Question

Consider the following reaction sequence.
${CH}_2={CH}_2\underset{dil.{KMnO}_4}{\overset{cold}{\to }}A\underset{{PCl}_5}{\overset{excess}{\to }}B$

The products $\left(A\right)$ and $\left(B\right)$ are, respectively:

• A. ${CH}_3{CH}_{2}OH$ and ${CH}_3{CH}_{2}Cl$
• B. ${CH}_{3}CHO$ and ${CH}_3{CHCl}_2$
• C. ${CH}_{2}OH{CH}_{2}OH$ and ${CH}_{2}Cl{CH}_{2}Cl$
• D. ${CH}_{2}OH{CH}_{2}OH$ and ${CH}_{2}OH{CH}_{2}Cl$

Answer 1

Answer: (C)

${CH}_{2}OH{CH}_{2}OH$ and ${CH}_{2}Cl{CH}_{2}Cl$

The products (A) and (B) are ${CH}_{2}OH{CH}_{2}OH$ and ${CH}_{2}Cl{CH}_{2}Cl$ respectively.

${CH}_2={CH}_2\underset{dil.{KMnO}_4}{\overset{cold}{\to }}\underset{A}{{CH}_{2}OH{CH}_{2}OH}\underset{{PCl}_5}{\overset{excess}{\to }}\underset{B}{{CH}_{2}Cl{CH}_{2}Cl}$

When ethylene is treated with cold dilute potassium permanganate solution, two $-OH$ groups are added to $C$ atoms joined by double bond to form a vicinal diol.

Thus, it is an example of dihydroxylation.

In the next step, on treatment with phosphorous pentachloride, the $-OH$ groups are replaced with $Cl$ atoms to form vicinal dichloride.

Option C is correct.