A) X = dilute aqueous \[NaOH,\]\[20{}^\circ C;\]Y = HBr/acetic acid, \[20{}^\circ C\]
B) X = concentrated alcoholic\[NaOH,\]\[80{}^\circ C;\]Y = HBr/acetic acid, \[20{}^\circ C\]
C) X = dilute aqueous \[NaOH,\] 20°C; Y =\[B{{r}_{2}}\text{/}CHC{{l}_{3}},\,0{}^\circ C\]
D) X = concentrated alcoholic\[NaOH,\]\[80{}^\circ C;\]Y = \[B{{r}_{2}}\text{/}CHC{{l}_{3}},\,0{}^\circ C\]
Correct Answer: B
Solution :
\[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}Br\xrightarrow[\text{dehydrobromination,}\,-HBr]{Conc.\,alc.\,NaOH,\,\,80{}^\circ C(x)}\] \[\begin{align} & \,\,\,\,\,\,C{{H}_{3}}CH=C{{H}_{2}} \\ & \,\,\,\,\,\underset{\vee }{\mathop{|}}\,HBr,\,acetic\,acid\,20{}^\circ C\,(Y) \\ & C{{H}_{3}}-CHBr-C{{H}_{3}} \\ \end{align}\]You need to login to perform this action.
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