Animated Solution for Chemistry - Organic Chemistry: The weight percentage of hydrogen in $\mathbf{Q}$, formed in the following reaction sequence, is ______. [Given : Atomic mass of $\text{H}=1$, $\text{C}=12$, $\text{N}=14$, $\text{O}=16$, $\text{S}=32$, $\text{Cl}=35$]

This problem is a beautiful journey through a classic multi-step organic synthesis, testing both your knowledge of reaction mechanisms and your precision in basic stoichiometry. Let's break down the transformation of chlorobenzene into our final product, $\mathbf{Q}$, and then calculate the required mass percentage.

Breaking the Tough C-Cl Bond

We start with chlorobenzene. Normally, aryl halides are notoriously unreactive towards nucleophilic aromatic substitution. The lone pairs on the chlorine atom delocalize into the benzene ring through resonance, giving the carbon-chlorine bond a partial double-bond character.

However, the first step employs Dow's Process. By using aqueous $\text{NaOH}$ under extreme conditions—a temperature of $623\text{ K}$ and a pressure of $300\text{ atm}$—we force the substitution to happen. The strong nucleophile ${\text{OH}}^{-}$ replaces the chlorine atom. But there is a catch! Because the reaction medium is highly basic, the acidic phenol formed immediately reacts with $\text{NaOH}$ to yield its salt, sodium phenoxide (${\text{C}}_6{\text{H}}_5\text{ONa}$).

Acidification and Exhaustive Nitration

In the second step, we introduce concentrated ${\text{H}}_2{\text{SO}}_4$ followed by concentrated ${\text{HNO}}_3$.

First, the sulfuric acid acts as a proton source, converting the phenoxide ion back into phenol.

Next, the mixture of concentrated ${\text{HNO}}_3$ and ${\text{H}}_2{\text{SO}}_4$ acts as a powerful nitrating agent, generating the electrophilic nitronium ion (${\text{NO}}_2^{+}$). The $-\text{OH}$ group on phenol is a strongly activating group due to its powerful $+M$ (mesomeric) effect. It pumps electron density into the ring, particularly at the ortho and para positions. Because the ring is so electron-rich, and we are using concentrated reagents, the nitration doesn't stop at mono-substitution. It undergoes exhaustive nitration to form 2,4,6-trinitrophenol, universally known as Picric acid. This is our major product, $\mathbf{Q}$.

To find the weight percentage of hydrogen in Picric acid, we must first determine its exact molecular formula.

Let's count the atoms carefully: - The benzene ring provides $6$ carbon atoms. - There are $3$ nitro ($-{\text{NO}}_2$) groups, contributing $3$ nitrogen and $6$ oxygen atoms. - The hydroxyl ($-\text{OH}$) group contributes $1$ oxygen and $1$ hydrogen atom. - Finally, there are only $2$ hydrogen atoms left on the benzene ring (at the meta positions).

Summing these up, the molecular formula is ${\text{C}}_6{\text{H}}_3{\text{N}}_3{\text{O}}_7$.

Now, let's calculate the molar mass ($M$) using the given atomic weights:

The total mass of hydrogen in one mole of Picric acid is $3\text{ g}$. The weight percentage of hydrogen is simply the mass of hydrogen divided by the total molar mass, multiplied by $100$:

Thus, the weight percentage of hydrogen in the final product is $1.31\%$.