A) \[{{H}_{2}}Pd/BaS{{O}_{4}}\]
B) \[{{H}_{2}},Pt{{O}_{2}}\]
C) \[NaB{{H}_{4}}\]
D) \[liq-N{{H}_{3}}/{{C}_{2}}{{H}_{5}}OH\]
Correct Answer: D
Solution :
\[{{H}_{2}}/Pd/BaS{{O}_{4}}\]reduces an alkyne to cis-alkene, \[{{H}_{2}}/Pt\]reduces it to alkane,\[NaB{{H}_{4}}\]. Does not reduce an alkyne. Reduction of an alkyne by active metal in liq. \[N{{H}_{3}}\] gives trans-alkene.
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