卤素化合物

A halogenoalkane 卤代烷 is an alkane with one or more halogen atoms in place of hydrogen (a C–X bond, where X is a halogen).

Making halogenoalkanes

• free-radical substitution 自由基取代 of an alkane with $\text{Cl}_2$ or $\text{Br}_2$ in ultraviolet light.
• electrophilic addition 亲电加成 of an alkene with a halogen $\text{X}_2$ or a hydrogen halide $\text{HX}$.
• substitution of an alcohol 醇, for example by $\text{HX}$, by $\text{PCl}_5$, by $\text{PCl}_3$ with heat, or by $\text{SOCl}_2$.

Three classes

A halogenoalkane is primary 伯, secondary 仲 or tertiary 叔, depending on how many carbon atoms are joined to the carbon that holds the halogen (one, two or three).

The C–X bond is polar, so the carbon is slightly positive. A nucleophilic substitution 亲核取代 happens when a nucleophile 亲核试剂 (a lone-pair species) attacks that carbon and replaces the halogen.

To identify the halogen, warm the halogenoalkane with silver nitrate 硝酸银 in ethanol. A silver halide precipitate 沉淀 forms, and its colour shows which halogen is present (white $\text{AgCl}$, cream $\text{AgBr}$, yellow $\text{AgI}$).

The same halogenoalkane can instead undergo elimination 消去 to form an alkene 烯烃. The conditions decide which reaction wins:

• $\text{NaOH}$ in water → nucleophilic substitution → an alcohol.
• $\text{NaOH}$ in ethanol, heated → elimination → an alkene.

Nucleophilic substitution can follow two routes:

• S$_\text{N}$2: one step. The nucleophile attacks at the same time as the halogen leaves, passing through a crowded transition state 过渡态 where both are half-bonded. The rate depends on both the halogenoalkane and the nucleophile.
• S$_\text{N}$1: two steps. First the C–X bond breaks to give a carbocation 碳正离子; then the nucleophile attacks it. The rate depends only on the halogenoalkane.

Drawing the mechanism for full marks. A curly arrow 弯箭头 shows a pair of electrons moving. It must start at a lone pair 孤对电子 or at a bond, and end where the new bond forms (or on the atom that takes the pair). For S$_\text{N}$2 the marks are usually:

1. the dipole 偶极 on the C–Br bond: $\delta+$ on C, $\delta-$ on Br.
2. a curly arrow from a lone pair on the O of $\text{OH}^-$ to the $\delta+$ carbon (draw the lone pair).
3. a curly arrow from the C–Br bond to the Br.
4. the transition state: square brackets, dotted HO···C and C···Br bonds, and an overall $-$ charge.
5. the products, including $\text{Br}^-$.

For S$_\text{N}$1 there are two steps, so two diagrams: the C–Br bond breaks first (arrow from the bond to Br) to give the carbocation with a full $+$ on the carbon, and then the lone pair of $\text{OH}^-$ attacks that $\text{C}^+$.

Why the rate equations differ. In S$_\text{N}$1 the slow step involves only the halogenoalkane, so rate $= k[\text{RBr}]$ (first order). In S$_\text{N}$2 both species meet in the one step, so rate $= k[\text{RBr}][\text{OH}^-]$ (second order). The "1" and "2" in the names count the species in the rate-determining step 决速步骤, not the number of steps.

Common mistakes.

• An arrow starting from the minus sign or from the H of $\text{OH}^-$. It must start at a lone pair on the oxygen.
• An arrow pointing from the Br to the carbon. Arrows show electrons moving away from the bond, into the Br.
• Forgetting the charge on the transition state, or drawing full bonds (not dotted ones) inside the brackets.
• Drawing a transition state for S$_\text{N}$1. S$_\text{N}$1 has a real intermediate 中间体 (the carbocation) instead.
• Saying tertiary halogenoalkanes use S$_\text{N}$1 "because they are more reactive". The reason is that three alkyl groups block the back of the carbon and stabilise the carbocation.

Alkyl groups push electrons towards the positive carbon (the inductive effect 诱导效应), so they stabilise the carbocation. This is why:

• primary halogenoalkanes mostly react by S$_\text{N}$2.
• tertiary halogenoalkanes mostly react by S$_\text{N}$1 (their carbocation is well stabilised).
• secondary halogenoalkanes use a mixture of the two.

How fast a halogenoalkane reacts depends on the strength of the C–X bond, measured by its bond energy 键能. The C–I bond is the weakest, so iodoalkanes react fastest; the C–Cl bond is the strongest of the three, so chloroalkanes react slowest. So when tested with silver nitrate, an iodoalkane gives its precipitate first — its higher reactivity 反应活性 comes from the weaker C–X bond.

Worked example. Predict the mechanism for the hydrolysis of 1-bromobutane and of 2-bromo-2-methylpropane. Classify the halogenoalkane first. 1-bromobutane is primary: the carbon carrying the $\text{Br}$ is barely shielded, so the nucleophile can attack the back of it and the mechanism is $\text{S}_\text{N}2$ - one step, with the rate depending on both the halogenoalkane and the nucleophile. 2-bromo-2-methylpropane is tertiary: three bulky methyl groups block that attack, but they also stabilise the carbocation formed once the $\text{Br}$ leaves, so it goes $\text{S}_\text{N}1$ - two steps, with the rate depending on the halogenoalkane only. Decide from the class (primary → $\text{S}_\text{N}2$, tertiary → $\text{S}_\text{N}1$), and note that the tertiary one hydrolyses faster despite being the more crowded.

• Draw nucleophilic substitution with a curly arrow from the nucleophile's lone pair to the $\delta+$ carbon and one from the $\text{C-X}$ bond.
• Match mechanism to class: $S_\text{N}1$ (tertiary, carbocation, two steps) vs $S_\text{N}2$ (primary, one step, transition state).
• Reactivity is set by bond enthalpy: $\text{C-I}$ is weakest, so iodoalkanes react fastest (not electronegativity).
• Elimination (hot, ethanolic KOH) vs substitution (warm, aqueous KOH) — the conditions decide the product; state them.