hello everybody and welcome to this a level chemistry video about Halla gino alkanes and in this video we will be having a look at the properties that allergy know alkanes have we will look at the nucleophilic substitution mechanism and then we'll have a look at the elimination mechanism and we'll finish up by having a look at a zone depletion and the role that halogen or alkanes played in the depletion of the ozone layer before we take a look at the properties and reactivity of halogen or alkanes it's definitely necessary to define what they are to do this let's take a look at an alkane and I'll pick the shortest one methane on here so this is methane and a halogen no alkane that is very similar to methane would be this one which we would call chloro methane because there is a chlorine group here attached to this carbon atom here and what is the rest of methane so essentially what's happened is a halogen has taken the place of a hydrogen atom in an alkane and that can happen once or more than once you can have halogen or alkanes that have got more than one halogen and different halogens in the same molecule halogen no alkanes don't really occur naturally but they are a really important synthetic compound they are used to make all sorts of things including refrigerants solvents pharmaceuticals and things like that they are the basis of PVC which is used in drain pipes teflon which is the nonstick coating on sauce pans and they are also important for a number of anesthetics and solvents now even though they have the same carbon skeleton as an alkane they are very definitely different in terms of their reactivity and the presence of that halogen and therefore the bond between the halogen and that carbon makes a hollow Gino alkane much more reactive than alkanes the general formula of halogen au alkanes is quite similar to the general formula of alkanes as you would expect if you have n carbon atoms where n could be any number the number of hydrogen atoms will be 2 n plus 1 and then there will be X where X is taken to be any halogen and it makes sense that it would be 2 n plus 1 instead of 2 n plus 2 because one hydrogen has been replaced by the halogen and the functional group of halogen of alkanes is often abbreviated to R X where R is a carbon chain of indeterminate length it can vary I've done a whole video about naming different functional groups and about isomers so I don't want to go into too much detail here but I think it is worth a quick recap about how we name halogen on alkanes so let's have this practice through some recap now when you name a halogen rlk you look for the longest on branched carbon chain which is a chain of two so this is going to be ethane and then you give the name the prefix depending on what the halogen is so in this case the halogen is fluorine so this is named fluoro ething and when you name halogen alkanes when you connect a prefix to the rest of the name you just run those two together and it is all technically one word fluoro ethane and the different prefixes are bromo chloro fluoro and i odo I've written them in that order because that's the order that they would appear a name if I change this floor Oh ethane and I take one of the hydrogen atoms out and I put a different halogen in instead what we have now is two halogens in the same allergy no alkane we have bromine and we have fluorine and since bromine is alphabetically before fluorine it needs to be bromo chloro ething and because the halogens could be both on the same carbon atom or on different carbon atoms we need numbers and so we call this 1 1 bromo because it's positioned on the first carbon and if the first carbon is the one that the bromine is attached to the carbon with the fluorine is to fluoro ethane and so the number shows which carbon has the bonded halogen in this case it is a little bit unclear whether it needs to be one bromo to fluoro or the other way around in fact it doesn't particularly matter and the convention is to go for one bromo to fluoro and have the numbers ascend a situation where the numbering is absolutely clear and it very definitely matters is shown in this next example here where we have a longest chain of three making this propane and we also have a chlorine atom attached to the second carbon here so there's our chlorine and it's on the second carbon one two one two doesn't matter which end you start numbering from so this needs chloro as the prefix and it needs a two because it's on the second carbon what about if you have multiple copies of the same halogen in different places in the molecule well if we take a look at this next example here we've got one carbon which makes it methane but you've got two iodine atoms and so this gets the prefix die for the two iodine's ioad o to show that it's his ID in that there are two of and then methane at the end so die iodoethane two iodine's attached to me thing this next example we've got the chain of two so this is definitely going to be Ethan but we've got three bromine atoms attached to it so this is going to be tri bromo be thane but this one is a little bit more complicated because we need to identify where the bromine are and so if we number this carbon number one and that carbon number two this one becomes 1 1 to try bromo ethane and that's very necessary because we could have 1 1 1 tri bromomethane or 1 1 to try bromo ethane and in general we try and keep the numbers as small as possible so it's 1 1 2 rather than 1 2 2 and the last prefix that we use is tetra so for instance if we had 4 chlorine atoms on this halogen or alkane that has got 4 carbons on it this would be tetra for the 4 chloro for the chlorine and then butane for the chain of 4 and then we'd need to identify the position and so the carbon atoms could be numbered 1 2 3 4 in which case this would be 2 3 4 4 etc chloro butane or alternatively we could number the carbon atoms from the other end 1 2 3 4 and this would be 1 1 2 3 tetra chloro butane and as I've said before when we have a choice between two different ways of doing something we want our numbers to be as small as possible so the green option is not the one that we go for and this would be 1 1 2 3 tetra chloro butane we're going to take a look now that's a few of the properties of halogen know alkanes and the first property is bond polarity now you should have come across polarity in the bonding and shapes topic and bond polarity is down to electronegativity and hopefully you'll remember that electronegativity is the ability of an atom to attract the electron density in a covalent bond so it's not just general ability to attract electrons is specific to the electron density or electrons in the pair from out of a covalent bond and hopefully you'll remember that the pattern in electronegativity is such that it increases as you go to the right in the periodic table and it increases as you go up the periodic table you don't need to memorize any of these electronegativities but you do need to remember that fluorine is the most electronegative of all of the atoms so if we take a look at these electronegativities what we're interested in is the bond between the carbon and the halogen so for all of the different potential halogen or alkanes we have a bond between the carbon and the halogen and this bond will be polar and the reason for the polarity is the comparative difference between carbon for instance and fluorine the difference in electronegativity is 1.5 fluorine is nearly twice as electronegative as carbon and what that means is the electron density in this covalent bond gets pulled towards the fluorine leaving the carbon electron deficient or slightly positive and the fluorine will be electron rich or slightly negative or we sometimes say Delta plus and Delta minus because that is the Greek letter Delta which means slightly positive or slightly negative if we move on to compare the carbon to chlorine bond the chlorine atom is less electronegative than fluorine but still significantly more electronegative than carbon so we are still going to get electron deficient and electron rich but the bond will not be quite so polar moving on to the bromine there is even a smaller difference still so I'm just representing the size of the dipole by writing the Delta plus and Delta minus more small and the iodine is even closer in electronegativity to the carbon so the dipole is going to be absolutely tiny on here but it will still be there so all of these bonds are polar and the bonds get more polar as you go up the group so C to F is the most polar and c2 iodine is the least polar moving on to a second property of halogen or alkanes which is solubility now solubility in water which is a polar solvent is determined by how strong the interactions are between the solvent water and the solute that we're trying to dissolve and even though the water solvent is polar and even though the halogen to carbon bond is polar the attraction is not strong enough to make the halogen or alkanes soluble in water so in fact we would say that howardino alkanes are insoluble in water and that's because the halogen alkanes have large regions the R group which are nonpolar and so the R group T the carbon chain so for instance this bit of my example skeletal formula appear is only able to form van der Waals forces these intermolecular forces are not the type of attractions that you would expect for something to be able to dissolve in water normally you'd be looking for something such as hydrogen bonding and so halogen or alkanes are in fact most soluble in hydrocarbons and that's because these are non-polar and actually they are so good at interacting with hydrocarbons that they're used in dry cleaning fluids to remove oily stains because oil is a mixture of hydrocarbons the final property we're going to discuss his boiling point and as with all families of organic chemicals the patent in general is a simple one the longer the chain the higher the boiling point and this is because the Van der Waals forces increase as the chain length increases because there are more electrons in the molecule again as with other organic families branching of the chain juices the boiling point and again this is down to the van der Waals forces this time they are weaker because of branching and this is because there is less surface area of contact meaning that even though there might be the same number of electrons there is less area for these intermolecular forces to act the next facts about boiling point is that it increases down the group now obviously you don't find halogen or alkanes in a group so to speak so what I mean by that is a halogen or alkane containing iodine will have a greater boiling point than the same halogen or alkane just having the iodine swapped for a bromine so I odo methane will have a greater boiling point than chloromethane for instance and this once again is down to the van der Waals forces the halogen is larger it has therefore got more electrons and therefore stronger van der Waals forces this outweighs the difference in polarity by the way so even though the carbons of fluorine bond is still very definitely more polar than the carbon so iodine bond it is more significant that the iodine atom is much bigger and so the polarity gets pushed to the side where the polarity is relevant halogen or alkanes to alkenes and so if we have halogen or alkanes and alkenes that have both got similar lengths so for instance chloro propane vs propane bond certainly has more significance but also it still comes back to Van der Waals forces because chlorine is significantly heavier than hydrogen it's got a relative atomic mass of 35 point five and it's got 17 electrons whereas the hydrogen that it's replaced has only got of one and one electron so so halogen no alkanes with more polarity and more electrons will have therefore a higher boiling point when halogen Amell came to react what happens almost every time is the carbon to halogen bond breaks in a level chemistry you need to be able to explain whether one halogen or alkane will react more quickly with another and the reactivity of halogen or alkanes ease with which the c2 X bond breaks if the bond breaks more easily it will be a more reactive halogen no alkane and the ease with which a bond breaks depends on two factors it depends on the polarity and it depends on the enthalpy if we just remind ourselves about the bond polarity and what we mean by that is the strength or size of this dipole carbons of fluorine is definitely the greatest then comes carbons chlorine then carbon to bromine and then carpenter iodine we've already explored this if we look at bond enthalpy bond enthalpy is how much energy is required to break a covalent bond and if we compare these bond enthalpies as we go down the group in the same way as looking at bond polarity we can see that without question the carbon to fluorine bond is significantly stronger than the carbon to iodine bond we can prove which of these two factors has the greatest effect on reactivity by doing a simple test tube reaction so if we got three test tubes with a different halogen no alkane in each one chloro butane bromo butane and iota butane and into each one of them we put aqueous acidified silver night great solution so AG no.3 is the formula for silver nitrate and so we put some of that into each of them and what happens is a hydrolysis reaction and what that means is the carbon to halogen bond breaks and the X minus ion is released into the solution and we make a precipitate of the silver halide which will be a white precipitate for the chloride butane and a cream precipitate for the bromo butane and a yellow precipitate for the iota butane and the crucial findings for this experiment is that the yellow precipitate forms first followed by the cream precipitate forming second and the white precipitate forming last and so therefore our conclusion is that reactivity increases as we go down the group in other words halogen or alkanes that contain iodine will react much faster than that same halogen or alkane that has chlorine instead and if we look back up at these two properties what that shows is that the significant property or the most significant property must be the bond enthalpy and the fact that I odo halogen Oh alkanes have the weakest bond and that outweighs the fact that the I odo have a Gino alkanes are the least polar the bond polarity once again is less significant just as it was for determining boiling point and solubility we're going to move on now to take a look at two reactions that have a gino alkanes undergo and we're going to look first at nucleophilic substitution and this type of reaction is called a substitution reaction because one atom or group of atoms is replaced by another and it's called nucleophilic substitution because a nucleophile causes the reaction to happen so what is a nucleophile well a nucleophile is something that has got a lone pair of electrons on an electronegative atom they are either negatively charged or the very least electron rich so Delta negative and how they act is they use their lone pair of electrons to form a new bond to a different atom and they're called nucleophiles because nucleo has got the same origin as the word nucleus and the nucleus of an atom is positive and file means loving or in this case seeking and so a nucleophile is taken to be positive seeking and so I've just mentioned that they use their lone pair of electrons to form a new bond they form a new bond with something that is electron deficient in other words with something that has got a delta plus charge such as the carbon atom in a halogen no alkane I'm going to keep this really simple and I'm going to have my hollow genome alkane as being chloromethane there are three nucleophiles that you need to be aware of for your course and they are the hydroxide ion which is negatively charged and has a lone pair on the oxygen which is how it is able to act as a nucleophile and when that happens the hydroxide ion takes the place of the halogen and we end up making alcohol in this case the alcohol is methanol the second nucleophile that you need to know of the cyanide ion and just like the hydroxide ion it is negatively charged and it has got a lone pair of electrons so it is able to act as a nucleophile and what we make in this case is we end up making something called a nitrile in this case the nitrile that we make has got two carbon atoms in it so this is called ethane nitrile and once again we've got our chloride ion being removed the final nucleophile that you need to know about is ammonia now ammonia has got a lone pair of electrons just like the other two but it doesn't have a negative charge and as a result of that the reaction is slightly different and because of this we actually need two ammonia molecules and what we end up making as a result of this nucleophilic substitution is we end up making an amine and this amine is called methyl amine and we end up making a slightly more complicated second product which is ammonium chloride unlike in the other two reactions where we make chloride ions which both have their own lone pair of electrons we make something slightly different here which is ammonium chloride and you need to be able to write equations such as these in exam situations not only do you need to be able to show this in exams you also need to be able to explain through something called mechanisms how these reactions are actually taking place and that's what we'll look at now and as I've said on the previous page the rate of substitution depends on the halogen with iodo being more reactive than bromo which is more reactive than chloro which is more reactive than fluoro mechanisms use something called curly arrows and what the curly arrows do is show the movement of electrons so if we have a look at the first of my examples from the previous page we've got our chloromethane and we've got our hydroxide ion notice by the way this I'm putting the negative on the left hand side near to the oxygen and think about what we know from the overall equation this chlorine is being replaced by this hydroxide ion so this is going to leave and so what happens is a new bond forms between the oxygen of hydroxide group and the electron deficient carbon so that carbon is Delta plus and that chlorine is Delta minus and then the second thing that happens is because this would leave the carbon with five bonds that's not stable so what has to happen is this carbon to chlorine bond breaks and so the chlorine leaves as the chloride ion and so as we showed on the previous page we produce our methanol and we produce our chloride ion what I should say is in an exam situation the only thing you would need to draw here is the red arrows my green is indicating extra explanation so this new bond forming and this bond breaking on here and in fact unless the language in the exam question says show clearly the structure of the organic product you in fact don't need to show the products at saw and this is likely to be a to mark question with one mark for each of the curly arrows showing the movement of electrons so in our second example very very similar things are going to happen the bond is going to form between the carbon in the cyanide nucleophile and the carbon in the halogen alkane and once again the carbon to chlorine bond in the halogen or alkane is going to break and then that's the end of the mechanism we're just going to make our products on here which once again I'm taking the time to show you because I think it's interesting seeing the triple bond in the nitrile part of this molecule and it also gives me an opportunity to remind you that I said on the previous page the halide ion this chloride ion is referred to as the leaving group the final mechanism that we need to look at for nucleophilic substitution is a little bit more complicated and it's down to the fact that the ammonia nucleophile is not charged the first two steps are absolutely identical we've got the new bond forming between the lone pair of electrons on the ammonia and the carbon that is electron deficient this curly arrow it's really important that it start the lone pair of electrons on the nucleophile and points to the carbon in the hollow Gino alkane you've got to be very specific with this and similarly the curly arrow that shows this bond breaking needs to very definitely start at the bond between the carbon and the halogen and then point to the halogen that is leaving doesn't have to really curl it just needs to have a really clear start and end point so the reason that this is more complicated should be apparent now because our nitrogen in this product has got four bonds instead of three and what that results in is it results in the nitrogen being positively charged and you must include this in the mechanism which is why I've drawn it as negative it's not immediately obvious why it should be positive but there's one really obvious reason the reactants over here have got no overall charge and we're producing a halide ion which is negative so that shows that the rest of this group here must be positive and the second reason is that the electrons at the nitrogen had complete ownership half before have been partially shared with the carbon this arrow remember is the movements of a pair of electrons and so nitrogen had full share of these two and now it's lost partial control with a bond to this carbon so it's overall is reduced in its electrons by 1 hopefully you'll remember that I said we need to ammonia's in this reaction and the reason that we need to ammonia is now going to become clear because this nitrogen is positive in order to become stable it needs to gain some electrons and it takes them from one of these bonds between itself and the hydrogen and that's the curly arrow there remember it needs to start at the bond and finish at the nitrogen and that restores nitrogen's charge back down to neutral and that's actually the end of the curly arrows that you need to show in the mechanism very rarely do you need to show this last one I'm going to show you just so you can appreciate and just so you can make sure you don't make the mistake nitrogen in this second ammonia comes along and it mops up that hydrogen that was going to be left by itself as h plus now H pluses can't exist by themselves they're too small they're too highly reactive something needs to react with them and it would be this second ammonia I'm gonna put it in green brackets now because it's unlikely you're going to be expected to show it in an exam but it is crucial that you know that it is the ammonia and it's not the chloride that gets involved the reason it's not the chloride is one really adjust probability if at the beginning of this we had a million molecules of our halogen of alkane and a million molecules of ammonia after this reaction has happened we will have one molecule with our nitrogen being positively charged and we'll have one chloride ion the chances of that chloride ion coming along and taking that hydrogen away are one in a million literally whereas the chances of an ammonia coming along and taking that hydrogen away are 999,999 out of a million and so obviously that is far more likely and that's why we need a second ammonia in this equation to give us this equation overall that I've showed you on the previous page now one last point of note is that the first ammonia that's involved in this mechanism acts as a nucleophile the second ammonia isn't a nucleophile it's accepting a proton and so it actually is acting as a vase returned briefly to the equations from before because I want to clarify something about the reactions first of all these are the equations that you would write if you've been asked to write the equations but if you've been asked to name the reagent that you would add to chloromethane you wouldn't say hydroxide ions and get the hydroxide ions from a solution of something like sodium hydroxide or potassium hydroxide dissolved in water so they would both be aqueous solutions you would need to warm it because otherwise this would happen too slowly and so to clarify we've got our reagent which is either of those two and we've got our conditions which is the warming the fact that it needs to be an aqueous solution halogen no alkanes don't mix particularly well with water and so it might be helpful to add a small amount of ethanol as well to help the chemicals mix so the ethanol is an additional solvent similarly to get the cyanide ions we would need to have a solution of potassium cyanide and this potassium cyanide is in a mixture of once again aqueous so we've got our water in there and we've got ethanol as well so sometimes confusingly called aqueous ethanolic so those are some of our conditions that is our reagent potassium cyanide and it also needs to be warmed to speed it up so reagents and conditions last of all the ammonia the reagents is nice and easy on this on the reagents is the ammonia the condition is that we need to have excess concentrated ammonia solution often the ammonia is considered to work better if it's dissolved in ethanol and we have quite a high pressure it's certainly considered to take place under pressure so to finish off nucleophilic substitution we have our halogen or alkane on the left hand side and the good news is it's going to be quite a predictable mechanism so in general where we've got mu for our nucleophile the nucleophile is going to attack the electron deficient carbon by moving its lone pair of electrons to form a new Bond X our halogen is going to leave by breaking the bond between the carbon and the halogen and the halogen is going to take both electrons remember and break that bond and so we're going to be left with our nucleophile attached to the carbon that was previously bonded to the halogen and the halogen has left over here as a halide ion remember that it is subtly different when ammonia is the nucleophile moving on to the second type of mechanism which is elimination we will see actually that this one is a little bit easier so in general we have got our halogen or alkane on the left-hand side and we're going to react it once again with potassium hydroxide or sodium hydroxide either is absolutely fine and we are going to produce an alkene potassium bromide and water this reaction is called elimination because a large molecule loses atoms or groups of atoms and I hope you can see that in this example the halogen or alkane has lost this bromine atom and it's lost one of the hydrogen's on an adjacent carbon atom on this occasion the hydroxide ion found in potassium hydroxide is not acting as a nucleophile it's acting as a base and remember a base removes a proton or an H+ ion the conditions for this reaction are different to the nucleophilic substitution conditions which is why we get a different product the potassium hydroxide or sodium hydroxide is dissolved in ethanol there is very definitely no water present and the mixture is heated rather than warmed since the reagent is identical and the conditions are quite similar it's important to know how you can achieve elimination rather than nucleophilic substitution so just to re-emphasize if we want to do the elimination we would heat rather than having it close to room temperature and we would have our solvent as ethanol whereas if we want a nucleophilic substitution to happen we would use water as a solvent and we just gently warm it that's just above room temperature you might have seen this reaction done in a lab we have our reaction mixture in a boiling tube and we heat it and the products form quickly and the products are gasses and so they travel through this delivery tube down here into a trough which has gotten upturned boiling tube or collection flask of some sort in there and the gaseous product rise up into here and they displace the water and the gas product forms here now the gas product should be an alkene and we can test for the alkene by adding bromine water and if we have indeed got an alkene the bromine water will go colorless because bromine water is yellow so more precisely it will go from yellow to colorless proving that we did indeed make an alkene the mechanism for the reaction in other words showing the movement of electrons by using curly arrows is I think really the most straightforward of all of them and the reason is that even though there are three curly arrows it is always the same three so we've got our hydroxide ion here that is acting as a base it uses its pair of electrons to attack this hydrogen and remove it then the pair of electrons that was holding the carbon to the hydrogen moves to between the two carbon atoms and finally the bromine to carbon bond once again breaks and so if I add in green we've got a new bond forming between the oxygen and the hydrogen the double bond is forming in here so that's another new bond and last of all the carbon some bromine bond is breaking and so we produce the bromide ion leaving group because it took the pair of electrons from the bond water and we make our alkene and it's better to show the alkyne having its bonds at 120 degree bond angles rather than at 90 degrees because that's far more realistic remember though that if you're asked to draw a mechanism you are only going to be expected to draw the left-hand side of this unless they say show the structure of the organic product and they're only really going to be asking you for those three curly arrows and of course and the atoms and charge that is in blue there are some slightly more complicated elimination processes and this is because sometimes you can end up make more than one different product for instance here with 2-bromobutane and the reason that you can make different products is that there are is that there are a variety of different hydrogen atoms that could be removed we could take this hydrogen atom or in fact any of these three hydrogen atoms they're all identical or we could take this hydrogen atom or this hydrogen atom those are the only ones that are options we can't take this one because it's on the same atom of carbon as the bromine is and so we couldn't form a double bond between this carbon and itself and we couldn't take one of these three hydrogen's because they are too far away from this carbon atom but we've got options for our green hydrogen atoms or our yellow and in fact both of these reaction mechanisms will occur because remember in mixtures of reactants we've got billions of molecules so all the possibilities will occur so if I show curly arrows for the green pathway first of all the hydroxide ion could remove this hydrogen atom and so that's what I'm going to show then the electrons in the carbon hydrogen bond will move to form the double bond here and the bromine atom will leave and will end up with the product being butte 1m and that is for the green pathway whether as the hydroxide ion could just as easily come along and remove this hydrogen when therefore it would be this pair of electrons that would move to form a double bond between the carbon atoms and this curlier would effectively still be the same but I'm drawing it as yellow and so this time we would get a different product and the product would be on this occasion buuuut - een so importantly we would get two different isomers I mentioned earlier that it was important to show the correct spatial arrangement around alkenes and the reason will become clear now because in fact butte 2e has got easy isomerism so we could make z butte - een or also II buttes oween which I won't bother drawing but I think you get the idea and so because the halogen oil cane was not symmetrical we end up with a mixture of isomeric products in reality even though you might try and emphasize one set of conditions over another you are likely to get a mixture of products when a halogen or alkane reacts with potassium hydroxide so just as a for instance if you've got to bromo propane you could if nucleophilic substitution happens so substitution nucleophilic you would make pro pan tool however elimination is likely to happen as well and so you are likely to make some protein too so you can heat and you can have ethanol as your solvent and you'll get more elimination happening and you can have aqueous potassium hydroxide and not get much above room temperature and you'll get more nucleophilic substitution happening but in reality you'll get a mixture of products and you would have to separate those by fractional distillation I want to finish this video by just looking at one type of halogen or alkane called chlorofluorocarbons or CFCs for short now these are halogen alkanes that contain chlorine and fluorine but no hydrogen CFCs are used as solvents and refrigerants but once the product they were being used in breaks the CFCs end up rising up to the ozone layer now I cover ozone and this reaction in my free-radical substitution video so to give you a just a brief summary zone is the molecule o3 and it decomposes in the upper atmosphere to make gin gas and the balanced equation is two ozone makes three oxygen and ozone is very beneficial because it absorbs ultraviolet light and ultraviolet light could cause things like cancer in humans but more profoundly life as we know it would not have evolved if the ozone layer hadn't been there the problem with CFCs was that when CFCs rose up to the ozone layer the ultraviolet light was the bond between the carbon and the chlorine to break resulting in the formation of very reactive chlorine free radicals this then caused a chain reaction where the chlorine free radical reacted with ozone to make the chlorine monoxide free radical and oxygen gas the chlorine monoxide free radical then reacts with a second ozone molecule to make two molecules of oxygen and another chlorine free radical and the key idea here is this will then repeat because this chlorine free radical has been regenerated because the chlorine is in effect acting as a catalyst as this free radical because it causes a reaction to happen in a different pathway but it doesn't get used up itself so overall if I just carefully strike through the things that are on the left-hand side and the right-hand side we can see that the overall equation is two molecules of ozone react together to form three molecules of oxygen which is the same overall equation as this decomposition that happens naturally except this happens much much faster because the chlorine is catalyzing the decomposition scientists spotted that chlorofluorocarbons were causing this problem of depleting the ozone layer and reported it to the government and eventually what happened is chemists carried out research and developed alternatives for refrigerants and coolants and these did not contain CFCs and then it became legislation passed by the government that we had to use CFC free solvents and to make sure that this problem was not going to get worse and whilst the issue has definitely improved we do still need to be mindful about holes in the ozone layer and we take particular notice of the skies above Antarctica but there are signs that there is a hole in the ozone layer developing over the Arctic rather than the Antarctic so it's really important that the use of harmful chemicals is regulated and monitored and we are vigilant for holes developing in the ozone layer okay that's everything for this video on halogen or alkanes I'll see you next time for alkenes bye bye