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.
The least acidic compound, among the following is
Choose the correct set of reagents for the following conversion:
The motion of an airplane is represented by the velocity-time graph as shown below. The distance covered by the airplane in the first 30.5 seconds is km.
A bead of mass \( m \) slides without friction on the wall of a vertical circular hoop of radius \( R \) as shown in figure. The bead moves under the combined action of gravity and a massless spring \( k \) attached to the bottom of the hoop. The equilibrium length of the spring is \( R \). If the bead is released from the top of the hoop with (negligible) zero initial speed, the velocity of the bead, when the length of spring becomes \( R \), would be (spring constant is \( k \), \( g \) is acceleration due to gravity):