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Haloalkanes and Haloarenes - Methods of Preparation

Grade 12CBSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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From Alcohols: Alcohols can be converted to haloalkanes using HX/ZnCl2HX/ZnCl_2 (Lucas reagent), PX3PX_3, PCl5PCl_5, or SOCl2SOCl_2. The reaction with thionyl chloride (SOCl2SOCl_2) is preferred because the by-products (SO2SO_2 and HClHCl) are gases and escape easily, leaving pure alkyl chloride.

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From Hydrocarbons (Addition of HXHX to Alkenes): Addition of HXHX to unsymmetrical alkenes follows Markovnikov's rule, where the negative part of the addendum adds to the carbon with fewer hydrogen atoms. In the presence of peroxides, HBrHBr (not HClHCl or HIHI) follows the Anti-Markovnikov rule (Kharasch effect).

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Halogen Exchange: This is used to prepare iodoalkanes and fluoroalkanes which are difficult to prepare by direct halogenation. The Finkelstein reaction uses NaINaI in dry acetone to produce R−IR-I. The Swarts reaction uses metallic fluorides like AgFAgF, Hg2F2Hg_2F_2, or SbF3SbF_3 to produce R−FR-F.

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From Diazonium Salts (Sandmeyer's Reaction): Primary aromatic amines react with NaNO2+HClNaNO_2 + HCl at 273−278K273-278 K to form benzene diazonium chloride. When treated with Cu2Cl2Cu_2Cl_2 or Cu2Br2Cu_2Br_2, the diazonium group is replaced by ClCl or BrBr.

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Free Radical Halogenation: Alkanes react with Cl2Cl_2 or Br2Br_2 in the presence of UV light or heat. This method usually results in a complex mixture of mono-, di-, and polyhaloalkanes, making it less suitable for laboratory synthesis of pure compounds.

📐Formulae

R−OH+HCl→ZnCl2R−Cl+H2OR-OH + HCl \xrightarrow{ZnCl_2} R-Cl + H_2O

R−OH+SOCl2→R−Cl+SO2↑+HCl↑R-OH + SOCl_2 \rightarrow R-Cl + SO_2 \uparrow + HCl \uparrow

CH3−CH=CH2+HBr→CH3−CH(Br)−CH3 (Markovnikov)CH_3-CH=CH_2 + HBr \rightarrow CH_3-CH(Br)-CH_3 \text{ (Markovnikov)}

CH3−CH=CH2+HBr→PeroxideCH3−CH2−CH2Br (Anti-Markovnikov)CH_3-CH=CH_2 + HBr \xrightarrow{\text{Peroxide}} CH_3-CH_2-CH_2Br \text{ (Anti-Markovnikov)}

R−X+NaI→Dry AcetoneR−I+NaX (Finkelstein Reaction)R-X + NaI \xrightarrow{\text{Dry Acetone}} R-I + NaX \text{ (Finkelstein Reaction)}

CH3−Br+AgF→CH3−F+AgBr (Swarts Reaction)CH_3-Br + AgF \rightarrow CH_3-F + AgBr \text{ (Swarts Reaction)}

C6H5N2+Cl−→Cu2Cl2/HClC6H5Cl+N2C_6H_5N_2^+Cl^- \xrightarrow{Cu_2Cl_2/HCl} C_6H_5Cl + N_2

💡Examples

Problem 1:

Identify the major product formed when propene (CH3−CH=CH2CH_3-CH=CH_2) reacts with HBrHBr in the presence of benzoyl peroxide.

Solution:

CH3−CH2−CH2BrCH_3-CH_2-CH_2Br (1-bromopropane)

Explanation:

In the presence of organic peroxides, the addition of HBrHBr to an unsymmetrical alkene follows the Anti-Markovnikov rule. The BrBr atom attaches to the primary carbon (the one with more hydrogen atoms) via a free radical mechanism.

Problem 2:

Write the chemical equation for the preparation of chloroethane from ethanol using thionyl chloride.

Solution:

C2H5OH+SOCl2→C2H5Cl+SO2+HClC_2H_5OH + SOCl_2 \rightarrow C_2H_5Cl + SO_2 + HCl

Explanation:

This is Darzen's process. It is the best method for preparing alkyl chlorides because the gaseous by-products SO2SO_2 and HClHCl escape from the reaction mixture.

Problem 3:

How is iodobenzene prepared from benzene diazonium chloride?

Solution:

C6H5N2+Cl−+KI→C6H5I+KCl+N2C_6H_5N_2^+Cl^- + KI \rightarrow C_6H_5I + KCl + N_2

Explanation:

Unlike the preparation of chlorobenzene or bromobenzene, the preparation of iodobenzene does not require a cuprous halide catalyst. Simply shaking the diazonium salt solution with potassium iodide (KIKI) is sufficient.