A) \[C{{H}_{3}}-C\equiv C-C{{H}_{3}}+Na/N{{H}_{3}}(liq.)\]
B) \[C{{H}_{3}}-CH=CH-C{{H}_{3}}+Li/N{{H}_{3}}(liq.)\]
C) \[C{{H}_{3}}-C=C-C{{H}_{3}}+Na/N{{H}_{4}}OH\]
D) \[C{{H}_{3}}-C\equiv C-C{{H}_{3}}+Pd/BaS{{O}_{4}}\]
Correct Answer: A
Solution :
When alkynes are reduced with Li or Na metal in \[\text{N}{{\text{H}}_{\text{3}}}\text{,}\]at low temperature, an anti-addition of hydrogen atoms takes place and traris or E product is obtained. \[{{H}_{3}}C-C\equiv C-C{{H}_{3}}+Na\xrightarrow{Liq.N{{H}_{3}}}\] \[{{H}_{3}}C\_\_\underset{\begin{align} & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,H \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2E-butene \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,or \\ & \,\,\,\,\,\,\,\,\,\,\,\,trans-2- \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,butene \\ \end{align}}{\mathop{\underset{|}{\mathop{C}}\,}}\,=\overset{H}{\mathop{\overset{|}{\mathop{C}}\,}}\,-C{{H}_{3}}\]
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