Question

Write short notes on the following: (i) Freons (ii) Sandmeyer reaction (iii) Friedel Craft acylation of haloarene. (2+1½+1½=5)
OR
What happens when (write chemical equation only): (i) Propene reacts with HI (ii) Ethyl bromide reacts with magnesium in the presence of dry ether (iii) Chlorobenzene is heated with sodium in dry ether (iv) The mixture of chlorobenzene and methyl chloride is heated with sodium in the presence of dry ether (v) Isopropyl alcohol reacts with conc. HCl and ZnCl₂. (1+1+1+1+1=5)

Hint:

Part 1: Freon-12 by Swarts reaction, Sandmeyer uses diazonium + cuprous halide, and the ring halogen is ortho/para directing in acylation. Part 2: apply Markovnikov (i), Grignard formation (ii), Fittig (iii) and Wurtz-Fittig (iv) coupling, and Lucas reagent (v).

Answer 1

Part 1 solved (short notes)

(i) Freons: The name Freon refers to a family of chlorofluorocarbon gases derived from methane/ethane by replacing the hydrogens with \(Cl\) and \(F\). Freon-12 (\(CCl_2F_2\)) is the industrially important example. Its preparation from \(CCl_4\) is an example of halogen exchange: \[CCl_4 + 2HF \xrightarrow{SbF_5} CCl_2F_2 + 2HCl\] Key uses are as refrigerant fluids and aerosol propellants, chosen because the gas is inert and non-flammable. The drawback is atmospheric: photolysis in the stratosphere liberates \(Cl^{\bullet}\) radicals that catalytically break down ozone, which is why CFCs are being phased out under the Montreal Protocol.

(ii) Sandmeyer reaction: Think of it as a substitution on the diazonium ion. Aniline is first diazotised (\(NaNO_2 + HCl\), 0–5°C) to \(C_6H_5N_2^+Cl^-\). A cuprous salt then supplies the nucleophile while \(N_2\) leaves as gas: \[C_6H_5N_2^+Cl^- \xrightarrow{Cu_2Cl_2 / HCl} C_6H_5Cl + N_2 \uparrow\] Using \(Cu_2Br_2/HBr\) gives bromobenzene and \(CuCN\) gives benzonitrile. Its usefulness lies in fixing a halogen (or \(CN\)) at the precise carbon that originally carried the amino group, which cannot be achieved by direct halogenation.

(iii) Friedel-Craft acylation of haloarene: Here an aryl halide behaves as the aromatic substrate. Acetyl chloride plus anhydrous \(AlCl_3\) produces the electrophilic acylium ion \(CH_3\overset{+}{C}=O\). The ring halogen, being an ortho/para director (though deactivating), steers the incoming acyl group to the para position mainly: \[C_6H_5Cl + CH_3COCl \xrightarrow{\text{anhyd. }AlCl_3} p\text{-}CH_3CO\text{-}C_6H_4\text{-}Cl + HCl\] Product: para-chloroacetophenone (a ketone). This is a classic electrophilic aromatic substitution.

Part 2 solved (OR — equations only)

(i) Propene + HI adds by Markovnikov rule, \(H\) to the carbon with more hydrogens: \[CH_3CH{=}CH_2 + HI \rightarrow CH_3CHICH_3 \ (\text{2-iodopropane})\] (ii) Ethyl bromide with magnesium in dry ether gives a Grignard reagent: \[C_2H_5Br + Mg \xrightarrow{\text{dry ether}} C_2H_5MgBr\] (iii) Chlorobenzene with sodium in dry ether couples two aryl groups (Fittig reaction): \[2C_6H_5Cl + 2Na \xrightarrow{\text{dry ether}} C_6H_5{-}C_6H_5 + 2NaCl\] (iv) Chlorobenzene + methyl chloride + sodium (Wurtz-Fittig) gives toluene: \[C_6H_5Cl + CH_3Cl + 2Na \xrightarrow{\text{dry ether}} C_6H_5CH_3 + 2NaCl\] (v) Isopropyl alcohol with conc. HCl and \(ZnCl_2\) (Lucas reagent) gives the alkyl chloride: \[(CH_3)_2CHOH + HCl \xrightarrow{ZnCl_2} (CH_3)_2CHCl + H_2O\]