A) \[C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ Cl \end{smallmatrix}}{\overset{\begin{smallmatrix} Cl \\ | \end{smallmatrix}}{\mathop{C}}}\,-C{{H}_{2}}-C{{H}_{2}}-C{{H}_{3}}\]
B) \[C{{H}_{3}}-C{{H}_{2}}-\underset{\begin{smallmatrix} | \\ Cl \end{smallmatrix}}{\overset{\begin{smallmatrix} Cl \\ | \end{smallmatrix}}{\mathop{C}}}\,-C{{H}_{2}}-C{{H}_{3}}\]
C) \[C{{H}_{3}}-C{{H}_{2}}-C{{H}_{2}}-CH{{ }_{2}}-\underset{\begin{smallmatrix} | \\ Cl \end{smallmatrix}}{\overset{\begin{smallmatrix} Cl \\ | \end{smallmatrix}}{\mathop{CH}}}\,\]
D) \[C{{H}_{3}}-\overset{\begin{smallmatrix} Cl \\ | \end{smallmatrix}}{\mathop{CH}}\,-\overset{\begin{smallmatrix} Cl \\ | \end{smallmatrix}}{\mathop{CH}}\,-C{{H}_{2}}-C{{H}_{3}}\]
Correct Answer: A
Solution :
A compound\[({{C}_{5}}{{H}_{10}}C{{l}_{2}})\]on hydrolysis gives \[{{C}_{5}}{{H}_{10}}O\]which reacts with hydroxylamine, so \[{{C}_{5}}{{H}_{10}}O\]must be aldehydic or ketonic group but it does not give Fehling test, so it must be a ketone and it form iodoform which is a characteristic test of\[C{{H}_{3}}CO-\]group. \[\underset{\begin{smallmatrix} ({{C}_{5}}{{H}_{10}}C{{l}_{2}}) \\ \,\,\,\,\,\,\,\,(A) \end{smallmatrix}}{\mathop{C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ Cl \end{smallmatrix}}{\overset{\begin{smallmatrix} Cl \\ | \end{smallmatrix}}{\mathop{C}}}\,-C{{H}_{2}}}}\,-C{{H}_{2}}-C{{H}_{3}}\xrightarrow[{}]{~hydrolysis}\] \[\underset{(unstable)}{\mathop{C{{H}_{3}}-\underset{\begin{smallmatrix} | \\ Cl \end{smallmatrix}}{\overset{\begin{smallmatrix} Cl \\ | \end{smallmatrix}}{\mathop{C}}}\,-C{{H}_{2}}}}\,-C{{H}_{2}}-C{{H}_{3}}\]You need to login to perform this action.
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