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 $\text{H}_2\text{SO}_4$ required to completely neutralize the ammonia evolved from 5.72 g of L in Kjeldahl's method of nitrogen estimation is ____ mL.

JEE Advanced · 2026 · 17 May 2026 · Paper 2 · Practical and Analytical Chemistry

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

H_{2} SO_{4}

required to completely neutralize the ammonia evolved from 5.72 g of L in Kjeldahl's method of nitrogen estimation is ____ mL.

Answer: 10

Solution

Follow the sequence to L.

$p$ -Xylene + $ClC H_{2} COCl$ / anhyd. $AlC l_{3}$ (Friedel–Crafts acylation) gives the chloromethyl ketone $Ar - CO - C H_{2} Cl$ ( $Ar$ = 2,5-dimethylphenyl).
$NaI$ , heat (Finkelstein) converts it to $Ar - CO - C H_{2} I$ ; the 3-nitrophenoxide then displaces iodide ( $S_{N} 2$ ) to give J $= Ar - CO - C H_{2} - O - C_{6} H_{4} N O_{2}$ .
$NaB H_{4}$ reduces the ketone to the secondary alcohol and $PB r_{3}$ converts that to the bromide K $= Ar - CH (Br) - C H_{2} - O - C_{6} H_{4} N O_{2}$ , molar mass 350 g mol $^{- 1}$ (given).
Excess $N H_{3}$ replaces Br by $N H_{2}$ : L $= Ar - CH (N H_{2}) - C H_{2} - O - C_{6} H_{4} N O_{2}$ . Its molar mass is $350 - 80 (Br) + 16 (N H_{2}) = 286$ g mol $^{- 1}$ . (M, from $PhSNa$ , is the sulphur analogue.)

Kjeldahl's method converts only amino/amide nitrogen to

N H_{3}

; the nitrogen of a nitro group is *not* estimated. So each mole of L gives one mole of

N H_{3}

(from the

N H_{2}

group).

Moles of L

= \frac{5.72}{286} = 0.02

mol

\Rightarrow

moles of

N H_{3} = 0.02

mol.

2 N H_{3} + H_{2} S O_{4} \to (N H_{4})_{2} S O_{4} \Rightarrow moles of H_{2} S O_{4} = \frac{0.02}{2} = 0.01 mol

Volume of 1 M

H_{2} S O_{4}

= \frac{0.01}{1} \times 1000 = 10

mL.

Answer: 10.00

JEE Advanced · 2026 · 17 May 2026 · Paper 2 · Numerical answer

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JEE Advanced · 2026 · 17 May 2026 · Paper 2 · Practical and Analytical Chemistry

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