Alkenes, Alkynes & Aromatic Hydrocarbons

Hydrogen gas will not react with an alkene under ordinary conditions — the H–H bond (436 kJ/mol) is strong and the reaction needs a catalyst to proceed at a useful rate. Platinum, palladium, or nickel metals adsorb both the H₂ and the alkene onto their surfaces, bringing both H atoms to the same face of the double bond simultaneously. Because both H atoms come from the same side, hydrogenation is a syn addition — both new C–H bonds form on the same face of the original C=C plane. This matters when the product can exist as cis or trans: syn addition to a disubstituted alkene gives the cis-disubstituted product.

R–CH=CH–R' + H₂ —Pt/Pd/Ni→ R–CH₂–CH₂–R'. Both H's add to the SAME face (syn addition) because they are delivered together from the catalyst surface.

When Br₂ (reddish-brown) is added to a solution containing an alkene, the solution decolourises instantly. This happens because the π electrons attack one Br of the Br–Br molecule, forming a three-membered bromonium ion intermediate that sits on top of the double bond. The bromide ion (Br⁻) then attacks from the bottom — the opposite face — opening the ring. The result is anti addition: the two bromine atoms end up on opposite faces of the original double bond. This is why the product of bromine addition to cyclopentene is trans-1,2-dibromocyclopentane, not cis.

CH₃CH=CH₂ + Br₂ → CH₃CHBr–CH₂Br. Anti addition via bromonium ion. Br₂ decolourisation is the classic unsaturation test — red-brown Br₂ → colourless vicinal dihalide.

When HCl or HBr adds to an unsymmetric alkene, there are two possible products: the H can go to C1 and the halide to C2, or vice versa. Vladimir Markovnikov noticed in 1870 that one product always predominates. His empirical rule: the H attaches to the carbon that already has more H's (the less substituted carbon). But why? The proton always attacks the alkene first, and it attacks the less substituted end because that creates the more substituted carbocation at the other carbon. A tertiary carbocation (3°) is far more stable than a secondary (2°) which is more stable than a primary (1°) — for exactly the same hyperconjugation reason as tertiary radicals are more stable. The halide then attacks the more stable carbocation. So Markovnikov’s rule is really just: the more stable carbocation forms, and that determines where the halide ends up.

Adding water across a double bond sounds simple, but water alone does nothing — alkenes are nonpolar and water is polar. The trick is to first protonate the alkene with H⁺ (from H₂SO₄ or HClO₄), forming a carbocation. That carbocation is then attacked by water (a nucleophile), forming an oxonium ion, which loses a proton to give the alcohol. Because the reaction goes through a carbocation intermediate, it follows Markovnikov’s rule: the OH group ends up on the more substituted carbon.

C=C + H₂O —H₂SO₄, heat→ alcohol (Markovnikov). e.g. 2-methylpropene + H₂O / 10% H₂SO₄ → tert-butanol (3° alcohol).

Water alone (no acid) doesn't react — alkenes are nonpolar; you need to protonate them first to generate a carbocation.

Potassium permanganate (KMnO₄) is a powerful oxidant, but its effect on a double bond depends entirely on how you use it. Cold and dilute: it gently adds two –OH groups across the double bond, leaving the carbon skeleton intact — the product is a diol, and the two OH groups end up on the same face (syn dihydroxylation). Hot and concentrated with acid: it destroys the double bond entirely, cleaving both the π and σ bonds between the two alkene carbons and oxidising each to its maximum state. Knowing which conditions give which product is a core Part V skill.

Ozonolysis is the forensic chemist’s best friend for structure determination. You bubble ozone (O₃) through a solution of the alkene; it cleaves the C=C and inserts an oxygen at each end. Then reductive work-up (Zn/H₃O⁺ or dimethylsulfide, Me₂S) liberates two carbonyl-containing fragments. The key rule: each alkene carbon becomes a C=O. If that carbon had an H attached, you get an aldehyde (R–CHO); if it had two C-substituents and no H, you get a ketone (R₂C=O). Because the products map back uniquely to the original double bond position, examiners love to give you ozonolysis products and ask you to deduce the alkene structure — or give you the alkene and ask for the products.

Benzene was discovered in 1825 and puzzled chemists for decades. Its molecular formula, C₆H₆, gives DU = 4 — suggesting three double bonds and one ring, like a cyclohexatriene. But benzene behaves nothing like a triene: it refuses to undergo addition reactions (which would destroy two double bonds at a time), it does not decolourise bromine water under ordinary conditions, and all six C–C bonds are the same length (139 pm — between single 154 pm and double 134 pm). Kekulé famously proposed the correct structure in 1865 — a regular hexagon with alternating double and single bonds — but even his structure failed to explain the equal bond lengths.

The modern picture resolves everything. All six C atoms in benzene are sp², so each contributes one unhybridised p orbital perpendicular to the ring plane. These six p orbitals overlap continuously around the entire ring, producing a π electron cloud that is delocalised over all six carbons — not fixed between alternating pairs. This complete delocalisation confers an extra resonance stabilisation energy of ~36 kcal/mol (the "aromatic stabilisation energy") compared with a hypothetical non-delocalised cyclohexatriene. This stabilisation is so large that benzene will not sacrifice it by undergoing addition — instead it reacts by substitution, which replaces one ring H but restores the aromatic π system.

All six C–C bond lengths in benzene are equal (139 pm) — between single (154 pm) and double (134 pm). DU = 4 but benzene undergoes substitution, not addition, because addition would destroy the 36 kcal/mol aromatic stabilisation.

Disubstituted benzenes: ortho (o-, 1,2), meta (m-, 1,3), para (p-, 1,4). When benzene is itself a substituent: phenyl (C₆H₅–).

Every aromatic substitution reaction follows the same two-step dance. First, an electrophile (E⁺) attacks the π cloud of the ring, forming a positively charged sigma complex (also called the arenium ion or Wheland intermediate). At this stage the aromatic system is temporarily disrupted — one ring carbon is sp³ and the positive charge is spread over the remaining three positions. The ring has given up its precious aromaticity for a moment, which is why this step is slow and rate-determining. Then, in the fast second step, a proton is lost from the sp³ carbon — restoring the aromatic π system. The net result is that one ring H has been replaced by E. The ring is regenerated, not destroyed. This is the mechanistic reason benzene prefers substitution over addition: addition would kill the aromaticity permanently; substitution restores it.

When benzene already carries one substituent and you add a second, you need to know two things: will the existing group speed up or slow down the ring toward further substitution (activation vs deactivation), and will the new group end up ortho/para or meta relative to the first? Both questions are answered by a single underlying principle: electron-donating groups activate the ring and direct ortho/para; electron-withdrawing groups deactivate and direct meta. The reason is the stability of the sigma-complex intermediate. Electron donors stabilise the positive charge in the sigma complex at the ortho and para positions (where the positive charge resides), making attack there faster. Electron withdrawers destabilise the positive charge at ortho and para, so attack at the meta position (where the EWG has less interaction with the positive charge) becomes relatively faster.