Preparation of haloalkanes

Route 1 — alkane to haloalkane (free-radical halogenation). Alkanes are unreactive to almost everything except halogens under UV light or heat, which trigger a free-radical chain reaction:

where: $h\nu$ denotes the light energy that homolyses $Cl_2$ into two $Cl^\bullet$ radicals, which then abstract an H from the alkane. Because every C-H bond in the molecule is a potential target, this route gives a mixture of mono-, di- and polyhalogenated products and is rarely used to make a single pure haloalkane on purpose, except when the starting alkane has only one kind of hydrogen (e.g. $CH_4$, or neopentane).

Route 2 — alkene to haloalkane (electrophilic addition of $HX$). Alkenes add $HX$ across the $C=C$ bond. For an unsymmetrical alkene, Markovnikov's rule decides which carbon gets the halogen: H adds to the carbon that already carries more hydrogens, and X adds to the more substituted carbon (because that pathway goes through the more stable carbocation intermediate). $HBr$ alone, in the presence of peroxides, reverses this outcome and goes anti-Markovnikov through a free-radical chain (the Kharasch effect) — this exception is specific to HBr and does not occur with HCl or HI, whose radical steps are respectively too slow or thermodynamically unfavourable.

Route 3 — alcohol to haloalkane. This is the workhorse route, because alcohols are cheap and widely available. Three reagents do the same overall job, $-OH \to -X$, but leave different by-products behind: $PX_3$ gives phosphorous acid, $PCl_5$ gives phosphoryl chloride and $HCl$ gas, and $SOCl_2$ (thionyl chloride, in the Darzens process) gives only gaseous $SO_2$ and $HCl$, both of which simply bubble out of the flask.

Route 4 — halide exchange. An existing haloalkane can be converted to a different haloalkane of the same carbon skeleton. The Finkelstein reaction swaps Cl or Br for I using $NaI$ in dry acetone; the Swarts reaction swaps Cl or Br for F using a metal fluoride such as $AgF$, $Hg_2F_2$, $CoF_2$ or $SbF_3$.

Route 5 — Hunsdiecker reaction. A silver salt of a carboxylic acid, refluxed with $Br_2$ in dry $CCl_4$, loses $CO_2$ and gains a bromine, shortening the chain by exactly one carbon:

where: $R\text{-}COOAg$ is the dry silver salt of the fatty acid, and the driving force is the escape of $CO_2$ gas together with the precipitation of insoluble $AgBr$, both of which pull the reaction to completion.

Every one of these five routes ends the same way — a carbon is left bonded to a halogen — but the paths differ because each starting material offers a different "leaving" event, and the reagent is picked to make that event as clean as possible. In alcohol-to-haloalkane conversion, the underlying step is substitution at the carbon bearing $-OH$: the reagent's job is to convert the poor leaving group $OH^-$ into a good leaving group before substitution happens. $PCl_5$ and $PX_3$ do this by forming an intermediate ester-like species that departs as phosphorus-containing by-products, while $SOCl_2$ forms a chlorosulfite intermediate that collapses to release $SO_2$ and $Cl^-$ directly at the carbon. The reason $SOCl_2$ is preferred in the laboratory is not that its mechanism is fundamentally different — it is that both of its by-products, $SO_2$ and $HCl$, are gases. They leave the flask on their own, so the alkyl chloride needs no separation from a solid or liquid contaminant, unlike the $POCl_3$ left behind by $PCl_5$ or the $H_3PO_3$ left behind by $PX_3$. In halide exchange, Finkelstein and Swarts are pulled forward by the same trick in opposite directions: Finkelstein removes $Cl^-$ or $Br^-$ from solution as an insoluble precipitate ($NaCl$ or $NaBr$) in dry acetone, so the equilibrium is dragged toward the iodide product even though iodide is not intrinsically a "stronger" nucleophile in every sense — it is simply that removing a product forces the reaction onward (Le Chatelier's principle applied to a substitution equilibrium). Swarts instead exploits the very strong $C-F$ bond that forms, which is thermodynamically more stable than the $C-Cl$ or $C-Br$ bond it replaces, so metal fluorides drive the exchange in the fluorine direction. In the Hunsdiecker reaction, the driving force is decarboxylation: the carboxylate radical formed after bromine attacks the silver salt is unstable and rapidly loses $CO_2$ to give a carbon-centred radical, which then abstracts a bromine atom from $Br_2$. Because the lost fragment is exactly the $-COOAg$ group, the alkyl halide product always has one carbon fewer than the starting acid — this chain-shortening is the diagnostic feature of the reaction, not an accident.

Starting from propan-1-ol, show how the same $-OH \to -Cl$ conversion looks with three different reagents, then use ethanoic acid to make bromomethane by the Hunsdiecker reaction, and state which alcohol route is cleanest and why.

Step 1. Convert propan-1-ol using $PCl_5$: $CH_3CH_2CH_2OH + PCl_5 \rightarrow CH_3CH_2CH_2Cl + POCl_3 + HCl$. Both $POCl_3$ (a liquid) and $HCl$ (a gas) form, so the product must be separated from residual $POCl_3$.

Step 2. Convert the same alcohol using $PCl_3$: $3\,CH_3CH_2CH_2OH + PCl_3 \rightarrow 3\,CH_3CH_2CH_2Cl + H_3PO_3$. Here $H_3PO_3$ is a solid/liquid by-product that again needs separation from the alkyl chloride.

Step 3. Convert the same alcohol using $SOCl_2$ with a trace of pyridine: $CH_3CH_2CH_2OH + SOCl_2 \rightarrow CH_3CH_2CH_2Cl + SO_2\uparrow + HCl\uparrow$. Both by-products are gases and escape the reaction vessel unaided, so the 1-chloropropane collected needs no further purification from solid or liquid contaminants — this is why $SOCl_2$ is the reagent of choice.

Step 4. Now start the Hunsdiecker sequence from ethanoic acid. First form the silver salt: $CH_3COOH + AgOH \rightarrow CH_3COOAg + H_2O$. Then reflux the dry salt with bromine in $CCl_4$: $CH_3COOAg + Br_2 \xrightarrow{CCl_4,\ \Delta} CH_3Br + CO_2\uparrow + AgBr\downarrow$. Ethanoic acid has two carbons; the product $CH_3Br$ has one, because the $-COOAg$ carbon leaves as $CO_2$ — exactly the one-carbon shortening rule for Hunsdiecker. Answer: of the three alcohol routes, $SOCl_2$ gives the purest 1-chloropropane because both its by-products are gases, and the Hunsdiecker sequence on ethanoic acid gives bromomethane, one carbon shorter than the starting acid.

• Applying Markovnikov's rule to $HBr$ addition without checking for peroxides — only $HBr$ (not $HCl$ or $HI$) switches to anti-Markovnikov via the Kharasch radical mechanism.
• Treating free-radical halogenation of alkanes as a clean, single-product method — it gives a mixture unless the alkane has only one type of hydrogen, unlike the alcohol routes.
• Swapping Finkelstein (halide $\to$ iodide, $NaI$/dry acetone) with Swarts (halide $\to$ fluoride, metal fluorides) — same reaction type, opposite target halogen.
• Miscounting carbons in a Hunsdiecker product — it always has one fewer carbon than the starting acid because the carboxyl carbon leaves as $CO_2$.
• Forgetting to dry the alcohol before adding $PCl_5$ or $SOCl_2$ — trace moisture reacts with these reagents first and lowers the yield of the alkyl halide.

Ask the tutor: Want to practice choosing the right reagent for a specific alcohol-to-haloalkane conversion, or work through more Hunsdiecker chain-shortening examples?