Difficulty: Intermediate | Prerequisites: CHM 255 Chapters 8 and 10, familiarity with bond dissociation energies, radical stability, and acid-base chemistry of organic compounds.
This set of reactions covers what happens when alkanes, alkenes, alcohols, and alkynes are treated with specific reagents for Exam 3. These are largely introductory transformations: radical halogenation for alkanes and alkenes, metal reduction of alcohols to alkoxides, and deprotonation of terminal alkynes. Each reaction type shows up again later in the course, so getting the fundamentals right here pays off.
Alkanes undergo radical halogenation (Br₂ or Cl₂ with heat or light) to give alkyl halides. Alkenes undergo allylic bromination via radicals. Alcohols react with Na metal to form alkoxides. Terminal alkynes are deprotonated by NaNH₂ to form acetylide anions, which are strong nucleophiles.
Radical (free radical)
A species with an unpaired electron. Radicals are highly reactive intermediates formed by homolytic cleavage of a bond (each atom gets one electron). Think of it as: a molecule that has been split down the middle and each half is desperate to find another electron.
Radical stability
3° > 2° > 1° > methyl. More substituted radicals are more stable due to hyperconjugation. The same trend as carbocation stability.
Allylic position
The carbon adjacent to a double bond. An allylic radical is resonance-stabilised because the unpaired electron can delocalise into the pi system.
Alkoxide
The conjugate base of an alcohol (RO⁻). Formed by treating an alcohol with a strong base or an alkali metal (e.g. Na). In simple terms, it is a deprotonated alcohol.
Acetylide anion
The conjugate base of a terminal alkyne (RC≡C⁻). Formed by treating a terminal alkyne with a very strong base such as NaNH₂. Acetylide anions are excellent nucleophiles and are used to form new C–C bonds in SN2 reactions.
Alkanes are largely unreactive, but they do undergo radical halogenation when treated with Br₂ or Cl₂ in the presence of heat (Δ) or light (hν).
The mechanism has three phases:
Initiation: Heat or light breaks the Br–Br (or Cl–Cl) bond homolytically, generating two halogen radicals (Br· + Br·)
Propagation: A bromine radical abstracts a hydrogen from the alkane, forming HBr and a carbon radical. The carbon radical then reacts with another Br₂ molecule, forming the alkyl bromide product and regenerating a Br· radical.
Termination: Any two radicals combine (Br· + Br·, R· + Br·, R· + R·), ending the chain.
Selectivity: Bromine is more selective than chlorine. Bromination preferentially occurs at the most substituted C–H bond (3° > 2° > 1°), because more substituted radicals are more stable. Chlorination is less selective and gives mixtures.
When an alkene is treated with Br₂ and heat, bromination can occur at the allylic position (the carbon next to the double bond) rather than across the double bond.
The allylic radical intermediate is resonance-stabilised. The unpaired electron can sit on either end of the allylic system, which is why this position is favoured.
The mechanism follows the same initiation/propagation/termination pattern as alkane radical halogenation:
Initiation: Br₂ → 2 Br· (heat)
Propagation: Br· abstracts the allylic H, forming a resonance-stabilised allylic radical. The allylic radical then reacts with Br₂ to form the allylic bromide product and regenerate Br·.
The product is an alkenyl bromide (the double bond is preserved, and Br is on the allylic carbon).
Alcohols (R–OH) react with sodium metal (Na) to form alkoxides (RO⁻ Na⁺) and hydrogen gas (H₂).
This is a simple acid-base reaction. The O–H bond is broken, and sodium donates an electron to reduce the proton to H₂. The result is the sodium salt of the alcohol.
For example: CH₃CH₂OH + Na → CH₃CH₂O⁻Na⁺ + ½ H₂
Alkoxides are strong bases and good nucleophiles. They appear frequently in later reactions, particularly in the Williamson ether synthesis (SN2 with an alkyl halide to form an ether).
Terminal alkynes (R–C≡C–H) have a relatively acidic proton (pKa ≈ 25) because the sp-hybridised carbon holds the electrons more tightly, stabilising the conjugate base.
Treating a terminal alkyne with sodium amide (NaNH₂) removes the terminal H to form an acetylide anion (R–C≡C⁻ Na⁺). NH₂⁻ is a very strong base (the conjugate acid, NH₃, has a pKa of about 38), so the equilibrium lies strongly to the right.
The acetylide anion is both a strong base and a strong nucleophile. It is commonly used in SN2 reactions with primary alkyl halides to form new C–C bonds, extending a carbon chain.
Students often confuse radical halogenation of alkenes (allylic position) with electrophilic addition of halogens across the double bond. Radical conditions (heat, light, no solvent) give allylic substitution. Electrophilic addition (Br₂ in CH₂Cl₂, no heat) adds across the double bond. The conditions determine the outcome.
Students sometimes think any base can deprotonate a terminal alkyne. You need a base whose conjugate acid has a higher pKa than the alkyne (~25). NaNH₂ works (pKa of NH₃ ≈ 38). NaOH does not (pKa of H₂O ≈ 15.7).
Students occasionally forget that radical stability follows the same order as carbocation stability (3° > 2° > 1° > methyl). This is why bromination of alkanes is selective for the most substituted position.
Students sometimes assume alkoxides are weak. They are strong bases and strong nucleophiles, fully capable of driving E2 elimination with secondary or tertiary substrates.
⚠️ Radical halogenation mechanisms (initiation, propagation, termination) are frequently tested. You may need to draw each step and identify the intermediates.
⚠️ Radical stability order (3° > 2° > 1° > methyl) determines selectivity in bromination. Expect a question asking you to predict which hydrogen is abstracted.
⚠️ Allylic bromination and resonance stabilisation of the allylic radical are commonly tested. Be ready to draw the resonance structures of the allylic radical.
⚠️ Acetylide formation with NaNH₂ is the setup for C–C bond-forming SN2 reactions. Exam questions may combine both steps: deprotonation then alkylation.
True or false: Chlorination of alkanes is more selective than bromination. (False. Bromination is more selective.)
Fill in the blank: Radical stability follows the order ______ > ______ > ______ > methyl. (3° > 2° > 1°)
True or false: NaOH is a strong enough base to deprotonate a terminal alkyne. (False. You need NaNH₂ or a similarly strong base.)
Fill in the blank: An allylic radical is stabilised by ______. (Resonance.)
True or false: Treating an alcohol with Na metal produces an alkoxide and water. (False. It produces an alkoxide and H₂ gas.)
Q: Propane is treated with Br₂ and light. What is the major product, and why?
A: 2-Bromopropane (isopropyl bromide). Bromine preferentially abstracts the secondary hydrogen because the resulting 2° radical is more stable than the 1° radical. Bromination is highly selective for the more substituted position.
Q: Write the product of the reaction: 1-butyne + NaNH₂, followed by CH₃Br.
A: Step 1 gives the sodium acetylide of 1-butyne (CH₃CH₂C≡C⁻ Na⁺). Step 2 is an SN2 reaction: the acetylide nucleophile attacks CH₃Br to form 2-pentyne (CH₃CH₂C≡CCH₃). A new C–C bond is formed.
Q: What is the difference between radical halogenation at the allylic position of an alkene and electrophilic addition of Br₂ across the double bond?
A: Radical conditions (Br₂ + heat or light, no solvent) abstract a hydrogen from the allylic position, preserving the double bond and placing Br at the allylic carbon. Electrophilic addition (Br₂ in an inert solvent like CH₂Cl₂) adds two Br atoms across the double bond, removing it. The conditions determine which reaction occurs.
Q: Ethanol is treated with sodium metal. Write the balanced equation.
A: 2 CH₃CH₂OH + 2 Na → 2 CH₃CH₂O⁻Na⁺ + H₂. The alcohol is deprotonated, forming sodium ethoxide and hydrogen gas.
Radical halogenation ties back to thermodynamics and bond dissociation energies from earlier chapters. The selectivity of bromination vs chlorination is explained by the Hammond postulate. Alkoxide formation connects directly to Williamson ether synthesis (SN2) and E2 elimination. Acetylide chemistry provides a bridge to carbon chain extension, which becomes central in synthesis problems later in the course.
Radical halogenation, free radical substitution, allylic bromination, allylic radical, resonance stabilisation, alkoxide, sodium alkoxide, Williamson ether synthesis, terminal alkyne, acetylide anion, NaNH₂, sodium amide, radical stability, initiation propagation termination, homolytic cleavage, bond dissociation energy, CHM 255, organic chemistry exam 3, Purdue organic chemistry, Chapter 8, Chapter 10