'X' is the isomer of $\text{C}_6\text{H}_{14}$. It has four primary carbons and two tertiary carbons. 'X' can be prepared from which of the following reaction?
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The Wurtz reaction ($2\text{R-X} + 2\text{Na} \to \text{R-R} + 2\text{NaX}$) is a powerful method for making symmetrical alkanes by coupling two alkyl halide molecules. To identify the reactant, mentally cleave the target symmetrical alkane at the middle bond.
$\text{Reaction with } \text{Ni}/\Delta \text{ and } \text{H}_2$ (Hydrogenation)
$\text{Reaction of a halide with } \text{Na}/\text{dry ether}$ (Wurtz Reaction)
$\text{Reaction of a halide with } \text{Zn}/\text{H}^+$ (Reduction)
$\text{Reaction of a halide with } \text{Na}/\text{dry ether}$ (Wurtz Reaction)
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The Correct Option isD
Solution and Explanation
Step 1: Determine the structure of the isomer X.
The molecular formula is $\text{C}_6\text{H}_{14}$, which is an alkane (general formula $\text{C}_n\text{H}_{2n+2}$).
The isomer X must have:
- 4 Primary carbons ($\text{C}1^\circ$, bonded to one other carbon).
- 2 Tertiary carbons ($\text{C}3^\circ$, bonded to three other carbons).
Let's try to sketch the structure. A molecule with two tertiary carbons must have a carbon backbone with three branches from each tertiary carbon. The only $\text{C}_6$ alkane that fits this description is 2,3-Dimethylbutane.
Structure of 2,3-Dimethylbutane: $\text{CH}_3-\text{CH}(\text{CH}_3)-\text{CH}(\text{CH}_3)-\text{CH}_3$.
The carbons at positions 1, 4, and the two methyl groups are $\text{C}1^\circ$ (4 primary carbons).
The carbons at positions 2 and 3 are $\text{C}3^\circ$ (2 tertiary carbons).
So, X is 2,3-Dimethylbutane.
Step 2: Analyze the given reactions for the preparation of X.
(A) $\text{Ni}/\text{H}_2$ is used for the hydrogenation of alkenes/alkynes. 2,3-Dimethylbutane can be formed from 2,3-Dimethyl-2-butene or 2,3-Dimethyl-1-butene, but this is a general reaction, not a specific preparation method.
(D) Wurtz reaction ($\text{R-X} + 2\text{Na} + \text{X-R} \to \text{R-R} + 2\text{NaX}$). This reaction is used to couple two alkyl halides to form a larger alkane.
To form 2,3-Dimethylbutane, the two R-groups must be identical and correspond to the 'half' of the molecule joined by the new bond.
$2,3-\text{Dimethylbutane}$ can be conceptually cleaved into two units: $(\text{CH}_3)_2\text{CH} - \text{CH}(\text{CH}_3)_2$.
The alkyl halide needed is Isopropyl halide, $\text{CH}_3-\text{CH}(\text{X})-\text{CH}_3$.
The reaction is: $2\text{CH}_3-\text{CH}(\text{I})-\text{CH}_3 + 2\text{Na} \xrightarrow{\text{dry ether}} \text{CH}_3-\text{CH}(\text{CH}_3)-\text{CH}(\text{CH}_3)-\text{CH}_3 + 2\text{NaI}$.
The Wurtz reaction of Isopropyl Halide successfully forms 2,3-Dimethylbutane.
Step 3: Conclude the final correct option.
The most plausible specific preparation method is the Wurtz reaction of isopropyl halide. Since the option lists the conditions for the Wurtz reaction, it is the correct choice.
