Preparation of Haloalkanes and Haloarenes

Route 1 — haloalkanes from alcohols (the main route). An alcohol's $-OH$ is replaced by $-X$ using one of three reagent families.

where: $R-OH$ is the alcohol, $HX$ is a hydrogen halide ($HI$, $HBr$, or $HCl$ with a $ZnCl_2$ catalyst), and $R-X$ is the haloalkane product. Reactivity of the acid follows $HI > HBr > HCl$, and reactivity of the alcohol follows $3° > 2° > 1°$ — this order is the chemistry behind the Lucas test used to distinguish alcohol classes. Phosphorus halides ($PCl_5$, or $PCl_3$/red phosphorus with $X_2$ for bromides and iodides) give cleaner products for many alcohols: $3R-OH + PCl_5 \rightarrow 3R-Cl + H_3PO_3$ (plus $POCl_3$ as a companion by-product). The preferred laboratory method, though, is thionyl chloride:

where: both by-products, $SO_2$ and $HCl$, escape as gases, leaving a pure alkyl chloride with no aqueous or solid residue to separate out (the Darzens procedure).

Route 2 — haloalkanes from hydrocarbons. Free-radical halogenation, $R-H + X_2 \xrightarrow{h\nu\ \text{or}\ \Delta} R-X + HX$, proceeds through initiation, propagation and termination steps but attacks carbons fairly indiscriminately, giving a mixture of products — useful only for simple, symmetric substrates. Alkenes can also add $HX$ (Markovnikov's rule, a Class 11 idea) or $X_2$ across the double bond to give a haloalkane directly.

Route 3 — haloarenes from arenes. Benzene undergoes electrophilic aromatic substitution with a halogen only in the presence of a Lewis acid catalyst:

where: the Lewis acid ($Fe$, $FeCl_3$, or $FeBr_3$) polarises the halogen molecule to generate the electrophile $X^+$, which the electron-rich benzene ring then attacks. Halogenated benzenes bearing an amino-derived substituent can also be made from diazonium salts (Sandmeyer and Balz–Schiemann reactions) — that route belongs to the Amines unit and is only named here.

There is no numerical formula to derive here — the topic is a set of substitution mechanisms — but the reactivity trends have a clean physical explanation worth walking through carefully.

Why tertiary alcohols react fastest with HX. The $R-OH + HX$ reaction proceeds by protonating the oxygen (turning it into the good leaving group $H_2O$), then losing water to form a carbocation, which the halide ion then captures. A tertiary carbocation, $R_3C^+$, is stabilised by three alkyl groups donating electron density and is comparatively long-lived; a primary carbocation is not stabilised at all and barely forms. This is exactly why tertiary alcohols react with plain HCl at room temperature while primary alcohols need the $ZnCl_2$ catalyst and heating — the catalyst helps generate a leaving group good enough to compensate for the unstable intermediate.

Why $SOCl_2$ is preferred over HX or $PCl_5$. A "clean" preparative reaction is one whose by-products leave the flask on their own. With $SOCl_2$, both by-products ($SO_2$ and $HCl$) are gases that bubble away, so the alkyl chloride needs no further purification. The $PCl_5$ route leaves behind $POCl_3$, a liquid that must be separated from the product; the HX route can leave water and excess acid mixed in. Chemists therefore reach for $SOCl_2$ whenever a pure primary or secondary chloride is the goal.

Why free-radical halogenation is a poor synthetic route. Once a chain reaction starts (initiation by light or heat homolysing $X_2$ into radicals), the radical can abstract a hydrogen from any similar carbon in the molecule, not just one preferred site. Every accessible C–H bond becomes a competing site, so the product is a statistical mixture — useful for teaching mechanism, but rarely useful for making one specific haloalkane in good yield.

Why haloarenes need a catalyst but haloalkanes from HX do not. Benzene's ring electrons are delocalised and only moderately reactive; $Cl_2$ or $Br_2$ alone cannot polarise enough to attack it. A Lewis acid catalyst grabs a lone pair from the halogen, pulling electron density away and leaving a strongly electrophilic $X^{\delta+}$ end that the ring can attack. Without the catalyst, the electrophile is simply too weak to react. Contrast this with phenol: its $-OH$ is directly attached to the ring, so the oxygen's lone pair conjugates into the ring, giving the C–O bond partial double-bond character. That strengthened bond resists the same substitution alcohols undergo, which is why phenols are not converted to haloarenes by HX or $PCl_5$ — direct ring halogenation or a diazonium-salt route is needed instead.

Problem. Starting from ethanol, $CH_3CH_2OH$, prepare ethyl chloride using thionyl chloride, then compare how this same alcohol would behave with plain HCl, and finally predict how tert-butyl alcohol would behave differently under the mildest of these conditions.

Step 1 — Write the $SOCl_2$ reaction.

Pyridine mops up the $HCl$ produced, pushing the reaction to completion and avoiding side reactions; both by-products then escape as gases, leaving pure ethyl chloride.

Step 2 — Predict the outcome with plain HCl instead. Ethanol is a primary alcohol, so its carbocation intermediate (were it to form by the $HX$ mechanism) would be highly unstable. In practice ethanol reacts only sluggishly with concentrated HCl and needs the $ZnCl_2$ catalyst (Lucas reagent) plus heating to go to completion — this is markedly slower and messier than the $SOCl_2$ route, which is why $SOCl_2$ is preferred whenever it is available.

Step 3 — Contrast with tert-butyl alcohol under the mildest conditions. $(CH_3)_3C-OH$ reacts with concentrated HCl alone, at room temperature, with no catalyst needed at all, because protonation and loss of water generates the relatively stable tertiary carbocation $(CH_3)_3C^+$, which chloride ion then captures:

Step 4 — State the conclusion. The reactivity order $3° > 2° > 1°$ for alcohols with HX, and the preference for $SOCl_2$ with primary/secondary alcohols, are two faces of the same idea: pick the reagent whose mechanism suits the stability of the intermediate (or by-product profile) for that class of alcohol. Ethanol gives clean ethyl chloride fastest via $SOCl_2$; tert-butyl alcohol gives tert-butyl chloride instantly with plain HCl alone.

• Treating phenols like alcohols. Phenol's C–OH bond has partial double-bond character from ring conjugation and resists HX/$PCl_5$ substitution; haloarenes come from direct ring halogenation or diazonium salts, never from phenol plus HX.
• Forgetting the Lewis acid catalyst for ring halogenation. Plain $Cl_2$ or $Br_2$ does not react with benzene; only $Fe$, $FeCl_3$, or $FeBr_3$ generates the electrophile $X^+$ needed.
• Applying one set of HX conditions to all alcohols. Primary alcohols typically need $ZnCl_2$/heat; tertiary alcohols react instantly at room temperature — the reactivity order $3° > 2° > 1°$ must be stated, not assumed away.
• Assuming free-radical halogenation is a good synthetic method. It gives a mixture of mono- and poly-halogenated isomers except for very simple, symmetric substrates, so it is rarely the preferred preparative route.

Ask the tutor: Want to walk through why each reagent's by-products determine whether it is the "preferred" method, or test yourself on predicting the right route for a given alcohol?