，因此 $\mathrm{OH^-}$ 在单一双分子步骤中从离去基团的背面直接进攻底物，得到 rate $= k[\mathrm{substrate}][\mathrm{OH^-}]$。

Q11HARDPaper 1HL3.4.2 Markovnikov Addition

Major product of propene $+$ $\mathrm{HBr}$?丙烯 $+$ $\mathrm{HBr}$ 的主要产物？

(A) 1-bromopropane.1-溴丙烷。
(B) 2-bromopropane.2-溴丙烷。
(C) 1,2-dibromopropane.1,2-二溴丙烷。
(D) propan-1-ol.丙-1-醇。

Answer:答案： (B)

The electrophile $\mathrm{H^+}$ (from polarized $\mathrm{HBr}$) adds first. Protonation at C1 (the terminal carbon, $\mathrm{CH_2}$) generates a secondary carbocation at C2, stabilized by two alkyl groups; protonation at C2 would generate a much less stable primary carbocation at C1. The reaction proceeds through the lower-energy secondary carbocation, so $\mathrm{Br^-}$ then bonds to C2, giving $\mathrm{CH_3CHBrCH_3}$ (2-bromopropane) as the major product. This is the essence of Markovnikov's rule: the halogen ends up on the more substituted carbon.亲电试剂 $\mathrm{H^+}$（来自极化的 $\mathrm{HBr}$）先加成。质子加在 C1（末端碳，$\mathrm{CH_2}$）上会在 C2 生成仲碳正离子，受两个烷基稳定；若质子加在 C2 上，则会在 C1 生成稳定性差得多的伯碳正离子。反应经由能量更低的仲碳正离子进行，随后 $\mathrm{Br^-}$ 与 C2 成键，生成主要产物 $\mathrm{CH_3CHBrCH_3}$（2-溴丙烷）。这正是马尔科夫尼科夫规则的本质：卤素最终连接在取代程度更高的碳上。

Q12HARDPaper 1HL3.2.5 E°cell & Spontaneity

$E^{\ominus}(\mathrm{Cu^{2+}/Cu}) = +0.34~\mathrm{V}$, $E^{\ominus}(\mathrm{Zn^{2+}/Zn}) = -0.76~\mathrm{V}$. $E^{\ominus}_\text{cell}$ and spontaneity for $\mathrm{Zn + Cu^{2+} \to Zn^{2+} + Cu}$?$E^{\ominus}(\mathrm{Cu^{2+}/Cu}) = +0.34~\mathrm{V}$，$E^{\ominus}(\mathrm{Zn^{2+}/Zn}) = -0.76~\mathrm{V}$。求 $\mathrm{Zn + Cu^{2+} \to Zn^{2+} + Cu}$ 的 $E^{\ominus}_\text{cell}$ 及自发性？

(A) Not spontaneous; $-1.10~\mathrm{V}$不自发；$-1.10~\mathrm{V}$
(B) Spontaneous; $+1.10~\mathrm{V}$自发；$+1.10~\mathrm{V}$
(C) Spontaneous; $+0.42~\mathrm{V}$自发；$+0.42~\mathrm{V}$
(D) Not spontaneous; $-0.42~\mathrm{V}$不自发；$-0.42~\mathrm{V}$

Answer:答案： (B)

$\mathrm{Cu^{2+}}$ is reduced (cathode) and Zn is oxidized (anode), so:$\mathrm{Cu^{2+}}$ 被还原（阴极），Zn 被氧化（阳极），故：

$$E^{\ominus}_\text{cell} = E^{\ominus}_\text{cathode} - E^{\ominus}_\text{anode} = 0.34 - (-0.76) = +1.10~\mathrm{V}$$

A positive $E^{\ominus}_\text{cell}$ corresponds to a negative $\Delta G^{\ominus}$ ($\Delta G^{\ominus} = -nFE^{\ominus}_\text{cell}$), so the reaction is spontaneous under standard conditions.正的 $E^{\ominus}_\text{cell}$ 对应负的 $\Delta G^{\ominus}$（$\Delta G^{\ominus} = -nFE^{\ominus}_\text{cell}$），故该反应在标准状态下自发。

SR 1MEDIUMPaper 2Redox Titration

$\mathrm{MnO_4^-(aq) + 8H^+(aq) + 5e^- \to Mn^{2+}(aq) + 4H_2O(l)}$; $\mathrm{Fe^{2+}(aq) \to Fe^{3+}(aq) + e^-}$.$\mathrm{MnO_4^-(aq) + 8H^+(aq) + 5e^- \to Mn^{2+}(aq) + 4H_2O(l)}$；$\mathrm{Fe^{2+}(aq) \to Fe^{3+}(aq) + e^-}$。

(a) Oxidation state of Mn in $\mathrm{MnO_4^-}$: let $x$ be its oxidation state; $x + 4(-2) = -1$, so $x = +7$. Oxidation state of Mn in $\mathrm{Mn^{2+}}$ is $+2$ (a monatomic ion equals its charge). Since Mn falls from $+7$ to $+2$, manganese is reduced.$\mathrm{MnO_4^-}$ 中 Mn 的氧化数：设为 $x$；$x + 4(-2) = -1$，故 $x = +7$。$\mathrm{Mn^{2+}}$ 中 Mn 的氧化数为 $+2$（单原子离子的氧化数等于其电荷）。Mn 从 $+7$ 降至 $+2$，故锰被还原。

(b) Multiply the Fe half-equation by 5 so that electrons cancel (5 electrons lost = 5 electrons gained), then add the two half-equations:将 Fe 半反应方程式乘以 5，使电子数相等（失去 5 个电子 = 得到 5 个电子），然后将两个半反应方程式相加：

$$\mathrm{MnO_4^-(aq) + 8H^+(aq) + 5Fe^{2+}(aq) \to Mn^{2+}(aq) + 4H_2O(l) + 5Fe^{3+}(aq)}$$

Check: charge on the left is $(-1) + 8(+1) + 5(+2) = +17$; charge on the right is $(+2) + 5(+3) = +17$. Charge and mass both balance.检验：左侧电荷为 $(-1) + 8(+1) + 5(+2) = +17$；右侧电荷为 $(+2) + 5(+3) = +17$。电荷与质量均已配平。

(c) $\mathrm{MnO_4^-}$ is reduced, so it is the oxidizing agent. $\mathrm{Fe^{2+}}$ is oxidized to $\mathrm{Fe^{3+}}$, so it is the reducing agent.$\mathrm{MnO_4^-}$ 被还原，故它是氧化剂。$\mathrm{Fe^{2+}}$ 被氧化为 $\mathrm{Fe^{3+}}$，故它是还原剂。

(d) Moles of $\mathrm{MnO_4^-}$ used:所用 $\mathrm{MnO_4^-}$ 的物质的量：

$$n(\mathrm{MnO_4^-}) = 0.0200~\mathrm{mol\,dm^{-3}} \times 0.02240~\mathrm{dm^3} = 4.48\times10^{-4}~\mathrm{mol}$$

From the 1:5 mole ratio in the equation above, moles of $\mathrm{Fe^{2+}}$:由上述方程式中 1:5 的摩尔比，$\mathrm{Fe^{2+}}$ 的物质的量：

$$n(\mathrm{Fe^{2+}}) = 5 \times 4.48\times10^{-4} = 2.24\times10^{-3}~\mathrm{mol}$$

Concentration of $\mathrm{FeSO_4}$:$\mathrm{FeSO_4}$ 的浓度：

$$c(\mathrm{FeSO_4}) = \dfrac{2.24\times10^{-3}~\mathrm{mol}}{0.0250~\mathrm{dm^3}} = 0.0896~\mathrm{mol\,dm^{-3}}$$

SR 2HARDPaper 2HLWeak Acids, Buffers & Titration Curves

$\mathrm{CH_3COOH}$, $K_a = 1.8\times10^{-5}~\mathrm{mol\,dm^{-3}}$ at 298 K.$\mathrm{CH_3COOH}$，298 K 下 $K_a = 1.8\times10^{-5}~\mathrm{mol\,dm^{-3}}$。

(a) pH of $0.150~\mathrm{mol\,dm^{-3}}$ $\mathrm{CH_3COOH}$. Since $\mathrm{CH_3COOH}$ is a weak acid with $K_a \ll C$, assume the dissociation is small enough that the equilibrium concentration of undissociated acid is approximately equal to the initial concentration:$0.150~\mathrm{mol\,dm^{-3}}$ $\mathrm{CH_3COOH}$ 的 pH。由于 $\mathrm{CH_3COOH}$ 是弱酸且 $K_a \ll C$，可假设电离程度很小，平衡时未电离酸的浓度近似等于初始浓度：

$$[\mathrm{H^+}] \approx \sqrt{K_a\,C} = \sqrt{(1.8\times10^{-5})(0.150)} = \sqrt{2.7\times10^{-6}} = 1.64\times10^{-3}~\mathrm{mol\,dm^{-3}}$$

$$\mathrm{pH} = -\log(1.64\times10^{-3}) = 2.78$$

(b) Buffer pH. Mixing equal volumes halves both concentrations, but the ratio $[\mathrm{A^-}]/[\mathrm{HA}]$ is unchanged: $0.100/0.150 = 0.667$. First find $pK_a$:缓冲溶液 pH。等体积混合使两个浓度都减半，但比值 $[\mathrm{A^-}]/[\mathrm{HA}]$ 不变：$0.100/0.150 = 0.667$。先求 $pK_a$：

$$pK_a = -\log(1.8\times10^{-5}) = 4.74$$

$$\mathrm{pH} = pK_a + \log\dfrac{[\mathrm{A^-}]}{[\mathrm{HA}]} = 4.74 + \log(0.667) = 4.74 - 0.18 = 4.57$$

(c) A strong acid to strong base titration would start at $\mathrm{pH} = 1.0$ for $0.100~\mathrm{mol\,dm^{-3}}$ acid, and its equivalence point would sit at exactly $\mathrm{pH} = 7$. Two features rule this out here: (1) the initial pH ($\approx 2.9$) is higher than the $\mathrm{pH} = 1.0$ expected for a fully dissociated $0.100~\mathrm{mol\,dm^{-3}}$ acid, showing the acid is only partially dissociated (weak). (2) the equivalence point lies above $\mathrm{pH} = 7$, because the salt formed, $\mathrm{CH_3COO^-}$, is the conjugate base of a weak acid and hydrolyzes water to give a mildly basic solution; this only happens for a weak acid to strong base titration.强酸滴强碱的滴定，$0.100~\mathrm{mol\,dm^{-3}}$ 的酸起始 $\mathrm{pH} = 1.0$，且等量点恰好位于 $\mathrm{pH} = 7$。此处有两项特征排除了这种情形：(1) 起始 pH（$\approx 2.9$）高于完全电离的 $0.100~\mathrm{mol\,dm^{-3}}$ 酸所预期的 $\mathrm{pH} = 1.0$，说明该酸只是部分电离（弱酸）。(2) 等量点位于 $\mathrm{pH} = 7$ 以上，因为生成的盐 $\mathrm{CH_3COO^-}$ 是弱酸的共轭碱，会使水解生成弱碱性溶液；这只在弱酸滴强碱时才会发生。

(d) At the half-equivalence point, $[\mathrm{HA}] = [\mathrm{A^-}]$, so $\log([\mathrm{A^-}]/[\mathrm{HA}]) = \log(1) = 0$ and $\mathrm{pH} = pK_a = 4.74$. The equivalence point of this titration is above $\mathrm{pH} = 7$ (around $\mathrm{pH} \approx 8.7$, from the hydrolysis of the acetate salt), so a suitable indicator must change color within that steep, high-pH region of the curve. Phenolphthalein (range 8.2 to 10.0) is the correct choice; methyl orange changes far too early (long before equivalence), and bromothymol blue's range sits below where the equivalence jump occurs.在半等量点，$[\mathrm{HA}] = [\mathrm{A^-}]$，故 $\log([\mathrm{A^-}]/[\mathrm{HA}]) = \log(1) = 0$，$\mathrm{pH} = pK_a = 4.74$。该滴定的等量点位于 $\mathrm{pH} = 7$ 以上（由乙酸根盐的水解可知约为 $\mathrm{pH} \approx 8.7$），因此合适的指示剂必须在曲线该陡峭的高 pH 区间内变色。正确选择是酚酞（范围 8.2 至 10.0）；甲基橙变色过早（远在等量点之前），溴百里酚蓝的范围则低于等量点跃升所在的区域。

SR 3HARDPaper 2HLElectrochemistry & Faraday's Law

$E^{\ominus}(\mathrm{Ag^+/Ag}) = +0.80~\mathrm{V}$; $E^{\ominus}(\mathrm{Pb^{2+}/Pb}) = -0.13~\mathrm{V}$.$E^{\ominus}(\mathrm{Ag^+/Ag}) = +0.80~\mathrm{V}$；$E^{\ominus}(\mathrm{Pb^{2+}/Pb}) = -0.13~\mathrm{V}$。

(a) Silver is the cathode (reduction) and lead is the anode (oxidation):银为阴极（还原），铅为阳极（氧化）：

$$E^{\ominus}_\text{cell} = E^{\ominus}_\text{cathode} - E^{\ominus}_\text{anode} = (+0.80) - (-0.13) = +0.93~\mathrm{V}$$

$E^{\ominus}_\text{cell}$ is positive, so the reaction is spontaneous under standard conditions.$E^{\ominus}_\text{cell}$ 为正，故该反应在标准状态下自发。

(b) In molten $\mathrm{PbBr_2}$, the only ions present are $\mathrm{Pb^{2+}}$ and $\mathrm{Br^-}$. $\mathrm{Pb^{2+}}$ is reduced at the cathode, and $\mathrm{Br^-}$ is oxidized at the anode:熔融 $\mathrm{PbBr_2}$ 中唯一存在的离子是 $\mathrm{Pb^{2+}}$ 与 $\mathrm{Br^-}$。$\mathrm{Pb^{2+}}$ 在阴极被还原，$\mathrm{Br^-}$ 在阳极被氧化：

$$\text{Cathode: } \mathrm{Pb^{2+}(l) + 2e^- \to Pb(l)}$$

$$\text{Anode: } \mathrm{2Br^-(l) \to Br_2(g) + 2e^-}$$

Products: liquid lead metal at the cathode, and bromine gas at the anode.产物：阴极生成液态铅金属，阳极生成溴气。

(c) Charge passed:通过的电荷量：

$$Q = It = (2.50~\mathrm{A})(40.0 \times 60~\mathrm{s}) = 2.50 \times 2400 = 6000~\mathrm{C}$$

Moles of electrons:电子的物质的量：

$$n(e^-) = \dfrac{Q}{F} = \dfrac{6000}{96\,500} = 0.0622~\mathrm{mol}$$

Since $\mathrm{Pb^{2+}}$ requires 2 electrons per Pb atom deposited:由于每沉积一个 Pb 原子需要 2 个电子：

$$n(\mathrm{Pb}) = \dfrac{n(e^-)}{2} = \dfrac{0.0622}{2} = 0.0311~\mathrm{mol}$$

$$m(\mathrm{Pb}) = n \times M = 0.0311~\mathrm{mol} \times 207.2~\mathrm{g\,mol^{-1}} = 6.44~\mathrm{g}$$

P3-1HARDPaper 3 HLRadical Substitution + SN1/SN2 + Addition + Nitration

This question surveys four organic reaction mechanisms from Reactivity 3.3 and 3.4.本题综合考查 Reactivity 3.3 与 3.4 中的四种有机反应机理。

(a) Free-radical substitution of ethane with chlorine.乙烷与氯气的自由基取代反应。

(i) Initiation(i) 引发

$$\mathrm{Cl_2 \xrightarrow{\;h\nu\;} 2Cl\bullet}$$

UV light supplies enough energy to break the weak $\mathrm{Cl-Cl}$ bond homolytically, each atom keeping one electron, giving two chlorine radicals.紫外光提供足够能量使较弱的 $\mathrm{Cl-Cl}$ 键均裂，每个原子各保留一个电子，生成两个氯自由基。

(ii) Propagation (two steps, producing chloroethane)(ii) 传播（两步，生成氯乙烷）

$$\mathrm{C_2H_6 + Cl\bullet \to C_2H_5\bullet + HCl}$$

$$\mathrm{C_2H_5\bullet + Cl_2 \to C_2H_5Cl + Cl\bullet}$$

Step 1: a chlorine radical abstracts a hydrogen atom from ethane, forming an ethyl radical and $\mathrm{HCl}$. Step 2: the ethyl radical reacts with a $\mathrm{Cl_2}$ molecule to give chloroethane and regenerate a chlorine radical, which can re-enter step 1. Adding the two steps gives the overall equation $\mathrm{C_2H_6 + Cl_2 \to C_2H_5Cl + HCl}$, with the chlorine radical unchanged, confirming the chain can propagate.第 1 步：氯自由基从乙烷夺取一个氢原子，生成乙基自由基与 $\mathrm{HCl}$。第 2 步：乙基自由基与一个 $\mathrm{Cl_2}$ 分子反应，生成氯乙烷并再生一个氯自由基，可重新进入第 1 步。将两步相加得到总方程式 $\mathrm{C_2H_6 + Cl_2 \to C_2H_5Cl + HCl}$，氯自由基数量不变，说明链反应可以持续传播。

(iii) Termination(iii) 终止

$$\mathrm{C_2H_5\bullet + Cl\bullet \to C_2H_5Cl}$$

Termination occurs whenever two radicals combine to form a stable molecule, removing radicals from the system. Other equally valid termination steps are $\mathrm{Cl\bullet + Cl\bullet \to Cl_2}$ and $\mathrm{C_2H_5\bullet + C_2H_5\bullet \to C_4H_{10}}$.只要两个自由基结合生成稳定分子，便发生终止，使自由基从体系中被移除。其他同样正确的终止步骤包括 $\mathrm{Cl\bullet + Cl\bullet \to Cl_2}$ 以及 $\mathrm{C_2H_5\bullet + C_2H_5\bullet \to C_4H_{10}}$。

(b) $S_N1$ (2-bromo-2-methylpropane) versus $S_N2$ (1-bromobutane).$S_N1$（2-溴-2-甲基丙烷）与 $S_N2$（1-溴丁烷）的比较。

Rate-determining step: for the tertiary substrate, the mechanism has two steps; the slow, rate-determining step is the unimolecular ionisation of the substrate, $\mathrm{(CH_3)_3C{-}Br \to (CH_3)_3C^+ + Br^-}$, which does not involve $\mathrm{OH^-}$ at all: the hydroxide only attacks the carbocation in a fast second step. For the primary substrate, there is a single concerted step in which $\mathrm{OH^-}$ attacks the carbon from the side opposite the leaving group as the $\mathrm{C-Br}$ bond breaks simultaneously.决速步：对叔底物而言，机理分两步；慢的决速步是底物的单分子电离，$\mathrm{(CH_3)_3C{-}Br \to (CH_3)_3C^+ + Br^-}$，完全不涉及 $\mathrm{OH^-}$：氢氧根仅在随后的快步中进攻碳正离子。对伯底物而言，是单一的协同步骤，$\mathrm{OH^-}$ 从离去基团的背面进攻碳原子，同时 $\mathrm{C-Br}$ 键断裂。

Molecularity: the $S_N1$ rate-determining step is unimolecular (only the substrate is involved), giving rate $= k[\mathrm{substrate}]$; the $S_N2$ step is bimolecular (both substrate and $\mathrm{OH^-}$ are involved in the single step), giving rate $= k[\mathrm{substrate}][\mathrm{OH^-}]$.分子数：$S_N1$ 的决速步是单分子反应（只涉及底物），速率 $= k[\mathrm{substrate}]$；$S_N2$ 是双分子反应（该单一步骤同时涉及底物与 $\mathrm{OH^-}$），速率 $= k[\mathrm{substrate}][\mathrm{OH^-}]$。

Stereochemistry at a chiral carbon: the $S_N1$ carbocation intermediate is planar ($sp^2$), so the nucleophile can attack from either face with roughly equal probability, giving a racemic mixture (racemisation) if the carbon were chiral. The $S_N2$ backside attack forces complete inversion of configuration at the carbon (Walden inversion), with no racemisation.手性碳处的立体化学：$S_N1$ 的碳正离子中间体是平面结构（$sp^2$），亲核试剂可从两面以大致相等的概率进攻，若该碳为手性碳则生成外消旋混合物（外消旋化）。$S_N2$ 的背面进攻会使该碳的构型发生完全的构型转化（瓦尔登转化），不发生外消旋化。

(c) Markovnikov addition of $\mathrm{HBr}$ to propene.$\mathrm{HBr}$ 对丙烯的马尔科夫尼科夫加成。

The electrophile is $\mathrm{H^+}$: the $\mathrm{H-Br}$ bond is polarised because $\mathrm{Br}$ is more electronegative than $\mathrm{H}$, so $\mathrm{H}$ carries a partial positive charge and is attacked first by the electron-rich $\mathrm{C{=}C}$ double bond.亲电试剂是 $\mathrm{H^+}$：$\mathrm{H-Br}$ 键因 $\mathrm{Br}$ 电负性大于 $\mathrm{H}$ 而极化，$\mathrm{H}$ 带部分正电荷，因而首先被富电子的 $\mathrm{C{=}C}$ 双键进攻。

Protonation at $\mathrm{C1}$ (the terminal $\mathrm{CH_2}$) leaves the positive charge on $\mathrm{C2}$, giving a secondary carbocation, $\mathrm{CH_3{-}CH^+{-}CH_3}$, stabilised by two alkyl groups through induction and hyperconjugation. Protonation at $\mathrm{C2}$ instead would leave the positive charge on $\mathrm{C1}$, giving a much less stable primary carbocation, $\mathrm{CH_3{-}CH_2{-}CH_2^+}$, stabilised by only one alkyl group.质子加在 $\mathrm{C1}$（末端 $\mathrm{CH_2}$）上，正电荷留在 $\mathrm{C2}$，生成仲碳正离子 $\mathrm{CH_3{-}CH^+{-}CH_3}$，受两个烷基通过诱导与超共轭效应稳定。若质子加在 $\mathrm{C2}$ 上，正电荷则留在 $\mathrm{C1}$，生成稳定性差得多的伯碳正离子 $\mathrm{CH_3{-}CH_2{-}CH_2^+}$，仅受一个烷基稳定。

The reaction proceeds through the lower-energy pathway leading to the more stable secondary carbocation. $\mathrm{Br^-}$ then bonds to $\mathrm{C2}$, giving $\mathrm{CH_3CHBrCH_3}$ (2-bromopropane) as the major product: the halogen ends up on the more substituted carbon, which is the essence of Markovnikov's rule.反应经由能量较低、通向更稳定仲碳正离子的路径进行。随后 $\mathrm{Br^-}$ 与 $\mathrm{C2}$ 成键，生成主要产物 $\mathrm{CH_3CHBrCH_3}$（2-溴丙烷）：卤素最终连接在取代程度更高的碳上，这正是马尔科夫尼科夫规则的本质。

(d) Electrophilic substitution: nitration of benzene.亲电取代：苯的硝化。

Concentrated $\mathrm{H_2SO_4}$ protonates concentrated $\mathrm{HNO_3}$, generating the electrophile, the nitronium ion:浓 $\mathrm{H_2SO_4}$ 质子化浓 $\mathrm{HNO_3}$，生成亲电试剂硝鎓离子：

$$\mathrm{HNO_3 + 2H_2SO_4 \to NO_2^+ + H_3O^+ + 2HSO_4^-}$$

The delocalised $\pi$-electron system of benzene attacks $\mathrm{NO_2^+}$, forming a non-aromatic, positively charged intermediate (the arenium ion, or sigma complex) in which the positive charge and the remaining four $\pi$ electrons are delocalised over five of the six ring carbons, while the carbon bonded to the incoming $\mathrm{NO_2}$ group becomes $sp^3$. A base ($\mathrm{HSO_4^-}$) then removes the $\mathrm{H^+}$ from that carbon, restoring the full delocalised aromatic ring and regenerating the $\mathrm{H_2SO_4}$ catalyst, giving nitrobenzene, $\mathrm{C_6H_5NO_2}$.苯环离域的 $\pi$ 电子体系进攻 $\mathrm{NO_2^+}$，生成非芳香的正电荷中间体（阳离子中间体，即 sigma 络合物），其中正电荷与剩余的四个 $\pi$ 电子离域在六个环碳中的五个上，而与新连接的 $\mathrm{NO_2}$ 基团成键的碳变为 $sp^3$ 杂化。随后碱（$\mathrm{HSO_4^-}$）从该碳上移去 $\mathrm{H^+}$，恢复完整的离域芳香环，并再生 $\mathrm{H_2SO_4}$ 催化剂，生成硝基苯 $\mathrm{C_6H_5NO_2}$。

Why substitution, not addition: benzene's ring has extra thermodynamic stability from its fully delocalised system of six $\pi$ electrons, so the ring undergoes substitution, which restores that aromatic stabilisation once the electrophile is incorporated, rather than addition, which would permanently destroy the delocalisation and give a far less stable, non-aromatic product, as would happen with an alkene's isolated, localised double bond.为何是取代而非加成：苯环因其完全离域的六个 $\pi$ 电子体系而具有额外的热力学稳定性，因此该环发生取代反应，使亲电试剂结合后仍能恢复芳香稳定性，而不是像烯烃中孤立、局域化的双键那样发生加成反应，因为加成会永久破坏离域体系，生成稳定性低得多的非芳香产物。

Mechanisms of Chemical Change: Solutions化学变化的机理：解析

Multiple Choice选择题

Structured Response结构化解答题

Organic Mechanisms (HL)有机反应机理（HL）