2,2-dichlorobutane on boiling with aqueous KOH gives

Options

A. butanal
B. 2-butanone
C. 2-butanol
D. butanoic acid

Answer

To solve the problem of what 2,2-dichlorobutane gives when boiled with aqueous KOH, we can follow these steps:

### Step 1: Identify the Reactants
We start with 2,2-dichlorobutane, which has the following structure:

```
Cl
|
CH3-CH-CH2-CH3
|
Cl
```

### Step 2: Understand the Reaction
When 2,2-dichlorobutane is boiled with aqueous KOH, it undergoes alkaline hydrolysis. This means that the chlorine atoms (Cl) will be replaced by hydroxyl groups (OH).

### Step 3: Replace Chlorine with Hydroxyl Groups
The reaction will replace both chlorine atoms with hydroxyl groups, leading to the formation of 2,2-dihydroxybutane:

```
OH
|
CH3-CH-CH2-CH3
|
OH
```

### Step 4: Elimination of Water
However, having two hydroxyl groups on the same carbon (the second carbon in this case) makes it unstable. Therefore, a dehydration reaction occurs where water (H2O) is eliminated from the molecule. This results in the formation of a double bond.

### Step 5: Formation of the Ketone
After the elimination of water, we end up with a double bond between the second and third carbon atoms, resulting in the formation of butan-2-one (a ketone):

```
O
||
CH3-C-CH2-CH3
```

### Step 6: Naming the Product
The final product is butan-2-one, which is a ketone with four carbon atoms. The name reflects the position of the carbonyl group (C=O) on the second carbon of the butane chain.

### Final Answer
When 2,2-dichlorobutane is boiled with aqueous KOH, it gives butan-2-one (a ketone).

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