Which of the following give para major product with aniline?

Options

A. `H_(2)SO_(4)`
B. `HNO_(3)//H_(2)SO_(4)`
C. `Br_(2)//H_(2)O`
D. `CH_(3)Cl//AlCl_(3)`

Answer

To determine which of the given options yields a para major product with aniline, we will analyze each option step by step.

### Step 1: Understanding Aniline and the Concept of Para Position
Aniline is an aromatic amine with the formula C6H5NH2. The para position refers to the position on the benzene ring that is opposite to the substituent. In this case, we are looking for reactions that predominantly yield products where the new substituent attaches at the para position relative to the NH2 group.

### Step 2: Analyzing Each Option

#### Option 1: Aniline with H2SO4 (Sulfonation)
- **Reaction**: Aniline reacts with sulfuric acid (H2SO4) to undergo sulfonation.
- **Product**: The sulfonic acid group (SO3H) will predominantly attach at the para position due to the activating effect of the NH2 group.
- **Conclusion**: This reaction gives a para major product.

#### Option 2: Aniline with HNO3 in H2SO4 (Nitration)
- **Reaction**: Aniline reacts with nitric acid (HNO3) in the presence of sulfuric acid.
- **Products**: This reaction produces three products: ortho-nitroaniline, meta-nitroaniline, and para-nitroaniline.
- **Distribution**: The distribution is approximately 2% ortho, 47% meta, and 51% para.
- **Conclusion**: The para product is the major product in this reaction.

#### Option 3: Aniline with Br2 in Water (Bromination)
- **Reaction**: Aniline reacts with bromine water (Br2 in H2O).
- **Products**: This reaction leads to the formation of 2,4,6-tribromoaniline, where bromine can attach at both the ortho and para positions.
- **Conclusion**: Since bromination produces significant amounts of both ortho and para products, it does not yield a para major product.

#### Option 4: Aniline with CHCl3 and AlCl3 (Friedel-Crafts Reaction)
- **Reaction**: Aniline reacts with chloroform (CHCl3) in the presence of aluminum chloride (AlCl3).
- **Conclusion**: This reaction does not proceed effectively because the AlCl3 forms a complex with the amine, preventing the Friedel-Crafts acylation or alkylation from occurring. Thus, it does not yield a para major product.

### Final Conclusion
From the analysis:
- **Option 1**: Para product (but not major)
- **Option 2**: Para major product
- **Option 3**: Not a para major product
- **Option 4**: Not a para major product

**The correct answer is Option 2, which gives a para major product with aniline.**