Question

Describe two methods of preparation of alkyl halides. Also give equations. Explain the following reactions of alkyl halides with examples: (i) Nucleophilic substitution reaction (ii) Elimination reaction.
OR
Write chemical equations of chlorination, nitration, sulphonation, Friedel-Crafts reaction and Wurtz (Fittig) reaction of chlorobenzene.

Hint:

For alkyl halides, recall preparation from alcohols or alkenes; a nucleophile swaps out the halide (SN1/SN2), while alcoholic KOH triggers beta-elimination to an alkene. For chlorobenzene, remember chlorine is an ortho/para director.

Answer 1

Option 1

Step 1: Two methods of preparation of alkyl halides.
(a) From alcohols: Alcohols react with halogen acids, phosphorus halides or thionyl chloride to replace the -OH group by -X.
\( R\text{-}OH + HX \rightarrow R\text{-}X + H_2O \) (X = Cl, Br, I; ZnCl₂ catalyst used with HCl)
\( 3\,R\text{-}OH + PCl_3 \rightarrow 3\,R\text{-}Cl + H_3PO_3 \)
\( R\text{-}OH + SOCl_2 \rightarrow R\text{-}Cl + SO_2 + HCl \)
Example: \( C_2H_5OH + HBr \rightarrow C_2H_5Br + H_2O \)
(b) From alkenes (addition of hydrogen halide): Alkenes add HX following Markovnikov's rule (the halogen goes to the more substituted carbon).
\( CH_2=CH_2 + HBr \rightarrow CH_3CH_2Br \)
\( CH_3CH=CH_2 + HBr \rightarrow CH_3CHBrCH_3 \)

Step 2: (i) Nucleophilic substitution reaction.
The carbon of the polar C-X bond bears a partial positive charge. A nucleophile (electron-rich species) attacks this carbon and replaces the halide, which departs as a good leaving group. It follows an SN1 route (two steps, through a carbocation) or an SN2 route (one concerted step, backside attack with inversion).
Example (hydrolysis): \( CH_3CH_2Br + KOH_{(aq)} \rightarrow CH_3CH_2OH + KBr \)
Example (cyanide): \( CH_3CH_2Br + KCN \rightarrow CH_3CH_2CN + KBr \)

Step 3: (ii) Elimination reaction.
On heating with alcoholic KOH, an alkyl halide loses a hydrogen from the beta-carbon and the halogen from the alpha-carbon as HX, forming a carbon-carbon double bond (dehydrohalogenation, beta-elimination). When more than one alkene is possible, the more substituted alkene is the major product (Saytzeff rule).
\( CH_3CH_2Br + KOH_{(alc)} \xrightarrow{\Delta} CH_2=CH_2 + KBr + H_2O \)
\( CH_3CHBrCH_2CH_3 + KOH_{(alc)} \rightarrow CH_3CH=CHCH_3 \text{ (major)} + HBr \)

Option 2: Reactions of chlorobenzene.
Chlorine is an ortho/para director, so electrophilic substitutions give mainly the para isomer.
Step 1: Chlorination: With anhydrous FeCl₃, chlorobenzene gives mainly p-dichlorobenzene (with some ortho).
\( C_6H_5Cl + Cl_2 \xrightarrow{FeCl_3} p\text{-}C_6H_4Cl_2 + HCl \)
Step 2: Nitration: With concentrated HNO₃ and concentrated H₂SO₄, it gives mainly p-nitrochlorobenzene.
\( C_6H_5Cl + HNO_3 \xrightarrow{H_2SO_4} p\text{-}O_2N\text{-}C_6H_4\text{-}Cl + H_2O \)
Step 3: Sulphonation: With concentrated (fuming) H₂SO₄ it gives mainly p-chlorobenzenesulphonic acid.
\( C_6H_5Cl + H_2SO_4 \rightarrow p\text{-}Cl\text{-}C_6H_4\text{-}SO_3H + H_2O \)
Step 4: Friedel-Crafts reaction: With CH₃Cl and anhydrous AlCl₃ (alkylation) it gives p-chlorotoluene (acylation with CH₃COCl gives p-chloroacetophenone).
\( C_6H_5Cl + CH_3Cl \xrightarrow{AlCl_3} p\text{-}CH_3\text{-}C_6H_4\text{-}Cl + HCl \)
Step 5: Wurtz (Fittig) reaction: Chlorobenzene with sodium metal in dry ether couples to give diphenyl (biphenyl).
\( 2\,C_6H_5Cl + 2Na \xrightarrow{dry\ ether} C_6H_5\text{-}C_6H_5 + 2NaCl \)
\[\boxed{\text{Both alternatives complete}}\]