A) Butyne-1
B) Butyne - 2
C) Butene - 1
D) Butene - 2.
Correct Answer: A
Solution :
: Since A\[({{C}_{4}}{{H}_{6}})\]adds two moles of\[B{{r}_{2}}\]to give B\[({{C}_{4}}{{H}_{6}}B{{r}_{4}})\]therefore, it is a butyne. Further since A forms a white ppt. with ammoniacal\[AgN{{O}_{3}}\] solution therefore, A is 1-butyne. \[C{{H}_{3}}C{{H}_{2}}C\equiv CH+AgN{{O}_{3}}+N{{H}_{4}}OH\xrightarrow[{}]{{}}\] \[\underset{(white\text{ }ppt.)}{\mathop{C{{H}_{3}}C{{H}_{2}}C}}\,\equiv CAg+N{{H}_{4}}N{{O}_{3}}+{{H}_{2}}O\]You need to login to perform this action.
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