`CH_(3)-overset(Cl)overset(|)CH-CH_(2)-CH_(3) overset("Pottasium tert. butoxide")to underset("Major")A` is

Options

A. `CH_(3)-CH=CH-CH_(3)`
B. `CH_(2)=CH-CH_(2)-CH_(3)`
C. `CH_(3)-overset(CH_(3))overset(|)C=CH_(2)`
D. Both (2) and (3)

Answer

To solve the problem, we need to analyze the reaction of the compound \( CH_3-CH(Cl)-CH_2-CH_3 \) with potassium tert-butoxide (KOC(CH_3)_3), which is a strong and hindered base. The goal is to determine the major product of this elimination reaction.

### Step-by-Step Solution:

1. **Identify the Structure:**
The compound \( CH_3-CH(Cl)-CH_2-CH_3 \) consists of a 4-carbon chain with a chlorine atom (Cl) attached to the second carbon. This makes it a secondary alkyl halide.

2. **Understand the Reagent:**
Potassium tert-butoxide (KOC(CH_3)_3) is a strong base and a hindered base. Hindered bases favor elimination reactions (E2 mechanism) over substitution reactions because they are less likely to approach the carbon atom directly bonded to the leaving group (Cl in this case).

3. **Mechanism of Reaction:**
In an E2 elimination, the base abstracts a proton (H) from a carbon adjacent to the carbon with the leaving group (Cl). The electrons from the C-H bond then form a double bond with the carbon that was attached to the leaving group, while the leaving group (Cl) departs.

4. **Identify Possible Protons to Remove:**
The compound has two adjacent carbons to the carbon with Cl:
- The carbon on the left (C1) has three hydrogens (H).
- The carbon on the right (C3) has two hydrogens (H).

5. **Determine the Major Pathway:**
Since potassium tert-butoxide is a hindered base, it will preferentially abstract the proton from the less hindered carbon (C3) rather than the more hindered carbon (C1).

6. **Elimination Reaction:**
- The base will abstract a hydrogen from C3, leading to the formation of a double bond between C2 and C3.
- The chlorine atom will leave, resulting in the formation of a double bond between C2 and C3.

7. **Final Product:**
The product of this elimination reaction will be \( CH_2=CH-CH_2-CH_3 \), which is butene. Given that the elimination occurs from the secondary carbon, the major product will be trans-2-butene or cis-2-butene, depending on the stereochemistry, but both are considered the same product in this context.

### Conclusion:
The major product of the reaction is butene, specifically \( CH_2=CH-CH_2-CH_3 \).