question_answer

4-bromo-l-butanol [A] can be converted into 5-hydroxy pentanoic acid [B] by following method [s]

\[I.\]\[A\xrightarrow[{}]{Mg/ether}\xrightarrow[(ii){{H}_{3}}{{O}^{+}}]{(i)C{{O}_{2}}}B\]

\[II.\]\[A\xrightarrow[{}]{KCN}\xrightarrow[{}]{{{H}_{3}}{{O}^{+}}}B\]

\[III.\]\[A\xrightarrow[{}]{Aq.\,KOH}\xrightarrow[{}]{KMn{{O}_{4}}/{{H}^{+}}}B\]

Select the correct alternate

A)  \[I,\text{ }II,\text{ }III\]           

B)  \[I,\text{ }II\]

C)  \[II,\text{ }III\]             

D)  \[II\]

Correct Answer: B

Solution :

 I. \[BrC{{H}_{2}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}OH\] [A] \[\xrightarrow[{}]{Mg/ether}Br\overset{s+}{\mathop{Mg}}\,\overset{s-}{\mathop{C{{H}_{2}}}}\,C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}OH\] \[\underset{5-hydroxy\text{ }pentanoic\text{ }acid}{\mathop{HO\overset{\begin{smallmatrix}  O \\  || \end{smallmatrix}}{\mathop{\underset{1}{\mathop{C}}\,}}\,\underset{2}{\mathop{C}}\,{{H}_{2}}\underset{3}{\mathop{C}}\,{{H}_{2}}\underset{4}{\mathop{C}}\,{{H}_{2}}\underset{5}{\mathop{C}}\,{{H}_{2}}OH}}\,\] II. \[\overset{(B)}{\mathop{BrC{{H}_{2}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}OH}}\,\] \[\xrightarrow[-KOH]{KCN}NCC{{H}_{2}}-C{{H}_{2}}-C{{H}_{2}}-C{{H}_{2}}-OH\] \[\downarrow {{H}_{3}}{{O}^{+}}\] \[\underset{5-hydroxy\text{ }pentanoic\text{ }acid}{\mathop{HCOO\overset{1}{\mathop{C}}\,-\overset{2}{\mathop{C}}\,{{H}_{2}}-\overset{3}{\mathop{C}}\,{{H}_{2}}-\overset{4}{\mathop{C}}\,{{H}_{2}}-\overset{5}{\mathop{C}}\,{{H}_{2}}-OH}}\,\] III. \[BrC{{H}_{2}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}OH\] \[\xrightarrow[-KBr]{Aq.KOH}HOC{{H}_{2}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}OH\]