Nematic Liquid Crystals

Question 1

Define a DSC and draw a heating and cooling DSC for a liquid crystalline material. Label the key features and describe how they relate to the properties of a liquid crystal.

Answer 1

Differential scanning calorimetry (DSC) measures the change in heat capacity of a material as a function of temperature.

Question 2

What is the general shape of a nematic liquid crystal and what causes the properties they are known for?

Answer 2

They have an elongated, rod-like shape (calamatic) and will allign with the molecules around it in a specific direction, the n director. They can also be disk shaped (discotic) This gives the material different properties across the material than down the material. This includes electron flow and optical properties.

Question 3

How do you measure molecular anisotropy?

Answer 3

The length/breadth ratio, b.

Question 4

Describe the liquid crystal phases, naming the transition points. What temperatures can these occur over?

Answer 4

The crystal state has all the molecules rigid in the solid structure, unable to rotate. Crystal → Nematic phase is the melt point, Nematic → Crystal is the recrystallisation point.

In the nematic phase, the molecules have an average direction, in the n direction, but each molecule has a different direction. Long range orientational ordering, no positional ordering. The incoming light is scattered when coming across the molecules. Nematic ⇔ Liquid is the clearing point.

In the isotropic (same properties in all directions) liquid, the molecules have no average direction and are free to rotate and tumble.

The temperatures of the transitions only depend on the molecular structure and can be over a huge range.

Question 5

Define birefringence and explain why it makes LCs opaque.

Answer 5

Birefringence is the property of a material to have two refractive indexs. This causes rotation of the planes of the light which causes the opaqueness of the material.

Question 6

Compare the rotations of the nematic molecules.

Answer 6

The relaxation times of the end over end rotations are much longer (10⁶ s) than the processions about the long axis (10¹¹ s).

Question 7

How is the range of angles of the molecules compared to the director shown? Give an equation and a typical values range.

Answer 7

The order parameter, S.

S = ½<3cos²(θ - 1)>

Where θ = angle made between the long axis of each molecule and the director. The <> brackets indicate this is averaged over a very large number of molecules.

S = 1 would indicate perfect order, S = 0 would indicate random allignment. Typical S values for LC are 0.4 < S < 0.7.

Question 8

Sketch a graph to indicate how the order parameter depends on temperature. List ways in which it can be measured.

Answer 8

If the molecular parameters are known, a macroscopic property can be measured such as: diamagnetism, birefringence, NMR and Raman scattering.

Question 9

Define and explain the cybotactic groups in the nematic phase.

Answer 9

Cybotactic groups are where there are regions of local order in a layer like form. These are known as smectic-A and C.

Smectic-A is where layers of the molecules form with the LC are perpendicular to the director.

Smectic-C is where director is not perpendicular to the layers, and locally the director appears to point in a different direction as the molecules are locally all alligned.

Question 10

Describe the x-ray diffraction of LCs in different directions.

Answer 10

The longitudinal distance between molecules, d, is greater than the lateral distance, a, so the diffraction due to d is at a smaller angle and is more intense. The diffraction due to a is at a wider angle as the gaps are smaller. The bands are broad and weak due to the movement of the molecules.

Question 11

Describe and give examples of the general structures of rod-like, nematic LC molecules.

Why would heteroatoms be added to the chain? What problems can they bring?

Answer 11

Composed of a series of rings (benzene, cyclohexane, pyrrole, thiophene, etc.), joined by linking groups (direct link, COO, CH₂CH₂, CH=CH, etc.) with small side groups on the ring (CN, F, Cl, Me, etc.).

At one end is a tail group which is normally an alkyl group or an alkoxy group. At the other end is a head group which is normally CN or F but can just be an alkyl chain.

Heteroatoms can add functionality and increase the rigidity but also attract ionic impurities.

Question 12

How do alkyl chain lengths affect the mixing of a nematic LC?

Answer 12

The ratio of aromatic to aliphatic determines how well layers form in the structure. The layers form as the aromatic regions want to be with the other aromatic regions and vice versa for the aliphatic regions.

Question 13

What is the quadropolar interaction and what effect does it have on the molecules?

Answer 13

The four dipoles formed when two molecules align with ends pointing in opposite directions. This means that their overall dipoles in the long axis (often influenced by the CN or F end group) are opposite and therefore attract. The attraction between the dipoles form new, perpendicular dipoles giving the four dipole structure.

This means molecules can form short lived dimers.

Question 14

How are dipole moments, μ, measured for nematic LC molecules?

Answer 14

The dipole moments for each group are given and can be summed by using trig and the angle of the shape (exterior angle = 360/n).

Question 15

For a simple LC with an alkyl or an alkoxy chain at the end, how does the carbon chain length affect the molecules phase transitions?

Answer 15

Alkyl: At short chain lengths the melting point of the crystal is very high due to the strong intermolecular interactions but these decrease very fast. At 5 carbons the nematic phase can form and both the clearing point and melting point increase at a similar rate, with a small range of the nematic phase temperatures.

Alkoxy: The trend for alkoxy end groups is similar, except the clearing point, while lower than the melting point, is much higher than the alkyl chain and maintains a higher temperature at longer chain lengths, even as the melting point decreases. This gives the nematic phase a much larger temperature range.

Question 16

Describe the effect of different rings on the clearing point of the nematic. Explain why this occurs.

Answer 16

The factors that affect the clearing point is the rigidity and the quadropolar interactions. Higher rigidity gives a higher transition, as does larger quadropolar interactions. However smaller overlap between molecules increases the clearing point. Ph = benzene, Ch = cyclohexane, Cch = cage cyclohexane.

R-Ph-Ch-CN has a very low cleraing point as sigma and pi regions intermix, interrupting quadropolar interactions whereas R-Ch-Ph-CN is much higher since the quadropolar is not interrupted and it has a smaller overlap. Swapping Ch of Cch increases rigidity as Cch cannot ring flip, increasing the clearing point.

Question 17

How do overall length to breadth ratios, polarisability and clearing point change with overlap of quadropolar regions.

Answer 17

Length to breadth ratios increase when the overlap is reduced and this increases the clearing point. However this decreases the polarisability of the LC.

Question 18

What is the effect of heteroatoms in the rings on the clearing point of LCs.

Answer 18

When two benzenes are adjacent, they have hydrogen overlap so the bond between the benzenes is twisted, reducing pi overlap. Adding heteroatoms (N or O) at the 2 and 6 postiions increases the clearing point.

Note that using mismatched size of heteroatoms reduces the clearing point (O and S).

Question 19

What are the factors in linking groups that affect the clearing point?

Answer 19

Interrupting sigma or pi regions greatly reduces the clearing point. Having rigid linking groups like an ester increases the clearing point.

Question 20

Give a brief overview of how boronic acid coupling occurs and what it achieves.

Answer 20

Boronic acid links aromatic groups by reacting a B(OH)₂ group on a ring to a bromine atom on a ring. This reaction must be done at approximately -78ºC to avoid side reactions. Iodine will react over bromine so if both are present, bromine can be used for later reactions.

The boron is introduced by a grignard reaction with B(^<em>i</em>Pro)₃, the coupling is Suzuki using a Pd(PPh₃)₄ catalyst, Na₂CO₃ and either EtOH or DME as a solvent.

Boronic acid can be used similarly to N₃ to introduce a range of groups onto a ring. The boron itself can also be introduced using a deprotonated ring.

Question 21

How do you introduce a CN group onto a ring?

Answer 21

Add a bromine atom by using Br₂, then substitute with CuCN.

Question 22

How is the stereochemistry of a cyclohexane ring linked to a benzene ring controlled? Give the reaction steps.

Answer 22

Using large groups such as a thioacetal during a reaction so that only the trans product (R group and Ph trans) are formed after reduction using Raney Ni/H₂.

Question 23

Give equations for birefringence, Δn, and dielectric anisotropy, Δε. What do these values depend on?

Answer 23

Birefringence is given by Δn = n_e - n_o where each n refers to the refractive index of a different plane.

Dielectric anisotropy is given by Δε = ε_|| - ε_|– where each is the permittivity across and along the long axis of the molecule. A +ve value means electrons are favoured to flow along the molecule.

Both values depend on the number of pi electrons. Large number of pi electrons gives a large Δn as the light is slowed due to the magnetic componant interacting with the electrons.

Question 24

How is average permittivity measured? How does this relate to the order parameter?

Answer 24

Average permittivity is measured by the parallel value plus two of the perpendicular values (for the 2 axis) divided by 3.

The order parameter is related to this by comparing the difference between the dielectric anisotropy and the average permittivity.

Question 25

For molecules with negative dielectric anisotropies, what effect do heteroatoms have on the value and what else affects this value?

Answer 25

When there are donor atoms like O, and electron withdrawing groups such as F, a larger perpendicular dipole is formed. When the dipole is closer to the donor atom the effect is larger.

Other factors are more polarisable pi electrons.

Question 26

Give the degrees of freedom for Gibbs phase diagram.

Answer 26

F = C (no. of componants) - P (phases) + 2

Question 27

How are mixtures of liquid crystals used to maximise the operating temperature range?

Answer 27

The eutectic point of a mixture of liquids is the point where both the solids are at an equlibrium with the liquid, not only one solid. This is the lowest temperature the liquid exists in and this means the temperature range of a nematic in this mixture is increased.

Question 28

Give some examples of what must be considered when considering molecules for LC componants.

What choices are typically made to address these requirements?

Answer 28

Stable to water, oxygen, light and electric fields, colourless for displays, wide temperature range, non-toxic, low viscosity, responds well to stimuli, very high purity and low cost.

Fluoro end groups, CN required too large of an electric field to respond. Eutectic binary mixtures used to have a larger temperature range.

Question 29

How do chiral molecules affect LC structure?

Answer 29

They form a heliacal shape around the director axis. There is no positional order but the orientation is ordered like a liquid crystal. Depending on the angle that you look at the sample, the light given off will change.

Question 30

How do you detect and represent optical activity?

Answer 30

Using a polarimeter.

[α]^T_λ which is the specific rotation.

Specific rotation = α/cl where α is the rotation of a plane of light down a tube of length l, containing concentration, c, of the sample.

Question 31

How can the changing reflection of colours from a chiral LC be represented?

Answer 31

Using a simplified Bragg equation and an average refractive index, with a changing angle of incidence. The width of the band is directly related to the birefringence of the LC, a low birefringence gives a thinner band of reflected light.

Question 32

Give two ways of introducing Br onto a molecule from an OH group.

Answer 32

Using PBr₃ and HBr/H₂SO₄.

Question 33

How can liquid crystals have chirality induced? What must be considered? Define the pitch of a chiral LC.

Answer 33

Using a chiral dopant, a chiral molecule. The compatability of the dopant must be considered, as well as the helical twisting power (HTP), β, of the dopant (1/P x C where P is change in pitch and c is concentration of dopant). Pitch is the distance for LC to undergo a 360 twist.

Question 34

Outline the two ways of measuring pitch in chiral, nematic LCs.

Answer 34

UV/vis spectroscopy can be used where increasing the dopant will decrease the pitch (twisting occurs over a shorter distance). This changes the absorption of the LC and this can be plotted to measure pitch.

Cano wedge cell places the LC in a cell with an angle of α. The pitch will be seperated into regions, seperated by defect lines. This is then used in the equation, P=2(p/2)tanα where p is the measured distance between defects.

Overall, UV is cheaper and easier.

Question 35

Give the conditions for stereoselectively converting a cyclohexene to a cyclohexane.

Answer 35

1. BF₃, DCM, THF and PCC to form carbonyl
2. BF₃.2AcOH and H₂S(CH₂)₂SH₂
3. Reduce with Raney Ni/H₂