`CH_(3)-underset(Cl)underset(|)(CH)-underset(CH_(3))underset(|)(CH)-CH_(3)+"Alcoholic KOH"overset(Delta)(to)`
(A) major + B (minor). A and B are:

Options

A. 3-methyl-but-1-ene, 2-methyl-but-2-ene
B. 2-methyl-but-2-ene, 3-methyl-but-1-ene
C. 2-methyl-but-1-ene, 3-methyl-but-1-ene
D. 2-methyl-but-2-ene, 2-methyl-but-1-ene

Answer

To solve the problem, we need to analyze the given haloalkane compound and determine the products formed when it is treated with alcoholic KOH and heated. The process involves elimination reactions, specifically beta elimination, which leads to the formation of alkenes.

### Step-by-Step Solution:

1. **Identify the Reactant**:
The given compound is a chloroalkane:
\[ CH_3-CH(Cl)-CH(CH_3)-CH_3 \]
This compound has a chlorine atom (Cl) attached to the second carbon (alpha position) and two beta positions (the first and third carbons).

**Hint**: Identify the positions of the halogen and the beta hydrogens for elimination.

2. **Understand the Reaction Conditions**:
The compound is treated with alcoholic KOH, which is a strong base. This will lead to an elimination reaction where a hydrogen atom from a beta carbon and the chlorine atom from the alpha carbon will be removed.

**Hint**: Remember that alcoholic KOH promotes elimination reactions rather than substitution.

3. **Identify Beta Positions**:
The beta positions are the carbons adjacent to the alpha carbon (where Cl is attached). In this case, the beta positions are:
- Beta 1: CH3 (first carbon)
- Beta 2: CH (third carbon)

**Hint**: Label the carbon atoms to clearly identify alpha and beta positions.

4. **Perform the Elimination**:
- **Major Product (A)**: When the hydrogen is removed from the beta 1 position (CH3) and the chlorine is removed from the alpha position, the major product formed is:
\[ CH_3-CH=CH-CH_3 \quad \text{(2-methylbut-2-ene)} \]
This product is more substituted and therefore more stable.

- **Minor Product (B)**: When the hydrogen is removed from the beta 2 position (CH) and the chlorine is removed from the alpha position, the minor product formed is:
\[ CH_2=CH-CH(CH_3)-CH_3 \quad \text{(3-methylbut-1-ene)} \]
This product is less substituted and therefore less stable.

**Hint**: Use the Zaitsev's rule to determine which product is major and which is minor based on substitution.

5. **Final Naming of Products**:
- Major Product (A): 2-methylbut-2-ene
- Minor Product (B): 3-methylbut-1-ene

**Hint**: Make sure to name the products correctly based on IUPAC nomenclature rules.

### Summary of Products:
- Major Product (A): 2-methylbut-2-ene
- Minor Product (B): 3-methylbut-1-ene

### Final Answer:
The correct option is:
- A (Major): 2-methylbut-2-ene
- B (Minor): 3-methylbut-1-ene