Two isomeric halo alkenes (A) and (B) have molecular formula `C_(5)H_(9)Cl`. (A) gives optically inactive compound whil (B) give optically active compound in hydrogenation. The two iosmers respectively are:

Options

A. 3-chloro-1-pentane and 4-chloro-2-pentene
B. 4-chloro-1-pentene and 2-chloro-2-pentene
C. 3-chloro-2-pentene and 1-chloro-2-pentene
D. 1-chloro-1-pentene and 5-chloro-1-pentene

Answer

To solve the problem of identifying the two isomeric halo alkenes (A) and (B) with the molecular formula C₅H₉Cl, we will follow these steps:

### Step 1: Understanding the Molecular Formula
The molecular formula C₅H₉Cl indicates that we have a compound with five carbon atoms, nine hydrogen atoms, and one chlorine atom. This suggests that we are dealing with alkenes (due to the presence of a double bond) and halides (due to the chlorine atom).

**Hint:** Look for structures that fit the molecular formula while considering the presence of a double bond and the chlorine atom.

### Step 2: Drawing Possible Isomers
We can draw different structural isomers of C₅H₉Cl. The two isomers we are interested in must differ in the position of the double bond and the chlorine atom.

1. **Isomer A:** Let's consider 3-chloro-1-pentene. This structure has the chlorine atom on the third carbon and a double bond between the first and second carbons.
2. **Isomer B:** Now consider 4-chloro-2-pentene. In this case, the double bond is between the second and third carbons, and the chlorine is on the fourth carbon.

**Hint:** Identify the positions of the chlorine and the double bond carefully to ensure they are different isomers.

### Step 3: Analyzing Optical Activity
Next, we need to determine the optical activity of the products formed upon hydrogenation of these isomers.

- **For Isomer A (3-chloro-1-pentene):**
When hydrogenated, it forms 3-chloro-pentane. This compound does not have any chiral centers (no carbon atom bonded to four different groups), making it optically inactive.

- **For Isomer B (4-chloro-2-pentene):**
Upon hydrogenation, this compound forms 4-chloro-pentane. This compound contains a chiral center (the fourth carbon has four different substituents: a chlorine, a hydrogen, a methyl group, and a propyl group), making it optically active.

**Hint:** Check for the presence of chiral centers in the products formed after hydrogenation to determine optical activity.

### Step 4: Conclusion
Based on the analysis:
- Isomer A is **3-chloro-1-pentene**, which gives an optically inactive compound upon hydrogenation.
- Isomer B is **4-chloro-2-pentene**, which gives an optically active compound upon hydrogenation.

Thus, the two isomers respectively are:
- (A) 3-chloro-1-pentene
- (B) 4-chloro-2-pentene

**Final Answer:** The two isomers are 3-chloro-1-pentene (A) and 4-chloro-2-pentene (B).