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.
For the thermal decomposition of \( N_2O_5(g) \) at constant volume, the following table can be formed, for the reaction mentioned below: \[ 2 N_2O_5(g) \rightarrow 2 N_2O_4(g) + O_2(g) \] Given: Rate constant for the reaction is \( 4.606 \times 10^{-2} \text{ s}^{-1} \).
A hydrocarbon which does not belong to the same homologous series of carbon compounds is
Let \[ I(x) = \int \frac{dx}{(x-11)^{\frac{11}{13}} (x+15)^{\frac{15}{13}}} \] If \[ I(37) - I(24) = \frac{1}{4} \left( b^{\frac{1}{13}} - c^{\frac{1}{13}} \right) \] where \( b, c \in \mathbb{N} \), then \[ 3(b + c) \] is equal to:
Let \( T_r \) be the \( r^{\text{th}} \) term of an A.P. If for some \( m \), \( T_m = \dfrac{1}{25} \), \( T_{25} = \dfrac{1}{20} \), and \( \displaystyle\sum_{r=1}^{25} T_r = 13 \), then \( 5m \displaystyle\sum_{r=m}^{2m} T_r \) is equal to: