Explain the reaction mechanism of dehydrohalogenation of ethyl bromide in the presence of KOH and ethanol.

Dear student!
The mechanism takes place as follows:

The given reaction is an elimination reaction where elimination of HBr occurs from ethyl bromide which takes place in presence of strong bases like alcoholic KOH etc.

The given reaction is an elimination reaction in which is as follows,

In first step, the ethyl bromide releases the bromide ion in the form of KBr to form a carbocation which further  releases an H⁺ ion to balance the positive charge and worms a water molecule.