Animated Solution for Chemistry - Organic Chemistry: Comprehension Passage

Welcome to a beautiful organic chemistry puzzle that perfectly blends multi-step synthesis with quantitative analysis! This problem is a classic JEE trap, testing not just your ability to predict products, but also your awareness of the subtle limitations of analytical methods. Let's decode this journey from $m$-xylene to our final products.

From $m$-Xylene to Compound K

We begin with $m$-xylene. The first step is a Friedel-Crafts Acylation using chloroacetyl chloride (${\text{Cl-CO-CH}}_2\text{-Cl}$) and anhydrous ${\text{AlCl}}_3$. This attaches the $-{\text{CO-CH}}_2\text{-Cl}$ group to the aromatic ring. Next, a Finkelstein reaction with $\text{NaI}$ swaps the chlorine atom for an iodine atom. Finally, an ${\text{S}}_{\text{N}}2$ attack by sodium $p$-nitrophenoxide replaces the iodine, yielding compound J.

Compound J is then subjected to reduction and bromination. Sodium borohydride (${\text{NaBH}}_4$) selectively reduces the ketone carbonyl ($\text{C=O}$) to a secondary alcohol ($\text{CH-OH}$). Following this, phosphorus tribromide (${\text{PBr}}_3$) converts the alcohol into an alkyl bromide ($\text{CH-Br}$), giving us compound K. The problem generously provides the molar mass of K as $350\text{ g/mol}$, which is our anchor for all subsequent calculations.

The Formation of L and M

From compound K, the reaction pathway splits into two parallel ${\text{S}}_{\text{N}}2$ reactions:

Path 1 (Formation of L): Reacting K with excess ammonia (${\text{NH}}_3$) replaces the bromine atom with an amino group ($-{\text{NH}}_2$). To find the molar mass of L, we simply take the mass of K, subtract the mass of bromine, and add the mass of the amino group:

Path 2 (Formation of M): Reacting K with sodium thiophenoxide ($\text{PhSNa}$) replaces the bromine atom with a phenylthio group ($-\text{SPh}$). The mass of the phenylthio group is $32+(6\times 12)+5=109$. Thus, the molar mass of M is:

Now we face the first quantitative challenge: estimating nitrogen in compound L using Kjeldahl's method. Compound L contains two nitrogen atoms—one in the newly added $-{\text{NH}}_2$ group, and one in the original $-{\text{NO}}_2$ group.

Here is the critical catch: Kjeldahl's method cannot estimate nitrogen present in nitro ($-{\text{NO}}_2$) groups, azo groups, or within aromatic rings (like pyridine). Under the reaction conditions, these nitrogens do not convert to ammonium sulfate. Therefore, only the amine nitrogen in compound L will evolve as ammonia (${\text{NH}}_3$).

This means $1\text{ mole}$ of L produces exactly $1\text{ mole}$ of ${\text{NH}}_3$. Given $5.72\text{ g}$ of L: This yields $0.02\text{ mol}$ of ${\text{NH}}_3$.

To neutralize this ammonia, we use sulfuric acid (${\text{H}}_2{\text{SO}}_4$). Since sulfuric acid is diprotic, the neutralization reaction is: Thus, $0.02\text{ mol}$ of ${\text{NH}}_3$ requires $0.01\text{ mol}$ of ${\text{H}}_2{\text{SO}}_4$. For a $1\text{ M}$ solution, the required volume is:

For the second question, we use the Carius method to estimate sulfur in compound M. In this method, all the sulfur in the organic compound is oxidized and precipitated as barium sulfate (${\text{BaSO}}_4$).

Compound M contains exactly one sulfur atom. Therefore, $1\text{ mole}$ of M will yield $1\text{ mole}$ of ${\text{BaSO}}_4$. Given $3.79\text{ g}$ of M: This will produce $0.01\text{ mol}$ of ${\text{BaSO}}_4$. The molar mass of ${\text{BaSO}}_4$ is $137+32+64=233\text{ g/mol}$.

This problem beautifully demonstrates why you must always be vigilant about the limitations of experimental methods. Keep practicing, and these conceptual traps will become your greatest strengths!