A) \[NaOH-B{{r}_{2}}\]
B) Sodalime
C) Hot conc\[{{H}_{2}}S{{O}_{4}}\]
D) \[PC{{l}_{5}}\]
Correct Answer: A
Solution :
Key Idea The reagent which can convert\[-\text{CON}{{\text{H}}_{\text{2}}}\]group into ?NH; group is used for this reaction. Among the given reagents only \[NaOH/B{{r}_{2}}\]converts \[-CON{{H}_{2}}\]group to \[-N{{H}_{2}}\] group, thus it is used for converting ace amide to methyl amine. This reaction is called Hofmann bromamide reaction. \[\underset{acetamide}{\mathop{C{{H}_{3}}CON{{H}_{2}}}}\,+NaOH+B{{r}_{2}}\xrightarrow[{}]{{}}\underset{methyl\,a\min e}{\mathop{C{{H}_{3}}N{{H}_{2}}}}\,\] \[+NaBr+N{{a}_{2}}C{{O}_{3}}+{{H}_{2}}O\]
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