hello so today i'll be reviewing as organic chemistry as fast as possible right i've divided the video into three sections the first part deals with an introduction to isomerism and some general terms that we need to know the second part deals with functional groups their production and the reactions they undergo including reactions reaction reagents and conditions and the last part consists of mechanisms involved in as chemistry so i'm gonna give timestamps in the description below so you can skip to them if you want all right so let's start so an organic compound typically contains carbon hydrogen and it may contain other elements as well so what is a hydrocarbon it is a compound made up of carbon and hydrogen only and what are isomers isomers are basically compounds with the same molecular formula but different structural formula or spatial arrangement of atoms they are of two types structural isomers and stereoisomers so structural isomers have the same molecular formula but different structural formula structure isomers are three types right chain isomers position isomers positional and functional group isomers so in chain isomers we basically juggle the position of the carbon atoms in the carbon skeleton right suppose we have the five carbon alkane in pentane an isomer is two methyl butane and another one is two two hyphen dimethyl propane so remember to give commas in between numbers and dashes have to be placed after before and after numbers all right that's the basic we these are the basics so anyway in position isomers it's kind of the same but here we're juggling the position the functional group around for example the five carbon alcohol pentane one all then an isomer exists spin ten tool and penton threol here the oh group is just changing positions between carbon number one two and three all right what about the last one for the last one we have functional group isomerism here the functional group completely changes they exist as pairs for example aldehydes are the functional group isomers of ketones esters are the functional group isomers of carboxylic acids and alcohols are functional group isomers of ethers ethers are not in our syllabus by the way so for example propanel is the functional group isomer of propanol italythenoid is the functional group isomer of butanoic acid the four carbon carboxylic acid and ethanol is the functional group isomer of dme dimethyl ether moving on to stereoisomers they are compounds with the same structural formula but different arrangement of atoms in space or different spatial arrangement of atoms there are of two types c strands or geometric isomers and on the other hand we have optical isomers so for c strands or geometric isomerism we require a carbon-carbon double bond an alkene basically sp2 carbon 120 degrees planar and we need two different atoms or groups of atoms attached to each carbon atom so for example if two hydrogens are attached to the same carbon atom we won't have cis trans isomerism it depends on like why do we get two different isomers because free rotation is not allowed right about the carbon-carbon double bond that is why so if the dipoles do cancel out that means if the heavy groups are on opposite sides we say that we have the trans isomer for example trans between is seen on the red and if the heavy groups are on one side and dipoles do not cancel out right we say that it's the cis isomer cis between now according to our syllabus something new has been added it may also be present in cyclic compounds how exactly for example if the heavy groups right if they're on the same side of the group same side of the ring i mean we say that we have the cis isomer cis3 methylcyclopentanol both heavy groups are on one side of the ring but what if the oh is above the ring and the ch3 is below the ring as you see here in the diagram ch3 is connected to the dotted line which represents that it's below the ring so if the one heavy group is above the ring and the other heavy group is below the ring it's basically a trans isomer okay so for optical isomerism we require a chiral carbon or an asymmetric asymmetric carbon it's a carbon atom attached to four different groups of atoms or atoms right and the mirror images are non-superimposable as you can see here you have to make sure that the images are naturally inverted look at how i drew the co h right it's laterally inverted make sure the weight shaped diagram is fine as well it has to be laterally inverted or you won't be getting marks okay we use an asterisk to denote the chiral carbon okay so each of the optical isomers are called enantiomers this has been added to the syllabus as well especially nato they rotate like physically they are the same physical they are the same physically and capably they have the same melting and boiling points they undergo the same reactions however they rotate to plane polarized light in opposite directions okay as you've seen in malice's law right the new addition to is physics plane polarized light only oscillates in one plane basically so each enantiomer basically rotates the plane rotates plane power is right in different directions for example the one on the right may rotate it clockwise dexter auditory or and the one on the left may rotate it anti-clockwise which is labor auditory right so if we have a mixture containing equal amounts of both enantiomers right it's called a racemic mixture if we pass plane positive polarized light through it it will not change its direction of oscillation all right and if you're asked what is an enantiomer it's basically one of those two one of the optical isomers that are formed okay so now we are dealing with some general terms fission for example fission means splitting up or breaking down it is of two types we have homolytic fission and heterolytic fission so homolytic fission means equal right homo equal and hetero means different so for homolytic fission it typically occurs in non-polar compounds like bromine chlorine we get free radicals right the covalent bond breaks and each each atom takes its own electron back basically homatic fission is a type of bond breaking in which the shared pair of electrons are equally divided between the two atoms molecules free radicals are formed right so what is a free radical it is a species with one or more unpaired electrons in its outer shell they are formed after homolytic fission they are extremely reactive because the unpaired electron has a tendency to become paired as soon as possible it wants the octet rate just like the chlorine free radical so we usually see homolytic fission in free radical substitution in all other mechanisms we are going to see heterolytic fission it typically occurs in a polar molecule so for example a b it breaks down to form the a anion and the b cation or vice versa so you show it in that way you see how i showed it with the blue blue pen right so that's heterolytic fission basically is taking both the electrons from the shared pair and remember that the more electronegative atom becomes the anion okay it is a type of bond breaking where one of the atoms or molecules takes both the electrons in the shared pair ions are formed after heterolytic fission classification of reagents regions are of two types in organic chemistry they are electrophiles or nucleophiles so electrophiles mean something that loves another electron so something that loves something negative right something that loves negative particles so it has to be positive right so electrophiles are positive in nature on the other hand nucleophiles are nucleus loving things that love the nucleus so if you want to love the nucleus which contains protons you need to be negative so i typically remember it this way like negative has an n right negative has an n so nucleophiles also have an n that's how i remember like it's faster to remember so next moving on an atom molecule can be a nucleophile electrophile without being charged okay for example water and ammonia they are nucleophiles because they have lone pairs okay or an alkene can be an uh can be a nucleophile because it has an electron rich pi bond right so the curly arrow must be from the nucleophile electrophile remember the electrophile can attack the nucleophile as we've seen in a2 chemistry where the nitronemia attacks the benzene ring for example uh you're gonna learn this next year like the attack can be from the electrophile to the nucleophile however the arrow has to be drawn from the nucleophile to the electrophile and the arrow must be curly the curly arrow represents transfer of electrons so next we need to know what a carbocation is it's an organic compound with a positively charged carbon atom it is of three types why are carbocations form so easily typically because carbon is on comparatively it's on the left hand side of the periodic table so it has low electronegativity so the other compounds except hydrogen which are attached to it have a tendency to take away its electron so carbocations are four types the methyl carbocation primary secondary and tertiary carbocations down this group right stability increases why is that because alkyl groups have a positive inductive effect what does that mean they have a tendency to donate electrons so think about it a carbocation is positive in nature it wants to react with something that is negative right so if other groups donate electrons towards it the intensity of the positive charge on the carbocation will actually decrease making it a weaker electrophile right so on the other hand the primary carbocation it only has one alkyl group so the intensity of the of the positive charge is not reduced that much right comparatively we say that electron-wasting groups have positive inductive effect alkyl groups are electron releasing positive inductive effect okay why is a tertiary carbocation the most stable it has three alkyl groups attached to the carbon atom alkyl groups are positive inductive effect so they stabilize the positive charge on the carbon atom which means reducing the intensity of the positive charge so you need to know this this is the relative stability of carbocations okay now we're moving on to the second section of the video where i'm talking about alkanes all the functional groups the reactions how we produce them so try to connect the dots okay so to produce alkanes we have two roots this is directly taken from the syllabus okay hopefully i haven't missed anything if i did please let me know so we can produce alkanes from alkenes by an addition reaction which is called hydrogenation using a nickel or platinum catalyst plus we require we also require heat [Music] okay so we also require heat and we're going to get an alkane there's another route we can use alumina catalyst along with heat for cracking we break down bigger hydrocarbons to form smaller ones we get an alkane and alkene and this is done to produce more economically favorable smaller alkanes and alkenes which are more useful for us right they have a few reactions they can undergo combustion uh which is of two types complete and incomplete in complete combustion we get carbon dioxide and water you can balance this it's pretty simple and in in complete combustion due to lack of oxygen we can either get carbon monoxide which is a harmful gas it's it binds to hemoglobin irreversibly or we can get suit just carbon when we have very less oxygen present okay what is the other reaction of alkanes free radical substitution uh basically alkanes react with bromine or chlorine in the presence of uv light and basically chlorine substitutes the substitutes one hydrogen in the alkane and this is a chain reaction it keeps on occurring until all hydrogens have been removed i've discussed the mechanism at the end of the video okay so alkenes all all alkanes are sp3 carbons by the way 109.5 degrees and tetrahedral in shape all alkenes are sp2 carbons they are 120 degrees bond angle and the shape is trigonal planar so to produce alkenes we have multiple roots we can either dehydrate an alcohol using aluminium oxide and heat alumina pumice or concentrate h2so4 plus heat or concentrate h3po4 plus heat in some mark schemes they show heat with h3po4 sometimes they don't you can either crack you can also crack alkanes red we saw that beforehand using alumina plus heat and we can also eliminate halogen alkanes using ethanolic naoh okay along with heat to get alkenes that's an elimination reaction these are the three roots so what are the reactions mainly they undergo electrophilic addition you will classify all of these as addition reactions okay in the marxism sometimes they don't accept hydrogenation it's an informal term so even though it's in the syllabus right so the first reaction is hydrogenation this is also addition we had hydrogen gas with the help of nickel or platinum catalyst and heat to get alkanes we can add water in the form of steam in the presence of heat and h3po for catalyst rate to get alcohols they also react with hydrogen halides hx and gaseous form in room temperature right under these conditions to form halogen alkanes no heat is required they also react with chlorine and bromine and halogens in general in uh no heat is required as well this occurs at room temperature to get a dibromo or a dichloro compound all right this is also the test for unsaturation because we know that when bromine is added to alkanes the um it gets decolorized right so these are the electrophilic addition reactions i want to talk about something else markovnikov addition what does this state like in this alkene over here propane if we add hbr to the compound we have two roots we have two roots basically we can either get one bromo propane or two bromo propane however we see that right so in markovnikov addition we have two possible routes either from propine we can we can get one bromopropane or two roma propane but two brown propane is the major product at around 70 to 80 percent why is that because the intermediates produced during the electrophilic addition mechanism are the one above and below respectively so the one produced above for one bromo propane is a primary carbocation and the one below is a secondary carbocation which has more inductive effect positive inductive effect due to the presence of two methyl groups that is why it's more stable that's why we get that at in a higher quantity okay so the next reaction is oxidation it can be done in two ways we can either use cold dilute camino for estified cold dilute camino for with alkenes to get um diols or we can use hot concentrated kmno4 which is acidified as well this causes oxidative cleavage or rupture of the alkene so there are three possibilities if there are two alkyl groups attached look at these hands red these are the alkyl groups if there are two alkyl groups attached to the carbon we are going to get ketones okay if there is one alkyl group we're gonna get carboxylic acids look there's only one hand over here regardless of the length of the chain right if there's one alkyl group we're going to get carboxylic acids what if what if there are zero alkyl groups so for example in ethene we basically get methanoic acid first which is unstable in the presence of um kmno4 right so further oxidative cleavage occurs you're going to see this in edo chemistry a lot though not in s um so we get carbon dioxide and water afterwards okay so there i would just want to say something there are two acids which are unstable in the presence of cheminophore methanolic acid and ethanolic acid these further oxidized to form carbon dioxide and water okay so these are the smallest carbon containing products all right so what is the last reaction of alkenes they undergo addition polymerization for example ethane can undergo polymerization to form polyethylene i'm showing you the repeat unit you have to show dangling bonds in the case of repeat units okay next we have halogen alkanes we can produce them mainly by free radical substitution of alkenes in the presence of uv light with chlorine or bromine we can add hydrogen headlights to alkenes in room temperature at room temperature okay so there are multiple ways we can produce them using alcohols we can use pcl5 and at room temperature we're going to get the halogen alkane more on this later we can use pcl3 along with heat to get the halogen alkane we can use thyrochloride to get the halogen alkane this also occurs at room temperature we can use a hydrogen helite we heat or heat under reflux okay um in this preparation is in situ in inside the reaction vessel basically or we can use uh halide salt like kbr and a concentrated acid like h2so4 or h3po4 this is also in situ preparation we need heat under reflux basically this is this is under the halogens chapter kbr first reacts with h2so4 to form hbr the hydrogen headlight then the hydrogen halide reacts with the alcohol to form the halogen alkane why do we do it in this way because the hydrogen head is typically gaseous right so it's very difficult to make it react directly with the alcohol so we need this in-situ preparation okay heat on the reflux okay what reactions do they undergo nucleophilic substitution mainly this is the characteristic reaction of halogen alkanes three nucleophilic substitution reactions are very important but before that remember that there are two types of nucleophilic substitution three degree hydrogen alkanes undergo sn1 while one degree halogen alkanes undergoes and two nucleophilic substitution reactions two degree can take part in both so if you're given a question with the two degree halogen alkane if you draw either one you're going to get full marks okay it's fine so the three nucleophilic substitution reactions are the first one is with naoh aqueous uh it does react with water also but it requires a lot of heat it's going to be a very slow reaction in the presence of nh aqueous it's going to be a very fast reaction this reaction is classified as hydrolysis the hydroxide basically replaces the chloride in the halogen alkyne we get an alcohol we can also react it with ethanolic ammonia and we have to apply heat in a sealed tube or heat under pressure that's what you need to write to get marks we're gonna get the amine so here's the thing this is a pretty hard one you need to understand that the amine can also act as a nucleophile why because it has a lone pair i haven't seen this in as yet but it was in the marsh feb 20 paper for a2 basically it can act as a nucleophile again because it has a lone pair so it can react with the halogen alkane again if it's in the same container so the reaction will keep on occurring until the nitrogen runs out of hydrogens okay to replace the chlorine or bromine in the halogen alkane okay it's a chain reaction kind of like free radical substitution what's the last nucleophilic substitution the halogen alkane can react with kc and ethanol the casein in the presence of heat to form nitriles okay so there is another reaction though this was mentioned in the syllabus um silver nitrate ethanol silver nitrate can be reacted with the halogen alkane to actually figure out which which halogen was present in the alkane right so if you get a white precipitate it's going to be silver chloride if it's cream it's silver bromide and if it's yellow it's silver iodide and it's important to mention that the silver iodide i mean the yellow precipitate will appear the fastest why because the cx the ci bond is the longest right it's the weakest and there's minimum orbital overlacking overlapping sorry so what's the next reaction halogen alkanes also undergo elimination in the presence of ethanolic naoh we get alkenes all right this is very important we get we get alkenes so the inorganic program ducks are actually um so each cell is removed from the halogen alkaline right so it actually reacts with nah afterwards to form nacl right plus h2o these are the inorganic products be careful okay so this follows besides the rule uh which states that if there are two possible elimination products we'll get the major part as the one which has more alkyl groups so for for example we had two chloropropane right so the product the major product will actually be a buta2 in rather than but one in because it has more alkyl groups it forms a more stable intermediate moving on to alcohols we can produce alcohols by hydration of alkenes steam plus h3po for catalyst oxidation of alkanes using coal dilute kmno4 right dials um hydrolysis of halogen alkanes using naoh equals plus heat okay and also we can reduce aldehydes and ketones using both li lh for n and nabh4 both work although li h4 is much stronger um these are the reduction equations using nascent nascent hydrogen okay so you can use this nice and hydrogen to show this this come quite often so you will only give a water on the right hand side if necessary for example i did not require water to balance the equation so i did not give water on the right hand side okay we can also reduce carboxylic acids using li lh4 only to get the primary alcohol okay so algae heads give us primary alcohols ketones give us secondary alcohols what about the hydrolysis of esters um they also give us alcohols this can be done in two ways either acidic hydrolysis or alkaline hydrolysis in the presence of heat we're going to get the alcohol here i broke down ethyl ethylene to form ethanol what reactions to alcohols undergo they will also undergo combustion to form carbon dioxide and water they react with active metals to form alkoxides for example ethanol reacts with sodium to form sodium ethoxide why the alcohol is slightly acidic what about this they also undergo nucleophilic substitution now i'm going to talk about these in details it reacts with pcl5 and so cl2 at room temperature to form the halogen alkane for pcl5 it will produce phosphoryl chloride pocl3 and steamy fumes of h7pc it will react with pcl3 in the presence of heat to form phosphorus acid it will also react with socl to at room temperature to form so2 and itself this is the best way to prepare the halogen alkane you'll find this in mcqs often because all the other products are gases so they will leave the container you'll only end up with the halogen alkane so i've missed two reactions here actually so pcl3 is quite stable but pbr3 and pi3 are not that stable so if you want to uh get the bromo alkane or the ideal kin you can opt to go for pbr3 and pi3 but they need to be made in c2 they need to be made in c2 for example we can use red phosphorus which is literally red in color it's not hot um red phosphorus and bromine right you can use this to make pbr3 in c2 then this can react with our alcohol ch3 ch2oh to form ch3ch2br right so this is one way of preparing um halogenobromides or iodides okay i mean bromo alkenes basically um they also react with hx in the presence of heat under reflux and you know the institute preparation kbr h2so4 to form hx and then following the reaction with alcohols to form halogen alkanes this has to be done in situ because hx is very unstable and it's very hard to react directly so we need to make it beforehand inside the vessel because the reaction temperature temperature is too high right it needs heat so the gaseous product would it would escape so we need to make it inside the container and then react it okay typically we heat under reflux when the temperature at which the reaction occurs is higher than the boiling points of any one of the reagents okay so we can also oxidize alcohols using k2cr27 st5 k2cr27 which is preferred or camino 4 estified primary alcohols are oxidized to carboxylic acids if they are heated under reflux for a long time for some time actually they can also be uh access to ld heads basically it's the root is like this primary alcohols are oxidized to ld heads followed by carboxy acids so you can get the aldehyde if you heat it and distill it almost immediately so distillation needs to take place almost immediately so secondary alcohols are oxidized to form ketones i've shown you the reactions so here i'm using nest and oxygen but like i'm using water right or else the equation wouldn't have been balanced that's why i need to use the h2 on the right hand side okay so tertiary alcohols cannot be oxidized you'll frequently get trick questions okay remember that got it so they also undergo dehydration we have three roots for dehydration alumina plus heat concentrate h2so4 plus heat and concentrate h3po4 you will see that this one does not require heat okay for dehydration uh we will get one molecule of water afterwards so for estrification uh alcohols react with carboxylic acids in the presence of a few drops of concentrated restores each other for and heat under reflux to form the ester so some secondary alcohols can be identified by the iodoform test we're gonna get a yellow precipitate more on this later when i discuss carbonyl so all carbonyl carbons are sp2 hybridized there are two types of carbonyl aldehydes or ketones aldehyde is cho and ketones are co r on both sides how to produce them so ld has a produced by oxidizing primary alcohols followed by distillation using k2cr27 acidified or kmno4 acidified i'm showing you the reaction here ketones are produced by oxidizing secondary alcohols so here uh we don't need distillation because that's the only product we're gonna get but to collect it we do need to distill it but we don't need to do it immediately okay where's the characteristic reaction of carbonyl they undergo nucleophilic addition reactions okay so remember the reagent is hcn and the catalyst is kcn or nscn what is the function of the catalyst basically hcn cannot provide a strong enough nucleophile so we use kcn or nscn to provide the cyanide ion cn minus in grade amounts okay which which gets regenerated at the end of the reaction so what do we get at the end of this reaction hydroxy nitriles are produced okay hydroxy nitriles that's the class of the compound produced so they also undergo reduction uh ld heads get reduced in the presence of you know either l lil h4 or nabh4 right to produce prime alcohols and secondary alcohols get reduced using the same reagents no heat required okay you'll see that in the book um nabh4 requires heat but in the market stream they don't mention that so secondary so ketones are reduced to secondary alcohols so the test for carbonyl is using two four dinitrophenyl hydrazine two for dnph this is a condensation reaction as you can see i have shown you here they react with ethanol is reacting with two for dna ph to form this orange precipitate and one molecule of water is liberated okay so what are the tests for aldehydes uh either the tolerance using the joint reagent or failings reagent tolerance reagent is a solution of ammonical silver nitrate a positive result is a silver mirror we need to warm the test tube containing tolerance reagent and the aldehyde or the sample and for fillings reagent this is basically bandix solution the copper 2 plus iron turns into cu plus and we get a brick red color of co2o as seen in biology if reducing sugar is present we also need to warm the container here so both of these are tests for aldehydes not ketones okay and the type of reaction is oxidation as the aldehyde is oxidized to form a carboxylic acid or to be honest a salt since the solution is alkaline the carboxy acid further reacts with the alkali in the solution to form a salt one important thing to note is that for you methanol acid also gives a positive result because hcoh even though it's a carboxylic acid it has a aldehyde group and it's easy to break i i taught you before hundred like uh it is oxidized in the presence of k104 and k207 in fact to form which just came out of camera for two from co2 and h2 okay so moving on we have a test for methyl ketones it's called the iodide from test using aqueous alkaline iodine the positive result is a yellow precipitate in the book they mentioned warming the testing but in the christian paper they don't mention warming okay so you don't need to say that the type of reaction is either oxidation or hydrolysis since the ch3 bond is broken it has two steps first we halogenate the methyl group followed by hydrolysis so who gives the positive methyl positive test for the iodo from test rate so there are two exceptions ethanol and ethanol and all methyl ketones and some secondary alcohols which have a methyl group at one side okay i've mentioned it here this is a typical reaction for the iodophone test we have butanone over here for carbon ketone it reacts with three molecules of iodine and four hydroxide as to form this basically hydrolysis occurs right of the ch3 bond we get chi3 from there and the ketone is actually oxidized to form a carboxylic acid or the salt basically since the solution is alkaline and some iodide ions and water molecules are liberated so the next thing on our list is carboxylic acid right these are all sp2 hybridized carbons 120 degrees so how do we produce carboxylic acids either by oxidizing uh one degree alcohols or aldehydes using k2cr27 sd5 or k104 plus heat under reflux so very few reactions require heat and reflux basically oxidation of alcohols and esterification in the other ones you can just apply heat it's going to be fine some nucleophilic substitution reactions do require heat under reflux like preparing some you know halogen alkanes from alcohols using hx right so you can also prepare them by hydrolyzing nitriles um sd cardiolysis h plus equals plus heat this is what you need to write to get marks in the questionnaire to get carboxylic acid the inorganic product is not important you can also hydrolyze it using an alkali but you need to estimate later on okay so it gives us the same thing it's fine you can also hydrolyze esters using h plus equals plus heat or alkali then acidifying you'll get the carboxylic acid so a carboxylic acid is an acid right so it's going to react with metals to form a salt plus hydrogen gas it's going to react with alkalis to form a salt plus water it's going to react with carbonates to form salt plus water plus carbon dioxide this is very important because alcohols do not react with sodium carbonate so it's important for as chemistry na2 you can differentiate between phenol and stop because alcohols will react with sodium so will carboxylic acids right but only carboxy acids will react with sodium carbonate remember that carbonates in general so they also undergo esterification they react with alcohols in the presence of a few drops of concentrated hgso4 and heat under reflex to form esters they can be reduced using li lh4 only we cannot use nabh4 because it's not strong enough um this is the reaction we're gonna get an alcohol primary alcohol okay so esters we can produce them by reacting alcohols and carboxylic acids and concentrated hgso4 plus heat under reflux to give us the ester ethyl ethanoid they can be hydrolyzed in two ways either in acidic conditions or alkaline acidic hydrolysis of esters gives us uh carboxylic acid and alcohol alkaline hydrolysis gives us a salt and alcohol because the carboxylic acid reacts with the alkali to form a salt so what about nitriles we can produce naturals in two ways either nucleophilic substitution of halogen alkanes using kc and ethernet casin heat and the halogen alkane to form um this nitrile or using nucleophilic addition of carbonyl compounds so this actually requires heat according to the syllabus although you won't see it in questionnaire at times it actually requires heat okay nucleophilic addition requires heat so ethanel um for example reacts with hcn the reagent in the presence of casein catalyst and heat to produce hydroxy nitriles why are these two reactions so import in organic chemistry because using them we can increase the number of carbon atoms okay so this is very important for drug synthesis okay so for uh the next one we can hydrolyze nitriles to form carboxylic acids alkaline hydrolysis also works but you need to acidify it to get the carboxylic acid and you can reduce naturals using allyl h4 or hydrogen plus nickel or platinum catalyst to get amines okay this isn't in the syllabus but i'm still gonna say this because you're gonna find this in paper one at times this is important for a2 so amites can be hydrolyzed in two ways acidic hydrolysis gives us the salt from the amine because the amine is basic due to the lone pair it reacts with the acid to form the salt and you're gonna get a carboxylic acid and alkaline hydrolysis is gonna give us a salt because the acid reacts with alkali to give us a salt and an amine okay we're moving on to the last section of the video where i talk about mechanisms we have four mechanisms in as and six mechanisms in a2 in a2 we have electrophilic substitution and addition elimination in as we have extra i mean and it needs we have free radical substitution electrophilic addition nucleophilic substitution and nucleophilic addition okay so starting with electrophilic i mean free radical substitution this is typically a reaction of alkanes okay um so arrows are not required in this mechanism but i saw it in the specimen 2022 paper for skim street paper too so you can learn it i'm going to show you we have three steps initiation propagation termination in initiation no free radicals are involved but we do get two due to the presence of uv light to the halogen breaks to form free radicals in propagation it actually occurs in multiple steps we have two steps here methane actually reacts with the chlorine free radical to give us the ch3 free radical and hcl then the ch3 free radical reacts with chlorine again to give us chloro methane and the free radical if you see this the chlorine free radical has been regenerated at the android so this is an example of catalysis and the breakdown of the ozone layer also occurs by free radical substitution that's why cfcs are bad they deplete those on there okay and in termination so the two free articles always remember in a termination reaction two free radicals will react to form a product with no free radicals okay and you can show these arrows as i've shown here i've seen this in the specimen paper okay the arrows can be shown here as well but i don't think they're required okay i saw this in the syllabus they're saying that arrows are not required so it's fine so electrophilic addition is a characteristic reaction of alkenes i've shown you two reactions here one with a non-polar bromine molecule when the bromine comes to close proximity to the uh electron rich pi bond the electron cloud is actually pushed away so it gets it becomes polar a dipole is induced so the attack is from the pi bond to the like the arrow is from the pi bond to the electro positive bromine over here and heterolytic fusion occurs so in the second step we actually get a carbocation and the bromide anion so the bromide and then will attack the carbon at the middle so we actually had a possibility of you know the carbocation being formed at the end the carbocation could also be formed here but it's just that the one being formed at the middle is much more likely because uh it's forming a two degree two degree carbocation which is much more stable so this is more common part b here um the alkene is reacting with hbr it already has an inherited dipole okay so heterolytic fusion is much more easier so in the first step we show the curly arrows the attack occurs and then in the second step the nucleophile is going to attack the um carbocation so why is it called electrophilic substitution because the electrophile right h plus with a partial positive charge is actually substituting or taking it's attacking the alkene essentially that's why it's called electrophilic addition okay we are referring to the first step so what about nucleophilic substitution this is a bit more complex nucleophilic substitution is of two types either sn1 or sn2 let's talk about sn1 first what is the uh why is it named sn1 because rate of reaction actually rate of reaction actually depends on the concentration rate of reaction actually depends on the concentration of the halogen alkane do you understand so uh in sn1 it's a double step it's a double star prediction in a multi-step reaction the slowest step is called the rate determining step so what happens here in the first step there's a heterolytic fusion of the halogen alkane we're gonna get a carbocation this is the slow step okay and in the second step the hydroxide actually attacks the carbocation and we get the um alcohol from the halogen alkane this is called hydrolysis of halogen alkanes so in a2 you're going to get a question like this if it was a chiral carbon how many products would we get like would we get only one uh isomer or both stereoisomers we would actually get both because the hydroxide can attack from any direction okay but in sn2 hear me out the hydroxide can only attack from one direction okay so we are only going to get one stereoisomer okay so in sn2 why is it called sn2 the two refers to the number of molecules that rate depends on okay one degree halogen alkanes undergo sn2 so rate depends on both the concentration of the halogen alkane as well as the hydroxide anion so both things occur simultaneously the hydroxide will attack i've actually missed the the dipole here this is del plus and this is del minus i missed it here as well this is del minus and this is del plus okay so the hydroxide actually attacks the electropositive carbon atom and heterolytic fusion occurs um then we get a hybrid compound which has not been we have not been able to um isolate this in the lab it's purely hypothetical which has a negative charge due to the hydroxide so basically these are dotted lines right what does this indicate this one is breaking and this one is being made so both occur simultaneously basically right and we get our final product so since there's only one step in the reaction and both the halogen alkane and hydroxide are taking part in it rate actually depends on the concentration of both of them what about the last mechanism that's nucleophilic addition uh it's for carbonyl compounds only it requires heat casein is the catalyst so in the first step the cn from the catalyst kcn or nscn actually attacks the electropositive carbon it's attached to a electronegative oxygen so a dipole is created okay and heterogeneity fission actually occurs of the pi bond so in the second step we actually get an o minus charge and the cn is attached to the carbon now because the carbon could not have five bonds right that's why we sacrificed one bond now what happens the o minus actually attacks an h plus where did we get this h plus from the hcn reagent that we added in the first place heterolytic fusion occurs in the hcn which actually gives us this h plus so o minus actually attacks that h plus and we get that we get our hydroxy uh hydroxy nitrile and the interesting part is the cn that we used initially has been regenerated that's why it's acting as a catalyst okay and the last one i've shown you the same reaction using a ketone okay we're getting a hydroxy nitrile here as well okay so um you can ask me any questions and drop a like and subscribe if you like the video i'm gonna do the a2 chemistry one as soon as possible and i'm gonna link it up here on the right maybe here when it's up alright see ya