Question

The major product (A) formed in the following reaction sequence is

• A.
• B.
• C.
• D.

Hint:

Follow the reaction sequence to determine the major product.

Answer 1

The Correct Option is B

1. Reduction of Nitro Group (Sn, HCl)

The first step involves the reduction of the nitro group (-NO2) to an amine group (-NH2) using Sn/HCl.

2. Acetylation of Amine Group (Ac2O, Pyridine)

The second step acetylates the amine group (-NH2) to form an amide (-NHCOCH3) using acetic anhydride (Ac2O) in the presence of pyridine. Pyridine acts as a base to neutralize the HCl produced during acetylation. This step is done to protect the amine group from reacting with bromine in the next step.

3. Bromination (Br2, AcOH)

The third step is bromination using Br2 in acetic acid (AcOH). The amide group (-NHCOCH3) is an ortho/para directing group, but it is also bulky. Therefore, bromination will occur primarily at the para position to the amide group due to steric factors.

4. Hydrolysis of Amide (NaOH(aq))

The fourth step is the hydrolysis of the amide group (-NHCOCH3) back to an amine group (-NH2) using NaOH(aq). This removes the protecting group.

The Final Product

The final product (A) will be a benzene ring with an amine group (-NH2) and a bromine atom (Br) at the para position to the amine.

Therefore, the structure of the final product (A) is 4-bromoaniline.

Answer 2

The given problem involves identifying the final product (A) from the reaction sequence:

\( \text{(i) Sn/HCl, (ii) Ac}_2\text{O, Pyridine, (iii) Br}_2\text{, AcOH, (iv) NaOH(aq)} \)

Concept Used:

This reaction sequence is a multi-step transformation starting from nitrobenzene. It involves reduction, protection, electrophilic substitution, and deprotection:

1. Reduction of the nitro group (–NO₂) to an amino group (–NH₂).
2. Acetylation of the –NH₂ group to form acetanilide, which directs electrophilic substitution to the ortho and para positions but reduces activation.
3. Bromination occurs at the para position (major) due to steric reasons.
4. Hydrolysis removes the acetyl protecting group, regenerating –NH₂.

Step-by-Step Solution:

Step 1: Reduction of nitrobenzene using tin and hydrochloric acid.

\[ \ce{C6H5NO2 ->[Sn/HCl] C6H5NH2} \]

The product is aniline.

Step 2: Acetylation of aniline with acetic anhydride in pyridine.

\[ \ce{C6H5NH2 ->[Ac2O, pyridine] C6H5NHCOCH3} \]

This forms acetanilide, which protects the amino group and moderates its activating effect.

Step 3: Bromination of acetanilide with bromine in acetic acid.

\[ \ce{C6H5NHCOCH3 ->[Br2, AcOH] 4-BrC6H4NHCOCH3} \]

Because the acetyl group directs substitution to ortho and para positions, the para isomer dominates due to less steric hindrance.

Step 4: Hydrolysis of the amide group in basic medium (NaOH, aqueous).

\[ \ce{4-BrC6H4NHCOCH3 ->[NaOH(aq)] 4-BrC6H4NH2 + CH3COONa} \]

The acetyl protecting group is removed, yielding p-bromoaniline.

Final Computation & Result:

The major product (A) formed is p-bromoaniline.

Correct Option: (2)