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 $171$.
The IUPAC name of the following compound is:
The compounds which give positive Fehling's test are:
Choose the CORRECT answer from the options given below:
The products formed in the following reaction sequence are: 
Consider the following sequence of reactions : 
Molar mass of the product formed (A) is ______ g mol\(^{-1}\).
If the mean and the variance of 6, 4, a, 8, b, 12, 10, 13 are 9 and 9.25 respectively, then \(a + b + ab\) is equal to:
Given three identical bags each containing 10 balls, whose colours are as follows:
| Bag I | 3 Red | 2 Blue | 5 Green |
| Bag II | 4 Red | 3 Blue | 3 Green |
| Bag III | 5 Red | 1 Blue | 4 Green |
A person chooses a bag at random and takes out a ball. If the ball is Red, the probability that it is from Bag I is $ p $ and if the ball is Green, the probability that it is from Bag III is $ q $, then the value of $ \frac{1}{p} + \frac{1}{q} $ is:
If \( \theta \in \left[ -\frac{7\pi}{6}, \frac{4\pi}{3} \right] \), then the number of solutions of \[ \sqrt{3} \csc^2 \theta - 2(\sqrt{3} - 1)\csc \theta - 4 = 0 \] is equal to ______.