Week 6 lecture 1
Week 6 lecture 1 Wednesday, September 30, 2015
1:28 PM
Resonance, EOC, pp. 179 ‐198 Resonance structures describe contribution to a molecule and not the real molecule So we can think of our resonance structures as dragons and unicorns (don’t really exist) but the real molecule/rhino has things that similar to both resonance structures
Note: when we talk about resonance we are talking about delocalized electrons Delocalized Electrons When we talk about resonance the electrons are spread over more than two atoms • This gives you extra stability When it is possible to draw more than one valid Lewis diagram for a molecule or ion, that species is said to have resonance. The electrons are DELOCALIZED i.e. not belonging to a single atom (lone pair) or confined to a bond between two atoms. The molecule or ion is said to be a RESONANCE HYBRID of the structures drawn. • So resonance hybrid is the 'actual molecule' For species with resonance, no single Lewis diagram is sufficient to describe them correctly Resonance structures may not have the same amount of energy Benzene You can think of the electrons as moving • So pi electrons moving around the benzene ring The reason why we can think of electrons as moving is because there is p orbitals on each of the carbon atoms, the p orbitals can overlap so we have cloud of electron density above and below the ring • And when we look at the electron density map, we see that the electrons are spread out evenly over all carbon atoms - So these electrons are delocalised Both resonance structures are equal, however the resonance hybrid is more stable than both of them Since both representations are equal in energy, each C‐C bond has a bond character in between a single and a double bond. • Measured bond length is: 1 39 Å shorter than a typical C‐C and longer than a typical C=C. - One of the consequences of resonances is that the bond length will be different CHEM1221 Page 1
- One of the consequences of resonances is that the bond length will be different Note: Ao = 1x10-10
Resonance also explain why we only get three isomers of benzene when we replace two of the hydrogens • So we are replacing two of the hydrogens If we didn’t have resonance there would be four isomers of benzene, if you had a di-substituted benzene • This is because A and B would be different In A there is a single bond between the two substitutes while in B there is a double bond • So if the electrons in a benzene molecule would LOCOLIZE it would result into two distinct (ortho) molecules but because both A and B are resonance structures, the real molecule has to be different to either of those which explain why we only have 1 ortho isomer Note: for your info, if benzene didn’t have resonance structures, the meta and para ones would still have one isomers
Note: from mod 1, substitutes immediate to each other are ortho (1,2), two spots way are meta (1,3) and three spots away are para (1,4) CHEM1221 Page 2
and three spots away are para (1,4) Cyclooctatetraene It is a cyclic carbon All of the carbons are sp2 hybridized which means it p orbitals could potentially overlap The problem is, it is not a planar molecule, and because of that the two ends point up (ends up in a boat shape) • This causes the p orbitals to be unable to overlap - This means the pi bonds are localized (no resonances) Alternating double bonds and single bonds in a ring system does not in all cases lead to resonance. Measured bond lengths are typical for alternating single (1.46 Å) and double (1.33 Å) bonds. • Since the electrons are localized in cyclooctatetraene the bond length alternate between 1.46 A (single bond) and 1.33 A (double bond) Another reason why this is not aromatic is due to the number of pi bonds… they are even • Or 4n+2=8 ==> n=not a whole number so it is not aromatic
Resonance Only electrons “move" . (Not really moving, but we think of them as if they were to help us find all the resonance structures) The only electrons that move are pi electrons or lone pair electrons. The total number of electrons can’t change. • So the total charge stays the same Electrons can be moved only between p orbitals. And only if they can be aligned properly • So that the Cyclooctatetraene example Examples Electrons can be moved only between p orbitals. • One way of doing that is to move pi electrons to sp2 hybridized carbon because sp2 hybridized carbon will have sp2 orbital + 1 p orbital - So we have p orbital in the sp2 hybridized carbon that we can more the pi electrons to Note: although in the resonance structures each of these carbons have a positive charge, in the real molecule, they have only a partial positive charge Note2: because the two resonance structures are symmetrical they contribute equally to the resonance hybrid If another sp3 carbon is getting in the way you cannot move the electrons because it will not obey the octet rule and also there is no p orbitals for the electrons as a sp3 hybrid has 3 sp orbitals and no p orbitals
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Note: the above example is an conjugated double bond where it goes double-single-double-single…etc. there are many of these in nature and are mainly responsible for the different colours of a lot molecules this is because the electrons are delocalized, it doesn't take that much energy to excite them so they observe visible light and you end up with different colours • Move lone-pair electrons to an sp2 carbon
Note: amides have two resonance contributors • Move electrons away from the away from the most electronegative atom
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OTHER EXAMPLES
Amide bonds are also stabilised by resonance. This affects the way that proteins adopt a three‐ dimensional shape, as the amide N atoms are also sp2‐ hybridised. (More on this in Module 3).
Ranking resonance structures CHEM1221 Page 5
Ranking resonance structures 1) Resonance structures with separated charges are less stable, and contribute less to the actual molecule, than resonance structures without separated charges. • Carboxylic acid is a good example of this - Here B is going to be less stable than A, and the A contributes a bit more to the actual structure
The left resonance is going to be more stable than the right resonance structure
E is going to be more stable than F
2) Having a negative charge on the more electronegative atom results in a more stable resonance structure, compared to the alternative. So if you had a choice, putting the charges on the more electronegative atoms gives you a more stable resonance structure • Because they are more electronegative they want to pull more electrons towards them… let them have the electrons - H is more stable than G so the real molecule will resemble H more than G
One more thing resonance is handy for doing is distributing charge • The right molecule is more stable than the left one because the negative charge is on the more electronegative atom CHEM1221 Page 6
electronegative atom However if we look at the charge distribution map, we can see that the electron density is spread • And remember the more you can spread electron density out, the more stable that is
Note: we talked about this The other thing resonance does is it can shorten and lengthen bonds As you can see in butane the c-c single bonds are 1.53A but in butadiene the c-c single bonds are 1.47 because of resonance • Because we have a little bit of resonance contributor Also the 1-butene c=c double bond is 1.319 and but the butadiene's c=c double bond is 1.322 • They are really close together, which means this tells you something about the resonances structures contribute
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Question: C2-c3 bond will have some double bond character so we expect it to be shorter than c5-c6
Stabilization of Charges Resonance can also stabilize charges So here in our benzyl cation, is much stable than the cyclohexane carbocation If you look at the distribution map you can see that in the carbocation the charge is stuck in the carbon while in the benzyl cation it is spread over the whole of the benzene ring so it is much more stable
How do you do it?
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You can only do that because all carbons are sp2 hybridize and are in one plane
Effect in a Reaction So if we did an electrophilic addition reaction like the one below…
• •
We would expect equally likely products according to markovnikov's rule According to his rule we should get a 50-50 ratio However when we do the reaction, only one product is detected This can best be explained by resonance stabilization of the carbocation intermediate. Because if we form the carbocation shown below (the one closer to the ring) it can be resonance stabilized as shown below - This makes that carbocation more stable, and it takes less energy to make that than the other one and that’s why you only get one type of product
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Relative stabilities Greater degree of substitution contributes to the stability of carbocation's (as was noted already in Module 1).
The more resonance structures you have the more stable your molecule
Note: experimentally we can look at the stability of carbocations and draw list of their relative stabilities
CHEM1221 Page 10
Essential skills – Module 2 Lecture 1
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1:28 PM
Resonance, EOC, pp. 179 ‐198 Resonance structures describe contribution to a molecule and not the real molecule So we can think of our resonance structures as dragons and unicorns (don’t really exist) but the real molecule/rhino has things that similar to both resonance structures
Note: when we talk about resonance we are talking about delocalized electrons Delocalized Electrons When we talk about resonance the electrons are spread over more than two atoms • This gives you extra stability When it is possible to draw more than one valid Lewis diagram for a molecule or ion, that species is said to have resonance. The electrons are DELOCALIZED i.e. not belonging to a single atom (lone pair) or confined to a bond between two atoms. The molecule or ion is said to be a RESONANCE HYBRID of the structures drawn. • So resonance hybrid is the 'actual molecule' For species with resonance, no single Lewis diagram is sufficient to describe them correctly Resonance structures may not have the same amount of energy Benzene You can think of the electrons as moving • So pi electrons moving around the benzene ring The reason why we can think of electrons as moving is because there is p orbitals on each of the carbon atoms, the p orbitals can overlap so we have cloud of electron density above and below the ring • And when we look at the electron density map, we see that the electrons are spread out evenly over all carbon atoms - So these electrons are delocalised Both resonance structures are equal, however the resonance hybrid is more stable than both of them Since both representations are equal in energy, each C‐C bond has a bond character in between a single and a double bond. • Measured bond length is: 1 39 Å shorter than a typical C‐C and longer than a typical C=C. - One of the consequences of resonances is that the bond length will be different CHEM1221 Page 1
- One of the consequences of resonances is that the bond length will be different Note: Ao = 1x10-10
Resonance also explain why we only get three isomers of benzene when we replace two of the hydrogens • So we are replacing two of the hydrogens If we didn’t have resonance there would be four isomers of benzene, if you had a di-substituted benzene • This is because A and B would be different In A there is a single bond between the two substitutes while in B there is a double bond • So if the electrons in a benzene molecule would LOCOLIZE it would result into two distinct (ortho) molecules but because both A and B are resonance structures, the real molecule has to be different to either of those which explain why we only have 1 ortho isomer Note: for your info, if benzene didn’t have resonance structures, the meta and para ones would still have one isomers
Note: from mod 1, substitutes immediate to each other are ortho (1,2), two spots way are meta (1,3) and three spots away are para (1,4) CHEM1221 Page 2
and three spots away are para (1,4) Cyclooctatetraene It is a cyclic carbon All of the carbons are sp2 hybridized which means it p orbitals could potentially overlap The problem is, it is not a planar molecule, and because of that the two ends point up (ends up in a boat shape) • This causes the p orbitals to be unable to overlap - This means the pi bonds are localized (no resonances) Alternating double bonds and single bonds in a ring system does not in all cases lead to resonance. Measured bond lengths are typical for alternating single (1.46 Å) and double (1.33 Å) bonds. • Since the electrons are localized in cyclooctatetraene the bond length alternate between 1.46 A (single bond) and 1.33 A (double bond) Another reason why this is not aromatic is due to the number of pi bonds… they are even • Or 4n+2=8 ==> n=not a whole number so it is not aromatic
Resonance Only electrons “move" . (Not really moving, but we think of them as if they were to help us find all the resonance structures) The only electrons that move are pi electrons or lone pair electrons. The total number of electrons can’t change. • So the total charge stays the same Electrons can be moved only between p orbitals. And only if they can be aligned properly • So that the Cyclooctatetraene example Examples Electrons can be moved only between p orbitals. • One way of doing that is to move pi electrons to sp2 hybridized carbon because sp2 hybridized carbon will have sp2 orbital + 1 p orbital - So we have p orbital in the sp2 hybridized carbon that we can more the pi electrons to Note: although in the resonance structures each of these carbons have a positive charge, in the real molecule, they have only a partial positive charge Note2: because the two resonance structures are symmetrical they contribute equally to the resonance hybrid If another sp3 carbon is getting in the way you cannot move the electrons because it will not obey the octet rule and also there is no p orbitals for the electrons as a sp3 hybrid has 3 sp orbitals and no p orbitals
CHEM1221 Page 3
Note: the above example is an conjugated double bond where it goes double-single-double-single…etc. there are many of these in nature and are mainly responsible for the different colours of a lot molecules this is because the electrons are delocalized, it doesn't take that much energy to excite them so they observe visible light and you end up with different colours • Move lone-pair electrons to an sp2 carbon
Note: amides have two resonance contributors • Move electrons away from the away from the most electronegative atom
CHEM1221 Page 4
OTHER EXAMPLES
Amide bonds are also stabilised by resonance. This affects the way that proteins adopt a three‐ dimensional shape, as the amide N atoms are also sp2‐ hybridised. (More on this in Module 3).
Ranking resonance structures CHEM1221 Page 5
Ranking resonance structures 1) Resonance structures with separated charges are less stable, and contribute less to the actual molecule, than resonance structures without separated charges. • Carboxylic acid is a good example of this - Here B is going to be less stable than A, and the A contributes a bit more to the actual structure
The left resonance is going to be more stable than the right resonance structure
E is going to be more stable than F
2) Having a negative charge on the more electronegative atom results in a more stable resonance structure, compared to the alternative. So if you had a choice, putting the charges on the more electronegative atoms gives you a more stable resonance structure • Because they are more electronegative they want to pull more electrons towards them… let them have the electrons - H is more stable than G so the real molecule will resemble H more than G
One more thing resonance is handy for doing is distributing charge • The right molecule is more stable than the left one because the negative charge is on the more electronegative atom CHEM1221 Page 6
electronegative atom However if we look at the charge distribution map, we can see that the electron density is spread • And remember the more you can spread electron density out, the more stable that is
Note: we talked about this The other thing resonance does is it can shorten and lengthen bonds As you can see in butane the c-c single bonds are 1.53A but in butadiene the c-c single bonds are 1.47 because of resonance • Because we have a little bit of resonance contributor Also the 1-butene c=c double bond is 1.319 and but the butadiene's c=c double bond is 1.322 • They are really close together, which means this tells you something about the resonances structures contribute
CHEM1221 Page 7
Question: C2-c3 bond will have some double bond character so we expect it to be shorter than c5-c6
Stabilization of Charges Resonance can also stabilize charges So here in our benzyl cation, is much stable than the cyclohexane carbocation If you look at the distribution map you can see that in the carbocation the charge is stuck in the carbon while in the benzyl cation it is spread over the whole of the benzene ring so it is much more stable
How do you do it?
CHEM1221 Page 8
You can only do that because all carbons are sp2 hybridize and are in one plane
Effect in a Reaction So if we did an electrophilic addition reaction like the one below…
• •
We would expect equally likely products according to markovnikov's rule According to his rule we should get a 50-50 ratio However when we do the reaction, only one product is detected This can best be explained by resonance stabilization of the carbocation intermediate. Because if we form the carbocation shown below (the one closer to the ring) it can be resonance stabilized as shown below - This makes that carbocation more stable, and it takes less energy to make that than the other one and that’s why you only get one type of product
CHEM1221 Page 9
Relative stabilities Greater degree of substitution contributes to the stability of carbocation's (as was noted already in Module 1).
The more resonance structures you have the more stable your molecule
Note: experimentally we can look at the stability of carbocations and draw list of their relative stabilities
CHEM1221 Page 10
Essential skills – Module 2 Lecture 1
CHEM1221 Page 11
