Hydrocarbons
The carbon-and-hydrogen backbone of organic chemistry — unreactive alkanes, electron-rich alkenes and alkynes, and the special stability of the benzene ring
- How hydrocarbons divide into alkanes, alkenes, alkynes and aromatics.
- Key preparations — Wurtz, decarboxylation, dehydrohalogenation, dehydration.
- The free-radical halogenation mechanism and electrophilic addition to alkenes.
- Markovnikov's rule and the anti-Markovnikov peroxide effect.
- The acidity of terminal alkynes and the chemistry of ozonolysis.
- Aromaticity (Hückel's rule), electrophilic aromatic substitution, and directing effects.
Classifying Hydrocarbons
Hydrocarbons contain only carbon and hydrogen. They split into saturated (alkanes, all single bonds), unsaturated (alkenes with \(\ce{C=C}\), alkynes with \(\ce{C#C}\)) and aromatic (benzene and its relatives). The degree of unsaturation governs reactivity: alkanes resist most reagents, while the electron-rich multiple bonds invite attack.
| Class | Bonding | General formula | Typical reaction |
|---|---|---|---|
| Alkane | C–C single | \(\ce{C_nH_{2n+2}}\) | free-radical substitution |
| Alkene | one \(\ce{C=C}\) | \(\ce{C_nH_{2n}}\) | electrophilic addition |
| Alkyne | one \(\ce{C#C}\) | \(\ce{C_nH_{2n-2}}\) | addition; terminal acidity |
| Aromatic | delocalised ring | (e.g. \(\ce{C6H6}\)) | electrophilic substitution |
Alkanes: Preparation
Alkanes — the "paraffins" (Latin parum affinis, little affinity) — are made by lengthening or shortening carbon chains, or by reducing unsaturation.
| Method | Reaction |
|---|---|
| Wurtz reaction | \(\ce{2R-X + 2Na -> R-R + 2NaX}\) |
| Hydrogenation | \(\ce{CH2=CH2 + H2 ->[Ni] CH3CH3}\) |
| Decarboxylation | \(\ce{CH3COONa + NaOH ->[\Delta] CH4 + Na2CO3}\) |
| Reduction of \(\ce{R-X}\) | \(\ce{R-X + 2[H] ->[Zn/HCl] R-H + HX}\) |
Alkanes: Free-Radical Halogenation
The signature alkane reaction is substitution by halogens in sunlight, which proceeds through a free-radical chain in three stages — initiation, propagation and termination.
Alkenes: Preparation
Alkenes are usually made by elimination — removing two groups from adjacent carbons to create the double bond. When more than one alkene is possible, Saytzeff's rule predicts the more substituted (more stable) alkene as the major product.
| Method | Reaction |
|---|---|
| Dehydrohalogenation | \(\ce{R-CH2-CH2-X ->[alc. KOH] R-CH=CH2 + HX}\) |
| Dehydration of alcohol | \(\ce{R-CH2-CH2-OH ->[conc. H2SO4][\Delta] R-CH=CH2 + H2O}\) |
| Dehalogenation | \(\ce{R-CHBr-CH2Br + Zn -> R-CH=CH2 + ZnBr2}\) |
Alkenes: Electrophilic Addition
The electron-rich \(\ce{C=C}\) bond is attacked by electrophiles, which add across it. With an unsymmetrical reagent like \(\ce{HX}\), Markovnikov's rule decides the orientation: the hydrogen adds to the carbon that already has more hydrogens, because that route passes through the more stable carbocation.
In the presence of peroxides, \(\ce{HBr}\) adds against Markovnikov (the Kharasch effect) because the mechanism switches to free radicals, forming the more stable radical instead of the more stable cation. This reversal works only for \(\ce{HBr}\), not \(\ce{HCl}\) or \(\ce{HI}\).
| Reagent | Product |
|---|---|
| \(\ce{H2}\) / Ni | alkane |
| \(\ce{X2}\) | vicinal dihalide |
| \(\ce{HX}\) | haloalkane (Markovnikov) |
| \(\ce{H2O}\) / \(\ce{H+}\) | alcohol (Markovnikov) |
Oxidation & Ozonolysis
Two oxidations of alkenes are classic exam tools. Cold dilute alkaline \(\ce{KMnO4}\) — Baeyer's reagent — gives a diol and decolourises, a quick test for unsaturation. Ozonolysis cleaves the double bond to two carbonyl fragments, revealing where the double bond was.
The \(\ce{C=C}\) is split, each carbon becoming a carbonyl. Working backwards from the aldehyde/ketone fragments tells you the structure of the original alkene — a favourite structure-determination trick.
Alkynes
Alkynes carry a triple bond and undergo addition like alkenes — but the terminal alkyne has a unique trick: its \(\ce{#C-H}\) hydrogen is weakly acidic. The \(sp\)-hybridised carbon has the most \(s\)-character, holding the bonding electrons tightly and stabilising the resulting carbanion.
Terminal alkynes react with sodium to form acetylides: \(\ce{HC#CH + Na -> HC#C-Na+ + \tfrac12 H2}\). With ammoniacal \(\ce{AgNO3}\) or \(\ce{Cu2Cl2}\) they give coloured precipitates — a test that distinguishes terminal from internal alkynes.
| Reaction | Reagent | Product |
|---|---|---|
| Hydration (Kucherov) | \(\ce{H2O}/\ce{H2SO4}/\ce{HgSO4}\) | acetaldehyde (from \(\ce{C2H2}\)) |
| Hydrohalogenation | \(\ce{HX}\) | vinyl halide → gem-dihalide |
| Preparation | \(\ce{CaC2 + 2H2O}\) | \(\ce{C2H2}\) (acetylene) |
Benzene & Aromaticity
Benzene (\(\ce{C6H6}\)) is a planar hexagonal ring with all six \(\ce{C-C}\) bonds of equal length, its six \(\pi\) electrons delocalised in a ring above and below the plane. This delocalisation gives an extra resonance stabilisation that makes aromatic compounds prefer substitution over addition.
Benzene has \(6\) \(\pi\) electrons (\(n=1\)), satisfying \(4n+2\). The same test flags aromatic ions and heterocycles; rings with \(4n\) \(\pi\) electrons are anti-aromatic and destabilised.
Electrophilic Aromatic Substitution
Because the ring is electron-rich, benzene reacts with electrophiles — but rather than add (which would destroy aromaticity), it substitutes, replacing a hydrogen and restoring the ring. Every such reaction follows the same three-step mechanism.
| Reaction | Reagent | Electrophile | Product |
|---|---|---|---|
| Nitration | conc. \(\ce{HNO3}/\ce{H2SO4}\) | \(\ce{NO2+}\) | nitrobenzene |
| Halogenation | \(\ce{Cl2}/\ce{FeCl3}\) | \(\ce{Cl+}\) | chlorobenzene |
| Sulphonation | oleum | \(\ce{SO3}\) | benzenesulphonic acid |
| F–C alkylation | \(\ce{RX}/\ce{AlCl3}\) | \(\ce{R+}\) | alkylbenzene |
| F–C acylation | \(\ce{RCOCl}/\ce{AlCl3}\) | \(\ce{RCO+}\) | aryl ketone |
Directing Effects
When benzene already carries a substituent, that group steers the incoming electrophile to particular positions and changes the ring's reactivity. Electron-donating groups activate the ring and direct ortho/para; electron-withdrawing groups deactivate and direct meta.
| Group | Effect on ring | Directs to |
|---|---|---|
| \(\ce{-OH},\ \ce{-NH2},\ \ce{-OR},\ \ce{-CH3}\) | activating | ortho / para |
| \(\ce{-NO2},\ \ce{-CN},\ \ce{-COOH},\ \ce{-CHO}\) | deactivating | meta |
| \(\ce{-Cl},\ \ce{-Br}\) (halogens) | deactivating | ortho / para |
Putting It to Work
Problem. What alkane forms when bromoethane undergoes the Wurtz reaction?
Solution. Two ethyl units couple:
Problem. Write a propagation step in the chlorination of methane.
Solution. A chlorine radical abstracts a hydrogen:
Problem. Give the major product of \(\ce{CH3-CH=CH2 + HBr}\) (no peroxide).
Solution. H adds to the terminal C; Br to the more substituted C (stable 2° cation):
Problem. What changes if the same reaction is run with \(\ce{HBr}\) and a peroxide?
Solution. Radical mechanism reverses the orientation (anti-Markovnikov):
Problem. Arrange \(\ce{CH3CH3},\ \ce{CH2=CH2},\ \ce{HC#CH}\) by increasing acidity and explain.
Solution. More \(s\)-character holds the lone pair more tightly:
Problem. Where does nitration of nitrobenzene occur, and is the ring more or less reactive than benzene?
Solution. \(\ce{-NO2}\) is a deactivating meta-director:
Chapter Summary
Alkanes (substitution), alkenes & alkynes (addition), aromatics (substitution).
Wurtz, decarboxylation, hydrogenation; free-radical halogenation chain.
Elimination preparations; electrophilic addition; Markovnikov & peroxide effect.
Terminal-H acidity (\(sp\)); acetylides; hydration to carbonyls.
Planar, cyclic, conjugated, \((4n+2)\) π electrons; benzene has 6.
Generate E⁺ → arenium ion → lose H⁺; o/p activators vs m deactivators.
Problems
For each item, first decide which class and reaction type it tests, then apply the relevant mechanism or rule. Difficulty rises down the list.
- Classify hydrocarbons and give the general formula of each class.
- Write the Wurtz reaction and explain why it suits only symmetrical alkanes.
- Give the three stages of the free-radical chlorination of methane with one equation each.
- State Saytzeff's rule and apply it to the dehydrohalogenation of 2-bromobutane.
- State Markovnikov's rule and predict the product of \(\ce{HBr}\) addition to propene.
- Explain the peroxide effect and why it operates only with \(\ce{HBr}\).
- What does ozonolysis reveal? Give the products from but-2-ene.
- Explain why terminal alkynes are acidic and arrange the acidity of ethane, ethene and ethyne.
- How would you distinguish a terminal alkyne from an internal one chemically?
- State Hückel's rule and verify that benzene is aromatic.
- Outline the three steps of electrophilic aromatic substitution.
- Predict the position and relative rate of nitration for (a) phenol and (b) nitrobenzene.