The reaction is `:`
`CH_(3)CHBr-CH_(2)Br+2KOH("alc.")overset(Delta)rarr CH_(3)-C equiv CH+2KBr+2H_(2)O`

Options

A. Dehalogenation
B. Dehydrohalogenation
C. Decarboxylation
D. Dehydration

Answer

To solve the problem, we need to analyze the given reaction step by step and identify the type of reaction taking place.

### Step 1: Identify the Reactants
The reactants in the given reaction are:
- CH₃CHBrCH₂Br (a dihalogenated alkane)
- 2 KOH (potassium hydroxide in alcoholic solution)

### Step 2: Understand the Reaction Conditions
The reaction is carried out in the presence of alcoholic KOH and heat (Δ), which indicates that it is likely to favor elimination reactions rather than substitution.

### Step 3: Analyze the Products
The products of the reaction are:
- CH₃C≡CH (an alkyne)
- 2 KBr (potassium bromide)
- 2 H₂O (water)

### Step 4: Determine the Type of Reaction
In the reaction:
- Both bromine atoms (halogens) are removed from the starting material.
- Additionally, hydrogen atoms are also removed.

This indicates that the reaction involves the elimination of both halogen atoms and hydrogen atoms, leading to the formation of a triple bond (alkyne).

### Step 5: Classify the Reaction
The removal of halogens and hydrogen suggests that this reaction is a **dehydrohalogenation** reaction.

### Conclusion
Thus, the type of reaction taking place in the given equation is **dehydrohalogenation**.

---