`CH_(3)CH_(2)CH_(2)OH underset(170^(@)C)overset(conc." "H_(2)SO_(4))to A underset(500^(@)C)overset(Cl_(2))toB`.
A and B. are

Options

A. `A=CH_(3)CH_(2)CH_(3),B=CH_(3)CH_(2)CH_(2)Cl`
B. `A=CH_(3)CH=CH_(2),B=CH_(2)ClCH=CH_(2)`
C. `A=CH_(2)=CH_(2),B=CH_(3)CH_(2)Cl`
D. `A=CH_(3)CH_(2)CH_(3),B=CH_(3)CH=CH_(2)`

Explanation

To solve the question, we need to identify the compounds A and B formed from the given reaction of propanol with concentrated sulfuric acid and then with chlorine gas.

### Step 1: Identify the starting compound
The starting compound is **propan-1-ol** (CH₃CH₂CH₂OH), which is a three-carbon alcohol.

### Step 2: Reaction with concentrated sulfuric acid
When propan-1-ol (CH₃CH₂CH₂OH) reacts with concentrated sulfuric acid (H₂SO₄) at 170°C, it undergoes dehydration. This means that a water molecule (H₂O) is eliminated from the alcohol.

The dehydration reaction can be represented as follows:
- The hydroxyl group (OH) from the alcohol and a hydrogen atom (H) from the adjacent carbon are removed, resulting in the formation of a double bond between the two carbon atoms.

The reaction can be summarized as:
\[ \text{CH}_3\text{CH}_2\text{CH}_2\text{OH} \xrightarrow{H_2SO_4, 170^\circ C} \text{CH}_3\text{CH}=\text{CH}_2 + \text{H}_2\text{O} \]

Thus, the product A is **propene (CH₃CH=CH₂)**.

### Step 3: Reaction with chlorine gas
Next, compound A (propene) reacts with chlorine gas (Cl₂) at 500°C. This reaction is known as allylic halogenation, where the chlorine atom adds to the allylic position (the carbon adjacent to the double bond).

The reaction can be represented as:
\[ \text{CH}_3\text{CH}=\text{CH}_2 + \text{Cl}_2 \xrightarrow{500^\circ C} \text{CH}_2\text{ClCH}=\text{CH}_2 \]

The product B formed is **allyl chloride (CH₂ClCH=CH₂)**.

### Conclusion
- Compound A is **propene (CH₃CH=CH₂)**.
- Compound B is **allyl chloride (CH₂ClCH=CH₂)**.

### Summary of the answer:
- A = Propene (C₃H₆)
- B = Allyl chloride (C₃H₅Cl)