Question

Consider the following reactions

$Meso-butan-2,3-diol$ is formed in:

• A. $\left(I\right),\left(II\right),\left(V\right)$
• B. $\left(III\right),\left(IV\right),\left(VI\right)$
• C. $\left(I\right),\left(II\right),\left(IV\right)$
• D. $\left(III\right),\left(IV\right),\left(V\right)$

Answer 1

The correct option is A $\left(I\right),\left(II\right),\left(V\right)$

$Meso-butan-2,3-diol$ is formed in (I) $KMn{O}_4$, (II) $Os{O}_4/NaHS{O}_3$ and (V) $H_2{O}_2+Os{O}_4$. The addition is syn addition. Two $-OH$ groups are added across carbon carbon double bond from the same side. $Meso-butan-2,3-diol$ has plane of symmetry.

In the cases (III) $HC{O}_{3}H/H_{2}O$, (IV) $MMPP+C_2{H}_5/H_{2}O$ and (VI) $H_2{O}_2+Se{O}_2$, trans addition results in $(\pm )$-butan-2,3-diol . Two $-OH$ groups are added across carbon carbon double bond from the opposite sides.