The major product (A) formed in the following reaction sequence is
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.
Identify the structure of the final product (D) in the following sequence of the reactions :
Total number of $ sp^2 $ hybridised carbon atoms in product D is _____.
Given below are two statements :
In the light of the above statements, choose the most appropriate answer from the options given below :
The number of optically active products obtained from the complete ozonolysis of the given compound is :
Match List-I with List-II
Choose the correct answer from the options given below :
Match List-I with List-II.
Choose the correct answer from the options given below :