A) Butyne-1
B) Butyne-2
C) Butene-1
D) Butene-2
Correct Answer: A
Solution :
\[\underset{\text{1}-\text{Butyne}}{\mathop{C{{H}_{3}}-C{{H}_{2}}-C}}\,\equiv CH+2B{{r}_{2}}\to C{{H}_{3}}-C{{H}_{2}}-\underset{\underset{\,\,Br}{\mathop{|}}\,}{\overset{\overset{\,\,Br}{\mathop{|}}\,}{\mathop{C}}}\,-\underset{\underset{\,\,Br}{\mathop{|}}\,}{\overset{\overset{\,\,Br}{\mathop{|}}\,}{\mathop{C}}}\,-H\] Since the molecule takes 2 moles of \[B{{r}_{2}}\]. Therefore it is alkyne. Also it gives white ppt with Tollen?s reagent therefore acidic H is present. Hence it is 1-Butyne.
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