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 :
Butyne-1,has the terminal hydrogen atom which reacts with anunoniacal silver nitrate easily and gives the white precipitate of silver acetylide, \[2C{{H}_{3}}C{{H}_{2}}C\equiv CH+A{{g}_{2}}O\xrightarrow[{}]{{}}\] \[\underset{white\text{ }ppt.}{\mathop{C{{H}_{3}}C{{H}_{2}}C\equiv CAg}}\,+{{H}_{2}}O\] But butyne-2 does not give this reaction. Hence, butyne-1 and butyne-2 are distinguished by\[AgN{{O}_{3}}+N{{H}_{4}}OH\]
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