An unknown alkyl halide $$(A)$$  reacts with alcoholic $$KOH$$  to produce a hydrocarbon $$\left( {{C_4}{H_8}} \right)$$  as the major product. Ozonolysis of the hydrocarbon affords one mole of propanaldehyde and one mole of formaldehyde. Suggest which organic compound among the following is the correct structure of the above alkyl halide $$(A)?$$

Solution :
\[\underset{\text{Propanaldehyde}}{\mathop{C{{H}_{3}}C{{H}_{2}}\underset{\begin{smallmatrix} | \\ H\, \end{smallmatrix}}{\mathop{C}}\,=}}\,\,\,\underset{\text{Formaldehyde}}{\mathop{O+O=C{{H}_{2}}}}\,\]      \[\xleftarrow[\frac{Zn}{{{H}_{2}}O}]{{{O}_{3}}}\underset{\begin{align} & \,\,\,\,\,\,\,\,\,\,\,\,{{C}_{4}}{{H}_{8}} \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\uparrow KOH\left( alc. \right) \\ & C{{H}_{3}}C{{H}_{2}}C{{H}_{2}}C{{H}_{2}}Br \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( A \right) \\ \end{align}}{\mathop{C{{H}_{3}}C{{H}_{2}}CH=C{{H}_{2}}}}\,\]