question_answer

Propyne and propene can be distinguished by:

A)  dil. \[KMn{{O}_{4}}\]

B)  conc. \[{{H}_{2}}S{{O}_{4}}\]

C)  \[AgN{{O}_{3}}\] in ammonia

D)  \[B{{r}_{2}}\] in \[CC{{l}_{4}}\]

Correct Answer: C

Solution :

Propyne and propene can be distinguished by ammonia cal \[AgN{{O}_{3}}\] solution. Propyne give white ppt. with it while propene has no effect. \[R-C\equiv CH+[Ag{{(N{{H}_{3}})}_{2}}]OH\to \]                              \[\underset{\begin{smallmatrix}  Silver\,alkylide \\  (white\,ppt.) \end{smallmatrix}}{\mathop{R-C\equiv C\,Ag\downarrow {{H}_{2}}O+N{{H}_{3}}}}\,\] Note: All terminal alkynes undergo this reaction. \[\underset{2-keto\,propanoic\,acis}{\mathop{\underset{\Pr opyne}{\mathop{C{{H}_{3}}C\equiv CH}}\,\xrightarrow[[O]]{Alk.\,KMn{{O}_{4}}}\,C{{H}_{3}}-\underset{O}{\mathop{\underset{||}{\mathop{C}}\,}}\,-COOH}}\,\]