In the given reaction `CH_3-CH_2-overset(Br)overset(|)(CH)-CH=CH_2overset("Alc.KOH//Delta)rarr(X)` X will be

Options

A. `CH_3-CH_2-CH=C=CH_2`
B. `CH_2=CH-CH_2-CH=CH_2`
C. `CH_3-CH=CH-CH=CH_2`
D. `CH_3-CH_2-CH_2-C-=CH`

Answer

To solve the given reaction step-by-step, we start with the reactants and analyze the reaction conditions:

### Step 1: Identify the Reactants
The reactants in the given reaction are:
- **CH₃-CH₂-CHBr-CH=CH₂** (1-bromo-3-butene)

### Step 2: Understand the Reaction Conditions
The reaction is treated with **alcoholic KOH** and heat. This indicates that we are likely dealing with an elimination reaction (dehydrohalogenation) rather than a substitution reaction. Alcoholic KOH is a strong base that can facilitate the elimination of hydrogen bromide (HBr) from the bromoalkane.

### Step 3: Elimination Reaction
In the presence of alcoholic KOH and heat, the bromine atom (Br) will be eliminated along with a hydrogen atom from the adjacent carbon. This results in the formation of a double bond.

### Step 4: Identify the Product
The elimination of HBr from **CH₃-CH₂-CHBr-CH=CH₂** will lead to the formation of a new double bond. The product will be:
- **CH₃-CH=CH-CH=CH₂** (1,3-pentadiene)

### Step 5: Final Product
Thus, the final product X is **CH₃-CH=CH-CH=CH₂**.

### Conclusion
The correct answer is that X will be **1,3-pentadiene**.