

MCQOPTIONS
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1. |
\[{{H}_{2}}C=CH-C{{H}_{2}}-C{{H}_{2}}-\underset{OH}{\mathop{\underset{|}{\mathop{C}}\,}}\,H-C{{H}_{3}}\] \[\xrightarrow[Pyridine]{SOC{{l}_{2}}}(A)\xrightarrow[({{H}_{2}}O)]{{{O}_{3}}/Zn}\underset{{{C}_{5}}{{H}_{9}}ClO}{\mathop{(B)}}\,\xrightarrow{NaB{{H}_{4}}}(C)\] Compound (C) is |
A. | \[C{{H}_{3}}-\overset{OH}{\mathop{\overset{|}{\mathop{C}}\,}}\,H-C{{H}_{2}}-\underset{Cl}{\mathop{\underset{|}{\mathop{C}}\,}}\,H-C{{H}_{3}}\] |
B. | \[HOC{{H}_{2}}-C{{H}_{2}}-C{{H}_{2}}-C{{H}_{2}}-C{{H}_{2}}-C{{H}_{2}}-Cl\] |
C. | \[HO-C{{H}_{2}}-C{{H}_{2}}-C{{H}_{2}}-\overset{Cl}{\mathop{\overset{|}{\mathop{C}}\,}}\,H-C{{H}_{3}}\] |
D. | \[HO-C{{H}_{2}}-C{{H}_{2}}-\underset{Cl}{\mathop{\underset{|}{\mathop{C}}\,}}\,H-C{{H}_{2}}-C{{H}_{3}}\] |
Answer» D. \[HO-C{{H}_{2}}-C{{H}_{2}}-\underset{Cl}{\mathop{\underset{|}{\mathop{C}}\,}}\,H-C{{H}_{2}}-C{{H}_{3}}\] | |