Topic 8: The Hydrocarbons

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Name of Homologous Series

General Formula



































































Saturated hydrocarbons (Molecules made of carbon and hydrogen only and only contain single bonds)

 ·        Have a functional group C-C (A functional group is a group that reoccurs throughout a homologous series. In this case, alkanes are characterized by single C-C bonds, which are its functional group. A homologous series is a family of compounds that have the same functional group and similar chemical properties.)

 ·         Have the suffix –ane. The only difference between one alkane and the next is the addition of CH2 onto the chain, hence we get the formula CnH2n+2.

 ·        Are relatively unreactive(saturated). Can only participate in combustion or substitution reactions.  The reason for their lack of reactivity is their strong, non-polar single bonds (only sigma bonds) that don’t break easily to participate in other reactions (Such as addition).

 ·        The first 4 are gases, then they become liquids (as more carbons are added, boiling points increase. This is because of the overall increase in Van der Waals forces.)

More detail on structure…


Alkanes are formed completely of single bonds (saturated). Thus, you always have a CH3 group capping each end of an alkane chain. σ-bonds allow rotation, thus, alkanes may have different orientations. This means alkanes cannot have geometric isomers.


As for bond angles, they can easily be derived by this example:





This picture may look a bit different to what you may usually see butane depicted as. Whenever the exam question asks you to draw a diagram showing the SHAPE of the molecule, always draw it like the above, with the correct bond symbols (These are described in topic 7 notes). Now, what about the bond angle?

ALL the bond angles in alkanes are 109.5°. You may have thought that the angle in C-C was 180°, but you can think about it this way.


You know that the CH3 at the end forms a tetrahedral shape, like methane. This is because you have four bond pairs of electrons around the carbon, therefore they repel as much as possible to form a tetrahedral shape, of bond angle 109.5°. The C-C bond is no different! Around each carbon in the chain, there are 4 bond pairs, thus, each bond angle is 109.5°.




The inertness of the alkanes can be easily demonstrated by the absence of reaction with bromine water, potassium permanganate, sulphuric acid and alkali. It does however undergo combustion, substitution and ‘cracking’.




Alkanes burn readily with a clean flame. The cleanliness of the flame distinguishes it from alkenes burning, as they generally burn with a sooty flame.


Under normal conditions, the alkane usually burns in complete combustion, forming Carbon dioxide and water:


Eg. CH4 + 2O2 --> CO2 + 2H2O


But if there is a lack of oxygen, it may undergo incomplete combustion. This produced toxic carbon monoxide.


CH4 + ?O2 --> ?CO2 + ?CO + ?H2O




Alkanes, in the presence of sunlight (UV light more specifically) can undergo a substitution reaction with halogens. There are three stages to this reaction:


-Chain initiation

-Chain propagation

-Chain termination


In chain initiation, a light particle (Photon) strikes a chlorine molecule. The chlorine molecule is split into two chlorine free radicals. A free radical is a species with an unpaired electron (odd number of electrons). They are highly reactive because they are electron deficient. The process by which they split is called homolytic (same) fusion. Both chlorine atoms are identical in terms of electron number. They are neutral, as they have the same number of protons as electrons.

The equation for initiation is:


Cl2 --> 2Cl•     


Since it is entirely made up of bond breaking, it is similar to atomisation and is endothermic.


The next step is chain propagation. This is composed of  2 reactions:


Cl• + CH4 --> HCl + •CH3


•CH3 + Cl2 --> CH3Cl + Cl•


Both reactions are related. The chlorine radical is produced by initiation, but the methyl radical is produced by the first propagation reaction. Because of the electron deficiency of the radicals, whatever they react with will also release another free radical. There aren’t enough electrons to distribute.


So, how does the reaction stop then? The termination reactions go as follows:


Cl• + Cl• --> Cl2


•CH3 + •CH3 --> CH3-CH3


Cl• + •CH3 --> CH3Cl


As you can see, the fact both are electron deficient means they just bond, without any bond breaking of some sort. This means the termination reactions are entirely exothermic.


A few things to notice:


-          The initiation step will have the same enthalpy change as the first termination step shown above, as it is the same bond that is formed and broken.


-          To recognise easily which is which, the initiation is normal --> free radical, propagation is normal + radical, and termination is radical + radical.




The cracking of alkanes is necessary to produce shorter chain alkanes or even alkenes. The heating (In the experiment we did, ONLY THE CATALYST IN THE TUBE IS HEATED) of the alkane over a catalyst (Aluminium oxide) can do two things:

-          Either release the corresponding alkene to the alkane and hydrogen gas

-          Or release a mixture of alkanes and alkenes.


That’s the best thing about cracking, you could make up your own equations, as long as they balance.



 ·        Unsaturated hydrocarbons (Have double bonds therefore enabling them to participate in many different reactions).

 ·        Have a functional group of C=C

 ·        Have a suffix of –ene

 ·        They start from ethene, as a minimum of two carbons are needed to make a double bond.

 ·        The first 3 are gases, then they become liquids.

More detail on structure…


Alkenes are different to alkanes in terms of bonding and structure in every possible way. The contain at least one double bond (Made of a sigma bond and a pi bond). A double bond does NOT allow rotation. Therefore, in some cases, you may have two isomers, that although may be structurally the same, may look different due to the fact the double bond won’t allow rotation. The best example is but-2-ene:


As you can see, the one on the left has both CH3 groups on the same side. This is called cis-but-2-ene. The other one has them on opposite sides, so is called trans-but-2-ene. These are geometric isomers. Try and draw but-1-ene out the same way as above, and then you will see why it doesn’t have any geometric isomers.


As for bond angles, I will use propene as an example:


As before, the bond angle in the CH3 is 109.5°. But, what about the rest? All the rest are 120°. The reason for it is that around each of the other two carbons, there are 3 bond centres, and no lone pairs. This means the electrons spread themselves out as much as possible, and this is forming a trigonal planar shape, ie. A separation of 120°.



Because of the double bond, alkenes can undergo quite a few more reactions than alkanes:




No different than the alkanes, except they burn to produce a sooty flame.




The oxidation of alkenes by potassium permanganate produces diols. The potassium permanganate turns colourless if acidified, or brown if alkaline. Diols are alcohols with two hydroxide groups (Check Topic 2 In the book). The hydroxides are bonded to the two carbons with the double bond.


-Bromine Water


In the absence of sunlight (To prevent any free radical substitution) a reaction occurs between halogens and the alkene:


C2H4 + X2 --> C2H4X2


This type of reaction is called electrophilic addition. An electrophile is responsible for the occurrence of this type of reaction.


It is an electron deficient species that forms a bond using electrons from the carbon it attacks. The puzzling thing in the above equation, is that the halogen molecule is not electron deficient! So how does electrophilic addition occur? When the molecule approaches the alkene, it is polarised by the pi-bond in the double bond. This means that one side of the molecule (halogen) will have more electrons than the other. The electron deficient side forms a covalent bond with the carbon, using electrons in the double bond. This removes the pi-bond, leaving a single bond.


It is better to explore electrophilic addition via the reaction between alkenes and sulphuric acid.



-Sulphuric Acid


Acids contain the most common electrophile: H+ ion. The ion, being electron deficient, attacks the double bond, and forms a bond with a carbon. The addition of a proton to the molecule means the entire molecule becomes slightly positive. It is now called a carbocation.  It is highly unstable, and bonds with the negative hydrogen sulphate ion from the sulphuric acid. This forms an alkyl-hydrogensulphate.