Question

The oxidation 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 oxidation 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. $\left(II\right)\to \left(P\right),\left(T\right)and\left(IV\right)\to \left(S\right)$
• B. $\left(II\right)\to \left(U\right)and\left(IV\right)\to \left(S\right)$
• C. $\left(II\right)\to \left(P\right),\left(U\right)and\left(IV\right)\to \left(Q\right),\left(R\right),\left(S\right)$
• D. $\left(II\right)\to \left(P\right)\left(U\right)and\left(IV\right)\to \left(R\right)$

Answer 1

The correct option is D $\left(II\right)\to \left(P\right)\left(U\right)and\left(IV\right)\to \left(R\right)$

(II) Oxidation of both cis and trans diol is possible using $Pb(OAc{)}_4$

(IV) The ring flipping is restricted due to bulkier group at equitorial position. Since the $Cl$ is not antiparallel to $OH$, so contraction of ring will take place through semi-pinacol rearrangement.