`CH_(3)-underset(Cl)underset(|)(C)H-underset(Cl)underset(|)(C)H-CH_(3)underset((ii)NaNH2)overset((i)Alc.KOH)(to)A` (major) `underset(Pd//CaCO_(3))overset(H_(2))(to)B`

Options

A. `CH_(3)CH_(2)-CH=CH_(2)`
B. `CH_(2)-CH_(2)-CH_(2)-CH_(3)`
C.
D.

Answer

To solve the given question, we need to follow the steps of the reactions described. Let's break it down step by step.

### Step 1: Identify the Initial Compound
The initial compound is given as:
\[ \text{CH}_3\text{CCl}(\text{CHCl})\text{CH}_3 \]

This compound has two chlorine atoms attached to the carbon chain.

### Step 2: Reaction with Alcoholic KOH
The first step involves the reaction of the compound with alcoholic KOH. Since alcoholic KOH is used, it promotes an elimination reaction (dehydrohalogenation).

1. **Elimination Reaction**:
- One chlorine atom (Cl) will be removed from the alpha carbon, and a hydrogen atom will be removed from the beta carbon.
- This results in the formation of a double bond between the two carbons.

The product after this step (let's call it A) will be:
\[ \text{CH}_3\text{C}=\text{C}\text{H}\text{CH}_3 \]
This represents a compound with a double bond between the two carbons.

### Step 3: Reaction with NaNH2
Next, the product A undergoes a reaction with sodium amide (NaNH2), which acts as a strong base.

2. **Formation of Alkyne**:
- NaNH2 will deprotonate the hydrogen adjacent to the double bond, leading to the formation of a triple bond.
- The product after this reaction (let's call it B) will be:
\[ \text{CH}_3\text{C} \equiv \text{C}\text{CH}_3 \]
This is a symmetrical alkyne.

### Step 4: Hydrogenation with Pd/CaCO3
The final step involves the hydrogenation of the alkyne using hydrogen gas (H2) in the presence of palladium on calcium carbonate (Pd/CaCO3), which acts as a Lindlar catalyst.

3. **Reduction of Alkyne to Alkene**:
- The alkyne will be partially hydrogenated to form a cis-alkene.
- The product after this step (let's call it C) will be:
\[ \text{C} \text{H}_3\text{C}=\text{C}\text{H}\text{CH}_3 \]
This is a cis-alkene.

### Conclusion
Thus, the final products A and B are:
- A: \[ \text{CH}_3\text{C} \equiv \text{C}\text{CH}_3 \] (an alkyne)
- B: \[ \text{C} \text{H}_3\text{C}=\text{C}\text{H}\text{CH}_3 \] (a cis-alkene)

The correct answer to the question is option 3.