The product(s) that are formed when the given compound is treated with $Br_2$ in the presence of $FeBr_3$ are:

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1. **Reagents and Reaction Type:** The reaction uses Bromine ($Br_2$) in the presence of Ferric Bromide ($FeBr_3$). $FeBr_3$ acts as a **Lewis acid** catalyst to generate the electrophile, the bromonium ion ($Br^+$) [NCERT 12th, Ch 10, Sec 10.5; NCERT 11th, Ch 13, Sec 13.5.5]. This indicates an **Electrophilic Aromatic Substitution** reaction.
   $$ Br_2 + FeBr_3 \rightarrow Br^+ + [FeBr_4]^- $$
2. **Substrate Analysis:** Although the specific structure of the 'given compound' is missing from the input, the reaction conditions and the probable answer format ('and', implying a mixture) suggest the substrate is a mono-substituted benzene ring containing an **ortho-para directing group** (e.g., $-CH_3$, $-OCH_3$, $-Cl$, etc.).
3. **Directive Influence:** Ortho-para directing groups increase electron density at the *ortho* (2, 6) and *para* (4) positions via resonance or inductive effects. Consequently, the electrophile ($Br^+$) attacks these positions, resulting in a mixture of **o-bromo** and **p-bromo** derivatives [NCERT 12th, Ch 10, Sec 10.5; NCERT 11th, Ch 13, Sec 13.5.6].
4. **Conclusion:** The products formed would be a mixture of the *ortho* and *para* isomers of the brominated compound. (e.g., if the reactant was Toluene, the products would be o-Bromotoluene and p-Bromotoluene).

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