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CBSE Class 12 Chemistry 2022 Term 2 Delhi Set 3

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Question : 3 of 12
Marks: +1, -0
An Organic compound (A) with molecular formula C3H7NO\mathrm{C}_3\mathrm{H}_7\mathrm{NO} on heating with Br2\mathrm{Br}_2 and KOH\mathrm{KOH} forms a compound (B). Compound (B) on heating with CHCl3\mathrm{CHCl}_3 and alcoholic KOH\mathrm{KOH} produces a foul smelling compound (C)(\mathrm{C}) and on reacting with C6H5SO2Cl\mathrm{C}_6\mathrm{H}_5\mathrm{SO}_2\mathrm{Cl} forms a compound (D) which is soluble in alkali. Write the structure of (A), (B), (C) and (D).
Solution:  
CH3CH2CONH2(A)+Br2+KOHCH3CH2NH2(B)\underset{(A)}{\mathrm{CH}_3\mathrm{CH}_2\mathrm{CONH}_2} + \mathrm{Br}_2 + \mathrm{KOH} \rightarrow \underset{(B)}{\mathrm{CH}_3\mathrm{CH}_2\mathrm{NH}_2}
CH3CH2NH2(B)+CHCl3+KOHCH3CH2NC(C)\underset{(B)}{\mathrm{CH}_3\mathrm{CH}_2\mathrm{NH}_2} + \mathrm{CHCl}_3 + \mathrm{KOH} \rightarrow \underset{(C)}{\mathrm{CH}_3\mathrm{CH}_2\mathrm{NC}} +KCl+H2O+ \mathrm{KCl} + \mathrm{H}_2\mathrm{O}
  A=  Propanaimde  \;A=\;\text{Propanaimde}\;
  B=  Ethyl amine  (1  amine  )\;B=\;\text{Ethyl amine}\; (1^{\circ} \;\text{amine}\;)
  C=  Ethyl isocyanide  \;C=\;\text{Ethyl isocyanide}\;
  D=  N-ethyl benzene sulphonamide  \;D=\;\text{N-ethyl benzene sulphonamide}\;
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