Question

Identify the major product of the following reaction:

• A. Benzoyl bromide
• B. Bromobenzene
• C. 4-bromo benzoic acid
• D. 3-bromo benzoic acid

Hint:

In electrophilic aromatic substitution, groups like –COOH direct incoming substituents to the meta position due to their electron-withdrawing nature. Always consider the directive influence of substituents before predicting products.

Answer 1

The Correct Option is D

Let us examine the given reaction step-by-step: \begin{itemize} \item The starting compound is **benzoic acid**, which has the structure: \[ \text{Benzoic acid: } \mathrm{C_6H_5COOH} \] \item The reagent used is **Br\textsubscript{2}/FeBr\textsubscript{3}**, a classic electrophilic bromination reagent that generates the **bromonium electrophile (Br\textsuperscript{+})** in situ via the reaction: \[ \mathrm{Br_2 + FeBr_3 \rightarrow Br^+ + FeBr_4^-} \] \item The type of reaction is **electrophilic aromatic substitution (EAS)**. \item Now, let’s consider the **substituent effect of –COOH**: \begin{itemize} \item The **–COOH (carboxylic acid)** group is an **electron-withdrawing group** due to both inductive $(-I)$ and resonance $(-R)$ effects. \item Electron-withdrawing groups **deactivate** the aromatic ring toward EAS and are **meta-directing**, i.e., they make electrophilic attack most favorable at the **meta position** relative to themselves. \end{itemize} \item So, in benzoic acid, when the Br\textsuperscript{+} electrophile approaches the ring, it prefers the **meta position** over ortho and para because: \begin{itemize} \item Ortho and para positions lead to unstable resonance intermediates due to direct competition with the withdrawing –COOH group. \item Meta position leads to a more stable arenium ion intermediate. \end{itemize} \item Therefore, the major product formed will be: \[ \text{3-bromo benzoic acid (Br at meta position w.r.t. –COOH)} \] \item Let’s draw the final product: \[ \begin{array}{c} \text{COOH group at position 1, Br at position 3}
\end{array} \] \item Incorrect options: \begin{itemize} \item (1) Benzoyl bromide is formed when benzoic acid reacts with SOCl\textsubscript{2} followed by reaction with Br\textsubscript{2}, not directly via bromination. \item (2) Bromobenzene would be formed from benzene, not benzoic acid. \item (3) 4-bromo benzoic acid is the **para isomer**, but para position is less favorable due to –COOH group being deactivating. \end{itemize} \end{itemize} Thus, the correct answer is: \[ \boxed{\text{(4) 3-bromo benzoic acid}} \]