2-chlorobutane is heated with alcoholic NaOH, the product formed in larger amount is

Options

A. 1-butene
B. 1-butyne
C. 2-butene
D. All of these

Explanation

To solve the problem of what product is formed in larger amounts when 2-chlorobutane is heated with alcoholic NaOH, we can follow these steps:

### Step 1: Identify the structure of 2-chlorobutane
2-chlorobutane is a four-carbon alkane with a chlorine atom attached to the second carbon. The structural formula can be represented as:
\[ \text{CH}_3 - \text{CH}(\text{Cl}) - \text{CH}_2 - \text{CH}_3 \]

### Step 2: Understand the reaction with alcoholic NaOH
When 2-chlorobutane is heated with alcoholic NaOH, a dehydrohalogenation reaction occurs. In this reaction, the alcoholic NaOH will help to eliminate the chlorine atom (Cl) and a hydrogen atom (H) from adjacent carbon atoms, leading to the formation of a double bond.

### Step 3: Elimination of Cl and H
The chlorine atom on the second carbon will be eliminated along with a hydrogen atom from the adjacent carbon (the third carbon). This results in the formation of a double bond between the second and third carbon atoms.

### Step 4: Formation of the product
After the elimination, the structure of the product will be:
\[ \text{CH}_3 - \text{CH} = \text{CH} - \text{CH}_3 \]
This compound is known as 2-butene.

### Step 5: Determine the major product
Since 2-butene can exist in two isomeric forms (cis and trans), the trans form is generally more stable and will be the major product formed in larger amounts.

### Conclusion
The product formed in larger amounts when 2-chlorobutane is heated with alcoholic NaOH is **2-butene**.