Delocalisation and Conjugation Flashcards
What does a reaction arrow represent
Used to represent a chemical reaction
The direction points in the direction of the reaction from reactants to products
Most common arrow used
What does an equilibrium arrow represent
is a double arrow and it represents a reversible reaction and the products may become reactants in the same process
What does a resonance arrow represent
Is a single double headed arrow, which shows the structures at both sides of the arrow which are resonance structures of another
What does a curly arrow represent
There are two types:
Double headed curly arrow: which represents the movements of an electron pair. The electron pair moves fro, the tail to the head (most widely used curly arrow
Single headed curly arrow: Which represents the movement of a single electron (used with radicals)
What is conjugation
Usually refers to a sequence of alternating double and single bonds
However, you can also get examples of conjugated systems that do not involve alternating double/single bonds
For example
What is delocalisation
Focuses on the molecular orbitals covering the conjugated system
Delocalised electrons are NOT confined to a single atom/bond between two atoms - they are shared by three or more atoms
How do conjugated systems relate to resonance forms
In a conjugated system, there is the possibility of having resonance forms/structures, that is several Lewis formulas of the same compound.
The resonance forms show that there is electron delocalisation
Benzene is a really familiar case of conjugation
Explain how?
All of carbons are sp² hybridised, and the ∏ electrons can delocalise around the ring
What is a resonance hybrid
is defined as the combination of all possible resonance forms for a given compound and it represents the overall delocalisation of electrons within the molecule
Curly arrows show the movement of electron from one resonance form to the other
What would the two resonance forms of benzene look like
Curly arrows show movement of electrons
Resonance arrow
What does the resonance hybrid of benzene look like
The Pi electrons are delocalised around the ring
Keep in mind that resonance hybrids are not always the best structure to represent molecules
Consider (E)-hexatriene
In its structure, some ‘single bonds’ are shorter than typical, and the central double bond is slightly longer than is typical
Why is this the case
This does apply for the C=C bond at the end of the molecule
This is due to the delocalisation of pi electrons
Because of this the central C=C bond can have single bond character and hence is longer than usual
And the C-C singl,e bonds have some double bond character and are hence shorter than usual
Can delocalisation occur in Propenal
Delocalisation of pi electrons, where the C=C double bond are conjugated to the C=O double bond
Is Arachidionic acid conjugated
Delocalisation of the pi electrons is not possible with the non-conjugated double bonds
This is due to the 4 C=C double bond which are seperated by two single bonds which are sp³ hybridised.
The sp³ hybridised carbon atoms have no p orbital to allow pi bonding
Is allene conjugated
Delocalisation of the pi electrons is not possible as the central carbon is sp hybridised rather than sp² hybridised
the two pi orbtials shown the the diagram are perpendicular to another and cannot overlap, and hence there cannot be conjugation between pi bonds on either side of the central carbon atom
There is different phasing opportunities with the pi bonds in 1,3-butadiene
(if we think of it like two ethane molecules bonded together) there are 4 pi orbitals in butadiene - we will get 4 pi molecular orbitals
What is the simplest organisation of phasing these orbitals
Having all 4 pi orbitals in phase
This will produce the lowest energy pi bonding molecular orbital
This has no nodes
What is the second lowest energy pi molecular orbital of butadiene
We can combine two ethane molecules, and combine the pi orbitals out of phase
This will produce one node between the central two carbon atoms (between carbon 2 and 3)
What is the second highest energy pi molecular orbital of butadiene
If we take two ethane molecules (where the pi orbitals on the ethane molecule are out of phase) and we combine the pi orbtials from the ethane molecules in phase, then we produce 4 pi orbtials
This will produce 2 nodes
What is the highest energy pi moleular orbital of butadiene
If we take two ethane molecules (where the pi orbitals on the ethane are out of phase) and we combine the pi orbitals out of phase, then we will produce 4 pi orbitals
This will produce 3 nodes
On the following diagram site which orbital arrangement is ψ1, ψ2, ψ3, ψ4
And which one is the HOMO and which one is the lumo
HOMO = highest occupied molecular orbital
LUMO = lowest energy unoccupied molecular orbital
On the following diagram site which orbital arrangement is ψ1, ψ2, ψ3, ψ4
And which one is the HOMO and which one is the lumo
HOMO = highest occupied molecular orbital ψ2
LUMO = lowest energy unoccupied molecular orbital ψ3
How many pi electrons does butadiene have
There are two electrons in each double bond
There are 2 double bonds = 4 pi electrons
These are put into the lowest energy MO ψ1 and ψ2
ψ1orientation of butadiene is fully delocalised
What does this mean
The electrons within this orbital can move all throughout the molecule
This is because all 4 pi orbitals are in phase, hence there is bonding interactions across all four carbon atoms This orbital has a +3 net bonding interactions - due to 3 in-phase interactions and no out-of-phase interactions
Why does Ψ2 have a net bonding interaction of +1
Ψ2 contains 1 node at the centre of the molecule
Thus we have a bonding interaction on the LHS and a bonding interaction on the RHS, but an antibonding interaction in the centre
Electrons in this orbital can bond on the left or the right hand side (like in isolated double bonds) but they cannot bond between carbon atoms 2 and 3
We describe this as 2 bonding interactions minus 1 antibonding interactions, so has a net bonding interaction of +1
We would also say that the double bond between carbons 2 and 3 is a partial double bond
How can the allyl anion be formed
The deprotonation of propene
Once deprotonated we have an anion with a negative charge next to a double carbon bond
How does this affect the movement of electrons
The p orbitals of the pi system (from the double carbon bond and the p orbital of the carboanion) are conjugated to another - the electrons can move between these orbtials
There are 3 different ways the molecular orbitals in an allyl anion can orientate themselves, which will affect the energy of the orbitals
describe them
Ψ1 - orbtials are all in phase - leading to a net bonding interaction of +2
Ψ2 - two orbitals are in phase and one out of phase - leading to a net bonding interaction of 0
Ψ3 - orbitals are all out of phase - leading to a net bonding interaction of -2
Which orientation of the 3 pi orbitals would be the HOMO and LUMO
The 4 pi electrons would occupy Ψ1 and Ψ2
Hence the LUMO would be Ψ3 and the HOMO would be Ψ2
Why in the case for the allyl ion is the resonance hybrud considered not as informative as the individual resonance forms
On the individual resonance forms, there can be a negative charge on either the right or left terminal carbon - but not on carbon 2
If we were to use the resonance hybrids, this information is lost because it appears from this structure that the negative charge is spread over all 3 carbon atoms (which is incorrect)
How is an allyl cation form
A negative leaving group (like Br) to produce a carbocation
Which oritentation of the molecular orbitals of the allyl cation results in the HOMO and LUMO
Because there are only two pi electron, they exist in the lowest energy conformation being Ψ1
Therefore the HOMO is Ψ1 and the LUMO is Ψ2
How could you form an allyl cation from cyclohexenol
If you treat cyclohexenol with a very strong acid, it will protonate the hydroxyl group and it will leave as water
How can we form an allyl anion form a carboxylic acid
If we treat a carboxylic acid with a base (e.g. a hydroxide ion) we can deprotonate it to produce the carboxylate anion and water as a by-product
The negative charge can delocalise between the two oxygen atoms
Suggest how the charge can delocalise in in the carboylate anion system
How can amides delocalise
The lone pair of electrons on the nitrogen can delocalise into the empty pi system of the carbonyl - this is because the nitrogen is sp² hybridised
This gives us N+ and O- as one of the resonance forms
This means we should expect partial double bond character between the carbonyl carbon and the nitrogen of the amide group
In Benzene, the oritentation of the molecular orbitals can affect the energy of the molecule
Describe the orientations of the different benzene molecular orbitals
Ψ1 - all molecular oribitals are in phase
Ψ2 - three in phase and 3 out of phase (two different ways to draw the node)
Ψ3 - two adjacent MOs in phase, then one out of phase (two different ways of drawing the node)
Ψ4 - every adjacent node is out of phase from another - 3 nodes
How would you draw a simplified version of the different molecular orbital oritentation of the benzene ring
Draw it as a hexagon standing on its point. Then at each carbon atom, we draw an orbital energy level and this gives us the correct pattern for the molecular orbitals in benezene
Ψ1 at the bottom being strongly bonding, then two Ψ2 forms being weaking bonding etc
