4.1.3 (a,b) Properties of Alkenes
Syllabus
(a) Alkenes as unsaturated hydrocarbons containing a C=C bond comprising a π-bond (sideways overlap of adjacent p-orbitals above and below the bonding C atoms) and a σ-bond (overlap of orbitals directly between the bonding atoms) (see also 4.1.2 a); restricted rotation of the π-bond
{Hybridisation is not required.}
{Use of model of orbital overlap to explain covalent bonding in organic compounds.}
(b) explanation of the trigonal planar shape and bond angle around each carbon in the C=C of Alkenes in terms of electron pair repulsion (see also 2.2.2 g–h, 4.1.2 b)
What does this mean?
The Basics
Alkenes are Hydrocarbons because they contain only Carbon and Hydrogen atoms.
They are unsaturated because they contain a C=C double-bond - this is what to write in an exam.
Alternatively, they are unsaturated because the Carbon atoms are not bonded to the maximum number of atoms - but don't write this in the exam.
The naming of Alkenes follows the Alkane model:- Meth, Eth, Prop, But, Pent, Hex, Hept, Oct, Non, Dec.
Although you can't have Methene because an Alkene must contain a C=C bond and Meth implies only one C atom.
Their general formula is CnH2n.
There are no possible isomers of Alkenes until we get to Butene.
Because the double bond can be at the end of the molecule or in the middle we can have But-1-ene (where the lowest numbered Carbon atom touched by the double bond is No 1) and But-2-ene (where the lowest numbered Carbon atom touched by the double bond is No 2)
But we could also move one of the Carbon atoms to form Methyl Propene.
Other isomers of C4H8 exist in Cyclobutane and Methylcyclopropane - but clearly these are not actually alkene because they don't contain a C=C bond
What is a Double Bond?
A C=C double bond contains a sigma bond and a pi bond.
The Sigma bond is much the same as in an Alkane, an orbital from one Carbon atom overlaps with a similar orbital in the other.
We can think of this as always being the overlap of s-orbitals (not strictly true) - which is where the name Sigma bond comes from, with Sigma being a Greek S.
Sigma bonds have a big overlap and are consequently very strong.
But not all orbitals are s-orbitals and these other orbitals can also overlap.
When p-orbitals overlap sideways it is called a pi bond after a Greek p.
Notice that the p-orbital shape means that they can overlap above and below the plane of the sigma bond - examiners will want you to write that in a definition.
Despite this picture the area of overlap is quite small compared to a sigma bond.
So pi bonds are not as strong as sigma bonds.
Looking at the bond enthalpy we can see that the difference between C-C and C+C is only 260 kJ/mol compared to the 350 kJ/mol for the sigma bond on its own.
Adding another pi bond to make a treble bond adds a similar amount of extra strength (230 kJ/mol).
When s-orbitals overlap they can freely rotate without ever breaking the overlap - which would require some energy.
Even if p, d or f orbitals overlapped head-on they could rotate without losing contact.
But in a pi bond the p-orbitals are overlapping sideways - above and below the plane of the sigma bond.
Any rotation would mean that contact is lost and the bond has broken.
But since breaking bonds needs energy pi bonds don't rotate in normal circumstances.
As a consequence many Alkenes have geometrical or stereo isomers (see later).
Bond Geometry
VSEPR tells us that double-bonds do not repel any more than single bonds.
So a Carbon atom in a double bond effectively has three bonds and no lone-pairs.
The furthest apart three bonds can get is 120o - which is trigonal planar.
So all alkenes are trigonal planar about the Carbon atoms in the C=C bond.
Although they are very likely to have other Carbon atoms in a tetrahedral arrangement because they have 4 bonds.
Videos
Exam-style Questions
1. (a) Four members of the homologous series of Alkenes are Ethene, Propene, But-1-ene and Pent-1-ene.
(i) Give one structural feature of the compounds that makes them members of the homologous series of Alkenes.
...................................................................................................................................................................................................................................
...............................................................................................................................................................................................................................(1)
(ii) State the trend in the boiling points from Ethene to Pent-1-ene.
...............................................................................................................................................................................................................................(1)
(b) There are five structural isomers of the molecular formula C5H10 which are Alkenes. The graphical formulae of two of these isomers are given.
Isomer 1 Isomer 2
Draw the graphical formulae of two of the remaining Alkene structural isomers.
Isomer 3 Isomer 4
(2)
(c) Draw the graphical formulae of three Alkenes which have the formula C4H8.
Isomer 1 Isomer 2 Isomer 3
(3)
(Total 7 marks)
Answers
1. (a) (i) the same general formula or CnH2n / the same functional group / a C=C / a double
bond / differ from their immediate neighbour by CH2 1
allow ‘all straight chain alkenes’
(ii) increases
not just ‘pent-l-ene highest’; allow ‘ethene lowest, pent-l-ene highest’ 1
(b) appropriate structure for pent-2-ene (1)
appropriate structure for 2-methylbut-2-ene (1)
appropriate structure for 3-methylbut-l-ene (1) max 2
accept structures with CH3– , –C2H5 and –CH2–
(c)
