A) \[C{{H}_{3}}COOH,C{{H}_{2}}=C{{H}_{2}}\]
B) \[C{{H}_{3}}CHO,\,C{{H}_{2}}=C{{H}_{2}}\]
C) \[{{C}_{2}}{{H}_{5}}OH,{{C}_{2}}{{H}_{5}}I\]
D) None of the above
Correct Answer: C
Solution :
Key Idea: When HI is present in less quantity and temperature is low, then two alkyl group separate and form one molecule of alcohol and other form alkyl halide. \[{{C}_{2}}{{H}_{5}}O{{C}_{2}}{{H}_{5}}+HI\xrightarrow{cold}\underset{\text{ethyl}\,\text{iodide}}{\mathop{{{C}_{2}}{{H}_{5}}I}}\,\] \[+\,\underset{\text{ethyl}\,\text{alcohol}}{\mathop{\,{{C}_{2}}{{H}_{5}}OH}}\,\]
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