Question

Toluene undergoes bromination in presence of iron followed by treatment with sodium in dry ether. The major product formed is \[ C_6H_5CH_3 \xrightarrow{Br_2/Fe} X \xrightarrow{2Na,\ dry\ ether} Y \]

• A. Biphenyl
• B. \(p,p'\)-Dimethylbiphenyl
• C. Diphenylmethane
• D. Ethylbenzene

Hint:

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.

Answer 1

The Correct Option is B

Step 1: Read the two-step plan.
Toluene is first brominated with $Br_2$ and iron, then the product is treated with sodium in dry ether. We must find the final product $Y$.

Step 2: Note the role of the methyl group.
In toluene the $-CH_3$ group activates the ring and sends new groups to the ortho and para spots.

Step 3: Do the bromination.
With $Br_2/Fe$ the bromine goes mainly to the para position, because there is less crowding there. This gives para-bromotoluene as the main product $X$. \[ C_6H_5CH_3 \xrightarrow{Br_2/Fe} p\text{-}BrC_6H_4CH_3 \]

Step 4: Do the sodium coupling.
Sodium in dry ether joins two aryl halide molecules by removing the bromine atoms. This is a Fittig-type coupling.

Step 5: Join the two rings.
Two para-bromotoluene molecules link through the carbons that held bromine. \[ 2\,p\text{-}BrC_6H_4CH_3 + 2Na \rightarrow CH_3C_6H_4\text{-}C_6H_4CH_3 + 2NaBr \] The two methyl groups end up at the para positions of the new biphenyl.

Step 6: Name the product.
The product is a biphenyl carrying a methyl group on each ring at the para spots, that is $p,p'$-dimethylbiphenyl.

Step 7: State the final answer.
The major product $Y$ is:
\[ \boxed{p,p'\text{-Dimethylbiphenyl}} \]