Which of the following products can be formed when 2 -chloro -2- methylpentane reacts with sodium methoxide in methanol?
1. `C_(2)H_(5)CH_(2)overset(CH_(2))overset("| ")underset(CH_(3))underset("| ")"C "-OCH_(3)`
2. `C_(2)H_(5)CH_(2)underset(CH_(3))underset("| ")"C "=CH_(2)`
3. `C_(2)H_(5)CH=underset(CH_(3))underset("| ")"C "-CH_(3)`

Options

A. 3 only
B. 1 and 2
C. 1 and 2
D. 1, 2 and 3

Answer

To solve the problem of determining which products can be 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:
- It is a tertiary alkyl halide (the carbon attached to the chlorine is bonded to three other carbons).
- The structure can be represented as:
```
CH3
|
CH3- C - CH2 - CH2 - CH3
|
Cl
```

### Step 2: Determine the Reaction Mechanism
Since we have a tertiary alkyl halide, the reaction can proceed via two mechanisms:
1. **Nucleophilic Substitution (SN1)**: The chloride ion (Cl-) leaves, forming a carbocation, followed by the nucleophile (methoxide ion, OMe-) attacking the carbocation.
2. **Elimination (E1)**: The chloride ion leaves, forming a carbocation, followed by the removal of a proton (H+) from a neighboring carbon to form a double bond.

### Step 3: Nucleophilic Substitution (SN1)
1. The Cl- leaves, forming a carbocation:
```
CH3
|
CH3- C+ - CH2 - CH2 - CH3
|
OMe
```
2. The methoxide ion (OMe-) attacks the carbocation:
- This results in the formation of the product:
```
CH3
|
CH3- C - OCH3
|
CH2 - CH2 - CH3
```
- This corresponds to the first product option given in the question.

### Step 4: Elimination (E1)
1. The Cl- leaves, forming the same carbocation as before:
```
CH3
|
CH3- C+ - CH2 - CH2 - CH3
|
```
2. A proton (H+) is removed from one of the adjacent carbons:
- If H is removed from the CH2 next to the carbocation, we get:
```
CH3
|
CH3- C = CH2
```
- If H is removed from the CH2 on the other side, we get:
```
CH2 = C - CH3
```
- Both of these products correspond to the second and third product options given in the question.

### Step 5: Conclusion
Since both the nucleophilic substitution and elimination pathways are possible, all three products can be formed:
1. Product 1: Ether (with OCH3)
2. Product 2: Alkene (C=CH2)
3. Product 3: Alkene (C=CH3)

Thus, the answer is that all three products can be formed.

### Final Answer
All three products can be formed when 2-chloro-2-methylpentane reacts with sodium methoxide in methanol.