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. \[\underset{methoxyethane}{\mathop{C{{H}_{3}}O{{C}_{2}}{{H}_{5}}}}\,+2HI\xrightarrow{{}}\underset{X}{\mathop{C{{H}_{3}}I}}\,+\underset{Y}{\mathop{{{C}_{2}}{{H}_{5}}OH}}\,\] \[{{C}_{2}}{{H}_{5}}OH+\underset{(aqueous)}{\mathop{NaOH}}\,+{{I}_{2}}\xrightarrow{{}}\underset{iodoform}{\mathop{CHI}}\,+HCOONa\] \[+{{H}_{2}}O+\] Nal
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