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Step-by-Step Solution
Step 1: Identify the Given Compound and Reaction
We are given an organic compound A with the formula $C_{4}H_{9}Cl$. It undergoes a Wurtz reaction with sodium metal in the presence of diethyl ether (represented as Na/diethyl ether), producing a hydrocarbon.
Step 2: Predict the Wurtz Reaction Product
In a Wurtz reaction, two alkyl halide molecules couple to form a hydrocarbon:
$2 \, C_{4}H_{9}Cl + 2 \, Na \longrightarrow C_{4}H_{9} - C_{4}H_{9} + 2 \, NaCl$
Thus, the resulting hydrocarbon from $C_{4}H_{9}Cl$ is an eight-carbon compound formed by joining two $C_{4}H_{9}$ groups.
Step 3: Analyze the Monochlorination Pattern
We are told that when this newly formed hydrocarbon undergoes monochlorination, it gives only one type of chloro-derivative. This means all the hydrogens in the hydrocarbon are equivalent (i.e., they occupy equivalent positions).
For all hydrogens to be equivalent, the hydrocarbon must be highly symmetrical. Among the possible $C_{4}H_{9}Cl$ isomers, t-butyl chloride ($\text{(CH}_3\text{)}_3\text{C-Cl}$) yields the coupling product 2,2-dimethylpropane ($\text{(CH}_3\text{)}_3\text{C-C(CH}_3\text{)}_3$) under Wurtz reaction conditions, which is also known as neopentane.
Step 4: Confirming the Structure and Final Answer
2,2-Dimethylpropane (neopentane) has only one type of hydrogen environment, so upon monochlorination, it produces only one type of chloro-derivative. Hence, the compound A must be t-butyl chloride.
Step 5: Conclusion
Therefore, given the requirement that the resultant hydrocarbon on monochlorination yields only one chloro derivative, t-butyl chloride is the correct answer.
