Weight Percentage of Hydrogen in Picric Acid Synthesis

To determine the weight percentage of hydrogen in product $\mathbf{Q}$, we analyze the given two-step reaction sequence:

1. Step 1: Conversion of Chlorobenzene to Phenol Chlorobenzene undergoes nucleophilic aromatic substitution when treated with $\text{NaOH}$ at high temperature ($623\text{ K}$) and high pressure ($300\text{ atm}$) (Dow's Process) to yield sodium phenoxide, which upon neutralization gives phenol ($\text{C}_6\text{H}_5\text{OH}$).

   $\text{C}_6\text{H}_5\text{Cl} \xrightarrow[\text{2. H}^+]{\text{1. NaOH, } 623\text{ K}, 300\text{ atm}} \text{C}_6\text{H}_5\text{OH}$

2. Step 2: Conversion of Phenol to Product $\mathbf{Q}$ When phenol is treated first with concentrated $\text{H}_2\text{SO}_4$, it undergoes sulfonation to form phenol-2,4-disulfonic acid. Subsequent treatment with concentrated $\text{HNO}_3$ results in nitration along with desulfonation to produce 2,4,6-trinitrophenol (commonly known as Picric acid) as the major product $\mathbf{Q}$.

   $\text{C}_6\text{H}_5\text{OH} \xrightarrow{\text{1. conc. H}_2\text{SO}_4, \text{ 2. conc. HNO}_3} \text{C}_6\text{H}_2(\text{NO}_2)_3\text{OH} \quad (\mathbf{Q})$

3. Molecular Mass Calculation for Compound $\mathbf{Q}$ ($\text{C}_6\text{H}_3\text{N}_3\text{O}_7$):

   • Carbon ($\text{C}$): $6 \times 12 = 72\text{ g/mol}$
   • Hydrogen ($\text{H}$): $3 \times 1 = 3\text{ g/mol}$
   • Nitrogen ($\text{N}$): $3 \times 14 = 42\text{ g/mol}$
   • Oxygen ($\text{O}$): $7 \times 16 = 112\text{ g/mol}$

   $\text{Molar mass of } \mathbf{Q} = 72 + 3 + 42 + 112 = 229\text{ g/mol}$

4. Weight Percentage of Hydrogen:

   $\text{Weight \% of H} = \left( \frac{\text{Mass of H}}{\text{Molar mass of } \mathbf{Q}} \right) \times 100$

   $\text{Weight \% of H} = \frac{3}{229} \times 100 \approx 1.31\%$

Thus, the weight percentage of hydrogen in $\mathbf{Q}$ is 1.31.