A primary alcohol, `C_(3)H_(8)O(W)` on heating with sulphuric acid undergoes dehydration to give an alkene , (X), which reacts with HCl to give (Y). (Y) on treatment with aqueous KOH gives compound (Z),`C_(3)H_(8)O`. Compounds (W) and (Z) are `:`

Options

A. functional isomers
B. position isomers
C. chain isomers
D. stereoisomers .

Explanation

To solve the problem step by step, we will identify the compounds involved and their transformations.

### Step 1: Identify the Primary Alcohol (W)
The primary alcohol with the formula \( C_3H_8O \) can be identified as **1-propanol** (CH3CH2CH2OH).

### Step 2: Dehydration Reaction
When 1-propanol is heated with sulfuric acid (H2SO4), it undergoes dehydration (loss of water) to form an alkene. The reaction can be represented as follows:

\[ \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} \xrightarrow{\text{H}_2\text{SO}_4, \text{heat}} \text{CH}_3\text{CH}=\text{CH}_2 + \text{H}_2\text{O} \]

The product formed here is **propene** (CH3CH=CH2), which we will denote as (X).

### Step 3: Reaction with HCl
The alkene (X), propene, reacts with hydrochloric acid (HCl) to form an alkyl halide. The reaction is as follows:

\[ \text{CH}_3\text{CH}=\text{CH}_2 + \text{HCl} \rightarrow \text{CH}_3\text{CHClCH}_3 \]

The product formed is **2-chloropropane** (CH3CHClCH3), which we will denote as (Y).

### Step 4: Reaction with Aqueous KOH
The alkyl halide (Y), 2-chloropropane, reacts with aqueous potassium hydroxide (KOH) to undergo hydrolysis, yielding an alcohol. The reaction can be represented as:

\[ \text{CH}_3\text{CHClCH}_3 + \text{KOH (aq)} \rightarrow \text{CH}_3\text{CHOHCH}_3 + \text{KCl} \]

The product formed here is **2-propanol** (CH3CHOHCH3), which we will denote as (Z).

### Step 5: Identify Compounds W and Z
- Compound (W) is **1-propanol** (C3H8O).
- Compound (Z) is **2-propanol** (C3H8O).

### Final Answer
The compounds (W) and (Z) are:
- (W) = 1-Propanol (C3H8O)
- (Z) = 2-Propanol (C3H8O)