18B.14-18B.17 - Condensation Polymers And Amino Acids

Why Condensation Polymers Form

An alkene can make an addition polymer because a $\ce{C=C}$ bond opens up. A dicarboxylic acid, a diol, a diamine or an amino acid usually has no $\ce{C=C}$ bond to open, so the chain has to grow by a different route: two functional groups react, a new link forms, and a small molecule is eliminated.

For a long chain to grow, the monomers must be bifunctional: each molecule needs two reactive ends so that it can link to two neighbours rather than stop the chain after one reaction.

The two link types in this lesson are:

The formation of a polyamide is therefore a condensation polymerisation reaction. Each amide link is made by joining a carboxyl-derived carbonyl carbon to a nitrogen atom and eliminating water from $\ce{-COOH}$ and $\ce{-NH2}$ ends.

Common-error contrast

Addition polymerisation keeps all atoms from the monomer in the polymer chain and uses up a $\ce{C=C}$ bond. Condensation polymerisation does not require a $\ce{C=C}$ bond and does lose small molecules. If a question asks for a condensation repeat unit, look for the link group formed, not for a double bond to open.

Drawing Condensation Repeat Units

Condensation repeat-unit questions are structure questions. The features to get right are the correct link group, the correct order of atoms in the link, both residues from the monomers, brackets, continuation bonds and n.

Use a general dicarboxylic acid and diol:

dicarboxylic acid $\ce{HOOC-R-COOH}$ and diol $\ce{HO-R'-OH}$

The acid $\ce{-OH}$ and alcohol $\ce{H}$ are removed as water, forming ester links. The repeat unit is:

Example:

• Monomers: $\ce{HOOC-CH2-CH2-COOH}$ and $\ce{HO-CH2-CH2-OH}$
• Repeat unit: $\left[\ce{-O-CH2-CH2-O-C(=O)-CH2-CH2-C(=O)-}\right]_{n}$

This is a polyester repeat unit because the chain contains ester links.

Route for a dicarboxylic acid and a diamine

Use a general dicarboxylic acid and diamine:

dicarboxylic acid $\ce{HOOC-R-COOH}$ and diamine $\ce{H2N-R'-NH2}$

The acid $\ce{-OH}$ and amine $\ce{H}$ are removed as water, forming amide links. The repeat unit is:

Example:

• Monomers: $\ce{HOOC-(CH2)4-COOH}$ and $\ce{H2N-(CH2)6-NH2}$
• Repeat unit: $\left[\ce{-NH-(CH2)6-NH-C(=O)-(CH2)4-C(=O)-}\right]_{n}$

This is a polyamide repeat unit. The nitrogen atom is part of the amide link; do not draw an ester oxygen in its place.

Route for amino acids

A 2-amino acid contains both functional groups needed for condensation:

$\ce{H2N-CH(R)-COOH}$

One molecule's $\ce{-COOH}$ group reacts with another molecule's $\ce{-NH2}$ group. The repeat unit is:

$\left[\ce{-NH-CH(R)-C(=O)-}\right]_{n}$

This is a polyamide because the peptide link is an amide link.

Worked example: choosing the right repeat unit

Question: Draw the repeat unit formed from $\ce{HOOC-(CH2)3-COOH}$ and $\ce{H2N-CH2-CH2-NH2}$.

Model route:

1. The first monomer is a dicarboxylic acid and the second is a diamine.
2. Carboxylic acid plus amine forms an amide link, $\ce{-C(=O)-NH-}$.
3. Keep the carbon skeletons: $\ce{(CH2)3}$ from the acid and $\ce{CH2-CH2}$ from the diamine.
4. Put the repeating section in brackets and add n.

Chemical interpretation: the answer shows two amide links per repeat cycle when the repeat units join together. It also shows that the small molecule eliminated during chain formation is not part of the repeat unit.

[DIAGRAM: asset_name: Polyamide repeat from hexanedioic acid and hexane-1,6-diamine, and an amino-acid repeat showing peptide linkage; asset_slug: edexcel_9ch0_polyamide_peptide; recommended_method: deterministic_chart; description: Polyamide repeat from hexanedioic acid and hexane-1,6-diamine, and an amino-acid repeat showing peptide linkage]

Zwitterions And Acid-Base Behaviour

2-amino acids have an amino group and a carboxylic acid group on the same carbon skeleton. In simple neutral shorthand an amino acid can be written as:

$\ce{H2N-CH(R)-COOH}$

In aqueous solution, simple amino acids near their isoelectric pH are represented predominantly as zwitterions:

$\ce{H3N+ - CH(R)-COO-}$

The carboxylic acid group donates a proton and becomes $\ce{-COO-}$; the amino group accepts a proton and becomes $\ce{-NH3+}$. This internal proton transfer explains why 2-amino acids can show both acidic and basic behaviour.

In acidic conditions, the carboxylate end can accept a proton:

In alkaline conditions, the ammonium end can lose a proton:

The important reasoning is not just "amino acids contain acid and amine groups". The zwitterion shows the two acid-base sites that exist together in solution, so amino acids can respond to both acids and alkalis.

Worked example: predicting the dominant charged form

Question: Glycine can be written as $\ce{H2N-CH2-COOH}$. Explain why glycine exists mainly as a zwitterion in aqueous solution near its isoelectric pH.

Model answer: the $\ce{-COOH}$ group can donate a proton and the $\ce{-NH2}$ group can accept a proton. Proton transfer gives $\ce{H3N+ - CH2-COO-}$. This species has a positive ammonium group and a negative carboxylate group, so it is a zwitterion with no overall charge.

Chemical interpretation: the molecule is not a simple neutral amine plus acid in water. Its internal acid-base reaction changes the charged groups present, which controls the acid-base behaviour and helps explain behaviour in separation methods.

[DIAGRAM: asset_name: Alanine cation, zwitterion and anion connected by acid-base reactions; asset_slug: edexcel_9ch0_alanine_charge; recommended_method: deterministic_chart; description: Alanine cation, zwitterion and anion connected by acid-base reactions]

Amino Acids And Plane-Polarised Light

Aqueous solutions of 2-amino acids can affect plane-polarised monochromatic light when the dissolved amino acid is optically active. The usual structural reason is a chiral centre at carbon-2: that carbon is bonded to four different groups.

For a typical 2-amino acid, the four groups around carbon-2 are:

• $\ce{NH2}$ (or $\ce{NH3+}$)
• $\ce{COOH}$ (or $\ce{COO-}$)
• $\ce{H}$
• $\ce{R}$

If these four groups are different, the molecule can exist as two optical isomers. A pure sample of one optical isomer rotates the plane of plane-polarised monochromatic light. The opposite optical isomer rotates it in the opposite direction under the same conditions.

Glycine is $\ce{H2N-CH2-COOH}$. Its carbon-2 is bonded to two hydrogen atoms, so it does not have four different groups and is not chiral. Glycine does not rotate plane-polarised monochromatic light.

A racemic mixture is also not optically active overall because equal amounts of two optical isomers rotate the plane in opposite directions and the rotations cancel. Therefore the careful claim is: many 2-amino acid solutions are optically active when they contain an excess of one optical isomer, but glycine or a racemic mixture will show no overall rotation.

Common-error contrast

Do not say "all amino acids rotate light because they contain an amino group". Optical activity comes from chirality, not from the presence of nitrogen. The structural check is whether carbon-2 has four different groups.