Kjeldahl estimation on L — JEE Advanced 2026 Paper 2 Q.17

01Original question — verbatim

JEE (Advanced) 2026, Paper 2, Chemistry, Section 4 — a question stem shared by Q.17 and Q.18. This simulation answers Q.17.

Question stem for question nos. 17 and 18

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

The scheme is redrawn here as live vector structures rather than reproduced from the paper, so the exam PDF itself is never published. Every structure is the same one the paper prints.

Given — atomic mass (amu)	Value	Needed for Q.17?
H	1	yes
C	12	yes
N	14	yes
O	16	yes
S	32	no — Q.18
Br	80	only to check the stem's M(K) = 350
Ba	137	no — Q.18

Q.17 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.

Two things have to be got right, and the second is the one that catches people. First the structure of L, which the stem lets you confirm rather than guess: whatever you propose for K has to weigh 350 g mol⁻¹. Second, which of L's nitrogens actually turns into ammonia — because L has two of them and Kjeldahl's digestion does not touch both.

Q.18, the mass of BaSO₄ from 3.79 g of M by the Carius method, is a separate question. This page builds M as far as identifying it and then leaves it alone.

02The Kjeldahl bench

The bench builds the compound first and weighs it to check itself against the stem's 350 g mol⁻¹, then asks you to commit to how many of L's nitrogens will come off as ammonia — and only then digests, distils and titrates.

00m-Xylene 01Build J 02Build K · weigh it 03Two routes 04Build L 05Nitrogen audit 06Digest · distil · titrate 07Answer waiting for your data

Bench — waiting for the atomic masses

t = 0.0 s

Nothing has been built yet. Check the data from the stem and press START EXPERIMENT.

The reagent shelf. Tap a bottle to see what it is for — the one in use lights up as the synthesis runs.

The atomic masses are pre-filled from the stem; the mass of L and the acid strength are the two quantities Q.17 gives.

03The nitrogen audit, and what the titre depends on

Kjeldahl's digestion turns nitrogen into ammonium sulphate — but only nitrogen it can reduce. Nitrogen in a nitro group, an azo group or an aromatic ring survives the digestion and is never counted. L happens to have one of each kind.

Explore a different mass of L 5.72 g

Your own weighing stays marked. The line is straight through the origin, which is why Kjeldahl works as a percentage-nitrogen method at any sample size.

mass explored

5.72 g

moles of L

— mol

titre needed

— mL

nitrogen by mass

— % N

04Record of the run

Each quantity appears only once the bench has established it, and every completed run is logged.

What this run established

Trial log

No runs yet. Change the mass or the acid strength and run it again — the titre follows both, and the structure of L never changes.

05Where this question goes wrong

Each card re-runs the bench with one mistaken assumption, on your own numbers.

06Step-by-step solution

Run the experiment and reveal the answer — the derivation opens with it.

The solution is still locked

The bench has not produced a titre yet. Run the experiment through to the end and press REVEAL ANSWER — this opens by itself at the same moment.

Step 1 — the only step that changes the carbon skeleton: Friedel–Crafts acylation

Chloroacetyl chloride with anhydrous AlCl₃ acylates m-xylene. Both methyls are o,p-directing, so the position that is ortho to one and para to the other is the one that reacts; the position between them is too hindered and the position meta to both is not activated. The product is a 2,4-dimethylphenacyl chloride.

m-C₆H₄(CH₃)₂ + ClCH₂COCl → Ar–CO–CH₂Cl (Ar = 2,4-dimethylphenyl)
Step 2 — Finkelstein exchange, then the phenoxide displaces the iodide

NaI with heat swaps chloride for iodide, which is the better leaving group for the step that follows. Sodium 3-nitrophenoxide then attacks the α-carbon in a plain S_N2 — an alkyl halide next to a carbonyl is unusually reactive towards substitution.

Ar–CO–CH₂Cl NaI Ar–CO–CH₂I NaO–C₆H₄–NO₂ J = Ar–CO–CH₂–O–C₆H₄–NO₂

J = C₁₆H₁₅NO₄, M = 192 + 15 + 14 + 64 = 285 g mol⁻¹
Step 3 — reduce, then brominate: and the stem hands you the check

NaBH₄ reduces the ketone to a secondary alcohol and leaves the nitro group and the ether alone. PBr₃ then replaces the hydroxyl with bromine.

J →_NaBH₄ Ar–CH(OH)–CH₂–O–Ar′ (C₁₆H₁₇NO₄, 287) →_PBr₃ K = Ar–CHBr–CH₂–O–Ar′

K = C₁₆H₁₆BrNO₃ : 192 + 16 + 80 + 14 + 48 = 350 g mol⁻¹ — exactly the molar mass the stem states, so the structure is confirmed rather than assumed
Step 4 — excess ammonia gives the amine, L

Ammonia displaces the bromide at the secondary benzylic carbon. The excess is what keeps it a primary amine instead of alkylating twice.

L = Ar–CH(NH₂)–CH₂–O–C₆H₄–NO₂ = C₁₆H₁₈N₂O₃, M = 192 + 18 + 28 + 48 = 286 g mol⁻¹

n(L) = 5.72 / 286 = 0.02 mol
Step 5 — the audit: L has two nitrogens and Kjeldahl counts one

Digestion with hot concentrated sulphuric acid converts nitrogen to ammonium sulphate, which is then made alkaline and distilled as ammonia. It only works on nitrogen that can be reduced to the −3 state under those conditions. Nitrogen in a nitro group — and equally in an azo group or an aromatic ring — is not converted, so it never appears as ammonia and is never estimated. L's amine nitrogen is counted; its nitro nitrogen is not.

n(NH₃) = n(L) × 1 = 0.02 mol (not 0.04)
Step 6 — the titration: sulphuric acid is dibasic

Two moles of ammonia are neutralised by one mole of H₂SO₄, because each sulphuric acid molecule supplies two protons.

2 NH₃ + H₂SO₄ → (NH₄)₂SO₄

n(H₂SO₄) = 0.02 / 2 = 0.01 mol ⟹ V = 0.01 / 1 = 0.01 L = 10 mL
What the two mistakes cost

Counting both nitrogens gives 20 mL. Forgetting that sulphuric acid is dibasic also gives 20 mL. Making both mistakes gives 40 mL — and, neatly, the two errors do not cancel, they compound. Only 10 mL comes from getting both right.

%N estimated = 1.4 × M_acid × V_mL × basicity ÷ mass = 1.4 × 1 × 10 × 2 / 5.72 = 4.90 % N — against 9.79 % nitrogen actually in the molecule

Independent verification

Derived first, checked afterwards. Every structure was built from the printed reagents alone — acylation, Finkelstein, S_N2, hydride reduction, bromination, ammonolysis — before any published solution was consulted.
The stem's own checksum. The independently constructed K, C₁₆H₁₆BrNO₃, comes to 350 g mol⁻¹ on the stem's atomic masses, which is exactly the figure the stem supplies. A wrong skeleton would not land on it, so this is a real structural confirmation and the simulation enforces it live.
Two more round numbers fall out. 5.72 g of L is 0.0200 mol exactly on M = 286, and — cross-checking the sibling question — 3.79 g of M (C₂₂H₂₁NO₃S, 379) is 0.0100 mol exactly, giving 2.33 g of BaSO₄. Three independent masses in the stem all resolve to clean amounts on these structures.
The Kjeldahl limitation, stated in the syllabus and load-bearing here. Nitrogen in nitro and azo groups and in rings is not converted to ammonium sulphate under Kjeldahl digestion. L was built with exactly one amine nitrogen and one nitro nitrogen, which is why the question carries the nitrophenoxide all the way through — and it is the single decision that separates 10 mL from 20 mL.
Stoichiometry checked both ways. 2 NH₃ + H₂SO₄ balances on N, H, S and O; equivalently the n-factor of ammonia is 1 and of sulphuric acid is 2, and equivalents match: 0.02 × 1 = 0.01 × 2.
Dimensions and scaling. mol ÷ (mol L⁻¹) = L, then ×1000 = mL. The titre is linear in the mass taken and inversely proportional to the acid strength: 11.44 g needs 20 mL, and 2 M acid needs 5 mL. The simulation reproduces both.
Distractor audit. Counting both nitrogens gives 20 mL; treating H₂SO₄ as monobasic gives 20 mL; doing both gives 40 mL; using M(K) = 350 in place of M(L) gives 8.17 mL; using M(J) = 285 gives 10.04 mL — close enough to 10 to be mistaken for it, which is why the reduction and bromination have to be tracked properly. None of them is 10 mL by the right route.
Independent numeric script (verify17.py, exact fractions): 23 of 23 checks pass, with every molar mass assembled atom by atom rather than recalled.
Headless-browser run of this page. The finished simulation was driven end to end in Chromium and asserted to reproduce 10 mL from the live model, to rebuild all four molar masses from the entered atomic masses, to flag it when the entered masses no longer reproduce the stem's 350, to keep the answer off the page until it is revealed, and to render with no console errors and no horizontal overflow at 390 px and 360 px.
Cross-check. Afterwards the answer was compared against a published worked solution for this paper, which identifies L with M = 286, states that one mole of L releases one mole of NH₃ under Kjeldahl's method, uses n-factors of 1 for ammonia and 2 for sulphuric acid, and reports the same 10 mL. IIT Roorkee's official key was not retrievable at the time of writing.

07What did we learn?

Four ideas carry this question. The cards fill their numbers in when the answer is revealed.

The four concepts this question turns on

CARD 01

A stated molar mass is a gift, not decoration

When a stem tells you a molar mass mid-sequence it is handing you a checksum. Build the structure your own way, add the atoms up, and if it matches you are on the right skeleton; if it does not, something earlier is wrong. Guessing is not required.

K = C₁₆H₁₆BrNO₃ ⟹ …

CARD 02

Kjeldahl does not see every nitrogen

Digestion reduces nitrogen to ammonium sulphate, so it only counts nitrogen it can reduce. Nitrogen in a nitro group, an azo group, or an aromatic ring comes through untouched and is never estimated. Put such a group in a molecule and the method under-reports on purpose.

L has 2 N, Kjeldahl counts 1 ⟹ …

CARD 03

Sulphuric acid is dibasic

Every H₂SO₄ molecule neutralises two ammonia molecules, so the moles of acid are half the moles of ammonia. In equivalents: n-factor 1 for NH₃, 2 for H₂SO₄, and equivalents must match.

2 NH₃ + H₂SO₄ → (NH₄)₂SO₄ ⟹ …

CARD 04

The whole chain, in one line

Mass to moles, moles to countable nitrogen, nitrogen to ammonia, ammonia to acid, acid to volume. Each arrow is one fact, and the only one that is not arithmetic is the second.

V = (w/M_L) × n_N,counted ÷ 2 ÷ c × 1000 ⟹ …

Formulas used

M(compound) = Σ (atomic mass × count) — every mass on this page is summed, not remembered

n(L) = w(L) / M(L)

n(NH₃) = n(L) × (nitrogens Kjeldahl can reduce)

2 NH₃ + H₂SO₄ → (NH₄)₂SO₄, so n(acid) = n(NH₃)/2

V = n(acid) / c, in litres; ×1000 for mL

%N = 1.4 × c × V(mL) × basicity ÷ w (the standard Kjeldahl form)

The 1.4 is 14 g mol⁻¹ × 100 ÷ 1000 — it is the molar mass of nitrogen hiding inside a percentage, not a magic constant.

Classification

Exam	JEE Advanced 2026 · Paper 2
Subject	Chemistry
Section	Section 4 · question stem for Q.17 and Q.18
Question	Q.17 — numerical value
Marking	+2 if correct, 0 otherwise
Chapter	Organic Chemistry — Some Basic Principles and Techniques
Topic	Kjeldahl estimation of nitrogen, on a compound built by a reaction sequence
Also draws on	Friedel–Crafts acylation · Finkelstein reaction · S_N2 · Hydride reduction · Mole concept · Acid–base titration
Difficulty	Hard
Answer	— run the experiment

01Original question — verbatim

Question stem for question nos. 17 and 18

02The Kjeldahl bench

Bench — waiting for the atomic masses

—

03The nitrogen audit, and what the titre depends on

04Record of the run

What this run established

Trial log

05Where this question goes wrong

06Step-by-step solution

The solution is still locked

Step 1 — the only step that changes the carbon skeleton: Friedel–Crafts acylation

Step 2 — Finkelstein exchange, then the phenoxide displaces the iodide

Step 3 — reduce, then brominate: and the stem hands you the check

Step 4 — excess ammonia gives the amine, L

Step 5 — the audit: L has two nitrogens and Kjeldahl counts one

Step 6 — the titration: sulphuric acid is dibasic

What the two mistakes cost

Independent verification

07What did we learn?

The four concepts this question turns on

A stated molar mass is a gift, not decoration

Kjeldahl does not see every nitrogen

Sulphuric acid is dibasic

The whole chain, in one line

Formulas used

Classification