An organic compound ‘A’ on treatment with NH3, gives ‘B’  which on heating gives ‘C’, ‘C’ when treated with Br2, in the  presence of KOH produces ethylamine. Compound ‘A’ is:

  1. A

    CH3COOH

  2. B

    CH3CH2CH2COOH

  3. C

     

  4. D

    CH3CH2COOH

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    Solution:

    Since, C when heated with Br2, in presence of KOH produces ethylamine, hence it must be propanamide and hence the organic compound (A) will be propanoic acid. The reactions follows. 
    CH3CH2COOHNH3(A)CH3CH2COONH4Δ(B)         CH3 -CH2- CONH2(C) Hoffmann  (C) bromamide  reaction KOH+Br2CH3CH2NH2(Ethylamine ) 

     

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