| Type | Bonding | General Formula |
|---|---|---|
| Alkanes | Only single C-C bonds (saturated) | CₙH₂ₙ₊₂ |
| Alkenes | At least one C=C double bond | CₙH₂ₙ |
| Alkynes | At least one C≡C triple bond | CₙH₂ₙ₋₂ |
| Aromatic | Delocalised ring system (e.g., benzene) | Varies (e.g., C₆H₆) |
Alkyl halides react with sodium metal in dry ether to give a higher alkane with double the number of carbons: $2\text{R-X} + 2\text{Na} \xrightarrow{\text{dry ether}} \text{R-R} + 2\text{NaX}$. Best suited for preparing symmetrical alkanes, since using two different alkyl halides gives a mixture of three possible products.
Electrolysis of an aqueous solution of the sodium/potassium salt of a carboxylic acid produces an alkane at the anode (with loss of CO₂ from each carboxylate, followed by combination of the resulting free radicals), along with H₂ at the cathode.
$$2\text{RCOO}^- \xrightarrow{\text{electrolysis}} \text{R-R} + 2\text{CO}_2 + 2e^-$$Predict the products (besides NaBr) when a mixture of methyl bromide and ethyl bromide is treated with sodium metal in dry ether (Wurtz reaction).
Three possible combinations occur: $\text{CH}_3-\text{CH}_3$ (ethane, from 2 methyl radicals), $\text{CH}_3-\text{CH}_2-\text{CH}_2-\text{CH}_3$ (n-butane, from 2 ethyl radicals), and $\text{CH}_3-\text{CH}_2-\text{CH}_3$ (propane, from one of each) — this mixture of products is exactly why Wurtz reaction with two different halides is not practically useful for a single pure product.
Alkanes react with halogens (Cl₂, Br₂) in the presence of sunlight or heat via a free radical chain mechanism, consisting of three stages: initiation, propagation, and termination.
Homolytic fission of Cl₂ under UV light generates chlorine free radicals.
The regenerated Cl• radical continues the chain, reacting with more methane.
Two radicals combine, consuming reactive species and stopping that chain.
Why does chlorination of methane give a mixture of products (CH₃Cl, CH₂Cl₂, CHCl₃, CCl₄), rather than stopping cleanly at one product?
Once some CH₃Cl forms, it is also present in the reaction mixture along with unreacted CH₄, and Cl• radicals can abstract a hydrogen from CH₃Cl just as readily as from CH₄, leading to further substitution (CH₂Cl₂, and so on). Since the chain reaction doesn't distinguish and stop after one substitution, a mixture of all possible chlorinated products is typically obtained.
Alkyl halides, when heated with alcoholic KOH, undergo elimination of HX to form an alkene. When more than one alkene is possible, the major product is usually the more substituted (more stable) alkene, as per Saytzeff's rule.
Alcohols, when heated with a strong acid (like concentrated H₂SO₄ or H₃PO₄), lose a water molecule to form an alkene — again generally following Saytzeff's rule for the major product.
Arises in alkenes where each carbon of the C=C double bond carries two different groups (restricted rotation around the double bond prevents interconversion). Cis isomer has similar/priority groups on the same side; trans has them on opposite sides.
Does but-2-ene (CH₃–CH=CH–CH₃) show geometrical isomerism? Explain.
Yes. Each double-bond carbon carries a methyl group and a hydrogen atom (two different groups on each carbon) — this satisfies the requirement for cis-trans isomerism. Cis-but-2-ene has both methyl groups on the same side; trans-but-2-ene has them on opposite sides.
When an unsymmetrical reagent like HX adds to an unsymmetrical alkene, the negative part (X) attaches to the carbon bearing fewer hydrogen atoms (the more substituted carbon), while H attaches to the carbon with more hydrogens. This happens because the reaction proceeds via the more stable (more substituted) carbocation intermediate.
Predict the major product when HBr adds to propene, CH₃–CH=CH₂, according to Markovnikov's rule.
H⁺ adds first to the terminal CH₂ carbon (which has more H atoms), generating the more stable secondary carbocation $\text{CH}_3-\overset{+}{\text{C}}\text{H}-\text{CH}_3$ rather than a less stable primary carbocation. Br⁻ then attacks this carbocation, giving 2-bromopropane ($\text{CH}_3\text{CHBrCH}_3$) as the major product.
In the presence of peroxides, HBr (specifically HBr, not HCl or HI) adds to unsymmetrical alkenes in the opposite orientation to Markovnikov's rule, via a free-radical mechanism rather than an ionic one — Br attaches to the carbon with more hydrogens.
Predict the major product when HBr adds to propene in the presence of peroxides.
The mechanism shifts to free-radical addition: a Br• radical adds first to the terminal carbon (generating the more stable secondary free radical, $\text{CH}_3-\overset{\bullet}{\text{C}}\text{H}-\text{CH}_2\text{Br}$), and this radical then abstracts H from another HBr molecule. The overall product is 1-bromopropane ($\text{CH}_3\text{CH}_2\text{CH}_2\text{Br}$) — the anti-Markovnikov product, opposite to normal HBr addition.
An alkene reacts with ozone to form an ozonide, which is then cleaved (in the presence of zinc dust and water, a reductive workup) to give two carbonyl compounds (aldehydes and/or ketones), with the C=C double bond completely broken. This reaction is extremely useful for determining the exact position of a double bond in an unknown alkene.
Ozonolysis of an alkene gives only acetone (CH₃-CO-CH₃) as the sole product. Deduce the structure of the original alkene.
Since ozonolysis gives only one product (not two different ones), the original alkene must have been symmetrical, with identical groups on both ends of the double bond. Acetone forming on both sides implies the original alkene was (CH₃)₂C=C(CH₃)₂ (2,3-dimethylbut-2-ene) — cleaving its central double bond gives two identical acetone molecules.
Alkenes react with cold, dilute, alkaline potassium permanganate to form a vicinal diol (glycol) — two -OH groups added on adjacent carbons where the double bond was. This decolourises the purple KMnO₄ solution and is used as a chemical test to detect the presence of unsaturation (Baeyer's test).
Calcium carbide reacts with water to give acetylene (ethyne), the simplest alkyne, industrially significant historically: $\text{CaC}_2 + 2\text{H}_2\text{O} \rightarrow \text{Ca(OH)}_2 + \text{C}_2\text{H}_2$.
A vicinal (adjacent-carbon) dihalide, when treated with excess alcoholic KOH, undergoes double dehydrohalogenation to form an alkyne.
Terminal alkynes (with a $\equiv\text{C-H}$ group) show weakly acidic character — the C-H bond can be broken by very strong bases (like sodamide, $\text{NaNH}_2$), since the resulting carbanion is stabilised by the high s-character (50%) of the sp-hybridised carbon, which holds the electron pair closer to the nucleus and more tightly.
Explain why ethyne is more acidic than ethene, which in turn is more acidic than ethane.
Acidity here relates to how stable the resulting carbanion is after losing H⁺, which depends on the s-character of the hybrid orbital holding the lone pair. Ethyne's carbon is sp hybridised (50% s-character), ethene's is sp² (33% s-character), and ethane's is sp³ (25% s-character). Higher s-character means the lone pair sits closer to the nucleus (more tightly held, more stable) — so ethyne's carbanion is the most stable, making ethyne the most acidic of the three, followed by ethene, then ethane.
Alkynes undergo addition reactions similar to alkenes, but can add reagents twice (across both pi bonds of the triple bond) if in excess, following Markovnikov's rule for unsymmetrical reagents at each step.
Predict the final product when excess HBr is added to propyne, CH₃-C≡CH.
First addition (Markovnikov): H adds to the terminal (≡CH) carbon, Br to the internal carbon, giving $\text{CH}_3-\text{CBr}=\text{CH}_2$ (2-bromopropene).
Second addition (Markovnikov again, on the new alkene): another HBr adds with Br going to the same (already Br-bearing, more substituted) carbon, giving the final product, 2,2-dibromopropane ($\text{CH}_3\text{CBr}_2\text{CH}_3$).
Benzene (C₆H₆) is a planar, hexagonal ring with all 6 carbons sp² hybridised. It's best represented as a resonance hybrid of two equivalent Kekulé structures (with alternating single/double bonds), giving all 6 C-C bonds equal length — intermediate between a typical single and double bond — due to complete delocalisation of the 6 π electrons above and below the ring plane.
For a compound to be aromatic, it must be: (1) cyclic, (2) planar, (3) fully conjugated (every ring atom has a p-orbital available), and (4) contain $(4n+2)$ π electrons in the delocalised system, where $n = 0, 1, 2, ...$. Benzene has 6 π electrons ($n=1$: $4(1)+2=6$), satisfying the rule.
Explain why all 6 carbon-carbon bond lengths in benzene are identical, even though the Kekulé structure suggests alternating single and double bonds.
Benzene's true structure is not a single Kekulé structure but a resonance hybrid of (at least) two equivalent Kekulé forms, where the 6 π electrons are completely delocalised around the entire ring rather than being fixed between specific pairs of carbons. Since every C-C bond has an equal share of this delocalised π electron density, all 6 bonds end up being identical in length and strength — intermediate between a pure single and pure double bond.
Benzene's electron-rich π system attracts electrophiles, initiating substitution reactions in three key steps: (1) generation of the electrophile, (2) attack of the electrophile on the ring, forming a resonance-stabilised carbocation intermediate (arenium ion/sigma complex), and (3) loss of a proton from this intermediate to restore aromaticity.
| Substituent Type | Directs Incoming Group To | Effect on Ring Reactivity | Examples |
|---|---|---|---|
| o,p-directing, activating | Ortho and para positions | Increases reactivity (more reactive than benzene) | −OH, −NH₂, −OR, alkyl groups |
| o,p-directing, deactivating | Ortho and para positions | Decreases reactivity (less reactive than benzene) | Halogens (−Cl, −Br) |
| Meta-directing, deactivating | Meta position | Decreases reactivity significantly | −NO₂, −CN, −COOH, −SO₃H |
Halogens are unusual: they direct incoming electrophiles to ortho/para positions (through electron donation via resonance, +M effect, when the intermediate carbocation forms) but still make the ring less reactive overall than benzene (due to their strong −I electron-withdrawing inductive effect dominating the overall electron density of the ring).
Predict the major product(s) when nitrobenzene undergoes further nitration (a second -NO₂ group is introduced).
The existing −NO₂ group is strongly meta-directing and deactivating. The second nitro group will therefore enter predominantly at the meta position relative to the first, giving 1,3-dinitrobenzene as the major product, and the overall reaction will proceed considerably more slowly than nitration of benzene itself (since −NO₂ deactivates the ring).