Difficulty: Intermediate | Prerequisites: Alkene stability trends, resonance structures, formal charge
This topic moves from molecular mechanics (force fields) to semi-empirical quantum mechanics (MOPAC/PM3). You are using quantum-mechanical calculations to compare the thermodynamic stability of alkene isomers and to examine how charge is distributed in resonance-stabilised ions. If you are not comfortable with the idea that more-substituted alkenes are more stable, or with drawing resonance structures for allylic systems, review those concepts first.
TL;DR: MOPAC PM3 calculations produce heats of formation (ΔHf) that let you rank alkene isomers by stability. The expected order is (E)-2-pentene > (Z)-2-pentene > 1-pentene (most negative ΔHf = most stable). Mulliken charge analysis of allylic cations and anions shows that charge is delocalised across the conjugated system, matching what resonance structures predict.
MOPAC
Molecular Orbital PACkage. A semi-empirical quantum chemistry program that solves an approximate version of the Schrodinger equation. Unlike MM2 (which treats atoms as classical balls and springs), MOPAC accounts for electronic structure, making it suitable for calculating properties like heats of formation and charge distributions. Think of it as a step up from molecular mechanics: it knows about electrons.
PM3 (Parameterised Model 3)
A specific semi-empirical method used within MOPAC. It uses experimental data to calibrate its approximations, giving reasonable accuracy for organic molecules at low computational cost.
Heat of Formation (ΔHf)
The enthalpy change when one mole of a compound is formed from its elements in their standard states. A more negative ΔHf means the compound is more thermodynamically stable. In simple terms, the lower the number, the more stable the molecule.
Mulliken Charge (Mulliken Population Analysis)
A method for assigning partial charges to individual atoms in a molecule based on how the electrons in the molecular orbitals are distributed. It divides the shared electron density equally between bonded atoms. Think of it as a bookkeeping system that tells you which atoms are slightly positive and which are slightly negative.
Allylic Cation
A carbocation where the positive charge is adjacent to a C=C double bond, allowing the charge to be delocalised by resonance across three atoms (C1=C2-C3+ becomes C1+-C2=C3). In simple terms, the positive charge is shared, making the ion more stable than an isolated carbocation.
Allylic Anion
A carbanion where the negative charge is adjacent to a C=C double bond, again allowing delocalisation. The lone pair on the terminal carbon overlaps with the pi system.
Resonance Delocalisation
The spreading of electron density (or positive charge) over multiple atoms through overlapping p orbitals. Resonance structures are the Lewis-structure way of representing this: the true structure is a weighted average of all valid resonance forms.
Alkene Substitution and Stability
More-substituted alkenes are more stable (lower in energy). The order is: tetrasubstituted > trisubstituted > disubstituted > monosubstituted > unsubstituted. This is due to hyperconjugation and reduced strain. For the pentene isomers: 2-pentene (disubstituted) is more stable than 1-pentene (monosubstituted), and the E isomer is more stable than the Z isomer because of reduced steric strain between the groups on the same side of the double bond.
The three isomers
1-pentene: CH2=CH-CH2-CH2-CH3 (monosubstituted, the double bond is at the end of the chain)
(E)-2-pentene: CH3-CH=CH-CH2-CH3, with the two larger groups on opposite sides of the double bond (trans)
(Z)-2-pentene: CH3-CH=CH-CH2-CH3, with the two larger groups on the same side (cis)
Expected stability order (most stable first)
(E)-2-pentene > (Z)-2-pentene > 1-pentene
This means (E)-2-pentene should have the most negative (lowest) ΔHf, and 1-pentene the least negative (highest)
Why this order holds
2-Pentene (both E and Z) is disubstituted at the double bond, while 1-pentene is only monosubstituted. More alkyl groups on the double bond stabilise it through hyperconjugation (overlap of adjacent C-H sigma bonds with the pi system)
Within the 2-pentene pair, the E isomer is more stable than the Z because in the Z form, the methyl and ethyl groups are on the same side of the double bond, causing a steric interaction that raises the energy slightly
What the MOPAC output confirms
The calculated ΔHf values should follow the predicted order
The difference between E and Z-2-pentene is small (typically around 1 kcal/mol), reflecting the modest steric clash in the Z form
The difference between 2-pentene and 1-pentene is larger, reflecting the more fundamental stability gain from increased substitution
The system
The structures are conjugated enone-type systems with numbered carbons C1 through C5 and an oxygen at C6
Structure (1) is the allylic cation (positive charge)
Structure (2) is the allylic anion (negative charge)
What Mulliken charges show
The charge is not concentrated on a single atom. Instead, it is spread across the conjugated system
For the allylic cation (1): expect significant positive charge on C1 and C3 (the terminal atoms of the allylic system), with C2 carrying less positive charge. The oxygen at C6 may also carry partial negative charge if it participates in resonance
For the allylic anion (2): expect significant negative charge on C1 and C3, with C2 less negative. The oxygen may bear additional negative charge
Correlation with resonance structures
Draw the resonance structures: for the cation, the positive charge can be placed on C1 or C3 (and potentially delocalised further to oxygen if the conjugation extends that far)
The Mulliken charges should mirror this: atoms that bear charge in one or more resonance structures will show a larger partial charge in the calculation
If the charges on C1 and C3 are approximately equal, this supports the idea that both resonance structures contribute equally
If one end carries more charge, the resonance structures are not equally weighted, and the dominant contributor is the one that places more charge on the atom with the larger calculated partial charge
Cation vs anion comparison
The charge distribution pattern should be roughly complementary: where the cation is most positive, the anion is most negative (at the same positions in the conjugated system)
This confirms that resonance delocalisation works the same way for both positive and negative charge in allylic systems
The oxygen's charge may differ between the two because its electronegativity affects how much it stabilises positive vs negative charge nearby
Heat of formation is reported directly by MOPAC as ΔHf in kcal/mol. No further calculation is needed to compare isomers: simply compare the numbers.
Stability ranking rule:
\text{More negative } \Delta H_f \implies \text{more stable}For Mulliken charges, the sum of all atomic charges in a neutral molecule equals zero. For a cation with charge +1, the charges sum to +1. For an anion with charge -1, they sum to -1.
\sum_{i} q_i = \text{total formal charge of the species}Resonance structures for the allylic cation:
H2C=CH-CH(+) ↔ (+)CH-CH=CH2
Both structures place the positive charge on a terminal carbon. The real structure is a blend, and the Mulliken charges quantify the blend.
Heat of formation calculations are used routinely in thermochemistry to predict whether a reaction is exothermic or endothermic. In industrial chemistry, knowing which alkene isomer is most stable helps predict the product distribution of catalytic cracking and isomerisation reactions in petroleum refining. Charge distribution calculations underpin computational drug design: knowing where partial charges sit on a molecule helps predict how it will bind to a protein target.
Students often confuse "more negative ΔHf" with "less stable." The opposite is true. A more negative heat of formation means the compound released more energy when formed from its elements, so it sits in a deeper energy well.
A common error is assuming (Z)-2-pentene is more stable than (E)-2-pentene because "Z" sounds like it should come first. Stability is determined by steric interactions, not alphabetical order. The E isomer has less steric strain because its larger groups are on opposite sides of the double bond.
Students sometimes think Mulliken charges represent formal charges. They do not. Mulliken charges are partial charges (fractions of an electron unit) that reflect how electron density is distributed across a molecule. Formal charges are integers assigned by a bookkeeping convention.
Another misconception is that resonance structures are real, interconvertible forms of the molecule. They are not. The molecule does not flip between resonance structures. The real structure is a single, fixed blend (the resonance hybrid), and Mulliken charges show you what that blend looks like numerically.
⚠️ You will likely be asked to rank alkene isomers by stability. Remember: more substituted > less substituted, and E > Z for the same substitution pattern.
⚠️ Expect a question asking you to interpret Mulliken charge data and explain whether it matches the predicted resonance structures. The answer is almost always yes, and you should point to specific atoms that carry partial charge consistent with where the resonance structures place formal charge.
⚠️ Know the difference between MM2 (molecular mechanics, calculates steric energy) and MOPAC/PM3 (semi-empirical quantum mechanics, calculates ΔHf and charge distributions). They answer different questions.
⚠️ Be prepared to explain why 1-pentene is less stable than 2-pentene in terms of hyperconjugation and degree of substitution at the double bond.
True or False: A more negative ΔHf means the compound is less stable. (False. More negative = more stable.)
Fill in the blank: (E)-2-pentene is more stable than (Z)-2-pentene because the larger groups are on ______ sides of the double bond, reducing steric strain. (Opposite.)
True or False: Mulliken charges on a neutral molecule sum to zero. (True.)
Fill in the blank: In an allylic cation, the positive charge is delocalised across ______ atoms by resonance. (Three, the two terminal carbons and the central carbon of the allyl system.)
True or False: MOPAC and MM2 calculate the same property. (False. MM2 calculates steric energy using molecular mechanics; MOPAC calculates heats of formation and electronic properties using semi-empirical quantum mechanics.)
Q: Rank 1-pentene, (E)-2-pentene, and (Z)-2-pentene from most stable to least stable. Explain your reasoning.
A: (E)-2-pentene > (Z)-2-pentene > 1-pentene. Both 2-pentene isomers are disubstituted at the double bond, while 1-pentene is monosubstituted. More substitution stabilises an alkene through hyperconjugation. Among the 2-pentenes, the E isomer avoids the steric clash that the Z isomer has between its cis-oriented methyl and ethyl groups.
Q: The MOPAC ΔHf for (E)-2-pentene is -8.2 kcal/mol and for 1-pentene is -5.3 kcal/mol. Which is more stable and by how much?
A: (E)-2-pentene is more stable by 2.9 kcal/mol (the difference between -8.2 and -5.3). The more negative value indicates greater stability.
Q: In the allylic cation, the Mulliken charges on C1 and C3 are both approximately +0.25, while C2 is approximately +0.05. What does this tell you about the resonance structures?
A: The nearly equal charges on C1 and C3 indicate that the two resonance structures (positive charge on C1, and positive charge on C3) contribute approximately equally to the resonance hybrid. The small positive charge on C2 is consistent with C2 never bearing a formal positive charge in any resonance structure but still being slightly electron-poor due to the adjacent delocalised positive charge.
Q: How does the charge distribution in the allylic anion compare to the allylic cation?
A: The pattern is complementary. In the anion, C1 and C3 carry the most negative partial charge (the positions where resonance structures place the formal negative charge), while C2 is less negative. The oxygen may carry more negative charge in the anion than in the cation because its electronegativity stabilises nearby negative charge more effectively than it stabilises nearby positive charge.
Q: Why does this lab use MOPAC/PM3 for heats of formation instead of MM2?
A: MM2 is a molecular mechanics method that models atoms as classical particles connected by springs. It calculates steric energy, not thermodynamic quantities like heats of formation. MOPAC/PM3 is a semi-empirical quantum mechanical method that includes electronic effects, making it capable of calculating ΔHf and charge distributions. These are properties that depend on electronic structure, which molecular mechanics cannot capture.
The alkene stability trends here are the same ones that govern Zaitsev's rule in elimination reactions (covered in Part 2 of these notes): the more substituted alkene is the thermodynamic product. Resonance and charge delocalisation come back in electrophilic aromatic substitution, where understanding partial charges on a ring helps predict directing effects. The jump from molecular mechanics to quantum mechanics in this lab foreshadows the more rigorous computational methods (Hartree-Fock, DFT) used in advanced courses and in research.
MOPAC, PM3, semi-empirical quantum mechanics, heat of formation, delta Hf, enthalpy of formation, Mulliken charge, Mulliken population analysis, partial charge, allylic cation, allylic anion, resonance delocalisation, resonance structures, conjugation, 1-pentene, E-2-pentene, Z-2-pentene, alkene stability, degree of substitution, hyperconjugation, E Z isomers, cis trans alkene, disubstituted alkene, monosubstituted alkene, charge distribution, CHEM 2311, computational chemistry organic