Methods of Preparation of Haloalkanes

Preparation from Alcohols

The most common way to make haloalkanes is from alcohols, replacing the -OH by -X. Several reagents do this:

(1) With hydrogen halides (HX): $$\text{R-OH} + \text{HX} \rightarrow \text{R-X} + \text{H}_2\text{O}$$ The reactivity order of the alcohols is 3° > 2° > 1°, and of the hydrogen halides is HI > HBr > HCl. (Lucas reagent - conc. HCl + ZnCl$_2$ - is used to distinguish 1°, 2°, 3° alcohols.)

(2) With phosphorus halides: $$\text{R-OH} + \text{PCl}_5 \rightarrow \text{R-Cl} + \text{POCl}_3 + \text{HCl}$$ $$3\text{R-OH} + \text{PCl}_3 \rightarrow 3\text{R-Cl} + \text{H}_3\text{PO}_3$$ (PBr$_3$ and PI$_3$, generated in situ from red P + Br$_2$/I$_2$, give bromides and iodides.)

(3) With thionyl chloride (SOCl$_2$) - the preferred lab method, because the by-products are gases that escape, leaving pure alkyl chloride: $$\text{R-OH} + \text{SOCl}_2 \rightarrow \text{R-Cl} + \text{SO}_2\uparrow + \text{HCl}\uparrow$$

Preparation from Hydrocarbons

(1) From alkanes — free-radical halogenation: Alkanes react with Cl$_2$ or Br$_2$ in the presence of UV light or heat to give a mixture of haloalkanes (substitution of H by X): $$\text{CH}_4 + \text{Cl}_2 \xrightarrow{\text{UV}} \text{CH}_3\text{Cl} + \text{HCl}$$ This gives mixtures (mono-, di-, poly-substituted), so it is of limited preparative use.

(2) From alkenes:

• Addition of HX (Markovnikov's rule - H adds to the carbon with more H's): $$\text{CH}_3\text{-CH=CH}_2 + \text{HBr} \rightarrow \text{CH}_3\text{-CHBr-CH}_3$$ With peroxides, HBr adds anti-Markovnikov (peroxide/Kharasch effect) giving CH$_3$-CH$_2$-CH$_2$Br.
• Addition of halogens (X$_2$) gives vicinal dihalides: $$\text{CH}_2\text{=CH}_2 + \text{Br}_2 \rightarrow \text{BrCH}_2\text{-CH}_2\text{Br}$$ (The decolourisation of bromine water is a test for unsaturation.)

Halogen Exchange Reactions

Two named reactions interconvert one halide for another:

(1) Finkelstein reaction — makes alkyl iodides from chlorides/bromides: $$\text{R-Cl (or R-Br)} + \text{NaI} \xrightarrow{\text{dry acetone}} \text{R-I} + \text{NaCl (or NaBr)}\downarrow$$ This works because NaCl and NaBr are insoluble in dry acetone but NaI is soluble - so the equilibrium is pulled toward the iodide as NaCl/NaBr precipitates.

(2) Swarts reaction - makes alkyl fluorides: $$\text{R-X} + \text{AgF (or Hg}_2\text{F}_2\text{, CoF}_2\text{, SbF}_3\text{)} \rightarrow \text{R-F} + \text{AgX}$$ Heating an alkyl chloride/bromide with a metallic fluoride replaces the halogen with fluorine.

Key Point — three named preparations to memorise:

• Finkelstein: R-X + NaI / dry acetone → R-I (makes iodides).
• Swarts: R-X + AgF/metal fluoride → R-F (makes fluorides).
• From alcohols: SOCl$_2$ is best (gaseous by-products → pure product).

[NEET Important] "Why dry acetone in the Finkelstein reaction?" — because NaI is soluble in acetone but NaCl/NaBr are not, so they precipitate and drive the reaction forward. This exact reasoning is frequently asked.

Solved Examples

Example 1: Finkelstein reaction

How would you convert 1-chlorobutane to 1-iodobutane?

Solution: Use the Finkelstein reaction — treat 1-chlorobutane with NaI in dry acetone: CH$_3$CH$_2$CH$_2$CH$_2$Cl + NaI → CH$_3$CH$_2$CH$_2$CH$_2$I + NaCl↓.

Example 2: Why dry acetone?

Why is dry acetone used as the solvent in the Finkelstein reaction?

Solution: NaI is soluble in acetone, but NaCl (and NaBr) are not. As the reaction proceeds, NaCl precipitates out, shifting the equilibrium toward the formation of the alkyl iodide.

Example 3: Swarts reaction

How is an alkyl fluoride prepared from an alkyl bromide?

Solution: By the Swarts reaction — heat the alkyl bromide with a metallic fluoride such as AgF, Hg$_2$F$_2$, CoF$_2$ or SbF$_3$: R-Br + AgF → R-F + AgBr.

Example 4: Alcohol with SOCl$_2$

Why is thionyl chloride preferred for converting an alcohol to an alkyl chloride?

Solution: R-OH + SOCl$_2$ → R-Cl + SO$_2$↑ + HCl↑. The by-products SO$_2$ and HCl are gases that escape, leaving a pure alkyl chloride with no contamination — making this the best laboratory method.

Example 5: Markovnikov addition

What is the major product when propene reacts with HBr (no peroxide)?

Solution: By Markovnikov's rule, H adds to the carbon with more hydrogens, so Br goes to the more substituted carbon: CH$_3$-CH=CH$_2$ + HBr → CH$_3$-CHBr-CH$_3$ (2-bromopropane).

Example 6: Peroxide effect

What is the product when propene reacts with HBr in the presence of peroxide?

Solution: The peroxide (Kharasch) effect gives anti-Markovnikov addition: CH$_3$-CH=CH$_2$ + HBr/peroxide → CH$_3$-CH$_2$-CH$_2$Br (1-bromopropane). (Note: only HBr shows the peroxide effect, not HCl or HI.)

Example 7: Reactivity order of alcohols with HX

Arrange 1°, 2° and 3° alcohols in order of increasing reactivity with a given haloacid.

Solution: 1° < 2° < 3° — tertiary alcohols react fastest with HX (they form the most stable carbocation).

Example 8: Vicinal dihalide

What is formed when ethene reacts with bromine?

Solution: A vicinal dihalide: CH$_2$=CH$_2$ + Br$_2$ → BrCH$_2$-CH$_2$Br (1,2-dibromoethane). The decolourisation of bromine confirms the C=C.

Example 9: Free-radical halogenation limitation

Why is the free-radical chlorination of an alkane not a good preparative method?

Solution: It gives a mixture of mono-, di- and polychlorinated products (plus different positional isomers) that are hard to separate, so it is of limited synthetic use.

Example 10: Preparation of 1-iodobutane

Write two ways to prepare 1-iodobutane from butan-1-ol.

Solution: (1) Direct: butan-1-ol + HI → 1-iodobutane + H$_2$O. (2) Two-step: butan-1-ol + SOCl$_2$ → 1-chlorobutane, then Finkelstein (NaI/dry acetone) → 1-iodobutane.