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When benzamide is treated with bromine and alkali gives compound A. Also when benzamide is reduced by LiAlH_4, compound B is formed. Find A and B. Write the equations. |
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Answer» Solution :(i) When benzamide react with bromine and ALKALI, Hoffmann reaction TAKE place and aniline `C_6H_5NH_2` is formed as (A). `{:(C_6H_5CONH_2 overset(Br_2"/"KOH)(to)C_6H_5NH_2+CO_2),(" BenzamideAniline(A) "):}` (ii) When Benzamide is reduced by `LiAlH_4`, Benzylamine `C_6H_5CH_2NH_2` is formed as (B). `{:(C_6H_5CONH_2 UNDERSET(4[H]) overset(Br_2"/"KOH)(to)C_6H_5NH_2+H_2O),("Benzylamine(B)"):}`. |
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