Question

Oxidation of ethene with cold alkaline $\text{KMn}{{\text{O}}_{\text{4}}}$ produces:
(A) Formaldehyde
(B) Glycol
(C) Ethylene glycol
(D) oxalic acid

Answer 1

Hint Alkenes are electron rich centres which mainly gives electrophilic addition reactions.
$\text{KMn}{{\text{O}}_{\text{4}}}$ Acts as a strong oxidizing agent but in low temperature or alkaline condition it causes imitated oxidation.
Ethene 1, 2- diol is known as glycol in which two $\text{(-OH)}$ groups are attached to the ethane in 1, 2 position.

Complete answer:
Cold dilute $\text{KMn}{{\text{O}}_{\text{4}}}$ is also known as Baeyer’s reagent, and decolourization of Baeyer’s reagent is the test of unsaturation. $\text{KMn}{{\text{O}}_{\text{4}}}$ Acts as a strong oxidizing agent which, causes oxidation of alkene At low temperature it causes partial oxidation of alkene into alcohol (vicinal diol), so the double bond of alkene will converted into two alcoholic groups.
On reacting ethene with Cold dilute $\text{KMn}{{\text{O}}_{\text{4}}}$ forms a cyclic paramagnetic intermediate and finally result in ethene 1, 2-diol or ethylene glycol.

So, option (C) is the correct answer.

$\text{3C}{{\text{H}}_{\text{2}}}\text{= C}{{\text{H}}_{\text{2}}}\,\text{+}\,\text{2KMn}{{\text{O}}_{\text{4}}}\,\text{+4}{{\text{H}}_{\text{2}}}\text{O}\,\,\to \,\text{3C}{{\text{H}}_{\text{2}}}\text{OHC}{{\text{H}}_{\text{2}}}\text{OH}\,\,\text{+}\,\,\text{2Mn}{{\text{O}}_{\text{2}}}\,\text{+2KOH}$

Additional information: Mild form of oxidizing agent ($\text{KMn}{{\text{O}}_{\text{4}}}$) causes less oxidation and ends up with a less oxidation product. Baeyer’s reagent is initially a coloured compound, after oxidation it will convert into a colourless compound. In this reaction solid potassium permanganate (which oxidation state is +7) will decompose into potassium manganate (with oxidation state +6), manganese oxide and oxygen. This liberated oxygen will oxidize alkene into alcohol.

Note: Hot $\text{KMn}{{\text{O}}_{\text{4}}}$ is a very strong oxidizing agent, if it reacts with ethene under high temperature and alkaline medium it will completely oxidized ethene into $\text{C}{{\text{O}}_{\text{2}}}$ and ${{\text{H}}_{\text{2}}}\text{O}$. The solution of $\text{KMn}{{\text{O}}_{\text{4}}}$ is made slightly alkaline by addition of sodium carbonate.
It is essential to make mild conditions for this reaction otherwise reaction will end up in some over oxidation products like acid or carbon dioxide.