The weight percentage of hydrogen in Q, formed in the following reaction sequence, is ________.

[Given: Atomic mass of H = 1, C = 12, N = 14, O = 16, S = 32, Cl = 35]

The correct answer is 1.31.

Step 1: Understand the reaction sequence

The image shows chlorobenzene undergoing a two-step reaction sequence:

1. Reaction with NaOH, 623 K, 300 atm (Dow's Process):
Chlorobenzene reacts with aqueous sodium hydroxide under high temperature and high pressure to undergo nucleophilic aromatic substitution, followed by acidification to yield phenol (C₆H₅OH).

2. Reaction with conc. H₂SO₄ and then conc. HNO₃:
Phenol is treated first with concentrated H₂SO₄ (sulfonation) and then with concentrated HNO₃ (nitration). This preparation avoids extensive oxidation and directly nitrates the phenol ring at ortho- and para-positions to form 2,4,6-trinitrophenol, commonly known as picric acid as the major product Q.

Step 2: Determine the molecular formula of product Q (Picric Acid)

Picric acid is 2,4,6-trinitrophenol.

Structure: A benzene ring with an -OH group at position 1 and nitro groups (-NO₂) at positions 2, 4, and 6.

Molecular formula of Q = C₆H₃N₃O₇

Step 3: Calculate the molecular weight of Q

Using the given atomic masses:

• H = 1

• C = 12

• N = 14

• O = 16

Molar mass of C₆H₃N₃O₇:

$\text{Molar Mass}=(6\×12)+(3\×1)+(3\×14)+(7\×16)$

$\text{Molar Mass}=72+3+42+112=229\text{\ g/mol}$

Step 4: Calculate the weight percentage of Hydrogen in Q

$\text{Weight \% of H}=\left(\frac{\text{Total mass of H}}{\text{Molar mass of Q}}\right)\×100$

$\text{Weight \% of H}=\left(\frac{3}{229}\right)\×100=\frac{300}{229}\≈1.31\text{\%}$

Hence, the weight percentage of hydrogen in product Q is 1.31.