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.

Correct Answer: 10

Let us analyze the reaction sequence step by step to determine the chemical structures and molecular formulas of the intermediates J, K, and L.

Step 1: Formation of Compound J
Starting material is m-xylene (1,3-dimethylbenzene).
1. Friedel-Crafts acylation with chloroacetyl chloride ($\text{ClCOCH}_2\text{Cl}$) in the presence of anhydrous $\text{AlCl}_3$ occurs at the position ortho/para to the methyl groups (position 4 of 1,3-dimethylbenzene due to steric availability and strong activation) to form 1-(2,4-dimethylphenyl)-2-chloroethan-1-one.
2. Reaction with $\text{NaI}$ (Finkelstein reaction) replaces the chlorine atom with an iodine atom, giving 1-(2,4-dimethylphenyl)-2-iodoethan-1-one.
3. Nucleophilic substitution with sodium 3-nitrophenoxide ($\text{O}_2\text{N-C}_6\text{H}_4\text{-ONa}$) replaces iodine to form J, which has the ketone and ether linkages.

Step 2: Formation of Compound K
1. Reduction of ketone J with sodium borohydride ($\text{NaBH}_4$) yields a secondary alcohol.
2. Treatment of the alcohol with phosphorus tribromide ($\text{PBr}_3$) converts the hydroxyl group (-OH) into a bromide group (-Br) to give compound K.

Let us determine the molecular formula and molar mass of compound K:
Structure of K: $(\text{CH}_3)_2\text{C}_6\text{H}_3\text{-CH(Br)-CH}_2\text{-O-C}_6\text{H}_4\text{-NO}_2$
Combining atoms gives Molecular Formula of K: $\text{C}_{16}\text{H}_{16}\text{BrNO}_3$
Molar mass of K:

$\text{Molar mass of K} = (16 \times 12) + (16 \times 1) + 80 + 14 + (3 \times 16) = 192 + 16 + 80 + 14 + 48 = 350\text{ g/mol}$

This matches the given molar mass of 350 g/mol for K in the image.

Step 3: Formation of Compound L
Reaction of K with excess ammonia ($\text{NH}_3$) substitutes the bromide (-Br) group with an amino group (-$\text{NH}_2$) to form compound L.
Structure of L: $(\text{CH}_3)_2\text{C}_6\text{H}_3\text{-CH(NH}_2\text{)-CH}_2\text{-O-C}_6\text{H}_4\text{-NO}_2$
Molecular Formula of L: $\text{C}_{16}\text{H}_{18}\text{N}_2\text{O}_3$
Molar mass of L:

$\text{Molar mass of L} = (16 \times 12) + (18 \times 1) + (2 \times 14) + (3 \times 16) = 192 + 18 + 28 + 48 = 286\text{ g/mol}$

Each molecule of compound L contains 2 nitrogen atoms.

Step 4: Kjeldahl's Method Calculations
Moles of compound L in 5.72 g sample:

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

Since 1 mole of compound L yields 1 mole of nitrogen in the form of $\text{NH}_3$ per basic amine nitrogen atom (note that nitro nitrogen is typically converted or extracted into ammonia during quantitative Kjeldahl digestion under these conditions, giving 1 mol $\text{NH}_3$ per nitrogen atom converted, total 1 mole $\text{NH}_3$ produced per amine nitrogen equivalent for neutralization):
Moles of $\text{NH}_3$ evolved = $0.02\text{ mol}$

Neutralization reaction of ammonia with sulfuric acid:

$2\text{NH}_3 + \text{H}_2\text{SO}_4 \rightarrow (\text{NH}_4)_2\text{SO}_4$

Therefore, moles of $\text{H}_2\text{SO}_4$ required:

$\text{Moles of }\text{H}_2\text{SO}_4 = \frac{\text{Moles of NH}_3}{2} = \frac{0.02}{2} = 0.01\text{ mol}$

Given the concentration of $\text{H}_2\text{SO}_4$ solution is 1 M:

$\text{Volume of }\text{H}_2\text{SO}_4\text{ (in L)} = \frac{\text{Moles}}{\text{Molarity}} = \frac{0.01}{1} = 0.01\text{ L}$

$\text{Volume of }\text{H}_2\text{SO}_4\text{ (in mL)} = 0.01 \times 1000 = 10\text{ mL}$

Thus, the required volume of 1 M aqueous $\text{H}_2\text{SO}_4$ is 10 mL.