2-Chloro-2-methylpentane on reaction with sodium methoxide in methanol yields :
(a)`C_2H_5CH_2undersetunderset(CH_3)(|)oversetoverset(CH_3)(|)C-OCH_3`
(b)`C_2H_5CH_2undersetunderset(CH_3)(|)C=CH_2`
(c )`C_2H_5CH=undersetunderset(CH_3)(|)C-CH_3`

Options

A. (a) and (c )
B. (c ) only
C. (a) and (b)
D. All of these

Answer

To solve the problem of what products are formed when 2-chloro-2-methylpentane reacts with sodium methoxide in methanol, we will analyze the reaction step by step.

### Step 1: Identify the Reactants
The reactant is 2-chloro-2-methylpentane, which has the following structure:
```
CH3
|
CH3-CH-CH2-CH2-Cl
|
CH3
```
We are reacting this with sodium methoxide (NaOCH3) in methanol (CH3OH).

### Step 2: Determine the Reaction Mechanism
Since 2-chloro-2-methylpentane is a tertiary alkyl halide, the reaction will proceed via the SN1 mechanism. In this mechanism, the first step involves the formation of a carbocation.

### Step 3: Formation of the Carbocation
When the chlorine atom leaves, it forms a stable tertiary carbocation:
```
CH3
|
CH3-C^+ -CH2-CH2
|
CH3
```
This carbocation is stabilized by the surrounding alkyl groups.

### Step 4: Nucleophilic Attack
The methoxide ion (OCH3-) will attack the carbocation:
```
CH3
|
CH3-C-OCH3
|
CH2-CH2
```
This product is formed through the SN1 pathway.

### Step 5: Elimination Reaction
In addition to the substitution reaction, an elimination reaction can also occur. The elimination will lead to the formation of alkenes. The elimination can happen in two ways, leading to two different alkenes:

1. **Major Product (more substituted alkene)**:
When a hydrogen is removed from the adjacent carbon, the more substituted alkene is formed:
```
CH3-C=CH-CH2-CH3
```
This is the major product.

2. **Minor Product (less substituted alkene)**:
When a hydrogen is removed from the other adjacent carbon, the less substituted alkene is formed:
```
CH3-CH=CH-CH2-CH3
```
This is the minor product.

### Step 6: Summary of Products
From the reaction, we can summarize the products:
- **Product A**: The substitution product (from SN1) is `C2H5CH(CH3)OCH3` (which corresponds to the structure of the ether).
- **Product B**: The minor alkene product is `C2H5CH=CH(CH3)`.
- **Product C**: The major alkene product is `C2H5C(CH3)=CH2`.

### Final Answer
The products formed from the reaction of 2-chloro-2-methylpentane with sodium methoxide in methanol are:
- (a) `C2H5CH(CH3)OCH3` (substitution product)
- (b) `C2H5CH=CH(CH3)` (minor alkene)
- (c) `C2H5C(CH3)=CH2` (major alkene)