Question

Which of the following alkyl halide is treated with sodium metal to obtain 2,2,3,3-tetramethyl butane?

• A. tert-Butyl bromide
• B. n-Propyl bromide
• C. sec-Butyl bromide
• D. n-Butyl bromide

Hint:

To solve a Wurtz reaction backwards, count the total number of carbons in the symmetrical product (here, 8 carbons) and divide by 2. This tells you the starting alkyl halide must have exactly 4 carbons, immediately narrowing down your choices to the butyl families.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The problem asks us to determine the starting alkyl halide that, when treated with sodium metal, yields a highly branched symmetric alkane (2,2,3,3-tetramethylbutane).

Step 2: Key Formula or Approach:
Treating an alkyl halide with sodium metal in the presence of dry ether is known as the

Wurtz reaction. This reaction couples two symmetrical alkyl fragments together by forming a carbon-carbon single bond: $$2\text{R-X} + 2\text{Na} \xrightarrow{\text{Dry Ether}} \text{R-R} + 2\text{NaX}$$ To find the required starting material, break the product molecule symmetrically in half down its central carbon-carbon bond.

Step 3: Detailed Explanation:
Let's look at the chemical structure of the target product, 2,2,3,3-tetramethylbutane: $$(\text{CH}_3)_3\text{C}-\text{C}(\text{CH}_3)_3$$ Breaking this highly symmetrical molecule directly in half across the central single bond reveals two identical alkyl fragments: $$(\text{CH}_3)_3\text{C}- \quad \text{and} \quad -\text{C}(\text{CH}_3)_3$$ This fragment represents a tertiary butyl group (tert-butyl). Therefore, the starting alkyl halide must be a tert-butyl halide, such as tert-butyl bromide: $$2(\text{CH}_3)_3\text{C-Br} + 2\text{Na} \xrightarrow{\text{Dry Ether}} (\text{CH}_3)_3\text{C}-\text{C}(\text{CH}_3)_3 + 2\text{NaBr}$$

Step 4: Final Answer:
The required starting material is tert-butyl bromide, which corresponds to option (A).