Identify Z in the following sequence of reactions: CH\(_3\)-CH\(_2\)-OH + PCl\(_3\) \(\xrightarrow{\text{alcohol}}\)\(X \xrightarrow{\text{KOH}}\)\(Y \xrightarrow{\text{conc. H\(_2\)SO\(_4\)}}\)\(Z \)
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Dehydrohalogenation with KOH leads to the formation of alkenes, and elimination reactions with concentrated sulfuric acid result in alkane formation.
Step 1: Understanding the reaction sequence.
- The first step involves the conversion of ethanol (CH\(_3\)-CH\(_2\)-OH) into an alkyl chloride, CH\(_3\)-CH\(_2\)-Cl, by reaction with PCl\(_3\).
- The second step uses alcoholic KOH, which leads to a dehydrohalogenation reaction, producing an alkene.
- The final step involves treating the alkene with concentrated sulfuric acid (H\(_2\)SO\(_4\)), which causes the alkene to undergo an elimination reaction, forming a hydrocarbon.
Step 2: Analyzing the options. (A) CH\(_3\)-CH\(_2\)-OH: This is the starting alcohol and does not fit the final product of the reaction. (B) (CH\(_3\))\(_2\)CH-CH\(_3\): This is incorrect as the final product is an alkane, not this structure. (C) CH\(_3\)-CH=CH\(_2\): This is an alkene intermediate, but the final product is an alkane. (D) CH\(_3\)-CH\(_3\)-OH: Correct — This is the alkane product formed after the elimination step.
Step 3: Conclusion.
The correct answer is (D) CH\(_3\)-CH\(_3\)-OH.
