Metal-Organic Boxes

Question 1

What is host-guest chemistry?

Answer 1

Many applications of metal-organic coordination compounds we want to design a structure with the metal ions further apart and a space or cavity in the middle
We can then bind or trap something in that space

Question 2

If in a square we have an interior bond angle of 90°
This could be achieved with four metals of the corners and four ligands for the edges, but which one would be best to choose?

Answer 2

Square planar metal ion would be ideal as it has ligand positions at 90° to another
E.g. Pd(II) and Pt(II)

Question 3

How would you classify ligands in a metal-organic box?

Answer 3

Ditopic monodentate ligands
e.g. 4,4’-bipyridine

Question 4

Why is it difficult to form discrete single boxes as shown here

Answer 4

• A square planar metal ion has four coordination sites, and this molecular square would only use two of them
• This would leave two coordination sites on each metal ion vacant
• Other ligands could then coordinate at other positions which could then lead to a wide variety of larger metal coordination structures (metal coordination polymers)

Question 5

How can we stop larger metal coordination polymers from forming and just the discrete squares?

Answer 5

This can be done with some of the bidentate chelating ligands
(But we cannot just mix the square planar metal ion, ditopic ligand, and bidentate blocking group all together due to combining in different ways)

Question 6

Instead we need to out the bidentate ligand on first in a separate step
How do we do this?

Answer 6

• You can use palladium (II) chloride (PdCl₂) which contain square planar Pd(II) centres bridged by chloride ions
• If we combine it with a bidentate ligand (such as en or bipy), the Pd(II) complexes are formed with one bidentate ligand only

Question 7

Why are only two of the chlorines displaced by the bidentate ligands to form these complexes

Answer 7

• Displacement of the first two chloride ions from the palladium centre by the bidentate ligand is easy as it binds more strongly do to the chelate effect
• Results in a mononuclear complex of palladium that is not charged
• As the bidentate ligands are neutral, they do not displace the final two chlorides (anions) due to electrostatics
• The chloride is a strongly coordinating anion to metals as hard to displace

Question 8

However, we cannot use these chloride complexes as starting materials for our square, as the 4,4’-bipyridine ligands will not be able to displace them either
How is this issue solved?

Answer 8

• This is solved by treating these complexes with a silver(I) salt of a weakly coordinating anion such as nitrate
  Ag(I) has a strong affinity for chloride and so pulls chlorides off the Pd(II) to form an insoluble precipiate of AgCl
• The vacant coordination sites on the Pd(II) are then replaced by NO₃⁻ anions which are bound through the oxygen

Question 9

NO₃⁻ are much more weakly coordinating and can be displaced by the 4,4’-bipyridine in the assembly
What happens when we react the formed Pd complex with more bipy

Answer 9

If these palladium units are combined with the ligand, then successful assembly of the square molecular box occurs
[M₄L₄]⁸⁺

Question 10

How can the following structure be analysed once synthesised

Answer 10

• Single crystal X-ray crystallography
• Mass spec
• ¹H NMR

Question 11

What would the ¹H NMR of the following compound look like?

Answer 11

• Due to the square being symmetric shapes and the ¹H NMR spectrum of the square only contains two signals from protons attached to the 4,4’bipyridine ligands as these are all equivalent
• (Pd is also low spin and diamagnetic making NMR straight forward)

Question 12

Even with these big blocking groups attached, it is possible to form a polymeric structure that repeatedly turns back on itself. All the bond angles in this structure would be 90° and so it would not be strained
Why does it not all self-assemble into polymers is due to…

Answer 12

Entropy
Forming molecular squares over a polymer leads to may more discrete species
These can all move around freely and so give more translational entropy of the system
Hence the molecular squares are favoured if the overall concentration is not too high (Le Chatelier principle)

Question 13

Although polymers can be avoided by working at an appropriate concentration, the system does not always behave exactly as we would design
What else can form

Answer 13

Molecular triangles are also observed with the same components and under the same reactions to make squares

Question 14

An equilateral triagnle has bond angles of 60°, so how is that possible with the Pd(II) metal centres preferring to be square planar?

Answer 14

• The ligands bend/bow outwards. THis allows the angle at the metal centre to be much closer to 90° so the distortion in metal geometry is within a tolerable range
• There is also some distortion in Ligand-Metal-Ligand bond angles at the Pd(II) centres ajd these angles are less than the ideal 90°

Question 15

What factors can influence the ratio of triangles to squares that form

Answer 15

• Concentration: This is due to entropy
• The bidentate blocking ligand
• Flexibility

Question 16

What would be the balanced equation for the equilibrium between triangle and squares?

Answer 16

4 triangles ⇌ 3 squares
The triangle is favoured at lower concentration and the square is favoured at higher overall concentrations

Question 17

How does the nature of the bidentate blocking ligand influence the ratio of triangles to squares that form?

Answer 17

As the blocking ligand becomes larger and more sterically demanding, it takes up more space around the Pd(II) centre, and forces the two pyridine ligands closer together, favouring more triangle

Question 18

How does ligand flexibility influence the ratio of triangles to squares that form

Answer 18

The more flexible the ligand, the eaiser it is to relieve strain around the metal centre, and hence more of the triangle is often fored if the ligand is longer and more flexible

Question 19

If we change the linear pyridine-based ligand for a bent ligand, then instead of molecular squares and triangles, we see the formation of molecular diamonds
These can be classified as?

Answer 19

[M₂L₂]⁴⁺

Question 20

What does the ¹H NMR spectrum of a diamond structure in low concentration look like?

Answer 20

• At low concentration, the spectrum was as expected
• There were only three signals from protons attached to aromatic rings (signals a and b from pyridine ring and signal c from benezene ring)
• This is consistent with the symmetry in the diamond

Question 21

What does the ¹H NMR spectrum look like for a molecular diamond for higher concentrations

Answer 21

• These signals from the low concentrations have disappeared, and a new set of signals grew
• This new set of signals had two signals for each ligand position (e.g. a’ and a’’ for the position which was a)
• This indicated that a new species of [2]catenane which is an interlocked structure that contains a mechanical bond which has lower overall symmetry

Question 22

Why has signal c’ moved significantly?

Answer 22

This is due to the π-π stacking interactions in the [2]catenane
This places this proton in the ring current shidling zone of the nearby aromatic rings and hence moves to a lower chemical shift

Question 23

Why is the [2]catenane structure preferred when interacting with water?

Answer 23

• The molecular diamond has an empty cavity that can bind hydrophobic molecules in water, as in the cavity they can get away from the water
• The ligands themselves are relatively hydrophobic and hence when they thread through each other in the [2]catenane, this is preferential to interacting with the water
• The π-π stacking between the ligands in the [2]catenane is also a favourable interaction

Question 24

What are other factors which affect the equilirbium of molecular diamonds and [2]catenane?

Answer 24

• Concentration
• Polarity of the solvent
• Competing guests

Question 25

How does concentration affect the equilirbium of molecular diamonds and [2]catenane?

Answer 25

Higher concentration favours the [2]catenane as this leads to fewer species being present
Due to entropy

Question 26

How does polarity of the solvent affect the equilirbium of molecular diamonds and [2]catenane?

Answer 26

• The [2]catenane has higher overall charge (8+) compared to the diamond (4+) and hence the [2]catenane is favoured in more polar solvents
• Solvent polarity can be decreased by adding methanol to water
• Solvent polarity can be increased by adding NaNO₃

Question 27

How does adding a competing guest of the solvent affect the equilibrium of molecular diamonds and [2]catenane?

Answer 27

• The [2] catenane does not have a cavity, so adding a competing guest that can bind in the cavity of the diamond but not in the [2]catenane, favours the diamond
• e.g. a benzoate anion due to being hydrophobic and negatively charged, so favourable electrostatic interactions with positively charged metal coordination structure

Question 28

Why are the palladium diamond and [2]catenane describe as labile?

Answer 28

With palladium(II), the [2]catenane forms when the overall concentration is increased because the M-L bonds are able to be broken again once formed
It is necessary to break at least one M-L bond before the structures can thread through each other

Question 29

With platinum(II), the M-L bonds are much stronger and much harder to break
This is due to…

Answer 29

Larger Δ with 5d metals platinum

Question 30

What are the conditions require to form a diamond with Pt(II)?
How would we describe the box onced formed?

Answer 30

Much harsher reaction conditions are used
100°C (compared to 25°C) due to bonds being much stronger
However, once formed, the Pt-ligand bonds do not break and we can describe this box as being locked

Question 31

Can [2]catenane also form from a Pt(II) diamond under normal conditions?

Answer 31

No formation of the [2]catenane is seen as threading needed cannot occur without the bonds breaking first
Therefore if the concentration in the solution is increased, there is no formation of [2]catenane

Question 32

What reaction conditions are required to unlock the Pt diamond to form [2]catenane?

Answer 32

• More forcing conditions are required
• This involves heating (to provide more energy) in concentrated NaNO₃ (to increase the polarity and favour dissociation)
• This must be performed at high concentrations to favour the [2]catenane
• Once formed, if these harsher conditions are removed (i.e., the NaNO₃ is removed and the temp is lowered) the [2]catenanse now also becomes locked in the catenane form

Question 33

What is a metal nanotube?

Answer 33

These are based on a cuboidal shape
One ligand occupies each long face of the cuboid
The smaller top and bottom face are empty, which is why this structure is referred to as a nanotube

Question 34

How can the length of a metal nanotube vary?

Answer 34

These structures can be made of varying lengths depending on the length of the ligand used, however will only form in the presence of a suitable guest molecule
These guest molecules act as a template and the nanotube forms around the guest - i.e. the larger the nanotube, the larger the optimal guest

Question 35

In general, what are the 3 features for the nanotube?

Answer 35

• hydrophobic
• able to form π-π stacking aromatic interactions with ligands
• negatively charged as favourable electrostatic interactions with positively charged nanotibes