Predict the product of the reaction `CH_3-underset(H)underset(|)overset(CH_3)overset(|)C- underset(Cl)underset(|)overset(H)overset|C-CH_2-CH=CH_2overset("alc. KOH)rarr`

Options

A. `(CH_3)_2C=CH-CH_2-CH=CH_2`
B. `(CH_3)_2CH-CH=CH-CH=CH_2`
C. `(CH_3)_2CH-CH=underset(CH_3)underset(|)C-CH=CH_2`
D. `(CH_3)_2C=CH-underset(CH_3)underset(|)C=CH_2`

Answer

To predict the product of the reaction of the compound with alcoholic KOH, we will follow these steps:

### Step 1: Identify the Starting Material
The starting material is a compound with the structure:
- CH₃-CH(CH₃)-C(Cl)-C(CH₂=CH₂)

This compound has a chlorine atom (Cl) attached to a carbon that is also connected to a carbon chain.

### Step 2: Understand the Reaction Conditions
The reaction is conducted in the presence of alcoholic KOH. Alcoholic KOH acts as a strong base and can facilitate elimination reactions (dehydrohalogenation) by removing a hydrogen atom and a halogen atom (in this case, Cl).

### Step 3: Identify the Acidic Hydrogen
In the presence of alcoholic KOH, the base will abstract an acidic hydrogen from the β-carbon (the carbon adjacent to the carbon with the Cl). In our structure, we have several hydrogens on the β-carbons.

### Step 4: Possible Elimination Pathways
There are two possible pathways for elimination:
1. **Pathway A**: Remove a hydrogen from the carbon adjacent to the carbon with Cl, leading to the formation of a double bond.
2. **Pathway B**: Remove a hydrogen from the other β-carbon.

### Step 5: Determine the Stability of the Resulting Alkenes
After the elimination, we will form alkenes:
- **From Pathway A**: This will yield a double bond between the two carbons adjacent to the carbon that originally had Cl.
- **From Pathway B**: This will yield a different alkene.

To predict the major product, we need to consider the stability of the resulting alkenes. Generally, more substituted alkenes are more stable due to hyperconjugation and the inductive effect.

### Step 6: Predict the Major Product
After analyzing the possible products from both pathways, we find that the product from Pathway A is more stable due to higher substitution. The resulting alkene will have the structure:
- CH₃-CH(CH₃)-C=CH₂

### Final Product
The final product of the reaction will be:
- **CH₃-CH(CH₃)-C=CH₂**