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. Decarboxytation
D. Dehydration

Answer

To solve the given reaction, we need to analyze the transformation of the reactant into the product and identify the type of reaction taking place. Here’s a step-by-step breakdown:

### Step 1: Identify the Reactant and Product
The reactant is `CH₃CHBrCH₂Br`, which has two bromine atoms attached to adjacent carbon atoms. The product is `CH₃C≡CH`, which is an alkyne with a triple bond between the second and third carbon atoms.

### Step 2: Understand the Role of KOH (alc.)
The presence of alcoholic KOH indicates that this is a dehydrohalogenation reaction. Alcoholic KOH is a strong base that facilitates the elimination of halogen atoms and hydrogen atoms from the carbon chain.

### Step 3: Mechanism of the Reaction
1. **First Elimination**: The first step involves the removal of one bromine atom (Br) from the carbon chain and a hydrogen atom (H) from the adjacent carbon. This results in the formation of a double bond (alkene).
- From `CH₃CHBrCH₂Br`, remove one `Br` and one `H`, leading to the formation of `CH₃C=CHBr`.

2. **Second Elimination**: In the second step, the remaining bromine atom (Br) is eliminated along with another hydrogen atom (H) from the adjacent carbon, resulting in the formation of a triple bond (alkyne).
- From `CH₃C=CHBr`, remove the second `Br` and `H`, yielding `CH₃C≡CH`.

### Step 4: Final Products
After both eliminations, the final products are:
- `CH₃C≡CH` (the alkyne)
- `2KBr` (from the two bromine atoms)
- `2H₂O` (from the hydroxide ions reacting with the eliminated hydrogen atoms)

### Conclusion
The overall reaction can be summarized as:
\[ CH₃CHBrCH₂Br + 2KOH \text{(alc.)} \xrightarrow{\Delta} CH₃C≡CH + 2KBr + 2H₂O \]

### Type of Reaction
This reaction is classified as **dehydrohalogenation** because it involves the removal of hydrogen and halogen atoms from the reactant.

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