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 \[C{{H}_{3}}O{{C}_{2}}{{H}_{5}}\] (methoxyethane). The reactions are as follows. \[\underset{methoxyethane}{\mathop{C{{H}_{3}}O{{C}_{2}}H}}\,+2HI\xrightarrow{{}}\underset{X}{\mathop{C{{H}_{3}}I}}\,+\underset{Y}{\mathop{{{C}_{2}}{{H}_{3}}OH}}\,\] \[{{C}_{2}}{{H}_{5}}OH+\underset{(aqueous)}{\mathop{NaOH}}\,+{{I}_{2}}\xrightarrow{{}}\underset{iodofrom}{\mathop{CH{{I}_{3}}}}\,+HCOONa\] \[+{{H}_{2}}O+NaI\]
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