Q12 / 125 · BITSAT 2016 · Chemistry

$\quad \mathrm{CH}_{3} \mathrm{C} \equiv \mathrm{CCH}_{3} \stackrel{\mathrm{H}_{2} / \mathrm{Pt}}{\longrightarrow} \mathrm{A} \stackrel{\mathrm{D}_{2} / \mathrm{Pt}}{\longrightarrow} \mathrm{B}$ The compounds $A$ and $B$, respectively are

  • A
    cis-butene-2 and $\mathrm{rac}-2,3$ -dideuterobutane
  • B
    trans-butene-2 and $\mathrm{rac}-2$, 3-dideuterobutane
  • Correct answer
    C
    cis-butene- 2 and meso-2, 3-dideuterobutane
  • D
    trans-butene-2 and meso-2, 3-dideuterobutane

Solution

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.

Hence, the correct answer is option: 'C'.

More Hydrocarbons Questions