A) propan-2-ol
B) propan-1-ol
C) ethoxyethane
D) methoxyethane
Correct Answer: D
Solution :
Molecular formula \[{{C}_{3}}{{H}_{8}}O({{C}_{n}}{{H}_{2n+2}}O)\] suggests that the organic compound, is either alcohol or ether. Since, the compound on reaction with\[HI\]gives two different compounds, it must be an unsymmetrical ether and its formula must be \[C{{H}_{3}}CO{{C}_{2}}{{H}_{5}}\](methoxyethane). The reactions are as follows \[\underset{methoxyethane}{\mathop{C{{H}_{3}}O{{C}_{2}}{{H}_{5}}}}\,+2HI\xrightarrow[{}]{{}}\underset{X}{\mathop{C{{H}_{3}}I}}\,+\underset{Y}{\mathop{{{C}_{2}}{{H}_{5}}OH}}\,\] \[{{C}_{2}}{{H}_{5}}OH+\underset{aqueous}{\mathop{NaOH}}\,+{{I}_{2}}\xrightarrow[{}]{{}}\underset{iodoform}{\mathop{CH{{I}_{3}}}}\,\]\[+HCOONa+{{H}_{2}}O+NaI\]You need to login to perform this action.
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