Structural Organisation and Self-assembly of Macromolecular Soft Materials

Question 1

What is the difference between hard and soft matter? Give examples in terms of plastics.

Answer 1

Hard matter has its interactions governed by atomic forces and quantum mechanics such as thermoplastics. Soft matter has forces with comparable strength to thermal energy at room temperature such as thermosetting plastics (form cross links after heating).

Question 2

Draw the phase transitions for a thermoplastic polymer.

Answer 2

Question 3

On a DSC plot for a thermoplastic, what are the 3 notable changes that may occur depending on the behaviour taken?

Answer 3

T_g, the glass transition point when going from the solids to the rubber associated with an increase in heat flow.

T_m, the melting point when transitioning from the semi-crystalline rubber to the liquid.

Crystallisation to a semi-crystalline rubber from the amorphous rubber through slow cooling.

Question 4

What are the 3 scales we look at polymers at? Give distances and observations.

Answer 4

The molecular scale, from nm to μm, looking at the atomic structure of the polymer.

The mesoscopic scale, 10 to several 100 μm, the structural organisation of molecules, either amorphous or ordered.

The macroscopic scale, normal viewing size, the bulk properties of the material.

Question 5

What are the 4 key functional groups of step-growth polymerisation?

Answer 5

Ester, amide, carbonate and urethane.

Question 6

Give the 3 types of addition polymerisation.

Answer 6

Radical, anionic, cationic.

Question 7

Describe the basic process of radical polymerisation and an example of a monomer that would undergo the reaction.

Answer 7

Initiation using a radical, propagation of the radical through alkenes and termination by combining radicals. Suitable for cyclisation polymerisation where a dialkene forms a ring by intermolecular reaction of a formed radical with a second alkene on a molecule.

Question 8

Describe the basic process of anionic polymerisation. Give possible reagents.

Answer 8

Initiation is done by molecules such as BuLi, KNH₂ or RMgX attacking an alkene which must be able to stabilise a negative charge. For example 2-cyanopropene where the cyano group stabilise the charge.

There is no formal termination step.

Question 9

Describe the basic process of cationic polymerisation.

Answer 9

An alkene which can stabilise a positive charge attacks an acid such as H₂SO₄, HClO₄, or a Lewis acid such as BF₃, AlCl₃. This then propagates to form the polymer and terminates by elimination of a hydrogen to form an alkene, or attack by a charged nucleophile.

Question 10

Describe the way that cyclic ethers polymerise.

Answer 10

All cyclic ethers can undergo cationic ring opening polymerisation, where the oxygen is protonated and another ring ether nucleophilicly attacks it from the oxygen to open the ring.

Epoxides (3 membered ring ethers) can also undergo anionic ring opening polymerisation because of the large partial positive charge on the carbons.

Question 11

Describe the ways to quantify the composition of a polymer. How can these be determined practically?

Answer 11

The molecular weight of polymers is a statistical average and can be analysed as such.

Number average: M_n = ΣnM/Σn where n = number of polymers of molecular weight M

Weighted average: M_w = ΣnM²/ΣnM

Polydispersity: PD = M_w/M_n

The molecular weights are determined by gel permeation chromotography.

Question 12

Describe how the molecular weight and polydispersity of a polymer affects its properties.

Answer 12

An increase in molecular weight strengthens the material, an increase in dispersity weakens the material, gives processing difficulties and has a large increase in viscosity.

Question 13

How does the rigidity of a polymer chain affect its properties?

What affect does substituting the chain for a Si atom and adding different side groups have (Me vs Cl)?

What is the effect of increasing the size of the side chain?

Answer 13

The greater the rigidity of the backbone, the higher the T_g and T_m values.

Si decreases the rigidity, side chains, especially organised ones, increase the rigidity and chlorine makes the chain more rigid than a methyl group.

Generally, increasing the side chain size makes packing works which decreases the temperatures.

Question 14

What effect does H bonding have on the T_g value and the likelihood of crystalline properties being shown?

Answer 14

H bonding pulls the chains together so T_g increases and the strong interactions between the chains makes crystal structures more likely.

Question 15

Describe the difference in strength between LDPE, HDPE and UHMWPE and why these properties arise.

Answer 15

LDPE has between 40-100 short branches between 1000 repeat units so it has a low density. This gives it a low crystallinity making it weak and flexible.

HDPE has 1-6 short branches per 1000 repeat units giving it a higher density and higher crystallinity, making the plastic hard.

UHMWPE or ultra high molecular weight HDPE is many HDPEs grafted together. The high molecular weight makes its stronger and suitable for use in bullet proof vests.

Question 16

Describe the differences in branching lengths between LDPE and HDPE and how they are prepared as such.

Answer 16

LDPE: many longer chains occur due to the loss of hydrogen to other chains which become deradicalised. Produced using a metal catalyst.

HDPE: very few, short branching chains along the polymer backbone due to a loss of hydrogen to the radical at the end of the same chain (rH in a ring). Produced using radical polymerisation.

Question 17

What properties are associated with cross linked polymers and how are they produced?

Answer 17

They are thermoset plastics, often insoluble (can swell into gels) and no liquid state.

For cross-link density: High - material is hard, Low - material is rubbery and highly stretchable.

Cross-links are made often with molecules specially designed with 2 or more reactive sites so reactions can occur from both sides, joining 2 molecules.

Question 18

List the types of copolymers.

Answer 18

1. Statistical copolymers - random distribution of the monomers in the chain.
2. Alternating copolymers - ABAB type structure
3. Block copolymers - substantial sequences/blocks of each monomer, composed by first making short homopolymers
4. Graft copolymers - blocks of one monomer are grafted onto the backbone of a homopolymer
5. Stereoblock copolymers - one monomer by blocks of different tacticitys
6. Elastomers - homopolymers linked by a cross linking group
7. Network polymers - cross linking groups are similar to main polymer chains.

Question 19

Give the different types of tacticity and the properties associated with each.

Answer 19

Isotactic - all subunits are alligned, high crystallinity, used in tough fibres and plastics.

Syndiotactic - alternating sides of side chains, slightly lower crystallinity, softer, used in cable insulation.

Atactic - random arrangement of side units, often no crystallinity, generally to soft for use.

Question 20

What are the 3 possible conformations of the polymer chain? How does maximising chain entropy affect the overall shape?

Answer 20

Staggered: Gauche(-), Trans and Gauche(+) rotating clockwise about the Newman projection.

The maximisation of chain entropy results in a coiled up shape of the chains.

Question 21

How do you predict the mean distance value between the ends of a polymer chain according to the random walk theory?

Answer 21

Use the root mean square formula:

{R_<em>n</em>²}^½ = ln^½

Where n is the number of repeat units and l is the average C-C bond length.

Question 22

When does the random walk theory breakdown and what can be done to account for this?

Answer 22

The distance predicted by the random walk theory can be incorrect due to branching, cross links and greater chain rigidity.

In solution, an alpha correction term is added with the value depending on the polymer itself or the solvent.

alpha > 1: good interactions with solvent, random coil expanded due to cross linking/branching, solvent interactions are favoured over the polymer interactions with itself.

alpha < 1: poor interactions with solvent, coil contracts to minimise contact with solvent.

Question 23

How do polymers crystallise? What factors influence this compared to typical crystallisation? How does crystallisation from different states affect the crystal products?

Answer 23

The chains fold and stack together to form a regular, vertically stretched sine wave type shape (hair pin). Polymer chains can stick out as well as loops at the point where there should be a turn. Kinetic factors have an important role in crystallisation due to the nature of the polymers.

Crystallisation from a dilute solution gives a high degree of crystallinity as plenty of time is given to neatly organise. The layers of lamella (crystal hairpin) stack together with so-called tie molecules joining them together. Length, l, is several μm, fold period, d, is about 10 nm.

Crystallisation from polymer melt gives a much lower degree of crystallinity as chain movement is slower and occurs from defect sites. Lamella form rapidally from defects to form spherulites. These have lamella branching radially outward from 0.1 μm to 1 mm, fold is perpendicular to the growth direction, there is an amorphous region between the fibrils and irregular chain folding with regular tie molecules.

Question 24

Describe how the molecular motion and the properties of polymers changes between states.

Answer 24

Below T_g: cooperative motion frozen, hard and brittle

Around T_g: gradual cooperative movement of segments but still slow

Semi-crystaline at T_g: cooperative motion still frozen until T_m

Just above T_g: random coils forming but movement is too slow to fully untangle, polymer is soft with some elastic recovery (rubbery)

Well above T_g: random coil conformations, some chain entanglement, no elastic recovery

Question 25

What factors affect the degree of crystallinity of a polymer in the structure and processing? How are the material properties affected by this?

Answer 25

Structural factors include: molecular weight distribution, chain stiffness, bulkiness and intermolecular forces and the regularity of the chain structure. Processing factors include: the liquid the crystals are grown from (solution or melt), cooling rate and the shear or strain placed onto the solution. Affected material properties include: transparency/opaqueness, elastic modulus (in the rubbery state) and brittleness (in the rubbery state)

Question 26

Describe how polymers mix and the thermodynamics explaining it.

Answer 26

Polymers mix extremely poorly, the larger the MW the less they will mix with even very similar polymers. This is because the entropy change due to mixing is much smaller when mixing polymers as much fewer microstates are possible (ΔS = *k*_BlnΩ where Ω is the number of microstates). The mixing of polymers is therefore determined by the enthalpic term; miscible polymer blends must have strong intermolecular forces between the two polymers such as hydrogen bonding, ionic interactions or electron donor-acceptor complexes.

Question 27

When can immicscible polymer blends be useful?

Answer 27

If large-scale phase separation can be prevented, there will be pockets of one polymer interwoven with the other. These have to be placed on a matrix with a phase mediator - often graft polymers. This creates impact-resistant plastics e.g HIPS.

Question 28

How will a diblock copolymer organise if the blocks are incompatible?

Answer 28

Microphase separation will occur and there will be 2 coiled phases of the polymer joined by the covalent bond. No macroscopic phase separation is possible due to the joining of the molecules.

Question 29

Name and describe 5 different star block copolymers and explain why block copolymers are useful.

Answer 29

A₂B - 3-Miktoarm: one polymer with another grafted onto the side ABC - 3-Miktoarm: three polymers all attached together at a single point A₂B₂ - 4-Miktoarm: 2 polymers attached together at a midpoint of both polymers A₃B - 4-Miktoarm: three A polymers and one B polymer attached together at a single point (AB)₄ - 4-Miktoarm: four diblock polymers attached at a single point Block copolymers are useful in nanotechnology as templates or due to their self-assembly.

Question 30

What is the Flory-Huggins interaction parameter, X? How is it related to temperature?

Answer 30

It describes the mixing enthalpy of block copolymers. If A-B interactions are unfavourable compared to A-A and B-B, then X \> 0 (usually the case) If A-B interactions are favourable compared to A-A and B-B, then X \< 0 X ~ 1/T with a higher temperature leading to faster separation.

Question 31

How does the chain length of diblock copolymers affect the predominant effect of mixing? What is the segregation product and what does it mean?

Answer 31

When the polymer chains are short (Degree of polymerisation, N, is small), the mixing entropy is large and is more important than the enthalpic term. When polymer chains are long (N is large) the entropic gain from a disordered arrangement is smaller so the enthalpic term is larger. The segregation product is XN where X is the Flory-Huggins interaction parameter and N is the degree of polymerisation (the sum of N for each block) which determines if microphase separation occurs. XN \> 10 is the weak segregation limit and produces a diffuse interface between the blocks. XN \> 50 is the strong segregation limit and produces sharp domains between phases. Given X, the degree of polymerisation required for segregation to occur can be determined.

Question 32

How does the diblock copolymer arrange on the macroscale?

Answer 32

A series of layers are formed with the connections between the polymers at the interface. The size of the layers is know as the domain width, d, and is directly related to N of each block. The value of d is typically 10-200 nm.

Question 33

How do you determine the volume fraction, f_A, for a diblock polymer? How does this affect the phases for linear polymers? What governs the phases formed?

Answer 33

f_A ≈ N_A / (N_A + N_B) When f_A ≈ 0.5, lamella form giving the alternating layers of polymer. When f_A ≈ 0.33 (0.18 - 0.35), the flat interface is unfavourable so cylinders form which stack together in 3D to form the cylindrical hexagonal phase with the A block on the inside of the cylinders. When f_A ≈ 0.25 or smaller, spheres form which pack together to form a BCC structure. Between the lamellar and hexagonal there is also a gyroid structure where one polymer is a network in the other. The entropy of penalty of extending the chains must be balanced with the enthalpic penalty of increasing the interface surface area.

Question 34

Sketch the phase diagram for diblock copolymer structures.

Answer 34

Question 35

How can the superlattices of block copolymers be distinguished and what are they used for?

Answer 35

Using TEM pictures, gyroid appears like a network of different polymers, hexagonal has a hexagon of holes around each hole which continues down through the material, cubic forms a square pattern. The matrix of holes can be used for nanolithography and selective incorporation of nanoparticles into a polymer.

Question 36

What differences between the polymer blocks can be incorporated?

Answer 36

Compositional contrasts such as hydrophilicity and polarity, conformational contrasts such as rigidity and length, morphological contrasts such as crystallinity and swelling and functionality contrast such as reactivity and biocompatibility.

Question 37

What are lyotropic hydrogels and how can they be used?

Answer 37

Polymers with a hydrophobic backbone and hydrophilic side chains which cause the polymer to swell. These can be adjusted to have permanent optical properties for use in contact lenses.

Question 38

What happens when you mix a lamella with hydrophobic and hydrophilic layers with water? How do these compare in size to biological systems?

Answer 38

The hydrophobic layers swell away from the junction, allowing the layer to curve back onto itself. This forms a vesicle, a spherical layer of the hydrophilic polymer which surrounds a hydrophilic sphere. This is called a polymersome and are often used for drug delivery. Polymersomes are ~23 nm in length and about 10x bigger than a phospholipid layer.

Question 39

What are functional polymers? Name and describe the processes of superglue and acrylates.

Answer 39

Functional polymers have in-built properties designed certain applications. Superglue, an acrylate - alkene and carbonyl connected by one bond, undergoes anionic polymerisation with OH^- ions from the atmosphere. The OH^- attacks the alkene, forming a negative charge which propagates the reaction. Acrylates can also undergo free-radical polymerisation by initiation using UV light or thermal energy.

Question 40

How is adhesion measured?

Answer 40

Cohesive bond strength is used. The work of cohesion, W_c, is the minimum work necessary to overcome the intermolecular attractive forces across a plane surface of 1 cm². W_c = 2γ_L where γ_L is the surface tension, often measured in dynes. Cohesion bond strength, σ_c = W_c / x where x is the thickness of the polymer.

Question 41

How can disc shaped nematic crystals be used in polymers?

Answer 41

When created into a polymer, the ordered structures have purposes such as giving displays wide viewing angles and as lubricants for disc drives.

Question 42

How and why is photopolymerisation in organised media done?

Answer 42

Macroscopic orientation is placed on the main chain polymers by drawing, extrusion, gel-drawing and spinning of melts and polymer blends of lyotropic LCs. Specific properties can be provided to the molecule such as mechanical, thermomechanical, optical, electro-optical. Photopolymerisation in the LC phase gives oriented, highly cross-linked network which is strong and brittle.

Question 43

How are polymer composites used for light shutters?

Answer 43

Liquid crystal droplets are suspended while acrylate is polymerised around it, meaning the droplets scatter light when they are randomly orientated, but transmit light when the droplets are alligned with the electric field.

Question 44

What properties mean that PMMA is useful for optical fibres?

Answer 44

PMMA is relatively transparent to light and can be used to split optical signals almost perfectly in 2 directions, working well up to fast internet speeds.

Question 45

How do epoxy resins work and why are they strong?

Answer 45

The epoxy part is an chloroepoxide which is combined with bisphenol-A and a hardener/catalyst to cause cross linking. The resin is strong as the biphenol-A coils up giving strength in all directions.

Question 46

Why would a polymer be preferential to a titanium rod for supporting bones?

Answer 46

Titanium is much stronger than bone so the new bones that regrow tend to be weaker. With polymers the mechanical stress can be controlled to match the body better.