Question

Identify the products C, D, and F formed in the following sets of reactions.

• A. C = p-nitrobromobenzene D = m-nitrobromobenzene F = p-nitrobromobenzene
• B. C = o-nitrobromobenzene D = m-nitrobromobenzene F = p-nitrobromobenzene
• C. C = o-nitrobromobenzene D = p-nitrobromobenzene F = o-nitrobromobenzene
• D. C = m-nitrobromobenzene D = p-nitrobromobenzene F = o-nitrobromobenzene

Hint:

In electrophilic aromatic substitution reactions, the position of substitution is directed by the nature of the substituent already present. Electron-withdrawing groups such as bromine direct the incoming group to the ortho and para positions.

Answer 1

The Correct Option is C

In this series of reactions, we first observe the halogenation of chlorobenzene with bromine in the presence of iron bromide (\(\text{FeBr}_3\)). This leads to the formation of bromobenzene (B). Next, we treat bromobenzene with concentrated nitric acid (\(\text{HNO}_3\)) and concentrated sulfuric acid (\(\text{H}_2\text{SO}_4\)) in a nitration reaction. The bromine group (-Br) is an electron-withdrawing group, which directs the incoming nitro group to the ortho and para positions relative to itself. This results in the formation of two products: o-nitrobromobenzene (C) and p-nitrobromobenzene (D). Finally, when the product mixture undergoes another halogenation reaction with bromine in the presence of iron bromide (\(\text{FeBr}_3\)), the nitration products (C and D) are both substituted with a second bromine atom. The outcome is the formation of o-nitrobromobenzene (F) and p-nitrobromobenzene (F).
Thus, the final products are: \[ C = \text{o-nitrobromobenzene}, \quad D = \text{p-nitrobromobenzene}, \quad F = \text{o-nitrobromobenzene}. \]