When n-propylbromide is treated with ethanolic KOH (potassium hydroxide dissolved in ethanol), it undergoes an elimination reaction known as dehydrohalogenation. This reaction removes a hydrogen atom from the beta-carbon (the carbon adjacent to the bromine) and the bromine atom itself, resulting in the formation of a double bond between the two adjacent carbon atoms.

Here is a detailed explanation of the reaction:

1. Initial Reaction:
The ethanolic KOH acts as a base and abstracts a hydrogen atom from the beta-carbon, creating a carbanion intermediate. The bromine atom simultaneously leaves, resulting in the formation of an alkoxide ion.

2. Formation of Alkene:
The alkoxide ion then reacts with another hydrogen atom, leading to the formation of an alkene. In this case, the alkene formed is propene (CH3-CH=CH2), which is the correct answer (option C). The alkene is formed due to the loss of a hydrogen atom from the beta-carbon and the elimination of the bromine atom.

Overall Reaction:
n-Propylbromide + Ethanolic KOH → Propene + Potassium Bromide + Water

The reaction mechanism involves the formation of an alkoxide ion intermediate, which further participates in the dehydrohalogenation reaction to form the alkene. The ethanolic KOH serves as a strong base to initiate the elimination reaction.

It is important to note that the reaction conditions, such as the presence of ethanolic KOH, are crucial for the dehydrohalogenation reaction to occur. In the absence of a strong base, other reactions, such as substitution or oxidation, may take place instead.