20.17
自由基卤化:热力学
反应的热力学有利性由吉布斯自由能 (ΔG) 的变化决定。 ΔG 有两个组成部分 - 焓 (ΔH) 和熵 (ΔS)。 对于烷烃卤化，熵分量可以忽略不计，因为反应物和产物分子的数量相等。 在这种情况下，ΔG 仅由焓分量决定。 决定 ΔH 的最关键因素是键的强度。 ΔH 可以通过比较断裂的键和形成的键之间…
烷烃的自由基卤化反应中，不同卤素的反应活性顺序不同。
这一差异可以从反应的热力学角度来理解。
对于烷烃卤化反应，由于反应物和产物的分子数相等，熵变项可忽略不计。因此，ΔG 等于 ΔH。
以甲烷的自由基氟化反应为例。该反应的焓变可通过被断裂键和新形成键的键解离能进行估算。
该反应的总ΔH为负值且绝对值较大，使得反应在热力学上有利，但具有高度爆炸性且不具实际可行性。
相反，对于碘化反应，正值的ΔH使其在热力学上不利，因而无法实现。
因此，仅氯化和溴化具有负的ΔH 数值在热力学上是有利的，并且在实践中是可行的。
两种反应的比较表明，溴化反应比氯化反应更慢。
仔细观察各个链增长步骤可知，溴化反应的第一个链增长步骤吸热更多，这会影响整体反应速率。
由于总反应仍然是放热的，溴化反应能够发生，但速度较慢。
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Q1: Why is radical fluorination of methane thermodynamically favorable but impractical?
Radical fluorination of methane has a large negative enthalpy change (ΔH = -431 kJ/mol), making it thermodynamically favorable. However, the reaction is highly exothermic and explosive, releasing extreme energy that makes it dangerous and unsuitable for laboratory or synthetic applications.
Q2: What determines whether radical halogenation of alkanes is thermodynamically favorable?
For alkane halogenation, entropy change is negligible because reactant and product molecule counts are equal, so Gibbs free energy (ΔG) depends solely on enthalpy (ΔH). The strength of bonds broken versus bonds formed determines ΔH. A negative ΔH indicates a thermodynamically favorable reaction.
Q3: Why is radical iodination of alkanes thermodynamically impossible?
Radical iodination has a positive enthalpy change (ΔH = +55 kJ/mol), resulting in a positive Gibbs free energy value. This positive ΔG makes the reaction thermodynamically unfavorable and prevents it from occurring under normal conditions, making iodination impossible to achieve.
Q4: How do chlorination and bromination compare in terms of reaction rate?
Bromination is slower than chlorination. The first propagation step (hydrogen abstraction) is exothermic for chlorination with low activation energy, but endothermic for bromination with high activation energy. Despite bromination's negative overall ΔH (-33 kJ/mol), the endothermic first step limits its rate.
Q5: What role does bond dissociation energy play in radical halogenation thermodynamics?
Bond dissociation energy determines the enthalpy change by comparing energy required to break bonds versus energy released when forming new bonds. Lower bond dissociation energies for bonds broken and higher energies for bonds formed produce more negative ΔH values, favoring thermodynamically favorable halogenation reactions.
Q6: Why are only chlorination and bromination practically feasible among halogenation reactions?
Chlorination (ΔH = -104 kJ/mol) and bromination (ΔH = -33 kJ/mol) both have negative enthalpy changes, making them thermodynamically favorable. Fluorination is too explosive, and iodination is thermodynamically unfavorable. This makes chlorination and bromination the only practical radical halogenation methods for alkanes.
Q7: How does the rate-determining step affect bromination versus chlorination?
The rate-determining step is hydrogen abstraction, the first propagation step. For chlorination, this step is exothermic with small activation energy, enabling fast reaction. For bromination, this step is endothermic with large activation energy, creating a kinetic barrier that slows the overall reaction despite favorable thermodynamics.