Question

Identify the major product from the following reaction sequence.

• A.

  

• B.

  

• C.

  

• D.

  

Hint:

\textbf{Oxidation of alkylbenzenes:} Alkyl groups attached to a benzene ring are oxidized to -COOH by strong oxidizing agents like KMnO$_4$/KOH followed by H$_3$O$^+$. (Requires at least one benzylic hydrogen). Toluene $\rightarrow$ Benzoic acid.
\textbf{Electrophilic Aromatic Substitution (EAS):} The -COOH group is:
Deactivating (makes the ring less reactive towards EAS than benzene).
Meta-directing (directs incoming electrophiles primarily to the meta positions).
\textbf{Bromination:} Br$_2$/FeBr$_3$ (or Br$_2$/Fe) is used for bromination of aromatic rings.
Combine the effects: Benzoic acid undergoes bromination at the meta position.

Answer 1

The Correct Option is A

The reaction sequence starts with toluene (methylbenzene). Step (i) KMnO$_4$ / KOH and Step (ii) H$_3$O$^+$, $\Delta$: KMnO$_4$ (potassium permanganate) in the presence of KOH (alkaline medium), followed by acidic workup (H$_3$O$^+$) and heating ($\Delta$), is a strong oxidizing agent for alkyl side chains on an aromatic ring. The methyl group (-CH$_3$) of toluene will be oxidized to a carboxylic acid group (-COOH). Regardless of the length of the alkyl side chain (as long as it has at least one benzylic hydrogen), it is oxidized to -COOH. So, toluene (C$_6$H$_5$-CH$_3$) is converted to benzoic acid (C$_6$H$_5$-COOH). The product after steps (i) and (ii) is benzoic acid. Step (iii) Br$_2$ / FeBr$_3$: This is an electrophilic aromatic substitution reaction: bromination of the benzene ring. FeBr$_3$ acts as a Lewis acid catalyst, polarizing Br$_2$ to generate the electrophile (Br$^+$ equivalent). The substituent already present on the benzene ring (-COOH in benzoic acid) directs the incoming electrophile (Br). The -COOH group (carboxylic acid) is an electron-withdrawing group and a meta-directing group for electrophilic aromatic substitution. It deactivates the ring, but substitution occurs primarily at the meta position relative to the -COOH group. So, bromination of benzoic acid will yield 3-bromobenzoic acid (or m-bromobenzoic acid). Structure of 3-bromobenzoic acid: The -COOH is at position 1. The Br is at position 3. This matches option (a). \[ \boxed{\text{3-Bromobenzoic acid}} \]