Preparation of alkyl halides — from alcohol, alkane, alkene routes

Question

What are the main methods of preparing alkyl halides from alcohols, alkanes, and alkenes? Which reagent and conditions should we pick for each route?

Solution — Step by Step

Alcohols are converted to alkyl halides by replacing the -OH group with a halogen. Three main reagent choices:

For R-Cl:

\text{R-OH} + \text{SOCl}_2 \xrightarrow{\text{pyridine}} \text{R-Cl} + \text{SO}_2 + \text{HCl}

Thionyl chloride ( $\text{SOCl}_2$ ) is the best reagent because both byproducts ( $\text{SO}_2$ and HCl) are gases — pure product without purification. This is called Darzen’s method.

For R-Br:

\text{3R-OH} + \text{PBr}_3 \rightarrow \text{3R-Br} + \text{H}_3\text{PO}_3

For R-I:

\text{R-OH} + \text{HI} \rightarrow \text{R-I} + \text{H}_2\text{O}

HI is the most reactive of the hydrogen halides (reactivity: HI > HBr > HCl).

With HCl + ZnCl $_2$ (Lucas reagent), the reaction rate depends on the type of alcohol:

Tertiary alcohol: immediate turbidity (instant reaction)
Secondary: turbidity in 5-10 minutes
Primary: no reaction at room temperature

This reactivity order ( $3^\circ > 2^\circ > 1^\circ$ ) follows carbocation stability.

\text{CH}_4 + \text{Cl}_2 \xrightarrow{h\nu \text{ or } \Delta} \text{CH}_3\text{Cl} + \text{HCl}

This is a free radical substitution reaction proceeding through initiation, propagation, and termination steps. Selectivity matters: Br $_2$ is more selective than Cl $_2$ . For alkanes with different types of H atoms, Br $_2$ preferentially attacks the most substituted carbon.

Reactivity of H atoms: $3^\circ > 2^\circ > 1^\circ$

\text{CH}_2\text{=CH}_2 + \text{HBr} \rightarrow \text{CH}_3\text{CH}_2\text{Br}

For unsymmetrical alkenes, Markovnikov’s rule applies: the halogen goes to the more substituted carbon (via the more stable carbocation intermediate).

Anti-Markovnikov addition (peroxide effect):

\text{CH}_3\text{CH=CH}_2 + \text{HBr} \xrightarrow{\text{peroxide}} \text{CH}_3\text{CH}_2\text{CH}_2\text{Br}

This works only with HBr (not HCl or HI) — called the Kharasch effect.

flowchart TD
    A["Need to prepare R-X"] --> B{"Starting material?"}
    B --> C["Alcohol R-OH"]
    B --> D["Alkane R-H"]
    B --> E["Alkene C=C"]
    C --> F["R-Cl: use SOCl2 + pyridine"]
    C --> G["R-Br: use PBr3"]
    C --> H["R-I: use HI or PI3"]
    D --> I["Free radical halogenation with Cl2/Br2 + hv"]
    E --> J["HX addition: Markovnikov"]
    E --> K["HBr + peroxide: Anti-Markovnikov"]

Why This Works

Each method targets a different type of bond: alcohol methods break the C-OH bond (nucleophilic substitution), alkane methods break the C-H bond (free radical), and alkene methods break the C=C pi bond (electrophilic addition). Choosing the right method depends on what functional group you are starting from and what regiochemistry you need in the product.

Alternative Method

For preparing aryl halides (not alkyl), electrophilic aromatic substitution using $\text{X}_2$ /Lewis acid is the standard route. Sandmeyer reaction from diazonium salts gives even better control over the position of halogen.

Common Mistake

Students often write “HCl + ZnCl $_2$ can convert any alcohol to an alkyl chloride.” This is wrong for primary alcohols — the Lucas reagent does NOT react with primary alcohols at room temperature. For primary R-Cl, always use SOCl $_2$ (Darzen’s method). NEET and JEE regularly give options that test whether you know this limitation.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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