2-Chloro-2-methylpropane on reaction with aqueous KOH gives X as the major product. X is

Options

A. but-2-ene
B. 2-methylbut-1-ene
C. 2-methylprop-1-ene
D. 2-methylbutan-2-ol

Explanation

To solve the question regarding the reaction of 2-chloro-2-methylpropane with aqueous KOH, we will follow these steps:

### Step 1: Identify the structure of 2-chloro-2-methylpropane
2-chloro-2-methylpropane is a tertiary haloalkane. Its structure can be represented as follows:

```
CH3
|
CH3-C-Cl
|
CH3
```

### Step 2: Understand the reaction conditions
When 2-chloro-2-methylpropane reacts with aqueous KOH, it undergoes an elimination reaction. Given that it is a tertiary halide, the reaction will favor an E2 elimination mechanism.

### Step 3: Identify the elimination mechanism
In the E2 mechanism, a base (in this case, KOH) abstracts a beta-hydrogen while the leaving group (Cl) departs simultaneously. The beta-hydrogens are those attached to the carbon atoms adjacent to the carbon bearing the chlorine atom.

### Step 4: Determine the beta-hydrogens
In the structure of 2-chloro-2-methylpropane, there are three beta-hydrogens available for elimination:

- The beta carbons are the ones adjacent to the carbon with the chlorine atom.

### Step 5: Perform the elimination
When a beta-hydrogen is removed, the electrons from the C-H bond will form a double bond between the alpha carbon (the one with the chlorine) and the beta carbon from which the hydrogen was removed. The chlorine will leave as Cl⁻.

### Step 6: Write the product
The major product formed from this elimination will be 2-methylpropene (or 2-methylprop-1-ene), which can be represented as:

```
CH3
|
CH3-C=CH2
```

### Step 7: Name the product
The product is named 2-methylprop-1-ene.

### Conclusion
Thus, the major product X formed from the reaction of 2-chloro-2-methylpropane with aqueous KOH is **2-methylprop-1-ene**.

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