A) \[C{{H}_{3}}-CH\left( N{{H}_{2}} \right)-C{{N}_{2}}N{{H}_{2}}\]
B) \[C{{H}_{3}}-CHOH-C{{H}_{2}}OH\]
C) \[C{{H}_{3}}-C\left( OH \right)=C{{H}_{2}}\]
D) \[C{{H}_{3}}-C\equiv CH\]
Correct Answer: D
Solution :
\[_{\text{5}}{{\text{Y}}^{\text{10}}}{{\text{=}}_{\text{5}}}{{\text{B}}^{\text{10}}}\] \[{{E}_{n}}=-\frac{13.6}{{{n}^{2}}}eV\] \[\Rightarrow \] \[34=-\frac{13.6}{{{n}^{2}}}\] Ale. KOH smoothly brings about dehydrobromination of y to give a mixture of vinyl bromide (A and B) while \[\Rightarrow \]being a strong base than ale. KOH readily brings about dehydrobromination of less reactive vinyl bromide to give propyne (Z).
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