Methods of Preparation of Haloalkanes

Haloalkanes are among the most versatile starting materials in organic chemistry, but they do not simply appear on their own. Chemists need reliable routes to build them from simpler, easily available compounds. Three major pathways exist: converting alcohols, modifying hydrocarbons, and swapping one halogen for another. Each route has its own strengths and limitations, and knowing when to use which method is a key practical skill.

Starting from Alcohols: Replacing $-OH$ with a Halogen

Alcohols are widely available and inexpensive, making them an obvious starting point. The core strategy is straightforward: remove the hydroxyl group ($-OH$) from the alcohol and put a halogen atom in its place. Several reagents can do this, and each brings different advantages.

Hydrogen Halides: The Most Direct Route

Reacting an alcohol with a concentrated hydrogen halide ($HCl$, $HBr$, or $HI$) is the simplest approach. The halide ion swaps places with the hydroxyl group, and water forms as a by-product:

$R-OH + HCl \xrightarrow{ZnCl_2} R-Cl + H_2O$

Not all alcohols respond the same way:

• Tertiary alcohols (3°) are the most reactive. All it takes is mixing the alcohol with concentrated $HCl$ at room temperature and giving the flask a shake. No catalyst is needed, and the reaction proceeds readily.
• Primary and secondary alcohols react much more slowly with $HCl$ and need anhydrous zinc chloride ($ZnCl_2$) as a catalyst to get things moving.

The overall reactivity order with any given haloacid is:

$3° > 2° > 1°$

For making alkyl chlorides specifically, you can either pass dry $HCl$ gas through a solution of the alcohol or heat the alcohol with concentrated aqueous hydrochloric acid.

For alkyl bromides, the standard method uses constant-boiling hydrobromic acid (48% $HBr$):

$R-OH + NaBr + H_2SO_4 \longrightarrow R-Br + NaHSO_4 + H_2O$

For alkyl iodides, heat the alcohol together with sodium iodide ($NaI$) or potassium iodide ($KI$) dissolved in 95% phosphoric acid ($H_3PO_4$). This combination delivers good yields of the desired iodide product.

Phosphorus Halides: Converting Three Alcohol Molecules at Once

Phosphorus trihalides ($PX_3$, where $X = Cl$ or $Br$) are efficient because one molecule converts three alcohol molecules in a single step:

$3R-OH + PX_3 \longrightarrow 3R-X + H_3PO_3 \qquad (X = Cl, Br)$

Phosphorus pentachloride ($PCl_5$) also works, though it handles only one alcohol molecule at a time and generates two extra by-products:

$R-OH + PCl_5 \longrightarrow R-Cl + POCl_3 + HCl$

There is a handy practical trick for preparing bromides and iodides. Rather than purchasing $PBr_3$ or $PI_3$ (which are unstable and difficult to store), chemists generate them in situ (directly inside the reaction flask) by mixing red phosphorus with bromine or iodine:

$R-OH \xrightarrow{red \ P/X_2} R-X \qquad (X_2 = Br_2, I_2)$

The red phosphorus reacts with the halogen to form $PBr_3$ or $PI_3$ on the spot. This freshly formed phosphorus halide then immediately converts the alcohol to the alkyl halide.

Thionyl Chloride: The Cleanest Route to Alkyl Chlorides

Of all the reagents for turning an alcohol into an alkyl chloride, thionyl chloride ($SOCl_2$) is the most popular choice in the laboratory:

$R-OH + SOCl_2 \longrightarrow R-Cl + SO_2 + HCl$

The reason is simple and elegant. Both by-products, sulfur dioxide ($SO_2$) and hydrogen chloride ($HCl$), are gases. They bubble out of the reaction mixture on their own, leaving behind only the pure alkyl chloride. No extra purification step is needed at all. This is a major practical advantage over the other methods, where liquid or solid by-products remain mixed with the product and must be separated.

Why These Methods Fail for Aryl Halides

A natural question arises: if replacing the $-OH$ works so well for alcohols, can the same approach convert phenols (compounds where $-OH$ is attached to an aromatic ring) into aryl halides? The answer is no.

In phenols, one of the lone pairs (unshared electron pairs) on the oxygen atom overlaps with the aromatic ring through resonance (sharing of electrons with the ring’s $\pi$ system). This overlap gives the $C-O$ bond a partial double-bond character, making it considerably stronger than the ordinary single $C-O$ bond found in alcohols. None of the reagents discussed above can overcome this extra bond strength, so different strategies are needed to prepare aryl halides.

Starting from Hydrocarbons

When alcohols are not convenient as starting materials, hydrocarbons themselves can serve as the source. Two broad strategies are available: replacing hydrogen atoms in alkanes and adding halogen atoms across double bonds in alkenes.

Free Radical Halogenation of Alkanes

Alkanes react with $Cl_2$ or $Br_2$ under UV light or heat through a free radical mechanism (a chain reaction driven by highly reactive species carrying unpaired electrons):

$CH_3CH_2CH_2CH_3 \xrightarrow{Cl_2 / UV} CH_3CH_2CH_2CH_2Cl + CH_3CH_2CHClCH_3$

The big downside of this approach is its lack of selectivity. The halogen radical can pull off any hydrogen atom in the molecule, so the reaction produces a messy mixture of different monochlorinated isomers alongside dichlorinated, trichlorinated, and more heavily substituted by-products. Isolating any one pure compound from this mixture is difficult, and the yield of each individual product remains low.

Because of this selectivity problem, free radical halogenation is rarely the first choice for preparing a specific haloalkane in the laboratory. On an industrial scale, however, the mixture can be separated economically, making this route practical for large-volume production.

Adding $HX$ Across Double Bonds: Markovnikov’s Rule in Action

Alkenes provide a much cleaner preparation route. When a hydrogen halide ($HCl$, $HBr$, or $HI$) adds across a $C=C$ double bond, an alkyl halide forms directly:

With a symmetrical alkene (where both carbons of the double bond carry identical groups), only one product is possible. But with an unsymmetrical alkene, two different products can form depending on which carbon receives the hydrogen and which gets the halogen.

This is where Markovnikov’s rule steps in: the hydrogen atom from $HX$ adds to the carbon of the double bond that already carries more hydrogen atoms, while the halogen goes to the carbon with fewer hydrogens. A helpful way to remember this is “the rich get richer”: the hydrogen-rich carbon gains yet another hydrogen.

For example, when propene reacts with $HI$:

$CH_3CH=CH_2 + HI \longrightarrow \underset{\text{minor}}{CH_3CH_2CH_2I} + \underset{\text{major}}{CH_3CHICH_3}$

The major product is 2-iodopropane because the hydrogen adds to the terminal $=CH_2$ (which already has more hydrogens), placing the iodine on the middle carbon.

Adding Halogens to Alkenes: The Bromine Colour Test

When bromine dissolved in carbon tetrachloride ($CCl_4$) is added to an alkene, the reddish-brown colour of bromine disappears. This colour change, known as the bromine decolourisation test, is one of the simplest and most reliable laboratory methods for detecting a $C=C$ double bond in any organic molecule.

The product of this addition is a vic-dibromide, short for vicinal (from the Latin vicinalis, meaning neighbouring) dibromide, which is colourless. The two bromine atoms end up on adjacent carbon atoms:

$H_2C=CH_2 + Br_2 \xrightarrow{CCl_4} BrCH_2-CH_2Br$

This is different from a gem-dihalide (from the Latin geminus, meaning twin), where both halogen atoms sit on the same carbon atom.

Worked Example 6.3: Finding All Monochloro Isomers

Problem: Identify all the possible monochloro structural isomers expected when $(CH_3)_2CHCH_2CH_3$ (2-methylbutane) undergoes free radical monochlorination.

Solution:

The strategy is to locate every structurally distinct type of hydrogen atom in the molecule. Each unique hydrogen type, when replaced by chlorine, gives a different structural isomer.

Drawing out 2-methylbutane and examining its structure reveals four distinct hydrogen environments:

1. The hydrogens on the two equivalent $-CH_3$ groups attached to the branching carbon
2. The single hydrogen on the branching $CH$ carbon itself
3. The hydrogens on the $-CH_2-$ group
4. The hydrogens on the terminal $-CH_3$ at the end of the chain

Replacing each type with a chlorine atom produces four monochloro isomers:

• $(CH_3)_2CHCH_2CH_2Cl$
• $(CH_3)_2CHCHClCH_3$
• $(CH_3)_2CClCH_2CH_3$
• $CH_3CH(CH_2Cl)CH_2CH_3$

Therefore, free radical monochlorination of 2-methylbutane gives four distinct monochloro structural isomers.

Swapping One Halogen for Another: Exchange Reactions

Sometimes you can easily introduce one halogen but actually need a different one. Halogen exchange reactions solve this problem by letting you replace the existing halogen with the one you want.

The Finkelstein Reaction: A Route to Alkyl Iodides

When an alkyl chloride or bromide is treated with sodium iodide ($NaI$) in dry acetone, the iodide ion displaces the existing halogen:

$R-X + NaI \xrightarrow{\text{dry acetone}} R-I + NaX \qquad (X = Cl, Br)$

This reaction, named after the German chemist Hans Finkelstein, works because of a clever solvent choice. Sodium iodide dissolves readily in acetone, but sodium chloride and sodium bromide are insoluble in acetone. As the reaction proceeds, $NaCl$ or $NaBr$ precipitates out of solution as a solid. According to Le Chatelier’s Principle, removing a product from the equilibrium shifts the reaction further toward the product side. This continuous removal of the by-product drives the reaction to completion.

The Swarts Reaction: The Best Way to Make Alkyl Fluorides

Alkyl fluorides are the hardest to prepare by direct methods. The Swarts reaction provides an elegant solution: heat an alkyl chloride or bromide with a metallic fluoride such as $AgF$, $Hg_2F_2$, $CoF_2$, or $SbF_3$:

$CH_3Br + AgF \longrightarrow CH_3F + AgBr$

The fluoride ion from the metal salt displaces the existing halogen, producing the desired alkyl fluoride. This remains the most dependable laboratory method for synthesising fluoroalkanes.