A) \[C{{H}_{3}}-C{{H}_{2}}-C{{H}_{2}}-C{{H}_{2}}-C{{H}_{2}}-OH\]
B) \[C{{H}_{3}}-C{{H}_{2}}-\underset{OH}{\mathop{\underset{\mathbf{|}}{\mathop{C{{H}_{2}}CH}}\,}}\,-C{{H}_{3}}\]
C) \[C{{H}_{3}}-C{{H}_{2}}-\underset{OH}{\mathop{\underset{\mathbf{|}}{\mathop{CH}}\,}}\,-C{{H}_{2}}-C{{H}_{3}}\]
D) \[C{{H}_{3}}-C{{H}_{2}}-\underset{C{{H}_{3}}}{\mathop{\underset{\mathbf{|}}{\mathop{CH}}\,}}\,-C{{H}_{2}}-OH\]
Correct Answer: C
Solution :
Since, the compound B gave a 2, 4-dinitrophenylhydrazine derivative but did not answer halo form test or silver mirror test, it must contains \[a>C=O\]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. Thus. A is\[\underset{(secondary\,alcohol)}{\mathop{C{{H}_{3}}C{{H}_{2}}-\underset{OH}{\mathop{\underset{\mathbf{|}}{\mathop{CH}}\,}}\,-C{{H}_{2}}-C{{H}_{3}}}}\,\xrightarrow[-\,{{H}_{2}}O]{[O]}\] \[\underset{\begin{smallmatrix} (do\,not\,answer\,haloform\, \\ test\,or\,silver\,mirror\,test) \end{smallmatrix}}{\mathop{C{{H}_{3}}C{{H}_{2}}\overset{O}{\mathop{\overset{\mathbf{||}}{\mathop{C}}\,}}\,-C{{H}_{2}}C{{H}_{3}}}}\,\]You need to login to perform this action.
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