got a video here with the essential information you need for the alkanes so we'll start with the general formula of the alkanes so in the table a you've got the name and the molecular formula for the first 10 alkanes and these are what we call aliphatic alkanes so they just have open carbon chains so what is the ratio the general ratio between carbons to hydrogens it's CN h2n + 2 where the n stands for the number of carbons you can also get cyclic alkanes so they're alkanes that have formed rings so there's three examples on the screen there so we've got cyclopropane so you can represent it like this or sceletal formulas like this that's got the molecular formula c3h6 6 and then cyclohexane C6 h12 and then this one here is methyle cyclohexene and that's C7 h14 and hopefully you can see the general formula is going to be cnh h2n this time alkanes are what we call saturated hydrocarbons so there's a couple of examples on the screen we've got ethane on the left with two carbons and butane on the right with the four carbons they're called saturated because all of the carbon carbon bonds are single and they're called hydrocarbons because they contain hydrogen and carbon only little bit about the shape now and the bond angle that you would get in aliphatic alkanes so these displayed formula are two dimensional and sometimes students think that because we draw it like this the bond angle there is 90 de well it's not it's actually a tetrahedral shape with a bond angle of 109.5° and the same goes for ethane as well so in effect we've got two tetrahedrons sort of stuck together and all alkanes have the tetrahedral Arrangement around the carbons with that 109.5 Dee Bond angle and that's because each carbon has four bonding pairs of electrons around it so we just look at the carbon in methane we got a pair of electrons in each of these Calum bonds so they will all repel each other equally and that gives you that tetrahedral shape and you get that Bond angle of 100 9.5 de we move on to Boiling Points of alkanes now so we're going to look at how chain length affects the boiling point of alkanes so we've got propane on the left with three carbons and we've got hexan on the right with six carbons and there's the boiling points of those two alkanes so it's minus 42° C for propane and 68° C for hexane and so you can see that the longer the chain the higher the boiling point so why is that it's because alkanes have what we call induced dipole dipole interactions or London forces we can call them between the molecules and remember alkanes are nonpolar molecules now these get stronger as the molecules get longer because there are more points of contact between the molecules I'm going to show you a a representation of that in a moment so hopefully that'll make it easier to visualize so because these intermolecular forces are getting stronger as the chain gets longer more energy is needed to overcome the forces of attraction between the molecules we're not breaking bonds that's such a common mistake stake we're not breaking bonds we're just overcoming forces between molecules so we've got this slide to try and explain that so on the left here we've actually got four propane molecules these are this this is the sceletal formula for propane and here's the intermolecular force so it's a weak force of attraction between the molecules and then between these four hexan molecules that's a much stronger intermolecular force so it's going to take more energy to break this intermolecular force than this one so as a sort of answer for an exam you could say stronger induced dipole forces or London forces between hexane molecules than propane molecules and therefore more energy is needed to overcome these forces and that gives it the higher boiling point obviously a longer chain still so this has got six carbons if you had a longer chain say with eight or 10 the boiling point just keeps going up and up and up now if we look at the effect of branching on the boiling point of alkanes so these are what we call structural isomers they've got identical carbons and hydrogens five carbons 10 hydrogens but the degree of branching is different so this has got no branching this has got one branch this methy Branch here and this one's got two branches the boiling points you can see the one with the greatest brunching has the lowest boiling point so why is that so again I've got sort of a visualization like that so hopefully from the previous um slide like this you can get your head around what's going on on here the intermolecular forces between the unbranched molecules are stronger than the ones with branching and the more branching that we've got the weaker the intermolecular force gets so how do we explain that well if I could move these on the screen I'd be able to push these really close together and what we can see is is these unbranched alkanes have the most points of contact or you could say they have the highest ability for what we call close packing pack really close together and that therefore gives them the strongest induced dipole or London intermolecular forces out of these three then if we move over to this one here on the right with the two branches so this example has the least points of contact these branches effectively stop the molecules getting too close together so they can't close Packers efficiently that gives them the weakest induced dipole or London forces and that's going to give it the lowest boiling point really easy to separate those and obviously this one here has the one branch so it has like an intermediate ability to close pack compared to these two so we move on to combustion of alkanes now so we'll start with complete combustion so you can see there alkanes combust completely in a plentiful supply of oxygen producing carbon dioxide and water so there's the example unbalanced at the moment for butane I just want to show you how I balance these I've got a set order that I use so I'll start with the carbons four carbons in the alkan so that means we'll get four c2s 10 hydrogens will give us five H2O when we double that five we get the 10 and then I go back and I work out the oxy so we've got 4 2 is 8 + 5 is 13 and I just always use an improper fraction 13 / 2 times that 2 gives us the 13 oxygen we we need incomplete combustion happens when you have a reduced supply of oxygen there's two scenarios you can get carbon monoxide and water or you can get carbon and water so we'll look at both examples again we'll use butane so this time we're getting Co not CO2 so let's use that method so we need four CO's five Waters and then count up the oxygen we've only got nine oxygen now so 9 over two multiplied by two gives us the nine ores we need and the example producing carbon now again four C's 5 H2O that's not changed but now we've only got five oxygen on the right so we need 5 over two o2s to balance the equation