The given compound is 3-bromo-3-cyclopentylhexane.

Dehydrobromination of this compound using alcoholic KOH will lead to the formation of alkenes.

The number of alkenes formed will depend on which hydrogen atoms are removed from the compound.

To determine the number of possible alkenes, we need to identify all the possible sets of hydrogen atoms that can be removed to form a double bond.

There are five sets of hydrogen atoms that can be removed to form alkenes:

1. The two hydrogen atoms on the same carbon atom as the bromine atom. This will form a double bond between the two adjacent carbons, resulting in 1,6-cyclopentadiene.

2. The two hydrogen atoms on the carbon atom adjacent to the bromine atom. This will form a double bond between the two adjacent carbons, resulting in 1,5-cyclopentadiene.

3. The two hydrogen atoms on the carbon atom adjacent to the carbon atom with the bromine atom. This will form a double bond between the two adjacent carbons, resulting in 1-cyclopentene.

4. The two hydrogen atoms on the carbon atom two carbons away from the bromine atom. This will form a double bond between the two adjacent carbons, resulting in 3-cyclopentene.

5. The two hydrogen atoms on the carbon atom three carbons away from the bromine atom. This will form a double bond between the two adjacent carbons, resulting in 4-cyclopentene.

Therefore, the total number of alkenes possible by dehydrobromination of 3-bromo-3-cyclopentylhexane using alcoholic KOH is 5.