10.6 Hybridization/LCAO & Stereochemistry

Hybridization/LCAO & Stereochemistry

Quick Answer: Hybridization explains molecular geometry by mixing atomic orbitals into hybrid orbitals; stereochemistry tracks the 3D arrangement of atoms that gives rise to isomers with identical connectivity but different shapes. The PA-CAT Bulletin of Information (rev. 20240815) groups these under the bond-properties / organic-I cluster (Table 5). Expect questions on hybrid state, R/S assignment, and E/Z naming.

Hybridization

Hybrid orbitals form by Linear Combination of Atomic Orbitals (LCAO) — mixing an s with p (and sometimes d) orbitals on the same atom to produce equivalent directional orbitals that overlap better with neighboring atoms.

How to identify hybridization from a structure: count σ bonds + lone pairs (the steric number) on the atom.

• Steric number 2 → sp (e.g., the carbons of C≡C, each with one σ to the other C and one σ to H).
• Steric number 3 → sp² (e.g., each carbon of C=C; the remaining p orbital forms the π bond).
• Steric number 4 → sp³ (e.g., saturated carbon, oxygen in H₂O, nitrogen in NH₃).

LCAO and Molecular Orbital Theory

When two atomic orbitals combine, they form two molecular orbitals: a lower-energy bonding MO (in-phase, constructive overlap) and a higher-energy antibonding MO (out-of-phase, node between nuclei). Electrons fill MOs from lowest energy up. Bond order = (bonding e⁻ − antibonding e⁻) / 2. For H₂: two electrons in σ₁s → bond order = 1. For He₂: σ₁s² σ₁s² → bond order = 0 (no bond). For O₂: the MO picture correctly predicts two unpaired electrons in π orbitals — paramagnetic, which a simple Lewis structure misses.

Stereochemistry: Chirality

A chiral center (stereocenter) is usually an sp³ carbon bonded to four different groups. A molecule with one chiral center is chiral and exists as two enantiomers — non-superimposable mirror images. Enantiomers share physical properties (melting point, solubility, BP) except for the direction they rotate plane-polarized light: (+) (dextrorotatory) rotates clockwise, (−) (levorotatory) rotates counterclockwise. A racemic mixture (±) is a 50:50 mix with zero net rotation.

With two or more stereocenters, diastereomers arise — stereoisomers that are not mirror images (e.g., cis/trans on a ring, or (2R,3R) vs (2S,3S) is enantiomeric, but (2R,3R) vs (2R,3S) is diastereomeric). The maximum number of stereoisomers is 2ⁿ for n stereocenters, reduced by symmetry (meso compounds).

A meso compound has chiral centers but is achiral overall because of an internal plane of symmetry (e.g., meso-tartaric acid). It is a diastereomer of the chiral enantiomeric pair.

R/S Configuration (Cahn-Ingold-Prelog)

1. Assign priority to each of the four groups on the stereocenter: higher atomic number = higher priority. For ties, compare the next set of atoms out until you break the tie.
2. Orient the molecule so the lowest-priority group points away from you.
3. Trace 1 → 2 → 3. Clockwise = R (rectus); counterclockwise = S (sinister).

If the lowest-priority group points toward you, the apparent direction is reversed: clockwise-appearing = S, counterclockwise-appearing = R.

Cis/Trans and E/Z Notation

Cis/trans labels relative position of substituents on a ring or double bond: cis = same side, trans = opposite sides. It fails when all four substituents on an alkene differ, so the E/Z system (CIP-based) is used:

• On each alkene carbon, pick the higher-priority substituent (CIP).
• If the two higher-priority substituents are on opposite sides → E (entgegen, opposite).
• If on the same side → Z (zusammen, together).

flowchart TD
    A[Stereocenter: sp3 C with 4 different groups] --> B{Mirror relationship?}
    B -->|Non-superimposable mirror images| C[Enantiomers: R vs S]
    B -->|Not mirror images| D[Diastereomers]
    D --> E[cis/trans on ring or alkene]
    D --> F[E/Z by CIP priority]
    C --> G[Same physical props, opposite optical rotation]

Worked Example — Assigning E/Z

Consider 2-bromo-2-butene drawn as CH₃ and Br on one carbon, H and CH₃ on the other. On C2: Br (Z=35) beats CH₃ (C, Z=6) → Br is higher priority. On C3: CH₃ beats H. If Br and the higher-priority CH₃ are on opposite sides of the double bond, the configuration is E; if on the same side, Z.

PA-CAT Tips

• Steric number = σ bonds + lone pairs → hybridization.
• R/S: lowest priority away, then 1→2→3 clockwise = R.
• E/Z: higher priority on each alkene carbon; opposite = E.
• A meso compound is achiral despite having stereocenters — internal mirror plane.
• Optical rotation sign (+/−) is experimentally determined and does not follow from R/S — (R)-glyceraldehyde is (+), but many R compounds are (−).