1-chlorobutane reacts with alcoholic `KOH` to from

Options

A. 1-butene
B. 1-butanol
C. 2-butene
D. 2-butanol

Answer

To solve the problem of what product is formed when 1-chlorobutane reacts with alcoholic KOH, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactant**:
The reactant is 1-chlorobutane, which has the molecular structure CH3-CH2-CH2-CH2Cl. This means it has a chlorine atom attached to the first carbon of the butane chain.

2. **Understand the Reaction Conditions**:
The reaction is carried out in the presence of alcoholic KOH. Alcoholic KOH is a strong base and acts as a dehydrohalogenating agent. This means it will facilitate the elimination of a hydrogen atom and a halogen atom (in this case, chlorine) from the alkyl halide.

3. **Identify the Mechanism**:
The reaction follows an elimination mechanism (specifically E2 mechanism), where a hydrogen atom from a beta carbon (the carbon adjacent to the carbon with the halogen) is removed along with the halogen atom.

4. **Locate the Beta Hydrogen**:
In 1-chlorobutane, the beta carbon is the second carbon in the chain (CH2). This carbon has two hydrogen atoms that can be eliminated.

5. **Perform the Elimination**:
When KOH removes the chlorine atom and one of the hydrogen atoms from the beta carbon, a double bond is formed between the first and second carbon atoms, resulting in the formation of but-1-ene (1-butene).

6. **Write the Reaction**:
The overall reaction can be represented as:
\[ \text{CH}_3\text{CH}_2\text{CH}_2\text{CH}_2\text{Cl} + \text{KOH} \rightarrow \text{CH}_3\text{CH}_2\text{CH}=\text{CH}_2 + \text{KCl} + \text{H}_2\text{O} \]
Here, but-1-ene is formed along with potassium chloride (KCl) and water (H2O) as byproducts.

7. **Conclude the Product**:
Therefore, the product formed from the reaction of 1-chlorobutane with alcoholic KOH is but-1-ene.

### Final Answer:
The correct option is **1-butene**.

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