Problem 54 - A Complete Resource Book in Chemistry for JEE Main

Question

Ethylene reacts with \(1 \%\) cold alkaline \(\mathrm{KMnO}_{4}\) to form (a) ethylene glycol (b) oxalic acid (c) \(\mathrm{HCHO}\) (d) ethyl alcohol

Step-by-Step Solution

Answer

(a) ethylene glycol

Step 1: Identify the Reaction

Ethylene ( ext{C}_2 ext{H}_4 ext{) is known to react with cold dilute alkaline potassium permanganate ( ext{KMnO}_4 ext{)} to form a product through an oxidation process known as 'syn' addition of hydroxyl groups.

Step 2: Determine the Product

Under these mild conditions, ext{KMnO}_4 ext{)} adds hydroxyl (OH) groups across the double bond of ethylene, converting it into a diol. This specific diol is ethylene glycol ( ext{C}_2 ext{H}_4(OH)_2 ext{)}.

Step 3: Verify the Answer

Since ethylene glycol is the only one among the given products that forms through the oxidation of ethylene with cold alkaline ext{KMnO}_4, the correct choice is option (a) ethylene glycol.

Key Concepts

Cold Alkaline KMnO4

Cold alkaline potassium permanganate (\(\text{KMnO}_4\)) is a powerful oxidizing agent. In chemistry, oxidizing agents are substances that gain electrons in a reaction, thereby oxidizing other substances. Potassium permanganate in alkaline conditions is often used to oxidize alkenes like ethylene into different products.
Cold and alkaline conditions specifically slow down the reaction, leading to gentler and more controlled oxidation.
When introduced to ethylene, the double bond provides the site for this reaction. \(\text{KMnO}_4\) oxidizes the ethylene without breaking the carbon skeleton entirely, which is possible in more extreme conditions.

• Gentle oxidation:
  Preserves more of the original molecule's structure.
• Cold conditions:
  Reduce the risk of over-oxidation.
• Alkaline medium:
  Helps stabilize the permanganate ions in the solution.

Understanding how cold alkaline \(\text{KMnO}_4\) functions provides clarity on its use in organic synthesis and reactions like ethylene oxidation.

Oxidation Reactions

Oxidation reactions involve the transfer of electrons from one substance to another. In simpler terms, oxidation often involves adding oxygen or removing hydrogen from a compound.
The reaction of ethylene with \(\text{KMnO}_4\) is an example of oxidation, where the compound adds oxygen to create a diol.This transformation highlights oxidation in action.
By reacting with the double bond in ethylene, the \(\text{KMnO}_4\) breaks the bond and adds a hydroxyl group (\(\text{OH}\)) to each of the carbon atoms involved in the bond.This process can vividly be seen as follows:

• The double bond:
  Serves as the reactive site for oxidizing agents.
• Formation of diols:
  Indicates successful oxidation by adding two hydroxyl groups.
• Soft conversion:
  Gentle oxidizing conditions prevent formation of unwanted side products.

Oxidation reactions like these are fundamental in creating complex molecules from simple ones, helping us understand chemical syntheses better.

Ethylene Glycol Formation

Ethylene glycol is a compound that emerges when ethylene (\(\text{C}_2\text{H}_4\)) undergoes a syn addition oxidation with cold alkaline \(\text{KMnO}_4\).
The syn addition process is where both hydroxyl groups are added simultaneously on the same side of the plane, thus maintaining the double bond's configuration.
This results in a molecule that functions as a diol, specifically ethylene glycol (\(\text{C}_2\text{H}_4(OH)_2\)), which has many practical uses, including being a key component in antifreeze formulations. Here’s how the formation occurs:

• Smooth oxidation:
  The reaction primarily targets the double bond for addition.
• Diol creation:
  Each carbon atom in the double bond receives a hydroxyl group.
• Single product type:
  Alleviates complexities often involved in radical or harsher conditions' reactions.

Understanding this formation illustrates how simple molecules can be transformed efficiently into more functional forms, which are incredibly useful in both industrial and chemical applications.