The total number of alkenes possible by dehydrobromination of 3-bromo-3-cyclopentylhexane using alcoholic `KOH` is

Answer

To determine the total number of alkenes possible by dehydrobromination of 3-bromo-3-cyclopentylhexane using alcoholic KOH, we will follow these steps:

### Step 1: Draw the Structure of 3-bromo-3-cyclopentylhexane
First, we need to visualize the structure of the compound. The compound consists of a hexane chain with a bromine atom attached to the third carbon, which also has a cyclopentyl group attached.

- **Structure**:
- The hexane chain is: CH3-CH2-CH(Br)-CH2-CH2-CH3
- The cyclopentyl group is attached to the third carbon, making it:
- CH3-CH2-C(Br)(C5H9)-CH2-CH3

### Step 2: Identify Beta Hydrogens
Next, we need to identify the beta hydrogens that can be eliminated along with the bromine atom. The beta carbons are the ones adjacent to the carbon that has the bromine.

- **Beta Carbons**:
- The carbon with Br (C3) has two beta carbons: C2 and C4.
- C2 has 2 hydrogens (attached to CH2).
- C4 has 2 hydrogens (attached to CH2).
- The cyclopentyl group also has hydrogens that can be considered.

### Step 3: Possible Eliminations
Now, we can perform the elimination reactions to form alkenes:

1. **Elimination from C2**:
- Removing Br and one hydrogen from C2 gives:
- Product: CH3-CH=CH-C5H9 (Cyclopentyl group remains intact)

2. **Elimination from C4**:
- Removing Br and one hydrogen from C4 gives:
- Product: CH3-CH2-CH=CH-C5H9

3. **Elimination from Cyclopentyl**:
- Removing Br and one hydrogen from the cyclopentyl group gives:
- Product: CH3-CH2-CH(C5H8)=C5H9 (double bond in the cyclopentyl ring)

### Step 4: Geometric Isomerism
Next, we need to consider if any of these products can exhibit geometric isomerism (cis/trans isomerism):

1. **First Product** (CH3-CH=CH-C5H9):
- This alkene can exhibit geometric isomerism because the double bond is between two different groups (alkyl and cyclopentyl).
- **Isomers**: 2 (cis and trans)

2. **Second Product** (CH3-CH2-CH=CH-C5H9):
- This alkene can also exhibit geometric isomerism.
- **Isomers**: 2 (cis and trans)

3. **Third Product** (with double bond in cyclopentyl):
- This product does not exhibit geometric isomerism because the two groups attached to the double bond are the same.
- **Isomers**: 1 (only one form)

### Step 5: Total Count of Alkenes
Now we can sum up the total number of unique alkenes:

- From the first product: 2 isomers
- From the second product: 2 isomers
- From the third product: 1 isomer

**Total number of alkenes = 2 + 2 + 1 = 5**

### Final Answer
The total number of alkenes possible by dehydrobromination of 3-bromo-3-cyclopentylhexane using alcoholic KOH is **5**.

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