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

We will use only one idea over and over: two ends meet, a small piece is thrown away, a new bond stays. Everything below is that idea, made visual.

Step 1 — Meet the two players and their "hands"

WHAT: We introduce the two monomers of nylon-6,6.

Hexamethylenediamine: $H_{2} N - (C H_{2})_{6} - N H_{2}$ . The two hands are $- N H_{2}$ groups (called amine groups). The middle $(C H_{2})_{6}$ is just a chain of 6 carbons — an inert "handle", no reaction happens there.
Adipic acid: $H O O C - (C H_{2})_{4} - C O O H$ . The two hands are $- C O O H$ groups (carboxylic acid groups).

WHY these two: To make a long chain we need an amine hand to shake a carboxylic-acid hand (that handshake makes an amide bond, coming in Step 3). Each molecule has two hands, so the chain can keep growing at both ends.

PICTURE: Look at the figure — the amine molecule is drawn in lavender with two mint hands ( $- N H_{2}$ ), the acid in coral with two butter hands ( $- C O O H$ ). Notice both middle sections are plain grey handles.

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

Step 2 — Zoom into ONE handshake: which atoms actually touch?

WHAT: We pick just one amine hand and one acid hand and look at the exact atoms that are about to react.

$acid carbon - C stays (= O) this O-H will leave - O - H + one H leaves H nitrogen stays - N - H$

Let me name each piece as it appears:

The acid carbon (the $C$ of $- C O O H$ ) — stays and becomes part of the new bond.
Its double-bond oxygen $= O$ — stays (it survives into the product).
Its $- O - H$ group — this leaves.
The amine nitrogen $N$ — stays and becomes part of the new bond.
One of the nitrogen's hydrogens $- H$ — this leaves.

WHY look this closely: because condensation polymers are defined by throwing away a small molecule. To see the small molecule appear, we must identify exactly which atoms depart. Here $- O H$ (from the acid) and $- H$ (from the amine) leave together.

PICTURE: The figure circles the departing $- O H$ in coral and the departing $- H$ in mint. The acid carbon and the nitrogen (which will bond) are marked with a dashed arrow showing them reaching toward each other.

Step 3 — The bond forms, water drops out (the amide link)

WHAT: Combine the departing $- O H$ and $- H$ into a water molecule, and let the $C$ and $N$ bond directly.

$acid hand - C O O H + amine hand H_{2} N - ⟶ amide link, stays - C O - N H - + thrown away H_{2} O$

Reading the equation term by term:

$- C O O H$ = the acid hand losing its $- O H$ .
$H_{2} N -$ = the amine hand losing one $- H$ .
$- C O - N H -$ = the new amide link; the $C$ (with its surviving $= O$ ) is now bonded straight to the $N$ (with its surviving $- H$ ).
$H_{2} O$ = the $- O H$ + $- H$ that left, combined — the small molecule that gives "condensation" its name.

WHY water and not something else: we removed exactly one $- O H$ and one $- H$ ; together those atoms ( $O$ , $H$ , $H$ ) are precisely $H_{2} O$ . Nothing is left over. This is why the Esterification reaction (acid + alcohol) also releases water — same $- O H + - H$ arithmetic.

PICTURE: The figure shows the two hands closing into a single amide link (drawn as a slate bar joining lavender and coral), while a small blue $H_{2} O$ droplet floats away.

Step 4 — But each monomer has TWO hands — so it grows both ways

WHAT: The molecule we just built still has an unused amine hand on the left and an unused acid hand on the right. Both can react again.

$H_{2} N - (C H_{2})_{6} - new amide N H - C O - (C H_{2})_{4} - C O O H$

Left end: $H_{2} N -$ is a free amine hand → can grab another adipic acid.
Right end: $- C O O H$ is a free acid hand → can grab another hexamethylenediamine.

WHY this matters: a long chain needs growth at both ends. If a monomer had only one hand, the chain would hit a dead end (a "cap") and stop short. Two hands each = endless threading. This is the step-growth idea: any two pieces (a monomer, a short chain, a long chain) can join wherever a free amine meets a free acid.

PICTURE: The figure shows the dimer in the centre with two glowing free hands (mint on the left, butter on the right), and dashed arrows pointing to fresh monomers waiting to attach on each side.

Step 5 — The pattern repeats: extract the repeat unit

WHAT: After many handshakes, the chain looks like an endlessly alternating ...amine–acid–amine–acid... sequence. We isolate the smallest block that repeats.

$[- N H - (C H_{2})_{6} - N H - C O - (C H_{2})_{4} - C O -]_{n}$

Reading the repeat unit left to right:

$- N H - (C H_{2})_{6} - N H -$ = the leftover of hexamethylenediamine after both its $- N H_{2}$ hands were used (each lost one $H$ ).
$- C O - (C H_{2})_{4} - C O -$ = the leftover of adipic acid after both its $- C O O H$ hands were used (each lost one $- O H$ ).
The two $- N H - C O -$ junctions inside = the two amide links per repeat unit.
$n$ = how many times this block repeats (thousands).

WHY the middle chains keep their exact carbon counts: the reactions only ever remove atoms from the hands ( $- O H$ and $- H$ ), never from the inert $(C H_{2})$ handles. So the diamine keeps all 6 of its carbons in $(C H_{2})_{6}$ , and adipic acid's 4 middle carbons stay as $(C H_{2})_{4}$ (its other 2 carbons are now inside the two $C O$ groups).

PICTURE: The figure boxes exactly one repeat unit out of a longer drawn chain, colour-coding the amine-derived half (lavender) and the acid-derived half (coral), with the two amide junctions flagged in slate.

Step 6 — Count the water: why $(2 n - 1)$ and not $2 n$

WHAT: We count exactly how many water molecules leave when we build a chain of $n$ repeat units.

Each repeat unit has 2 amide links, and each amide link releases 1 water. Naively that suggests $2 n$ waters. But look carefully at the two ends of the whole chain:

$one leftover hand H_{2} N - \dots\dots one leftover hand - C O O H$

WHY it's $(2 n - 1)$ , one less: a chain of $n$ repeat units contains $2 n$ monomers total, and $2 n$ beads joined in a line need only $2 n - 1$ bonds between them (think of 4 beads → 3 links). The very ends keep one unused amine hand and one unused acid hand — those two hands never reacted, so their water was never released.

$n H_{2} N (C H_{2})_{6} N H_{2} + n H O O C (C H_{2})_{4} C O O H ⟶ [- N H (C H_{2})_{6} N H - C O (C H_{2})_{4} C O -]_{n} + (2 n - 1) H_{2} O$

$(2 n - 1)$ = (2 links per unit) $\times$ ( $n$ units) $-$ 1 missing end-bond.

PICTURE: The figure lays out a short chain with $n = 3$ units (6 monomer beads), draws the 5 internal bonds each dropping a water droplet, and marks the two end hands as "unused → no water here". Count: $2 (3) - 1 = 5$ droplets.

Step 7 — Edge case: what if a monomer had only ONE hand?

WHAT: Suppose we accidentally mixed in a molecule with only one $- N H_{2}$ (a monoamine, one hand).

WHY show this: it reveals why bi-functionality is non-negotiable. A one-handed molecule can shake once and then has nothing left to react — it caps the chain and stops growth right there.

$\dots - C O - dead end, no second hand N H - R$

$N H - R$ = amide link formed, but $R$ has no further reactive group → the chain is terminated.

This is also the flip side of Step 5 in thermoset chemistry: a molecule with three hands doesn't cap — it branches into a 3-D net (that's bakelite/melamine, a different story). Nylon uses exactly two hands per monomer → clean linear chains → meltable, drawable fibres.

PICTURE: Two panels. Left: a one-handed monomer joins and freezes the chain (a red "STOP"). Right: for contrast, the normal two-handed monomer keeps a free hand glowing "GO".

The one-picture summary

WHAT: The whole nylon-6,6 story compressed: two two-handed monomers, alternating handshakes, one water per handshake, an extracted repeat unit with two amide links, and $(2 n - 1)$ waters total.

Recall Feynman: retell the whole walkthrough

Picture two kinds of beads. Purple beads have a sticky hand ( $- N H_{2}$ ) on each end; orange beads have a different sticky hand ( $- C O O H$ ) on each end. A purple hand and an orange hand want to shake — but to do it, the purple hand must throw away one little $H$ and the orange hand must throw away its $- O H$ ; those two scraps fly off together as a drop of water, and the two beads lock into an "amide" bond. Because every bead has two hands, once you make one bond you still have free hands on both ends, so the beads keep threading: purple-orange-purple-orange, forever. The smallest repeating chunk is one purple plus one orange, holding two amide bonds — that's the repeat unit. If you count the water: line up $2 n$ beads and you need $2 n - 1$ bonds to link them (four beads need three links), so you drop $2 n - 1$ water molecules, with two lonely hands left unused at the very ends. And the moment a bead has only one hand, it shakes once and stops the chain dead — which is exactly why nylon insists every monomer bring two hands to the table.