Alkenes

Each carbon in ethene is sp² hybridised (trigonal planar, 120°). Head-on overlap of sp² orbitals gives the sigma bond, and the unhybridised p orbitals overlap sideways to give one pi bond, with electron density above and below the plane. Two sigma bonds cannot form between the same two atoms, and sp³ describes alkane carbons. The pi bond is a single bond with two lobes, not two pi bonds. The pi bond is weaker than the sigma bond (less effective overlap), which is why alkenes undergo addition: the pi bond breaks while the sigma bond stays.

Hydrogenation adds H₂ across the C=C: C₂H₄ + H₂ → C₂H₆, using a finely divided nickel catalyst and heat (about 150 °C). Hydrogenating some C=C bonds in vegetable oils raises their melting point to make margarine. Steam with H₃PO₄ (about 300 °C, 60 atm) adds H₂O, giving ethanol, not ethane. Ethanolic NaOH removes HX from halogenoalkanes to make alkenes; it does not react with ethene. UV light starts radical reactions with halogens, not addition of hydrogen.

Dehydration of an alcohol removes H₂O: CH₃CH₂CH₂OH → C₃H₆ + H₂O, using heated Al₂O₃ or hot concentrated H₂SO₄ or H₃PO₄. Alkenes also form by cracking and by elimination of HBr from a bromoalkane with ethanolic NaOH; but in aqueous NaOH, substitution happens instead and propan-1-ol is formed. Propane is already saturated, so it does not react with hydrogen. Acidified KMnO₄ oxidises primary alcohols (towards aldehydes and carboxylic acids); it does not make alkenes.

Cold dilute acidified KMnO₄ adds an OH group to each carbon of the C=C, giving the diol ethane-1,2-diol; MnO₄⁻ is reduced, so the purple colour fades. Carbon dioxide and water come from splitting the C=C with hot concentrated KMnO₄ (each =CH₂ end gives CO₂). Orange to green is the colour change of dichromate(VI), not manganate(VII), and adding OH to only one carbon (ethanol) would need hydration with steam. Ethene does react: decolourising KMnO₄ is a test for C=C, like decolourising bromine water.

In addition polymerisation the pi bond opens and the two C=C carbons bond to neighbouring monomers, so the backbone carries the two carbons of the double bond and the CH₃ hangs off as a branch: –CH₂–CH(CH₃)–. Putting all three carbons in the backbone is wrong; the methyl carbon was never part of the C=C. The polymer is saturated, so the repeat unit cannot keep a C=C. –CH₂–CH₂– is the repeat unit of poly(ethene).

The C=C breaks and each carbon becomes C=O. The left carbon carries two CH₃ groups (=CR₂), so it becomes the ketone propanone, (CH₃)₂CO. The right carbon carries H and CH₃ (=CHR), so it first forms an aldehyde, which hot KMnO₄ oxidises further to ethanoic acid, CH₃COOH. A diol forms only with cold dilute KMnO₄; claiming hot KMnO₄ gives a diol is a classic error. Ethanal would not survive these vigorous oxidising conditions. CO₂ comes only from a terminal =CH₂ group, which this alkene does not have, and the carbon chain is split, so a four-carbon ketone cannot form.

A ketone comes from a C=C carbon bearing two alkyl groups. A single product means the alkene is symmetrical, and joining two (CH₃)₂C= units gives (CH₃)₂C=C(CH₃)₂, C₆H₁₂. Hex-3-ene is symmetrical but each C=C carbon carries an H, so it gives two molecules of propanoic acid, not a ketone. 2-methylpent-2-ene gives propanone from one end but propanoic acid from the =CH-C₂H₅ end. Hex-1-ene gives CO₂ from the =CH₂ end and pentanoic acid.

n(C₂H₄) = 2.80 / 28.0 = 0.100 mol. Ethene has one C=C, so it adds one Br₂ in a 1 : 1 ratio. M_r(Br₂) = 159.8, so mass = 0.100 × 159.8 = 15.98 g = 16.0 g. 7.99 g uses the A_r of a single Br atom instead of Br₂. 32.0 g assumes two Br₂ add, which would need two C=C bonds. 17.2 g uses M_r 26.0 for ethene, the value for ethyne, C₂H₂.