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Structure and Bonding

1. Hybridization

1.1 sp3^3 Hybridization

Definition 1 (sp3^3 Hybridization): One ss and three pp orbitals combine to form four equivalent sp3^3 hybrid orbitals, arranged tetrahedrally with bond angles of 109.5°.

ψsp3=12(ψs+ψpx+ψpy+ψpz)\psi_{sp^3} = \frac{1}{2}(\psi_s + \psi_{p_x} + \psi_{p_y} + \psi_{p_z})

Each sp3^3 orbital has 25% ss character and 75% pp character. Examples: methane (CH4_4), ethane, water (bent due to lone pairs).

1.2 sp2^2 Hybridization

Definition 2 (sp2^2 Hybridization): One ss and two pp orbitals combine to form three equivalent sp2^2 hybrid orbitals in a trigonal planar arrangement (120°). The remaining unhybridized pzp_z orbital forms π\pi bonds.

ψsp2=13ψs+23ψp\psi_{sp^2} = \frac{1}{\sqrt{3}}\psi_s + \sqrt{\frac{2}{3}}\psi_p

Each sp2^2 orbital has 33.3% ss character. Examples: ethylene (C2_2H4_4), formaldehyde, benzene.

1.3 sp Hybridization

Definition 3 (sp Hybridization): One ss and one pp orbital combine to form two sp hybrid orbitals in a linear arrangement (180°). Two unhybridized pp orbitals form two perpendicular π\pi bonds.

ψsp=12(ψs+ψp)\psi_{sp} = \frac{1}{\sqrt{2}}(\psi_s + \psi_p)

Each sp orbital has 50% ss character. Examples: acetylene (C2_2H2_2), CO2_2, HCN.

1.4 Bond Strength and Hybridization

More ss character     \implies shorter, stronger bonds:

Bond length: sp3>sp2>sp\text{Bond length: } \text{sp}^3 > \text{sp}^2 > \text{sp}

Bond strength: sp3<sp2<sp\text{Bond strength: } \text{sp}^3 < \text{sp}^2 < \text{sp}

Hybridizationss CharacterC–C Bond Length (pm)C–C Bond Energy (kJ/mol)
sp3^325%154347
sp2^233.3%133614
sp50%120839

2. VSEPR Theory

2.1 Electron Pair Repulsion

Theorem 1 (VSEPR Theory): Electron pairs around a central atom arrange to minimize repulsion: lone pair–lone pair > lone pair–bond pair > bond pair–bond pair.

2.2 Common Geometries

Steric NumberGeometryBond AngleExample
2Linear180°CO2_2, BeCl2_2
3Trigonal planar120°BF3_3, CO32_3^{2-}
4Tetrahedral109.5°CH4_4
4Trigonal pyramidal< 109.5°NH3_3
4Bent< 109.5°H2_2O
5Trigonal bipyramidal90°, 120°, 180°PCl5_5
6Octahedral90°, 180°SF6_6

2.3 Effect of Lone Pairs and Multiple Bonds

  • Lone pairs occupy more space than bonding pairs, compressing bond angles.
  • Double and triple bonds repel more than single bonds.
  • In trigonal bipyramidal geometry, lone pairs preferentially occupy equatorial positions (less crowding: 2 neighbors at 90° vs 3 for axial).

3. Molecular Orbital Theory for Organics

3.1 Sigma (σ\sigma) and Pi (π\pi) Bonds

Definition 4 (σ\sigma Bond): A bond formed by head-on overlap of orbitals along the internuclear axis. Electron density is symmetric about the bond axis.

Definition 5 (π\pi Bond): A bond formed by lateral overlap of parallel pp orbitals, perpendicular to the internuclear axis. Electron density is above and below (or around) the bond axis.

3.2 Bonding in Ethylene

The C=C double bond consists of:

  • One σ\sigma bond (sp2^2–sp2^2 overlap).
  • One π\pi bond (p–p lateral overlap).

Rotation about a π\pi bond requires breaking it (270\sim 270 kJ/mol), explaining the planarity of alkenes.

3.3 Bonding in Acetylene

The C≡C triple bond:

  • One σ\sigma bond (sp–sp overlap).
  • Two π\pi bonds (two perpendicular p–p overlaps).

4. Conjugation and Resonance

4.1 Conjugated Systems

Definition 6 (Conjugation): Alternating single and double bonds allow pp orbitals to overlap across multiple atoms, creating a delocalized π\pi system.

Examples: 1,3-butadiene, α,β\alpha,\beta-unsaturated carbonyls, benzene.

4.2 Resonance

Definition 7 (Resonance): When a molecule or ion can be represented by two or more valid Lewis structures (resonance forms), the actual structure is a hybrid — a weighted average.

Theorem 2 (Resonance Rules):

  1. Resonance forms differ only in electron placement; nuclei do not move.
  2. All resonance forms must have the same number of unpaired electrons.
  3. The resonance hybrid is more stable than any individual form.
  4. More stable resonance forms contribute more to the hybrid.

Stability ranking of resonance forms:

  • Octet rule satisfied > octet rule violated
  • Fewer formal charges > more formal charges
  • Negative charge on electronegative atoms > on electropositive atoms
  • Charge separation minimized > charge separation maximized

Example 1: The nitrate ion NO3_3^- has three equivalent resonance structures, each with one N=O and two N–O bonds. The actual N–O bond order is 4/34/3.

\blacksquare

4.3 Resonance Energy

Definition 8 (Resonance Energy): The extra stabilization of a conjugated system compared to a hypothetical system with localized bonds.

Benzene: ΔHhyd\Delta H_{\text{hyd}} for 3 localized double bonds = 3×(120)=3603 \times (-120) = -360 kJ/mol. Experimental ΔHhyd\Delta H_{\text{hyd}} = 208-208 kJ/mol. Resonance energy = 152 kJ/mol.

5. Aromaticity

5.1 Huckel”s Rule

Theorem 3 (Huckel’s Rule): A planar, cyclic, fully conjugated system with (4n+2)(4n + 2) π\pi electrons is aromatic (exceptionally stable). Systems with 4n4n π\pi electrons are antiaromatic (destabilized).

π\pi ElectronsnnAromaticityExample
20AromaticCyclopropenyl cation
41AntiaromaticCyclobutadiene
61AromaticBenzene
82AntiaromaticCyclooctatetraene (tub)
102AromaticNaphthalene

5.2 Criteria for Aromaticity

  1. Cyclic — the π\pi system must form a closed loop.
  2. Planar — all pp orbitals must be parallel for effective overlap.
  3. Fully conjugated — every atom in the ring must have a pp orbital (no sp3^3 atoms in the ring).
  4. (4n+2)(4n + 2) π\pi electrons — Huckel’s rule.

5.3 Aromatic Heterocycles

Pyridine: 6 π\pi electrons from the C=N ring; the nitrogen lone pair is in an sp2^2 orbital perpendicular to the π\pi system and does not participate.

Pyrrole: 6 π\pi electrons; the nitrogen lone pair (in a pp orbital) contributes to the π\pi system.

Furan and Thiophene: 6 π\pi electrons; the heteroatom lone pair contributes to the π\pi system.

5.4 Anti-Aromaticity and Non-Aromaticity

Antiaromatic: Meets all criteria except has 4n4n π\pi electrons. Highly destabilized; distorts geometry to escape antiaromaticity (e.g., cyclooctatetraene adopts a tub-shaped non-planar conformation).

Non-aromatic: Fails one or more criteria (not cyclic, not planar, not fully conjugated, or wrong electron count but not 4n4n).

6. Stereochemistry

6.1 Chirality

Definition 9 (Chirality): A molecule is chiral if it is not superimposable on its mirror image. A chiral center (stereocenter) is a carbon atom bonded to four different substituents.

Theorem 4 (Chirality): A molecule with a single stereocenter exists as a pair of enantiomers (non-superimposable mirror images) that are chemically identical in an achiral environment but rotate plane-polarized light in opposite directions.

6.2 The R/S Naming System

Definition 10 (Cahn-Ingold-Prelog Rules):

  1. Assign priority to substituents based on atomic number (higher = higher priority).
  2. Orient the molecule so the lowest-priority group is pointing away.
  3. Read the remaining three in order of decreasing priority:
    • Clockwise → RR (Rectus)
    • Counterclockwise → SS (Sinister)

For double bonds (E/Z):

E (Entgegen): higher priority groups on opposite sides\text{E (Entgegen): } \text{higher priority groups on opposite sides}

Z (Zusammen): higher priority groups on same side\text{Z (Zusammen): } \text{higher priority groups on same side}

6.3 Optical Activity

Definition 11 (Optical Activity): Enantiomers rotate plane-polarized light. The specific rotation:

[α]=αobscl[\alpha] = \frac{\alpha_{\text{obs}}}{c \cdot l}

where αobs\alpha_{\text{obs}} is the observed rotation (degrees), cc is concentration (g/mL), and ll is path length (dm).

An enantiomeric mixture: ee=[major][minor][major]+[minor]×100%ee = \frac{[\text{major}] - [\text{minor}]}{[\text{major}] + [\text{minor}]} \times 100\%

6.4 Diastereomers

Definition 12 (Diastereomers): Stereoisomers that are not mirror images. They have different physical and chemical properties.

  • Molecules with 2 or more stereocenters can have diastereomeric relationships.
  • Meso compounds: Molecules with stereocenters that are achiral overall due to an internal plane of symmetry.

Example 2: Tartaric acid has three stereoisomers: (R,R)(R,R), (S,S)(S,S) (enantiomers), and meso (internally compensated, [R,S][R,S] with a symmetry plane).

\blacksquare

6.5 Fischer Projections

Definition 13 (Fischer Projection): A 2D representation of a 3D molecule with:

  • Horizontal bonds projecting toward the viewer.
  • Vertical bonds projecting away from the viewer.

To interchange two substituents: swap any two groups → invert stereochemistry.

7. Conformational Analysis

7.1 Newman Projections

Definition 14 (Newman Projection): View along a C–C bond. The front carbon is represented by the point where bonds meet; the back carbon by a circle.

7.2 Conformations of Ethane

The dihedral angle ϕ\phi between H atoms on adjacent carbons determines the energy:

E(ϕ)=V02(1+cos3ϕ)E(\phi) = \frac{V_0}{2}(1 + \cos 3\phi)

  • Staggered (ϕ=60°,180°,300°\phi = 60°, 180°, 300°): minimum energy.
  • Eclipsed (ϕ=0°,120°,240°\phi = 0°, 120°, 240°): maximum energy (12\sim 12 kJ/mol above staggered).

7.3 Butane Conformations

ConformationDihedral AngleRelative Energy (kJ/mol)
Anti180°0
Gauche±60°+3.8
Eclipsed (Me–H)±120°+16
Eclipsed (Me–Me)+19

The anti conformation is most stable due to minimal steric hindrance.

7.4 Cyclohexane

Definition 15 (Chair Conformation): The most stable conformation of cyclohexane, with bond angles of 109.5° and zero angle strain.

  • Axial bonds: Point alternately up and down, roughly parallel to the ring axis.
  • Equatorial bonds: Point outward, roughly in the plane of the ring.

At room temperature, cyclohexane undergoes rapid ring flipping, interconverting axial and equatorial positions.

Theorem 5 (A-Value): The conformational free energy difference between axial and equatorial:

SubstituentA-Value (kJ/mol)Preferred Position
CH3_37.3Equatorial
C2_2H5_57.4Equatorial
t-Bu> 20Equatorial (locks)
OH4.0Equatorial
Cl2.1Equatorial
Br1.8Equatorial

7.5 Disubstituted Cyclohexanes

  • 1,2-diaxial interactions: Substituents in the 1,2-positions on the same side experience steric repulsion (gauche butane interactions).
  • cis-1,3-diaxial: Severe repulsion if both groups are axial.
  • trans-decalin: Rigid (no ring flip); cis-decalin: more flexible.

7.6 Cyclohexane Conformational Equilibrium

For a monosubstituted cyclohexane:

Keq=[equatorial][axial]=eΔG/RTK_{\text{eq}} = \frac{[\text{equatorial}]}{[\text{axial}]} = e^{-\Delta G/RT}

Example 3: For methylcyclohexane at 298 K with ΔG=7.3\Delta G = 7.3 kJ/mol:

Keq=e7300/(8.314×298)=e2.9519K_{\text{eq}} = e^{-7300/(8.314 \times 298)} = e^{-2.95} \approx 19

The equatorial conformer is favored ~95%.

\blacksquare

8. Physical Organic Chemistry Concepts

8.1 Inductive Effects

Definition 16 (Inductive Effect): Electron withdrawal or donation through σ\sigma bonds, decreasing with distance.

  • Electron-withdrawing groups (EWG): -NO2_2, -CN, -C=O, halogens (at short range).
  • Electron-donating groups (EDG): Alkyl groups, -O^-, -NH2_2.

8.2 Hyperconjugation

Definition 17 (Hyperconjugation): Delocalization of σ\sigma electrons (in most cases C–H) into an adjacent empty or partially filled pp or π\pi orbital.

This stabilizes carbocations, explains the preference for staggered conformations, and contributes to the stability of alkenes (more alkyl substituents = more hyperconjugation = more stable).

8.3 Field Effects

Definition 18 (Field Effect): Electrostatic interaction through space (not through bonds), important for polar substituents near a reaction center.

Common Pitfalls

  1. Confusing hybridization and geometry. sp2^2 hybridized carbon is trigonal planar (3 groups), but sp3^3 nitrogen with a lone pair is also trigonal planar (σ=3\sigma = 3). Fix: Steric number (bonded atoms + lone pairs) determines geometry, not hybridization alone.
  2. Misidentifying aromatic vs non-aromatic systems. Cyclooctatetraene is non-aromatic (tub-shaped, not planar), not antiaromatic. Fix: Check all four criteria: cyclic, planar, fully conjugated, and electron count.
  3. Wrong R/S assignment from Fischer projections. Interchanging any two groups in a Fischer projection inverts stereochemistry. Fix: Place the lowest-priority group at top or bottom (vertical) for a valid R/S assignment.
  4. Ignoring hyperconjugation in carbocation stability. Tertiary carbocations are more stable not just because of inductive effects but also because of more C–H hyperconjugation. Fix: Count the number of adjacent C–H bonds that can hyperconjugate.
  5. Confusing E/Z with R/S. E/Z refers to double-bond geometry (alkenes); R/S refers to chirality at stereocenters. Fix: Use CIP priorities for both, but apply to different structural features.
  6. Wrong axial/equatorial assignments in chair conformations. Axial bonds alternate up/down around the ring. Fix: Draw the chair carefully; remember that ring flip converts all axial to equatorial and vice versa.
  7. Overcounting π\pi electrons in aromatic heterocycles. The nitrogen lone pair in pyridine does not contribute to the π\pi system; in pyrrole it does. Fix: Check whether the lone pair is in a pp orbital (contributes) or an sp2^2 orbital (does not).

Summary

  • Hybridization: sp3^3 (tetrahedral), sp2^2 (trigonal planar + pπp_\pi), sp (linear + 2 pπp_\pi).
  • VSEPR: Electron pairs arrange to minimize repulsion; bond angles deviate from ideal values due to lone pairs and multiple bonds.
  • Conjugation and resonance: Delocalized π\pi systems; resonance hybrids are more stable than individual forms.
  • Aromaticity: Huckel’s rule (4n+2)(4n + 2) π\pi electrons; must be cyclic, planar, and fully conjugated.
  • Stereochemistry: R/S system for chiral centers; E/Z for double bonds; enantiomers vs diastereomers.
  • Conformational analysis: Newman projections; cyclohexane chair conformations; A-values determine substituent preferences.

Worked Examples

Example 1: Determining Molecular Geometry

Problem: Predict the geometry and bond angle around the central carbon in CH3NO2 (nitromethane). Solution: The central carbon is sp3 hybridised (tetrahedral geometry for the C-H bonds, approximately 109.5 degrees). The nitrogen is sp2 hybridised with a formal positive charge. The two N-O bonds are equivalent due to resonance. The C-N bond length is shorter than a typical C-N single bond due to partial double bond character from resonance delocalisation.

Example 2: Analysing Conformational Energy

Problem: For methylcyclohexane, determine which conformation is more stable (equatorial vs axial methyl) and calculate the energy difference. Solution: The equatorial conformation is more stable. The A-value for a methyl group is 1.7 kcal/mol. At room temperature (kT = 0.6 kcal/mol), the equilibrium constant K = exp(-Delta G/RT) = exp(-1.7/0.6) = exp(-2.83) = 0.059. The equatorial:axial ratio is approximately 17:1, meaning about 94% of molecules are in the equatorial conformation at room temperature.

Cross-References

TopicSiteLink
Reaction MechanismsWyattsNotesView
SpectroscopyWyattsNotesView
Quantum ChemistryWyattsNotesView
Organic Chemistry — MIT 5.12MIT OCWView