Question

The oxidative cleavage of 1,2-diol by a periodic acid or periodic salt to give corresponding carbonyls is known as Malaprade oxidation.

Vicinal diols can also be oxidized to form aldehydes and ketones by lead tetraacetate. This oxidative cleavage is analogous to malaprade oxidation and is known as Criegee oxidation.

When Halohydrin is treated with $A{g}_{2}O$ or Silver salt, it gives pinacol-pinacolone type rearrangement called semi-pinacol rearrangement.
List-I contains reactions and List-II contains products that can be formed.



Which of the following options has the correct combination considering reaction in List-I and their product(s) in List-II?

• A. $\left(I\right)\to \left(T\right),\left(U\right)and\left(III\right)\to \left(Q\right)$
• B. $\left(I\right)\to \left(P\right),\left(T\right),\left(U\right)and\left(III\right)\to \left(Q\right)$
• C. $\left(I\right)\to \left(T\right),\left(U\right)and\left(III\right)\to \left(R\right)$
• D. $\left(I\right)\to \left(P\right),\left(U\right)and\left(III\right)\to \left(S\right)$

Answer 1

The correct option is A $\left(I\right)\to \left(T\right),\left(U\right)and\left(III\right)\to \left(Q\right)$

(I) There is no reaction on anti-diol attached with ring by $HI{O}_4$ due to incapability of formation of cyclic ester.

(III) Due to presence of bulky group at equitorial position, flipping is restricted. Since $Cl$ is antiperiplanar to $OH$, epoxide will be formed.