hi am so glad you're here let's jump right into covering all the reactions covered in organic chemistry 2 one of the first topics typically covered is diin specifically we need to know how to make them and then how they react when presented with electrophiles in order to generate conjugated diin we need to begin by doing an elimination reaction in certain circumstances we have a single halide or we have a dihalide species in both of these examples you can do an elimination reaction to form your conjugated dine and in the presence of potassium tur butoxide both of these reagents would produce a conjugated dine when reacting a conjugated dine with hbr you typically always produce a one two adduct of the addition of the bromine and the hydrogen so this would be the one two adduct because the hydrogen and the bromine go at positions one and two on the carbon chain and in addition to that you also generally produce a 14 adduct so in this case you would produce a hydrogen at this location at the one position and a bromine at the four we have called these previously the kinetic product and the thermodynamic product so the thermodynamic product is going to be your 14 addition and your kinetic product is going to be your one two addition in addition we can add bromine gas or br2 to a conjugate a dine and we would get the same one two addu and 14 addu specifically this is going to generate a bromine at the one position and the two position this would be the kinetic product or the one two addition or the 14 product which is going to be the thermodynamic product and importantly depending on the conditions of the reaction that is going to generate the major product as being one or the other so for example for kinetic products to be formed you generally want low temperatures like at 0° c and a very short reaction time however for something like the thermodynamic product you generally want higher temperatures and a longer Reaction Time pause the video try these problems independently then resume the video to check your answers in all deal Alder reactions you always start with a conjugated dine reacting with what is called a dienophile and these electrons in the pi system are delocalized such that they can form a cyclic structure in this case you end up with a cyclohexene ring where the X group is coming off the side that is not part of a pi system importantly the new bonds formed are located here and here and the pi electrons now IDE in the carbons that previously were just a sigma Bond so that is where this Pi bond is located when performing deals Alder reactions on Dino files that are either Cy or trans substituents we need to consider the stereochemistry involved in the product formation so again we end up with the cyclohexene ring and since the reactant dienophile was CIS for the substituents coming off the alkine they need to remain CIS on the product so in this case we would draw both both of them as wedges since we formed new stereo centers we need to indicate the stereo chemistry and we do so by drawing a wedge for both indicating that both are going in the same direction for the trans dienophile where the substituents are going in opposite directions this gives rise to a cyclohex ring where now we need to indicate that both of them are going in opposite directions so we would draw one as a wedge and one as a dash to indicate that they are trans to one another as opposed to if they are cyst to one another importantly deals ERS required that the dine be in the S Cy confirmation in order to proceed so for that reason any time you have a cyclic system that locks the position into the sis confirmation indicating that around this Sigma Bond the alkenes are cyst to one another this is actually going to speed up the reaction now once we have our dine in the sis confirmation it can react with any Dina file that contains a pi Bond specifically ones that have electron withdrawing groups like Esters for example or Nitro groups or cyano groups anything that will activate the dienophile will make this reaction proceed more rapidly and since there are two Pi bonds present in this Dina file this means that only one of them is being used in the deal's Alder reaction so only one set of those electrons are being used so in our cyclohexene confirmation we need to indicate that fact that there's still going to remain a pi Bond as well as on the opposite side of the cyclohexene ring so we end up with a system with two Pi bonds now importantly this also means since these carbons are SP2 hybridized we do not need to indicate wedges or dashes indicate that that they are stereocenters because they are SP2 hybridized and thus planer for that reason the the Esters that are present would actually just come off as straight lines so we need to indicate that with a straight line since this is an SP2 hybridized carbon and they are trigonal planer and do not contain stereo centers now for this example in addition to this ring confirmation this cycle pentadine being locked into the sis confirmation we also need to consider the fact that our Dina file contains two Pi Bonds in these aldhy so remember an aldhy is CH but you can also draw it like this both of these are representing an aldah and these Pi electrons present in this electron withdrawing group are going to be stabilized in the Endo position so for that reason now that we have this fused ring cyclic system where we had our cyclopentene which contained five carbon we need to still contain those same five carbons so those are positions 1 2 3 4 and five whereas our new cyclohexene ring is a part of this boat confirmation and then we need to consider the Endo rule for these Pi containing electron withdrawing groups on the dino file so both of these substituents would be going in the downwards Direction and again this is called the Endo rule where the substituents would be pointing towards the larger ring as opposed to the smaller fused ring up at the top for the electrocyclic Rea reactions remember these Pi electrons are allowed to move in such a way that you can actually close this ring and end up with a system where you have moved the pi electrons and this is called an electrocyclic reaction remember that both of these reactions can either proceed either through Heat or the presence of light and it's important to remember that the conditions that are used and determine the stereo chemistry that results in this closed ring system because the orbitals will either rotate dis rotatory or con rotatory depending on which conditions heat or light were used in the system importantly the same is true for 4 Pi systems as it is for six Pi systems or even 8 Pi systems in the case of this four Pi system we were going to end up with a cycline ring and notice that the equilibrium favors the products in the six Pi system but the reactants in the four Pi system largely this comes from The Ring strain that would would result in this cyob buttine ring for sigmatropic rearrangements we have cazin reactions and cope rearrangements and remember sigmatropic just means that the sigma bond is changing locations importantly for a sigmatropic rearrangement to occur this is going to move the pi electrons and end up breaking this Sigma Bond and moving it here so we are breaking this Sigma Bond and we are making a new one at this location this is going to result in a new product in the presence of heat where you end up moving that Sigma bond in addition to moving the pi electrons so now this would be the reaction that occurs to create this product and again remember the location of our Sigma bond is changing places which is why we call it a sigmatropic rearrangement cazin rearrangements also undergo a very similar process and notably this is going to lead to the cause of a sigma Bond being changed from locating between the oxygen and the carbon on this side to being present on the other side so in the presence of heat this is going to give us a new product where you have a carbon oxygen double bond as well as another Pi Bond on this side and notice that our Sigma bond has moved importantly all of these rearrangements are three three rearrangements because the carbons involved in this movement and where the breaking and the making of bonds occur there are three atoms up top and three atoms below so we often call those a 33 sigmatropic rearrangement now let's try some reactions at benzilic positions and reductions of benzene pause the video try these problems independently then resume the video to check your answers while Benzene itself is typically very unreactive the benzilic carbon position meaning the carbon position next to the Benzene ring typically is very reactive for that reason it can undergo things like oxidation using either of these conditions which would completely oxidize off the rest of the remaining carbon chain no matter what it is as long as there is a carbon hydrogen bond into a carboxilic acid so for this reason this product leads to the formation of a Caro oxyc acid at that Benzel position additionally since there is a carbon hydrogen bond at that Benzel position here we can do things like brominate using halogenation using a free radical specifically NBS is what allows us to produce that bromine radical therefore this product would be the bromination at that position through a mechanism that follows a radical formation in addition if there are already substituents at that benzilic position like for example a benzilic bromide we can do things like substitu reactions in this case it would be an sn1 reaction replacing the bromine with something like an alcohol and this would be an sn1 reaction because it is a quinary carbon and therefore it can't undergo a sn2 reaction where everything happens at the same time however in the next case we can follow an sn2 reaction where everything happens at the same time and the water would attack nucleophilically in order to replace that benzol bromide in order to give us a benzol alcohol we can also do things like elimination reactions specifically using h2so4 concentrated sulfuric acid you can eliminate this alcohol in order to produce an alken so an alken will now be present at that benzilic position the same is true if you use a strong base like sodium methoxide in the presence of a benzil bromide so for example this would produce the same product where you have one of the substituents as a methyl group and one as an alken also if you're getting value from this video make sure to subscribe to my channel for more chemistry content now let's get back to talking about the reduction of benzene now typically Benzene is going to be incredibly stable because it's an aromatic compound with a fully conjugated Pi system with a cyclic structure it satisfies hul's rule of 4N plus 2 pi electrons now importantly though under extremely forcing conditions in the presence of hydrogen gas rainy nickel at high temperatures and high pressures you can actually do a full reduction of benzene this would lead to the formation of cyclohexane where all of the pi bonds are completely reduced and replaced with carbon hydrogen bonds at every position in order to fully saturate this carbon the same is true for Benzene that contains positions that are either electron donating or electron withdrawing now importantly when you have an electron donating group this is going to lead to reduction at the other carbons so what I mean by that is that you're going to end up with two Pi bonds instead of three and importantly the reduction is going to happen at the carbons that are not attached to that electron donating group now importantly the opposite is true for when you have an electron withdrawing group in this case you would still end up with two electrons however they the reduction would actually happen at this position that contains this electron withdrawing Group which we know since it has this carbonal group therefore notice this is the position that was reduced as well as this carbon however in the presence of an electron donating group it's not the carbon attached to that substituent but in fact the ones adjacent to that so for that reason this is going to be the reduction with an electron donating group and this would be the reduction with an electron withdrawing group Benzene can also undergo a variety of what are called electrophilic aromatic substitution reactions when Benzene is introduced to bromine gas in the presence of a Lis acid of iron Tri bromine this is going to do a bromination an electrophilic aromatic substitution at one of the positions to place a bromine on the Benzene ring similarly chlorination can happen in the presence of chlorine gas and a different leis acid in aluminum Tri chloride so this would place a chlorine substituent on that Benzene ring nitration occurs in the presence of nitric acid and sulfuric acid where you can place a Nitro group onto the Benzene ring which is important because this can undergo a variety of subsequent reactions like reduction to a primary amine or an analine specifically in the presence of a metal like either iron or zinc in addition to hydrochloric acid followed by step two which introduces a base this produces a reduction of NO2 to nh2 so therefore we are able to produce an primary amine in this way creating the analine Benzene derivative adding fuming sulfuric acid by itself in the presence of benzene is actually going to place an S so3h group or a sulfination at the position where now we can do what's called a sulfination now importantly this reaction is one of the only ones that are reversible meaning you can go back to Benzene just by adding a little bit more of dilute sulfuric acid so dilute h2so4 will allow you to go backwards to bend meaning that it's one of the functional groups that can be installed to Benzene and subsequently removed the next two reactions are freal crafts reactions specifically isolation isolation and alkal so alkal and isolation are always done in the presence of aluminum Tri chloride where you can place either an asil group meaning a carbonal compound can be added in order to place for example this group or you can do an alkal just just to add a single alkal chain to your Benzene ring now importantly in this case we were using methyl chloride but you need to consider the fact that in the case of fredal crafts alkal if you're producing a primary carbocation through this process or this mechanism then you can get a rearrangement in these reactions for that reason if you wanted to install a linear alkal chain one would likely use freal crafts isolation followed by reduction of the carbonal group in order to produce a linear alkal chain we can form a linear alkal chain if we reduce this Ketone into an alkal group using for example zinc in the presence of mercury and HCL this would allow us to do the reduction of a ketone into an alkal chain now importantly if you do a fral crafts alkal you are creating a Benzel position or a Benzel carbon hydrogen bond with which you can do all of those reactions that we just learned about for example radical bromination at the Benzel position or oxidation into carboxilic acids now let's look at ketones and aldhy pause the video try these problems independently then resume the video to check your answers ketones are an incredibly versatile functional group that are capable of tremendous amounts of chemical reactions specifically those listed on the screen specifically we've talked about Reactions where you can form hydrates by replacing two of the alcohols or the waters that are present in the addition of a carbonal carbon which is called nucleophilic addition reactions other types of nucleophilic addition reactions are things like forming acetal so instead of having the hydrate formed instead you would have two ethers formed which would be called an acetal in the presence of a molecule that contains two alcohols which is called a diol you can make cyclic acetes as well which are an incredibly useful protecting group for things like ketones when undergoing multi-step synthesis additionally in the presence of an acid and a primary amine we can form things that are called amines so this is going to be a carbon nitrogen double bond and it is going to contain a single R Group coming off and this is what happens when you use a primary amine when using a secondary amine in the same reaction with a catalytic amount of acid you will form what are called enamines where you have the nitrogen to carbon single Bond the nitrogen is now a tertiary nitrogen and there is an alken at this POS at this carbon position now if you're using a catalytic amount of acid and hydren which is nh2 nh2 this is going to form what is called a hydrone so a nitrogen to nitrogen Bond is formed as part of this Iman structure i m i n and this can undergo subsequent reactions Like the Wolf kishner reduction where you can actually completely remove this to just create two brand new carbon hydrogen atoms and this is done in the presence of a strong base like sodium hydroxide typically water and also in the presence of heat in addition to these nucleophiles adding at the carbonal carbon position we can consider several others specifically nucleophilic carbons so for example a grenard reagent creates what's is effectively an R minus group or a nucleophilic carbon which will do a nucleophilic addition at that carbonal carbon position following acidic workup will allow us to create a new carbon carbon bond in addition to now creating that alcohol under basic conditions cyano groups or nitriles will also add nucleophilically at that carbonio carbon position creating what are called cyano hydrant so a cyano hydron is a carbon that contains both an alcohol and a nitr at the same carbon position we've also learned about using iic reagents which are going to be a carbon to phosphorus ID that means that you have a compound that is overall neutral but the adjacent hetero atoms are oppositely charged so remember a resonance structure for this would make the carbon nucleophilic and negatively charged whereas the phosphorus would be positively charged this is going to allow us to create new alken bonds or carbon carbon double bonds specifically replacing the oxygen in order to create a brand new alken at this position so this would be our product of our viig reaction and again that is viig VI TTI the last one is using an hydric acid in order to create an Esther so in this case we would create a brand new Esther compound which is a carbonal carbon where you have another carbon oxygen bond in addition to some R group that was present here now let's talk about carboxilic acids pause the video try these problems independently then resume the video to check your answers largely there are a few different ways to produce carboxilic acids we've already seen the case where you have a benzilic carbon hydrogen bond that can undergo oxidation with potassium permanganate in order to make a benzo carboxilic acid additionally we can make carboxilic acids by using alal bromides specifically if we have an alal bromide that can undergo a substitution reaction with sodium cyanide this will produce first a cyano group at the position where you had the bromade so for example this would produce an alkal cyano group from here remember that cyano groups can undergo oxidation to produce carox IC acids via the addition of an acid and heat therefore the subsequent transformation for this would be a carboxilic acid where now the carboxilic acid is formed at this position additionally we can form carboxilic acids via a grenard reagent so the first step in this reaction would be to create your grenard reagent where the Magnesium would insert into the alkal bromide Bond following the formation of the grard reagent the grard reagent would react with CO2 and after acidic workup would produce a carb carboxilic acid so this was another method to form carboxilic acids of alkal bromides finally it's important to remember that carboxilic acids can subsequently be reduced via lithium aluminum hydride or some Boron Ester species each of these would produce primary alcohols where you are reducing the carbonal oxygen so therefore you're leaving behind two brand new CH bonds now let's talk about the preparation and reactions of acid chlorides pause the video try these problems independently then resume to video to check your answers acid chlorides are going to be the most reactive carboxilic acid derivative in fact they can be turned into all of the other carboxilic acid derivatives so we just learned about how to turn a carboxilic acid into an acid chloride using Thal chloride and we can actually turn the acid chloride into a carboxilic acid simply by the addition of water we can also turn acid chlorides into Esters through the addition of of an alcohol so R where the r is any alcohol could be ethanol methanol Etc and pyodine so Pine is often the solvent used in this process and we add an alcohol to create a brand new Esther we can turn acid chlorides into amids so an amid is remember is going to be your carbonal that contains a nitrogen containing group on the other Carbon on the other side of the carbon just by adding excess amounts of an amine so ammonia for example and excess amount meaning XS ammonia would allow us to create a primary amid but we could also do something like create by using excess secondary amines to create a brand new tertiary amid so this is a way that we could add different types of R groups to our amid we can also use acid chlorides to create ketones so remember a ketone is going to be an R Group off coming off the carbonal by using an organometallic reagent which we'll learn about more in this course so R2 coer lithium it's called a Gilman reagent acid chlorides also undergo reactions with grard reagents so rmgbr for example is what's called a grard reagent and when you also have an acid in your system so h3o plus just meant to depict any sort of acid we can actually create a brand new alcohol species that contains two different R groups so we can create brand new tetrahedral carbons with an alcohol on them by adding an excess amount of these grenard reagents in the presence of acid we can create primary alcohols by using lithium aluminum hydride which will allow us to create a primary alcohol where the other R groups would just be hydrogen atoms and then finally we can also create alahh Tides so remember an aldah is going to be that carbonal with a hydrogen coming off the carbon by using this reagent so lithium aluminum O3 in the presence of an acid like acid chlorides and hydrides can be converted back to carboxilic acids simply with the presence of H2O or water again like acid chlorides and hydrides can be converted into Esters through the addition of an alcohol so an Esther can be created through the addition of an alcohol again like acid chlorides we can create primary amids simply by adding different amines to create brand new amid bonds so a primary amine could be formed through the addition of excess ammonia or if we wanted to make a secondary or tertiary amid we would simply add different types of amines to our system so excess amines can be used to create brand new amids from anhydrides again like acid chlorides we can actually create ketones by simply adding that Gil reagent which again was R2 copper lithium can be added to anhydrides to make ketones alcohols can be made by adding excess amounts of grenard reagents so again that's some R Group with magnesium bromide for example and some acid h3o+ we can create brand new alcohols as well and we can create ones that have a stereo Center potentially because we're creating uh tetrahedral alcohols just like acid chlorides we can also make primary alcohols by using something like excess lithium aluminum hydride as well as water or acid to create that primary alcohol so remember a primary alcohol is going to be attached to a carbon with two other hydrogen atoms attached to it and again like acid chlorides we can use lithium aluminum ethers in order to create brand new aldah es so this allows us to create alides which again is carbon to hydrogen with the carbonal functional group one of the simplest ways to make Esters is going to be through deproteination of a carboxilic acid followed by an sn2 or substitution reaction to create a brand new Esther Bond a more common way that people make Esters though is called fiser esterification where you use an alcohol and some catalytic acid and you can convert carboxilic acids very readily into brand new Esters and again this is called Fisher esterification so Fisher esterification is the most common way that these types of ERS are made in Industry the mechanism for Fisher esterification follows a pathway that you might predict where the catalytic amount of acid is allowed to protonate that carbonal oxygen to create a brand new positively charged species which will allow for this carbonal carbon to be a attacked by things like our alcohol so once we have made this more electrophilic the lone pair on an oxygen from alcohol will come up kick up those electrons to allow us to create a brand new tetrahedral alcohol and allow us to further undergo the transformation to an Esther through a series of proton transfers so remember we still have plenty of our alcohol present which has lone pairs on the oxygen which will deprotonate that o group allowing us to create a neutral species so now we have a fully neutral compound where we have two alcohols and an O group from here one of the alcohols can be protonated by the brand new acid that has been formed allowing us to create a good leaving group which was going to be an O2 derivative so we have our alcohol we also have a brand new o 2 that is positively charged and our o group from here the electrons can come down from one of the oxygen atoms and kick off that good leaving group allowing us to create a brand new protonated Esther where now this proton that is on the carbonal oxygen can subsequently be deprotonated by for example the alcohol that is still present in our compound allowing us to finally generate our brand new Esther nitrs are commonly made through substitution reactions with alal halides for example an alal bromide will undergo a substitution reaction with sodium cyanide to create a brand new nitr additionally nitriles can be made from amid bonds so remember an amid is going to be a carbonal compound that contains a nitrogen containing group on the carbonal carbon by adding again that Thal chloride that we've learned about in the preparation of acid chlorides in order to create our brain BR new nitr and the byproducts of this are going to be that sodium dioxide and hydrochloric acid now nitriles and amids are unique in that they really only undergo two different types of reactions so they are the least reactive of the carboxilic acid derivatives and therefore they typically only create one other type of functional group transformation and that is going to be the creation of a primary amine so a primary amine can be created from both of these from nitriles we just add lithium aluminum hydride in the presence of water to create our primary amine and we use the exact same reagent and actually the lithium aluminum hydride can reduce this carbonal oxygen to again create that primary amine now let's talk about enol and enolate reactions pause the video try these problems independently then resume the video to check your answers via a process that proceeds through enols and enolates we are able to do alpha Al halogenation or different types of reactions at Alpha carbons of carbonal compounds remember the alpha carbon is going to be near the carbonal carbon which is electrophilic and the alpha carbon is going to be the carbon that is directly adjacent to that so in this Ketone example there are two alpha carbons which contain very acidic carbon to hydrogen atoms of both ketones and carboxilic acids we can do an alpha halogenation meaning that at the alpha position we can place a hallogen in this case bromine under acidic conditions we can add a bromine atom to that Alpha carbon importantly for carboxilic acids there's only a single Alpha carbon since the other side contains the hydroxy portion of a carboxilic acid meaning that the carboxilic acid would be halogenated at that alpha position only so we would place a bromine at this position the hofor reaction is actually a two-step reaction in which you can take a ketone and turn one of the r groups into a carboxilic acid so this is going to place a hydroxy group here via a two-step process where first you brominate and then subsequently replace that bromine with the hydroxy group to make this carboxilic acid additionally enols and enolates allow us to create brand new carbon carbon bonds through what is called the alol reaction like alol addition or alol condensations in this first example an aldol addition is going to allow us to create a beta hydroxy Ketone where we can create a brand new carbon carbon bond that adjoins the these two species we call a beta hydroxy alahh because at the beta position is where the hydroxy group is again this is called an aldol addition importantly though through the addition of a tremendous amount of heat we can actually do what is called an aldol condensation where the entire hydroxy group is actually removed and what you are left with is a alpha and Alpha Beta unsaturated aldhy so now you have the alpha beta because at the alpha and beta positions is where you can find this unsaturated alken we can also do crossed alol additions where you have two different types of carbonal compounds that react to form a new carbon carbon bond between the two that's called a crossed aldog reaction and in the presence of a strong base and heat we can do this cross alol addition allowing us to generate that Alpha Beta unsaturated aldah so in this case we would create this Alpha Beta unsaturated alahh and again the name unsaturated comes from the fact that there is an alen present in here and again this is at the alpha and beta positions additionally it's possible to do a cyclisation aldol reaction in where you deprotonate the alpha hydrogen at one of these positions and it can react at a different carbonal functional group on the same molecule this is called an intr molecular aldol condensation and the product of this reaction is going to contain the same number of carbon atoms in which case this is going to be six but importantly the reaction happen happens at the alpha position which would be here and to the five position so our ring would actually be a five membered ring with a methyl group coming off of it so this final product should look like a five membered Brain still remaining our Ketone and it's going to be unsaturated and un at that unsaturated position is where that methyl group is coming off for Esters we can also do what are called cazin condensations in a cazin condensation we are again forming that enolate between the alpha carbon and the carbonal group of an Esther using the deprotonation of a strong base like sodium ethoxide or LDA we can do this claz in condensation in order to produce a brand new carbon carbon Bond where you will be left with a product where you have formed this new carbon carbon Bond at this position you can also do a crossed cazin condensation between two different Esters as long as one of them does not contain a free Alpha carbon that contains a hydrogen bonding that would be deproteinated in order to get a single type of product for example this one is the only one that contains an acidic Alpha carbon hydrogen which would be deprotonated by the strong base like sodium ethoxide or LDA generating a carboncarbon bond from this position to here so that is going to allow us to make our product which again is going to contain that brand new carbon carbon Bond and you would lose the a oxide portion thus allowing us to generate this product where you are left with this Esther on this side and then just like for alol reactions we can also do an intr molecular meaning on the same molecule condensation which would allow you to produce this new five membered ring where the Ketone is here and the new carbon carbon bond is formed at this position leaving behind our Esther now let's talk about enol and enolate reactions pause the video try these problems independently then resume the video to check your answers via enate ions we can also alalade at the alpha carbon position of different types of molecules importantly depending on which base you use like sodium hydride or LDA you will get a different isomer or a different constitutional isomer that is for form Med at the alkal position of each Alpha hydrogen via enolate ions you can also alkal Alpha carbons depending on which base you use at different locations so for example for sodium hydride since it is a smaller base it would deprotonate the alpha carbon at this position the more substituted or LDA would instead deprotonate the Alpac carbon hydrogen that is at the less substituted position allowing us to create an enolate that will allow us to form two different types of alkal positions the R Group will undergo a substitution reaction at those deprotonated positions so as I mentioned for the sodium hydride example the alkal would take place at the more substituted position whereas for LDA the less substituted position would be the one that gets alculated so for that reason that product would be alkal here importantly you can also do Michel additions once you have formed enols or enolates through stabilized carbon nucleophiles like Gilman reagents or other Michel acccept ctors so in this first example we would alkal at this position because the Gilman reagent is less hard or is a softer leis acid allowing us to alkal at this position so the product that's that would be formed is going to be alkal at that position allowing us to generate a product that has a new R Group at this position additionally we can do Reactions where we would deprotonate this Alpha hydrogen which is going to be incredibly acidic due to its position between these two carbonal groups allowing us to alkal with this Michael acceptor again at this position which is going to allow us to make a product that looks like this and where we still have these two carbonal groups and the alkal position is here so this is going to be our final product now let's talk about the preparation of amines pause the video try these problems independently then resume the video to check your answers for synthesizing am meines I've grouped these three reactions below because they follow a similar pathway first transform a functional group into something else and then subsequently reduce that functional group which likely contains the nitrogen in it into a primary amine for example the first step is going to extend our carbon chain by adding a CN group to the position where the hydro where the bromine used to be doing a substitution reaction specifically an sn2 reaction now from here lithium aluminum hydride followed by the addition of water allows us to reduce that position and extend our carbon chain to make this primary amine notice that this product contains one additional carbon to our starting material which is what sodium cyanide does by extending the carbon chain conversely using sodium Aid you would still do a substitution reaction however importantly the carbon chain is not extended because this Aid molecule will do a substitution and not extend the carbon chain however once you have produce this azid complex this alkal Aid you can do reduction using either hydrogen over platinum or again that lithium aluminum hydride followed by the addition of water in order to reduce this to the primary amine which is going to now be nh2 the next example uses carboxilic acids to First convert to acid chlorides using Thal chloride and then subsequently create the amid Bond using an excess amount of the amine so our product here is going to be an amid once we produce this amid we can reduce the carbonal functional group using again lithium aluminum hydride followed by the addition of water to create our primary amine now importantly all of these different methods allow us to produce primary amines additionally we know of the gabrial synthesis and reductive amination which allow us to synthesize Ames specifically using this phthalamide molecule we can make am meanes first by deprotonating the hydrogen can attach to the nitrogen making a very nucleophilic nitrogen which would undergo a substitution reaction of alal bromides and then using hydren to liberate the thalomid molecule allowing us to leave behind the primary amine which in this case is going to contain a product which looks like this for reductive amination we are always doing a process that proceeds through an imine so first we proteinate the carbonal oxygen then the amine will attack the carbonal carbon position and then subsequently sodium cyanoborohydride will do a reduction of an amine in order to make an amine using ammonia this is how we make primary amines so in this case this would be our product using a primary amine as our starting material we would make a secondary amine where you have a nitrogen attached to two R groups and starting with a secondary amine we can make a tertiary amine so in this case there are going to be three R groups coming off of that nitrogen making it a tertiary amine now let's talk about the reactions of aerody diazonium salts pause the video try these problems independently then resume the video to check your answers one of the most popular reactions for this are called sandmire Reactions where you take copper salts like for example copper chloride and you can exchange the substituent to chloride from the diazonium salt similarly we can do this for copper iodide and copper bromide to generate the new Aero halide species in exchange for the diazonium salt the sandmire reaction also works for Cyanide so you can use copper cyanide to exchange this position for a new cyano group and importantly remember that we have already learned about Reactions where you can transform cyano groups into for example carox IC acids through oxidation so this is a powerful way to install brand new functional groups along an aeral compound there are other examples as well that are incredibly powerful one of those such examples are fluorination where you can actually exchange the diazonium salt for a fluoride another example that I want you to consider that may seem relatively simple or straightforward is going to be reduction through adding something like phosphoric acid we can actually exchange that Di I onium salt for just a simple hydrogen and I want you to keep that one in mind because we can actually use diazonium salts to direct functional groups and direct synthesis and then subsequently reduce the diazonium salt to get rid of it so it may be a powerful way for you to utilize this tool in organic synthesis and then the final example that I want you to consider is the fact that we can exchange just diazonium salt for an alcohol and we do this by using the conditions of using sulfuric acid h2so4 in water with a little bit of heat which I'll indicate with the Delta symbol now let's talk about some reactions involving Organo metallic compounds pause the video try these problems independently then resume the video to check your answers harnessing the power of Organo metallic chemistry allows us to do incredible Transformations like form new carbon carbon bonds importantly the first step then is going to be to create your Organo metallic reagents which are anything that has a metal attached to a carbon the way that we make lithium reagents is by taking an alkoh halide adding two two equivalents of lithium and forming our new organolithium species similarly we form grenard reagents by starting with an aloh halide adding magnesium turnings in ether and forming our grenard reagent which inserts itself into the aloh halide position allowing us to create our grenard reagent we can also create what are Gilman reagents through a two-step process which Begins by making an organolithium species same as in the previous reaction however following that up with a copper salt allowing us to create a brand new Gilman reagent which contains a copper and lithium in it we also learned about Suzuki couplings which allow us to take any vinyl or aerob borinate eser species and couple it with an alkal or Aero halide to create a brand new carbon carbon bond by attaching the new species at this position where you would split the alalh halide and the organoboronate Esther species forming the new carbon carbon Bond at that location this allows us to make brand new carbon carbon bonds which contain a brand new carbon carbon Bond formed at that position and again this is called Suzuki coupling and that covers all the reactions of organic chemistry 2 if you want to dive deeper into any of these reactions I would encourage you to check out this playlist for more in-depth videos if you found this content useful I would encourage you to subscribe to my channel to check out more chemistry content I'll see you in the next video