While alkanes and alkenes are both hydrocarbons, the primary difference is that alkanes are saturated molecules, containing only single covalent bonds (σ-bonds) between the carbon atoms whereas alkenes are unsaturated molecules containing a double covalent bond (combination of a π-bond and a σ-bond). This leads to them having different properties, and different chemistry.
Alkanes
The family of alkane molecules has similar chemistry to one another because they are a homologous series: a series of organic compounds having the same functional groups, each successive member differing by –CH2– Their general formula is CnH2n+2 where n is the number of carbon atoms in the alkane.
Alkanes can be in the form of straight or branched chains, as illustrated here by the various alkanes with molecular formula C6H14.
We can also have cyclic alkanes called cycloalkanes. In these molecules there are 2 less hydrogen atoms, so the general formula is CnH2n:
Physical Properties of Alkanes
The graph shows that there is a trend in the melting and boiling points of the straight chain alkanes with increasing number of carbon atoms in the chain. Room temperature (298K) is shown with a green dashed line, from which we can see that:
- Alkanes with 4 carbons or less both melt and boil below room temperature. They are therefore gases at room temperature.
- Alkanes with between 5 and 17 carbon atoms melt below room temperature but boil above room temperature. They are therefore liquids at room temperature,
- Alkanes with greater than 18 carbon atoms melt and boil above room temperature and are therefore solids at room temperature.
This is easily explained. Alkanes are simple molecules, so the only things that holds one molecule to another are weak intermolecular forces. There is not a significant difference in electronegativity between C and H atoms, so these are not polar molecules. The only intermolecular forces are London forces. The strength of London forces increases with the number of electrons in the molecule, so increasing chain length gives rise to increasing London forces, which require more energy to overcome when melting or boiling.
Volatility is the ease with which a substance turns into a vapour. It can be seen that the volatility of the alkenes decreases with increasing chain length.
Branching of the alkane molecule also has an effect on melting and boiling points. When comparing to alkanes which are isomers, and hence isoelectronic (containing the same number of electrons), it might be expected that the London forces would be the same strength and hence the melting and boiling points would be the same. In fact the more branched the alkane is, the lower its melting and boiling point. This is explained in terms of the area of surface contact between the molecules. Since London forces get weaker with distance apart, the greater the area of surface contact, the stronger the London forces will be.
Shapes of Molecules and Electron Density of Alkanes
The shape around every carbon atom in an alkane is tetrahedral, with 109.5º bond angles. This is because each carbon atom forms four bonding pairs, leaving no lone pairs.
We have previously considered electrons as being located in atomic orbitals, within the subshells and shells of atoms. When electrons are shared between atoms to form covalent bonds they are no longer located in the orbitals of each atom, but in newly-created molecular orbitals. The shared pair of electrons that form a single covalent bond are found in an ellipsoid-shaped orbital called a sigma (σ-orbital). This can be though of as having been formed by the direct overlap of the s- or p- orbitals in the bonded atoms. Here is an illustration of the molecular orbitals for an ethene molecule, showing how they form single covalent bonds referred to as σ-bonds.
Chemical Properties of Alkanes
As already discussed, alkanes are remarkably inert, showing no reaction with acids or alkalis for example. There are two reasons for this:
- very low polarity of C-H and C-C bonds (see electron density maps). C-C and C-H bonds are non polar, consequently the bonds in alkanes are not susceptible to attack by most common chemical reagents
- high bond enthalpies of C-H and C-C bonds. These bonds require a significant amount of energy to break, so reactions have a high activation energy.
For some uses, the chemical inertness of the alkanes is their greatest asset. Compounds that are non- corrosive to metals (lubricating oils), harmless to our skin (vaseline), and safe in contact with foods (polyethene), are enormously useful to us.
1: Combustion
When an excess of oxygen is available, alkanes combust completely giving carbon dioxide and water as the only products. Alkanes make excellent fuels for domestic, industrial and transport use because they react with oxygen exothermically (a fuel is defined as a source of useful chemical energy). They have low toxicity, and short chain alkanes are easily ignited.
e.g. CH4 + 2O2 → CO2 + 2H2O ΔcHθ = -890 kJmol-1
When only limited oxygen is available, combustion is incomplete and carbon monoxide, or even carbon particles (soot) may be formed. The presence of carbon particles can be detected when hydrocarbons burn because the flame becomes yellow and smoky rather than pale blue.
e.g. 2CH4 + 3O2 → 2CO + 4H2O or 2CH4 + 2½O2 → CO + C + 4H2O
Dangers of incomplete combustion arise because CO binds permanently to haemoglobin, preventing O2 and CO2 transport. CO is therefore toxic, and is particularly dangerous because it is a colourless and odourless gas so hard for people to detect.
2: Substitution reaction of alkanes with halogens
Hydrogen atoms in an alkane can be substituted by halogen atoms. The mechanism involves the formation of free radicals, so it is referred to as a radical substitution mechanism.
Conditions for the reaction: ultraviolet light is required (the amount present in sunlight is sufficient), so this is an example of a photochemical reaction.
Example: when chlorine is mixed with methane and exposed to sunlight, chloromethane is formed and hydrogen chloride gas is evolved. CH4(g) + Cl2(g) → CH3Cl(g) + HCl(g)
Mechanism: the example here is for the reaction of methane with chlorine, but other halogens and other alkanes react in the same way and can be substituted for each other in this mechanism.
INITIATION STEP The energy of UV light is enough to break the Cl-Cl bond in the Cl2 molecule. Absorption of light energy to break a bond is called photodissociation. The type of bond fission (breaking) is homolytic, resulting in two Cl• radicals being formed.
Cl-Cl(g) → Cl•(g) + Cl•(g) conditions: uv
Radicals are very reactive, and react rapidly with other molecules or chemical species…
PROPAGATION STEPS A chlorine radical reacts with methane to produce hydrogen chloride and a methyl radical
Cl• + H-CH3 → Cl-H + •CH3
The methyl radical is then able to react with a chlorine molecule to produce chloromethane and regenerate a chlorine radical.
Cl-Cl + •CH3 → CH3Cl + Cl•
Forming a new chlorine radical means these two propagation steps can happen again and again – a chain reaction which needs no more UV to keep going. Measurements show these steps can occur some 10,000 times for one UV photon initiation.
TERMINATION STEP The reaction only stops when two radicals combine to form a new molecule. This removes radicals from the reaction mixture, stopping the chain reaction.
There are several possibilities:
Cl• + Cl• → Cl2
Cl• + •CH3 → CH3Cl
•CH3 + •CH3 → C2H6 so a small amount of ethane is also formed.
Formation of a mixture of products
The mechanism of radical substitution does not produce a single pure product. A mixture of products is formed, so there are limitations on the use of this reaction for synthesising a specific desired product.
There are three ways in which more than one product might be formed: e.g. if we want to use this reaction to make 1-bromopropane from propane:
CH3CH2CH3 + Br2 → CH3CH2CH2Br + HBr
1: The bromine radical can attack any position in the alkane chain, so 2-bromopropane can also be formed by this pair of propagation steps:
CH3CH2CH3 + Br• → CH3•CHCH3 + HBr
CH3•CHCH3 + Br2 → CH3CHBrCH3 + Br•
2: One possible termination step is the reaction of two alkyl radicals, forming an alkane of twice the length of the original alkanes, so two propyl radicals can react to form hexane:
•CH2CH2CH3 + •CH2CH2CH3 → CH3(CH2)4CH3
And because there are two types of propyl radical that can be formed, depending on which hydrogen atom was removed from the propane molecule, 2-methylpentane and 2,3-dimethylbutane are also possible products that can be formed in termination steps.
3: There is no reason why any of the products formed in any of the reactions cannot themselves react with bromine radicals in further radical substitution reactions, e.g. 1-bromopropane formed in the original reaction could react again to form 1,2-dibromopropane in these propagation steps:
CH3CH2CH2Br + Br• → CH3CH•CH2Br + HBr
CH3CH•CH2Br + Br2 → CH3CHBrCH2Br + Br•
Alkenes
Like alkanes, alkenes can be straight chain or branched chain. The double bond between the carbon atoms can be anywhere in the chain. It is also possible to have cyclic alkenes (with or without branches off the ring). Note: benzene is sometimes drawn as a hexagon with three alternating double bonds, making it look like an cyclic alkene. IT IS NOT AN ALKENE.
The alkenes form a homologous series with general formula CnH2n (for cycloalkenes it would be CnH2n-2). Each member of the series differs by a -CH2– unit.
Physical Properties of Alkenes
The melting and boiling points of alkenes follow the same trends as those of alkanes, for the same reasons. Because the have a 120º bond angles at the carbons with the double bond and 109.5º bond angles elsewhere alkene molecules don’t pack quite so neatly and closely together, and as a result the London forces between alkenes are a little weaker than those of correspondingly sized alkanes. This leads to melting and boiling points that are a little lower than those of the alkanes.
e.g. (boiling points in Kelvin) ethene 169K ethane 185K
propene 226K propane 231K
Shapes of Molecules and Electron Density of Alkenes
The two carbon atoms with a double covalent bond between them have the equivalent of 3 bonding pairs and no lone pairs of electrons, so the shape at these carbon atoms is trigonal planar and the bond angles are 120º. Elsewhere in the alkene the carbon atoms are just like those in alkanes, so the bond angles are 109.5º and the shape at these carbon atoms is tetrahedral.
The shape around the C=C bond and the fact that there is restricted rotation about this bond is a result of how the double bond is formed. The first of the two covalent bonds between the two carbon atoms is a σ-bond formed exactly in the way discussed for alkanes. Each of the two carbon atoms therefore has three σ-bonds. This leaves a p-orbital on each carbon atom containing one electron. These p-orbitals are at 90º to the plane of the molecule, but by sideways overlap of the p-orbitals a new molecular orbital, referred to as a π-orbital is formed between the two carbon atoms, providing a the second bond in the double bond, which is referred to as a π-bond.
Unlike in alkanes, where we saw there was little variation in the electron density around the molecule, the double bond in an alkene localises two bonding pairs of electrons in a small region of space, and so the double bond is a region of high electron density which is susceptible to attack by reagents that are attracted to regions of negative charge. We call such reagents electrophiles (electron-loving). The electron density map for ethene, below, shows the region of high electron density (red) clearly, and can be compared with the electron density map for alkanes to see why alkenes are reactive while alkanes are inert, and why the attacking species approaches the alkene side-on.
Chemical Properties of Alkenes
Most alkene reactions involve breaking the π-bond, which has a lower bond enthalpy than the σ-bond and is therefore weaker. This leaves the single covalent bond (σ-bond) intact allowing the attacking species to be attached to the carbon atoms where the double bond was, to form a saturated molecule. These are addition reactions.
With lots of negative charge in the area of high electron density between the two C=C carbons, species wanting to accept an electron pair (often positively charged species, but not always) are attracted here where they can attack the double bond. These species are electrophiles. Thus these reactions are termed electrophilic addition reactions.
1: Reaction of alkenes with hydrogen
Reagent: H2
Conditions: Ni catalyst, temperature 150ºC
Example: 
Notes: This reaction is used for ‘hardening’ vegetable oils, which are unsaturated molecules, by making them saturated. This has the effect of raising the melting points so they are more solid for use in margarines. The process is called hydrogenation.
2: Reaction of alkenes with halogens
Reagents: Cl2, Br2, I2
Conditions: mix at room temperature
Example:
Notes:
- This reaction is usually carried out with bromine as a test for alkenes. The orange bromine water is decolourised by the alkene.
- Fluorine is too reactive to carry out this reaction with alkenes. It tends to ignite them.
3: Reaction of alkenes with hydrogen halides
Reagents: HCl, HBr, HI
Conditions: mix gases at room temperature
Example:
Notes:
- reactivity increases from HF to HI. HF will react with an alkene only under pressure.
- addition of a hydrogen halide to an unsymmetrical alkene can result in two possible isomeric products, depending on which way round the H-Br molecule adds across the double bond:
4: Reaction of alkenes with steam
Reagent: H2O(g)
Conditions: acid catalyst such as H3PO4, 300°C temperature and 6MPa pressure
Example:
Notes:
- This is the industrial route to alcohols for use a solvents, not for drinking – methylated spirits, industrial alcohol etc.
- Where water is added to a molecule, as here, we call this a hydration reaction. This is not to be confused with a hydrolysis, which is breaking a molecule up due to a reaction with water forming two products.
- With an unsymmetrical alkene two isomeric products are possible:
Mechanism of Electrophilic Addition
These reactions of alkenes all happen by a mechanism in which an electrophile attacks the electron-rich double bond, breaking it and resulting in a saturated molecule in which the electrophile has been added in. This mechanism is electrophilic addition. The steps are drawn as follows (remember that a curly arrow represents movement of a PAIR of electrons):
After this stage we are left with a negative ion, and an alkane with a positive charge on one of the carbon atoms that was in the double bond. This happens because each C atom donated one electron to the original π-bond, but these were both used to make a covalent bond between the electrophile and one the carbon atoms. The other C atom is therefore one electron missing, giving it a + charge. We call it a carbocation.
In this way the final product molecule, bromoethane in this case, is formed. Putting the stages together, the full mechanism should be written as shown here:
It should be relatively easy to see that water acts as an electrophile in the above mechanism exactly in the same way that H-Br does. We simply replace the H-Br with H-OH, and in the second stage we have a hydroxide ion, OH–.
What is less obvious is how H2 or halogens such as Br2 can act as electrophiles in this mechanism as both atoms in these molecules have the same electronegativity, and so there is no δ+ part to be attracted. What happens here is that we get an induced dipole. The bond in H-H or Br-Br is not normally polar, but when one of these molecules gets near the high electron density in the C=C double bond of the alkene, the negative charge repels the bonding pair of electrons in the H-H or Br-Br bond. These electrons are pushed away from the alkene, leaving the atom nearest to the alkene δ+ and the one furthest away δ-. Thus, the mechanism for the reaction with bromine can be written as follows:
Major and Minor Addition Products
We noted how when hydrogen halides or steam are added to an unsymmetrical alkene, we get a mixture of two possible products, depending on which way round the molecule adds to the carbon atoms of the C=C. In practice we get one product formed in greater quantity than the other one. Markownikoff’s rule allows us to predict which will be the major product.
Markownikoff’s Rule: The hydrogen atom of the electrophile, H-X, is added to the carbon of the C=C which has most hydrogen atoms/least carbon atoms directly bonded to it.
Thus if we react propene with HCl we can predict that the major product will be 2-chloropropane and the minor product will be 1-chloropropane:
We need to understand and be able to explain why this happens, in terms of the stability of the intermediate carbocations formed in the electrophilic addition mechanism. Having a carbon atom bonded to the carbon with the + charge makes is more stable than having a hydrogen atom bonded to it. Thus the more carbon atoms (less hydrogen atoms) there are bonded to the carbon with the + charge the more stable this intermediate carbocation will be. The most stable carbocation will last longer and have more chance of reacting with the negative ion in Step 2 to form the final product. We don’t have to be able to explain why having carbon rather than hydrogen atoms bonded to it makes the carbon with the + charge more stable, but at a simple level we can accept that carbon atoms have more electrons than hydrogen atoms, and the extra electrons around help to stabilise the + charge.
We can classify the carbocations depending on how many carbon atoms are bonded to the carbon atom with the + charge. A primary carbocation will have one carbon bonded to it; a secondary carbocation will have two; a tertiary carbocation will have three. Thus we can see that a tertiary carbocation will be most stable leading to the major product, and a primary carbocation will be least stable leading to the minor product.
Chemistry 