1.

Reaction \[C{{H}_{3}}CON{{H}_{2}}\xrightarrow{NaOBr}\] gives [CPMT 1983, 93, 97]

A.                 \[C{{H}_{3}}Br\]               
B.                 \[C{{H}_{4}}\]
C.                 \[C{{H}_{3}}COBr\]         
D.                 \[C{{H}_{3}}N{{H}_{2}}\]
Answer» E.


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