Primary alkyl halide `C_4H_9Br` (X) reacts with alc. KOH to give compound (Y). (Y) reacts with HBr to give compound (Z) which is an isomer of (X). When (X) reacts with Na metal it gives compound (P). (X), (Y), (Z) and (P) are

Options

A. `{:(" "X," "Y," "Z," "P),(CH_3-underset(CH_3)underset(|)(CH)-CH_2Br,CH_3-underset(CH_3)underset(|)C=CH_2,CH_3-underset(CH_3)underset(|)overset(Br)overset(|)C-CH_3,C_8P_18):}`
B. `{:(" "X," "Y," "Z," "P),(CH_3CH_2CH_2CH_2Br,CH_3CH=CHCH_3,CH_3-underset(Br)underset(|)(CH)-CH_2CH_3,C_5H_10):}`
C. `{:(" "X," "Y," "Z," "P),(CH_3-underset(Br)underset(|)(CH)-CH_2CH_3,CH_3CH=CHCH_3,CH_3CH_2CH_2CH_2Br,C_7H_14):}`
D. `{:(" "X," "Y," "Z," "P),(CH_3-underset(CH_3)underset(|)(CH)-CH_2Br,CH_3CH=CHCH_3,CH_3-underset(Br)underset(|)(CH)-CH_2CH_3,C_6H_12):}`

Explanation

To solve the question systematically, let's break down the reactions step by step.

### Step 1: Identify Compound X
Given that compound X is a primary alkyl halide with the formula `C4H9Br`, we can deduce that it is likely to be 1-bromobutane (CH3-CH2-CH2-CH2-Br).

### Step 2: Reaction of X with Alcoholic KOH to form Y
When 1-bromobutane (X) reacts with alcoholic KOH, it undergoes an elimination reaction (dehydrohalogenation) to form an alkene. In this case, the reaction will lead to the formation of but-1-ene (Y):
\[ \text{C4H9Br} + \text{alc. KOH} \rightarrow \text{C4H8} + \text{HBr} \]
Thus, compound Y is but-1-ene (CH3-CH2-CH=CH2).

### Step 3: Reaction of Y with HBr to form Z
When but-1-ene (Y) reacts with HBr, it undergoes an electrophilic addition reaction. The HBr adds across the double bond, leading to the formation of 2-bromobutane (Z):
\[ \text{C4H8} + \text{HBr} \rightarrow \text{C4H9Br} \]
Here, Z is 2-bromobutane, which is an isomer of X (1-bromobutane).

### Step 4: Reaction of X with Na Metal to form P
When 1-bromobutane (X) reacts with sodium metal, it undergoes the Wurtz reaction, leading to the formation of an alkane. In this case, two molecules of 1-bromobutane can couple to form octane (P):
\[ 2 \text{C4H9Br} + 2 \text{Na} \rightarrow \text{C8H18} + 2 \text{NaBr} \]
Thus, compound P is octane (C8H18).

### Summary of Compounds
- **X**: 1-bromobutane (C4H9Br)
- **Y**: But-1-ene (C4H8)
- **Z**: 2-bromobutane (C4H9Br)
- **P**: Octane (C8H18)

### Final Answer
X, Y, Z, and P are:
- X: 1-bromobutane (C4H9Br)
- Y: But-1-ene (C4H8)
- Z: 2-bromobutane (C4H9Br)
- P: Octane (C8H18)