Everything from the 4 lessons of this topic on a few pages: the key idea of each lesson, the diagram to remember, the points to learn and the traps examiners set. Read it, cover it, test yourself.
Alkanes are saturated hydrocarbons, CnH2n+2, with only single bonds and tetrahedral carbons. Their C–C and C–H bonds are strong and almost non-polar, so they are attacked by neither nucleophiles nor electrophiles, which makes them generally unreactive. They come from crude oil by fractional distillation and cracking. They burn: complete combustion gives CO2 and H2O; in limited oxygen, incomplete combustion gives toxic CO and soot. Boiling points rise with chain length and fall with branching.
When balancing a combustion equation, balance C first, then H, then O last; if O needs a half, double everything. Alkanes do not decolourise bromine water in the dark: they react with halogens only by free-radical substitution in UV light.
Alkanes react with chlorine or bromine in ultraviolet light by free radical substitution. Initiation: UV light breaks Cl–Cl by homolytic fission into two chlorine radicals. Propagation: a radical reacts with a molecule to give a product and a new radical, so the chain continues. Termination: two radicals combine and the chain stops. Because radicals are so reactive, further substitution occurs and a mixture of products forms (CH3Cl, CH2Cl2, CHCl3, CCl4, and some ethane).
Radicals form by homolytic fission and carry an unpaired electron, shown by a dot. Every propagation step has one radical on each side; every termination step has two radicals in and none out. A step that makes HCl and a methyl radical is propagation, not termination.
Alkenes are unsaturated, CnH2n, with a C=C made of a σ and a π bond. The π electrons are a region of high electron density above and below the plane, so alkenes undergo addition: with H2 (Ni catalyst) to alkanes, with steam (H3PO4 catalyst) to alcohols, with HX to halogenoalkanes and with X2 to dihalogenoalkanes. Bromine water turns from orange to colourless. Cold, dilute acidified KMnO4 gives a diol; hot, concentrated acidified KMnO4 splits the C=C, and the products reveal where the double bond was.
Cold, dilute KMnO4 adds two OH groups (a diol); hot, concentrated KMnO4 breaks the C=C. Do not mix them up. In the cleavage products, each carbon of the original C=C becomes a C=O carbon: =CH2 goes all the way to CO2.
In electrophilic addition the π electrons of the C=C attack an electrophile. With Br2, the approaching π cloud induces a dipole in Br–Br; the π pair forms a bond to the Brδ+ atom while Br–Br breaks heterolytically, giving a carbocation and Br−, which then attacks the positive carbon. With HBr, Hδ+ is the electrophile. Alkyl groups push electron density towards a positive carbon (the positive inductive effect), so carbocation stability is tertiary > secondary > primary. With an unsymmetrical alkene the major product comes from the more stable carbocation: propene + HBr gives mainly 2-bromopropane.
The first curly arrow must start from the C=C double bond (the π electrons) and point to the electrophile, not start at the bromine. Draw the carbocation with a full + charge on carbon, and the bromide ion with a lone pair and a − charge. The major product comes from the more substituted carbocation.