A) \[{{C}_{2}}{{H}_{5}}O{{C}_{2}}{{H}_{5}}\]
B) \[C{{H}_{3}}\underset{\begin{smallmatrix} | \\ OC{{H}_{3}} \end{smallmatrix}}{\mathop{CH}}\,C{{H}_{3}}\]
C) \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}OC{{H}_{3}}\]
D) \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}OH\]
Correct Answer: A
Solution :
\[{{C}_{4}}{{H}_{10}}O\]can have two structures: \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}OH\]and\[{{C}_{2}}{{H}_{5}}O{{C}_{2}}{{H}_{5}}\]. Since it does not react with sodium metal, it can not be an alcohol. \[{{C}_{4}}{{H}_{5}}O{{C}_{2}}\underset{(excess)}{\mathop{Hl}}\,\to 2{{C}_{2}}{{H}_{5}}l+{{H}_{2}}O\]
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