A) \[C{{H}_{3}}Br\]
B) \[C{{H}_{3}}C{{H}_{2}}OH\]
C) \[C{{H}_{3}}OH\]
D) \[C{{H}_{3}}COC{{H}_{3}}\]
Correct Answer: A
Solution :
When \[C{{H}_{3}}Br\]reacts with \[LiAl{{H}_{4}}\] and \[Na,\] it gives \[C{{H}_{4}}\]and \[{{C}_{2}}{{H}_{6}}\] both. \[C{{H}_{3}}Br\xrightarrow{LiAl{{H}_{4}}}\underset{Methane}{\mathop{C{{H}_{4}}+H-Br}}\,\] \[C{{H}_{3}}Br+2Na+BrC{{H}_{3}}\xrightarrow{{}}\] \[\underset{Ethane}{\mathop{C{{H}_{3}}-C{{H}_{3}}}}\,+2NaBr\]
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