Question

Which among the following is used to prepare $CH_3 - CH_2 - CH_2 - CH_3$ from $CH_3 - CH_2 - Cl$?

• A. Na / dry ether
• B. $Zn-Cu$ couple / ethanol
• C. $LiAlH_{4}$ / ether
• D. $CH_{3}-COCl$ / $AlCl_{3}$

Hint:

Chemistry Tip: Wurtz Reaction is best for preparing symmetrical alkanes with an even number of carbon atoms. If you see the carbon count double, think Sodium/Ether!

Answer 1

The Correct Option is A

Concept: Chemistry (Organic Chemistry) - Wurtz Reaction.

Step 1: Analyze the reactant and the product. The reactant is ethyl chloride ($CH_3-CH_2-Cl$), which is a 2-carbon alkyl halide. The product is n-butane ($CH_3-CH_2-CH_2-CH_3$), which is a 4-carbon alkane.

Step 2: Observe the change in the carbon chain. The number of carbon atoms in the product is exactly double the number of carbon atoms in the starting alkyl halide ($2 \times 2 = 4$). The product is also a symmetrical alkane.

Step 3: Recall the named reaction for coupling alkyl halides. The specific reaction used to prepare higher symmetrical alkanes from alkyl halides by doubling the carbon chain is known as the Wurtz reaction.

Step 4: Identify the reagents for the Wurtz reaction. The Wurtz reaction requires treating an alkyl halide with metallic sodium ($Na$) in the presence of dry ether as a solvent. The ether must be dry (anhydrous) to prevent the highly reactive sodium from reacting with water.

Step 5: Verify the reaction equation. $2 CH_3CH_2Cl + 2 Na \xrightarrow{\text{dry ether}} CH_3CH_2-CH_2CH_3 + 2 NaCl$. Since the reagents $Na / \text{dry ether}$ perfectly facilitate this doubling transformation, Option A is the correct choice. $$ \therefore \text{The reagent used is Na / dry ether.} $$