In Carius method of estimation of halogen, 0.172 g of an organic compound showed presence of 0.08 g of bromine. Which…

Step 1: Calculate %Br from the Given Data

Given:

• Mass of compound = 0.172 g
• Mass of Br = 0.08 g

$\%Br = \frac{0.08}{0.172} \times 100 = 46.51\%$

Step 2: Calculate Molar Mass from %Br

If the compound has one Br atom (Br = 80 g/mol): $\text{Molar mass} = \frac{80}{0.4651} = 172 \text{ g/mol}$

If the compound has two Br atoms (2 × Br = 160 g/mol): $\text{Molar mass} = \frac{160}{0.4651} = 344 \text{ g/mol}$

Step 3: Match with Given Structures

(a) Bromoethane $\ce{CH3CH2Br}$: MW = 109 g/mol → %Br = 80/109 = 73.4% ❌

(b) Unknown: Cannot evaluate without structure.

(c) 2,4-Dibromoaniline $\ce{C6H3(NH2)Br2}$: MW = 250.9 g/mol → %Br = 160/250.9 = 63.8% ❌

(d) 4-Bromoaniline $\ce{C6H4(NH2)Br}$: MW = 172 g/mol → %Br = 80/172 = 46.51% ✅

Step 4: Conclusion

Molar mass = 172 g/mol with one Br atom → 4-bromoaniline ($\ce{C6H4BrNH2}$, MW = 6(12) + 4(1) + 14 + 1 + 80 = 72 + 4 + 14 + 1 + 80 = 171 ≈ 172 g/mol)

Answer: (d) 4-bromoaniline

Key Points to Remember:

• First calculate %Br, then deduce MW, then match with options
• For one Br: $\text{MW} = \frac{80}{\%Br/100}$
• 4-Bromoaniline: $\ce{C6H4(Br)(NH2)}$, MW = 12(6) + 4 + 14 + 1 + 80 = 172 g/mol ✓
• This reverse-calculation approach is powerful for structure determination from Carius data