`Ph-CH_(2)-CH_(2)-C-=C-CH_(3)overset(Na//NH_(3))to A overset(Cl_(2)//hv)to B`
Identify B

Options

A.
B.
C. `Ph -CH_(2)-CH(Cl)-CH=CH-CH_(3)`
D. None of these

Explanation

To solve the problem step by step, we will analyze the reaction sequence provided in the question:

### Step 1: Identify the starting compound
The starting compound is given as `Ph-CH2-CH2-C≡C-CH3`. This structure can be represented as:
- A phenyl group (Ph) attached to a chain that includes a propyl group and a terminal alkyne (C≡C).

### Step 2: Perform the Birch reduction
The first reaction involves sodium (Na) in liquid ammonia (NH3). This is known as the Birch reduction, which typically reduces alkynes and alkenes to form cis-alkenes.

In this case, the terminal alkyne (C≡C) will be reduced to a cis-alkene. The structure after Birch reduction (let's denote it as A) can be represented as:
- `Ph-CH2-CH2-CH=CH-CH3` (where the double bond is in the cis configuration).

### Step 3: Identify the product A
From the Birch reduction, we have:
- A = `Ph-CH2-CH2-CH=CH-CH3` (cis isomer).

### Step 4: Perform chlorination
Next, the compound A undergoes chlorination in the presence of light (Cl2/hv). This reaction typically involves free radical substitution, where chlorine radicals will abstract hydrogen atoms from the alkene.

In this case, since we started with a cis-alkene, the chlorination will lead to the formation of a trans product. The product after chlorination (let's denote it as B) can be represented as:
- `Ph-CH2-CH2-CH=CH-CH2Cl` (where the double bond is now in the trans configuration).

### Step 5: Identify the product B
Thus, the final product B after chlorination is:
- B = `Ph-CH2-CH2-CH=CH-CH2Cl` (trans isomer).

### Conclusion
The final answer is that compound B is the trans isomer of the chlorinated product derived from the original compound.