Z' in the following sequence of reactions is
`C_(6)H_(6)overset(HNO_(3)//H_(2)SO_(4))underset(Delta)rarr W overset(Zn//HCl)rarr X overset(NaNO_(2))underset(HCl)rarr Y overset(H_(2)O//H_(3)PO_(2))rarr Z`

Options

A.
B.
C.
D.

Explanation

To find the compound Z in the given sequence of reactions, we will analyze each step carefully.

### Step-by-step Solution:

1. **Starting Compound**: The reaction begins with benzene, which is represented by the formula \( C_6H_6 \).

2. **Nitration of Benzene**: The first step involves the nitration of benzene using a mixture of nitric acid (\( HNO_3 \)) and sulfuric acid (\( H_2SO_4 \)). This reaction generates a nitronium ion (\( NO_2^+ \)), which acts as an electrophile. Benzene, being a nucleophile due to its pi electrons, undergoes electrophilic substitution to form nitrobenzene (\( C_6H_5NO_2 \)).
- **Product W**: Nitrobenzene (\( C_6H_5NO_2 \))

3. **Reduction of Nitrobenzene**: The next step involves the reduction of nitrobenzene using zinc and hydrochloric acid (\( Zn/HCl \)). This reaction reduces the nitro group (\( -NO_2 \)) to an amino group (\( -NH_2 \)), resulting in the formation of aniline.
- **Product X**: Aniline (\( C_6H_5NH_2 \))

4. **Diazotization of Aniline**: The aniline is then treated with sodium nitrite (\( NaNO_2 \)) in the presence of hydrochloric acid (\( HCl \)) at low temperatures (0 to 5 degrees Celsius). This process is known as diazotization, where the amino group is converted into a diazonium salt (\( C_6H_5N_2^+Cl^- \)).
- **Product Y**: Diazonium salt (\( C_6H_5N_2^+Cl^- \))

5. **Reduction of Diazonium Salt**: Finally, the diazonium salt is treated with water and phosphorous acid (\( H_3PO_2 \)). This reaction reduces the diazonium group, effectively removing it and regenerating benzene.
- **Product Z**: Benzene (\( C_6H_6 \))

### Final Answer:
Thus, the compound Z in the sequence of reactions is **Benzene** (\( C_6H_6 \)).

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