Organic Chemistry Tutoring session Exam Notes: Key Concepts and Problem Solving
The third molecule (likely the most linear) has the highest boiling point.
Linear molecules like straight chains have higher boiling points due to increased van der Waals forces.
Highly branched molecules are less stable, leading to lower boiling points.
Example (Ramen): Adding sauce to ramen water can make it boil faster. This is attributed to the sauce's composition (acting like a branched molecule or affecting the intermolecular forces of water), lowering the overall boiling point of the mixture, thus causing the water to boil faster at a lower temperature.
The molecule with the most branches will have the lowest boiling point.
Dipole Moments and Polarity
A molecule possesses a dipole moment if there is a net separation of charge, making it polar.
Nonpolar Molecules (No Dipole Moment):
Molecule A: Pulls are equal in all directions, canceling each other out, resulting in no net dipole moment.
Molecule D: Both electronegativities are the same, and electron pulls are in exact opposite directions, canceling out.
Molecule C (example with a hexagon): If one fluorine is going 'up' and another is going 'down' in a 3D hexagonal structure, their individual dipole moments oppose and cancel each other out, resulting in no net dipole moment.
Polar Molecules (With Dipole Moment):
Molecule B: If substituents are pulling in directions that do not cancel (e.g., both 'up' or at an angle that results in a net pull), there will be a net dipole moment.
Water (H_2O): Water is a polar molecule because its bent molecular geometry ensures that the individual dipole moments of the O-H bonds do not cancel out, resulting in a net dipole moment pointing towards the oxygen.
Solubility Rule: "Like dissolves like." Polar solvents dissolve polar solutes, and nonpolar solvents dissolve nonpolar solutes. This is why water (polar) and oil (nonpolar) do not mix.
Reaction Mechanisms and Formal Charge
Step 1: Electron Movement: Identify the flow of electrons using arrows. Typically, electrons move from areas of high electron density to areas of low electron density (or towards more electronegative atoms).
Identifying Charges: Determine which atom is more electronegative; it will attract electrons, potentially gaining a negative charge, while the other atom will lose electrons, potentially gaining a positive charge.
Intermediate Formation: When a bond transfers to an atom, that atom may gain a negative charge. For example, if hydrogen loses its electron to oxygen, oxygen acquires a negative charge.
Formal Charge Calculation: The formal charge is calculated using the formula:
Formal Charge = (\text{Valence electrons}) - (\text{Lone pair electrons}) - (1/2 \times \text{Bonding electrons})Example Application:
Oxygen with a negative charge: If oxygen has 6 valence electrons, 6 lone pair electrons, and forms 1 bond (so 2 bonding electrons), its formal charge is 6 - 6 - (1/2 \times 2) = 6 - 6 - 1 = -1. This confirms the negative charge on oxygen.
Nitrogen in NH2 connected to CH3: Nitrogen has 5 valence electrons. If it forms 4 bonds (e.g., to 2 H, 1 C, and 1 H, or if referring to primary amine where it's connected to 2 H and 1 C and forms 4 bonds in total and has no lone pairs) and has 0 lone pairs, its formal charge is 5 - 0 - (1/2 \times 8) = 5 - 4 = +1. This explains a positive charge on nitrogen.
Crucial Point: Always include and correctly calculate the charges in your reaction mechanisms, as missing them can lead to incorrect answers.
pKa and Acidity/Reactivity
Fundamental Concept: Understanding the relationship between pKa and acidity/basicity is essential.
Reactivity Rule: An acid-base reaction will significantly proceed if the pKa of the conjugate acid formed is at least 3 units greater than the pKa of the reacting acid. So, ( \text{pKa}{\text{conjugate acid}}) - (\text{pKa}{\text{reacting acid}}) > 3
Example 1: Reaction with Hydroxide Ion (OH^-)
The conjugate acid of hydroxide is water (H_2O), with a pKa of 16.
Molecule a (pKa 11): 16 - 11 = 5. Since 5 > 3, molecule 'a' can react with hydroxide.
Molecule b (pKa 19): Since 19 > 16 (meaning the conjugate acid of molecule b would be stronger than water, thus molecule b is a weaker acid than water), molecule 'b' cannot react with hydroxide.
Molecule c (pKa 5.3): 16 - 5.3 = 10.7. Since 10.7 > 3, molecule 'c' can react with hydroxide.
Molecule d (pKa 5.2): 16 - 5.2 = 10.8. Since 10.8 > 3, molecule 'd' can react with hydroxide.
Example 2: Reaction with Bicarbonate Ion (HCO_3^-)
The conjugate acid corresponding to bicarbonate in this context is implied to have a pKa around 10.2 (could be referring to carbonic acid's second dissociation constant). The reactivity rule (difference >3) is applied.
For molecule a, it cannot react (assuming its conjugate acid pKa is not >3 units higher than 10.2).
For molecule b, it cannot react.
For molecule c, the speaker states it cannot react (despite 10.2 - 5.3 = 4.9 which is >3 - follow the explicit instruction given by the speaker).
For molecule d, it can react (since 10.2 - 5.2 = 5 which is >3).
Resonance Structures
Significance: This is a critically important concept for OChem 1 and 2.
Drawing Rules:
The chemical formula of the molecule must remain the same across all resonance structures.
The net charge of the molecule must remain the same across all resonance structures (e.g., if the initial structure has two negative and one positive charge, all resonance forms must maintain this overall charge).
Drawing Strategy: Move double bonds to different positions in the molecule, or move lone pairs to form double bonds, or convert double bonds into lone pairs. Pay attention to charges that form as a result of electron movement.
Challenge: The difficulty often lies in identifying and drawing all possible resonance structures without drawing duplicates or missing any. Sometimes, professors require drawing all electron-pushing arrows to show the movement of electrons. Typically, 2 to 4 arrows are involved per resonance step.
Formal Charge of Radicals
Radical: A species containing an unpaired electron.
Formal Charge Calculation with Radicals: The formal charge formula still applies.
Example (Carbon Radical): Carbon has 4 valence electrons. If it has one unpaired electron (radical) and forms 3 bonds (e.g., to 3 hydrogens), it has 1 lone electron (from the radical) and 6 bonding electrons (3 \times 2). So, the formal charge is 4 - 1 - (1/2 \times 6) = 4 - 1 - 3 = 0. Therefore, a carbon with 3 bonds and a radical (total of 7 electrons around it) has a formal charge of 0.
Do not be confused by the presence of a radical; remember to account for the single unpaired electron in the formal charge calculation.
Types of Carbons
Tertiary Carbon (3^\text{o} Carbon): A carbon atom that is directly bonded to three other carbon atoms. When identifying, sometimes hydrogens are ignored, and only the number of carbon connections is considered.
Quaternary Carbon (4^\text{o} Carbon): A carbon atom that is directly bonded to four other carbon atoms.
Skeleton Diagrams (Chairs): In chair conformations, "solid" lines indicate a substituent is pointing 'up', and "dashed" lines indicate a substituent is pointing 'down'. Axial positions are generally the easiest to identify.
Chair Conformations and Energy Calculations
Chair Conformations: Represent the 3D structure of cyclohexane and similar rings. Substituents can occupy axial (perpendicular to the ring) or equatorial (roughly parallel to the ring) positions, and they can be directed 'up' or 'down' relative to the ring plane.
Stability and Energy (A-Values): Different substituents and their positions (axial vs. equatorial, and their relative positions to other substituents) result in different energy levels due to gauche and eclipsed interactions.
Gauche Interaction: Occurs when two substituents are not directly aligned but are on adjacent carbons, often resulting in a stable, lower-energy conformation (e.g., 2.1 kJ/mol for an OH-H interaction, 3.8 kJ/mol for a CH3-H gauche interaction).
Eclipsed Interaction: Occurs when two substituents are directly aligned, leading to higher energy and less stable conformations (e.g., 6 for CH3-H eclipsed, 11 for CH3-CH3 eclipsed).
Calculation Procedure for Conformational Energy (A-value):
Draw All Hydrogens: This is often the most challenging step. Explicitly drawing all hydrogens helps visualize interactions.
Identify Interactions: Determine which substituents are interacting with each other. For example, an axial hydroxyl group on carbon 1 will interact with axial hydrogens on carbons 3 and 5.
Sum Interaction Energies: Use a provided table of A-values (energy increments) for specific types of interactions (e.g., OH-H gauche, CH3-H eclipsed). Sum these values for all relevant interactions in a given conformer.
Example: For an axial OH, it interacts with 2 axial hydrogens. If the OH-H interaction energy is 2.1 kJ/mol, the total energy from these interactions is 2 \times 2.1 = 4.2 kJ/mol.
For CH3 groups, different values apply: 3.8 for gauche CH3-H, 6 for eclipsed CH3-H, 11 for eclipsed CH3-CH3.
Hydrogen-hydrogen interactions are generally considered to have 0 energy difference as they are too small to cause significant strain.
Calculate % of More Stable Conformer:
Determine the energy difference (\Delta G) between the less stable and more stable conformer: \Delta G = \text{Energy}{\text{less stable}} - \text{Energy}{\text{more stable}}. Note: \Delta G is always a positive number for this calculation; always subtract the smaller energy from the larger one.
Use a provided graph (x-axis: \Delta G, y-axis: % more stable conformer) to find the percentage of the more stable conformer present in the equilibrium mixture.
Example: If \Delta G = 7.2 kJ/mol, locate 7.2 on the x-axis and project up to the curve to find the corresponding percentage on the y-axis, which might be around 90\text{%}.
Lewis Acids and Bases
Lewis Base: An electron pair donor. These typically have lone pairs of electrons or a negative charge (e.g., oxygen atoms in water or alcohols, anions).
Lewis Acid: An electron pair acceptor. These are typically electron-deficient, having a positive charge or incomplete octets (e.g., carbocations, metal ions, compounds like BH_3).
Identification Tip: If a species has a positive charge or lacks lone pairs, it is likely a Lewis acid. If it has lone pairs or a negative charge, it is likely a Lewis base.
Functional Groups
Importance: Essential to memorize and recognize common functional groups.
Carboxylic Acid: Characterized by a -COOH group (a carbonyl group, C=O, bonded to a hydroxyl group, -OH).
If the hydrogen is removed (e.g., -COO^-), it becomes a carboxylate ion, which is still derived from a carboxylic acid.
Ester: Characterized by a -COOR group (a carbonyl group, C=O, bonded to an oxygen with an alkyl or aryl group, -OR).
Ether: Characterized by a -ROR' group (an oxygen atom single-bonded to two alkyl or aryl groups). It lacks a carbonyl group.
Common Confusion (Ester vs. Ether): Esters have two oxygens (one in a carbonyl, one in an ether linkage), while ethers only have one oxygen (in a C-O-C linkage).
Energy Differences and Strain
Energy Differences in Conformers (Cont.):
Gauche: Interactions like CH3-H in a gauche conformation lead to lower energy and higher stability (e.g., 3.8).
Eclipsed: Interactions like CH3-H in an eclipsed conformation lead to higher energy and lower stability (e.g., 6). CH3-CH3 eclipsed is even higher energy (e.g., 11).
Small groups like hydrogen have negligible interaction energy with other groups, often considered 0.
Types of Strain:
Steric Strain: Arises from repulsive interactions when two large functional groups are forced into close proximity, causing their electron clouds to overlap.
Example: Cis-1,2-dimethylcyclobutene exhibits steric strain between the two methyl groups if they are close.
Angle Strain: Occurs when bond angles deviate significantly from their ideal (VSEPR-predicted) values. This is common in small rings (e.g., cyclobutane has significant angle strain because its ideal bond angles are 90^\text{o} instead of 109.5^\text{o}).
Torsional Strain: Arises from repulsive interactions between bonding electron pairs or substituents on adjacent atoms when they are eclipsed. This generally occurs in cyclic compounds, making them less stable.
Exception (Cyclohexane): The chair conformation of cyclohexane is a stable, strain-free structure because it minimizes both angle and torsional strain, making it the most stable cyclic conformation.
Constitutional Isomers: Molecules that have the same molecular formula but different connectivity (different arrangement of atoms).
Drawing Constitutional Isomers: Start with the base structure (e.g., cyclopropane). Then, move substituents to different carbon positions to create new connectivity while maintaining the same total number of carbons and hydrogens and adhering to the naming conventions (e.g., "dimethyl"). Do not break the cyclic structure. For example, for trans-1,2-dimethylcyclopropane, you would keep the cyclopropane ring and move the methyl groups to different ring carbons (e.g., 1,1-dimethylcyclopropane, 1,2-dimethylcyclopropane with cis). Make sure the same molecular formula is maintained.
This comprehensive note covers fundamental organic chemistry concepts, beginning with a discussion on boiling points, where linear molecules exhibit higher boiling points due to stronger van der Waals forces, while branched molecules have lower boiling points. It then delves into dipole moments and polarity, defining polar molecules by a net charge separation and nonpolar ones by canceling pulls, and introduces the