The reaction of t - butyl chloride and sodium ethoxide gives mainly

Options

A. t - butyl ethyl ether
B. 2,2 - dimethylbutane
C. 2 - methylprop -1 - ene
D. isopropyl n- propyl ether

Answer

To solve the question regarding the reaction of t-butyl chloride with sodium ethoxide, we will follow these steps:

### Step 1: Identify the Reactants
The reactants in this reaction are:
- Tertiary butyl chloride (t-butyl chloride), which has the structure:
\[ \text{(CH}_3\text{)}_3\text{CCl} \]
- Sodium ethoxide (NaOC\(_2\)H\(_5\)), where the ethoxide ion (OC\(_2\)H\(_5\)) acts as a strong base.

### Step 2: Understand the Mechanism
Tertiary butyl chloride is a tertiary alkyl halide, which means it has three alkyl groups attached to the carbon that is bonded to the chlorine atom. Due to this steric hindrance, the reaction with sodium ethoxide will favor elimination over substitution.

### Step 3: Elimination Reaction
In the presence of a strong base like sodium ethoxide, the reaction will proceed via an elimination mechanism (E2 mechanism). The ethoxide ion will abstract a proton (H\(^+\)) from a β-carbon (the carbon adjacent to the carbon bearing the chlorine), leading to the formation of a double bond.

### Step 4: Identify the Products
The elimination of HCl will result in the formation of an alkene. The major product of this reaction will be:
\[ \text{2-methylprop-1-ene} \]
The reaction can be summarized as follows:
\[ \text{(CH}_3\text{)}_3\text{CCl} + \text{NaOC}_2\text{H}_5 \rightarrow \text{C}_3\text{H}_6 + \text{NaCl} + \text{C}_2\text{H}_5\text{OH} \]
Where:
- The major product is 2-methylprop-1-ene (C\(_3\)H\(_6\)).
- Byproducts include sodium chloride (NaCl) and ethanol (C\(_2\)H\(_5\)OH).

### Step 5: Conclusion
Thus, the main product formed from the reaction of t-butyl chloride with sodium ethoxide is **2-methylprop-1-ene**.

### Final Answer
The correct option is **2-methylprop-1-ene**.

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