When the all-cis isomer of `C_6H_6Cl_6` (1,2,3,4,5,6 -Hexachlorohexane) is heated with alc. KOH, the most probable product is :

Options

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D.

Explanation

To solve the problem of what product is formed when the all-cis isomer of C6H6Cl6 (1,2,3,4,5,6-hexachlorohexane) is heated with alcoholic KOH, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Structure of the Compound**:
- The compound is 1,2,3,4,5,6-hexachlorohexane, which has all chlorine atoms in the same plane (all-cis isomer). This means that all the Cl atoms are either above or below the plane of the carbon chain.

2. **Understand the Reaction Conditions**:
- The reaction involves heating with alcoholic KOH. KOH in alcohol acts as a strong base, which will facilitate elimination reactions.

3. **Mechanism of Reaction**:
- The mechanism involved here is E1cb (Elimination Unimolecular via Conjugate Base). In this mechanism, the base (OH-) abstracts a hydrogen atom from the β-carbon (the carbon adjacent to the carbon bearing the leaving group, Cl in this case).

4. **Identify β-Hydrogens**:
- In the structure of 1,2,3,4,5,6-hexachlorohexane, each carbon has a hydrogen atom that can be abstracted by OH-. Since all Cl atoms are in the same plane, the elimination will occur in a way that allows for the formation of a double bond.

5. **Elimination of Chlorine Atoms**:
- The elimination will proceed with the loss of Cl atoms. As OH- abstracts a hydrogen, a Cl atom will leave, forming a double bond. This process will continue, leading to the formation of a product with multiple double bonds.

6. **Final Product Formation**:
- After the elimination of three Cl atoms, the resulting product will be 1,3,5-trichlorobenzene. The remaining structure will have alternating double bonds due to the elimination of Cl atoms.

### Conclusion:
The most probable product when the all-cis isomer of C6H6Cl6 is heated with alcoholic KOH is **1,3,5-trichlorobenzene**.