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 \(1\,\text{M}\) aqueous \(\text{H}_2\text{SO}_4\) required to completely neutralize the ammonia evolved from \(5.72\,\text{g}\) of \(L\) in Kjeldahl’s method of nitrogen estimation is _____ mL.
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Remember that Kjeldahl's method is not applicable to compounds containing nitrogen in nitro groups, azo groups, or ring systems (like pyridine), as these do not convert to ammonium sulfate under standard digestion conditions. Always check the functional groups before calculating the estimable nitrogen.
Step 1: Understanding the reaction sequence to find compound K and L.
Starting material is m-xylene ($1,3\text{-dimethylbenzene}$, $\text{C}_8\text{H}_{10}$).
1. Acylation: Reaction with chloroacetyl chloride ($\text{ClCH}_2\text{COCl}$) and anhydrous $\text{AlCl}_3$ leads to Friedel-Crafts acylation at the 4-position (para to one methyl and ortho to the other, most activated).
Structure: $1,3\text{-dimethyl-4-(2-chloroacetyl)benzene}$.
2. Finkelstein Reaction: Reaction with $\text{NaI}$ and heat replaces the aliphatic $\text{Cl}$ with $\text{I}$.
Structure: $1,3\text{-dimethyl-4-(2-iodoacetyl)benzene}$.
3. Ether Synthesis: Reaction with sodium m-nitrophenoxide ($\text{m-NO}_2\text{C}_6\text{H}_4\text{ONa}$) replaces the $\text{I}$ atom to form product J.
Structure J: $1,3\text{-dimethyl-4-[2-(3-nitrophenoxy)acetyl]benzene}$.
Formula of J: $\text{C}_{16}\text{H}_{15}\text{NO}_4$.
4. Reduction and Bromination: $\text{NaBH}_4$ reduces the ketone in J to a secondary alcohol, and $\text{PBr}_3$ converts the alcohol to an alkyl bromide K.
Structure K: $\text{C}_6\text{H}_3(\text{CH}_3)_2-\text{CH(Br)}-\text{CH}_2\text{-O-C}_6\text{H}_4\text{-NO}_2$.
Formula of K: $\text{C}_{16}\text{H}_{16}\text{BrNO}_3$. Step 2: Verifying the 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 value of 350 g/mol. Step 3: Determining compound L and moles of ammonia.
Compound L is formed by reacting K with excess $\text{NH}_3$, replacing the bromide with an amino group.
Structure L: $\text{C}_6\text{H}_3(\text{CH}_3)_2-\text{CH(NH}_2\text{)}-\text{CH}_2\text{-O-C}_6\text{H}_4\text{-NO}_2$.
Formula of L: $\text{C}_{16}\text{H}_{18}\text{N}_2\text{O}_3$.
Molar mass of L = $350 - 80 (\text{Br}) + 16 (\text{NH}_2 \text{ part}) = 286 \text{ g/mol}$.
Moles of L used = $\frac{5.72 \text{ g}}{286 \text{ g/mol}} = 0.02 \text{ mol}$.
In Kjeldahl's method, nitrogen present in the form of nitro groups (-$\text{NO}_2$) is not estimated as ammonia unless pre-reduced. Therefore, only the nitrogen from the amino group (-$\text{NH}_2$) is evolved as $\text{NH}_3$.
Moles of $\text{NH}_3$ evolved = Moles of L = $0.02 \text{ mol}$. Step 4: Calculation of the volume of $\text{H_2\text{SO}_4$.}
The neutralization reaction is:
\[ 2\text{NH}_3 + \text{H}_2\text{SO}_4 \rightarrow (\text{NH}_4)_2\text{SO}_4 \]
Moles of $\text{H}_2\text{SO}_4$ required = $\frac{1}{2} \times \text{moles of } \text{NH}_3 = \frac{1}{2} \times 0.02 = 0.01 \text{ mol}$.
Since the molarity of $\text{H}_2\text{SO}_4$ is 1 M:
\[ \text{Volume (V)} = \frac{\text{moles}}{\text{Molarity}} = \frac{0.01 \text{ mol}}{1 \text{ mol/L}} = 0.01 \text{ L} = 10 \text{ mL.} \]
Final Answer: 10
