A) methanol and bromoethane
B) ethyl hydrogen sulphate and alcoholic\[KOH\]
C) ethyl hydrogen sulphate and aqueous \[KOH\]
D) ethanol and alcoholic \[KOH\]
Correct Answer: D
Solution :
\[C{{H}_{2}}=C{{H}_{2}}\xrightarrow{Conc.{{H}_{2}}S{{O}_{4}}}C{{H}_{3}}-\underset{(A)}{\mathop{C{{H}_{2}}}}\,-HS{{O}_{4}}\] (addition reaction) \[C{{H}_{3}}-C{{H}_{2}}-HS{{O}_{4}}\xrightarrow{\Delta ,{{H}_{2}}O}C{{H}_{3}}-\underset{(B)}{\mathop{C{{H}_{2}}}}\,-OH\] (hydrolysis) \[C{{H}_{3}}-C{{H}_{2}}-OH\xrightarrow{PB{{r}_{3}}}C{{H}_{3}}-C{{H}_{2}}-Br\] (bromination) \[C{{H}_{3}}-C{{H}_{2}}-Br\xrightarrow[(\Delta )]{Alc.\,KOH(\Delta )}C{{H}_{2}}=C{{H}_{2}}\] (dehydrohalogenation) Hence, B and D in the above given reactions are \[C{{H}_{3}}C{{H}_{2}}OH\](ethanol) and ale. \[KOH\].
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