Organic Chemistry
Bonds, isomerism, alkanes, alkenes, aromatics, alkyl halides, oxygenated compounds, acids, amines, biomolecules and polymers. Theory, examples, and solved exercises.
Complete Theory
4Carbon has atomic number and electron configuration . The four valence electrons () can form up to four covalent bonds, a property that makes it the fundamental element of organic chemistry, capable of building chains, branches, and rings of unlimited complexity.
The atomic orbitals of carbon in the valence shell are:
A sigma bond () forms by head-on overlap along the bond axis; a pi bond () forms by lateral overlap above and below the molecular plane, and is weaker and more reactive than a bond.
Real-world applications: carbon's versatility is the basis of life and organic materials: plastics, pharmaceuticals, fuels, textile fibers, polymers. The ability to form stable C—C bonds is what distinguishes organic chemistry from inorganic chemistry.
The atomic orbitals of carbon in the valence shell are:
- One 2s orbital (spherical, lower energy)
- Three 2p orbitals (2px, 2py, 2pz, dumbbell-shaped, oriented along the Cartesian axes, slightly higher energy)
A sigma bond () forms by head-on overlap along the bond axis; a pi bond () forms by lateral overlap above and below the molecular plane, and is weaker and more reactive than a bond.
Real-world applications: carbon's versatility is the basis of life and organic materials: plastics, pharmaceuticals, fuels, textile fibers, polymers. The ability to form stable C—C bonds is what distinguishes organic chemistry from inorganic chemistry.
Hybridization is the linear combination of atomic orbitals to form equivalent hybrid orbitals, with shape and energy intermediate between the starting orbitals. For carbon, three fundamental types exist:
sp³ Hybridization: combination of one 2s orbital + three 2p orbitals → four equivalent sp³ hybrid orbitals.
Real-world applications: hybridization determines reactivity: sp carbons (acetylenes) are more acidic (pKₐ ≈ 25) than sp³ (pKₐ ≈ 50), enabling selective deprotonation reactions.
sp³ Hybridization: combination of one 2s orbital + three 2p orbitals → four equivalent sp³ hybrid orbitals.
- Geometry: tetrahedral, bond angles
- C—C bond length: (single bond)
- Examples: (methane), (ethane), (propane)
- All bonds are of type
- Geometry: trigonal planar, bond angles
- C=C bond length: (double bond: )
- Examples: (ethylene), (propylene)
- Geometry: linear, bond angles
- C≡C bond length: (triple bond: )
- Examples: (acetylene), (hydrogen cyanide)
Real-world applications: hybridization determines reactivity: sp carbons (acetylenes) are more acidic (pKₐ ≈ 25) than sp³ (pKₐ ≈ 50), enabling selective deprotonation reactions.
Organic molecules can be represented with different types of formulas, each with a different level of detail:
For molecules with multiple possible Lewis structures, resonance structures are used (e.g., benzene, carbonate ion). The real structure is a hybrid of all limiting structures.
Real-world applications: the skeletal formula is the universal standard in organic chemistry (scientific publications, CAS databases, pharmacology manuals). Every aspiring chemist must be able to read and write it fluently.
- Lewis formula: shows all atoms, bonds (lines), and lone pairs of electrons (dots). Example: shows C with 4 single C—H bonds, no lone pairs. Essential for understanding electronic structure and formal charge distribution.
- Condensed formula: writes atoms in sequence, grouping hydrogens bonded to each carbon. Example: for propane, or for hexane. It is the most common notation in chemical reactions for compactness.
- Skeletal (line-angle) formula: each vertex and line ending represents a carbon atom; hydrogens bonded to carbon are implied (counted by difference, knowing C forms 4 bonds). Single bonds are simple lines, double bonds are parallel lines, triple bonds are three parallel lines. Heteroatom symbols (O, N, Cl, etc.) are explicit.
Example: n-butane is a zig-zag line of 4 segments; cyclohexane is a hexagon; benzene is a hexagon with a circle or three alternating double bonds.
For molecules with multiple possible Lewis structures, resonance structures are used (e.g., benzene, carbonate ion). The real structure is a hybrid of all limiting structures.
Real-world applications: the skeletal formula is the universal standard in organic chemistry (scientific publications, CAS databases, pharmacology manuals). Every aspiring chemist must be able to read and write it fluently.
VSEPR theory (Valence Shell Electron Pair Repulsion) predicts molecular geometry based on repulsion between electron pairs (bonding and lone pairs) in the valence shell. In organic chemistry, it applies mainly to carbon, nitrogen, and oxygen atoms.
Carbon geometries by hybridization:
Real-world applications: molecular geometry determines the three-dimensional shape of drugs, affecting their recognition by biological receptors (key-lock principle). Ring strain in cyclopropane makes it useful as an anesthetic gas (rapid ring opening).
Carbon geometries by hybridization:
- sp³ Carbon (4 bonds): regular tetrahedron, angles . Examples: alkanes, cycloalkanes. The tetrahedral arrangement minimizes repulsion between the 4 bonding electron pairs.
- sp² Carbon (3 bonds + 1 ): trigonal planar, angles . Rotation around the double bond is prevented (the bond would break), leading to geometric cis-trans isomerism.
- sp Carbon (2 bonds + 2 ): linear, angles . Examples: alkynes, nitriles.
- sp³ Nitrogen (3 bonds + 1 lone pair): trigonal pyramidal geometry, angles (e.g., ammonia, amines). The lone pair occupies more space than a bond, compressing the angles.
- sp³ Oxygen (2 bonds + 2 lone pairs): bent geometry, angles (e.g., water, alcohols, ethers). The two lone pairs further compress the angle.
Real-world applications: molecular geometry determines the three-dimensional shape of drugs, affecting their recognition by biological receptors (key-lock principle). Ring strain in cyclopropane makes it useful as an anesthetic gas (rapid ring opening).
Worked Examples
2Example 1Carbon Hybridization in Methane, Ethylene, and Acetylene
Given
Methane CH₄ (4 single C—H bonds)
Ethylene C₂H₄ (C=C double bond)
Acetylene C₂H₂ (C≡C triple bond)
Find
Type of hybridization for each carbon
Geometry and bond angles
Types of bonds present
Step-by-step solution
1Methane (CH₄): carbon forms 4 identical bonds with 4 hydrogen atoms. To have 4 equivalent orbitals, the 2s orbital combines with the three 2p orbitals, generating 4 sp³ hybrid orbitals. Tetrahedral geometry with angles of . All C—H bonds are bonds formed by head-on overlap of C sp³ orbitals with H 1s orbitals.
2Ethylene (C₂H₄): each carbon forms 3 bonds (2 with H and 1 with the other C) and 1 bond. Hybridization is sp²: 3 sp² hybrid orbitals (for bonds) + 1 pure 2p orbital (for the bond). Trigonal planar geometry with angles . The C=C double bond consists of one bond (head-on sp²-sp² overlap) and one bond (lateral 2p-2p overlap). Rotation around the double bond is prevented.
3Acetylene (C₂H₂): each carbon forms 2 bonds (1 with H and 1 with the other C) and 2 bonds. Hybridization is sp: 2 sp hybrid orbitals (for bonds) + 2 pure 2p orbitals (for the two bonds). Linear geometry with angles . The C≡C triple bond has one bond and two bonds perpendicular to each other.
✓ Final result: CH₄: sp³ tetrahedral 109.5°; C₂H₄: sp² trigonal planar 120°; C₂H₂: sp linear 180°
Example 2Writing Lewis and Skeletal Formulas
Given
Molecular formula: C₃H₈O (isopropanol)
Atoms: 3 C, 8 H, 1 O
Find
Complete Lewis formula
Condensed formula
Skeletal formula
Step-by-step solution
1For the Lewis formula, arrange the atoms: C—C—C with O bonded to the central carbon (C₂). Each C must have 4 bonds, O 2 bonds + 2 lone pairs. The structure is: CH₃—CH(OH)—CH₃. Draw all C—H bonds explicitly and the lone pairs on oxygen (two pairs). Formal charge is zero on all atoms.
2Condensed formula: CH₃CH(OH)CH₃ or (CH₃)₂CHOH. The parentheses indicate the OH group bonded to the central carbon, and (CH₃)₂ indicates two methyl groups bonded to the same carbon.
3Skeletal formula: a zig-zag line of 3 segments (the 3 carbons). An —OH (explicit) is bonded to the central carbon. All hydrogens are implied. The left terminal carbon has 3 H (counted by difference: 4 – 1 bond = 3 H), the central carbon has 1 H (4 – 3 bonds = 1 H), the right terminal carbon has 3 H.
✓ Final result: Lewis: CH₃—CH(OH)—CH₃ with 2 lone pairs on O; Condensed: (CH₃)₂CHOH; Skeletal: 3-C zig-zag with —OH
Exercises with Solutions
4Exercise 1Hybridization and geometryMedium
Problem to solve
Indicate the hybridization, geometry, and bond angles for each carbon in the following molecules: (a) propane C₃H₈, (b) propene C₃H₆, (c) propyne C₃H₄.
Given data
C₃H₈ (alkane, only single bonds)C₃H₆ (alkene, one double bond)C₃H₄ (alkyne, one triple bond)
Step-by-step solution
1In propane (C₃H₈, CH₃—CH₂—CH₃) all three carbons form only bonds, so sp³ hybridization for all, tetrahedral geometry (angles ). The terminal carbons are bonded to 3 H, the central one to 2 H.
2In propene (C₃H₆, CH₂=CH—CH₃) the C₁ and C₂ of the double bond have sp² hybridization (trigonal planar geometry, ), while C₃ (methyl) is sp³ (tetrahedral, ).
3In propyne (C₃H₄, CH≡C—CH₃) the C₁ and C₂ of the triple bond have sp hybridization (linear geometry, ), while C₃ is sp³ (tetrahedral, ).
✓ Final answer: Propane: all sp³ (109.5°); Propene: C₁,C₂ sp² (120°), C₃ sp³ (109.5°); Propyne: C₁,C₂ sp (180°), C₃ sp³ (109.5°)
Exercise 2Representation formulasMedium
Problem to solve
Convert the condensed formula (CH₃)₂CHCH₂OH into a skeletal formula and a Lewis formula.
Given data
(CH₃)₂CHCH₂OH (3-methyl-1-butanol)
Step-by-step solution
1The condensed formula tells us: a central carbon (CH) bonded to two methyls (CH₃)₂, to a CH₂, and to an OH. The main chain is C—C—C—OH with a methyl branch on C₂.
2Skeletal formula: zig-zag line of 4 carbons (C₁—C₂—C₃—C₄). On C₂ (second vertex), a segment goes up or down (the branched CH₃). On C₄ (last carbon), the —OH group is explicitly attached. All H are implied.
3Lewis formula: draw all C—C, C—H, and C—O bonds explicitly, with two lone pairs on the oxygen of OH. Verify each C has 4 bonds, O has 2 bonds, each H has 1 bond.
✓ Final answer: Skeletal: 4-C zig-zag with —OH at the end and —CH₃ branch on C₂
Exercise 3Formal charge and resonanceHard
Problem to solve
For the nitrite ion NO₂⁻, determine: (a) the resonance structures, (b) the formal charge on each atom, (c) the hybridization of nitrogen.
Given data
NO₂⁻ (nitrite ion)
Step-by-step solution
1Nitrogen (group 15) has 5 valence electrons, oxygen 6. Charge −1: total electrons = 5 + 2×6 + 1 = 18 e⁻ (9 pairs). Skeleton: N central, two O outer.
2First resonance structure: N bonded with double bond to O₁ (N=O) and single to O₂ (N—O⁻). Nitrogen has one lone pair. Second structure: N=O₂ and N—O₁⁻. The two structures are equivalent and symmetric.
3Formal charge on first structure: N: 5 – 2 – ½(6) = 0; O₁(=): 6 – 4 – ½(4) = 0; O₂(−): 6 – 6 – ½(2) = −1. On the second structure, the charges swap between O₁ and O₂.
4Nitrogen has sp² hybridization: 3 electron domains (2 bonds + 1 lone pair), bent geometry (about 115°). The N—O bond has order 1.5 due to resonance.
✓ Final answer: Two equivalent resonance structures; N charge 0, O with charge −1 alternating; N sp² hybrid
Exercise 4VSEPR geometry in organic moleculesVery Hard
Problem to solve
Predict the geometry around each carbon, nitrogen, and oxygen atom in the alanine molecule (CH₃CH(NH₂)COOH), an amino acid. Indicate hybridization and angles.
Given data
Alanine: CH₃—CH(NH₂)—COOH
Step-by-step solution
1C₁ (CH₃, methyl): 4 bonds → sp³, tetrahedral (109.5°). C₂ (CH, alpha carbon): 4 bonds (C—CH₃, C—NH₂, C—COOH, C—H) → sp³, tetrahedral (109.5°).
2C₃ (COOH, carboxyl group): the carbon has a C=O double bond (sp² with carbonyl oxygen) and two single bonds C—OH and C—C → sp², trigonal planar (120°).
3N (NH₂, amino group): 3 bonds (2 N—H + 1 N—C) + 1 lone pair → sp³, trigonal pyramidal geometry (107°).
4O (OH, hydroxyl group): 2 bonds (O—H + O—C) + 2 lone pairs → sp³, bent geometry (104.5°). O (C=O): double bond, sp² hybrid, linear with respect to carbon (120° between C=O and C—OH).
✓ Final answer: Aliphatic C: sp³ tetrahedral; Carboxyl C: sp² trigonal planar; N: sp³ pyramidal; O(OH): sp³ bent
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