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

Intuition How to use this page

Cover every solution. Try the problem yourself first — pencil moving, guessing allowed — then reveal. Each level climbs one rung: from naming what you see, to building repeat units, to analysing structure, to inventing new polymers, to mastering the whole logic. The one trap that catches most students at each level is dissected right after.

Two notation habits used everywhere below, read these first:

::: in the self-tests means "cover the right-hand side". Everything left of ::: is the question; everything right is the answer you should try to say before uncovering. It is just a flash-card separator.
HOOC– is the carboxyl group written backwards. A carboxylic acid group is $- C O O H$ : a carbon double-bonded to one oxygen and single-bonded to an $- O H$ . When that same group sits at the left end of a molecule, chemists mirror-write it as $H O O C -$ so the reactive $- O H$ still points outward toward whatever it will react with. So $H O O C - (C H_{2})_{4} - C O O H$ is the same carboxyl on both ends — one flipped for the left, one normal on the right. Nothing new, just a left-hand spelling.

For the mass arithmetic below we use these atomic masses (g/mol): $H = 1, C = 12, N = 14, O = 16$ .

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

Look at the figure: the blue monomer and yellow monomer each lose an atom (the red water that pops out), and what is left — the block still holding one $- C O - N H -$ link — is bracketed in green. That green bracket is the repeat unit. Every worked solution that writes $[\dots]_{n}$ is naming exactly this green tile.

Level 1 — Recognition

L1·Q1

Classify each as polyamide, polyester, or 3-D cross-linked thermoset: nylon-6,6, PET, bakelite, Kevlar, melamine–formaldehyde, nylon-6.

Recall Solution

Nylon-6,6 → polyamide (amine + acid → amide link).
PET (terylene) → polyester (alcohol + acid → ester link).
Bakelite → 3-D cross-linked thermoset (phenol has 3 reactive sites).
Kevlar → polyamide (aromatic → aramid, but still $- C O - N H -$ ).
Melamine–formaldehyde → 3-D cross-linked thermoset (3 $- N H_{2}$ groups).
Nylon-6 → polyamide (from one monomer, caprolactam).

Mnemonic check: "PET is the polyEster; rest of the fibres are Amides; Bakelite + Melamine are the big meshes."

L1·Q2

For each linkage below, name it and name the two functional groups that made it: $- C O - N H - - C O - O -$

Recall Solution

$- C O - N H -$ is the amide link, from a carboxylic acid $(- C O O H)$ + an amine $(- N H_{2})$ . See Amide bond formation.
$- C O - O -$ is the ester link, from a carboxylic acid $(- C O O H)$ + an alcohol $(- O H)$ . See Esterification. In both, the small molecule spat out is $H_{2} O$ . The picture at the top of the page (the red water popping out) is exactly this event.

Level 2 — Application

L2·Q1

Adipic acid is $H O O C - (C H_{2})_{4} - C O O H$ . How many carbon atoms does it contain, and why does the name "nylon-6,6" use a 6 for it?

Recall Solution

Count all carbons: the two carboxyl carbons (one in each $C O O H$ — remember $H O O C -$ is just a flipped $- C O O H$ , so it has a carbon too) plus the four $C H_{2}$ carbons. $2 (from the two C O O H) + 4 (the (C H_{2})_{4}) = 6.$ So adipic acid is a 6-carbon diacid. The other monomer, hexamethylenediamine $H_{2} N (C H_{2})_{6} N H_{2}$ , is a 6-carbon diamine. Hence nylon-6,6 = (6-C diamine, 6-C diacid).

L2·Q2

Write the repeat unit of PET (terylene) from ethylene glycol $H O - C H_{2} C H_{2} - O H$ and terephthalic acid $H O O C - C_{6} H_{4} - C O O H$ , and state how many water molecules leave per repeat unit.

Recall Solution

Each ester link forms by acid- $O H$ + alcohol- $H$ leaving as water. One repeat unit has two ester links, so 2 water molecules leave per repeat unit: $[- O - C H_{2} C H_{2} - O - C O - C_{6} H_{4} - C O -]_{n}$ Read the skeletal diagram below rather than the flat text: the blue glycol block $(- O - C H_{2} C H_{2} - O -)$ and the yellow terephthalic block $(- C O - C_{6} H_{4} - C O -)$ alternate, joined at the two green ester links. The diol supplies the two $- O -$ oxygens; the diacid supplies the two $- C O -$ carbonyls. See Esterification.

Level 3 — Analysis

L3·Q1

When $n$ molecules of hexamethylenediamine react with $n$ molecules of adipic acid to make one linear nylon-6,6 chain, how many water molecules are released? Explain the " $- 1$ ".

Recall Solution

Set up the exact mapping first (the analogy made rigorous):

1 bead = 1 monomer. We have $n$ diamine beads + $n$ diacid beads $= 2 n$ beads total.
1 thread = 1 condensation reaction, and every condensation reaction makes exactly one amide link and ejects exactly one $H_{2} O$ . So threads = links = waters, one-for-one.

To string $2 n$ beads into one open line, count the gaps between neighbours: with $2 n$ beads in a row there are $2 n - 1$ gaps, hence $2 n - 1$ threads. Therefore: $waters = links = gaps = 2 n - 1 \Rightarrow (2 n - 1) H_{2} O$ The " $- 1$ " appears because the two chain ends stay unreacted — the beads at each end have one free hand (one un-threaded side). See the figure: 4 beads → 3 threads → 3 waters, and you can literally count the two free ends.

Why $2 n - 1 \approx 2 n$ for large $n$ (the Carothers idea): the degree of polymerisation is governed by the Carothers equation $\overset{ˉ}{X}_{n} = \frac{1}{1 - p},$ where $p$ is the fraction of functional groups that have actually reacted (the extent of conversion, $0 \leq p < 1$ ). Two lessons drop out:

Because real polymerisation is an equilibrium and never reaches $p = 1$ , a few end groups always stay free — that is the physical origin of the " $- 1$ ": open chains have unreacted ends. Only as $p \to 1$ do the ends become a negligible fraction, i.e. $2 n - 1 \to 2 n$ .
Chain length is brutally sensitive to conversion: $p = 0.99$ gives $\overset{ˉ}{X}_{n} = 100$ , but $p = 0.999$ gives $\overset{ˉ}{X}_{n} = 1000$ . To get long, strong nylon you must drive off the water hard (Le Chatelier) so more links form.

L3·Q2

Kevlar and nylon-6,6 are both polyamides with the same $- C O - N H -$ link. Yet Kevlar is far stiffer and stronger. Give two structural reasons, naming the concepts.

Recall Solution

Aromatic backbone. Kevlar's monomers (terephthalic acid + p-phenylenediamine) are built on benzene rings, which are flat and rigid. The chain is a straight rod, not a floppy $(C H_{2})_{6}$ string as in nylon → far less able to coil or stretch.
Extensive inter-chain Hydrogen bonding. Every amide $N - H$ donates a hydrogen bond to the $C = O$ of a neighbouring chain, and because the rigid rods pack into flat sheets, these H-bonds line up in dense, regular arrays. Many aligned H-bonds = very high tensile strength (stronger than steel per unit weight). Nylon-6,6 also H-bonds, but its flexible chains coil, so the bonds are fewer and less aligned.

Level 4 — Synthesis

L4·Q1

You are handed succinic acid $H O O C - (C H_{2})_{2} - C O O H$ (a 4-carbon diacid) and ethylenediamine $H_{2} N - (C H_{2})_{2} - N H_{2}$ (a 2-carbon diamine). Design the polymer: draw its repeat unit, name its family, and give it a "nylon-x,y" style name.

Recall Solution

Both are bi-functional (two reactive ends each) → they can alternate into a long chain. Acid + amine → amide link + water, so this is a polyamide (a nylon). Repeat unit (amine block, then acid block): $[- N H - (C H_{2})_{2} - N H - C O - (C H_{2})_{2} - C O -]_{n}$ The nylon-x,y naming rule, made explicit and visualised: by universal convention the first number = carbons in the diamine, the second number = carbons in the diacid, in that fixed order (amine, then acid). It is an ordered pair, like $(x, y)$ coordinates — swapping them would name a different polymer. The figure shows the two counted blocks side by side with the count arrows. Carbon counts: ethylenediamine has 2 carbons; succinic acid has $(C H_{2})_{2} + 2 C O O H$ carbons $= 2 + 2 = 4$ . Amine first, acid second → nylon-2,4. Two waters leave per repeat unit.

L4·Q2

Explain, using the idea of functionality (number of reactive groups per monomer), why swapping phenol (3 reactive sites) for a mono-substituted phenol with only 1 free reactive position would fail to give bakelite. What would you get instead?

Recall Solution

Bakelite's hardness comes from a 3-D cross-linked network, which requires monomers that can branch — i.e. functionality $\geq 3$ . Phenol offers three reactive ring positions (2 ortho + 1 para), so each phenol can bond to three neighbours → a net forms. With only one free reactive position, each such monomer can make at most one bridge — it acts as a chain-stopper (dead end). You cannot form a network; at best you'd get tiny molecules or short, non-cross-linked pieces — no infusible thermoset. This is the same reason monomers must be at least bi-functional to chain at all, and tri-functional to make thermosets.

Level 5 — Mastery

L5·Q1 (mass loss on polymerisation)

One repeat unit of nylon-6,6 is $[- N H (C H_{2})_{6} N H - C O (C H_{2})_{4} C O -]$ . Compute: (a) the molar mass of the two free monomers added together, (b) the molar mass of the repeat unit, (c) confirm the difference equals the mass of the water eliminated.

Recall Solution

Monomer masses. Hexamethylenediamine $H_{2} N (C H_{2})_{6} N H_{2} = C_{6} H_{16} N_{2}$ : $6 (12) + 16 (1) + 2 (14) = 72 + 16 + 28 = 116 g/mol .$ Adipic acid $H O O C (C H_{2})_{4} C O O H = C_{6} H_{10} O_{4}$ : $6 (12) + 10 (1) + 4 (16) = 72 + 10 + 64 = 146 g/mol .$ (a) Sum of free monomers $= 116 + 146 = 262 g/mol$ .

Repeat unit $C_{12} H_{22} N_{2} O_{2}$ (two waters, i.e. 2 $H_{2} O$ = $4 H + 2 O$ , are gone from the 262): $12 (12) + 22 (1) + 2 (14) + 2 (16) = 144 + 22 + 28 + 32 = 226 g/mol .$ (b) Repeat unit $= 226$ g/mol.

Difference $= 262 - 226 = 36$ g/mol $= 2 \times 18$ = two water molecules. ✔ This is the numerical fingerprint of a condensation polymer: repeat unit is lighter than its monomers by exactly the water eliminated. (Contrast with Addition polymers, where nothing is lost and the repeat unit equals the monomer mass.)

L5·Q2 (degree of polymerisation)

A nylon-6,6 sample has average molar mass $\approx 45, 200$ g/mol. Using the repeat-unit mass from Q1, estimate $n$ , the number of repeat units per chain.

Recall Solution

Each repeat unit weighs $226$ g/mol (the end-group waters are negligible for a long chain). $n \approx \frac{45200}{226} = 200.$ So on average $\approx 200$ repeat units per chain — meaning about $400$ monomers strung together. Large $n$ is exactly why the $(2 n - 1)$ from L3 is essentially $2 n$ : the two dangling ends are a rounding error. Via the Carothers equation $\overset{ˉ}{X}_{n} = 1/ (1 - p)$ , $n \approx 200$ repeat units ( $\approx 400$ reacted monomer units) corresponds to conversion $p$ so high that $1 - p \approx 1/400$ , i.e. $p \approx 0.9975$ — the water was driven off almost completely.

L5·Q3 (full-logic classification)

Melamine is $C_{3} N_{3} (N H_{2})_{3}$ . State (i) its functionality, (ii) the polymer family with formaldehyde, (iii) the physical consequence, and (iv) why it, like bakelite, will char rather than melt.

Recall Solution

(i) Functionality = 3 (three $- N H_{2}$ groups). (ii) Melamine + formaldehyde → a cross-linked network polymer (same architecture family as bakelite). (iii) A 3-D network is one giant covalently-bonded molecule; chains cannot slide past each other → hard, rigid, "unbreakable crockery", infusible. (iv) On heating, a thermoset can't melt because melting requires chains to flow, and a cross-linked net has no independent chains to flow. Instead of softening, the covalent network eventually breaks down chemically → it chars. This is the defining behaviour of thermosetting polymers.