Question

The compound A (major product) is

• A.
• B.
• C.
• D.

Answer 1

The Correct Option is B

To solve the problem, we need to determine the major product formed when the given compound undergoes a reaction with $ \text{Br}_2 $ under UV light. The reaction involves the substitution of a hydrogen atom in the alkyl group by bromine ($ \text{Br} $).

1. Understanding the Reaction:
The given compound is:
$$ \text{O}_{2}\text{N}-\underset{\text{CH}_{3}-\text{CH}_{3}}{\underbrace{\text{C}_{6}\text{H}_{4}}} $$
This is a benzene ring substituted with a nitro group ($ \text{NO}_2 $) and an ethyl group ($ \text{CH}_2\text{CH}_3 $).

When this compound reacts with $ \text{Br}_2 $ under UV light, the reaction proceeds via a radical mechanism. UV light provides the necessary energy to initiate the reaction by generating free radicals.

2. Mechanism of the Reaction:
Under UV light:
1. The bromine molecule ($ \text{Br}_2 $) absorbs energy and dissociates into two bromine atoms ($ \text{Br}^\bullet $):
$$ \text{Br}_2 \xrightarrow{\text{UV Light}} 2\text{Br}^\bullet $$
2. One of the bromine atoms ($ \text{Br}^\bullet $) abstracts a hydrogen atom from the ethyl group ($ \text{CH}_2\text{CH}_3 $), forming a methyl radical ($ \text{CH}_3\text{CH}_2^\bullet $) and $ \text{HBr} $:
$$ \text{CH}_2\text{CH}_3 + \text{Br}^\bullet \rightarrow \text{CH}_3\text{CH}_2^\bullet + \text{HBr} $$
3. The methyl radical ($ \text{CH}_3\text{CH}_2^\bullet $) then reacts with another bromine atom ($ \text{Br}^\bullet $) to form the brominated product:
$$ \text{CH}_3\text{CH}_2^\bullet + \text{Br}^\bullet \rightarrow \text{CH}_3\text{CH}_2\text{Br} $$

3. Identifying the Major Product:
The reaction results in the substitution of one of the hydrogen atoms in the ethyl group ($ \text{CH}_2\text{CH}_3 $) by bromine ($ \text{Br} $). The major product will have the bromine atom attached to one of the carbon atoms in the ethyl group.

Given the options:
- (B) shows bromination at the end of the ethyl chain ($ \text{CH}_2\text{CH}_2\text{Br} $).
- (A) shows bromination at the middle of the ethyl chain ($ \text{CH}_2\text{CHBrCH}_3 $).
- (C) shows bromination directly on the benzene ring.
- (D) shows bromination at the middle of the ethyl chain ($ \text{CH}_2\text{CHBrCH}_3 $).

In radical halogenation, the more stable radical (tertiary or secondary) is favored. However, in this case, both primary and secondary positions are possible, but the primary position (end of the chain) is slightly more favorable due to steric factors.

4. Conclusion:
The major product is formed by bromination at the end of the ethyl chain, resulting in the structure:
$$ \text{O}_{2}\text{N}-\underset{\text{CH}_{2}-\text{CH}_{2}\text{-Br}}{\underbrace{\text{C}_{6}\text{H}_{4}}} $$

Final Answer:
(B) $ \text{O}_{2}\text{N}-\underset{\text{CH}_{2}-\text{CH}_{2}\text{-Br}}{\underbrace{\text{C}_{6}\text{H}_{4}}} $

Answer 2

To solve this problem, we need to understand the mechanism of free radical substitution in the presence of UV light and bromine (Br₂) for the given compound.

1. Analyzing the Reaction:
The given compound is a nitrobenzene derivative with a methyl group (-CH₃) attached to the benzene ring. The reaction involves the addition of bromine (Br₂) under UV light, which induces a free radical substitution reaction. Under these conditions, the UV light provides the energy required to break the Br₂ bond homolytically, generating two bromine radicals (Br•).

2. Free Radical Substitution Mechanism:
The reaction proceeds through the following steps:

• Initiation: The UV light causes the homolytic cleavage of Br₂, generating two Br• radicals.
• Propagation: One Br• radical abstracts a hydrogen atom from the methyl group (-CH₃) on the benzene ring, forming a methyl radical (-CH₂•). The resulting methyl radical then reacts with another molecule of Br₂ to form the major product.
• Termination: The reaction ends when two radicals combine to form a stable product, but this step is less relevant to the identification of the major product.

3. Product Formation:
The major product of this reaction will have a bromine atom substituted at the position where the methyl group was originally attached. Since the reaction is under UV light, it follows the mechanism of free radical substitution rather than electrophilic aromatic substitution. Therefore, the bromine will substitute the hydrogen atom attached to the methyl group (-CH₃) on the benzene ring, resulting in the formation of a bromoalkyl group.

4. Identifying the Major Product:
In the options: - Option (B) shows a bromoalkyl group (Br-CH₂-CH₃) attached to the benzene ring. This corresponds to the substitution of the hydrogen on the methyl group by a bromine atom. 

- Option (A) shows a bromine attached directly to the benzene ring, but this does not reflect the substitution occurring at the methyl group. 

- Option (C) shows a bromine attached to the benzene ring with a -CH₃ group, but this is not the expected outcome under UV light and Br₂ conditions. 

- Option (D) shows a bromine attached to the position adjacent to the nitro group, but this is unlikely due to the selectivity of the free radical mechanism.

Final Answer:
The major product is shown in option (B), where the bromoalkyl group is attached to the benzene ring at the position where the methyl group was originally attached.