Volume of Sulfuric Acid Required in Kjeldahl Estimation of Nitrogen

To determine the volume of $1\text{ M}$ aqueous $\text{H}_2\text{SO}_4$ required to neutralize the ammonia evolved from compound L, we first deduce the chemical structures and molecular formulas of the compounds in the reaction sequence.

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Step 1: Identification of Reaction Sequence Products

1. Formation of Compound J:

   • Friedel-Crafts Acylation: Reaction of $m$-xylene ($1,3$-dimethylbenzene) with chloroacetyl chloride ($\text{Cl-CH}_2-\text{CO-Cl}$) in the presence of anhydrous $\text{AlCl}_3$ yields $2$-chloro-$1$-($2,4$-dimethylphenyl)ethan-$1$-one as the major electrophilic substitution product at the $4$-position.
   • Finkelstein Reaction: Treatment with $\text{NaI}$ and heat substitutes the chlorine atom with iodine to yield $2$-iodo-$1$-($2,4$-dimethylphenyl)ethan-$1$-one.
   • Nucleophilic Substitution: Reaction with sodium $3$-nitrophenoxide ($\text{NaO}-\text{C}_6\text{H}_4-3\text{-NO}_2$) substitutes the iodine to give product J: $\mathbf{J} = (2,4\text{-(CH}_3)_2\text{C}_6\text{H}_3)-\text{C}(=\text{O})-\text{CH}_2-\text{O}-\text{C}_6\text{H}_4(3\text{-NO}_2)$

2. Formation of Compound K:

   • Reduction of the carbonyl group of J using $\text{NaBH}_4$ forms a secondary alcohol.
   • Treatment with $\text{PBr}_3$ converts the alcohol into a benzylic bromide, yielding K: $\mathbf{K} = (2,4\text{-(CH}_3)_2\text{C}_6\text{H}_3)-\text{CH}(\text{Br})-\text{CH}_2-\text{O}-\text{C}_6\text{H}_4(3\text{-NO}_2)$

   Verification of Molar Mass of K:

   • Molecular formula of $\mathbf{K}$: $\text{C}_{16}\text{H}_{16}\text{BrNO}_3$
   • Molar Mass of $\mathbf{K} = (16 \times 12) + (16 \times 1) + 80 + 14 + (3 \times 16) = 192 + 16 + 80 + 14 + 48 = 350\text{ g/mol}$ (Matches the given molar mass).

3. Formation of Compound L:

   • Nucleophilic substitution of the bromide group in K with excess ammonia ($\text{NH}_3$) yields the primary amine L: $\mathbf{L} = (2,4\text{-(CH}_3)_2\text{C}_6\text{H}_3)-\text{CH}(\text{NH}_2)-\text{CH}_2-\text{O}-\text{C}_6\text{H}_4(3\text{-NO}_2)$

   Molar Mass of L:

   • Molecular formula of $\mathbf{L}$: $\text{C}_{16}\text{H}_{18}\text{N}_2\text{O}_3$
   • Molar Mass of $\mathbf{L} = (16 \times 12) + (18 \times 1) + (2 \times 14) + (3 \times 16) = 192 + 18 + 28 + 48 = 286\text{ g/mol}$

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Step 2: Estimation of Ammonia Evolved via Kjeldahl's Method

In Kjeldahl's method:

• Nitrogen present in nitro groups ($-\text{NO}_2$) is not converted into ammonium sulfate ($\text{NH}_3$) under standard digestion conditions.
• Only the amine nitrogen ($-\text{NH}_2$) is quantitatively converted to ammonia ($\text{NH}_3$).

Thus, $1\text{ mole}$ of compound L yields $1\text{ mole}$ of $\text{NH}_3$.

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

$\text{Moles of } \text{NH}_3 \text{ evolved} = 0.02\text{ mol}$

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Step 3: Neutralization with $\text{H}_2\text{SO}_4$

The neutralization reaction between ammonia and sulfuric acid is given by: $2\text{NH}_3 + \text{H}_2\text{SO}_4 \rightarrow (\text{NH}_4)_2\text{SO}_4$

• Moles of $\text{H}_2\text{SO}_4$ required: $\text{Moles of } \text{H}_2\text{SO}_4 = \frac{1}{2} \times \text{Moles of } \text{NH}_3 = \frac{0.02}{2} = 0.01\text{ mol}$

• Volume of $1\text{ M}$ $\text{H}_2\text{SO}_4$ required: $\text{Volume (in L)} = \frac{\text{Moles}}{\text{Molarity}} = \frac{0.01\text{ mol}}{1\text{ M}} = 0.01\text{ L} = 10\text{ mL}$

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Final Answer

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