2-chloro-2-methylpentane on reaction with sodium methoxide in methanol yields:
(a) `C_(2)H_(5)CH_(2)overset(CH_(3))overset(|)underset(CH_(3))underset(|)(C)-OCH_(3)` (b) `C_(2)H_(5)CH_(2)underset(CH_(3))underset(|)(C)=CH_(2)` (c) `C_(2)H_(5)CH_(2)=underset(CH_(3))underset(|)(C)-CH_(3)`

Options

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

Answer

To solve the problem, we need to analyze the reaction of 2-chloro-2-methylpentane with sodium methoxide in methanol.

### Step-by-Step Solution:

1. **Identify the Reactant**:
- The reactant is 2-chloro-2-methylpentane. Its structure can be represented as:
```
CH3
|
CH3- C - CH2 - CH3
|
Cl
```

2. **Understand the Reaction Conditions**:
- Sodium methoxide (NaOCH3) is a strong nucleophile and is used in a polar protic solvent (methanol). This combination typically favors elimination reactions over substitution reactions.

3. **Determine the Possible Mechanisms**:
- The reaction can proceed via two mechanisms:
- **SN1 Mechanism**: This is likely due to the tertiary carbon (the carbon with the chlorine attached) which can stabilize a carbocation.
- **E2 Mechanism**: This mechanism can also occur due to the strong base (sodium methoxide) leading to elimination.

4. **Substitution Product**:
- If the reaction proceeds via SN1, we can expect a substitution product. The product would be:
```
CH3
|
CH3- C - CH2 - OCH3
|
NaCl
```
- This corresponds to one of the possible products.

5. **Elimination Products**:
- For elimination, we can have two possible products:
- **Path A**: The formation of an alkene by removing a hydrogen from the adjacent carbon:
```
CH3
|
CH3- C = CH2
|
CH2
```
- **Path B**: A more substituted alkene:
```
CH3
|
CH2 = C - CH3
|
CH3
```

6. **Analyze the Products**:
- The products formed from the elimination reactions are:
- Product A: `C2H5CH2=C(CH3)CH3`
- Product B: `C2H5CH2=C(CH3)CH2`
- The substitution product is less favored but still possible.

7. **Conclusion**:
- All three products (substitution and elimination) can be formed, but the elimination products will dominate due to the reaction conditions favoring elimination.

### Final Answer:
The answer is that all products (A, B, and C) are possible.

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