Alkane = All single bonds (saturated). Alkene = has a double bond. Alkyne = has a triple bond. Memory: All Linear, Alkane Double = Doesn't apply, It's sIngle only.
⚗️ Hydrocarbons
CnH(2n+2)
Alkane Molecular Formula
Alkanes follow CnH(2n+2). Methane=CH4, Ethane=C2H6, Propane=C3H8. Each carbon adds 2 H's plus the 2 end H's.
⚗️ Hydrocarbons
MEMO: 1-4 gases, 5-17 liquids, 18+ solids
Alkane Physical States
At room temperature: C1–C4 alkanes are gases (methane, ethane, propane, butane). C5–C17 are liquids. C18+ are solids (wax, tar).
⚗️ Hydrocarbons
E/Z = Examine substituents on each carbon of double bond
E/Z Isomerism
E (from German 'entgegen' = opposite): higher priority groups on opposite sides. Z (from German 'zusammen' = together): higher priority groups on same side. Use Cahn-Ingold-Prelog priority rules.
⚗️ Hydrocarbons
'More H = more saturated = less reactive'
Degree of Unsaturation
Degrees of unsaturation (DoU) = (2C + 2 + N - H - X) / 2. Each ring or double bond = 1 DoU. Triple bond = 2 DoU. Benzene ring = 4 DoU.
⚗️ Hydrocarbons
Markovnikov's rule: H adds to less substituted carbon (more H already there)
Markovnikov's Rule
When HX (or H₂O) adds to an alkene, H adds to the carbon that already has MORE hydrogens. The result: more substituted carbocation forms (more stable). Memory: 'The rich get richer' — carbon with more H gets the extra H. Modern explanation: more substituted carbocation (3° > 2° > 1°) is more stable. Anti-Markovnikov: radical addition (ROOR + HBr, H goes to more substituted carbon, Br to less).
Markovnikov
H to C with more H → more substituted carbocation
Anti-Markovnikov
Radical conditions (ROOR/HBr) → Br to more substituted C
Why?
More substituted carbocation is more stable
Exception
Hydroboration: B to less substituted C (syn addition)
⚗️ Hydrocarbons
Carbocation stability: 3° > 2° > 1° > methyl — more substitution = more stable
Carbocation Stability
Tertiary carbocations (3°) are most stable — three alkyl groups donate electrons by hyperconjugation and induction. Stability order: 3° > 2° > 1° > methyl. Allylic and benzylic carbocations are especially stable (resonance delocalization). Vinyl and aryl carbocations are very unstable. Rearrangements occur: hydride or methyl shift from adjacent C can form a more stable carbocation (Wagner-Meerwein).
Most stable
Tertiary (3°) — 3 alkyl groups donate electrons
Resonance stabilized
Allylic (R₂C=CR-CR₂⁺) and benzylic (PhCH₂⁺)
Unstable
Primary, vinyl, aryl
Rearrangements
1,2-hydride or methyl shift → more stable carbocation
Alkanes react with halogens under UV light (or heat) via radical chain mechanism. Three stages: Initiation: X₂ → 2X• (UV breaks X-X bond). Propagation: X• + R-H → HX + R•; R• + X₂ → RX + X•. Termination: any two radicals combine. Selectivity: F₂ (most reactive, least selective, explosive) > Cl₂ (selective) > Br₂ (most selective — most discriminating) > I₂ (too slow, endothermic). Bromine prefers most substituted C (more stable radical).
Initiation
X₂ + hν → 2X• (homolytic cleavage)
Propagation 1
X• + R-H → HX + R•
Propagation 2
R• + X₂ → RX + X•
Selectivity
Br₂ most selective (3° >> 2° >> 1°)
F₂
Too reactive, not selective, dangerous
⚗️ Hydrocarbons
Hydroboration-oxidation: anti-Markovnikov, syn addition of H and OH across alkene
Hydroboration-Oxidation
BH₃/THF adds across alkene syn (both B and H add to same face). B goes to less hindered carbon (anti-Markovnikov). Oxidation with H₂O₂/NaOH replaces B with OH (retention of configuration). Overall: anti-Markovnikov, syn addition of OH. Net result: OH on less substituted carbon. Compare to acid-catalyzed hydration (Markovnikov, OH on more substituted). Key reagents: (1) BH₃·THF, (2) H₂O₂, NaOH.
Dihydroxylation: OsO₄ → syn diol; mCPBA then H₂O → anti diol
Dihydroxylation of Alkenes
Two ways to add two OH groups across a double bond: Syn dihydroxylation: OsO₄ (catalytic with N-oxide co-oxidant) → syn-diol (both OH on same face). Also: KMnO₄ (cold, dilute) → syn-diol. Anti dihydroxylation: (1) mCPBA → epoxide, (2) H₂O/acid or base → anti-diol (ring opens with inversion). Sharpless asymmetric dihydroxylation uses chiral ligands with OsO₄ → enantioselective syn diol.
⚗️ Hydrocarbons
Alkyne acidity: terminal R-C≡C-H, pKa ~25 — deprotonated by NaNH₂ or n-BuLi
Terminal Alkyne Reactions
Terminal alkynes (R-C≡C-H) have unusually acidic C-H bonds (pKa ~25) because the sp-hybridized carbon holds electrons closer to the nucleus (more s-character = more electronegative). Deprotonated by NaNH₂ (sodium amide) or n-BuLi to give acetylide anion (R-C≡C⁻). Acetylides are strong bases AND good nucleophiles — react with alkyl halides (SN2) and carbonyl compounds. Key carbon-carbon bond forming reaction.
Why acidic
sp carbon has 50% s-character — holds electrons close
pKa comparison
Alkyne ~25 < alkene ~44 < alkane ~50
Deprotonation
NaNH₂ (pKa ~38) or n-BuLi (pKa ~50)
Use
Acetylide + R'X → R-C≡C-R' (alkylation)
⚗️ Hydrocarbons
Conformational analysis: cyclohexane chair — equatorial preferred over axial
Conformational Analysis of Cyclohexane
Cyclohexane adopts chair conformation (most stable). Each carbon: one axial bond (up or down, alternating) and one equatorial bond (nearly horizontal). Ring flip interconverts axial and equatorial. Large substituents prefer equatorial (less steric strain — 1,3-diaxial interactions avoided). A-value = energy cost of putting group axial vs equatorial. tert-Butyl group locks ring (too large to go axial). Trans-diequatorial is most stable disubstituted isomer.
Chair
Most stable cyclohexane conformation
Equatorial
Preferred for large groups — less 1,3-diaxial strain
Ring flip
Axial becomes equatorial and vice versa
tert-Butyl
Too large axial — locks chair conformation
Trans-1,2
Can be diequatorial (most stable)
🎓 Common Exam Questions
Q: What is the molecular formula for alkanes, and what are the first four alkanes?
A: Alkanes follow CₙH₂ₙ₊₂. C1: methane (CH₄). C2: ethane (C₂H₆). C3: propane (C₃H₈). C4: butane (C₄H₁₀). Each additional carbon adds CH₂. Alkanes are saturated (only single bonds) and are the least reactive class of organic compounds.
Q: What are the physical states of alkanes at room temperature?
A: C1–C4 alkanes are gases at room temperature (methane, ethane, propane, butane). C5–C17 are liquids. C18+ are solids (wax, tar, asphalt). As chain length increases, van der Waals forces increase, raising the boiling point. Branching lowers the boiling point compared to straight-chain isomers.
Q: How do you assign E/Z configuration to alkenes?
A: Use Cahn-Ingold-Prelog (CIP) priority rules to rank substituents on each carbon of the double bond. Higher priority group = higher atomic number (or higher sum for ties). E (entgegen = opposite): higher priority groups on opposite sides of the double bond. Z (zusammen = together): higher priority groups on the same side. E is like 'trans' but based on priority, not size.
Q: How do you calculate degrees of unsaturation (DoU)?
A: DoU = (2C + 2 + N – H – X) / 2. Where C = carbons, N = nitrogens, H = hydrogens, X = halogens. Each ring = 1 DoU. Each double bond = 1 DoU. Each triple bond = 2 DoU. Benzene ring = 4 DoU (3 double bonds + 1 ring). DoU = 0 means fully saturated. DoU ≥ 4 with C6 suggests aromatic ring.
Q: What is the difference between alkanes, alkenes, and alkynes in terms of bonding and reactivity?
A: Alkanes: all single bonds, sp³ carbons, saturated — least reactive (mainly free radical halogenation). Alkenes: one C=C double bond, sp² carbons — undergo electrophilic addition (HX, H₂O, Br₂, H₂). Alkynes: one C≡C triple bond, sp carbons — also undergo addition reactions and can be deprotonated (terminal alkynes, pKa ~25). Reactivity: alkyne > alkene > alkane.
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