In the following sequence of reactions, B is
`C_(6)H_(6) underset(FeCl_(3))overset(Cl_(2))to A overset(Na)to B`

Options

A. chlorobenzene
B. benzyl chloride
C. diphenyl
D. chlorophenylmethane

Explanation

To solve the problem, we need to analyze the given sequence of reactions step by step.

### Step 1: Identify the starting material
The starting material is benzene, which has the molecular formula \( C_6H_6 \).

### Step 2: First reaction with \( Cl_2 \) and \( FeCl_3 \)
When benzene reacts with chlorine gas (\( Cl_2 \)) in the presence of a Lewis acid catalyst like \( FeCl_3 \), a substitution reaction occurs. This reaction leads to the formation of chlorobenzene (compound A). The reaction can be summarized as follows:

\[ C_6H_6 + Cl_2 \xrightarrow{FeCl_3} C_6H_5Cl + HCl \]

Here, one hydrogen atom from benzene is replaced by a chlorine atom, resulting in chlorobenzene (\( C_6H_5Cl \)) and hydrochloric acid (\( HCl \)) as a byproduct.

### Step 3: Second reaction with sodium in dry ether
The next step involves the reaction of chlorobenzene (A) with sodium (\( Na \)) in dry ether. This reaction is a Wurtz coupling reaction, where two molecules of chlorobenzene react with sodium to form biphenyl (compound B) and sodium chloride (\( NaCl \)). The reaction can be summarized as follows:

\[ 2 C_6H_5Cl + 2 Na \xrightarrow{dry\ ether} C_6H_5-C_6H_5 + 2 NaCl \]

In this reaction, two chlorobenzene molecules couple together, leading to the formation of biphenyl (\( C_{12}H_{10} \)).

### Conclusion
Thus, the final product B formed from the sequence of reactions is biphenyl.

\[ \text{B is } C_{12}H_{10} \text{ (biphenyl).} \]

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