Mass of a condensation copolymer — JEE Advanced 2026 Paper 2 Q.14

01The question — verbatim

Section 3, numerical value — +4 for the correct value, 0 otherwise.

In the following reaction sequence, major products X and Y are acyclic monomers.

CH₃I —(1) KCN (2) H₃O⁺, Δ (3) Red P, Br₂ (4) NH₃ (excess)→ X
Caprolactam —H₃O⁺, Δ→ Y

500 mol of X completely reacts with 500 mol of Y to give 1 mol of a single biodegradable acyclic copolymer Z as the only product. The amount of Z formed in grams is ______.

Given — Atomic mass (in amu): H : 1, C : 12, N : 14, O : 16, Br : 80

Two words carry the whole question. Acyclic means the chain has two loose ends, so joining n units needs n − 1 links, not n. And a single copolymer means all 1000 units end up in one molecule — which is what makes the count 999 rather than 1000.

What each part of the question is for

What you are given	Value	What it is for
The four-step sequence from CH₃I	—	identifies X, and with it the molar mass 75
Caprolactam + H₃O⁺, Δ	—	identifies Y, and with it the molar mass 131
Amount of X	500 mol	half the monomer units
Amount of Y	500 mol	the other half
Amount of polymer formed	1 mol	how many chains the 1000 units are shared between — this is what fixes the number of links
Atomic masses	H 1, C 12, N 14, O 16	build every molar mass on the page
Atomic mass of Br	80	only for the bromo intermediate — no bromine survives into X, so it never touches the answer

02The experiment

Enter the amounts, then watch the chemistry happen: X is built from CH₃I one reagent at a time, caprolactam's ring is opened to give Y, and the two are polymerised while the water is counted off. The answer comes last.

1Enter the data 2Discover X 3Discover Y 4Polymerise 5The mass waiting for your data

Fill in the three amounts to arm the reactor.

03The two monomers

Filled in from your run. Every molar mass is added up from the atomic masses you entered — nothing here is a remembered number.

04Where it goes wrong

Four answers this question produces every year. Each is one wrong move, and the reactor above will reproduce all of them if you set it up that way.

85000 — losing 1000 waters

The commonest slip by far: one water per monomer instead of one per link. A chain of 1000 beads has 999 strings. Set the polymer amount to 0 chains — the reactor refuses, because a chain has to exist for its ends to be loose.

103000 — forgetting the water

Adding the monomer masses and stopping. Condensation polymerisation ejects a small molecule at every link; the polymer is always lighter than the monomers that made it.

94000 — using caprolactam as Y

Caprolactam is C₆H₁₁NO, M = 113. It is the cyclic starting material, and the question says the monomers are acyclic. Hydrolysis adds a water and opens the ring to give M = 131.

stuck — hunting for bromine

Br = 80 is given, so it feels needed. It is not: bromoacetic acid is an intermediate and every bromine leaves at the NH₃ step. Not every number in a question is a number you use.

05The record of the run

Filled in from your own run, quantity by quantity, as each one is established.

06Step-by-step solution

Run the experiment and the derivation opens by itself.

The solution is still locked

Enter the amounts, run the reactor to the end and reveal the answer — the derivation and the answer box unlock with it.

Follow the carbon to find X

Each reagent does one thing, and only one step changes the carbon count:

CH₃I ⟶ KCN CH₃CN → H₃O⁺, Δ CH₃COOH → red P / Br₂ BrCH₂COOH → NH₃ excess H₂N–CH₂–COOH

Cyanide displaces iodide and brings a carbon with it; hydrolysis takes the nitrile all the way to the acid; red P / Br₂ is Hell–Volhard–Zelinsky, which brominates the α carbon; excess ammonia then displaces that bromide.

X = glycine, C₂H₅NO₂
M = 2(12) + 5(1) + 14 + 2(16) = 75 g mol⁻¹
Open the ring to find Y

Caprolactam is a cyclic amide — a seven-membered ring of N, a carbonyl carbon and five CH₂ groups. Acidic hydrolysis cuts the C–N bond and adds the elements of water across it, and the ring falls open into a straight chain.

C₆H₁₁NO + H₂O → H₂N–(CH₂)₅–COOH

Y = 6-aminohexanoic acid, C₆H₁₃NO₂
M = 6(12) + 13(1) + 14 + 2(16) = 131 g mol⁻¹ (113 + 18, as it must be)
Count the links, not the units

1000 mol of monomer units end up in 1 mol of chains, so every chain is 1000 units long. A chain with two loose ends has one fewer link than it has units:

links = 1000 − 1 = 999 → 999 mol of H₂O eliminated

Every amide link is made by an –NH₂ and a –COOH condensing, and each one ejects exactly one water. A cyclic polymer would have lost 1000 — the word acyclic in the question is what makes it 999.
Weigh what went in and take off what left

mass in = 500 × 75 + 500 × 131 = 37 500 + 65 500 = 103 000 g

water out = 999 × 18 = 17 982 g

mass of Z = 103 000 − 17 982 = 85 018 g
Check it atom by atom

The same answer without any subtraction: count the atoms in 1 mol of chain.

C: 500(2) + 500(6) = 4000 · H: 500(5) + 500(13) − 999(2) = 7002
N: 500 + 500 = 1000 · O: 500(2) + 500(2) − 999 = 1001

4000(12) + 7002(1) + 1000(14) + 1001(16) = 85 018 g

Independent verification

Derived before anything was looked up. Both structures, both molar masses, the link count and the arithmetic were fixed from first principles, and only then compared with a published solution.
The carbon count tracked through the whole sequence. CH₃I has one carbon and glycine has two; the only step that adds carbon is the cyanide displacement. That alone forces X to be a two-carbon acid, independently of recognising the reagents.
Hydrolysis checked atom by atom, not just by name: C₆H₁₁NO + H₂O = C₆H₁₃NO₂ balances on every element, and 113 + 18 = 131 on the masses.
The answer computed twice, two different ways. Subtracting the water from the monomer mass gives 85 018 g; counting the atoms in one mole of chain (C₄₀₀₀H₇₀₀₂N₁₀₀₀O₁₀₀₁) and multiplying by the atomic masses gives 85 018 g to machine precision.
Conservation of mass checked explicitly: 103 000 g of monomer = 85 018 g of polymer + 17 982 g of water, exactly.
The model tested away from the paper's numbers. With one unit per chain no links form and no water is lost, and the mass reduces to the monomer mass — the formula degrades correctly at the boundary rather than by luck.
The distractors reproduced deliberately: 1000 waters gives 85 000; no water gives 103 000; caprolactam in place of Y gives 94 000. None of them is 85 018.
Answer key cross-check. A published worked solution for this paper identifies the same X and Y, uses the same 999 waters and reports the same 85 018. IIT Roorkee’s official key was not retrievable at the time of writing.
Headless-browser run of this page, driving the reactor from the empty panel to the reveal, asserting that the answer is nowhere on the page until it is revealed, that the molar masses are computed from the entered atomic masses rather than hard-coded, that impossible amounts are refused, and that the page never scrolls sideways at 390 px.

07Concepts and classification

What the question is made of, and where it sits in the syllabus.

What you need to know

R–X + CN⁻ → R–CN (one carbon longer)

R–CN ⟶{H₃O⁺} R–COOH

red P / Br₂ = Hell–Volhard–Zelinsky: α-bromination of an acid

α-bromo acid + NH₃ (excess) → α-amino acid

lactam + H₂O → amino acid (the ring opens)

n units in one acyclic chain → n − 1 condensations

mass of polymer = Σ (monomer masses) − (links) × 18

Classification

Exam	JEE (Advanced) 2026, Paper 2
Question	Q.14, Section 3 — numerical value
Marking	+4 for the correct value, 0 otherwise, no negative marking
Subject	Organic Chemistry
Chapter	Polymers
Topic	Condensation copolymer from a reaction sequence
Also needs	Nitrile route, HVZ reaction, lactam hydrolysis, mole arithmetic
Difficulty	Moderate–Hard — three separate places to slip
Answer	— run the experiment

01The question — verbatim

02The experiment

03The two monomers

04Where it goes wrong

05The record of the run

06Step-by-step solution

The solution is still locked

Follow the carbon to find X

Open the ring to find Y

Count the links, not the units

Weigh what went in and take off what left

Check it atom by atom

Independent verification

07Concepts and classification

What you need to know

Classification