A) \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}OH\]
B) \[C{{H}_{3}}CH(OH)C{{H}_{2}}C{{H}_{3}}\]
C) \[{{(C{{H}_{3}})}_{3}}COH\]
D) \[C{{H}_{3}}C{{H}_{2}}-O-C{{H}_{2}}C{{H}_{3}}\]
Correct Answer: B
Solution :
\[C{{H}_{3}}CH(OH)C{{H}_{2}}C{{H}_{3}}\xrightarrow{\text{Conc}\text{.}\,{{H}_{2}}S{{O}_{4}}}\]\[\underset{({{C}_{4}}{{H}_{8}})}{\mathop{C{{H}_{3}}CH=CHC{{H}_{3}}}}\,\] \[\underset{2-\text{Butanol}}{\mathop{C{{H}_{3}}CHOHC{{H}_{2}}C{{H}_{3}}}}\,\xrightarrow{[O]}\underset{\text{Butanone}}{\mathop{C{{H}_{3}}COC{{H}_{2}}C{{H}_{3}}}}\,\] Butanone gives both an oxime and positive iodoform test, therefore, the original compound is 2-butanol.
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