Complete the following reactions: C6H5−CONH2Br2/KOH−−−−−−→AKOH−−−−→CHCl3B−−−−−→Sn/HClC
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Solution
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When amide reacts with Br2 in presence of KOH, an amine is formed with and the product has one carbon less than the parent substrate, it is called Hoffmann bromamide reaction and when aniline (a primary amine) is made to react with KOH/CHCl3, a carbylamine is formed. Sn/HCl is used for the reduction of many carbon compounds.