2-chloro-2-methylpentane on reaction with sodium methoxide in methanol yields (A) `C_2H_5CH_2overset(CH_3)overset(|)underset(CH_3)underset(|)C-OCH_3`
(B) `C_2H_5 CH_2underset(CH_3)underset(|)C=CH_2`
(C ) `C_2H_5 CH=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 of what product is formed when 2-chloro-2-methylpentane reacts with sodium methoxide in methanol, we can follow these steps:

### Step 1: Understand the Reactants
We have 2-chloro-2-methylpentane (C5H11Cl) and sodium methoxide (NaOCH3) in methanol (CH3OH). Sodium methoxide acts as a base in this reaction.

**Hint:** Identify the structure of 2-chloro-2-methylpentane. It has a chlorine atom attached to the second carbon of a pentane chain, which is also branched with a methyl group.

### Step 2: Reaction Mechanism
The reaction mechanism involves an elimination reaction (E2 mechanism). Sodium methoxide will abstract a proton (H+) from a β-carbon (adjacent to the carbon with the leaving group Cl) leading to the formation of a double bond.

**Hint:** Remember that in E2 reactions, the base abstracts a proton while the leaving group departs simultaneously.

### Step 3: Identify β-Hydrogens
In 2-chloro-2-methylpentane, the β-carbons are the carbons adjacent to the carbon with the Cl group. We need to identify the β-hydrogens available for elimination. The structure allows for multiple β-hydrogens.

**Hint:** Count the number of β-hydrogens on the adjacent carbons to determine where the elimination can occur.

### Step 4: Forming the Alkene
When the base abstracts a proton from one of the β-carbons, the chlorine leaves, forming a double bond between the α-carbon (the carbon with the Cl) and the β-carbon from which the proton was removed.

**Hint:** Visualize the resulting structure after elimination to see which alkene is formed.

### Step 5: Determine the Stability of the Alkene
The stability of the formed alkene is determined by the number of alkyl groups attached to the double bond. More substituted alkenes are generally more stable.

**Hint:** Use Zaitsev's rule, which states that the more substituted alkene is usually the major product in elimination reactions.

### Step 6: Identify the Correct Product
After analyzing the possible products, we find that the most stable alkene formed from the elimination reaction is 2-methyl-2-pentene. This corresponds to option (C) in the provided choices.

**Hint:** Compare the structures of the products given in the options to confirm which one matches your derived product.

### Final Answer
The product formed when 2-chloro-2-methylpentane reacts with sodium methoxide in methanol is **(C) `C2H5 CH=CH(CH3) CH3`**.