Condensation polymers — nylon-6,6, nylon-6, terylene (PET), bakelite, melamine, Kevlar: Worked examples

This is the practice companion to the parent topic. Here we do not learn new theory — we exhaust it. We will find every kind of question this topic can ask, then solve one of each so that no exam scenario can surprise you.

Before we begin, one tool we will lean on constantly is mass bookkeeping. When two molecules join and spit out water, mass is conserved: the mass that goes in equals the mass of the polymer chain plus the mass of every water molecule that left. That single idea (nothing vanishes, it just leaves as $H_{2} O$ ) is the engine behind half of these examples.

The scenario matrix

Every question in this topic falls into one of these cells. Our examples below are labelled with the cell they hit, and together they touch every row.

Cell	Case class	What makes it tricky	Covered by
A	Identify linkage from structure	amide vs ester look alike	Ex 1
B	Count carbons / decode the name	the "6,6" trap	Ex 2
C	Water-molecules bookkeeping	is it $n$ or $2 n - 1$ ?	Ex 3
D	Mass / degree-of-polymerisation calc	subtract the water!	Ex 4
E	Bi- vs tri-functional → chain vs 3-D net	why some melt, some don't	Ex 5
F	Degenerate / limiting input	mono-functional cap, single monomer	Ex 6
G	Real-world word problem	pick the right polymer	Ex 7
H	Exam-style twist / odd-one-out	mixed classification	Ex 8

Example 1 — Cell A: read the linkage off a structure

Forecast: guess before reading — which one is the ester?

Look at the atom sitting after the carbonyl $C = O$ . Why this step? The whole difference between the two families is what the acid's carbon bonded to. If it bonded to nitrogen ( $N$ ) it is an amide; if to oxygen ( $O$ ) it is an ester.
Fragment $- C O - N H -$ : carbon → nitrogen → amide link. Amides come from carboxylic acid + amine (see Amide bond formation).
Fragment $- C O - O -$ : carbon → oxygen → ester link. Esters come from carboxylic acid + alcohol (see Esterification).

Verify: count what left. In both cases the acid donated $- O H$ and the partner donated $- H$ , so both released $H_{2} O$ . The only difference is the partner's other atom ( $N$ vs $O$ ) — exactly what we read off. ✓

Example 2 — Cell B: decode "nylon-6,6" without falling in the trap

Forecast: the tempting wrong answer is "4". Why is it wrong?

Write every carbon. Why this step? You cannot count what you don't see. Adipic acid = $H O O C - (C H_{2})_{4} - C O O H$ . The two $H O O C$ groups each carry one carbon (the carboxyl carbon), and $(C H_{2})_{4}$ holds four.
Total $= 1 + 4 + 1 = 6$ carbons. Why this matters: the "4" only counts the middle $C H_{2}$ units; the two carboxyl carbons are still carbons.
Hexamethylenediamine $H_{2} N (C H_{2})_{6} N H_{2}$ has $(C H_{2})_{6} = 6$ carbons.
So nylon-6,6 = (6 carbons in the diamine, 6 carbons in the diacid). The "6"s count carbons per monomer, not amide links.

Verify: the two numbers are equal (6 and 6), matching the symmetric name "6,6". If the acid part were truly 4 carbons the polymer would be called nylon-6,4 — it is not. ✓

Example 3 — Cell C: how many water molecules for $n$ repeat units?

Forecast: is it $n$ , $2 n$ , or $2 n - 1$ ?

Count the amide links being made. Why this step? Each condensation link releases exactly one water. So (water molecules) = (links formed).
Each repeat unit contains two amide links (one diamine-to-diacid on each side). $n$ repeat units, if joined into one closed loop, would need $2 n$ links.
But the chain has two open ends (it is a line, not a ring). A line of $n$ repeat units has one fewer junction than a closed loop. Why this matters: the two end groups stay unreacted ( $- N H_{2}$ and $- C O O H$ dangling), so one link fewer is made.
Water released $= 2 n - 1$ .

Verify: test $n = 1$ . One repeat unit = one diamine + one diacid making $2 (1) - 1 = 1$ water. Check: joining them at one end forms one amide, releases one water, leaving a dimer with free ends. ✓ Matches the parent-note formula $(2 n - 1) H_{2} O$ .

Example 4 — Cell D: mass bookkeeping (subtract the water!)

Forecast: will it be exactly $100 \times (116 + 146)$ , or less?

Add the raw monomer masses. Why this step? Everything that enters the chain came from monomers. $100 \times 116 + 100 \times 146 = 11600 + 14600 = 26200 g/mol$ .
Count water lost. Why this step? Condensation removes mass as water; the chain must weigh less than the sum. Links formed $= 2 n - 1 = 2 (100) - 1 = 199$ waters.
Subtract. Mass of water lost $= 199 \times 18 = 3582 g/mol$ .
Chain mass $= 26200 - 3582 = 22618 g/mol$ .

Verify: the answer must be smaller than 26200 — and it is (22618). Sanity ratio: water lost is $3582/26200 \approx 13.7%$ of the raw mass, believable for small monomers. This "lighter than the sum of monomers" is exactly the condensation signature the parent note names. ✓

Example 5 — Cell E: two holes vs three holes → melts or not

Forecast: which one will you never be able to melt?

Figure — Condensation polymers — nylon-6,6, nylon-6, terylene (PET), bakelite, melamine, Kevlar

Give P two "hands". Why this step? Two reactive ends let a monomer bond on both sides — like a bead with two holes threading a string (left panel of the figure). This grows a linear chain.
A linear chain can slide past its neighbours when heated → it softens and melts → thermoplastic (see Thermosetting vs thermoplastic polymers). Nylons and PET live here.
Give Q three "hands". Why this step? A third bonding site lets the chain branch sideways and stitch to neighbouring chains, building a 3-D cross-linked net (right panel).
A single giant net has no separate chains to slide → it cannot melt; it chars → thermosetting. Bakelite and melamine live here.

Verify: cross-check against the parent note — it lists phenol (3 reactive sites) → bakelite and melamine (three $- N H_{2}$ ) → both thermosets. Both have functionality $\geq 3$ , exactly our rule. Bi-functional nylons/PET are thermoplastic. ✓

Example 6 — Cell F: degenerate inputs (a cap, and a single monomer)

Forecast: does one reactive group help or hurt growth?

(a) A mono-functional molecule has only one hand. Why this step? Once it bonds, its single site is used up — it becomes a dead end (a chain cap). Limiting behaviour: the more caps you add, the shorter the average chain, because growth stops wherever a cap lands. So mono-functional = chain terminator; you cannot build long chains from them alone.
(b) Nylon-6's monomer is caprolactam, a 7-membered ring already holding both an $- N H$ and a $C = O$ . Why this step? A single monomer can polymerise if it carries both kinds of reactive group at once — one on each "end". It doesn't need a partner.
The ring opens and self-links head-to-tail: the $- N H$ of one attacks the $C = O$ of the next. This is ring-opening polymerisation, giving $[- N H - (C H_{2})_{5} - C O -]_{n}$ .

Verify: count carbons in the repeat unit $- N H - (C H_{2})_{5} - C O -$ : five from $(C H_{2})_{5}$ plus one carbonyl carbon $= 6$ . This matches the "6" in nylon-6. ✓ And degenerate case (a) matches real industry: mono-functional acids are used deliberately to control (cap) molecular weight.

Example 7 — Cell G: real-world word problem

Forecast: two of these are polyamides — but only one is a thermoset. Which?

(i) Plates need hardness + heat resistance. Why this step? Plates get hot and dropped, so we need a rigid 3-D net (won't melt, won't crack). → Melamine–formaldehyde (three $- N H_{2}$ → cross-linked thermoset).
(ii) Vest needs extreme tensile strength per weight. Why this step? Stopping a bullet needs stiff, hydrogen-bonded chains. → Kevlar — rigid aromatic polyamide with extensive inter-chain Hydrogen bonding.
(iii) Bottle needs a clear, drawable, recyclable thermoplastic. Why this step? Bottles are blow-moulded from a meltable clear polyester. → PET (terylene), the polyester of ethylene glycol + terephthalic acid.

Verify: each choice matches the parent note's stated uses — melamine → unbreakable crockery, Kevlar → bulletproof vests, PET → plastic bottles. Classification check: (i) thermoset, (ii) & (iii) thermoplastic; (ii) amide, (iii) ester. All internally consistent. ✓

Example 8 — Cell H: exam twist (odd-one-out + classification)

Forecast: three separate sorts — don't mix the axes.

(a) Polyester = ester link = acid + alcohol. Why this step? Only PET pairs an alcohol (ethylene glycol) with an acid. → PET (the odd one out among these polyamides/nets).
(b) Thermoset = tri-functional → 3-D net. Why this step? Only phenol (3 sites) and melamine (3 $- N H_{2}$ ) cross-link. → Bakelite and Melamine.
(c) Electrophilic aromatic substitution. Why this step? Bakelite's first step attaches $- C H_{2} O H$ onto phenol's ring at ortho/para positions — a textbook Electrophilic aromatic substitution (formaldehyde acts as electrophile). → Bakelite.

Verify: cross-tabulate — PET (ester, thermoplastic), Nylon-6,6 & Kevlar (amide, thermoplastic), Bakelite & Melamine (network, thermoset). No polymer lands in two conflicting boxes; the three questions probe three independent axes and each has a clean answer. ✓

Recall Quick self-test (cover the answers)

Water released for $n = 3$ repeat units of a two-link-per-unit polyamide? ::: $2 (3) - 1 = 5$ .
Carbons in hexamethylenediamine? ::: 6.
Is a mono-functional monomer a chain-builder or a chain-capper? ::: Capper (dead end).
The only polyester among nylon/PET/Kevlar/bakelite/melamine? ::: PET.
Which forms via ring-opening from one monomer? ::: Nylon-6 (caprolactam).