NEETChemistryHaloalkanes and Haloarenes
Match List-I with List-II. List-I (Reaction Sequence) List-II (Major Final Product) (A) Bromoethane KCN LiAlH ₄ (I) Propan-1-amine (B) Bromoethane AgNO ₂ Sn/HCl (II) Ethanamine (C) Bromoethane KCN H ₃ O ⁺/ (III) Propanoic acid (D) Bromoethane KNO ₂ (IV) Ethyl nitrite Choose the correct answer from the options given below:
Options
- A(A)-(II), (B)-(I), (C)-(III), (D)-(IV)
- B(A)-(I), (B)-(II), (C)-(III), (D)-(IV)
- C(A)-(I), (B)-(IV), (C)-(III), (D)-(II)
- D(A)-(III), (B)-(II), (C)-(I), (D)-(IV)
Correct answer
B. (A)-(I), (B)-(II), (C)-(III), (D)-(IV)
Step-by-step solution
(A) Bromoethane reacts with KCN (an ambident nucleophile) to form propanenitrile as the major product. Reduction of propanenitrile with LiAlH ₄ yields propan-1-amine. Thus, (A) matches with (I). (B) Bromoethane reacts with AgNO ₂ to form nitroethane because the silver-oxygen bond is covalent, making nitrogen the attacking center. Reduction of nitroethane with Sn/HCl yields ethanamine. Thus, (B) matches with (II). (C) Bromoethane reacts with KCN to form propanenitrile, which on complete hydrolysis with acidic water