`CH_3-undersetunderset(H)(|)oversetoverset(CH_3)(|)C-CH_2Br underset(Delta)overset(Al KOH) to ` Alkene,
Alkene is -

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Answer

To solve the problem of determining the alkene formed from the reaction of 2-bromobutane with alcoholic KOH upon heating, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactant:**
The compound given is 2-bromobutane, which can be represented as:
\[ CH_3-CH(Br)-CH_2-CH_3 \]

2. **Understand the Reaction Conditions:**
The reaction is conducted with alcoholic KOH and involves heating. Alcoholic KOH promotes elimination reactions, specifically dehydrohalogenation.

3. **Dehydrohalogenation Mechanism:**
In this reaction, we will eliminate a hydrogen atom (H) and a bromine atom (Br) from adjacent carbon atoms:
- The bromine (Br) is a leaving group.
- A hydrogen atom from the adjacent carbon will also be removed.

4. **Identify the Atoms to be Eliminated:**
- From the 2-bromobutane, we can eliminate the Br from the second carbon and an H from the adjacent (first or third) carbon.
- For example, we can remove H from the first carbon (CH3) and Br from the second carbon.

5. **Formation of the Alkene:**
After the elimination of H and Br, a double bond will form between the first and second carbon atoms:
\[ CH_3-CH=CH-CH_3 \]
This results in the formation of 1-butene.

6. **Final Structure of the Alkene:**
The alkene formed can be represented as:
\[ CH_2=CH-CH_2-CH_3 \]

7. **Conclusion:**
The alkene produced from the reaction of 2-bromobutane with alcoholic KOH is 1-butene.

### Final Answer:
The alkene formed is 1-butene.