Alkanes Alkanes are aliphatic saturated hydrocarbons (no = double bonds, and atoms only). They are identified by having a ane name ending. The alkanes have 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 2 - Their general formula is n 2n+2 where n is the number of carbon atoms in the alkane. We need to recognise the first 10 alkanes by name and formula, and be able to recognise chains attached as branches to an organic molecule as being alkyl groups. Alkane Molecular formula Alkyl group Structural formula e.g. methane 4 methyl- 3 - pentane 5 12 pentyl- 3 ( 2 ) 4 - The cycloalkanes are alicyclic saturated hydrocarbons, having two less atoms, and a ring rather than a chain structure. e.g. cyclohexane 6 12 compared to hexane 6 14 Physical Properties: Temperature (K) 650 600 550 500 450 400 350 300 250 200 150 100 Variation of boiling and melting point of alkanes with chain length Room temperature For the straight chain alkanes: Low molecular weight alkanes ( 1 to 4 are gases at room temperature A little heavier molecular weight alkanes ( 5 17 ) are liquids at room temperature igher molecular weight alkanes ( 18 and above) are solids at room temperature. 50 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 hain length (-atoms) Trend in melting and boiling points: As the number of carbons increases, the melting and boiling points get higher. We say that the volatility of the alkanes decreases with increasing number of carbons in the alkane molecule. [volatility: the ease with which a substance turns into a vapour]. Explanation: The lack of polarity of alkane molecules means that only weak induced dipole-dipole forces (London forces) are present between molecules. There are no permanent dipole-dipole interactions or hydrogen bonds. As the number of electrons in the molecule increases (with increasing number of atoms) the strength of the London forces also increases. More energy is needed to separate the molecules, so melting and boiling points rise. Page 1
Effect of branching: For isoelectronic alkanes, the melting and boiling points decrease as the chain gets more branched. Explanation: As the molecule gets more ball shaped and less long and thin the molecules can t get so close together. There are fewer points of contact and so the intermolecular forces are weaker. The vapour is formed more easily, and so the boiling point gets lower. Shapes of molecules: The atoms in alkanes are held together by σ-molecular orbitals which lie along the axis of each bond, as shown here for an ethene molecule. The molecular orbitals can be considered as the overlap of two atomic orbitals. The two electrons in each σ-orbital attract both nuclei binding them together in a σ- bond. We need to be able to draw 3D displayed formulae for alkanes, so we used dashed bonds to represent bonds going into the plane of the paper and wedged bonds to show those coming out of the plane of the paper: Each atom in an alkane has a tetrahedral arrangement of 4 σ-bonds around it, so it has:! 109.5 o bond angles! free rotation about bonds! which makes alkane chains are very flexible heck your understanding: i) List all the physical properties of pentane you can, based on the information so far in this section. hemical Properties: General: alkanes are remarkably inert, showing no reaction with acids or alkalis for example. There are two reasons for this: very low polarity of - and - bonds as and have similar electronegativities, - and - bonds are non polar. onsequently the bonds in alkanes are not susceptible to attack by most common chemical reagents high bond enthalpies of - and - 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. ompounds that are noncorrosive to metals (lubricating oils), harmless to our skin (vaseline), and safe in contact with foods (poly(ethene)), are enormously useful to us. Reactions: 1: ombustion: 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. 4 + 2O 2 " O 2 + 2 2 O Δ c θ = -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. 2 4 + 3O 2 2O + 4 2 O or 2 4 + 2½ O 2 " O + + 4 2 O Page 2
Dangers of incomplete combustion arise because O binds permanently to haemoglobin, preventing O 2 and O 2 transport. O is therefore toxic, and is particularly dangerous because it is a colourless and odourless gas so hard for people to detect. heck your understanding: ii) Write a balanced symbol equation (including state symbols) for the complete combustion of hexane. iii) Write another equation to show combustion with a limited supply of oxygen, where equal volumes of carbon monoxide and carbon dioxide are produced. 2: Substitution reaction of alkanes with halogens: ydrogen 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 (free) radical substitution mechanism. onditions for the reaction: ultraviolet light is required (the amount present in sunlight is sufficient), so this is an example of a photochemical reaction. Application When chlorine is mixed with methane and exposed to sunlight, chloromethane is formed and hydrogen chloride gas is evolved: 4(g) + l 2(g) 3 l (g) + l (g) Mechanism of Radical Substitution: 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 l-l bond in the l 2 molecule. Absorption of light energy to break a bond is called photodissociation. The type of bond fission (breaking) is homolytic, resulting in two l radicals being formed: l-l (g) " l (g) + l (g) 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 l + - 3 l- + 3 The methyl radical is able to react with a chlorine molecule to produce chloromethane and regenerate a chlorine radical: l-l + 3 3 l + l Forming a new chlorine radical means the reaction 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 STEPS 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: l + l " l 2 l + 3 " 3 l 3 + 3 " 2 6 so a small amount of ethane is also formed Page 3
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: 3 2 3 + Br 2 " 3 2 2 Br + Br The bromine radical can attack any position in the alkane chain, so 2-bromopropane can also be formed by this pair of propagation steps: 3 2 3 + Br " 3 3 + Br 3 3 + Br 2 " 3 Br 3 + Br 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: 3 2 2 + 3 2 2 " 3 ( 2 ) 4 3 But because there are two types of propyl radical that can be formed, depending on which position in the chain was attacked, 2-methylpentane and 2,3-dimethylbutane are also possible products that can be formed in termination steps. 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: 3 2 2 Br + Br " 3 2 Br + Br 3 2 Br + Br 2 " 3 Br 2 Br + Br heck your understanding: iv) hlorine reacts with cyclohexane in the presence of light. Show the mechanism steps in this reaction. Bromine reacts with ethane to produce bromoethane when illuminated with light. v) Show the initiation step in the mechanism for this reaction vi) Show the propagation steps in the mechanism for this reaction vii) Show a termination step in which bromoethane is produced and suggest two reasons why bromoethane is not the only organic product in this reaction Page 4
Answers to heck Your Understanding questions: i) List all the physical properties of pentane you can, based on the information given: - liquid at room temperature (but very close to its boiling point) - higher boiling point than similar isoelectronic alkanes - fairly volatile - insoluble in water - flexible chain, can flex into different conformations - 109.5 bond angles throughout ii) Write a balanced symbol equation (including state symbols) for the complete combustion of hexane: 6 14(l) + 9½ O 2(g) " 6 O 2(g) + 7 2 O (l) - or quantities all x 2 iii) Write another equation to show combustion with a limited supply of oxygen, where equal volumes of carbon monoxide and carbon dioxide are produced: Equal volumes of O and O 2 mean equal numbers of moles of each gas produced therefore three atoms in each hexane make O and the other three atoms make O 2 6 14(l) + 8O 2(g) " 3 O (g) + 3 O 2(g) + 7 2 O (l) iv) Initiation l 2 " l + l Propagation 6 12 + l 2 " 6 11 + l l 2 + 6 11 " 6 11 l + l Termination e.g. l + l " l 2 v) Show the initiation step in the mechanism for this reaction Br 2 " Br + Br vi) Show the propagation steps in the mechanism for this reaction 3-3 + Br " 2 3 + Br Br-Br + 2 3 " Br + 3 2 Br vii) Show a termination step in which bromoethane is produced and suggest two reasons why bromoethane is not the only organic product in this reaction 3 2 + Br " 3 2 Br Ethyl radicals can combine to produce butane 3 2 + 2 3 " 3 2 2 3 Bromoethane can itself react by the same mechanism to make e.g. 1,1-dibromoethane 3 2 Br + Br 2 " 3 Br 2 + Br (N.B formation of 1,2-dibromoethane would be an equally acceptable answer) Page 5