Question

A student decided to prepare ethane-1,2-diol from ethene in two steps. What are the reagent required to carry out conversion in Step I and Step II?


Select one:
A.
Step I
Br2 in ultraviolet light
Step II
Cold alkaline KMnO4

B.
Step I
Cl2 in CCl4
Step II
Aqueous KOH

C.
Step I
Aqueous Cl2
Step II
KOH in ethanol

D.
Step I
Br2 in CCl4
Step II
Hot acidified KMnO4

Answer 1

Answer:

Option B is the answer

Explanation:

In option A, step I is occurring under ultraviolet condition. So, Br2 will go under homolytic cleavage. It will form radical and will follow radical mechanism. The alkene would not undergo radical mechanism.

In option C, when alkene is treated with aq Cl2 addition product will be have hydroxy and Cl on adjacent carbons.

Now when, the product will be treated with alcoholic KOH, the product might go elimination reaction giving you back vinylic alcohol

In option D

when alkene is treated with Br2 CCl4 it will form di bromo alkane. When treated with hot acidified KmnO4 the alkene can undergo oxidation.

Option B seems to be the most appropriate answer because, when alkene is treated with Cl2 In CCl4 it will give dichloro alkane. The product can undergo substitution with hydroxy and give you diol