What would be the produt when neopentyl chloride reacts with sodium ethoxide

Options

A. 2-Methyl-2butanol1
B. Neopentylalcoho1
C. both a and b
D. 2-Methyl-2-butene

Explanation

To determine the product formed when neopentyl chloride reacts with sodium ethoxide, we can follow these steps:

### Step 1: Identify the Reactants
Neopentyl chloride (C5H11Cl) has the structure where one carbon atom is bonded to four methyl groups (C(CH3)4CH2Cl). Sodium ethoxide (C2H5O-) is a strong base.

### Step 2: Understand the Reaction Mechanism
The reaction will proceed via an E2 elimination mechanism due to the presence of a strong base (sodium ethoxide). In this mechanism, the leaving group (Cl) will be eliminated along with a hydrogen atom from a β-carbon, leading to the formation of a double bond.

### Step 3: Formation of Carbocation
When neopentyl chloride reacts with sodium ethoxide, the chloride ion (Cl-) will leave, forming a carbocation. The structure of the carbocation will be a primary carbocation at the end of the neopentyl chain.

### Step 4: Carbocation Rearrangement
Carbocations can rearrange to form more stable structures. In this case, a methyl group can shift to the carbocation, resulting in a more stable tertiary carbocation.

### Step 5: Elimination to Form the Alkene
Once the more stable carbocation is formed, elimination will occur. A hydrogen atom from a β-carbon will be removed, and a double bond will be formed between the α-carbon (the one that was attached to the leaving group) and the β-carbon.

### Step 6: Identify the Product
The final product will be a more substituted alkene due to the elimination process. The resulting compound will be 2-methylbut-2-ene.

### Final Product
The product formed from the reaction of neopentyl chloride with sodium ethoxide is **2-methylbut-2-ene**.

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