`CH_(3)CH_(2)CHCl_(2) overset(Aq.KOH)rarr`?

Answer

To solve the reaction of geminal dihalide \( CH_3CH_2CHCl_2 \) with aqueous KOH, we will follow these steps:

### Step 1: Identify the Reactants
We have a geminal dihalide, \( CH_3CH_2CHCl_2 \), which contains two chlorine atoms attached to the same carbon atom (the second carbon in this case).

### Step 2: Understand the Reaction Conditions
The reaction is performed in the presence of aqueous KOH. KOH dissociates in water to give \( K^+ \) and \( OH^- \). The hydroxide ion (\( OH^- \)) will act as a nucleophile.

### Step 3: First Substitution Reaction
In the first step, one of the chlorine atoms will be substituted by the hydroxide ion. The reaction can be represented as follows:

\[ CH_3CH_2CCl_2 + KOH \rightarrow CH_3CH_2CCl(OH) + KCl \]

Here, one \( Cl \) is replaced by \( OH \), resulting in the formation of a mono-hydroxy compound.

### Step 4: Second Substitution Reaction
Next, the remaining chlorine atom on the same carbon will also be substituted by another hydroxide ion. This can be represented as:

\[ CH_3CH_2CCl(OH) + KOH \rightarrow CH_3CH_2C(OH)(OH) + KCl \]

Now, we have a compound with two hydroxyl groups attached to the same carbon atom.

### Step 5: Elimination of Water
The compound \( CH_3CH_2C(OH)(OH) \) is unstable due to the presence of two hydroxyl groups on the same carbon. Therefore, it undergoes dehydration (loss of water) to form a carbonyl compound (an aldehyde):

\[ CH_3CH_2C(OH)(OH) \rightarrow CH_3CH_2C=O + H_2O \]

This results in the formation of the final product, which is an aldehyde:

\[ CH_3CH_2C=O \quad \text{(butanal)} \]

### Final Product
The final product of the reaction is \( CH_3CH_2C=O \) (butanal).

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