\[{\rm{C}}{{\rm{H}}_{\rm{3}}} - {\rm{C}}{{\rm{H}}_{\rm{2}}} - {\rm{Cl}} \overset{Alc. KOH}{\rightarrow} {\rm{A}}\], the product is
(A) \[{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{C}}{{\rm{H}}_{\rm{2}}}{\rm{OK}}\]
(B) \[{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{CHO}}\]
(C) \[{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{C}}{{\rm{H}}_{\rm{2}}}{\rm{OC}}{{\rm{H}}_{\rm{2}}}{\rm{C}}{{\rm{H}}_{\rm{3}}}\]
(D) \[{\rm{C}}{{\rm{H}}_{\rm{2}}} = {\rm{C}}{{\rm{H}}_{\rm{2}}}\]

Hint: Here, the reaction of an alkyl halide in presence of alcoholic KOH is given and we have to identify the product obtained in the reaction. We know that the reagent alcoholic KOH gives the product through the process of elimination.

Complete Step by Step Answer:
Let's first discuss the dehydrohalogenation reaction. In this reaction, a molecule of hydrogen atom and a halogen atom are removed from the compound in presence of alcoholic KOH. This reaction forms the alkene as the product. And this process is also termed \[\beta \] elimination. We know that a \[\beta \] carbon indicates that carbon atom that is next to the halogen bonded carbon atom. And, the \[\beta \]elimination says that the removal of the hydrogen atom takes place from the \[\beta \]carbon.

Let's understand the detailed elimination reaction of the given alkyl halide in the presence of potassium hydroxide of alcoholic nature. In this reaction, the elimination of hydrogen atoms from the \[\beta \]carbon occurs and the formation of a water molecule occurs. And the halogen atom leaves the group and alkene forms.

Therefore, the reaction of ethyl chloride with alcoholic KOH gives ethene as the product.
Hence, option D is right.

Note: It is to be noted that only alcoholic KOH gives an elimination reaction because of the strong nature of the base. Due to its strong basic nature, it can pull an atom of hydrogen to give a molecule of water and thus forms a double bond. Also, the high temperature favours the formation of an elimination product.