Heating 1,2 dibromocyclohexane with 2 equivalents of alcoholic KOH yields

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Answer

To solve the question of what product is formed when heating 1,2-dibromocyclohexane with 2 equivalents of alcoholic KOH, we can follow these steps:

### Step 1: Identify the Reactants
We start with 1,2-dibromocyclohexane, which has two bromine atoms attached to the first and second carbon of the cyclohexane ring.

### Step 2: Understand the Reaction Conditions
We are using 2 equivalents of alcoholic KOH. Alcoholic KOH is a strong base and promotes elimination reactions rather than substitution reactions.

### Step 3: Determine the Type of Reaction
Since we are using alcoholic KOH, we can expect an elimination reaction (specifically, an E2 mechanism) to occur. In this reaction, we will remove a bromine atom and a hydrogen atom to form a double bond.

### Step 4: Elimination of Bromine and Hydrogen
1. The first equivalent of KOH will remove one bromine atom (Br) and a hydrogen atom (H) from adjacent carbons (C1 and C2).
2. This will form a double bond between C1 and C2, resulting in the formation of a cyclohexene derivative.

### Step 5: Consider the Stability of the Product
Since we have two equivalents of KOH, we can perform another elimination. The second equivalent of KOH can remove another bromine atom (if present) and a hydrogen atom from the adjacent carbon to create a more stable alkene.

In this case, the major product will be a conjugated diene due to the stability of the double bonds formed. The most stable product will be 1,3-hexadiene, as it allows for conjugation.

### Final Product
The final product of heating 1,2-dibromocyclohexane with 2 equivalents of alcoholic KOH will be **1,3-hexadiene**.

### Summary of Steps:
1. Identify the reactants (1,2-dibromocyclohexane).
2. Recognize the reaction conditions (alcoholic KOH).
3. Determine the type of reaction (elimination).
4. Perform the elimination to form a double bond.
5. Consider the stability of the product to find the major product (1,3-hexadiene).