4.1.2 (a,b,c) Properties of Alkanes

Syllabus

(a) Alkanes as saturated Hydrocarbons containing single C–C and C–H bonds as σ-bonds (overlap of orbitals directly between the bonding atoms); free rotation of the σ-bond

{Hybridisation not required.}

{Use of model of orbital overlap to explain covalent bonding in organic compounds.}

(b) explanation of the tetrahedral shape and bond angle around each Carbon atom in Alkanes in terms of electron pair repulsion (see also 2.2.2 g–h)

{Learners should be able to draw 3-D diagrams.}

(c) explanation of the variations in boiling points of Alkanes with different Carbon-chain length and branching, in terms of induced dipole–dipole interactions (London forces) (see also 2.2.2 k)

What does this mean?

The simplest homologous series in Organic Chemistry are the Alkanes.

All other Organic substances are named according to the naming system for Alkanes.

Alkanes are saturated because each Carbon atom is bonded to the maximum number of other atoms (4).

Although, if you're asked for a definition of saturated you should say that it contains only singleC-C bonds (no double or treble bonds).

VSEPR theory tells us that if an atom has 4 charge centres (electron pairs) and all are bonds (no lone pairs) then they will repel to minimum repulsion at 109.5^o.

This is the tetrahedral bond angle.

So the bonding around every Crabon atom is tetrahedral.

This means that our usual 2-D representation of Alkanes is not very accurate.

A better representation might look like the diagram of Butane (right).

But clearly a truly 3-D representation of an Alkane isn't easy to draw on a 2-D medium such as paper.

Though this isn't strictly the case we can assume that all bonding with Alkanes is made up of overlapping s-orbitals in Hydrogen and Carbon atoms - meaning that every bond is a σ-bond.

If we apply our prior knowledge we can see that there are no polar bonds in Alkanes because C-C bonds cannot be polar by definition, and C-H bonds do not have a big enough difference in electronegativity to count as polar.

So no Alkanes are polar molecules.

This leaves us only van der Waals intermolecular forces (strictly speaking London Dispersion forces since van der Waals can cover other forces too) to account for their boiling/melting points.

We know that van der Waals increase in strength with the number of electrons, which increase with molecular size.

So it should be no surprise that the trend in Alkane boiling points is that it increase with the number of Carbon atoms in the chain.

The graph shows us that Methylpropane and Butane have different boiling points despite having the same mass and both being C₄H₁₀.

Butane is a straight molecule with no branches.

Methylpropane is branched.

This stops methylpropane molecules approaching each other as closely.

Forces become weaker over greater distances (as with magnets)

So the van der Waals forces in branched molecules are always weaker than in non-branched molecules and the boiling points are consequently lower.