Match List-I with List-II
Choose the correct answer from the options given below: |
(A)-(I), (B)-(IV), (C)-(II), (D)-(III) (A)-(I), (B)-(III), (C)-(II), (D)-(IV) (A)-(I), (B)-(II), (C)-(IV), (D)-(III) (A)-(III), (B)-(IV), (C)-(II), (D)-(I) |
(A)-(I), (B)-(IV), (C)-(II), (D)-(III) |
The correct answer is Option (1) → (A)-(I), (B)-(IV), (C)-(II), (D)-(III)
Hydrolysis of Ethyl Chloride C₂H₅Cl + OH⁻ → C₂H₅OH + Cl⁻ Here, the Cl atom is replaced by OH⁻, so this is a substitution reaction. Thus, Hydrolysis of Ethyl Chloride → Substitution reaction (I) 2-Propanol with Concentrated H₂SO₄ When 2-propanol is heated with concentrated sulphuric acid, dehydration occurs forming propene. CH₃–CHOH–CH₃ → CH₃–CH=CH₂ + H₂O Since a molecule of water is removed, this is an elimination reaction. Thus, 2-Propanol + Conc. H₂SO₄ → Elimination reaction (IV) Reaction of HCN with Acetone Acetone reacts with HCN to form cyanohydrin. (CH₃)₂CO + HCN → (CH₃)₂C(OH)CN Here, HCN adds across the C=O double bond, so it is an addition reaction. Thus, HCN + Acetone → Addition reaction (II) Acetamide with NaOH and Br₂ Acetamide reacts with bromine and NaOH in Hofmann bromamide reaction, producing a primary amine with one carbon less. CH₃CONH₂ → CH₃NH₂ Since the carbon chain length decreases, this is a degradation reaction. Thus, Acetamide + NaOH + Br₂ → Degradation reaction (III) |