A) \[{{H}_{2}}S{{O}_{4}}\]
B) \[AgN{{O}_{3}}+N{{H}_{4}}OH\]
C) \[{{K}_{2}}C{{r}_{2}}{{O}_{7}}\]
D) \[N{{H}_{4}}OH\]
Correct Answer: B
Solution :
Key Idea: Only terminal alkynes react with ammoniacal silver nitrate to produce white ppt. of silver acetylide. Butyne-1 is terminal alkyne and butyne-2 is not. \[\underset{Butyne-1}{\mathop{2C{{H}_{3}}CH\equiv CH}}\,+\underset{\begin{align} & \downarrow \\ & C{{H}_{3}}C{{H}_{2}}C\equiv CAg \\ & Silver\text{ }acetylide \\ & \,\,\,\,\,\,(White\text{ }ppt.) \\ \end{align}}{\mathop{N{{H}_{4}}OH}}\,+AgN{{O}_{3}}\] \[\underset{Butyne-2}{\mathop{C{{H}_{3}}CH\equiv CHC{{H}_{3}}}}\,+N{{H}_{4}}OH+AgN{{O}_{3}}\xrightarrow[{}]{{}}\] No reaction \[\therefore \]\[AgN{{O}_{3}}+N{{H}_{4}}OH\]is used to distinguish butyne-1 and butyne-2
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