Formal Charge Sample Question:
Assign all formal charges in the following molecules:
Answer Key:
C1: (4 electrons) – (4 electrons) = 0
4 bonds = 4 electrons, 1 electron/bond
C2: (4 electrons) – (4 electrons) = 0
4 bonds = 4 electrons, 1 electron/bond
O1: (6 electrons) – (6 electrons) = 0
2 bonds = 2 electrons, 1 electron/bond
2 lone pairs = 4 electrons, 2 electrons/lone pair
O2: (6 electrons) – (7 electrons) = 0
1 bond = 1 electron
3 lone pairs = 6 electrons, 2 electrons/lone pair
Partial Charge Sample Question:
Assign all significant partial charges and dipoles in the following molecules:
Answer Key:
Oxygen is a FON element, so it will always create a polar bond unless it is bonded to itself. The dipole created by oxygen is going to oxygen, away from the carbon it is bonded to due to oxygen being significantly more electronegative than carbon.
Oxygen delocalizes electron density away from the carbon it is bonded to, making carbon partially positive and oxygen partially negative.
Nitrogen is a FON element, so it will always create a polar bond unless it is bonded to itself. The dipole created by nitrogen is going to nitrogen, away from the carbon it is bonded to due to nitrogen being significantly more electronegative than carbon.
The carbon-carbon bonds and implicit hydrogen-carbon bonds do not have any dipoles shown due to those bonds being non-polar.
Relationship between Formal Charge and Partial Charge Sample Question:
For each pair of molecules given, draw any dipoles or partial charges that are present. Compare the two molecules and explain any differences in dipoles and/or partial charges. Consider factors such as electronegativity and formal charges in your explanation.
Answer Key:
In both molecules there are dipoles towards the oxygen atoms, away from the atoms they’re bonded to. This is because oxygen is a FON element, so it will always create polar bonds unless it is bonded to itself.
However, the dipoles towards the formally positive oxygen are larger than the dipoles towards the neutral oxygen. This is because oxygen will always get itself to a partially negative state due to its extremely high electronegativity. Therefore, the formally positive oxygen has to pull harder to delocalize more electron density away from the atoms it is bonded to. This is due to the formally positive oxygen starting in an electron deficient state. The formally neutral oxygen does not need to delocalize as much electron density to become partially negative.
Due to the dipoles towards the formally positive oxygen being larger, the hydrogens and carbon it is bonded to will be more partially positive than the hydrogen and carbon bonded to the formally neutral oxygen.
Sigma and Pi Bonds Sample Question:
For each molecule provided, determine whether each bond is a sigma (σ) bond or a pi (π) bond. Clearly label each bond as sigma or pi directly on the molecular structure.
Answer Key:
The 3 double bonds in the ring are made of 1 sigma bond and 1 pi bond. Every bond has 1 sigma bond, however a bond can never have more than 1 sigma bond. That means the second bond in the double bonds must be a pi bond.
The single bond connected to the benzene ring is a sigma bond because all single bonds are always sigma bonds.
The two terminal single bonds in the substituent on the ring are sigma bonds because all single bonds are always sigma bonds.
The double bond in the substituent on the ring is made of 1 sigma bond and 1 pi bond. Every bond has 1 sigma bond, however a bond can never have more than 1 sigma bond. That means the second bond in the double bond must be a pi bond.
Hybridization Sample Question:
Clearly label the hybridization of each non-hydrogen atom and lone pair directly on each molecular structure:
Answer Key:
The terminal carbon on the left side of the molecule is sp3 hybridized because it has 4 sigma bonds that are all counted in the hybridization. Note that there are 3 implicit hydrogens not shown on that carbon.
The two carbons in the double bond are sp2 hybridized because they have 3 sigma bonds that are counted in the hybridization, and 1 pi bond that is not counted in the hybridization.
The carbon single bonded to the oxygen is sp3 hybridized because it has 4 sigma bonds that are all counted in the hybridization. Note that there are 2 implicit hydrogens not shown on that carbon.
The oxygen is sp2 hybridized. It has 2 sigma bonds which are both counted in the hybridization. One lone pair on the oxygen is in a hybridized orbital and is therefore counted in the hybridization, however the second lone pair does not hybridize and is in a p-orbital. This is because only 1 lone pair can ever be in a hybridized orbital. The hybridized lone pair has the same hybridization as the oxygen its on with an about (~) sign included. The non-hybridized lone pair is in a p-orbital. Note that the bond between the hydrogen and the oxygen is not shown in this example.
*Note that the ~ sign being added to the hybridization of a hybridized lone pair is technically correct, however some notations will not include it. Either way, a hybridized lone pair will always have the same hybridization as the atom its on, and a non-hybridized lone pair will always be in a p-orbital.
Resonance Structures:
Determine and draw all good resonance structures for the following molecule:
Answer Key:
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 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.
Pi-System: The pi-system is an average of the 3 resonance structures. There is a dotted line across carbons 1-5 which 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.
Course Description:
Key Features:
- Formal Charge:
- Practice labeling all formal charges in molecules.
- 20 questions
- Partial Charge:
- Practice identifying all significant partial charges and dipoles in molecules.
- 20 questions
- Relationship between Formal Charge and Partial Charge:
- Compare molecules and analyze the effects of formal charges on dipoles and partial charges.
- 10 questions
- Sigma and Pi Bonds:
- Practice identifying sigma and pi bonds in molecules.
- 10 questions
- Hybridization:
- Practice determining the hybridization of atoms and lone pairs in diverse molecules.
- 20 questions
- Resonance Structures:
- Practice drawing resonance structures and pi-systems for different molecules.
- 3o questions
- Detailed Answer Keys: Access detailed answer keys for each worksheet, featuring:
- Step-by-step worked solutions for every problem.
- Comprehensive explanations that clarify key concepts and problem-solving strategies.
Enhanced Learning Experience:
- Consolidate your knowledge and skills with structured practice that mirrors the content and complexity of Module 1.
Recommended Prerequisite:
- Completion of the instructional videos from Module 1 is highly recommended before engaging with this practice module. This ensures familiarity with the core concepts and prepares you for effective practice and application.
Who Should Engage with This Practice Module:
- This practice module is designed for learners seeking extensive practice and reinforcement in fundamental organic chemistry concepts. Whether you are preparing for exams, reviewing key concepts, or aiming to enhance your problem-solving skills, this module provides essential practice and support.
Benefits:
- Access to a substantial number of practice problems for each topic, ensuring thorough concept mastery.
- Detailed answer keys with step-by-step worked solutions and explanations for each practice set.
- Reinforce understanding through extensive practice and application of organic chemistry principles.
It turned out to be great, I learned alot