Consider the following reaction sequence in which J, K, L and M are the major products.

Given:
Atomic mass (in amu): H : 1, C : 12, N : 14, O : 16, S : 32, Br : 80, Ba : 137

The volume of 1 M aqueous H₂SO₄ required to completely neutralize the ammonia evolved from 5.72 g of L in Kjeldahl’s method of nitrogen estimation is _____ mL.

The correct answer is 20 (or 20 mL).

Step 1: Determine the structures of J, K, and L from the given reaction sequence:

1. Formation of J:
Starting material is m-xylene (1,3-dimethylbenzene).
- Step 1: Friedel-Crafts acylation with 2-chloroacetyl chloride (Cl-CH₂-CO-Cl) in the presence of anhydrous AlCl₃ inserts an acyl group (-CO-CH₂-Cl) onto the aromatic ring (electrophilic substitution at the position ortho to one methyl and para to the other, position 4 of 1,3-dimethylbenzene).
- Step 2: Finkelstein reaction with NaI replaces -Cl with -I to form -CO-CH₂-I.
- Step 3: Nucleophilic substitution with sodium 3-nitrophenoxide (3-O₂N-C₆H₄-ONa) replaces -I with the phenoxy group -O-C₆H₄-3-NO₂.
Thus, J is 2-(3-nitrophenoxy)-1-(2,4-dimethylphenyl)ethan-1-one.

2. Formation of K:
- Step 1: Reduction of the carbonyl group in J using NaBH₄ yields the secondary alcohol (-CH(OH)-CH₂-O-C₆H₄-3-NO₂).
- Step 2: Reaction with PBr₃ replaces the hydroxyl group (-OH) with a bromine atom (-Br).
Thus, K is 4-(1-bromo-2-(3-nitrophenoxy)ethyl)-1,3-dimethylbenzene.
- Chemical Formula of K: C₁₆H₁₆BrNO₃
- Molar mass of K = 16(12) + 16(1) + 80 + 14 + 3(16) = 192 + 16 + 80 + 14 + 48 = 350 g/mol (which matches the value given in the diagram).

3. Formation of L:
- Reaction of K with excess ammonia (NH₃) undergoes nucleophilic substitution where the bromine atom (-Br) is replaced by an amino group (-NH₂).
Thus, L is 2-(2,4-dimethylphenyl)-2-(3-nitrophenoxy)ethan-1-amine (or its isomer having -NH₂ group replacing -Br).
- Chemical Formula of L: C₁₆H₁₈N₂O₃
- Molar Mass of L = 16(12) + 18(1) + 2(14) + 3(16) = 192 + 18 + 28 + 48 = 286 g/mol.

Step 2: Kjeldahl's method and Stoichiometry:

Each molecule of L contains 2 nitrogen atoms (one from the nitro group and one from the amine group). In standard Kjeldahl's method, nitrogen in organic compounds is converted quantitatively into ammonia gas (NH₃).
Therefore, 1 mole of compound L yields 2 moles of NH₃.

Moles of compound L in 5.72 g:

$\text{Moles of L}=\frac{5.72\text{g}}{286\text{g/mol}}=0.02\text{mol}$

Moles of NH₃ evolved:

$\text{Moles of}{\text{NH}}_3=2\times 0.02=0.04\text{mol}$

Step 3: Neutralization with H₂SO₄:

The neutralization reaction of ammonia with sulfuric acid is:

$2{\text{NH}}_3+{\text{H}}_2{\text{SO}}_4\to {\left({\text{NH}}_4\right)}_2{\text{SO}}_4$

From the reaction stoichiometry, 2 moles of NH₃ require 1 mole of H₂SO₄ for complete neutralization.

$\text{Moles of}{\text{H}}_2{\text{SO}}_4\text{required}=\frac{1}{2}\times \text{Moles of}{\text{NH}}_3=\frac{0.04}{2}=0.02\text{mol}$

Given the concentration of H₂SO₄ solution is 1 M:

$\text{Volume of 1 M}{\text{H}}_2{\text{SO}}_4\text{(in L)}=\frac{0.02\text{mol}}{1\text{M}}=0.02\text{L}$

Converting the volume into mL:

$\text{Volume (in mL)}=0.02\times 1000=20\text{mL}$