Question

How m-chloroaniline can be prepared from benzene?

• A. (i) Cone. $ HN{{O}_{3}}+ $ conc. $ {{H}_{2}}S{{O}_{4}} $ (ii) $ C{{l}_{2}}/Fe $ (iii) $ Sn/HCl $
• B. (i) Cone. $ HN{{O}_{3}}+ $ cone. $ {{H}_{2}}S{{O}_{4}} $ (ii) $ NaOH $ (iii) Dil. $ {{H}_{2}}S{{O}_{4}}+C{{l}_{2}} $
• C. (i) Cone. $ HN{{O}_{3}}+ $ cone. $ {{H}_{2}}S{{O}_{4}} $ (ii) $ Sn/HCl $
• D. (i) Cone. $ HN{{O}_{3}}+ $ cone $ {{H}_{2}}S{{O}_{4}} $ (ii) $ Sn/HCl $

Answer 1

Answer: (A)

Correct answer is (a) (i) Cone. $ HN{{O}_{3}}+ $ conc. $ {{H}_{2}}S{{O}_{4}} $ (ii) $ C{{l}_{2}}/Fe $ (iii) $ Sn/HCl $