Question

Convert Benzene \(\rightarrow\) 3-Bromophenol by choosing appropriate reagents (i) to (v) in the correct sequence.
\[ \text{(i) } NaNO_2/HCl \, (0^\circ C) \] \[ \text{(ii) Conc. } HNO_3/H_2SO_4 \] \[ \text{(iii) } H_2O/283 \, K \] \[ \text{(iv) } Fe/HCl \] \[ \text{(v) } Br_2/Fe \]

• A. \(x, ii, iii, iv\)
• B. \(ii, v, iv, i, iii\)
• C. \(iii, iv, v, ii, i\)
• D. \(v, ii, i, iv, iii\)

Hint:

To prepare meta-substituted phenols, first introduce a meta-directing group like \(-NO_2\), then convert it into \(-NH_2\), form diazonium salt, and finally replace it with \(-OH\).

Answer 1

The Correct Option is B

Step 1: Understand the target compound.
The target compound is \(3\)-bromophenol, which means bromine and hydroxyl group are present at meta positions on the benzene ring.
So, we need to introduce \(-Br\) and \(-OH\) in such a way that they finally become meta to each other.
\[ \text{Target compound} = m\text{-bromophenol} \]

Step 2: Introduce nitro group on benzene.
First, benzene is treated with concentrated \(HNO_3/H_2SO_4\).
This is nitration of benzene and gives nitrobenzene.
\[ C_6H_6 \xrightarrow{Conc. \, HNO_3/H_2SO_4} C_6H_5NO_2 \] Therefore, the first reagent is (ii).

Step 3: Bromination of nitrobenzene.
The nitro group \((-NO_2)\) is a strong electron-withdrawing group and is meta-directing.
So, when nitrobenzene reacts with \(Br_2/Fe\), bromine enters at the meta position.
\[ C_6H_5NO_2 \xrightarrow{Br_2/Fe} m\text{-bromonitrobenzene} \] Therefore, the second reagent is (v).

Step 4: Reduce nitro group to amino group.
Now, \(m\)-bromonitrobenzene is reduced using \(Fe/HCl\).
The \(-NO_2\) group is converted into \(-NH_2\), while the bromine remains at the meta position.
\[ m\text{-bromonitrobenzene} \xrightarrow{Fe/HCl} m\text{-bromoaniline} \] Therefore, the third reagent is (iv).

Step 5: Convert amino group into diazonium salt.
The amino group \((-NH_2)\) is converted into diazonium salt by treating with \(NaNO_2/HCl\) at \(0^\circ C\).
This reaction is called diazotization.
\[ m\text{-bromoaniline} \xrightarrow{NaNO_2/HCl \, (0^\circ C)} m\text{-bromobenzenediazonium chloride} \] Therefore, the fourth reagent is (i).

Step 6: Hydrolysis of diazonium salt to phenol.
Finally, diazonium salt is treated with water at \(283 \, K\).
The diazonium group \((-N_2^+Cl^-)\) is replaced by \(-OH\), forming \(3\)-bromophenol.
\[ m\text{-bromobenzenediazonium chloride} \xrightarrow{H_2O/283K} m\text{-bromophenol} \] Thus, the correct sequence is:
\[ ii \rightarrow v \rightarrow iv \rightarrow i \rightarrow iii \] Final Answer:
The correct sequence of reagents is:
\[ \boxed{ii, \, v, \, iv, \, i, \, iii} \]