JEE Main 2025 · 22 January · morning · Numerical answer · +4 / −1
Consider the following sequence of reactions :
Molar mass of the product formed (A) is __________ g mol⁻¹.
Answer and solution
Answer 154
Solution
Chemistry ma’am · 4 steps
Follow the nitrogen out of the ring. Then Na in ether joins two aryl halides into biphenyl.
- Reduce, then diazotise. Sn + HCl reduces nitrobenzene to aniline, C₆H₅NH₂. NaNO₂ and HCl at 0 °C turn aniline into benzenediazonium chloride, C₆H₅N₂⁺Cl⁻.
- Sandmeyer reaction. Cu₂Cl₂ replaces the -N₂⁺ group with Cl and releases N₂ gas. You now have chlorobenzene, C₆H₅Cl.
- Fittig reaction. Sodium in dry ether removes Cl from two chlorobenzene molecules and joins the two phenyl rings directly. The product A is biphenyl, C₆H₅-C₆H₅.
2C₆H₅Cl + 2Na →(dry ether) C₆H₅-C₆H₅ + 2NaCl
- So. Biphenyl is C₁₂H₁₀, with molar mass 12 × 12 + 10 × 1 = 154 g mol⁻¹. The answer is 154.
The common slip: 112.5. Stopping at chlorobenzene gives 112.5. Step (iv), Na in ether, still has to couple two of those molecules into biphenyl.
The idea to keep. Diazonium salts let you swap -NH₂ for a halogen. Na in dry ether with an aryl halide means Fittig coupling, so two rings join and the carbon count doubles.
Answer from the NTA final answer key. Solution by PrepWiser.