A) \[C{{H}_{4}}\]
B) \[{{C}_{2}}{{H}_{6}}\]
C) \[{{C}_{3}}{{H}_{8}}\]
D) \[{{C}_{4}}{{H}_{10}}\]
Correct Answer: A
Solution :
\[{{C}_{2}}{{H}_{5}}OH\] is a protic solvent. Thus, when Grignard reagent \[(RMgBr)\] reacts with it, it results in the formation of \[R-H\]. \[\overset{-\delta }{\mathop{C{{H}_{3}}}}\,-\overset{+\delta }{\mathop{Mg}}\,-Br+{{C}_{2}}{{H}_{5}}\,\,\,\,\overset{-\delta +\delta }{\mathop{OH}}\,\xrightarrow{{}}\]\[C{{H}_{4}}+Mg(O{{C}_{2}}{{H}_{5}})Br\]
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