So hang in there, keep watching these videos. I'm going to do my best to show you as many memorization techniques or understandings as possible in order to remember this very much a memorization unit. These are the objectives for 10.2.
Going over the writing conventions, the single arrow represents just a single electron that's been transferred. So you can see here And here we have just the single electrons. The double arrow means that both electrons have moved. Now always draw the movement of electrons.
Don't draw the movement of positive. Draw the movement of the negative electrons. This here does not represent an electron. This big dot here represents a radical.
It actually looks like this, which just means the valence shell isn't full. This is a revision from topic two, so be aware of... That as the complex. That's the charge of the complex and these are the ligands.
And lastly, these dashed lines mean it's not a complete bond, but there is a shared bond. They're delocalised electrons. This is better actually written like this.
It means it's going out into the page. This one's coming out towards you, so it's a much clearer, thicker line. This will help you give better 3D diagrams.
Alkane reactions now. Just some terminology. Catenation means that an element can form bonds with itself.
This is very important in organic chemistry because the carbon and the carbon-hydrogen bonds are very stable. They're non-polar, and they don't react. So this gives us the strong, stable base with which to form all our compounds. We will refer to the data booklet later, but this will give you very confident information here about. the results so you can see the stability of the molecule there.
It takes 348 kilojoules per mole to break it which shows you how stable the carbon-carbon bond. If we look at the carbon-hydrogen bond here that's also very high 414. So it's important to go back to your topic 2 and talk about the orbital. So there's large these large atoms have shielding so they the positive nuclear charge Hazen doesn't have a good ability to attract itself to the bonding electrons here. That's why this silicon has a much weaker bond enthalpy.
Combustion and incomplete combustion. Hopefully this is revision from before IB. Complete combustion is CO2 and incomplete combustion is carbon monoxide, which will bind much more strongly than O2 to your hemoglobin and so cause poisoning.
More terminology. Heterolytic means it's uneven, so that sort of splitting will cause one to be positive here and one to be negative. Homolytic splitting causes these dots here which means that they are radicals and they have an equal breakage here at the double bond.
Moving on to some alkane reactions now. Take special note of these red notes up here. This means that you must know the mechanism. So when you see these drawings, you must be able to draw that.
So that's the half. One, the half arrow here. So UV has come in and broken these bonds, and you can see this is chlorine here. If you look that up in your data book, that's a much weaker.
So if UV attacks a molecule, it's going to break off the chlorine because it's got the weakest bond enthalpy. And so you can draw it like this, and that's homolytic fission. So when you're drawing these mechanisms, you also need to know this terminology. Initiation creates the radical.
Propagation. means that there is still a radical created here, so that there will be more reactions. And termination means the radicals have disappeared. Now, the radicals themselves are broken by UV light in this particular instance, and the reason that they don't react with each other and go back to being how they were is because they're in a solution or an environment where there's heaps and heaps of these things.
So it's just reaction kinetics. The chances are this thing, when it's created, isn't the next thing it's going to bump into is something else and not itself. The second thing here is you can basically draw, because these are unstable, they don't have full valence shell electrons, they will pretty much react with anything. So you could come up with some more ideas here about other things that will react with anything it comes in contact with. So that will help you sort of work out.
Well, I don't have to draw it like this. I could draw something else. As long as it hits, whatever it hits with, it will react with.
Now, to terminate, just grab two radicals and give them full electron shells. And that finishes it off. In reality, you're going to get...
All of these organic chemistry is a mess. That's why we have so many separation techniques. So again, the mechanism needs to be known here.
The bond is only going to break with UV light. Once you get these radicals, it can then react with this methane molecule here. And so you draw the...
electrons here and you draw a line going here to make the attack. Alright, so that's Bromo, so you will see the Bromine here. The Bromine is brown with UV light that will...
then go colourless. Alken reactions now. Okay, you can see here that the double bonds are not as stable as the single bonds.
So the single bond is 3, 4, 8, and the double bond is 6, 1, 12. So if you divide that by... 2 you get 306 so per bonding pair it's weaker I think the main thing is there's a high degree of electronegativity here so if there's lots of negatives here it's going to be very easy for something positive to come in and do something there and so with any of these reactions all you're doing you don't have to know the mechanism here but all you're doing is you're breaking these bonds here and so you're taking an electron over here and one over here and some other electrons forming from some other thing. So in this case it's hydrogen.
So you break those form radicals and that's where you get the new bonds from. Memorization, nickel catalyst 150 degrees. All right so you need to probably do a list of things that you can't work out and work out ways to memorize that.
I don't have a way to memorize that one I'm afraid. It usually makes sense for heat to help reactions usually and some sort of catalyst. Again you can see you just always just break off the double bond and that's how you get your product.
Now alkenes do the same as alkanes except you don't need UV light so if you mix some bromine water which is brown and it probably goes off really quickly if you've done this in the classroom so I don't know if it's worked if your bromine's gone off and it will go colourless. The alkane itself will also do this if you have UV light remember. So here is such as just a mechanism for how it works both electrons moving over here. That's making this positive, which will be attracted to this electrodense region.
That will form a bond there, and then this will be a positive charge here to form this reaction. Again, the addition of water. So you can, H2O, you can put the H on one side, the OH on the other.
This one's a rather cruel one. If they ask you for the mechanism, which they haven't said in the syllabus, but this is how the mechanism is suggested. So concentrated H2SO4 on heat. If you're in trouble and you can't guess what the catalyst is, go for H2SO4 because it's the king of chemicals and needed for so many organic reactions.
Lastly, polymerization. This is just it reacting with itself. So this is the monomer here, the polymer here.
So I know how to do the brackets with the N there. Again, I'm assuming this is revision. So one goes over here, one goes over here, and it joins with itself.
as an addition polymerization. So in summary, here we have these here. What I suggest is you look at this and try and memorize it, cover it up and then draw it. And you need to draw everything perfectly so you need to pay attention to the reaction conditions here and the products.
And if you get one wrong, have a look at what the answer is. Get a new piece of paper, cover it up and write it again. Moving to alcohol reactions now.
Now again, this is combustion that we've covered in the past. This one here is complete combustion. And you can see that the more carbon atoms there are, more carbon bonds there are, of course, there's more energy stored in the molecule. The question is, per carbon, is that the most efficient? So what you'd need to do is divide these numbers by this by 8 and this by 7, this by 6, and have a comparison to see which one's the most efficient.
Moving on to the oxidation of our carbon. Now what I do is, you can see here with the tertiary alcohol you just need to realise that there's no way for this to have access to this because it's blocked off steric hindrance. So it's very difficult for that to form anything else that's more oxidised, you can't form a double bond there and you can't...
you can't break the carbon-carbon bond because it's too strong, so you can't do oxidation that way. So there's five bonds here, so that doesn't work. So what I do is I grab the molecule, say this one here, and I think, what other ways can I add more oxygens without destroying it? And so I can see... I could get rid of this here, alright, I could add oxygens here and the reason we don't add an extra oxygen on top of there is because that would then break, that would then have five bonds.
So we're not left with many options, we're left with a double bond O and that gives us an aldehyde and the carboxylic acid has slightly more oxygen in there so that's how you know the carboxylic acid comes second. This one here is the ketone and then we run out of bonds. If you look at the bonding involved here, this is dipole and so that will have weaker bonding then here we have a dipole as well and we have hydrogen bonding here. So this will have the weaker melting point. So if you're making this compound here, this one's going to come off first, and it's going to evaporate.
You're not going to get any of this. So what you do is you have this. inside a reflux.
So if you just want everything to an aldehyde, here, 23 degrees Celsius, as opposed to the hydrogen bonding here as ethanoic acid. So if you just want that, then you can just let it all evaporate off and do distillation so it'll just come off here. But if you want it to get further oxidized, what you have to do is invert it here which is a reflux, let it cool and drop back down and then further react to get more oxidized. Now you have to know the names of the oxidization reagents. So here these things get reduced.
So this is the memorization bit here. The orange to green. 6 to 3. Again I don't know a trick for remembering those so put that in your memorization ones, those ones there.
Condensation reactions now, now we've covered these a bit now, so any sort of small molecule that comes off is a condensation that condensates, gets rid of a small product. So a carboxylicate and an alcohol, which one of those three can you get? And you can see that that one is really the only option.
There's no nitrogen and there's no chlorine. So you join those two, concentrated sulfuric acid, generally if you can't remember what the conditions are, and so that's an esterification reaction that's forming an ester. Halogen and alkane reactions now. A nucleophile, that means it loves positive.
That must mean itself is negative. Compare that to an electrophile. That would be negative.
That would love negative, so itself, it would be positive. So a nucleophile is something that's negative, so that would be something like a H-, it's negative. It's going to attack a positive carbon, most likely.
The electrophile, it's positive, so it loves something negative. So that proton would be an electrophile. So just for standard level you're going to have to learn these reactions.
I think it would be helpful to look at the high level SN1, SN2. How do you know it's OH-? How do you know it's X?
Just as a pre-reference. Pre-emptive on that. Have a look at the carbon-oxygen bond and the carbon-halogen bond. If you can't remember which one it is, it's doing the substitution. If you look at your data booklet, you can see here that the carbon-oxygen bond is 358. it's quite strong.
Whereas if you look at your halogens, 3, 2, 4, 2, 8, 5, 2, 2, 8, not including fluorine, they're much weaker. So that's how I know which one's going to do which if I've forgotten which one is it going to be OH or a halogen that comes in that's going to do the substitution. So apart from that it's just a simple swap over which is what the word substitution means. Benzene reactions now.
Now specifically they've told you don't need to memorize anything here. Just know that the electrophile is going to come in and replace the H+. So here's the So here we have some examples, just grabbing the electrophile here.
Now this doesn't look like an electrophile, but if you're getting H2 coming off, so the O2-is joined with the H+, if you remove an O2-from that, that becomes positive in a more complex mechanism. Same with this, HCl. If you're removing a Cl-from there, that would be heteroelectric fission, so that would create in the more complex mechanism somewhere.
this will have a positive charge when it attacks the benzene.