Transcript for:
Aromatic Substitution Reactions Overview

first let me give you an overview of what we're going to cover in this video we're going to talk about orthop Pera and metad directors uh which groups are activating deactivating uh electrophilic aromatic substitution and the difference of that with nucleophilic aromatic substitution reactions we're going to talk about how to rank compounds in terms of their reactivity we're going to go over a list of reactions that you need to know for your tests we're going to go over some synthesis problems mechanism problems and then towards the end we're going to focus on nucleophilic aromatic substitution that's going to be like the last 5 10% of this video but the other 80 90% will be focused on electrophilic aromatic substitution so let's begin on the left side we have the orthop parad directors and on the right side we have the metad directors just so you know Ortho is one two and Pera is 14 meta is 13 I'll explain how that will apply soon uh shortly now notice that all of the activating groups except the weekly activating ones they have lone peers on the first atom aka the the aiming group has a lone paer the O3 group has a lone pair even the Amite group has a lone pair when you see that it's strongly activated and it's usually an orthopair director however if you look at the Helens they have loone pairs too now they're weekly deactivating but because of the presence of the lone Pairs and how they can donate electron density by residents the entire left side is uh is orthop pair directors The Only Exception is the r groups even though they don't have loone pairs they are weekly activating because they donate electron density to considering um by means of the inductive effect rather than resonance now if we look at the groups on the right these are the metad directors and notice that in the first atom that's attached to the ring the Benzene ring is like right here there's no lone pairs on the first atom connected to the ring in fact all of those atoms are partially positive this carbon atom is partially positive because it's attached to oxygen which is partially negative the oxygen atom is pulling electron density away from the carbon and that creates a partial positive charge on the carbon same thing here too now these groups are strongly deactivated if you look at the sulfur it's being pulled by three oxygen atoms so it has a a very powerful partial positive charge same thing with the Nitra this carbon uh um it's being pulled by the nitrogen the nitrogen pulls the electrons toward itself leaving a partial positive charge on the carbon atom now if you look at the uh groups on the left the orthop par directors this nitrogen has a partial negative charge same thing is true with this nitrogen and that oxygen so the benzing ring it's attached to an atom that has a partial negative charge and so those um Bening Rings tend to be activated because they can donate electron density to the uh to the Benz ring now even though Florine is well Florin also has a partial negative charge so it's also an orthopair director now notice the difference between nh2 which has a partial negative charge and NH3 even though in both cases nitrogen is Electro negative here we have a positive formal charge and it converts from a strongly activating group to a deactivating group group even the Nitro group the nitrogen has a positive charge one of the oxygen atoms has a negative charge but you can see a pattern here all the metad directors the first element has a partial positive charge or a positive formal charge and for the orthopair directors the first element usually has a partial negative charge between carbon and hydrogen carbon is more electr negative than hydrogen so carbon has a partial negative charge hydrogen is partially positive with respect to carbon so do you see that Trend all of these first atoms for the orthopair directors have a partial negative charge and on the right for meta directors they have a partial positive charge now the next thing we're going to go over is uh we're going to make a distinction between electrophilic aromatic substitution reactions and nucleophilic aromatic substitution reactions so EAS that's going to stand for electrophilic aromatics substitution reactions in this reaction a hydrogen will be replaced with an electrophile so therefore it's called electrophilic substitution we replace a hydrogen atom with an electrophile and it's aromatic because we're dealing with aromatic Rings like benzing Nas stands for nucleophilic aromatic substitution for these types of reactions we're going to replace like a a bromine atom with a nucleophile in this case the nucleophile could be like hydroxide we're going to talk about this towards the end of the video so that would be an example of a nucleophilic aromatic substitution reaction for EAS this is the electrophile therefore the Benzene ring is the nucleophile and nucleophiles tend to be electron so the more electrons you add to a nucleophile the more nucleophilic it will be so think of the electron donating groups that we talked about in the last slide such as like the the O group or the nh2 group if we put an O group here it can donate electron density to the ring making it even more nucleophilic and that's why um whenever you have a lone pair here it can make the ring more nucleophilic and therefore it's usually activating with the exception of the halogens if you add like an NO2 group um those are electron withdrawn groups the metad directors and so they deactivate the ring because they make the ring less nucleophilic for nucleophilic aromatic substitution is the other way around the alcohol the O group would be a deactivator because in a nucleophilic aromatic substitution the benzing ring act as the electrophile instead of the nucleophile so an o or an nh2 group they would deactivate the ring because they make it more nucleophilic and less electrophilic however um groups such as NO2 and uh like Carbono groups s so3h the metad directors they would make the ring more reactive towards nucleophilic aromatic subtitution because they make it more electrophilic they pull electron density away from the ring so I just want to go over that so let's say if you get a question and let's say it asks you to rank the following compounds in order of increasing reactivity towards EAS electrophilic aromatic substitution so let me give you a list of compounds feel free to pause the video and try these questions yourself as well so towards EAS we're going to say that um group number one is the most reactive and number four is the least reactive so for electrophilic aromatic substitution the benzing ring is the nucleophile so the activating groups the orthop par directors they make the ring more nucleophilic so phenol will be an example that would be number one number four is the Nitro group it's a powerful strongly deactivated metad director number three is the alahh it's a moderately deactivated metad director number two is the floring group because it's weakly deactivated and it's an orthop par director but this is number one the most reactive now if we were to rank the reactivity of these compounds towards nucleophilic aromatic substitution it's going to be the other way around in nucleophilic aromatic substitution reactions the Benzene ring is the electrophile so electron withdrawing groups such as NO2 will make the ring more reactive so this would be number one this will be number two and that's number three phenol would be the worst for nucleophilic aromatic substitution okay now let's talk about why um phenol activates the ring towards electrophilic aromatic substitution so there's phenol it has two lone pairs now and let's say here's an electrophile so this is an EAS reaction and therefore the Benzene ring is a nucleophile phenol well the alcohol is an electron donating group and it activates the ring because it makes the ring more nucleophilic if we draw the resonance structure it can donate a pair of electrons and that would make this ring have a negative charge as we can see it has a carban instead of a carbo cation so the ring is more nucle philic because of the negative charge it's more electron now let's look at NO2 let's see why it deactivates the ring towards an EAS reaction this oxygen has three lone Pairs and that one has two that oxygen has a negative charge and nitrogen has a positive formal charge so here's the electrophile which has a a positive charge and we're still considering we're still analyzing an EAS reaction so the Nitro group can withdraw electron density from the ring so we call it an electron withdrawn group it can pull this double bond toward itself and this is the uh reson structure that we're about to draw and so now the ring has a positive charge and this positively charged ring does not want to react with the electrophile because like charges they repel each other and that's why NO2 deactivates the ring it puts a positive charge on the ring and it doesn't want to react with an electrophile it wants to react with a nucleophile a nucleophile would definitely react with that carboon so that's why uh electron with drawn groups work better for nucleophilic aromatic substitution they don't work very well for EAS and that's why they're deactivating groups or they they deactivate the ring towards EAS reactions but the phenol like we considered before it created a carbo a carban which had a negative charge which is attracted to an electrophile which has a positive charge opposite charges attract and that's why um phenol is an activator towards EAS okay so now that we covered that topic let's talk about the reactions that you need to know for this test so I'm going to go over a list of reactions and for now just feel free to write these down and just make sure you know them and then I'll show you how to apply it soon so the first reaction we're going to talk about is nitration now you can look up the mechanism but I'm just going to give you the end result of the product in this reaction a hydrogen atom is going to be replaced with an NO2 group and that's nitration the next reaction is bromination br2 and f br3 f br3 is the Catalyst that makes this reaction work it helps generate a an a bromine electrophile basically a bromine with a positive charge it replaces that hydrogen with one bromine atom but here's a question for you let's say if you add bromine to Nitro Benzene this is called bromo by the way and whenever you see an NO2 group it's a Nitro group what's going to happen if you add bromine to Nitro Bing now whenever you have a substituent on a ring it's going to tell the other group where to go now remember we talked about orthop and meta directors NO2 is a metad director and it's a deactivator so this ring is less reactive than the original ring however it can still work now we said that metad directors are one three directors so let's say if this is carbon 1 the NO2 will direct the bromine to go towards carbon 3 we can also count it this way this is also the meta position now these two positions are equal so if I put a bromine atom on both on either um we're going to get the same product so I'm just going to put it on one of them so that's what metad directors do they direct any new electrophile to the metap position so that's the product of that reaction now let's say if we perform nitration on uh bromobenzene where is the NO2 group going to go now bromine it's it's a weakly deactivating group but it's still an orthop paric group Ortho is in the second position relative to the uh bromine atom it's one two is Ortho 14 is per so the NO2 group can be directed in this position or on carbon 4 so we can get a mixture of products so the first product I'm going to draw is the ortho product which looks like this so if you want to name that compound it's um Ortho bromo nitrobenzene and let's see if I can fit the other product I'm running out of space here so this product is called Paran Nitro um parab bromon Nitro bending we have to alphabetize it B comes before n so here you get a mixture of products and that's why you need to know all of that information on the last page because you need to know which groups are orthopair directon and which ones are metad directon but I just want to give you a preview of how to apply um those Direct in effects but let's go over some more reactions that you need to know the next reaction that you want to add uh to your list that you need to memorize is chlorination if we add chlorine with the aluminum chloride leis acid Catalyst we're going to replace the hydrogen atom with a chlorine atom the next reaction that you need to add to your list is iodination so if we add I2 with nitric acid it simply replaces a hydrogen atom with an electrophile which is iodine the next reaction you need to know is sulfination if we had SO3 with h2so4 or just h2so4 with heat um you're going to get the same product which is basically s so3h now keep in mind if you have two molecules of h2so4 um actually even just one molecule of h2so4 if you add a lot of heat to the reaction H2 s04 can decompose into water and sulfur trioxide so pure sulfuric acid could generate the sulfur trioxide electrophile and that reaction could work so you need heat to make it work however this reaction is reversible for example if you add h3o+ this will convert back to the Benzene form so that process is desulfination the next reaction we're going to consider is the fro Craft's alation reaction and we're going to spend a little time talking about this one so let's say if we add a methyl chloride with a Lewis acid Catalyst basically the ls acid catalyst takes away the chlorine atom and then you're going to get the ch3 with a plus charge and the Benzene ring is going to attack that electrophile so the electrophile in this reaction is this it's a methylation and this is going to be your product now you can add pottassium panate in the presence of an acid Catalyst and this is going to convert it well maybe that's not a catalyst because it's water is going to be consumed in the reaction but this will be converted to a carboxilic acid so that's how you can make benzoic acid from Benzene you can add methyl chloride then oxidize it with potassium PR magnate now instead of adding meth ethylchloride excuse me methyl chloride you can add eth chloride with an aluminum Catalyst and then you'll get C2 ch3 instead and then following that let me start on a new page so once you add your Etho group you can add NBS NBS is a radical reaction and it doesn't react with the Benzene ring it reacts with the benzilic hydrogen outside of the Bening ring and this is useful because you can create a lot of different products with this process what's going to happen is it's going to replace a hydrogen atom with a bromine atom then after that reaction you can add sodium hydroxide or any other strong base and what's going to happen is you're going to have an E2 elimination reaction hydroxide is going to grab this hydrogen this is going to form a bond well those electrons will be used to form that double bond and the br will be kicked out and so you're going to get styrene basically it's like a a double bond outside the benine ring at least I believe that styring sometimes my memory of compounds uh fails me but I think that's what it is but you can get that product too and from styrene you can make so many other products you can do any type of alen addition reaction you can add hbr you can add H2O Plus turn into an alcohol bh3 thf hydroboration oxy mercuration um all sorts of reactions you can perform with an alkan but you're going to have to follow these steps to get to styrene first now let's say let's go back to Fredo craft calculation what's going to happen if we add instead of methylchloride or ethyl chloride what if we add propy chloride with an Alum an aluminum acid Catalyst here's what we will not get we won't get this product instead we're going to get uh this one here's why the aluminum acid Catalyst it takes away chlorine atom and when chlorine leaves it creates a positive um carbo cat but it's primary and primary carbo cation are not stable especially if it's adjacent to a secondary carbon what's going to happen is you're going to have a hydride shift and so your plus charge is on a secondary uh carbon and so the Benzene ring it attacks there so that's why um it's attached to the secondary carbon well right now it's tertiary but before it was it um before it's connected to the Ben ring it was secondary now the methyl cine that we formed before that that can't rearrange and the ethoca won't rearrange either because both carbons are primary so you don't have that issue with methyl and ethol chloride with the fetra ulation but when you start having like propo chloride or buto chloride you need to watch out for rearrangements the Benzene ring is going to attach to the more substituted carbon atom of that structure so now let's talk about the Fredo Craft's isolation reaction or acation reaction so here we have an acid chloride and our aluminum chloride acid catalyst so the mechanism is very similar but we're not going to go into detail like too much detail into it but what I want you to know is that the Catalyst takes away the chlorine and then the Ben ring is going to connect to this carbon so this is going to be the product of this reaction so this turns into a keto next you can use the clemon reduction which is an alloy of zinc and Mercury with hydrochloric acid and it's going to reduce the Ketone to an alkane so notice this is how you can make propo Benzene you don't want to use um propo chloride with aluminum chloride you won't get propo benzene instead um you're going to get this you get isopropylbenzene but if you want to make propo Benzene use the Fredo crafts isolation reaction instead of the alation reaction because this one won't rearrange now you can also add a turbo chloride now because the chlorine on a tertiary carbon um it's not going to rearranged so you can get Turbo Benzene but now let's talk about oxidation of benzene Rings let's go back to a K4 with HD+ let's say we have a turb group a methyl group an ethyl group and this long chain um isopropyl group if you add potassium paginate with h3o+ all of the substituents all of the carbon atoms will be oxidized except this one this carbon has no benzilic hydrogens it's coronary this carbon has benzilic hydrogen this one too and this one too all of the other structures will be oxidized so this is going to be the product of this reaction now we talked about this reaction before we use methyl chloride then once we add potassium magnate it turned into a carbic acid so every carbon structure outside of the benzing Ring except for the tbua group will turn into a carboxilic acid regardless of how many carbon atoms it has attached to on a chain this entire carbon structure will be cleaved same thing here too it doesn't matter what other carbon atoms is attached to it it's going to turn into um benzoic Ain or a carboxilic acid functional group the next reaction we're going to talk about the Gamin Coast reaction I believe that's how you pronounce it um let's say if you had carbon monoxide and hydrochloric acid with an aluminum chloride Catalyst and copper chloride basically your intermediate is an acid chloride which looks like this however this intermediate is not stable and you can't really store it so you have to make it in the process of reacting it with a Benz ring but it works just like the Fredo isolation reaction where um pretty much we're going to take off the chlorine and add this group to the Benz ring so this is a good way of making Benz alahh um but you need like the reactions conditions are intense like you need high pressure and high temperature to make this reaction work but there are other ways of making balide though but this is one way to do it so now let's talk about grin reactions uh with Benzene rings so first let's add uh bromine with febr3 to make bromo Benzene now what do you think is going to happen if we add magnesium if you add magnesium uh to the structure magnesium inserts itself between the carbon and the Brom atom so now we have a grard and we can react as this is we can react as gred with many other reagents by the way this is called Phenom magnesium bromide so let's say if we start with Phenom magnesium bromide and let's say we react it with water if we react it with water um you need to be aware that this carbon has a partial negative charge so it's nucleophilic it's going to grab a hydrogen and expel hydroxide so basically this is going to convert back to Benzene at this point the Magnesium which is basically a plus two charge is an ion and bromide and hydroxide they're just going to form a salt so MGB r o but you don't have to worry about that because that's going to be somewhere in the solution the product of inches is Benzene so that's how you can convert bromo Benzene into Benzene turn it into a Griner agent and then add water um if you add Dum which is D2 um d2o like basically uh deuterated water the mechanism will work the same way but instead of getting uh Benzene well this is going to be similar to Benzene hydrogen andum or isotopes of each other but you can get this product but let's continue with thanom magnesium bromide we can also react with carbon dioxide if we do we're going to get um benzoate and if we add H2O plus we can pronate the carboxilate ion and so now we have benzoic acid so that's another way you can make benzoic acid is by reacting a grard with uh carbon dioxide let's say if we react with ethylene oxide or this two carbon oxide this will extend the carbon structure by two carbons and so this oxygen will have negative charge and then if we add h3o+ this will turn into a primary alcohol now what we can also do is using a grard we can add it to an aloh haly so this would be an sn2 rea action so this is another way to alcate the benzing ring if you don't want to use feto crafts alation or isolation and as you can see there's hardly any rearrangements now for this to work you need to use um a primary aloh haly if you use a secondary or tertiary alcohal you could get a E2 reaction and um basically you'll just get Benzene plus an Aline so make sure you use a methyl or primary aloh helite for that reaction to work now let's say if we add it to like Benz alide Or Another carbon compound the Gren could attack this carbon and then that can open so now we have like two Benzene Rings which looks like this and then add an h2+ we could turn that into a secondary alcohol now let's talk about some other reactions that you need to know so let's um use nitration and let's make nitro Benzene now from Nitro benzene you can make analine by using any one of the following reactions you can use uh a metal an active metal with an acid such as iron metal and hydrochloric acid you can use zinc metal with hydrochloric acid or you can use a tin metal with hydrochloric acid you can also use hydrogen gas with a platinum or padium cadus what's going to happen is the Nitro group will be reduced to um an aming group or an nh2 group so right now this was a strongly deactivating group the Nitro group and now we have a strongly activating group The aiming group at this point if we add sodium nitrite with hydrochloric acid under very cold conditions um we can get a diazonium salt this reaction is called D diazotization I believe but it looks like this this nitrogen has a positive formal charge and it forms an ion pair or salt with a chloride ion so once we have that diazonium group we can create a lot of different products so we can add copper bromide and that's going to put a br on it we can add copper chloride and that's going to put a CL you might be thinking okay I know how to put a bromine on a ring and a chlorine on a ring so what's the point of this reaction later in this video I'm going to give you a problem and you're going to see why this reaction is important by the way whenever you react to daium salt with with let's say a copper One Salt like copper bromide copper chloride or copper cyanide that is referred to as the sandmire reaction or it's known as a semi reaction the N2 group is a very good leaving group because it leaves his nitrogen gas and nitrogen gas is very stable here we're going to replace the N2 group with a cyanide now there's some more vactions that I want to talk about let's see if I can fit it in here if we add um hbf 4 which is basically H+ and BF4 minus we can replace the N2 group with a floring group and there's two more we need to go over if we add h3o+ we can completely get rid of the N2 group so basically actually no we can get rid of it but we're going to replace it with uh an alcohol so that's how you can get phenol the last one is uh H3 P2 and that's how you can go back to Benzene it's going to replace the N2 group with a hydrogen so as you can see there's a lot of reactions that you need to know for this test okay so let's focus on applying some of these reactions so let's say if you started with Benzene what's the product of this reaction if you add uh SO3 with h204 this is sulfination and the product is simply so3h we can put it anywhere on the Benz ring now let's say if we continued and we added uh methylchloride with al3 this is the Fredo Craft's alation reaction and the SO3 group will determine where that metal group is going to go so now we have to go back to the very beginning of what we talked about so3h is it an activator or a deactivator and is it an orthopair director or a metad director so3h is strongly deactivated and it's a meta director and meta is basically on the third carbon from the directing group so this is carbon one that's two this is carbon 3 this is also carbon 3 so the methyl group can go here or here but because of symmetry um those two positions are identical so we simply have to put it on one of them so we're going to put the method group here now let's say if we continued and um let's say we add let's say if we choose to add chlorine with al3 where is the chlorine going to go so let's look at the uh so3h group it's a metad director so it can only direct the chlorine to this position now the ch3 group is an orthop parad director Ortho is one two so this is Caron if that's number one this is number two so the ch3 group could direct the chlorine atoms to go here or here the par of position is 14 so this is number three this is number four the methyl can also direct the chlorine group to go here as well so chlorine is being directed to all four positions which position will it go to well for one thing it's doesn't really want to go here because of these two bulky relatively bulky groups there's too much uh steric factors over here so it's going to approach the benzing ring from the uh from a site that's more accessible which is the left side the left side is more accessible than the right side so we're not going to consider this position it's too sterically hindered now both the SO3 H group and the methyl group are directing it in different locations however who who is stronger is the methyl group stronger than the uh sulfonic group or is it the other way around so now we got to go back to that chart the methyl group is an R Group which is weekly activated the S so3h group is a strongly deactivated group so methyl activates the ring more than the so3h group so methyl is going to direct it it's going to tell the chlorine where to go because it's more activating than the so3h group so chlorine will go either here or here so we're going to get a mixture of products so whenever you have to decide like where or which group is going to like determine where the new group is going to go look at which one is more activated in this case methyl is more activated so it's going to tell where the chlorine should go or it's going to direct the chlorine uh where to go on the ring so this is these are the two products that we're going to get for this reaction so let's try another example like this so starting with Benzene what is the product of this first step if we had nitric acid and suric acid by now you know this is nitration and we're simply going to add an NO2 group but let's put it on the top we could put it anywhere on a ring now let's say if we add um a three carbon acid chloride so this is the froc Craft's illation reaction where is the NO2 going to direct uh this group we know NO2 is a metad director so it's going to tell the acid chloride to go here and the leis acid catalyst is going to remove the chlorine atom so then this is going to be our product we're going to get a ketone so now let's say if we add uh bromine with with febr3 which is the catalyst so this is bromination where is the bromy going to go on which part of the Ring now NO2 is a metad director so N2 will want the bromine atoms to go on the on this position which is on the third carbon with respect to NO2 now the Ketone is also a metad director is moderately deactivating if this is carbon one this is two this is three the k will also tell the bromine atom to go here and both groups are directing it towards the same position so that's where bromine's going to is going to go so this is our only product or this is the only product of the this reaction try this example so what if we have two Bening rings and let's say if we add a br2 with f br3 where is the bromine atom going to go and on which ring so go ahead and feel free to pause this video and try this uh yourself so what we need to find out or what we need to know is which ring is more activated is it the ring on the left or the ring on the right what would you say the ring on the left is deactivated because this ring sees a carbonal group a carbon group is a deactivating a moderately deactivating group that carbon is partially positive both of these oxygens are pulling electron density away from the carbon and so this Benz ring SCS a partial positive carbon and that partially positive carbon makes the Benz ring um less nucleophilic so it's less reactive towards electrophilic aromatic substitution which is what we're dealing with right now however it's more reactive towards nucleophilic aromatic substitution so this ring is deactivated so it's not going to do anything this oxygen has a lone paer and it's partially negative so it makes the ring more nucleophilic towards electrophilic aromatic substitution so this is the ring in which bromine is going to go on now keep in mind I'm just going to draw the resar structures this ring is activated because the oxygen can donate electron density to it as you can see it can put a negative charge on that carbon and this ring is deactivated because the carbonal removes electron density away from that ring and as you can see the ring has a positive charge instead of a negative charge and in E the ring has to be the nucleophile and right now this ring is more nucleophilic than this one because it has a partial negative charge so the bromine atom is going to go on the ring that's on the right the one that's activated so whenever it's attached to um like an oxygen atom directly that has a lone pair that's the activated ring now where is the bromine atom going to go on that ring now this oxygen is moderately activating group because it's also attached to this carbonal group which kind of weakens the oxygen but it's still an orthopair director which means that broi can go here or here but this is a relatively big group so Ortho is not as favored as perah even though it still can occur but we're just going to draw the par position or the parag group for now because this side is more accessible um than KNE sides however statistically we have this there's twice we have a probability of the bromine atom is two times more likely to go in the oropos than a paros if it wasn't for steric factors because there's two orthop positions and one paros so you also have to take that into consideration but because this group is bulky we're just going to put the BR on a PA of position sometimes I'm at a loss for words you probably haven't noticed that by now okay so now we have this product and uh let's say if we add one more electrophile in this case ethyl chloride with aluminum chloride where is the ethyl group going to go we know the catalyst is going to take away chlorine so let's look at the direction effects this oxygen will direct the eth group to go in the ortho position which are those two which is the same but the bromine atom is also orthop parad dirction and it's going to tell the ethyl group to go here or here the top is the same as the bottom because of symmetry so which is going to win bromine or oxygen well we need to know which of these two atoms is more activated it turns out the oxygen this oxygen is moderately Activa and bromine is weakly deactivating so uh the oxygen has priority so then this is going to be our product now I do want to mention something because I don't think I've mentioned it before we said that phenol is strongly activated and um an Esther which looks like this is moderately deactivated or at least we I showed it to you in the beginning of this video but I probably didn't talked about it as much now the reason why phenol is strongly activating is because all it can do is simply donate electron density to the ring now this oxygen can do that as well however it's weakened by the carbonal group that's next to it this carbonal group is an electron withdrawing group but the oxygen is an electron donating group just like the the um phenol however the electron with draw group pulls on these oxygen atoms so the benzing ring doesn't always receive electrons from that oxygen because of the resonance structure so therefore that's why um this group is moderately deactivating instead of strongly activating however this is the major resonance form between uh these two so this is the minor one so this oxygen with a positive formal charge doesn't really deactivate the ring that much because this is the major resonance contributor the ring usually sees an oxygen with a partial negative charge instead of a positive formal charge so that's why overall this is still considered an electron donating group because the electron donating group is attached directly to the ring if the ring was attached directly to the Carbono group overall it would be electron withdrawn instead of electron donated so it's really dependent on which group is directly attached to the Bening ring so now that we've covered direct and effects orthop meta stuff um reactivity the reactions that you need to know let's talk about synthesis and how to make products so let's say if um you're given this problem on a test and you want to start from Benzene well starting from Benzene you want to make this product how would you do it so feel free to pause the video and using the reactions that you now know go ahead and list the reagents that you need to make this product now the real question is should we add the sulfur group first or the methyl group first should we add s so3h or ch3 well SO3 3 H is the metad director and let's say if we perform the reaction in this order this is not the correct order by the way let's say if we use sulfination followed by FOC Craft's ulation so the first thing that's going to happen is we're going to get the so3h group and then once we add the methyl chloride the S so3h group is the metad director and it's going to tell it to go here and so we won't get the product that we want so for these type of questions the order matters the order in which you choose your reagents is very important so we what we have to do is make this number one and this number two now let's say if we add the methyl chloride group first so we're going to get ch3 or just ch3 the chlorine is going to leave and then we can um add sulfur trioxide with sulfuric acid now the ch3 group is an orthopair director so it can direct the SO3 group here or here these two are the same now because the methyl group is not very B the ortho will be the major product instead of the per if we had a bulky group here um per is going to be the major product but if your substituent is methyl I remember seeing in the textbook it's like 60% Ortho 40% per if you have like an Etho group it's like 5050 but if you have like a tur buo group excuse me um it's predominantly per because tur buo groups are very bulky but 40% per is still significant though so I would just write both but since we're only concerned with making this product we're going to focus on the oropos so the way you would write your answer is step one you would use methyl chloride and aluminum chloride and step two um you need to use sulfur trioxide and h204 so here's another one let's say if you're starting from Benzene and you wish to make this product so for each of these synthesis problems feel free to pause the video and try it yourself and then unpause it and see if you have the right answer okay so let's plan a mental let's make a mental outline of uh how to make this product if we add the methyl group first we won't get the answer we want because methyl is orthop parad dant and right now these two groups are meta with respect to each other if this is carbon one then this is carbon 3 so we need a metad director now nh2 is an orthopair director so if we put the nh2 first on the ring it's going to tell the methyl group to go in the orthopair position so that's not going to work either so what can we do in this reaction what we need to know is that there's an intermediate um before we make the nh2 group to make nh2 we have to make NO2 first and NO2 is a metad Direction so here's what we're going to do step one we're going to use nitration with sulfuric acid and that's going to add the NO2 group and before we convert that to nh2 we're going to add the methyl group because the nl2 group is going to tell it to go in a metap position and then after that we can reduce NO2 to nh2 using um let's say tin with hydrochloric acid so step one was nitration so here's the NO2 group step two FOC crafts ulation actually let me put the N2 group here so it's going to direct the meth group to to go on a meta position which is here and then step three reduction the NO2 will be converted to an nh2 group once we use tin metal and hydrochloric acid starting from Benzene how can you make this product now be careful with this one I haven't talked about it yet but I think now it's a good time to talk about it go ahead and try that one so we apparently don't want to add this group first because the Ketone is a metad director and uh these two groups are pair with respect to each other PA is4 and we don't want to add NO2 first well we don't want to add NO2 and then the Ketone because NO2 is a metad director we need to use the nh2 group because it's a parad director so it appears as if we need to add hn3 first with sulfuric acid and then add reduce it this time we'll use like Fe and HDL and then add the as chloride with the Catalyst but there's an issue with this reaction that you have to be aware of by the way that's not the right answer that it seems like it's the right answer but it's not and I'll explain why so the first step is nitration so now we have our NO2 group and then once we add iron metal with hydrochloric acid it reduces into an nh2 group now here's the issue the nh2 has a l here and if we react it with the acid chloride it's going to become deactivated this is the powerful leis acid Catalyst and aluminum has a positive three charge because chloride has a minus one charge and because nitrogen has a partial charge it's attracted to the aluminum catalst so it's going to uh combine with it and once that happens nitrogen will no longer have a partial negative well it's going to have a positive formal charge I should say whenever nitrogen has four bonds it has a positive formal charge and whenever it has a positive formal charge it's no longer an orthopair director which is what we want it's now a strongly deactivated metad director so this Ben doesn't want to even react anymore so pretty much this reaction stops if we do this so you don't want to add the Catalyst in the presence of an Aiming group instead what you want to do is you want to use the aiming group and you want to react it with an acid chloride however this this is the important part you want to react it with the acid chloride without the al3 catalyst so I'm going to write it here I'm going to take it out so don't add this with the acid um chloride so react the aming with the acid chloride but without the acid catalyst so what's going to happen is these two are going to combine and we're going to remove HCL the amine is going to be converted into an amide if you take away HDL we're going to have a nit um that has a hydrogen one hydrogen is gone one um is still there and notice that it's an am now basically what we have what we just did is we added a protectant group to the aing we turn into an amide and this process is reversible if we had H2O plus with heat we can hydrolyze the Amite and get back the uh aiming group but we don't want to do that yet so now that we have the amide we can react it with another leis acid another acid chloride and because of the lone pair on this nitrogen it's still an orthopair Direction it's no longer strongly activated but it's moderately activated the Carbono group weaken the uh electron donating effects of the nh2 group but that's a good thing because we can now perform this reaction because the um the nitrogen group is no longer as active as it was before um it won't react with the acid chloride and here's why the Carbono group can pull electron density away from the nitrogen and so it has a resonance form so notice the partially the positive formal charge of the resonance form now this positive charge doesn't make the ring deactivated because this resonance form is the major contributor this one is the minor contributor so for the most most of its time the ring sees this lone here and so it's still orthopair um directing but it's not as um nucleophilic as the nh2 groups so this Catalyst doesn't react with this nitrogen so this nitrogen doesn't PA up with the Catalyst which is good that means that this reaction can work now so the Catalyst can take away chlorine and the Benzene ring is going to react with the acid chloride so whenever you have an nh2 group and you want to perform some sort of electrophilic aromatic substitution reaction uh be careful because if you use a powerful Catalyst like al3 or febr3 it's going to deactivate the nh2 group and so the reaction won't work so what you need to do is add this protecting group then perform the reaction and then remove the protecting group so we want the par of product which is going to be over here all right so now I'm going to redraw it but I'm going to rotate it a little so I'm going to put the NIT back on the top instead of the top right position so now we have this our last step is to add H2O plus and that's going to basically hydrolyze this Bond it's not going to affect a ketone so now we have our aming group and now we have our Ketone so whenever you have an Aiming and if you have like powerful C cdic conditions right now should be H3 NH3 plus by the way because we have a strong acid but if anytime you add like a leis acid to it or even H2O plus the nh2 will turn into like a NH3 group or a nitrogen with a plus charge if you add al3 and it's going to deactivate the ring and it's going to be a meta director instead of an orthop par director so watch out for that so right now by the way we need heat to make this work if we add another group to this this right now is a metad director but here it was an orthop paric director so that's why this group still went over here um but we need to convert that back to nh2 so what we need to do is um add a dilute solution of water and sodium hydroxide sodium hydroxide will deprotonate the extra hydrogen so now we have nh2 and we have our Ketone so that's it for that example I know it was a long discussion but um it was a necessary one so real quick how do you make benzoic acid from Benzene do you remember this one CU chances are this will be on your test now we've covered this one already the first step was to add methyl chloride with the al3 Catalyst and then oxidize it with potassium magnate and h2+ now do you remember how to make styrene from Benzene we've also covered this earlier step one was to add ethyl chloride so we get the two carbons that we need step two was to add NBS to add a bromine on the uh benzilla carbon which is right here and then step three was to use an E2 reaction we can use sodium hydroxide or some other strong base like sodium methoxide sodium ethoxide or even tur oxide and then we can get styrene but instead of styrene instead of a double bond on the outside how can you make a triple bond instead how would you do that well those same three reactions that we covered we want to um use those three reactions to make styrene now once you have styrene what you want to do is um add bromine and you can use D chloromethane which is the non-polar solvent well slightly polar but relatively non-polar so what this is going to do is going to add two bromine atoms across the double bond and there's a hydrogen here and there's a hydrogen there so now what you want to use is um sodium hydroxide and it's going to convert to the alkine so here's what's going to happen first hydroxide is going to grab one of the hydrogen atoms and then this this is going to form a double bond and it's going to expel the bromine atom so right now what we have is I need to start on another page we have a double bond um we still have a bromine atom on this carbon and we still have a hydrogen here so another hydroxide molecule could grab this hydrogen form a triple bond and expel the bromine atom so now we have an kind but now let's say if you want to start from Benzene and you want to extend a carbon chain used an alkine how would you do it now keep in mind there's a hydrogen here and so what we want to use is um sodium amide N2 the amide ion is a very powerful base and it can remove this hydrogen and put a negative charge on that carbon so now you have an alkine with a negative charge and notice that we need to add um one two three four carbons so what you need is a four carbon aloh halide and then perform your sn2 reaction so this carbon is going to attack that carbon and it's going to expel the bromine atom and then you're going to get this product so now let's say if you have um Benzene and you want to make uh benzo alcohol how would you do it now there's many ways of doing this the first is we can use the gadman Kos reaction which is carbon monoxide HCL alcl3 and CCL we talked about this earlier in this video it creates an acid chloride but a one carbon acid chloride which is very unstable and so the L acid Catalyst removes the acid chloride or removes the chlorine from the acid chloride and uh we're going to get a one carbon alahh group now you can use a reducing agent like nabh4 or lithium aluminum hydride and it's going to reduce the alahh to a primary alcohol that's one way you can do it another way is you can add bromine with febr3 that's going to create bromobenzene and Then followed by magnesium so you can create a grer reagent once you have your Griner re agent you can use carbon dioxide which will turn it into a and benzoate group and then you can reduce it using lithium aluminum hydde which can reduce carboxilic acids into alcohols and Then followed by protonation or you can call Aquis workup so starting from Benzene try this example how would you make uh this product how can you make a four carbon alcohol with the alcohol being on the first Carbon on the benzilla carbon so what we need is a Gren reagent so we're going to add um we're going to perform bromination and then we're going to add magnesium in the Second Step so initially we going to get this uh mgbr group so once we have Phenom magnesium bromide we can react it with an alide so basically we need a four carbon alide so 1 2 3 4 so what's going to happen is this grer is going to attack the carbon and that P Bond's going to break so initially we have an Al coxide ion and Then followed by h2+ or even water water can work this is going to grab a hydrogen expel hydroxide and so now we have this product how can you convert Benzene into phenol how would you do it now this is another reaction that we've covered already but feel free to pause the video and see if you can figure it out so the first step is nitration hno3 with h2so4 that's the first step so that's going to add an NO2 group to the Benzene ring step two is to reduce it using um zinc metal with hydrochloric acid so now we have an nh2 group step three is to use sodium nitrite with HCL sometimes you may see this as hoo H2 or H2 this combination makes uh nitrous acid and nitrous acid converts the Amin into a doonium sot which looks like this and this nitrogen has a plus charge now from here we can make a lot of different products but if you want to make make phenol simply add h2+ this is a very good leaving group it leaves as nitrogen gas now how do you make anol from Benzene this is anasol it has an oc3 group how would you do it so first we're going to use the same four steps that we just covered and uh we're going to make phenol first so after the nitration followed by reduction followed by diazotization followed by h2+ we're going to make phenol and then once you have phenol you can simply there's a lot of things you can do you can react it with an alcohol using an acid Catalyst and so you can remove H2O and it's going to turn into an ether that's one way you can do it or you can use the Williamson's the Williamson's ether synthesis reaction which I think we should go over so in this reaction the first step is detination so phenol is a relatively acidic alcohol most alcohols have a PK of like 16 or 18 but because of resonance fenol has a PK of 10 so hydroxide is a suitable base um that we can use to remove the uh acidic hydrogen so now we have phenoxide by the way um this ring is even more activated than phenol because of the negative charge on the oxygen so once you have phenoxide we can now react it with an aloh halide like methyl chloride so this oxygen is going to attack the methyl expel the chlorine group and that's how we can make an ether simply deprotonate the hydrogen and then add an aloh halide to it now we can make different types of ethers all we got to do is just change number of carbons in that structure so that's how you can make an AOL now something you should be aware of when you're dealing with phenol when you deprotonate when you remove that hydrogen if carbon dioxide is present if this reaction is exposed to the air it can react with carbon dioxide this ring is strongly activated because of the negative charge of the phenoxide ion so this lone pair can form a double bond and cause this double bond to attack carbon dioxide this is the Kobe Schmid reaction by the way and now there's a hydrogen here so if we have pH oxide chances are we have a small amount of hydroxide in a solution because this reaction is occurring under basic conditions hydroxide could also react with carbon dioxide to create carbonate by the way now hydroxide could remove the hydrogen these electrons can form a double bond and um we're going to get the phenoxide ion again so now the benzoate group deactivates The Ring well it's not really completely deactivated yet but it weakens it because it's uh it's a deactivating group but the phenoxide is still activating so overall this Benzene ring should still be activating but it wasn't as active as it was before because this carbonal group is a deactivator so we can say it's less active activating but still activating overall so this could react with another CO2 molecule but if we had h2+ then we can get this product and if I remember correctly I believe this is called salicyclic acid and um it's one step closer to making aeto cocyclic acid which is aspirin so now let's say if you have the Benzene ring and um if you want to make uh let's say if you want an Esther so we covered how to make phenol how to make an ether but what if you want to make an Esther how would you do it I ran out of space so first you want to make phenol using the same four steps nitr reduction using zinc and HCL diation and then h2+ once you have phenol then you can react it with a carboxylic acid using an acid catalyst so all you got to do is remove water and then connect the two groups together and that's how you can make an Esther so in the case of aspirin once we have a um this compound once we have phena with benzoic acid all we got to do is add another carbic acid and this oxygen is more nucleophilic than this hydroxy group so this is going to react with the carboxilic acid and so now we have aceto cocyclic acid which is aspirin and water was removed now here's another uh synthesis problem for you let's say if you have Benzene and you want to create 135 tribromo Benzene how would you do it now you might be wondering okay I could simply add bromine with febr3 three times if you add it the first time nothing's wrong with that you're going to get a bromine atom here but notice that this bromine atom is not metad direct it's orthop parad Direction it's going to direct the second bromine atom to go here or here or here which is what we don't want all of these bromine atoms are meta with respect to each other so how can make this work if we add more bromine atoms you won't get that product so here's what you need to do the key is not to add bromine at the beginning starting with Benzene you want to add uh nitration you want to add a Nitro group so you want to perform nitration but you want to do it three times not just once but three times so initially the first Nitro group is going to add very quickly to the ring because the Benzene ring is not activated or deactivated and the Nitro group is a very powerful electrophile now adding the second Nitro group is going to be it's going to take longer because once the first Nitro group is on the Benzene Ring The Ring is deactivated so you may have to like heat up the reaction mixture and increase the reaction time but it can work though by the way NO2 is a meta director so it directed here now that we have two NO2 groups they're going to both direct the third NO2 group to go to that position but now the ring is even more deactivated so you probably have to add more heat more reaction time to make it work but you can still get the job done you just got to add more energy to it so now this ring is fully nitrated so now what we want to do is reduce it using Fe and HL so we're going to convert all of the NO2 groups into an nh2 group now you might be wondering okay where am I going with this and you'll see in a minute just hanging there so now we're going to use sodium nitrite with hydrochloric acid which can also be written as ho o or simply H2 it's going to convert the nh2 groups into um the diazonium groups now keep in mind there's a chloride ion for each N2 group so our next step is to add copper bromide excess copper bromide this is the sandmire reaction and it's going to replace the N group with bromine atoms and that's how you can make 135 tribromo benzene now here's a question for you let's say if you have a Nitro group how can you convert it back to Benzene how would you do it to convert it back to Benzene we need to replace it with a hydrogen and we can do that by converting it to an nh2 group we can use iron metal with HCL and then once we have the nh2 group we can use hono or sodium nitrate with HCL so now we have our diazonium salt and then the last step we need to do is um add H3 po2 this reaction is going to replace this group with a hydrogen so that's how you can convert Nitro Benzene back to benzene but now let's say if you have Benzene what can you do to make um floral Benzene how would you do it so first we need to use the same three steps nitration reduction and then diazotization to make the N2 group once we have the N2 group then we could simply add h B F4 which is basically H+ and Boron with three Florine atoms the boron has a negative formal charge and it's going to replace this group with a floring atom here's another question for you let's say if you have Benzene how can you make excuse me how can you make uh two Bening rings attached to each other so how can you make basically D benzy think about this one and see if you can figure it out feel free to pause the video do some research now I want to show you something here's a hint that's going to help you out let's say if you have phenol attached to or next to par choron Nitro Benzene actually let me redraw it I want to draw the Nitro group like this what do you think is going to happen in this reaction well what you need to realize is that if we want to com if you want to combine two Benzene rings together you need an activated ring and a deactivated ring this ring is activated because of the electron donating group and this ring is deactivated because of the electron with drawn group so basically this ring is the nucleophile because it's electron rich and this ring is an electrophile in organic chemistry the nucleophile always reacts with the electrophile and let's talk about why this oxygen has some lone Pairs and it can donate electron density uh to the this double bond can move here and this double bond can put a negative charge on the par of position now the Nitro group is an electron withdrawn group and it can pull electron density away from the ring so notice the two structures that we now have so we have an oxygen that has a double bond and it's attached to the ring and we now have a negative charge on a parac carbon and this oxygen has a positive formal charge and if we drew the resonance structure or if we draw the resonance structure for on this group this oxygen now has a negative charge and that one already had a negative charge to begin with now there's a chlorine atom here and this carbon has a positive charge but chlorine does have some lone pairs so it can stabilize that positive charge if it wants to donate a pair of electrons we're not going to do that though but the fact is that this carbon has a partial positive charge and this carbon is partially negative these are the minor uh resonance contributor forms of the reactions that we started with but the fact is that this ring is nucleophilic and it's going to attack the ring that's electrophilic so those two benzing rings can combine now the chlorine group is crucial to making this reaction work because we need a leave group you'll see why soon in the beginning of this video we talked about electrophilic aromatic substitution reactions and nucleophilic aromatic substitution reactions um for electrophilic aromatic substitution reactions the electrophile um replaces the hydrogen atom and in the beginning we talked about for Nas nucleophilic aromatic substitution reactions the nucleophile replaces a Le group in this case the chlorine so this is the nucleophile and right now for this Benzene ring it's undergoing an EAS reaction the hydrogen atom that's right here will be replaced with the electrophile which is this entire benzing ring this whole structure and for this benzing ring on the right side it's undergoing an N reaction because it's reacting with a nucleophile which is this and in such reaction the nucleophile kicks out the leaving group which is the chlorine group so notice what's going to happen so we have a double bond here a double bond here we still have a hydrogen we still have the chlorine which is going to leave soon and here we have a double bond so this oxygen with a negative charge is going to reform a double bond this double bond is going to go back here where it was and this double bond is going to move here and it's going to expel the chlorine leaving group and since I'm out of space I need to go back to the drawn board and make a new page so now this ring is aromatic again and so we have our normal Nitro group the chlorine has been expelled but now we have to remove a hydrogen and this oxygen still has a positive formal charge so we need to use the base the base could be the solvent if this reaction occurs under water like or with water as a solvent we can use water as a solvent or something else I haven't really specified the reaction conditions but for the sake of keeping things simple let's just use water as dissolvent so water is going to grab a hydrogen and this bond has two electrons which those electrons will be used to form a double bond that double bond is forced to move over here and this Bond breaks and oxygen pulls those electrons toward itself so that oxygen pulls all of those electrons toward itself excuse me so now we have our second aromatic ring so that's how you can convert you can put two benzing rings together but in the example that we talked about this is just a Hint by the way all of this discussion this whole discussion was a hint the issue with this one though is we can't get rid of the hydroxy group I mean we don't have a reaction where we can replace that with a hydrogen maybe there is one I don't know what it is but I do know that we can replace the NO2 group with a hydrogen we need to remove all groups and replace them with hydrogen if we want to make dibenzene the Nitro group we can convert it to nh2 group and we convert that into a N2 group then add H3 P2 and get the Benzene ring so first let's start with Benzene and let's make the reagents that we need so first we can add chlorine cl2 with fe3 that's going to put the CL on the first carbon and Then followed by nitration now we're going to get this group okay so that was the electrophile that we had when we want to connect the two Benzene rings together to make the nucleophile here's what we need to do we're going to um add nitric acid again now following nitration we're going to reduce it to an Aiming group so we're going to use zinc and htl now the nh2 group can act just like the phenol group because it's a powerful electron donating group just like phenol so once we isolate the nh2 group and make sure it doesn't have any positive charge this is going to be our nucleophile and now we have the electrophile and using the same mechanism that we did last time we're going to get this product the chlorine is going to leave so basically this is a cant reaction now what we could do next is reduce this reduce the NO2 group to an nh2 group using fcl so now we have two nh2 groups next we can react this with um sodium nitrite and HCL to make a diazonium salt now there may be some side reactions in this process because chances are doesn't the N two groups don't form at the same time because let's say if one side has an nh2 and the other side has an N2 group the ring with the nh2 could react with the other ring that has the N2 group I'll give an example of that but those would be side products let's focus on the product that we want to get so this reaction won't occur with 100% yield so just keep that in mind that side products will occur but at this point what we can do is add H3 P2 and then that's going to replace the N2 group with a hydrogen so now we have made dibenzene there's probably other ways of making dibenzene but this is just one of many ways of doing it and it's probably not the most efficient way but this is just a a learning exercise that's how you can make it though now here's a question for you let's say if you have phenol or let's say even um analine nh2 a Benzene ring and you add it to this so this is another example of a couping reaction and this is why I was saying that you can have some side products because the this was h2n which we had in the last example it can react with um this group as well so what's going to happen if we put these two uh these two groups together so this oxygen is going to form a double bond that double Bond's going to move here and then this double bond is going to attack the nitrogen group and then the triple bond is going to go down to a double so now we have two Lan piers and um we have a double bond here another double bond this oxygen now has a positive charge and one Lum here and there still a hydrogen here so now we got to remove the hydrogen so whatever the base is in this reaction maybe it's a chloride ion it's going to grab the hydrogen these electrons will form a double bond that double Bond's going to move here and then we're going to get alcohol again so that's another way you can couple uh two Benzene rings together just keep in mind when you do so the triple bond is now reduced to a double bond nh2 could have done the same thing and in that last example that we covered when we had the D Benzene ring chances are that the N2 groups won't be converted to um the nh2 groups won't be converted to N2 groups at the same time so it's possible that this nitrogen could attack this nitrogen and this could go here so now notice the structure that we now have so this could form a polymer it can keep growing so to speak wait hold on this this had to be connected to uh this group there we go so this is one of the side products that can occur when I talked about making that dzo group and this polymer could keep growing this could react with another molecule at this end and it just could form a long chain polymer so in chemistry you can always get side products depending on what what you're reacting with okay but let's continue so now we've covered all the synthesis reactions that actually wait this one more I want to talk about let's say if you have Benzene how would you make this compound so how would you connect two benzing rings together by means of an eser so feel free to pause the video brainstorm and try to figure out this one what you need to do is first make a carbic acid add methyl chloride with aluminum chloride so now you have your methyl group you can use ethylchloride or propo chloride doesn't matter and then use potassium manganate with H2O plus and this will convert it to a benzoic acid now the next thing you need to do is make phenol so you need to use nitration followed by um reduction fcl first you will get the NO2 group then the nh2 group and then step three sodium nitrite with HCL to make the N2 group and then step four h3o+ and you want to make phenol whenever you add an alcohol and a carboxilic acid you can get Esther so now what you want to do is combine the carboxilic acid and the alcohol together and using an acid Catalyst this reaction is reversible by the way remove water so water is going to be a side product if you add heat to it you can drive the reaction forward because water is going to escape from the reaction mixture and then that's how you can make this particular compound so now let's go over a mechanism problems so let's say you have a test and uh you get this um let's say you're get given this reactant how can you propose the mechanism for the formation of this compound how would you do it so hopefully you um have reviewed the mechanism for the FOC Craft's ulation reaction but the first thing that happens is the chloride reacts with the catalyst this chlorine is partially it has a partial negative charge and aluminum has a plus three charge in it structure so these two are attracted to each other so chlorine binds with the aluminum so that's the first step so now aluminum has a negative formal charge and chlorine has a positive formal charge now this Bond can break chlorine can take these electrons with it and so what we now have is an acinum carbo or cation and we have the al4 ion somewhere in a solution now this carbocation resonate stabilized this lone pair this oxygen can use one of its lone pairs to form a triple bond and so then you you get this uh carbon with a triple bonded oxygen in which case the oxygen now has the positive formal charge so you have those two resonance forms but we're going to use this resonance form to draw the product now notice that this molecule can bend and it can bend in such a way that it can form another ring so we have four carbons here one two three four so one 2 3 4 and on carbon 4 we have a double bond o and is a plus charge here so this Benzene ring can use a double bond to attack that carbocation whenever a double bond attacks a carbocation it creates another uh carboon but now the ring is closed but the question is where is the other carboon it's usually one of these two carbon atoms of the double bond that broke now this carbon atom it regained a bond but this one it lost the bond and it didn't get it back so the plus charge is here which means that there's a hydrogen here so now by the way keep in mind that al4 minus is in equilibrium with alcl3 and cl minus however um this form is more stable we can use chloride or we can use al4 minus as a weak base to grab a hydrogen but I'm not going to get into the particulars about which base to use or I'm just going to keep it simple so chloride is going to grab a hydrogen and this carbon hydrogen bond is going to break regenerating the aromatic ring and so we get this product so that's the mechanism for that reaction here's another example so let's say you have an acid chloride and a Griner reagent immediately these two will react let me draw this that's a chlorine propose a mechanism for the formation of this product so what we need to realize is that this carbon atom is partially negative because magnesium has a positive two charge and this carbon atom is partially positive so the partially negative carbon atom is attracted to the partially positive carbon atom and so it attacks it and then when that happens the pi Bond breaks but the ring closes so we have a one two three four five six carbon ring now when this carbon attacks the Carbono group the Magnesium bromide um leaves and it's in a solution somewhere so we don't have to worry about it anymore right now we have an oxygen with a negative charge and we still have a chlorine attached to this atom so then this oxygen uh reforms the double bond that tetraedro intermediate collapses and it expels the chlorine group and so now we have our product which looks like that try this example propose a mechanism for the formation of this product so the first thing that's going to happen is the chlorine atom is going to react with the Catalyst let's start with that one so this aluminum has a negative formal charge and chlorine has a POS POS formal charge now this chlorine could leave forming a carbocation now keep in mind there's another chlorine atom here and uh it could combine with the carboon 1 two 3 4 we can get a five membered ring so I'm just simply drawing the resonance form and that process is reversible so we could get this intermediate in in that example but we're going to focus on uh this resonance form because our goal is simply to propose a mechanism so now the Bening ring is going to react with this carbo and we still have the other chlorine atom here so we're going to call this carbon 1 2 3 4 so carbon so here's carbon 1 2 3 4 Carbon one has a meth group and carbon 4 has a methyl group and it has a chlorine atom now keep in mind that since we one of these carbon atoms will form the plus charge but since this carbon atom was used to create the connection the plus charge is now on that chlorine atom now that chlorine atom can close and form a ring but we're not really concerned about that particular resonance form so we have like this SE seven member ring but this process is reversible it can open and close right now it's not going to stay that way because we need to regenerate our aromatic ring so let's go back to this resonance 4 now keep in mind the other chloride ion which is combined with alcl4 some of it is dissolve in the solution but most of it is attached to the uh aluminum chloride Catalyst and the part that's still disolved we can use that in um as a weak base to grab this hydrogen and this bond is going to break put in a double bond there so once we regenerate our atic ring this chlorine atom will no longer be attracted to that caroet which means it's going to be free to react with the uh Lis acid Catalyst because right now it's not free to react because it can form that seven member ring but that's a reversible process so once we have this now that chloride ion is free to react with the catalyst so now chlorine has a positive formal charge and then it could leave but it can also to come back so most of these reactions are reversible but once we have this caroet then the Benzene ring will attack it and then it's going to close so now we have this structure but we need to find out where the plus the plus charge is now because we um we form this bond that means that the plus charge is on this carbon and that means that we still have a hydrogen atom here so we're going to use a chloride ion from the solution which comes from the al4 minus ion and we're going to use that to regenerate our Benzene ring so now we have our product which looks like that try this one so Benzene in the presence of another Aline using HF as a catalyst and it can make this product this reaction is very similar to a Fredo Craft's alation reaction but without the aluminum chloride Catalyst but go ahead and propose a mechanism for this process but let's get started the first thing that happens is the alen reacts with HF this double bond grabs the hydrogen expels the fluoride so now we have a carbocation and we have a fluoride ion in a solution the hydrogen goes on the uh less substitute carbon atom which is a secondary one so we can get a tertiary carbo now the fluoride could react with a carbo and that could be a side product of this reaction but also the fluoride well not the fluoride also the benzing ring could react with the carbocadon as well so this won't be your only product um it's definitely going to be one of the products but you could also have this product in this reaction but now our goal is simply to propose a mechanism for the formation of that product so we're going to combine these two so the benzing ring is going to attack the carbo canadine now keep in mind that there's two hydrogen here so this is carbon 1 two three so one two three on carbon one we have two methyl groups and we still have a hydrogen here and another hydrogen here so this carbon atom lost a bond so it has the positive charge and we're going to use floride as a weak base to grab this hydrogen and these electrons will be used to regenerate the benzing ring and so now we have our product so that reaction wasn't too bad so now for our last mechanism well this time we're going to propose two mechanisms actually but with one reaction so propose a mechanism for the formation of this product and the formation of that product so feel free to pause the video and and try that try those examples so what do you think the first step for this reaction is let's start with this one so starting with styrene which you know how to make now we talked about it earlier we're going to react with the acid catalyst so this double bond is going to react with the hydrogen and in the process it's going to create a plus charge the hydrogen is going to go on the less Subs to the carbon so the plus charge is on the secondary benzilla carbon and it's stabilized by the benzing ring because we can draw a lot of resonance structures but now once we have that carbocation it can react with another styrene molecule but here's a question for you which double bond is more reactive the double bond on the outside or in the Benzene ring now the bonds and the Bening rings are very stable because of the aromatic ring and so they're not very reactive unless there's a a powerful electrophile in front of it they could react with a plus charge but this double bond is far more reactive and it's more likely to react with the double bond with the plus charge so we're going to say this is carbon one which connects to carbon two and that's carbon 3 so one two 3 carbon one has this methyl Group which I'm going to draw on top and so right now we have this structure the question is where's the plus charge carbon one gained a bond and carbon 2 lost a bond but got it back carbon 3 lost the bond and it didn't get it back so the plus charge is here and now this is a hydrogen here now we really didn't specify what type of uh solvent but let's say we chose a H3 Plus or actually this stying is nonpolar so we need to use a different solent all right so we're going to describe this solvent as like a base whatever it is and this base is going to grab a hydrogen and that carbon hydrogen bond it's going to break and form a double bond it's always easy to use it's always easy to use water as a solvent but since styrene is nonpolar and can't dissolve in water or doesn't dissolve very well um we can't use water as an example but that's how you can show the mechanism for the formation of uh that product so make sure you know what your solving is if you're on a test if not if they if your teacher simply gives you an acid Catalyst use some generic base to remove the hydrogen but now let's talk about the formation of the other product now we're going to start back with this particular carbocation that we had because the first two steps are the same but what we're going to do is redraw the structure in a way that it's easy to see what's going to happen so this is number one two three so here's carbon one carbon 2 carbon 3 carbon one has a methyl group carbon 3 has a Bening ring but carbon 3 has a plus charge and the Benzene ring could react with that plus charge and that's what's going to happen here and that's how it closes to form the five membered ring so mechanisms is something that you just have to practice with and as you practice you get better um at predicting what the mechanism is going to be based on what like you have so now we need to find out where the plus charge is it's usually one of these two carbon atoms now this carbon re it formed a bond even though it lost double bond this one lost a bond but didn't get it back so um the plus charge is now on that carbon but we still have a hydrogen here our last step is to use the base to grab the hydrogen and break the carbon hydrogen bond use those electrons to regenerate the Benzene ring and so now we have our product now you might be wondering this is a long video and this chapter has a lot of information we've covered most of it the only thing we didn't cover is there's a few details we need to talk about on nucleophilic aromatic substitution we made a comparison between EAS and Nas but let's review that again so for electrophilic aromatic substitution a hydrogen was replaced with an electrophile so that's why it's called electrophilic substitution but what we're now going to discuss is nucleophilic aromatic substitution in which a leaving group let's say like a bromine atom is replaced by some sort of nucleophile the nucleophiles that we're going to use would be like hydroxide or amide nh2 minus now keep in mind um alcohols which has lone pairs nh2 which has a lone paer those are the activators for the EAS reaction the activators for the nucleophilic aromatic substitution reactions are these are electron donating groups by the way but the activators are now the electron withdrawn groups which are like the NO2 groups um s so3h those groups they activate the rink towards nucleophilic aromatic substitution so here's an example we's say we have a bromine atom and we have an electron withdrawn group there's two types of reactions within the nucleic aromatic substitution category and there's one reaction which is called the addition elimination mechanism which proceeds via the me the meisenheimer complex and then there's the other reaction which proceeds via the benzine intermediate what you need to know is if you're performing a nucleophilic aromatic stion reaction where you're add in like a a hydroxide nucleophile if you see an electron withdrawn group like an NO2 group you need to know that this reaction will proceed via the addition elimination mechanism so we're going to add a group first and then we're going to eliminate something we're going to remove something so we're going to add hydroxide and then eliminate bromide the end result is that bromine will be replaced with a nucleophile hydroxide this is a good way to make phenol by the way but you have to get rid of the NO2 group when you're done so now let's talk about the mechanism this time I'm going to draw the NO2 group like this so the first thing that hydroxide will do is it's going to attack this carbon because that carbon is attached to a bromine atom bromine is electronegative and it withdraws electron density from that carbon by Me by means of the inductive effect so that carbon is partially positive and hydroxide having a negative charge is attract to that carbon So It Begins by attacking that carbon and this Bond breaks it can put a negative charge on that carbon so now we're going to get this structure which looks like this and so now we have a negative charge now we can draw another resonance form this negative charge can move here and put a negative charge in that carbon so the negative charge can jump um by two carbons so now we have a double bond here and one here and now that negative charge could form this carban can form a double bond and that double Bond's going to break so now we have another resonance structure which looks like this now this is a fourth resonance structure sure but I'm out of space so I'm probably not going to draw it but the lone paer will be right here let me just show you the arrows this lone pair can form let me change the color this Lan pair could form a double bond and this double bond can put another carbon there that's the other resonance form so all of these uh resonance forms combined is known as the meisenheimer comp complex now to get the product one of these oxygen can reform a double bond this double bond can form another double bond and that double bond can move here expel in the bromine atom so now we have this an alcohol um our Benzene ring which is now aromatic again and a Nitro group so that's the addition elimination mechanism we add hydroxide and then we eliminate bromide the next one is the benzine intermediate so let's say if we do not have an electron withd Drawing Group so if there's no NO2 group in a ring and let's say if we add a very powerful base like nh2 minus hydroxide if we use hydroxide this reaction won't work very well because the ring is not activated towards nucleophilic aromatic substitution but if we were to use hydroxide to make it work we would have to use some very high temperature like over 300° C to make it work but sodium amide is a very powerful base and this reaction can work under um normal room temperature conditions now there's a hydrogen here and so what sodium amide does is uh the amide ion grabs the hydrogen these electrons are used to form a triple bond and then the Brom ion is expelled so now um we have a benzine intermediate as you can see it's like a Benzene ring but with an alkine so they call it uh benzine now the nh2 minus some of it has been converted to NH3 but you still have some nh2 minus left over in the uh solution it can attack the benzine from this position or from this position so we can get two we can get up to two products dependent on um or even three products depending on what substituents are already present in the ring but let's say for the attacks there the triple bond will be converted to a double bond and it's going to put a carban on a ring so here we have our nh2 group we have a double bond again and and we have a negative charge now notice how we regenerated NH3 or we created NH3 so this carbon can grab a hydrogen from NH3 regenerating the nh2 minus ion so the net result was that we replaced the bromide Ion with nh2 by means of the benzine intermediate so because there's no other substituents we can only get one product so even if we put the nh2 on this position these two products are the same but now let's say if there was a substituent present now we can't use an electron withdrawn group we have to use an electron donating group basically like a weak electron donating group so that we can still have the benzine intermediate if we use a powerful electron withdrawing group then it's going to go the reaction will proceed by means of the addition elimination mechanism okay so if we had amide we can form the benzine on the right side or we could form it on the left side so that means that the nh2 group can go on three different positions it can go directly where the bromine atom was or it can go on the right side or it can go on the left side I meant to put that on the right side so notice that if we have one substituent we only get two products these products are the same they're identical so they're considered as one product and this one is different so we get a total of two products in this reaction now let's say if we have two different substituents let's say we have a methyl and we have an eth and we have a bromine atom here now if we use nh2 minus or sodium amide now we can get all three different products because there's no symmetry anymore so this is one product here here's the other product and then here's the last product so as you can see all of these products are different um if we name it we'll get different names