16.1 Hydrocarbons: Alkanes, Alkenes, Alkynes & Aromatic Systems

16.1 Hydrocarbons: Alkanes, Alkenes, Alkynes & Aromatic Systems

Quick Summary: Hydrocarbons contain solely carbon and hydrogen, divided into saturated (alkanes, cycloalkanes) and unsaturated classes (alkenes, alkynes, aromatics). Molecular geometry follows carbon hybridization: alkanes are sp3 tetrahedral (109.5°), alkenes are sp2 trigonal planar (120°), and alkynes are sp linear (180°). Cyclohexane eliminates ring strain via its chair conformation. Alkenes undergo electrophilic additions governed by Markovnikov's rule. Terminal alkynes exhibit weak acidity due to high s-character. Benzene possesses six delocalized π-electrons, conferring ~150 kJ/mol resonance stabilization that favors electrophilic substitution over addition.

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1. Classification of Hydrocarbons

Hydrocarbons divide into two primary categories:

• Saturated Hydrocarbons: Contain only carbon-carbon single bonds (σ-bonds), maximizing hydrogen content: acyclic alkanes and cyclic cycloalkanes.
• Unsaturated Hydrocarbons: Contain carbon-carbon multiple bonds (π-bonds) or aromatic rings: alkenes (C=C), alkynes (C≡C), and aromatic hydrocarbons (arenes).

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2. Alkanes and Cycloalkanes

Acyclic alkanes (CnH2n+2) possess sp3 carbons with tetrahedral geometry (109.5°).

• Properties & Conformations: Nonpolar molecules held by London dispersion forces; boiling points increase with chain length but decrease with branching. Cylindrical σ-symmetry allows rapid C-C rotation, favoring staggered over eclipsed conformations to minimize torsional strain.
• Reactions: Chemically inert toward acids and bases, alkanes undergo combustion and radical halogenation (Cl2, Br2 with light) via initiation, propagation, and termination.
• Cycloalkanes & Ring Strain: Ring closure introduces angle strain (deviation from 109.5°) and torsional strain (eclipsing). Cyclopropane (60°) and cyclobutane (88°) exhibit severe strain (>110 kJ/mol). Conversely, cyclohexane (C6H12) puckers into a chair conformation with ideal 109.5° angles and staggered bonds, eliminating strain (0 kJ/mol). Substituents prefer equatorial positions to minimize 1,3-diaxial strain.

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3. Alkenes & Electrophilic Additions

Alkenes (CnH2n) possess a C=C double bond (one σ, one π) with sp2 trigonal planar geometry (120°).

• Restricted Rotation: Lateral 2p overlap restricts rotation (260 kJ/mol barrier), producing cis-trans isomerism when each alkene carbon bears two distinct groups.
• Electrophilic Addition: The nucleophilic π-cloud attacks electrophiles:
  • Hydrogenation: Catalytic syn-addition of H2 over Pd/C yields alkanes.
  • Halogenation: Addition of Br2 or Cl2 yields vicinal dihalides via a cyclic halonium ion (anti-addition); bromine decolorization confirms unsaturation.
  • Markovnikov's Rule: In additions of HX or H2O/H+ to unsymmetrical alkenes, H+ adds to the carbon with more hydrogens, forming the more stable carbocation (3° > 2° > 1°) prior to nucleophilic capture.

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4. Alkynes: Geometry & Terminal Acidity

Alkynes (CnH2n-2) contain a C≡C triple bond (one σ, two orthogonal π-bonds) with sp linear geometry (180°).

• The short (1.20 Å) triple bond adds up to two equivalents of H2, X2, or HX.
• Terminal Alkyne Acidity: Terminal alkynes (R-C≡C-H) exhibit weak acidity (pKa ≈ 25) relative to alkenes (pKa ≈ 44) and alkanes (pKa ≈ 50). Because an sp orbital has 50% s-character, electrons reside closer to the nucleus, stabilizing the acetylide conjugate base (R-C≡C-) and permitting deprotonation by strong bases like NaNH2.

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5. Aromatic Hydrocarbons: Benzene & Hückel's Rule

Benzene (C6H6) is a planar regular hexagon with uniform C-C bond lengths (1.39 Å). Each sp2 carbon contributes a 2p orbital to a delocalized cyclic π-system.

• Hückel's Rule: A cyclic, planar, conjugated system is aromatic if it contains 4n + 2 π-electrons. Benzene contains six π-electrons (n = 1), yielding ~150 kJ/mol resonance energy.
• Electrophilic Aromatic Substitution (EAS): Benzene resists addition to preserve aromaticity, undergoing substitution instead:
  • Halogenation: C6H6 + Br2 —(FeBr3)→ C6H5Br + HBr
  • Nitration: C6H6 + HNO3 —(H2SO4)→ C6H5NO2 + H2O (via NO2+)

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6. Comparative Reference Tables

Hydrocarbon Classes Comparison

Hydrocarbon Reaction Profiles

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7. Worked Example: Markovnikov Addition

Problem: Predict the major product when 2-methyl-2-butene reacts with anhydrous HCl. Justify using carbocation stability.

Step 1: Analyze alkene structure In (CH3)2C=CHCH3, C2 bears two methyl groups (zero hydrogens); C3 bears one methyl and one hydrogen.

Step 2: Compare intermediates Protonation yields two possible carbocations:

• Pathway A (Proton adds to C3): Yields a tertiary (3°) carbocation, (CH3)2C+-CH2CH3.
• Pathway B (Proton adds to C2): Yields a secondary (2°) carbocation, (CH3)2CH-C+HCH3.

Step 3: Evaluate stability The 3° carbocation is far more stable due to hyperconjugation and inductive donation from three alkyl groups, lowering activation energy.

Step 4: Nucleophilic attack Chloride attacks the carbocation at C2: (CH3)2C+-CH2CH3 + Cl- → (CH3)2C(Cl)-CH2CH3 Conclusion: The major product is 2-chloro-2-methylbutane, strictly obeying Markovnikov's rule.