A) \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}OH\]
B) \[C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}\underset{\begin{smallmatrix} | \\ OH \end{smallmatrix}}{\mathop{CH}}\,C{{H}_{3}}\]
C) \[C{{H}_{3}}C{{H}_{2}}\underset{\begin{smallmatrix} | \\ OH \end{smallmatrix}}{\mathop{CH}}\,C{{H}_{2}}C{{H}_{3}}\]
D) \[C{{H}_{3}}C{{H}_{2}}\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\mathop{CH}}\,C{{H}_{2}}OH\]
E) \[C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ C{{H}_{3}} \end{smallmatrix}}{\overset{\begin{smallmatrix} C{{H}_{3}} \\ | \end{smallmatrix}}{\mathop{C}}}\,-C{{H}_{2}}-OH\]
Correct Answer: C
Solution :
Since, the compound B gave a 2, 4-dinitrophenylhydrazine derivative but did not answer halogen test or silver mirror test, it must contains a group, but it is neither a methyl ketone nor an aldehyde. Moreover compound B is obtained by the oxidation of compound A, having molecular formula\[{{C}_{5}}{{H}_{12}}O,\]so the compound A must be a secondary alcohol. \[C{{H}_{3}}-C{{H}_{2}}\underset{\begin{smallmatrix} \,\,\,\,\,\,\,2{}^\circ \text{ }alcohol \\ (compound\text{ }A) \end{smallmatrix}}{\mathop{\underset{\begin{smallmatrix} | \\ OH \end{smallmatrix}}{\mathop{CH}}\,}}\,-C{{H}_{2}}-C{{H}_{3}}\xrightarrow[-{{H}_{2}}O]{[O]}\] \[\underset{\begin{smallmatrix} \,\,\,\,\,\,\,\,\,\,\,ketone \\ (compound\text{ }B) \end{smallmatrix}}{\mathop{C{{H}_{3}}-C{{H}_{2}}-\underset{\begin{smallmatrix} || \\ O \end{smallmatrix}}{\mathop{C}}\,-C{{H}_{2}}-C{{H}_{3}}}}\,\]You need to login to perform this action.
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