The following compound is obtained when `CH_(3)CH_(2)CHBr_(2)` is boiled with aqueous caustic potash:

Options

A. `CH_(3)-C -= CH`
B. `CH_(3)-CH=CHBr`
C. `CH_(3)CH_(2)CHO`
D. None of these

Explanation

To solve the problem of what compound is obtained when `CH3CH2CHBr2` is boiled with aqueous caustic potash (potassium hydroxide, KOH), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactant**: The compound given is `CH3CH2CHBr2`, which is 1-bromo-2-bromopropane. It has two bromine atoms attached to the second carbon of the propane chain.

2. **Understand the Reagent**: Aqueous caustic potash (KOH) provides hydroxide ions (OH⁻), which act as nucleophiles in the reaction.

3. **Nucleophilic Substitution Reaction**: The hydroxide ion will attack the carbon atom bonded to the bromine atoms. Since we have two bromine atoms, we can expect two nucleophilic substitution reactions to occur.

4. **First Nucleophilic Attack**: The first hydroxide ion will attack one of the bromine-bearing carbon atoms (the second carbon). This will lead to the displacement of one bromine atom, forming an intermediate compound:
\[ CH3CH2C(OH)Br \]

5. **Second Nucleophilic Attack**: The second hydroxide ion will then attack the same carbon atom, leading to the displacement of the second bromine atom. This will yield:
\[ CH3CH2C(OH)H \]
which can be simplified to:
\[ CH3CH2C(OH)H \rightarrow CH3CH2C(OH)H \]

6. **Dehydration to Form an Aldehyde**: The compound `CH3CH2C(OH)H` is not stable as it has two hydroxyl groups on one carbon. Therefore, a dehydration reaction occurs, where water is eliminated, leading to the formation of an aldehyde:
\[ CH3CH2CHO \]
This is propanal, which is an aldehyde.

7. **Final Product**: The final product obtained from the reaction of `CH3CH2CHBr2` with aqueous caustic potash is propanal (CH3CH2CHO).

### Final Answer:
The compound obtained is propanal (CH3CH2CHO).

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