Question

p-Nitrotoluene with \(Br_2\) (heat)

Hint:

Br\(_2\) + heat → side chain bromination (free radical substitution).

Answer 1

Bromination of aromatic compounds in the presence of heat typically leads to side-chain bromination, where a bromine atom attaches to the alkyl side chain of the molecule rather than directly to the aromatic ring.
Reaction Overview:
In the case of \(p\)-nitrotoluene, bromination occurs at the benzylic (side-chain) position when treated with bromine (\(Br_2\)) and heat. This results in the substitution of a hydrogen atom from the methyl group of the toluene with a bromine atom, forming \(p\)-nitrobenzyl bromide. The reaction is as follows: \[ p\text{-Nitrotoluene} \xrightarrow{Br_2,\ heat} p\text{-Nitrobenzyl bromide} \]
Step 1: Mechanism of Side-Chain Bromination.
In the presence of heat, \(Br_2\) generates a bromine radical (\(Br·\)), which is highly reactive. The methyl group (\(-CH_3\)) attached to the benzene ring undergoes homolytic cleavage, generating a benzyl radical at the benzylic position. This radical then reacts with the bromine radical to form the final product, \(p\)-nitrobenzyl bromide.
Step 2: Product Formed.
The product of this reaction is \(p\)-nitrobenzyl bromide, where the bromine atom is substituted at the benzylic carbon (the methyl group of toluene), creating a bromoalkyl group attached to the aromatic ring.
Final Answer:
The product formed in the bromination of \(p\)-nitrotoluene in the presence of heat is \(p\)-nitrobenzyl bromide.