A) \[C{{H}_{3}}-CH=C{{H}_{2}}\]
B) \[C{{H}_{3}}-C\equiv CH\]
C) \[C{{H}_{3}}C{{H}_{2}}CH\backslash /N{{H}_{2}}N{{H}_{2}}\]
D) \[C{{H}_{3}}C{{H}_{2}}CH\backslash /ClN{{H}_{2}}\]
Correct Answer: B
Solution :
\[C{{H}_{3}}-C{{H}_{2}}-CHC{{l}_{2}}\underset{\Delta }{\mathop{\xrightarrow{NaN{{H}_{2}}}}}\,\] \[C{{H}_{3}}-CH=CHCl\underset{\Delta }{\mathop{\xrightarrow{NaN{{H}_{2}}}}}\,\] \[\underset{Final\,product}{\mathop{C{{H}_{3}}-C\equiv CH}}\,\]
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