`CH_(3)CH_(2)CH=CH_(2) overset(NBs)to X overset(alc. KOH)to Y`.
The compound Y is

Options

A. But-2-en-2-ol
B. But-3-en-1-ol
C. 1,3-Butadience
D. 1,2-Butadiene

Explanation

To solve the given problem step by step, we will analyze the reactions involving the compound CH₃CH₂CH=CH₂ (butene) with N-bromosuccinimide (NBS) and alcoholic KOH.

### Step 1: Reaction with NBS
1. **Identify the starting compound**: The starting compound is CH₃CH₂CH=CH₂, which is 1-butene.
2. **Understand the role of NBS**: N-bromosuccinimide (NBS) is used for allylic bromination. It will brominate the hydrogen atom at the allylic position (the carbon adjacent to the double bond).
3. **Identify the allylic position**: In 1-butene, the allylic position is the carbon atom next to the double bond. Therefore, the hydrogen on the second carbon (CH₂) will be replaced by a bromine atom.
4. **Write the product of the reaction**: The product after bromination will be CH₃CHBrCH=CH₂ (3-bromobutene).

### Step 2: Reaction with Alcoholic KOH
1. **Identify the product from Step 1**: The product from the first step is CH₃CHBrCH=CH₂.
2. **Understand the role of alcoholic KOH**: Alcoholic KOH is a strong base that promotes dehydrohalogenation, which means it will remove a hydrogen atom and a halogen atom (bromine in this case) to form a double bond.
3. **Identify the hydrogen and bromine to be removed**: In CH₃CHBrCH=CH₂, we can remove the hydrogen from the carbon adjacent to the bromine (the second carbon) and the bromine itself.
4. **Write the product of the reaction**: After removing H and Br, we will form a new double bond between the second and third carbons, resulting in CH₂=CH-CH=CH₂, which is 1,3-butadiene.

### Final Answer
The compound Y is **1,3-butadiene**.

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