Conversion of Chlorobenzene to Biphenyl:
The conversion of chlorobenzene (C₆H₅Cl) to biphenyl (C₆H₅-C₆H₅) involves a reaction known as the **Wurtz-Fittig Reaction**. This reaction is a type of cross-coupling reaction where two aryl halides are coupled in the presence of sodium metal.
Reaction:
2 C₆H₅Cl + Na → C₆H₅-C₆H₅ + 2 NaCl
Mechanism:
1. The reaction is carried out in dry ether and requires heat. Sodium metal donates electrons to the chlorobenzene, generating a phenyl radical (C₆H₅•) and sodium chloride (NaCl).
2. The phenyl radicals couple together to form biphenyl (C₆H₅-C₆H₅).
Conditions:
- Sodium metal (Na) is used as a reducing agent.
- The reaction takes place in the presence of anhydrous ether to prevent moisture from interfering with the reaction.
Conclusion:
By heating chlorobenzene with sodium metal in ether, chlorobenzene undergoes a Wurtz-Fittig reaction to form biphenyl.
Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.
(i) (CH3 )2CHNH2 (ii) CH3 (CH2 )2NH2 (iii) CH3NHCH(CH3 )2
(iv) (CH3 )3CNH2 (v) C6H5NHCH3 (vi) (CH3CH2 )2NCH3 (vii) m–BrC6H4NH2
Give one chemical test to distinguish between the following pairs of compounds.
(i) Methylamine and dimethylamine
(ii) Secondary and tertiary amines
(iii) Ethylamine and aniline
(iv) Aniline and benzylamine
(v) Aniline and N-methylaniline
Account for the following:
(i) pKb of aniline is more than that of methylamine.
(ii) Ethylamine is soluble in water whereas aniline is not.
(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.
(iv) Although amino group is o– and p– directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline.
(v) Aniline does not undergo Friedel-Crafts reaction.
(vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines. (vii) Gabriel phthalimide synthesis is preferred for synthesising primary amines.