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π-Molecular Orbitals and Conjugation

From Isolated $p$ Orbitals to $\pi$ Molecular Orbitals

Hybridization explains the $\sigma$ skeleton: $sp^2$ carbons in ethene are trigonal planar, and the leftover $p$ orbital on each carbon stands perpendicular to that plane. Lewis structures then draw a second line between the carbons and call it a $\pi$ bond. That second line is another convenient lie. The $\pi$ bond is not a second stick. It is what you get when those two $p$ orbitals combine into molecular orbitals.

Two $p$ orbitals do not stay "two $p$ orbitals." They mix to give two $\pi$ molecular orbitals:

  • A bonding combination ($\pi$) — the $p$ lobes on the two carbons have the same phase (same color on the same face of the molecule). Electron density piles up between the carbons, above and below the plane. This orbital is occupied. It is the HOMO.
  • An antibonding combination ($\pi^*$) — the $p$ lobes have opposite phase, so a nodal plane appears between the carbons. This orbital is empty. It is the LUMO.

Red and blue in the models below are phase, not charge. A color change from red to blue (or blue to red) marks a node: a place where the wavefunction is zero. The molecular plane is a node for every $\pi$ orbital (that is just the $p$ orbital's built-in node). Extra nodes that cut across the carbon chain are the ones that raise the energy.

Action: ethene
  • Start with the isolated $p$ orbitals. Confirm that the $p$ lobes stand perpendicular to the molecular plane. Drag to rotate if you want a more edge-on view of the $\mathrm{C=C}$ bond.
  • Switch to $\pi$ (HOMO). The two $p$ orbitals have merged into one orbital that is bonding: same color on the same face, density between the carbons. This is where ethene's $\pi$ electrons live, and it is why alkenes behave as nucleophiles.
  • Switch to $\pi^*$ (LUMO). Count the new node between the two carbons. Empty orbitals with a node between nuclei are antibonding and high in energy.

 

Red / blue = phase. A color swap marks a node. The HOMO holds the $\pi$ electrons that electrophiles attack.
Take Note
  • The $\sigma$ framework (the $sp^2$ bonds in the plane) is still there. We are looking only at the $\pi$ system built from leftover $p$ orbitals.
  • $\pi$ has zero nodes between carbons — bonding, occupied, HOMO, nucleophilic.
  • $\pi^*$ has one node between carbons — antibonding, empty, LUMO. Nucleophiles would donate into this orbital if the reaction called for it; ordinary electrophilic addition to an alkene uses the HOMO instead.
  • Energy tracks nodes: more internuclear nodes $\Rightarrow$ higher energy.

Conjugation: four $p$ orbitals, four $\pi$ MOs

1,3-Butadiene is two alkenes that share a single bond between them: $\mathrm{C=C-C=C}$. If those four $p$ orbitals line up in the same plane they are conjugated. They no longer make two separate $\pi$ bonds. They mix into four molecular orbitals that span the whole chain.

For a linear polyene with $n$ conjugated $p$ orbitals you always get $n$ $\pi$ MOs. Butadiene has $n = 4$, so four levels. Each level $k$ has $k-1$ internuclear nodes:

MO Internuclear nodes Occupied? Role
$\psi_1$ ($\pi_1$) 0 — bonding everywhere yes (2 electrons) lowest-energy $\pi$ MO
$\psi_2$ ($\pi_2$) 1 — node through the central $\mathrm{C-C}$ yes (2 electrons) HOMO
$\psi_3$ ($\pi_3^*$) 2 no LUMO
$\psi_4$ ($\pi_4^*$) 3 — antibonding everywhere no highest-energy $\pi$ MO

Four $\pi$ electrons fill $\psi_1$ and $\psi_2$. The HOMO is therefore $\psi_2$, not $\psi_1$. That matters for reactivity: an electrophile does not care about the lowest occupied orbital. It cares about the highest occupied one, because those electrons are the most loosely held.

Action: 1,3-butadiene
  • Turn on the isolated $p$ orbitals first. All four carbons have a $p$ lobe. That is the conjugated array before the orbitals mix.
  • Step through $\psi_1 \rightarrow \psi_4$. At each step, count how many times the color flips as you walk from $\mathrm{C1}$ to $\mathrm{C4}$. That is the node count.
  • Stop on the HOMO ($\psi_2$). Where is the amplitude largest? Those carbons are the nucleophilic sites (the ends of the diene).
  • Stop on the LUMO ($\psi_3$). Empty, more nodes, higher energy. This is the orbital a nucleophile would use if it added to the diene.

 

Walk $\mathrm{C1}$ to $\mathrm{C4}$ and count color flips. Nodes increase with energy. The HOMO is $\psi_2$, not $\psi_1$.

What the HOMO is trying to tell you

Once you can see the orbitals, three reactivity rules stop being slogans:

  1. Nucleophiles donate from the HOMO. Ethene's nucleophilic identity is the $\pi$ HOMO sitting above and below the $\mathrm{C=C}$ bond. That is why electrophiles add to the $\pi$ face, not to the $\sigma$ skeleton. In butadiene the HOMO ($\psi_2$) has its largest lobes on the terminal carbons, so electrophiles attack $\mathrm{C1}$ or $\mathrm{C4}$.
  2. Conjugation raises the HOMO and lowers the LUMO relative to an isolated alkene. Butadiene is a better nucleophile and a better electrophile than ethene for that reason. The HOMO–LUMO gap also shrinks, which is why conjugated polyenes absorb visible light (think lycopene in tomatoes) while ethene does not.
  3. Nodes are a ladder. $\psi_1$ is bonding all along the chain (no internuclear nodes). Each step up the ladder inserts one more node. You do not have to memorize which MO is the HOMO if you can count electrons and count nodes: four $\pi$ electrons fill the two lowest MOs, so the second one is the HOMO.
Pitfall
  • Do not call $\psi_1$ the HOMO just because it is "the $\pi$ orbital." In a conjugated system several $\pi$ MOs are occupied. The HOMO is the highest of those.
  • Red and blue are not $\delta+$ / $\delta-$. They are the sign of the wavefunction. Same color on neighboring atoms is bonding; a color swap is a node.
  • The node in the molecular plane does not count when you rank $\pi$ MOs against each other. Every $\pi$ MO has that node. Count the additional nodes that cut the chain.

Questions

Question: In ethene, which orbital is the HOMO and why does that orbital make ethene a nucleophile?

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Question: As you click from $\psi_1$ to $\psi_4$ in butadiene, what happens to the number of internuclear nodes, and which orbital is the HOMO?

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Question: Why do electrophiles attack the ends of 1,3-butadiene rather than the internal carbons? Use the HOMO, not a resonance cartoon, in your answer.

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Question: How does conjugation change nucleophilicity compared with an isolated alkene such as ethene?

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