Resonance Structures Practice Worksheet Answer Key
Resonance Structures Practice Worksheet Answer Key:
Determine and draw all good resonance structures for the following molecules:
1. 
Hybridization: The hybridization of each atom in the molecule needs to be determined before drawing resonance structures. Only sp and sp2 hybridized atoms can participate in resonance, so we need to find which atoms have those hybridizations. This will tell us what parts of the molecule we can use in the resonance structures. The hybridizations of all atoms and lone pairs in the following molecule are shown in the image below:
Based on the hybridizations shown in the previous image, we now know what parts of the molecule can participate in resonance. The atoms that are sp and sp2 hybridized, and the lone pairs that are in a p orbital can participate in resonance and are highlighted in the image below:
Structure #1: Starting molecule
Structure #2: The second resonance structure is derived from structure #1. It is formed by taking 1 lone pair on oxygen 1 and moving it down to create a pi bond between oxygen 1 and the nitrogen. At the same time, the pi bond between oxygen 2 and the central nitrogen is moved up to become a lone pair. This results in oxygen 1 becoming formally neutral due to oxygen 1 losing 1 electron from the lone pair that is now a pi bond. Oxygen 2 is also formally negative due to gaining 1 electron from the lone pair that was a pi bond.
Structure #3: The third resonance structure is derived from structure #2. It is formed by taking 1 lone pair on oxygen 3 and turning it into a pi bond between oxygen 3 and the nitrogen. At the same time, the pi bond between the central nitrogen and oxygen 1 is moved back up to become a lone pair while keeping oxygen 2 negatively charged. This results in oxygen 3 becoming formally positive due to oxygen 3 losing 1 electron from the lone pair that is now a pi bond. It also results in oxygen 1 becoming formally negative due to oxygen 1 gaining 1 electron from the lone pair that was a pi bond.
Structure #4: The fourth structure is not a good resonance structure. This is because the nitrogen is formally neutral while oxygen 3 is formally positive. Oxygen is more electronegative than nitrogen, so it is not as energetically favorable to have a positive oxygen when the nitrogen could be positive. Therefore, we do not form this structure.
Pi-System: The pi-system is an average of the 3 resonance structures. Each oxygen around the central nitrogen is single bonded in some structures, and double bonded in other structures. This is represented by dotted lines in addition to the single bonds, showing an average of a single bond and a double bond. The partial charges in the pi-system are a reflection of the formal charges in each resonance structure.
2. 
Hybridization: The hybridization of each atom in the molecule needs to be determined before drawing resonance structures. Only sp and sp2 hybridized atoms can participate in resonance, so we need to find which atoms have those hybridizations. This will tell us what parts of the molecule we can use in the resonance structures. The hybridizations of all atoms and lone pairs in the following molecule are shown in the image below:
Based on the hybridizations shown in the previous image, we now know what parts of the molecule can participate in resonance. The atoms that are sp and sp2 hybridized, and the lone pairs that are in a p orbital can participate in resonance and are highlighted in the image below:
Structure #1: Starting structure
Structure #2: The second resonance structure is derived from structure #1. It is formed by moving the pi bond between oxygen and carbon 1 up to the oxygen in the form of a lone pair. This creates a formal negative charge on the oxygen due to the oxygen gaining 1 electron from the lone pair that was a pi bond, and a formal positive charge on carbon 1 due to carbon 1 losing 1 electron from the pi bond that moved.
Structure #3: The third resonance structure is derived from structure #2. It is formed by moving the pi bond on carbons 2 and 3 over to carbons 1 and 2. This results in the formal positive charge moving to carbon 3 due to carbon 3 losing 1 electron from the pi bond that moved.
Pi-System: The pi-system is an average of the 3 resonance structures. There is a dotted line across the oxygen and carbons 1-3. This represents the pi electron density being spread across the oxygen and those carbons. The partial charges in the pi-system are a reflection of the formal charges in each resonance structure.
3.
Hybridization: The hybridization of each atom in the molecule needs to be determined before drawing resonance structures. Only sp and sp2 hybridized atoms can participate in resonance, so we need to find which atoms have those hybridizations. This will tell us what parts of the molecule we can use in the resonance structures. The hybridizations of all atoms and lone pairs in the following molecule are shown in the image below:
Based on the hybridizations shown in the previous image, we now know what parts of the molecule can participate in resonance. The atoms that are sp and sp2 hybridized, and the lone pairs that are in a p orbital can participate in resonance and are highlighted in the image below:
Structure #1: Starting structure
Structure #2: The second resonance structure is derived from structure #1. It is formed by moving the pi bond between carbon 1 and oxygen up to the oxygen up to become a lone pair. This creates a formal negative charge on the oxygen due to the oxygen gaining 1 electron from the lone pair that was a pi bond, and a formal positive charge on carbon 1 due to carbon 1 losing 1 electron from the pi bond that moved.
Structure #3: The third resonance structure is derived from structure #2. It is formed by moving the pi bond between carbons 2 and 3 up to carbons 1 and 2. This moves the formal positive charge from carbon 1 to carbon 3 due to carbon 3 losing 1 electron from the pi bond that moved.
Structure #4: The fourth resonance structure is derived from structure #3. It is formed by moving the pi bond between carbons 4 and 5 to carbons 3 and 4. This moves the formal positive charge from carbon 3 to carbon 5 due to carbon 5 losing 1 electron from the pi bond that moved.
Pi-System: The pi-system is an average of the 4 resonance structures. There is a dotted line across the oxygen and carbons 1-5 because the oxygen and carbons 1-5 were involved with a pi bond in at least one resonance structure. This represents the pi electron density being spread across the oxygen and those carbons. The partial charges in the pi-system are a reflection of the formal charges in each resonance structure.
4.
Hybridization: The hybridization of each atom in the molecule needs to be determined before drawing resonance structures. Only sp and sp2 hybridized atoms can participate in resonance, so we need to find which atoms have those hybridizations. This will tell us what parts of the molecule we can use in the resonance structures. The hybridizations of all atoms and lone pairs in the following molecule are shown in the image below:
Based on the hybridizations shown in the previous image, we now know what parts of the molecule can participate in resonance. The atoms that are sp and sp2 hybridized, and the lone pairs that are in a p orbital can participate in resonance and are highlighted in the image below:
Structure #1: Starting structure
Structure #2: The second resonance structure is derived from structure #1. It is formed by moving the pi bond between carbons 2 and 3 over to carbons 1 and 2. This moves the formal positive charge to carbon 3 due to carbon 3 losing 1 electron from the pi bond that moved.
Structure #3: The third resonance structure is derived from structure #2. It is formed by moving the pi bond between carbons 4 and 5 over to carbons 3 and 4. This moves the formal positive charge to carbon 5 due to carbon 5 losing 1 electron from the pi bond that moved.
Pi-System: The pi-system is an average of the 3 resonance structures. There is a dotted line across carbons 1-5 because carbons 1-5 were involved with a pi bond in at least one resonance structure. This represents the pi electron density being spread across those carbons. The partial charges in the pi-system are a reflection of the formal charges in each resonance structure.
5. 
Hybridization: The hybridization of each atom in the molecule needs to be determined before drawing resonance structures. Only sp and sp2 hybridized atoms can participate in resonance, so we need to find which atoms have those hybridizations. This will tell us what parts of the molecule we can use in the resonance structures. The hybridizations of all atoms and lone pairs in the following molecule are shown in the image below:
Based on the hybridizations shown in the previous image, we now know what parts of the molecule can participate in resonance. The atoms that are sp and sp2 hybridized, and the lone pairs that are in a p orbital can participate in resonance and are highlighted in the image below:
Structure #1: Starting structure
Structure #2: The second resonance structure is derived from structure #1. It is formed by moving the pi bonds around the benzene ring.
Structure #3: The third resonance structure is derived from structure #2. It is formed by moving the pi bond between carbons 2 and 3 out to carbons 1 and 2. This moves the positive charge into the ring and onto carbon 3 due to carbon 3 losing 1 electron from the pi bond that moved.
Structure #4: The fourth resonance structure is derived from structure #3. It is formed by moving the pi bond between carbons 4 and 5 to carbons 3 and 4. This moves the positive charge onto carbon 5 due to carbon 5 losing 1 electron from the pi bond that moved.
Structure #5: The fifth resonance structure is derived from structure #4. It is formed by moving the pi bond between carbons 6 and 7 to carbons 5 and 6. This moves the positive charge onto carbon 7 due to carbon 7 losing 1 electron from the pi bond that moved.
Structure #6: The sixth resonance structure is derived from structure #4. It results from a lone pair on the oxygen moving down to form a pi bond between the oxygen and carbon 5. The formal positive charge moves from carbon 5 to the oxygen due to oxygen losing 1 electron from the lone pair that is now a pi bond.
Pi-System: The pi-system is an average of the 6 resonance structures. There is a dotted line across the oxygen and carbons 1-7. This represents the pi electron density being spread across the oxygen and those carbons. The partial charges in the pi-system are a reflection of the formal charges in each resonance structure.
6.
Hybridization: The hybridization of each atom in the molecule needs to be determined before drawing resonance structures. Only sp and sp2 hybridized atoms can participate in resonance, so we need to find which atoms have those hybridizations. This will tell us what parts of the molecule we can use in the resonance structures. The hybridizations of all atoms and lone pairs in the following molecule are shown in the image below:
Based on the hybridizations shown in the previous image, we now know what parts of the molecule can participate in resonance. The atoms that are sp and sp2 hybridized, and the lone pairs that are in a p orbital can participate in resonance and are highlighted in the image below:
Structure #1: Starting structure
Structure #2: The second resonance structure derived from structure #1. It is formed by moving the pi bond between carbon 1 and oxygen 1 up to oxygen 1 in the form of a lone pair. This results in a formal positive charge on carbon 1 due to carbon 1 losing 1 electron from the pi bond that moved. It also results in oxygen 1 having a formal negative charge due to oxygen 1 gaining 1 electron from the pi bond that is now a lone pair.
Structure #3: The third resonance structure is derived from structure #2. It is formed by moving the pi bond between carbons 2 and 3 to carbons 1 and 2. The formal positive charge moves to carbon 3 due to carbon 3 losing 1 electron from the pi bond that moved.
Structure #4: The fourth resonance structure is derived from structure #3. It is formed by moving a lone pair on oxygen 2 down to form a pi bond between oxygen 2 and carbon 3. The formal positive charge moves to oxygen 2 due to oxygen 2 losing 1 electron from the lone pair that is now a pi bond.
Pi-System: The pi-system is an average of the 4 resonance structures. There is a dotted line across oxygens 1 and 2 and carbons 1-3. This represents the pi electron density being spread across those oxygens and carbons. The partial charges in the pi-system are a reflection of the formal charges in each resonance structure.
7.
Hybridization: The hybridization of each atom in the molecule needs to be determined before drawing resonance structures. Only sp and sp2 hybridized atoms can participate in resonance, so we need to find which atoms have those hybridizations. This will tell us what parts of the molecule we can use in the resonance structures. The hybridizations of all atoms and lone pairs in the following molecule are shown in the image below:
Based on the hybridizations shown in the previous image, we now know what parts of the molecule can participate in resonance. The atoms that are sp and sp2 hybridized, and the lone pairs that are in a p orbital can participate in resonance and are highlighted in the image below:
Structure #1: Starting structure
Structure #2: The second resonance structure derived from structure #1. It is formed by moving the pi bond between carbons 1 and oxygen 1 up to oxygen 1 in the form of a lone pair. This results in a formal positive charge on carbon 1 due to carbon 1 losing 1 electron from the pi bond that moved. It also results in oxygen 1 having a formal negative charge due to oxygen 1 gaining 1 electron from the lone pair that was a pi bond.
Structure #3: The third resonance structure is derived from structure #2. It is formed by moving the pi bond between carbons 2 and 3 to carbons 1 and 2. The formal positive charge moves to carbon 3 due to carbon 3 losing 1 electron from the pi bond that moved.
Structure #4: The fourth resonance structure is derived from structure #3. It is formed by moving the bottom pi bond between carbons 4 and 5 to carbons 3 and 4. The formal positive charge moves to carbon 5 due to carbon 5 losing 1 electron from the pi bond that moved.
Structure #5: The fifth resonance structure is derived from structure #1. It is formed by moving 1 lone pair on oxygen 2 to form a pi bond between oxygen 2 and carbon 2. At the same time, the pi bond between carbons 2 and 3 is pushed onto carbon 3 as a lone pair. This creates a formal positive charge on oxygen 2 due to oxygen 2 losing 1 electron from the lone pair that is now a pi bond, and a formal negative charge on carbon 3 due to carbon 3 gaining 1 electron from the lone pair that was a pi bond.
Structure #6: The sixth resonance structure is derived from structure #5. It is formed by moving the lone pair on carbon 3 to form a pi bond between carbons 3 and 4. This pushes the pi bond between carbons 4 and 5 out onto carbon 5 in the form of a lone pair. The formal negative charge is moved to carbon 5 due to carbon 5 gaining 1 electron from the lone pair.
Pi-System: The pi-system is an average of the 6 resonance structures. There is a dotted line across oxygens 1 and 2 and carbons 1-5. This represents the pi electron density being spread across the oxygen and those carbons. The partial charges in the pi-system are a reflection of the formal charges in each resonance structure. Note that carbons 3 and 5 do not have partial charges shown. This is because each carbon has both formal positive and negative charges in some of the resonance structures. When a formal positive and formal negative charge are averaged out, it comes to no charge.
8.
Hybridization: The hybridization of each atom in the molecule needs to be determined before drawing resonance structures. Only sp and sp2 hybridized atoms can participate in resonance, so we need to find which atoms have those hybridizations. This will tell us what parts of the molecule we can use in the resonance structures. The hybridizations of all atoms and lone pairs in the following molecule are shown in the image below:
Based on the hybridizations shown in the previous image, we now know what parts of the molecule can participate in resonance. The atoms that are sp and sp2 hybridized, and the lone pairs that are in a p orbital can participate in resonance and are highlighted in the image below:
Structure #1: Starting structure
Structure #2: The second resonance structure is derived from structure #1. It is formed by moving the lone pair on the nitrogen down to form a pi bond between the nitrogen and carbon 1. At the same time, the pi bond between carbons 1 and 2 is pushed onto carbon 2 in the form of a lone pair. This results in a formal positive charge on the nitrogen due to the nitrogen losing 1 electron from the lone pair that is now a pi bond. This also results in a formal negative charge on carbon 2 due to carbon 2 gaining 1 electron from the pi bond that is now a lone pair.
Structure #3: The third resonance structure is derived from structure #2. It is formed by moving the lone pair on carbon 2 over to form a pi bond between carbons 2 and 3. At the same time, the pi bond between carbons 3 and 4 is pushed onto carbon 4 in the form of a lone pair. This results in the formal negative charge moving to carbon 4 due to carbon 4 gaining 1 electron from the pi bond that is now a lone pair.
Pi-System: The pi-system is an average of the 3 resonance structures. There is a dotted line across the nitrogen and carbons 1-4. This represents the pi electron density being spread across the nitrogen and those carbons. The partial charges in the pi-system are a reflection of the formal charges in the resonance structures.
9.
Hybridization: The hybridization of each atom in the molecule needs to be determined before drawing resonance structures. Only sp and sp2 hybridized atoms can participate in resonance, so we need to find which atoms have those hybridizations. This will tell us what parts of the molecule we can use in the resonance structures. The hybridizations of all atoms and lone pairs in the following molecule are shown in the image below:
Based on the hybridizations shown in the previous image, we now know what parts of the molecule can participate in resonance. The atoms that are sp and sp2 hybridized, and the lone pairs that are in a p orbital can participate in resonance and are highlighted in the image below:
Structure #1: Starting structure
Structure #2: The second resonance structure is derived from structure #1. It is formed by moving the pi bonds around the benzene ring.
Structure #3: The third resonance structure is derived from structure #2. It is formed by moving a lone pair on the oxygen down to form a pi bond between the oxygen and carbon 1. At the same time, the pi bond between carbon 1 and 2 is pushed onto carbon 2 in the form a lone pair. This results in a formal positive charge on oxygen due to oxygen losing 1 electron from the lone pair that is now a pi bond. Carbon 2 also has a formal negative charge due to carbon 2 gaining 1 electron from the pi bond that is now a lone pair. Note that the electronegativity of oxygen is greater than the hydrogen it is bonded to (EN X>Y), so the alcohol is an electron donating group to the ring.
Structure #4: The fourth resonance structure is derived from structure #3. It is formed by moving the lone pair on carbon 2 to form a pi bond between carbons 2 and 3. At the same time, the pi bond between carbons 3 and 4 is pushed onto carbon 4 in the form of a lone pair. This results in the formal negative charge moving to carbon 4 due to carbon 4 gaining 1 electron from the pi bond that is now a lone pair.
Structure #5: The fifth resonance structure is derived from structure #4. It is formed by moving the lone pair on carbon 4 to form a pi bond between carbons 4 and 5. At the same time, the pi bond between carbons 5 and 6 is pushed onto carbon 6 as a lone pair. This results in the formal negative charge moving to carbon 6 due to carbon 6 gaining 1 electron form the pi bond that is now a lone pair.
Pi-System: The pi-system is an average of the 5 resonance structures. There is a dotted line across the oxygen and carbons 1-6. This represents the pi electron density being spread across the oxygen and those carbons. The partial charges in the pi-system are a reflection of the formal charges in each resonance structure.
10.
Hybridization: The hybridization of each atom in the molecule needs to be determined before drawing resonance structures. Only sp and sp2 hybridized atoms can participate in resonance, so we need to find which atoms have those hybridizations. This will tell us what parts of the molecule we can use in the resonance structures. The hybridizations of all atoms and lone pairs in the following molecule are shown in the image below:
Based on the hybridizations shown in the previous image, we now know what parts of the molecule can participate in resonance. The atoms that are sp and sp2 hybridized, and the lone pairs that are in a p orbital can participate in resonance and are highlighted in the image below:
Structure #1: Starting structure
Structure #2: The second resonance structure is derived from structure #1. It results from moving the pi bonds around the benzene ring.
Structure #3: The third resonance structure is derived from structure #2. It is formed by moving the pi bond between carbon 1 and the oxygen up to the oxygen in the form of a lone pair. At the same time, the pi bond in the benzene ring between carbons 2 and 3 is moved to carbons 1 and 2. This results in a formal negative charge on the oxygen due to oxygen gaining 1 electron from the pi bond that is now a lone pair. Carbon 3 also has a formal positive charge due to carbon 3 losing 1 electron from the pi bond that moved. Note that the electronegativity of carbon 1 is less than the oxygen it is bonded to (EN Y>X), so the ketone is an electron withdrawing group from the ring.
Structure #4: The fourth resonance structure is derived from structure #3. It is formed by moving the pi bond between carbons 4 and 5 to carbons 3 and 4. This results in the formal positive charge moving to carbon 5 due to carbon 5 losing 1 electron from the pi bond that moved.
Structure #5: The fifth resonance structure is derived from structure #4. It is formed by moving the pi bond between carbons 6 and 7 to carbons 5 and 6. This moves the formal positive charge to carbon 7 due to carbon 7 losing 1 electron from the pi bond that moved.
Pi-System: The pi-system is an average of the 5 resonance structures. There is a dotted line across the oxygen and carbons 1-7. This represents the pi electron density being spread across the oxygen and those carbons. The partial charges in the pi-system are a reflection of the formal charges in each resonance structure.
11.
Hybridization: The hybridization of each atom in the molecule needs to be determined before drawing resonance structures. Only sp and sp2 hybridized atoms can participate in resonance, so we need to find which atoms have those hybridizations. This will tell us what parts of the molecule we can use in the resonance structures. The hybridizations of all atoms and lone pairs in the following molecule are shown in the image below:
Based on the hybridizations shown in the previous image, we now know what parts of the molecule can participate in resonance. The atoms that are sp and sp2 hybridized, and the lone pairs that are in a p orbital can participate in resonance and are highlighted in the image below:
Structure #1: Starting structure
Structure #2: The second resonance structure is derived from structure #1. It is formed by moving the pi bonds around the benzene ring.
Structure #3: The third resonance structure is derived from structure #2. It is formed by moving a lone pair on the sulfur to form a pi bond between the sulfur and carbon 1. At the same time the pi bond between carbons 1 and 2 is pushed onto carbon 2 in the form of a lone pair. This results in a formal positive charge on the sulfur due to sulfur losing 1 electron from the lone pair that is now a pi bond. Carbon 2 also has a formal negative charge due to carbon 2 gaining 1 electron from the pi bond that is now a lone pair.
Structure #4: The fourth resonance structure is derived from structure #3. It is formed by moving the lone pair on carbon 2 to form a pi bond between carbons 2 and 3. At the same time the pi bond between carbons 3 and 4 is pushed onto carbon 4 as a lone pair. This moves the formal negative charge to carbon 4 due to carbon 4 gaining 1 electron from the pi bond that is now a lone pair.
Structure #5: The fifth resonance structure is derived from structure #4. It is formed by moving the lone pair on carbon 4 to form a pi bond between carbons 4 and 5. At the same time the pi bond between carbons 5 and 6 is pushed onto carbon 6 as a lone pair. This moves the formal negative charge to carbon 6 due to carbon 6 gaining 1 electron from the pi bond that is now a lone pair.
Structure #6: The sixth resonance structure is derived from structure #1. It is formed by moving a lone pair on the sulfur to form a pi bond between the sulfur and carbon 7. At the same time the pi bond between carbons 7 and 8 is pushed onto carbon 8 as a lone pair. This results in a formal positive charge on the sulfur due to sulfur losing 1 electron from the lone pair that is now a pi bond. Carbon 8 also has a formal negative charge due to carbon 8 gaining 1 electron from the pi bond that is now a lone pair.
*note: all resonance structures with a benzene ring have a twin structure with the pi bonds rotated in the ring, however not all of those structures are shown. The only example of this shown in the answers is structures #1 and #2, but an additional structure with rotated pi bonds could also be shown for structure #6.
Pi-System: The pi-system is an average of the 6 resonance structures. There is a dotted line across the sulfur and carbons 1-8. This represents the pi electron density being spread across the oxygen and those carbons. The partial charges in the pi-system are a reflection of the formal charges in each resonance structure.
12.
Hybridization: The hybridization of each atom in the molecule needs to be determined before drawing resonance structures. Only sp and sp2 hybridized atoms can participate in resonance, so we need to find which atoms have those hybridizations. This will tell us what parts of the molecule we can use in the resonance structures. The hybridizations of all atoms and lone pairs in the following molecule are shown in the image below:
Based on the hybridizations shown in the previous image, we now know what parts of the molecule can participate in resonance. The atoms that are sp and sp2 hybridized, and the lone pairs that are in a p orbital can participate in resonance and are highlighted in the image below:
Structure #1: Starting structure
Structure #2: The second resonance structure is derived from structure #1. It is formed by moving the lone pair on carbon 1 to form a pi bond between carbons 1 and 2. At the same time the pi bond between carbon 2 and oxygen 1 is moved up to to oxygen 1 in the form of a lone pair. This results in the formal negative charge moving to oxygen 1 due to oxygen 1 gaining 1 electron from the pi bond that is now a lone pair.
Structure #3: The third resonance structure is derived from structure #1. It is formed by moving the lone pair on carbon 1 to form a pi bond between carbons 1 and 3. At the same time the pi bond between carbon 3 and oxygen 2 is moved up to oxygen 2 in the form of a lone pair. This results in the formal negative charge moving to oxygen 2 due to oxygen 2 gaining 1 electron from the pi bond that is now a lone pair.
Pi-System: The pi-system is an average of the 3 resonance structures. There is a dotted line across oxygens 1-2 and carbons 1-3. This represents the pi electron density being spread across the oxygen and those carbons. The partial charges in the pi-system are a reflection of the formal charges in each resonance structure.
13.
Hybridization: The hybridization of each atom in the molecule needs to be determined before drawing resonance structures. Only sp and sp2 hybridized atoms can participate in resonance, so we need to find which atoms have those hybridizations. This will tell us what parts of the molecule we can use in the resonance structures. The hybridizations of all atoms and lone pairs in the following molecule are shown in the image below:
Based on the hybridizations shown in the previous image, we now know what parts of the molecule can participate in resonance. The atoms that are sp and sp2 hybridized, and the lone pairs that are in a p orbital can participate in resonance and are highlighted in the image below:
Structure #1: Starting structure
Structure #2: The second resonance structure is derived from structure #1. It is formed by moving the pi bonds around the ring.
Structure #3: The third resonance structure is derived from structure #2. It is formed by moving the pi bond between the nitrogen and carbon 1 to the nitrogen in the form of a lone pair. This results in a formal negative charge on the nitrogen due to the nitrogen gaining 1 electron from the pi bond that is now a lone pair. It also results in a formal positive charge on carbon 1 due to carbon 1 losing 1 electron from the pi bond that moved.
Structure #4: The fourth resonance structure is derived from structure #3. It is formed by the pi bond between carbons 2 and 3 moving to carbons 1 and 2. This results in the formal positive charge moving to carbon 3 due to carbon 3 losing 1 electron from the pi bond that moved.
Structure #5: The fifth resonance structure is derived from structure #4. It is formed by the pi bond between carbons 4 and 5 moving to carbons 3 and 4. This results in the formal positive charge moving to carbon 5 due to carbon 5 losing 1 electron from the pi bond that moved.
Pi-System: The pi-system is an average of the 5 resonance structures. There is a dotted line across the nitrogen and carbons 1-5. This represents the pi electron density being spread across the nitrogen and those carbons. The partial charges in the pi-system are a reflection of the formal charges in each resonance structure.
14.
Hybridization: The hybridization of each atom in the molecule needs to be determined before drawing resonance structures. Only sp and sp2 hybridized atoms can participate in resonance, so we need to find which atoms have those hybridizations. This will tell us what parts of the molecule we can use in the resonance structures. The hybridizations of all atoms and lone pairs in the following molecule are shown in the image below:
Based on the hybridizations shown in the previous image, we now know what parts of the molecule can participate in resonance. The atoms that are sp and sp2 hybridized, and the lone pairs that are in a p orbital can participate in resonance and are highlighted in the image below:
Structure #1: Starting structure
Structure #2: The second resonance structure is derived from structure #1. It is formed by the pi bond between the oxygen and carbon 1 moving to the oxygen in the form of a lone pair. At the same time, the lone pair on the nitrogen moves to form a pi bond between the nitrogen and carbon 1. This results in a formal negative charge on oxygen 1 due to oxygen gaining 1 electron from the pi bond that is now a lone pair. It also results in a formal positive charge on the nitrogen due to the nitrogen losing 1 electron from the lone pair that is now a pi bond.
Structure #3: The third resonance structure is derived from structure #2. It is formed by the pi bond between the nitrogen and carbon 1 moving back to the nitrogen in the form of a lone pair. This results in the formal positive charge moving to carbon 1 due to carbon 1 losing 1 electron from the pi bond that moved. Note that structure #3 is a minor resonance contributor because having a formal positive carbon adjacent to a highly electronegative nitrogen is very unstable.
Pi-System: The pi-system is an average of the 3 resonance structures. There is a dotted line across the oxygen, nitrogen, and carbon 1. This represents the pi electron density being spread across the oxygen and those carbons. The partial charges in the pi-system are a reflection of the formal charges in each resonance structure.
15.
Hybridization: The hybridization of each atom in the molecule needs to be determined before drawing resonance structures. Only sp and sp2 hybridized atoms can participate in resonance, so we need to find which atoms have those hybridizations. This will tell us what parts of the molecule we can use in the resonance structures. The hybridizations of all atoms and lone pairs in the following molecule are shown in the image below:
Based on the hybridizations shown in the previous image, we now know what parts of the molecule can participate in resonance. The atoms that are sp and sp2 hybridized, and the lone pairs that are in a p orbital can participate in resonance and are highlighted in the image below:
Structure #1: Starting structure
Structure #2: The second resonance structure is derived from structure #1. It is formed by the pi bond between carbons 2 and 3 moving to carbons 1 and 2. This results in the formal positive charge moving to carbon 3 due to carbon 3 losing 1 electron from the pi bond that moved.
Structure #3: The third resonance structure is derived from structure #2. It is formed by the pi bond between carbons 4 and 5 moving to carbons 3 and 4. This results in the formal positive charge moving to carbon 5 due to carbon 5 losing 1 electron from the pi bond that moved.
Structure #4: The fourth resonance structure is derived from structure #3. It is formed by the pi bond between carbons 6 and 7 moving to carbons 5 and 6. This results in the formal positive charge moving to carbon 7 due to carbon 7 losing 1 electron from the pi bond that moved.
Structure #5: The fifth resonance structure is derived from structure #4. It is formed by a lone pair on the oxygen moving to form a pi bond between the oxygen and carbon 7. This results in the formal positive charge moving to the oxygen due to the oxygen losing 1 electron from the lone pair that is now a pi bond.
Pi-System: The pi-system is an average of the 5 resonance structures. There is a dotted line across the oxygen and carbons 1-7. This represents the pi electron density being spread across the oxygen and those carbons. The partial charges in the pi-system are a reflection of the formal charges in each resonance structure.
16.
Hybridization: The hybridization of each atom in the molecule needs to be determined before drawing resonance structures. Only sp and sp2 hybridized atoms can participate in resonance, so we need to find which atoms have those hybridizations. This will tell us what parts of the molecule we can use in the resonance structures. The hybridizations of all atoms and lone pairs in the following molecule are shown in the image below:
Based on the hybridizations shown in the previous image, we now know what parts of the molecule can participate in resonance. Those parts are highlighted in the image below:
Structure #1: Starting structure
Structure #2: The second resonance structure is derived from structure #1. It is formed by moving the pi bond between carbon 1 and the oxygen up to the oxygen in the form of a lone pair. This results in a formal negative charge on the oxygen due to the oxygen gaining 1 electron from the pi bond that is now a lone pair, and formal positive charge on carbon 1 due to carbon 1 losing 1 electron from the pi bond that moved.
Structure #3: The third resonance structure is derived from structure #2. It is formed by moving the pi bond between carbons 2 and 3 to carbons 1 and 2. This results in the formal positive charge moving to carbon 3 due to carbon 3 losing 1 electron from the pi bond that moved.
Structure #4: The fourth resonance structure is derived from structure #3. It is formed by moving the lone pair from the nitrogen to form a pi bond between the nitrogen and carbon 3. This moves the formal positive charge onto the nitrogen due to nitrogen losing 1 electron from the lone pair that is now a pi bond.
Pi-System: The pi-system is an average of the 4 resonance structures. There is a dotted line across the nitrogen and carbons 1-3. This represents the pi electron density being spread across the oxygen and those carbons. The partial charges in the pi-system are a reflection of the formal charges in each resonance structure.
17.
Hybridization: The hybridization of each atom in the molecule needs to be determined before drawing resonance structures. Only sp and sp2 hybridized atoms can participate in resonance, so we need to find which atoms have those hybridizations. This will tell us what parts of the molecule we can use in the resonance structures. The hybridizations of all atoms and lone pairs in the following molecule are shown in the image below:
Based on the hybridizations shown in the previous image, we now know what parts of the molecule can participate in resonance. Those parts are highlighted in the image below:
Structure #1: Starting structure
Structure #2: The second resonance structure is derived from structure #1. It is formed by moving the pi bond between the oxygen and carbon 1 to the oxygen in the form of a lone pair. This results in a formal negative charge on the oxygen due to the oxygen gaining 1 electron from the pi bond that is now a lone pair. It also results in a formal positive charge on carbon 1 due to carbon 1 losing 1 electron from the pi bond that moved.
Structure #3: The third resonance structure is derived from structure #2. It is formed by the pi bond between carbons 2 and 3 moving to carbons 1 and 2. This results in the formal positive charge moving to carbon 3 due to carbon 3 losing 1 electron from the pi bond that moved.
Structure #4: The fourth resonance structure is derived from structure #3. It is formed by the pi bond between carbons 4 and 5 moving to carbons 3 and 4. This results in the formal positive charge moving to carbon 5 due to carbon 5 losing 1 electron from the pi bond that moved.
Structure #5: The fifth resonance structure is derived from structure #4. It is formed by the pi bond between carbons 6 and 7 moving to carbons 5 and 6. This results in the formal positive charge moving to carbon 7 due to carbon 7 losing 1 electron from the pi bond that moved.
Structure #6: The sixth resonance structure is derived from structure #5. It is formed by the lone pair on the nitrogen moving to form a pi bond between the nitrogen and carbon 7. This results in the formal positive charge moving to the nitrogen due to nitrogen losing 1 electron from the lone pair that is now a pi bond.
Pi-System: The pi-system is an average of the 6 resonance structures. There is a dotted line across the oxygen, nitrogen, and carbons 1-7. This represents the pi electron density being spread across the oxygen, nitrogen, and carbons 1-7. The partial charges in the pi-system are a reflection of the formal charges in each resonance structure.
18.
Hybridization: The hybridization of each atom in the molecule needs to be determined before drawing resonance structures. Only sp and sp2 hybridized atoms can participate in resonance, so we need to find which atoms have those hybridizations. This will tell us what parts of the molecule we can use in the resonance structures. The hybridizations of all atoms and lone pairs in the following molecule are shown in the image below:
Based on the hybridizations shown in the previous image, we now know what parts of the molecule can participate in resonance. Those parts are highlighted in the image below:
Structure #1: Starting structure
Structure #2: The second resonance structure is derived from structure #1. It is formed by the pi bond between carbon 1 and the oxygen moving up to the oxygen in the form of a lone pair. This results in a formal negative charge on the oxygen due to the oxygen gaining 1 electron from the lone pair that was the pi bond. It also results in a formal positive charge on carbon 1 due to carbon 1 losing 1 electron from the pi bond that moved.
Structure #3: The third resonance structure is derived from structure #2. It is formed by the pi bond between carbon 2 and 3 moving to carbons 1 and 2. This results in the formal positive charge moving to carbon 3 due to carbon 3 losing 1 electron from the pi bond that moved.
Structure #4: The fourth resonance structure is derived from structure #3. It is formed by the pi bond between carbons 4 and 5 moving to carbons 3 and 4. This results in the formal positive charge moving to carbon 5 due to carbon 5 losing 1 electron from the pi bond that moved.
Structure #5: The fifth resonance structure is derived from structure #4. It is formed by the pi bond between carbons 6 and 7 moving to carbons 5 and 6. This results in the formal positive charge moving to carbon 7 due to carbon 7 losing 1 electron from the pi bond that moved.
Pi-System: The pi-system is an average of the 5 resonance structures. There is a dotted line across the oxygen and carbons 1-7. This represents the pi electron density being spread across the oxygen and those carbons. The partial charges in the pi-system are a reflection of the formal charges in each resonance structure.
19.
Hybridization: The hybridization of each atom in the molecule needs to be determined before drawing resonance structures. Only sp and sp2 hybridized atoms can participate in resonance, so we need to find which atoms have those hybridizations. This will tell us what parts of the molecule we can use in the resonance structures. The hybridizations of all atoms and lone pairs in the following molecule are shown in the image below:
Based on the hybridizations shown in the previous image, we now know what parts of the molecule can participate in resonance. Those parts are highlighted in the image below:
Structure #1: Starting structure
Structure #2: The second resonance structure is derived from structure #1. It is formed by moving the pi bonds around the benzene ring.
Structure #3: The third resonance structure is derived from structure #2. It is formed by moving the the pi bond between carbon 1 and oxygen up to the oxygen in the form of a lone pair. This results in a formal negative charge on the oxygen due to the oxygen gaining 1 electron from the pi bond that is now a lone pair, and a formal positive charge on the carbon that lost the pi bond.
Structure #4: The fourth resonance structure is derived from structure #3. It is formed by moving the pi bond between carbons 2 and 3 over to carbons 1 and 2. This results in the formal positive charge moving to carbon 3 due to carbon 3 losing 1 electron from the pi bond that moved.
*note: all resonance structures with a benzene ring have a twin structure with the pi bonds rotated in the ring, however not all of those structures are shown. The only example of this shown in the answers is structures #1 and #2, but additional structures with rotated pi bonds could also be shown for structure #3 and structure #4.
Structure #5: The fifth resonance structure is derived from structure #4. It is formed by moving the pi bond from carbons 4 and 5 in the ring to carbons 3 and 4. This results in the formal positive charge moving to carbon 5 due to carbon 5 losing 1 electron from the pi bond that moved.
Structure #6: The sixth resonance structure is derived from structure #5. It is formed by moving the pi bond from carbons 6 and 7 to carbons 5 and 6. This results in the formal positive charge moving to carbon 7 due to carbon 7 losing 1 electron from the pi bond that moved.
Structure #7: The seventh resonance structure is derived from structure #6. It is formed by moving the pi bond from carbons 8 and 9 to carbons 7 and 8. This results in the formal positive charge moving to carbon 9 due to carbon 9 losing 1 electron from the pi bond that moved.
Pi-System: The pi-system is an average of the 7 resonance structures. There is a dotted line across the oxygen and carbons 1-9. This represents the pi electron density being spread across the oxygen and those carbons. The partial charges in the pi-system are a reflection of the formal charges in each resonance structure.
20. 
Hybridization: The hybridization of each atom in the molecule needs to be determined before drawing resonance structures. Only sp and sp2 hybridized atoms can participate in resonance, so we need to find which atoms have those hybridizations. This will tell us what parts of the molecule we can use in the resonance structures. The hybridizations of all atoms and lone pairs in the following molecule are shown in the image below:
Based on the hybridizations shown in the previous image, we now know what parts of the molecule can participate in resonance. Those parts are highlighted in the image below:
Structure #1: Starting structure
Structure #2: The second resonance structure is derived from structure #1. It is formed by moving the pi bonds around the benzene ring.
Structure #3: The third resonance structure is derived from structure #2. It is formed by moving the lone pair on nitrogen 1 down to form a pi bond between nitrogen 1 and carbon 1. At the same time, the pi bond between carbons 1 and 2 is pushed onto carbon 2 in the form of a lone pair. This results in a formal positive charge on the nitrogen due to the nitrogen losing 1 electron from the lone pair that is now a pi bond. Carbon 2 also has a formal negative charge due to gaining 1 electron from the pi bond that is now a lone pair.
*note: all resonance structures with a benzene ring have a twin structure with the pi bonds rotated in the ring, however not all of those structures are shown. The only example of this shown in the answers is structures #1 and #2, but an additional structure with rotated pi bonds could also be shown for structure #3.
Structure #4: The fourth resonance structure is derived from structure #3. It is formed by moving the lone pair on carbon 2 to form a pi bond between carbons 2 and 3. At the same time, the pi bond between carbons 3 and 4 is pushed onto carbon 4 in the form of a lone pair. This results in the formal negative charge moving to carbon 4 due to carbon 4 gaining 1 electron from the pi bond that is now a lone pair.
Structure #5: The fifth resonance structure is derived from structure #4. It is formed by moving the lone pair on carbon 4 to form a pi bond between carbons 4 and 5. At the same time, the pi bond between carbons 5 and 6 is pushed onto carbon 6 in the form of a lone pair. This results in the formal negative charge moving to carbon 6 due to carbon 6 gaining 1 electron from the pi bond that is now a lone pair.
Structure #6: The sixth resonance structure is derived from structure #5. It is formed by moving the lone pair on carbon 6 to form a pi bond between carbons 6 and 7. At the same time, the pi bond between carbons 7 and 8 is pushed onto carbon 8 in the form of a lone pair. This results in the formal negative charge moving to carbon 8 due to carbon 8 gaining 1 electron from the pi bond that is now a lone pair.
Pi-System: The pi-system is an average of the 6 resonance structures. There is a dotted line across the nitrogen and carbons 1-8. This represents the pi electron density being spread across the oxygen and those carbons. The partial charges in the pi-system are a reflection of the formal charges in each resonance structure.
21.
Hybridization: The hybridization of each atom in the molecule needs to be determined before drawing resonance structures. Only sp and sp2 hybridized atoms can participate in resonance, so we need to find which atoms have those hybridizations. This will tell us what parts of the molecule we can use in the resonance structures. The hybridizations of all atoms and lone pairs in the following molecule are shown in the image below:
Based on the hybridizations shown in the previous image, we now know what parts of the molecule can participate in resonance. Those parts are highlighted in the image below:
Structure #1: Starting structure
Structure #2: The second resonance structure is derived from structure #1. It is formed by the pi bond between the oxygen and carbon 1 moving up to the oxygen in the form of a lone pair. This results in a formal negative charge on the oxygen due to the oxygen gaining 1 electron from the lone pair that was a pi bond. Carbon 1 also has a formal positive charge because it lost 1 electron from the pi bond that moved.
Structure #3: The third resonance structure is derived from structure #2. It is formed by the pi bond between carbons 2 and 3 moving to carbons 1 and 2. This results in the formal positive charge moving to carbon 3 due to carbon 3 losing 1 electron from the pi bond that moved.
Pi-System: The pi-system is an average of the 3 resonance structures. There is a dotted line across the oxygen and carbons 1-5. This represents the pi electron density being spread across the oxygen and those carbons. The partial charges in the pi-system are a reflection of the formal charges in each resonance structure.
22.
Hybridization: The hybridization of each atom in the molecule needs to be determined before drawing resonance structures. Only sp and sp2 hybridized atoms can participate in resonance, so we need to find which atoms have those hybridizations. This will tell us what parts of the molecule we can use in the resonance structures. The hybridizations of all atoms and lone pairs in the following molecule are shown in the image below:
Based on the hybridizations shown in the previous image, we now know what parts of the molecule can participate in resonance. Those parts are highlighted in the image below:
Structure #1: Starting structure
Structure #2: The second resonance structure is derived from structure #1. It is formed by moving the pi bond between the nitrogen and carbon 1 onto the nitrogen in the form of a lone pair. This makes the nitrogen formally neutral due to the nitrogen gaining 1 electron from the lone pair that was a pi bond. Carbon 1 also becomes formally positive because it lost 1 electron from the pi bond.
Structure #3: The third resonance structure is derived from structure #2. It is formed by moving the pi bond from carbons 2 and 3 to carbons 1 and 2. This moves the formal positive charge to carbon 3 due to carbon 3 losing 1 electron from the pi bond that moved.
Structure #4: The fourth resonance structure is derived from structure #3. It is formed by moving the pi bond from carbons 4 and 5 to carbons 3 and 4. This moves the formal positive charge to carbon 5 due to carbon 5 losing 1 electron from the pi bond that moved.
Pi-System: The pi-system is an average of the 4 resonance structures. There is a dotted line across the nitrogen and carbons 1-5. This represents the pi electron density being spread across the oxygen and those carbons. The partial charges in the pi-system are a reflection of the formal charges in each resonance structure.
23.
Hybridization: The hybridization of each atom in the molecule needs to be determined before drawing resonance structures. Only sp and sp2 hybridized atoms can participate in resonance, so we need to find which atoms have those hybridizations. This will tell us what parts of the molecule we can use in the resonance structures. The hybridizations of all atoms and lone pairs in the following molecule are shown in the image below:
Based on the hybridizations shown in the previous image, we now know what parts of the molecule can participate in resonance. Those parts are highlighted in the image below:
Structure #1: Starting structure
Structure #2: The second resonance structure is derived from structure #1. It is formed by moving 1 of the pi bonds between carbon 1 and nitrogen 1 to nitrogen 1 in the form of a lone pair. This creates a formal negative charge on nitrogen 1 due to nitrogen 1 gaining 1 electron from the lone pair that was a pi bond. Carbon 1 also has a formal positive charge due to losing 1 electron from the pi bond that moved.
Structure #3: The third resonance structure is derived from structure #2. It is formed by the pi bond between carbons 2 and 3 moving to carbons 1 and 2. This results in the formal positive charge moving to carbon 3 due to carbon 3 losing 1 electron from the pi bond that moved.
Structure #4: The fourth resonance structure is derived from structure #3. It is formed by the lone pair on nitrogen 2 moving to form a pi bond between nitrogen 2 and carbon 3. This results in the formal positive charge moving to nitrogen 2 due to nitrogen 2 losing 1 electron from the lone pair that is now a pi bond.
Pi-System: The pi-system is an average of the 4 resonance structures. There is a dotted line across both nitrogens and carbons 1-3. This represents the pi electron density being spread across the oxygen and those carbons. The partial charges in the pi-system are a reflection of the formal charges in each resonance structure.
24.
Hybridization: The hybridization of each atom in the molecule needs to be determined before drawing resonance structures. Only sp and sp2 hybridized atoms can participate in resonance, so we need to find which atoms have those hybridizations. This will tell us what parts of the molecule we can use in the resonance structures. The hybridizations of all atoms and lone pairs in the following molecule are shown in the image below:
Based on the hybridizations shown in the previous image, we now know what parts of the molecule can participate in resonance. Those parts are highlighted in the image below:
Structure #1: Starting structure
Structure #2: The second resonance structure is derived from structure #1. It is formed by moving the pi bonds around the benzene ring.
Structure #3: The third resonance structure is derived from structure #2. It is formed by 1 lone pair on the bromine moving down to form a pi bond between the bromine and carbon 1. At the same time, the pi bond between carbons 1 and 2 is pushed onto carbon 2 in the form of a lone pair. This results in carbon 2 having a formal negative charge due to carbon 2 gaining 1 electron from the lone pair that was a pi bond. The bromine also has a formal positive charge due to losing 1 electron from the lone pair that is now a pi bond.
Structure #4: The fourth resonance structure is derived from structure #3. It is formed by the lone pair on carbon 2 moving to form a pi bond between carbons 2 and 3. At the same time, the pi bond between carbons 3 and 4 is pushed onto carbon 4 in the form of a lone pair. This results in the formal negative charge moving to carbon 4 due to carbon 4 gaining 1 electron from the lone pair that was a pi bond.
Structure #5: The fifth resonance structure is derived from structure #4. It is formed by the lone pair on carbon 4 moving to form a pi bond between carbons 4 and 5. At the same time, the pi bond between carbons 5 and 6 is pushed onto carbon 6 in the form of a lone pair. This results in the formal negative charge moving to carbon 6 due to carbon 6 gaining 1 electron from the lone pair that was a pi bond.
Pi-System: The pi-system is an average of the 5 resonance structures. There is a dotted line across the bromine and carbons 1-6. This represents the pi electron density being spread across the oxygen and those carbons. The partial charges in the pi-system are a reflection of the formal charges in each resonance structure.
25. 
Hybridization: The hybridization of each atom in the molecule needs to be determined before drawing resonance structures. Only sp and sp2 hybridized atoms can participate in resonance, so we need to find which atoms have those hybridizations. This will tell us what parts of the molecule we can use in the resonance structures. The hybridizations of all atoms and lone pairs in the following molecule are shown in the image below:
Based on the hybridizations shown in the previous image, we now know what parts of the molecule can participate in resonance. Those parts are highlighted in the image below:
Structure #1: Starting structure
Structure #2: The second resonance structure is derived from structure #1. The pi bond between the nitrogen and carbon 1 moves to the nitrogen in the form of a lone pair. This results in a formal negative charge on the nitrogen due to the nitrogen gaining 1 electron from the lone pair that was a pi bond. Carbon 1 also has a formal positive charge due to losing 1 electron from the pi bond that moved.
Structure #3: The third resonance structure is derived from structure #1. It is formed by 1 lone pair on the oxygen moving to form a pi bond between oxygen and carbon 6. At the same time, the pi bond between carbons 6 and 7 is pushed onto carbon 7 in the form of a lone pair. This results in a formal positive charge on the oxygen due to the oxygen losing 1 electron from the lone pair that is now a pi bond. Carbon 7 also has a formal negative charge due to gaining 1 electron from the lone pair that was a pi bond.
Structure #4: The fourth resonance structure is derived from structure #1. It is formed by 1 lone pair on the oxygen moving to form a pi bond between the oxygen and carbon 8. At the same time, the pi bond between carbons 8 and 9 is pushed onto carbon 9 in the form of a lone pair. This results in a formal positive charge on the oxygen due to the oxygen losing 1 electron from the lone pair that is now a pi bond. Carbon 9 also has a formal negative charge due to gaining 1 electron from the lone pair that was a pi bond.
Structure #5: The fifth resonance structure is derived from structure #2. It is formed by the pi bond between carbons 2 and 3 moving to carbons 1 and 2. This results in the formal positive charge moving to carbon 3 due to carbon 3 losing 1 electron from the pi bond that moved.
Structure #6: The sixth resonance structure is derived from structure #2. It is formed by the pi bond between carbons 4 and 5 moving to carbons 1 and 4. This results in the formal positive charge moving to carbon 5 due to carbon 5 losing 1 electron from the pi bond that moved.
Pi-System: The pi-system is an average of the 6 resonance structures. There is a dotted line across the oxygen, nitrogen, and carbons 1-9. This represents the pi electron density being spread across the oxygen and those carbons. The partial charges in the pi-system are a reflection of the formal charges in each resonance structure.
26.
Hybridization: The hybridization of each atom in the molecule needs to be determined before drawing resonance structures. Only sp and sp2 hybridized atoms can participate in resonance, so we need to find which atoms have those hybridizations. This will tell us what parts of the molecule we can use in the resonance structures. The hybridizations of all atoms and lone pairs in the following molecule are shown in the image below:
Based on the hybridizations shown in the previous image, we now know what parts of the molecule can participate in resonance. Those parts are highlighted in the image below:
Structure #1: Starting structure
Structure #2: The second resonance structure is derived from structure #1. It is formed by moving the lone pair on carbon 1 to form a pi bond between carbons 1 and 2. At the same time, the pi bond between carbons 2 and 3 is pushed onto carbon 3 in the form of a lone pair. This results in the formal negative charge moving to carbon 3 due to carbon 3 gaining 1 electron from the lone pair that was a pi bond.
Structure #3: The third resonance structure is derived from structure #2. It is formed by moving the lone pair on carbon 3 to form a pi bond between carbons 3 and 4. At the same time, the pi bond between carbon 4 and the oxygen moves to the oxygen in the form of a lone pair. This results in the formal negative charge moving to the oxygen due to the oxygen gaining 1 electron from the pi bond that is now a lone pair.
Structure #4: The fourth resonance structure is derived from structure #1. It is formed by moving the lone pair on carbon 1 to form a pi bond between carbons 1 and 5. At the same time, the pi bond between carbons 5 and 6 moves to carbon 6 in the form of a lone pair. This results in the formal negative charge moving to carbon 6 due to carbon 6 gaining 1 electron from the pi bond that is now a lone pair.
Pi-System: The pi-system is an average of the 4 resonance structures. There is a dotted line across the oxygen and carbons 1-6. This represents the pi electron density being spread across the oxygen and those carbons. The partial charges in the pi-system are a reflection of the formal charges in each resonance structure.
27.
Hybridization: The hybridization of each atom in the molecule needs to be determined before drawing resonance structures. Only sp and sp2 hybridized atoms can participate in resonance, so we need to find which atoms have those hybridizations. This will tell us what parts of the molecule we can use in the resonance structures. The hybridizations of all atoms and lone pairs in the following molecule are shown in the image below:
Based on the hybridizations shown in the previous image, we now know what parts of the molecule can participate in resonance. Those parts are highlighted in the image below:
Structure #1: Starting structure
Structure #2: The second resonance structure is derived from structure #1. It is formed by moving the lone pair on carbon 1 to form a pi bond between carbons 1 and 2. At the same time, the pi bond between carbons 2 and 3 is pushed onto carbon 3 in the form of a lone pair. This results in the formal negative charge moving to carbon 3 due to carbon 3 gaining 1 electron from the lone pair that was a pi bond.
Structure #3: The third resonance structure is derived from structure #2. It is formed by moving the lone pair on carbon 3 to form a pi bond between carbons 3 and 4. At the same time, the pi bond between carbons 4 and 5 is pushed onto carbon 5 in the form of a lone pair. This results in the formal negative charge moving to carbon 5 due to carbon 5 gaining 1 electron from the lone pair that was a pi bond.
Structure #4: The fourth resonance structure is derived from structure #1. It is formed by moving the lone pair on carbon 1 to form a pi bond between carbons 1 and 6. At the same time, the pi bond between carbon 6 and the oxygen moves up to form a lone pair on the oxygen. This results in the formal negative charge moving to the oxygen due to the oxygen gaining 1 electron from the lone pair that was a pi bond.
Structure #5: The fifth resonance structure is derived from structure #1. It is formed by moving the lone pair on the nitrogen to form a pi bond between the nitrogen and carbon 6. At the same time, the pi bond between carbon 6 and the oxygen moves up to the oxygen. This results in a formal negative charge being created on the oxygen due to the oxygen gaining 1 electron from the lone pair that was a pi bond. The nitrogen also has a formal positive charge due to the nitrogen losing 1 electron from the lone pair that is now a pi bond.
Structure #6: The sixth resonance structure is derived from structure #5. It is formed by moving the lone pair on carbon 1 to form a pi bond between carbons 1 and 2. At the same time, the pi bond between carbons 2 and 3 is pushed onto carbon 3 in the form of a lone pair. This results in the formal negative charge moving from carbon 1 to carbon 3 due to carbon 3 gaining 1 electron from the lone pair that was a pi bond.
Structure #7: The seventh resonance structure is derived from structure #6. It is formed by moving the lone pair on carbon 3 to form a pi bond between carbons 3 and 4. At the same time, the pi bond between carbons 4 and 5 is pushed onto carbon 5 in the form of a lone pair. This results in the formal negative charge moving from carbon 3 to carbon 5 due to carbon 5 gaining 1 electron from the lone pair that was a pi bond.
Pi-System: The pi-system is an average of the 7 resonance structures. There is a dotted line across the oxygen, nitrogen, and carbons 1-6. This represents the pi electron density being spread across the oxygen and those carbons. The partial charges in the pi-system are a reflection of the formal charges in each resonance structure.
28.
Hybridization: The hybridization of each atom in the molecule needs to be determined before drawing resonance structures. Only sp and sp2 hybridized atoms can participate in resonance, so we need to find which atoms have those hybridizations. This will tell us what parts of the molecule we can use in the resonance structures. The hybridizations of all atoms and lone pairs in the following molecule are shown in the image below:
Based on the hybridizations shown in the previous image, we now know what parts of the molecule can participate in resonance. Those parts are highlighted in the image below:
Structure #1: Starting structure
Structure #2: The second resonance structure is derived from structure #1. It is formed by moving the pi bond in the ring between carbons 2 and 3 to carbons 1 and 2. This results in the formal positive charge moving to carbon 3 due to carbon 3 losing 1 electron from the pi bond that moved.
Structure #3: The third resonance structure is derived from structure #2. It is formed by moving the pi bond between carbons 4 and 5 to carbons 3 and 4. This results in the formal positive charge moving to carbon 5 due to carbon 5 losing 1 electron from the pi bond that moved.
Structure #4: The fourth resonance structure is derived from structure #3. It is formed by moving the pi bond between carbons 6 and 7 to carbons 5 and 6. This results in the formal positive charge moving to carbon 7 due to carbon 7 losing 1 electron from the pi bond that moved.
Structure #5: The fifth resonance structure is derived from structure #4. It is formed by moving the pi bond between carbons 8 and 9 to carbons 1 and 8. This results in the formal positive charge moving to carbon 9 due to carbon 9 losing 1 electron from the pi bond that moved.
Structure #6: The sixth resonance structure is derived from structure #5. It is formed by moving the pi bond between carbons 10 and 11 to carbons 9 and 10. This results in the formal positive charge moving to carbon 11 due to carbon 11 losing 1 electron from the pi bond that moved.
*note: all resonance structures with a benzene ring have a twin structure with the pi bonds rotated in the ring, however not all of those structures are shown. Additional resonance structures with rotated pi bonds could be shown for structure #1, structure#5, and structure #6.
Pi-System: The pi-system is an average of the 6 resonance structures. There is a dotted line across carbons 1-11. This represents the pi electron density being spread across the oxygen and those carbons. The partial charges in the pi-system are a reflection of the formal charges in each resonance structure.
29.
Hybridization: The hybridization of each atom in the molecule needs to be determined before drawing resonance structures. Only sp and sp2 hybridized atoms can participate in resonance, so we need to find which atoms have those hybridizations. This will tell us what parts of the molecule we can use in the resonance structures. The hybridizations of all atoms and lone pairs in the following molecule are shown in the image below:
Based on the hybridizations shown in the previous image, we now know what parts of the molecule can participate in resonance. Those parts are highlighted in the image below:
Structure #1: Starting Structure
Structure #2: The second resonance structure is derived from structure #1. It is formed by moving the pi bonds around the benzene ring.
Structure #3: The third resonance structure is derived from structure #1. It is formed by moving the radical to form a pi bond between carbons 1 and 2 with 1 electron from the pi bond between carbons 2 and 3. The other electron in the pi bond between carbons 2 and 3 moves to carbon 3 and forms a radical.
Structure #4: The fourth resonance structure is derived from structure #3. It is formed by moving the radical to form a pi bond between carbons 3 and 4 with 1 electron from the pi bond between carbons 4 and 5. The other electron in the pi bond between carbons 4 and 5 moves to carbon 5 and forms a radical.
Structure #5: The fifth resonance structure is derived from structure #4. It is formed by moving the radical to form a pi bond between carbons 6 and 7 with 1 electron from the pi bond between carbons 5 and 6. The other electron in the pi bond between carbons 6 and 7 moves to carbon 7 and forms a radical.
Structure #6: The sixth resonance structure is derived from structure #4. It is formed by moving the radical to form a pi bond between carbons 8 and 9 with 1 electron from the pi bond between carbons 5 and 8. The other electron in the pi bond between carbons 8 and 9 moves to carbon 9 and forms a radical.
Pi-System: The pi-system is an average of the 9 resonance structures. There is a dotted line across carbons 1-9. This represents the pi electron density being spread across the oxygen and those carbons. There are no partial charges due to there not being any formal charges in any of the resonance structures.
30.
Hybridization: The hybridization of each atom in the molecule needs to be determined before drawing resonance structures. Only sp and sp2 hybridized atoms can participate in resonance, so we need to find which atoms have those hybridizations. This will tell us what parts of the molecule we can use in the resonance structures. The hybridizations of all atoms and lone pairs in the following molecule are shown in the image below:
Based on the hybridizations shown in the previous image, we now know what parts of the molecule can participate in resonance. Those parts are highlighted in the image below:
Structure #1: Starting structure
Structure #2: The second resonance structure is derived from structure #1. It is formed by moving the pi bond between the nitrogen and carbon 1 to the nitrogen in the form of a lone pair. This results in a formal negative charge on the nitrogen due to the nitrogen gaining 1 electron from the lone pair that was a pi bond. Carbon 1 also has a formal positive charge due to losing 1 electron from the pi bond that moved.
Structure #3: The third resonance structure is derived from structure #2. It is formed by moving the pi bond between carbons 2 and 3 to carbons 1 and 2. This results in the formal positive charge moving to carbon 3 due to carbon 3 losing 1 electron from the pi bond that moved.
Structure #4: The fourth resonance structure is derived from structure #3. It is formed by moving the pi bond between carbons 4 and 5 to carbons 3 and 4. This results in the formal positive charge moving to carbon 5 due to carbon 5 losing 1 electron from the pi bond that moved.
Structure #5: The fifth resonance structure is derived from structure #2. It is formed by moving the pi bond between carbons 6 and 7 to carbons 1 and 6. This results in the formal positive charge moving to carbon 7 due to carbon 7 losing 1 electron from the pi bond that moved.
Structure #6: The sixth resonance structure is derived from structure #5. It is formed by moving the pi bond between carbons 8 and 9 to carbons 7 and 8. This results in the formal positive charge moving to carbon 9 due to carbon 9 losing 1 electron from the pi bond that moved.
Structure #7: The seventh resonance structure is derived from structure #6. It is formed by moving the pi bond between carbons 10 and 11 to carbons 9 and 10. This results in the formal positive charge moving to carbon 11 due to carbon 11 losing 1 electron from the pi bond that moved.
Pi-System: The pi-system is an average of the 7 resonance structures. There is a dotted line across carbons 1-11. This represents the pi electron density being spread across the oxygen and those carbons. The partial charges in the pi-system are a reflection of the formal charges in each resonance structure.
