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Question: Answered & Verified by Expert
$\quad \mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{CCH}_{3} \stackrel{\mathrm{H}_{2} / \mathrm{Pt}}{\longrightarrow} \mathrm{A} \stackrel{\mathrm{D}_{2} / \mathrm{Pt}}{\longrightarrow} \mathrm{B}$

The compounds $A$ and $B$, respectively are
ChemistryHydrocarbonsBITSATBITSAT 2016
Options:
  • A cis-butene-2 and $\mathrm{rac}-2,3$ -dideuterobutane
  • B trans-butene-2 and $\mathrm{rac}-2$, 3-dideuterobutane
  • C cis-butene- 2 and meso-2, 3-dideuterobutane
  • D trans-butene-2 and meso-2, 3-dideuterobutane
Solution:
2048 Upvotes Verified Answer
The correct answer is: cis-butene- 2 and meso-2, 3-dideuterobutane
Catalytic hydrogenation of alkynes gives cis-alkene which in turn adds deuterium atoms in presence of $\mathrm{H}_{2}$ again in $\mathrm{ci}$ -manner forming meso -2,3 -dideuterobutane.

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