4.1.2 Alkanes

Alkanes are saturated hydrocarbons: they contain only carbon and hydrogen, and all the carbon–carbon bonds are single bonds. Each C–C and C–H bond is a σ bond formed by end-on overlap.

Because a single σ bond is cylindrically symmetrical around the bond axis, you can rotate around a C–C single bond without breaking it. That becomes an important contrast with the restricted rotation of C=C in alkenes.

Around a carbon with four single bonds, there are four bonding regions. They repel one another as far apart as possible, giving a tetrahedral arrangement with bond angles of about 109.5°.

When you draw a displayed formula flat on the page, do not let the drawing fool you into saying 90° or 120°. A carbon with four single bonds has a three-dimensional tetrahedral arrangement.

As an alkane chain gets longer, it has more electrons and a larger contact surface, so London forces between molecules become stronger. You therefore need more energy to separate the molecules and the boiling point generally rises.

Branching works the other way for isomers: a more branched molecule is more compact, with less effective surface contact between molecules, so the London forces are weaker and the boiling point is usually lower.

Alkanes are fairly unreactive because their C–C and C–H bonds have relatively high bond enthalpies and are only weakly polar. There is no strongly δ+ centre for a nucleophile to attack and no electron-rich functional group for an electrophile.

If you are explaining the low reactivity, give both ideas: the bonds are hard to break and their low polarity gives attacking reagents little electrostatic reason to react.

With plenty of oxygen, complete combustion of an alkane gives CO₂ and H₂O. With limited oxygen, incomplete combustion can give CO and/or carbon as well as water.

Carbon monoxide is especially dangerous because it binds strongly to haemoglobin and reduces the blood’s ability to carry oxygen. When you balance a combustion equation, I usually balance C first, H second and O last.

When you halogenate an alkane, UV light starts a free-radical substitution chain reaction. In initiation, the halogen bond breaks homolytically. Propagation then keeps the chain going: one radical is used and another radical is produced. Termination happens when two radicals combine.

For methane and chlorine, I would know the two propagation steps as a pair because they make the chain logic obvious.

Radical substitution is difficult to stop neatly after one H has been replaced. The first haloalkane can react again, so further substitution gives di-, tri- and more highly substituted products.

With larger alkanes, hydrogen atoms can also occupy different environments, so substitution at different carbon positions gives structural isomers. If you need one pure product, this lack of selectivity is a real disadvantage of the reaction.