Consider the following sequence of reactions to produce major product (A):
The molar mass of the product (A) is g mol−1. (Given molar mass in g mol−1 of C: 12,
H: 1, O: 16, Br: 80, N: 14, P: 31)
Let's break down the reaction sequence:
1. Bromination (Br2, Fe):
The starting material is 3-nitrotoluene. Bromination occurs ortho and para to the methyl group, but meta to the nitro group. Therefore, the major product is 4-bromo-3-nitrotoluene.
2. Reduction (Sn, HCl):
The nitro group (-NO2) is reduced to an amino group (-NH2). So, we now have 4-bromo-3-aminotoluene.
3. Diazotization (NaNO2, HCl, 273 K):
The amino group is converted to a diazonium salt. So, we get 4-bromo-3-tolyldiazonium chloride.
4. Reduction (H3PO2, H2O):
The diazonium salt is replaced by a hydrogen atom. Thus, the amino group gets replaced with hydrogen. Therefore, we obtain 4-bromotoluene.
Final Product Analysis:
The final product (A) is 4-bromotoluene (C7H7Br).
Molar Mass Calculation:
Molar mass = 7(12) + 7(1) + 1(80) = 84 + 7 + 80 = 171 g/mol.
Final Answer:
The final answer is $171$.
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 \( T_r \) 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: