Question:
Toluene undergoes bromination in presence of iron followed by treatment with sodium in dry ether. The major product formed is
\[
C_6H_5CH_3
\xrightarrow{Br_2/Fe}
X
\xrightarrow{2Na,\ dry\ ether}
Y
\]
Toluene undergoes bromination in presence of iron followed by treatment with sodium in dry ether. The major product formed is
\[
C_6H_5CH_3
\xrightarrow{Br_2/Fe}
X
\xrightarrow{2Na,\ dry\ ether}
Y
\]
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Remember that a methyl group is an ortho/para director. In bromination of toluene, the para product is generally the major product due to lower steric hindrance. Subsequent Wurtz coupling joins two aromatic rings together.
Updated On: Jun 10, 2026
- Biphenyl
- \(p,p'\)-Dimethylbiphenyl
- Diphenylmethane
- Ethylbenzene
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The Correct Option is B
Solution and Explanation
Concept:
The reaction sequence involves:
• Electrophilic aromatic substitution (bromination of toluene).
• Wurtz-Fittig type coupling in presence of sodium metal and dry ether.
The methyl group is an activating and ortho/para-directing group. Therefore bromination occurs predominantly at the para position due to lower steric hindrance.
Step 1: Bromination of toluene The methyl group activates the benzene ring and directs the incoming bromine atom to the ortho and para positions. \[ C_6H_5CH_3 \xrightarrow{Br_2/Fe} o\text{-bromotoluene} + p\text{-bromotoluene} \] The para isomer is the major product because it experiences less steric crowding. Thus, \[ X=p\text{-bromotoluene} \]
Step 2: Wurtz coupling reaction Two molecules of \(p\)-bromotoluene react with sodium metal in dry ether. \[ 2\,p\text{-BrC}_6H_4CH_3 + 2Na \rightarrow p\text{-CH}_3C_6H_4-C_6H_4CH_3 + 2NaBr \] A carbon-carbon bond is formed between the two aromatic rings.
Step 3: Identify the final product The product contains two benzene rings joined together with methyl groups at para positions. Therefore the product is \[ \boxed{p,p'\text{-Dimethylbiphenyl}} \]
Final Answer \[ \boxed{\text{Option (B)}} \]
• Electrophilic aromatic substitution (bromination of toluene).
• Wurtz-Fittig type coupling in presence of sodium metal and dry ether.
The methyl group is an activating and ortho/para-directing group. Therefore bromination occurs predominantly at the para position due to lower steric hindrance.
Step 1: Bromination of toluene The methyl group activates the benzene ring and directs the incoming bromine atom to the ortho and para positions. \[ C_6H_5CH_3 \xrightarrow{Br_2/Fe} o\text{-bromotoluene} + p\text{-bromotoluene} \] The para isomer is the major product because it experiences less steric crowding. Thus, \[ X=p\text{-bromotoluene} \]
Step 2: Wurtz coupling reaction Two molecules of \(p\)-bromotoluene react with sodium metal in dry ether. \[ 2\,p\text{-BrC}_6H_4CH_3 + 2Na \rightarrow p\text{-CH}_3C_6H_4-C_6H_4CH_3 + 2NaBr \] A carbon-carbon bond is formed between the two aromatic rings.
Step 3: Identify the final product The product contains two benzene rings joined together with methyl groups at para positions. Therefore the product is \[ \boxed{p,p'\text{-Dimethylbiphenyl}} \]
Final Answer \[ \boxed{\text{Option (B)}} \]
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