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Question: Product in the reaction is: \( C{{H}_{3}}-CH=CH-C{{H}_{3}}\xrightarrow[{{H}^{+}}]{KMn{{O}_{4}}} \) ...

Product in the reaction is: CH3−CH=CH−CH3→H+KMnO4C{{H}_{3}}-CH=CH-C{{H}_{3}}\xrightarrow[{{H}^{+}}]{KMn{{O}_{4}}}
(A) CH3−CH(OH)−CH(OH)−CH3C{{H}_{3}}-CH\left( OH \right)-CH\left( OH \right)-C{{H}_{3}}
(B) 2CH3CHO2C{{H}_{3}}CHO
(C) 2CH3COOOH2C{{H}_{3}}COOOH
(D) CH3CHO+CH3COOHC{{H}_{3}}CHO+C{{H}_{3}}COOH

Explanation

Solution

Hint : We know that during hydroxylation of alkenes, the carbon-hydrogen bond in the alkene oxidizes to a carbon-hydroxyl bond. Hydroxylation of alkenes is an oxidation reaction. A reagent that increases the oxidation number is called an oxidizing agent. Both potassium permanganate and potassium dichromate are oxidizing agents but the hydroxylation of alkenes takes place in particular conditions.

Complete Step By Step Answer:
First, let us know about the hydroxylation reactions of alkenes. Hydroxylation of alkenes is an oxidation reaction where a carbon-carbon double bond converts to a carbon-hydroxyl bond. Converting an alkene to a glycol requires adding a hydroxyl group to each end of the double bond. This addition is called hydroxylation (or hydroxylation) of the double bond. Hydroxylation of Alkenes is the oxidation of an organic compound. When oxygen is added to the C−HC-H group and forms a bond, there is a generation of −OH-OH groups creating COH.COH. As oxygen reacts very slowly, there is a need for catalysts or enzymes that are also termed as Hydroxylases
The main methods of effecting cis-hydroxylation are by reaction with potassium permanganate, with osmium tetroxide alone or as a catalyst, or with silver iodoacetate according to Woodward procedure. The most basic method of trans-hydroxylation is undoubtedly the reaction with peracids.
CH3−CH=CH−CH3→H+KMnO4CH3−CH(OH)−CH(OH)−CH3 diol formation C{{H}_{3}}-CH=CH-C{{H}_{3}}\xrightarrow[{{H}^{+}}]{KMn{{O}_{4}}}\underset{\text{ }diol\text{ }formation}{\mathop{C{{H}_{3}}-CH\left( OH \right)-CH\left( OH \right)-C{{H}_{3}}}}\,
Therefore, the correct answer is option A.

Note :
Remember that the potassium manganate (VII)\left( VII \right) solution is a strong oxidizing agent. So if we are not using cold dilute conditions then product formation will not stop at the diol. The manganate (VII)\left( VII \right) ions can further oxidize the diol if we are using concentrated acidic conditions.