**75.** Consider the following sequence of reactions: Starting compound: Nitrobenzene (\ce{C6H5NO2}) **Reagents:** 1. Sn + HCl 2. NaNO₂, HCl (0°C) 3. Cu₂Cl₂ 4. Na, Ether The final product formed is \( A \). **Question:** Molar mass of the product formed (A) is ______ g mol\(^{-1}\).

Key Concepts Needed

1. Reduction of Nitro Group
   $\text{C}_6\text{H}_5\text{NO}_2 \xrightarrow{\text{Sn / HCl}} \text{C}_6\text{H}_5\text{NH}_2$
   Tin in acid supplies electrons and protons, converting the strongly electron-withdrawing $\ce{NO_2}$ to the nucleophilic amine $\ce{NH_2}$.

2. Diazotisation
   $\text{C}_6\text{H}_5\text{NH}_2 + \text{NaNO}_2 + \text{HCl} \; (0^\circ\text{C}) \rightarrow \text{C}_6\text{H}_5\text{N}_2^+\,\text{Cl}^- + 2\,\text{H}_2\text{O}$
   Amines react with nitrous acid (generated in situ) below 5 °C to form unstable but highly versatile diazonium salts.

3. Sandmeyer Reaction
   $\text{C}_6\text{H}_5\text{N}_2^+\,\text{Cl}^- \xrightarrow{\ce{Cu2Cl2}} \text{C}_6\text{H}_5\text{Cl} + \ce{N_2}\uparrow$
   Copper(I) chloride catalytically replaces the diazonium group by chlorine.

4. Wurtz-Fittig (Aryl Wurtz) Coupling
   $2\,\text{C}_6\text{H}_5\text{Cl} + 2\,\text{Na} \xrightarrow{\text{dry ether}} \text{C}_{12}\text{H}_{10} + 2\,\text{NaCl}$
   Although less efficient with aryl halides, sodium can couple two chlorobenzenes to give biphenyl.

5. Molar-Mass Calculation
   Biphenyl formula: $\ce{C_{12}H_{10}}$
   $M = 12(12.01) + 10(1.008) \approx 144.12 + 10.08 = 154.20\;\text{g mol}^{-1}$

What’s happening, kiddo?

Imagine you have a Lego® car (nitrobenzene) and you keep breaking off one type of Lego brick and snapping on another until you get a bigger, cooler car (biphenyl).

1. First swap: You remove the nitro brick and pop on an amino brick (–NH₂).
2. Second swap: You quickly turn that –NH₂ into a magic sticker called a diazonium tag. It’s super reactive but only if you keep it ice-cold!
3. Third swap: You peel off the tag and stick on a chlorine brick.
4. Fourth swap: Put two of these chlorine-cars together in a garage filled with sodium metal and they clamp onto each other. Now you’ve built a long two-car train called biphenyl.

The weight of this two-car train is about 154 grams for every mole of trains.

Step-by-Step Solution

1. Reduction of Nitrobenzene
   $\ce{C6H5NO2}\;\xrightarrow[HCl]{Sn}\;\ce{C6H5NH2}$
   Product: Aniline.

2. Diazotisation of Aniline (0 °C)
   $\ce{C6H5NH2 + NaNO2 + 2HCl \;\to\; C6H5N2+ Cl- + 2 H2O + NaCl}$
   Product: Benzenediazonium chloride.

3. Sandmeyer Chlorination
   $\ce{C6H5N2+ Cl- \xrightarrow{Cu2Cl2}\; C6H5Cl + N2}\uparrow$
   Product: Chlorobenzene.

4. Aryl Wurtz Coupling
   $\ce{2 C6H5Cl + 2 Na \xrightarrow{dry\ ether}\ C12H10 + 2 NaCl}$
   Product: Biphenyl ($\ce{C12H10}$).

5. Molar Mass Calculation
   Carbon: $12 \times 12.01 = 144.12$ g mol⁻¹
   Hydrogen: $10 \times 1.008 = 10.08$ g mol⁻¹
   $M_{\text{biphenyl}} = 144.12 + 10.08 \approx 154.20\,\text{g mol}^{-1}$

Answer: $\boxed{\approx 154\ \text{g mol}^{-1}}$