1. Tardigrade
  2. Question
  3. Chemistry
  4. CH 3 C ≡ CCH 3 xrightarrowH 2 / P t A xrightarrowD2 / P t B The compounds A and B, respectively are

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Q. $CH _{3} C \equiv CCH _{3} \xrightarrow{H 2 / P t} A \xrightarrow{D_{2} / P t} B$
The compounds $A$ and $B$, respectively are

BITSATBITSAT 2016

Solution:

Catalytic hydrogenation of alkynes gives cis-alkene which in turn adds deuterium atoms in presence of $H_2$ again in cis-manner forming meso-$2, 3$-dideuterobutane.