Alkanes and Cycloalkanes
IUPAC nomenclature of alkanes and cycloalkanes, physical properties (boiling point as a function of chain length), conformations (eclipsed/staggered, chair/boat of cyclohexane), radical halogenation.
Complete Theory
4IUPAC nomenclature (International Union of Pure and Applied Chemistry) provides systematic rules for assigning a unique name to every organic compound. For linear alkanes, numerical prefixes are used: meth- (1C), eth- (2C), prop- (3C), but- (4C), pent- (5C), hex- (6C), hept- (7C), oct- (8C), non- (9C), dec- (10C) + suffix -ane.
For branched alkanes the rules are:
Real-world applications: IUPAC nomenclature is the international standard for scientific communication, chemical databases (CAS, PubChem), and regulatory documentation (REACH). An IUPAC name uniquely identifies a substance, avoiding ambiguity of common names.
For branched alkanes the rules are:
- Identify the longest main chain (maximum number of C atoms). If two chains have the same length, choose the one with more branches.
- Number the main chain starting from the end closest to the first substituent (first point of difference rule).
- Identify and name the substituents (alkyl groups: —CH₃ methyl, —CH₂CH₃ ethyl, —CH(CH₃)₂ isopropyl, —C(CH₃)₃ tert-butyl, etc.).
- Write the name: position(multiplier prefix)-substituent + main chain name. Substituents in alphabetical order (ignoring multiplier prefixes di-, tri-).
- Separate numbers and letters with hyphens, numbers from each other with commas.
Real-world applications: IUPAC nomenclature is the international standard for scientific communication, chemical databases (CAS, PubChem), and regulatory documentation (REACH). An IUPAC name uniquely identifies a substance, avoiding ambiguity of common names.
The physical properties of alkanes depend mainly on London forces (dispersion), which are weak intermolecular forces due to instantaneous dipoles. The main characteristics:
- Boiling point: increases with increasing molecular mass (more electrons → larger dipole fluctuations). For isomers of equal mass, the more linear chain has a higher boiling point (greater surface area contact). n-pentane (C₅) boils at 36°C, 2,2-dimethylpropane (neopentane) at 9.5°C. Rule: for each added CH₂, the boiling point increases by about 20–30°C up to C₁₅, then gradually less.
- Solubility: alkanes are nonpolar and soluble only in nonpolar solvents (CCl₄, benzene, ether) — "like dissolves like". They are insoluble in water (polar solvent).
- Density: alkanes have density , lower than water (1.0 g/mL), so they float (think of oil slicks on water).
- Physical state: C₁-C₄ gases, C₅-C₁₇ liquids, C₁₈⁺ solids (waxes). Methane (C₁) boils at −161°C, propane (C₃) at −42°C (used as liquefied fuel), octane (C₈) at 126°C.
Conformations are different spatial arrangements of atoms obtained by rotation around single bonds (C—C). They are not isomers, but conformers that interconvert rapidly at room temperature.
Ethane (CH₃—CH₃): rotation around the C—C bond produces:
Ethane (CH₃—CH₃): rotation around the C—C bond produces:
- Eclipsed conformation: the H atoms of the two carbons are aligned (viewed along the C—C axis). Higher energy due to steric repulsion (torsional strain). Energy barrier: .
- Staggered conformation: the H atoms are alternated (each H of one C falls between two H of the other C). Minimum energy (most stable). It is the preferred conformation at room temperature.
- Anti: the two CH₃ groups are at 180° → minimum energy
- Gauche: the two CH₃ groups are at 60° → energy higher than anti
- Eclipsed: CH₃ groups at 0° (fully eclipsed) or 120° (partially eclipsed) → maximum energy
- 6 axial (a) bonds: vertical, alternating up/down
- 6 equatorial (e) bonds: tilted outward from the ring
- Ring flip exchanges axial and equatorial positions. Bulky substituents prefer the equatorial position (less steric hindrance).
Radical halogenation is a substitution reaction in which a hydrogen atom of an alkane is replaced by a halogen (typically Cl₂ or Br₂) through a free radical mechanism. It requires activation by UV light or heat (≥ 250°C for chlorination).
Three-step mechanism:
Bromination is much more selective than chlorination because the reaction is less exothermic, so the energy barrier more closely reflects radical stability (Hammond postulate). Bromination occurs almost exclusively at the tertiary position, while chlorination gives mixtures of products.
Real-world applications: radical chlorination of methane produces methyl chloride, dichloromethane, chloroform, and carbon tetrachloride (depending on the Cl₂:CH₄ ratio), all important industrial solvents. Selective bromination is useful in organic synthesis for introducing bromine at a specific position.
Three-step mechanism:
- Initiation: the halogen molecule splits into two radicals by homolytic bond cleavage: $mathrm{Cl_2 \xrightarrow{h u} 2,Clcdot}Delta H = 242\,\mathrm{kJ/mol}$; Br₂ requires less, 192 kJ/mol).
- Propagation (2 stages): (a) the Cl• radical abstracts an H from the alkane, forming HCl and an alkyl radical R•: . (b) The R• radical reacts with Cl₂ to form RCl and regenerate Cl•: . Each propagation cycle produces one molecule of alkyl halide and regenerates the radical.
- Termination: two radicals combine (R• + Cl• → RCl, R• + R'• → R—R', Cl• + Cl• → Cl₂), consuming the radicals and stopping the reaction.
Bromination is much more selective than chlorination because the reaction is less exothermic, so the energy barrier more closely reflects radical stability (Hammond postulate). Bromination occurs almost exclusively at the tertiary position, while chlorination gives mixtures of products.
Real-world applications: radical chlorination of methane produces methyl chloride, dichloromethane, chloroform, and carbon tetrachloride (depending on the Cl₂:CH₄ ratio), all important industrial solvents. Selective bromination is useful in organic synthesis for introducing bromine at a specific position.
Worked Examples
2Example 1Nomenclature of 3-methylhexane
Given
Structural formula: CH₃CH₂CH(CH₃)CH₂CH₂CH₃
Find
IUPAC name of the compound
Step-by-step solution
1Identify the longest continuous chain of carbon atoms. Starting from the left: C₁—C₂—C₃(C₇)—C₄—C₅—C₆. The chain is 6 carbons (not 7: the branched CH₃ carbon is attached to C₃ but is not on the main chain). So the base name is hexane.
2Number the main chain from left to right. The methyl substituent (—CH₃) is bonded to C₃. If we numbered from the right, the methyl would be at C₄ (first point of difference: 3 < 4). So the correct numbering is from the left: 3-methylhexane.
3Full name: 3-methylhexane. No other substituents. Verify the molecular formula: C₇H₁₆ (branched heptane).
✓ Final result: 3-methylhexane (main chain: hexane, methyl at position 3)
Example 2Boiling Point of C₅H₁₂ Isomers
Given
n-pentane (linear): Bp = 36°C
2-methylbutane (isopentane): Bp = 28°C
2,2-dimethylpropane (neopentane): Bp = 9.5°C
Find
Explain the difference in boiling point
Step-by-step solution
1The dominant intermolecular forces in alkanes are London forces (dispersion), which depend on the contact surface area between molecules. The larger the surface area, the greater the number of temporary instantaneous dipole-dipole interactions.
2n-pentane has a linear chain that allows close contact along the entire length between adjacent molecules (maximum surface area). Isopentane (branched) has a smaller surface area. Neopentane (nearly spherical) has the smallest contact surface area.
3Consequently, the boiling point decreases as branching increases, even with the same molecular mass (72 g/mol for all three). This principle is important in oil refining: branched isomers have better anti-knock properties (higher octane number).
✓ Final result: Bp: n-pentane (36°C) > isopentane (28°C) > neopentane (9.5°C) — decreases with branching
Exercises with Solutions
4Exercise 1IUPAC nomenclatureMedium
Problem to solve
Assign the IUPAC name to the compound CH₃CH₂CH(CH₃)CH(CH₃)CH₂CH₃.
Given data
CH₃CH₂CH(CH₃)CH(CH₃)CH₂CH₃
Step-by-step solution
1Longest chain: 6 carbons (hexane). The chain has two methyl substituents.
2Numbering from the left: CH₃(1)—CH₂(2)—CH(3, with CH₃)—CH(4, with CH₃)—CH₂(5)—CH₃(6). Substituents at positions 3 and 4.
3Numbering from the right gives the same positions (3 and 4). Alphabetical order: "ethyl" comes before "methyl", but we only have methyls.
4Name: 3,4-dimethylhexane.
✓ Final answer: 3,4-dimethylhexane
Exercise 2Physical propertiesMedium
Problem to solve
Arrange in order of increasing boiling point: pentane, 2-methylbutane, hexane, 2,2-dimethylpropane.
Given data
C₅H₁₂ (pentane)C₅H₁₂ (2-methylbutane)C₆H₁₄ (hexane)C₅H₁₂ (2,2-dimethylpropane)
Step-by-step solution
1Separate by molecular mass: hexane (C₆, M=86) has higher mass and therefore higher boiling point (69°C) than all C₅ compounds (M=72).
2Among the C₅ isomers, order by increasing branching (smaller surface → weaker London forces → lower Bp).
3Order: 2,2-dimethylpropane (9.5°C) < 2-methylbutane (28°C) < pentane (36°C) < hexane (69°C).
✓ Final answer: 2,2-dimethylpropane (9.5°C) < 2-methylbutane (28°C) < pentane (36°C) < hexane (69°C)
Exercise 3Cyclohexane conformationsHard
Problem to solve
For methylcyclohexane, determine the most stable conformation between chair and boat. Compare the stability of the axial and equatorial positions of the methyl group.
Given data
Methylcyclohexane: cyclohexane with CH₃ on C₁
Step-by-step solution
1The chair conformation has no ring strain (angles about 109.5°) and is the most stable. The boat conformation has steric strain between flag-pole H atoms and is about 30 kJ/mol less stable.
2In the chair conformation, the methyl can occupy axial (vertical) or equatorial (tilted) positions. In the axial position, CH₃ interacts with axial H atoms at positions 3,5 (1,3-diaxial interactions). This steric repulsion makes the axial conformation about 7.6 kJ/mol less stable.
3The chair conformation with the methyl equatorial (CH₃ "lying" outward) is therefore the most stable. At room temperature, about 95% of molecules have the methyl in the equatorial position (shifted conformational equilibrium).
✓ Final answer: Chair with equatorial methyl (95% at 25°C) — 1,3-diaxial interactions disfavor the axial position
Exercise 4Radical halogenationVery Hard
Problem to solve
Consider the chlorination and bromination of 2-methylbutane. Predict the main product in each case and explain the difference in selectivity.
Given data
2-methylbutane: CH₃CH(CH₃)CH₂CH₃
Step-by-step solution
12-methylbutane has three types of hydrogens: 9 primary (1°) H on the three CH₃, 2 secondary (2°) H on CH₂, 1 tertiary (3°) H on CH. The corresponding radicals have stability: 3° > 2° > 1°.
2In chlorination (poorly selective), products form in proportions that reflect both radical stability and statistical abundance of hydrogens. Approximate statistical prediction: 1° products ≈ 60–65%, 2° ≈ 20–25%, 3° ≈ 15–20%. A mixture of chlorides is obtained.
3In bromination (highly selective), the main product is almost exclusively the tertiary bromide (>99%). Bromine is less reactive than chlorine, and the reaction is endothermic (slow step: H abstraction). According to the Hammond postulate, the transition state resembles the radical, so the most stable radical (3°) forms preferentially.
✓ Final answer: Chlorination: mixture (1° > 2° > 3°). Bromination: almost only 3° (>99%) — selectivity reflects radical stability (Hammond postulate)
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