Question

Identify substrate 'S' in the following reaction.
\(\text{S}\overset{\text{Na / dry ether}\,}{\rightarrow }\) 3, 4 - diethyl - 3, 4 - dimethyl hexane

• A. 3 - chloro - 2 - methylpentane
• B. 2 - chloro - 3 - methylpentane
• C. 3 - chloro - 3 - methylpentane
• D. 2 - chloro - 2 - methylpentane

Hint:

Wurtz joins two alkyl groups. Split the product in half to find the alkyl halide.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
In the Wurtz reaction, two molecules of an alkyl halide react with sodium in dry ether and the two alkyl groups join:
\[ 2\text{R-X} + 2\text{Na} \to \text{R-R} + 2\text{NaX} \]
So the product is made of two identical R groups.

Step 2: Key Formula or Approach:
Break the product at its central C-C bond to get R, then attach X.

Step 3: Detailed Explanation:
The product is 3,4-diethyl-3,4-dimethylhexane. The central bond is between C3 and C4. Each of these carbons carries a methyl, an ethyl, and the rest of the hexane chain (which is also an ethyl: C2-C1 on one side, C5-C6 on the other).
So each half is a carbon bearing one methyl and two ethyl groups: \(\text{C(CH}_3)(\text{C}_2\text{H}_5)_2\).
Its longest chain is five carbons with a methyl on C3, so the radical is 3-methylpentan-3-yl.
Attaching chlorine gives 3-chloro-3-methylpentane. Options (A) and (B) are secondary chlorides that would give a different coupled product. Option (D) would give 2,3-dimethyl-type chain products.

Final Answer:
The substrate is 3-chloro-3-methylpentane, option (C). \[ \boxed{\text{3-chloro-3-methylpentane (C)}} \]