Question

Chlorobenzene on heating with concentrated $\text{HNO}_3$ in presence of concentrated $\text{H}_2\text{SO}_4$ gives

• A.

  

• B.

  

• C.

  

• D.

  

Hint:

Halogens attached to a benzene ring are unique: they are deactivating due to their strong inductive effect, but they are always ortho/para-directing due to resonance.
The para-product is almost always the major product because it minimizes steric clash between adjacent functional groups.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks for the product formed when chlorobenzene undergoes a reaction with a nitrating mixture consisting of concentrated $\text{HNO}_3$ and concentrated $\text{H}_2\text{SO}_4$ under heating conditions.
This is a classic electrophilic aromatic substitution reaction, specifically the nitration of a haloarene.

Step 2: Key Formula or Approach:
In chlorobenzene, the chlorine atom has lone pairs of electrons that undergo $+R$ (resonance) effect with the benzene ring, increasing electron density at the ortho and para positions.
Although chlorine is deactivating due to its strong $-I$ (inductive) effect, it is strictly an ortho/para-directing group.
The nitrating mixture generates the nitronium ion ($\text{NO}_2^+$) as the active electrophile, which will attack the ortho and para positions of chlorobenzene.

Step 3: Detailed Explanation:
When chlorobenzene reacts with concentrated $\text{HNO}_3$ and concentrated $\text{H}_2\text{SO}_4$, nitration takes place.
The electrophile $\text{NO}_2^+$ replaces a hydrogen atom at either the ortho or para position relative to the chlorine atom.
This yields a mixture of two structural isomers: 1-chloro-2-nitrobenzene (ortho product) and 1-chloro-4-nitrobenzene (para product).
Due to steric hindrance between the bulky chlorine atom and the incoming nitro group at the ortho position, the ortho isomer is formed in smaller amounts as the minor product.
Conversely, the para position experiences minimal steric repulsion, making 1-chloro-4-nitrobenzene the highly stable and major product of the reaction.
Option (A) accurately depicts this mixture or the predominant major/minor combination.

Step 4: Final Answer:
The reaction yields a mixture of ortho and para substituted products, with the para-substituted product being the major component, corresponding to option (A).