Class 15: Introduction to Molecular Orbitals and Carbonyl Compounds Flashcards
Draw a molecular orbital diagram of a pi-bond system (e.g. C=O, C=C)
- Draw the participating atomic p orbitals
- Combine them in-phase to get the pi bonding MO (π)
- Combine them out-of-phase to get the pi antibonding MO (π*)
- Arrange the MO energy levels vertically with π lower than π*
- Show the distribution of electrons in the MOs
- Fill lower energy π first, then π*
- Follow Aufbau principle and Hund’s rules
- Indicate bond order as (# π electrons - # π* electrons)/2
Some key points:
- π MOs are delocalized over the whole π system
- More nodes in π* lead to higher antibonding energy
- Bond order > 0 means bonding capacity
- Higher bond order indicates stronger π bond
Explain the origins of directionality of an addition into a carbonyl system.
- Carbonyl carbon (C=O) is electrophilic due to polar bond
- Electron density is pulled towards more electronegative oxygen
- This makes the carbonyl carbon relatively electron-deficient
- Nucleophiles attack the partially positive carbonyl carbon
- The π electrons from C=O bond act as the nucleophile
- Two possible resonance forms result after nucleophilic attack
- An oxyanion and a carbocation
- The carbocation form is more stable/favored
- So the preferred direction minimizes carbocation character
- Avoids highly unstable/reactive carbocations
- Favors tertiary > secondary > primary carbocations
- Steric factors can also influence the preferred direction
- Nucleophile attacks less hindered side
- Electron-withdrawing groups stabilize carbocation character
- Can reverse the typical addition direction
So the electrophilicity of the carbonyl carbon, resonance stabilization of carbocation character, and steric factors dictate the preferred direction.
Predict the relative order of Lewis acidity or electrophilicity based on the carbonyl compound’s structure.
- More electronegative substituents increase electrophilicity
- e.g. F > Cl > Br > alkyl
- Due to electron-withdrawing inductive effects
- Resonance effects also influence electrophilicity
- Aryl > alkyl (resonance stabilization)
- α,β-unsaturated > saturated alkyl
- Hybridization of the carbonyl carbon
- sp2 (aldehyde, ketone) > sp (CO2, carbonates)
- sp allows better orbital overlap with nucleophile
- Steric effects play a role
- More hindered carbonyls are less electrophilic
- Due to reduced accessibility to nucleophile
- Typical order of increasing electrophilicity:
Aldehydes < Ketones < Esters < Acid chlorides < Anhydrides < Acid halides - Exceptions occur with very electronegative substituents
- e.g. CF3CO is more electrophilic than acetyl chloride
So the key factors are inductive effects, resonance, hybridization, sterics and substituents on the carbonyl carbon.
Explain why a carbonyl compound can be electrophilic.
- Carbonyl carbon (C=O) has a partial positive charge
- Due to greater electronegativity of oxygen
- Pulls electron density away from carbon
- This makes the carbonyl carbon electron-deficient
- And susceptible to nucleophilic attack
- The C=O pi bond is polarized
- Carbon side is electron-poor (electrophilic)
- Oxygen side is electron-rich (nucleophilic)
- Resonance effects can further increase electrophilicity
- e.g. Aryl ketones are stabilized by resonance
- Makes the carbonyl carbon more electrophilic
- Electron-withdrawing substituents enhance electrophilicity
- By inductive electron withdrawal
- e.g. Acyl halides, acid anhydrides
- Hybridization also affects electrophilic character
- sp2 (aldehydes, ketones) < sp (acid chlorides)
- Lower hybridization allows better orbital overlap
So the polar nature of the C=O bond, resonance effects, inductive effects of substituents, and carbonyl carbon hybridization all contribute to the electrophilic reactivity.
Use your newly gained vocabulary, structures, and curved arrows in the context of reactions: mechanisms, reaction pathway, and transition state.
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Molecular Orbitals (MOs):
- MO theory explains bonding in molecules using atomic orbitals.
- Combination of atomic orbitals form molecular orbitals, which can be bonding, antibonding, or nonbonding.
- Bonding MOs result from constructive interference of atomic orbitals, stabilizing the molecule.
- Antibonding MOs result from destructive interference, destabilizing the molecule.
- Nonbonding MOs have no net effect on bond strength.
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Carbonyl Compounds:
- Contain a carbon-oxygen double bond (C=O).
- Highly polar due to the electronegativity difference between C and O.
- Common examples include aldehydes, ketones, carboxylic acids, and esters.
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Reactions: Mechanisms, Pathways, and Transition States:
- Mechanisms: Describes the step-by-step process by which reactants are converted into products.
- Reaction Pathway: Sequence of steps leading from reactants to products.
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Transition State: High-energy, unstable state in which bonds are being broken and formed during a reaction.
- Represents the highest energy point along the reaction pathway.
- Determines the rate of the reaction.
- Often represented as a transition state diagram or using transition state theory.
Draw a molecular orbital diagram of a pi-bond system (e.g. C=O, C=C)
STRATEGY:
1. Determine the bonds and orbitals interacting in those bonds
EX: pi bond with p orbitals, sigma bond with sp2 or sp3
2. Draw the y axis with delta G and then add your orbitals that haven’t bonded yet (DEPENDS ON ATOM) – ex: O will be lower energy than C due to Zeff. Add electrons.
P orbitals will be higher energy than the hybrid orbitals due to the hybrid orbitals S character
3. Add sigma bonding and antibonding (one bond, one line, 2 bonds, 2 lines)—connect them to the sp2 or sp3 orbitals
4. Add pi bonding above sigma bonding and pi antibonding below sigma antibonding—connect to the p orbitals
5. Identify where the electrons are in the molecule
Are they lone pairs?
Are they bonding to Hydrogen?
Are they in the bond we’re drawing?
6. fill in electrons to bonding and antibonding
Bond order=(bonding-antibonding)/2
LUMO (lowest unoccupied molecular orbital)
HOMO (highest occupied molecular orbital)
This is because of the different orbital overlap: stronger bonds are with atoms that are closer to each other on the periodic table, and therefore have more similar overlap
Explain the origins of directionality of an addition into a carbonyl system.
There is open bonding orbitals that are sp2 and become sp3 when they are in the tetrahedral intermediate
Use your newly gained vocabulary, structures, and curved arrows in the context of reactions: mechanisms, reaction pathway, and transition state.
Curved arrows strategy:
Draw them from the ELECTRONS to the molecule that you intend to use
Acyl-Substitution reactions:
Simple reactions
1. Electrophilic addition (attack)–> tetrahedral intermediate
2. Elimination
TS: the pathway from the reactants to the tetrahedral intermediate, and the pathway from the intermediate to the products or to another intermediate
Predict the relative order of Lewis acidity or electrophilicity based on the carbonyl compound’s structure.
Carbonyls are lewis acids
Determine how electrophilic and acidic the compound is depending on its leaving group as well as the dipole present on the oxygen
When protonated, the C becomes more electrophilic because the H pulls electrons from the O which causes the O to pull even more electrons from the C to try to stabilize its positive charge
This makes the carbon EVEN MORE partially positive and therefore electron loving
OVERALL:
The only reactions possible are:
Bronsted-lowry acid/base reactions (protonation, deprotonation)
Nucleophilic addition
Elimination
Give a claim, evidence, and reasoning for which carbonyl compound is the stronger electrophile:
an acid chloride or an amide?
Claim: The acid chloride is the stronger electrophile.
Evidence and Reasoning: Both compounds can be stabilized by the lone pair of Cl or N via
resonance. However, due to differences in size and electronegativity between N and Cl they
do not stabilize equally.
The acid chloride has second resonance structure with a carbon-to-chloride double bond
whose π bond is weak due to poor orbital overlap between C 2p and Cl 3p orbitals. Therefore,
this second resonance structure does not contribute significantly to the resonance hybrid and
the acyl chloride is not significantly stabilized by electron density donated by Cl.
The amide has a second resonance structure with a carbon-to-nitrogen double bond whose π
bond is much stronger than in the acid chloride due to the similarity in sizes between the
atoms. Therefore, the second resonance structure contributes significantly to the resonance
hybrid and the carbonyl group is stabilized by electron density donated by N.
Another factor to consider is the difference in electronegativity between N and Cl. Cl (3.16 on
Pauling scale) is more electronegative than N (3.04). The carbonyl C of an acid chloride will have a
greater partial positive charge on the electrophilic C.
b. Based on your answer to a), which of the two is lower in free energy?
Amide
