AS Organic Chemistry Principles Overview

Hello, so today I'll be reviewing AS organic chemistry as fast as possible, right? I'm dividing 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 going to give timestamps in the description below so you can skip to them if you want. Right. So let's start. So an organic compound typically contains carbon, hydrogen and 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. Structural isomers are of 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 5-carbon alkane in pentane. An isomer is 2-methylbutane and another one is 2,2-dimethylpropane. So remember to give commas in between numbers and dashes have to be placed after, before and after numbers, alright? That's the basic. These are the basics. So anyway, in positional isomers, it's kind of the same, but here we're juggling the functional group around. For example, the 5-carbon alcohol. pentane 1l then an isomer exists pentane 2l and pentane 3l 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 the um carboxylic acids and alcohols are functional group isomers of ethers. Ethers are not in our syllabus by the way. So for example propanol is the functional group isomer of propanone. Ethyl ethanoid 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. They are of two types, cis-trans or geometrical isomers, and on the other hand, we have optical isomers. So for cis-trans or geometrical 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 right 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 you 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, cis-3-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 laterally inverted. Look at how I drew the COOH, it's laterally inverted. Make sure the wedge shaped diagram is fine as well. it has to be laterally inverted or you won't be getting marks okay uh 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 in a2 they rotate like physically they are the same Physically, they're the same physically and capillary. 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 his physics plane polarized light only oscillates in one plane basically so each an enchemer basically rotates the plane rotates plane polarized light in different directions For example, the one on the right may rotate it clockwise, dextro-rotatory, and the one on the left may rotate it anti-clockwise, which is lever-rotatory. So if we have a mixture containing equal amounts of both enantiomers, it's called a racemic mixture. If we pass plane polarized light through it, it will not change its direction of oscillation. 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. Homolytic 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 right 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 heterodic fission. It typically occurs in a polar molecule. So for example, AB, 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 pen, right? So that's heterodic fission basically. A 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 electrons in the shared pair. Ions are formed after heterolytic fission. Classification of reagents. Reagents 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, epitrophile without being charged, okay? For example, water and ammonia, they are nucleophiles because they have lone pairs. Or an alkene can be a nucleophile because it has an electron-rich pi bond. So the curly arrow must be from the nucleophile to electrophile. Remember, the electrophile can attack the nucleophile, as we've seen in A2 chemistry, where the nitronium ion attacks the benzene ring, for example. You're going to learn this next year. 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 formed so easily? Typically because carbon is on comparatively it's on the left hand side of the periodic table. So it has low electron negativity. So the other compounds except hydrogen which are attached to it have a tendency to take away its electron. So carbocations are of 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. So, on the other hand, a primary carbocation, it only has one alkyl group. So, the intensity of the positive charge is not reduced that much, right, comparatively. We say that electron-reducing 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 have 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 function groups, their 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, right? by an addition reaction which is called hydrogenation using a nickel or platinum catalyst plus we require we also require heat 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 an alkene and this is done to produce more economically favorable smaller alkenes and alkenes which are more useful for us right they have a few reactions they can undergo combustion 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 incomplete 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 soot 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 uh 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 So, alkenes. All alkenes 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 a trigonal planar. So to produce alkenes, we have multiple routes. 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, right? 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 routes. So what are the reactions? Mainly they undergo electrophilic addition. You will classify all of these as addition reactions, okay? In the Marx scheme, sometimes they don't accept hydrogenation. It's an informal term. So even though it's in the syllabus, right? So the first fraction is hydrogenation. This is also addition. We add 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 H3PO4 catalyst, right? To get alcohols. They also react with hydrogen halides, HX 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 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, propene, if we add HBr to the compound, we have two roots. We have two roots. Basically, we can either get one bromopropane or two bromopropane. However, we see that right so in Markovnikov addition we have two possible routes Either from propene we can we can get one bromopropane or two bromopropane But two bromopropane 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 bromopropane 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 caminophore acidified cold dilute caminophore with alkanes to get um diols or we can use hot concentrated k-amino4 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 right 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 going to 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 there are zero alkyl groups? So, for example, in ethene, we basically get methanoic acid first, which is unstable in the presence of KM4, right? So, further oxidative cleavage occurs. You're going to see this in A2 chemistry a lot though. not in ace 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 chemnophore methanic acid and ethanedioic acid these further oxidize 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 ethene can undergo polymerization to form polyethene 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 alkanes in the presence of uv light with chlorine or bromine we can add hydrogen halides to alkenes in room temperature at room temperature okay so there are multiple ways we can produce them using alcohols we can use pcl5 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 thionyl chloride to get the halogen alkene. This also occurs at room temperature. We can use a hydrogen halide. We need heat or heat under reflux. Okay. In this preparation is in-situ, inside the reaction vessel basically. Or we can use 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 halide 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 under 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 halogen alkanes undergo sn1 while one degree halogen alkanes undergo sn2 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 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 equals 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 NaOH 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 alkyl. 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 going to 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 March-Feb 2022 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 KCN, ethanolic KCN, in the presence of heat to form nitriles. Okay, so there is another reaction though. This was mentioned in the syllabus. Silver nitrate, ethanolic silver nitrate can be reacted with the halogen alkane to actually figure out which halogen was present in the alkane, right? So if we 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 C-X, the C-I bond is the longest, right? It's the weakest and there's minimum orbital overlapping. So what's the next reaction? Halogen alkanes also undergo elimination in the presence of ethanolic NaOH. we get alkenes right this is very important we we get alkenes so the inorganic progra products are actually um so hcl is removed from the halogen right so it actually reacts with noh afterwards to form na cl right plus h2o these are the inorganic products be careful okay so this follows the sides of rule Which states that if there are two possible elimination products, we'll get the major product as the one which has more alkyl groups. So for example, we had two chloropropane, right? So the product, the major product will actually be but-2-in rather than but-1-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 h3po4 catalyst oxidation of alkenes using cold dilute k1o4 right diols um hydrolysis of halogen alkenes using naoh aqueous plus heat okay and also we can reduce aldehydes and ketones using both li alh4 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 nascent hydrogen to show this. These 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 LiLH4 only to get the primary alcohol. Okay. So aldehydes give us primary alcohols. Ketones give us secondary alcohols. What about the hydrolysis of esters? 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 ethanoate to form ethanol. What reactions do 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 SOCl2 at room temperature to form the halogen alkane. For PCl5, it will produce phosphoryl chloride POCl3 and steamy fumes of HCl, Abicene. It will react with PCl3 in the presence of heat to form phosphorous acid. It will also react with SOCl2 at room temperature to form SO2 and HCl. This is the best way to prepare the halogen alkane. You'll find this in MCQs often. because all the other products are gaseous 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 get the bromo alkane or the iodalkane, 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 these 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 halogen bromides or iodides okay i mean bromalkins basically They also react with HX in the presence of heat under reflux and you know the in-situ preparation KBr plus 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 is too high, right? It needs heat. So the gaseous product 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 K2Cr2O7, acidified K2Cr2O7, which is preferred, or KMnO4, acidified. 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 oxidized to aldehydes. Basically, the route is like this. Primary alcohols are oxidized to aldehydes followed by carboxylic 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 nitrogen 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 routes for dehydration, alumina plus heat, concentrated H2SO4 plus heat, and concentrated H3PO4. You will see that this one does not require heat, okay, for dehydration. We will get one molecule of water afterwards. So, for esterification, alcohols react with carboxylic acids in the presence of a few drops of concentrated H2SO4 and heat under reflux to form the ester. So some secondary alcohols can be identified by the Iodoform test. We're going to 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 aldehydes are produced by oxidizing primary alcohols followed by distillation using K2Cr2O7 acidified or KMnO4 acidified. I'm showing you the reaction here. Ketones are produced by oxidizing secondary alcohols. So here we don't need distillation because that's the only product we're going to get. But to collect it, we do need to distill it, but we don't need to do it immediately. Where is the characteristic reaction of carbonyl? They undergo nucleophilic addition reactions. OK, so remember, the reagent is HCN and the catalyst is KCN or NACN. What is the function of the catalyst? Basically, HCN cannot provide a strong enough nucleophile. So we use KCN or NACN to provide the Cyanide ion Cn-in great amounts, okay, which which gets regenerated at the end of the reaction So what do we get at the end of this reaction? Hydroxynitrile are produced okay hydroxynitrile That's the class of the compound produced so they also undergo reduction Aldehydes get reduced in the presence of you know either Lial H4 or NABH4, right? to produce primary 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 marks scheme 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 four dnph this is a condensation reaction as you can see i have shown you here they react with Ethanol is reacting with 2,4-DNPH 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 tolerance reagent or failings reagent. Tolerance reagent is a solution of ammoniacal 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 failings reagent, this is basically a Bendix solution. The copper 2 plus ion turns into Cu plus. and we get a brick red color of co2o as seen in biology if a 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 this 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 a2 methanoic 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 taught you beforehand, right? Like it is oxidized in the presence of K104 and K2CO2O7, in fact, to form. with just k-104 right k-104 to form co2 and h2o okay so moving on we have a test for methyl ketones it's called the iodoform test using aqueous alkaline iodine the positive result is a yellow precipitate in the book they mention warming the test tube but in the question 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 test for the iodoform test? 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 iodoform test. We have butanone over here, 4-carbon ketone. It reacts with 3 molecules of iodine and 4 hydroxyl ions 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 iodine 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 one degree alcohols or aldehydes using k2cr2o7 acidified or k1o4 plus heat under reflux so very few reactions require heat under 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 i'm halogen alkanes from alcohols using hx right so you can also prepare them by hydrolyzing nitrals um acidic hydrolysis h plus equals plus heat this is what you need to write to get marks in the question paper to get carboxylic acid the inorganic product is not important you can also hydrolyze it using an alkali but you need to acidify it 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 and A2. You can differentiate between phenol and stuff 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 h2so4 and heat under reflex to form esters they can be reduced using lil h4 only we cannot use nabh4 because it's not strong enough um this is the reaction we're going to get an alcohol primary alcohol okay so esters we can produce them by react alcohols and carboxylic acids and constant H2SO4 plus heat under reflux to give us the ester ethyl ethanoate. They can be hydrolyzed in two ways, either in acidic conditions or alkaline. Acidic hydrolysis of esters gives us 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 nitriles in two ways, either nucleophilic substitution of halogen alkanes using KC and ethanolic KC and heat. and the halogen alkane to form this nitrile or using nucleic addition of carbonyl compounds so this actually requires heat according to the syllabus although you won't see it in questionary period times it actually requires heat okay nucleic addition requires heat so Ethan L for example reacts with HCN the reagent in the presence of casein catalyst and heat to produce hydroxyl nitrile why are these two reactions so important 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 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 nitriles using allyl H4 or hydrogen plus nickel or platinum catalyst to get amines. Okay. This isn't in the syllabus, but I'm still going to say this because you're going to find this in paper one at times. This is important for A2. So, MRIs can be hydrolyzed in two ways. Acetic 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 going to get a carboxylic acid. And alkaline hydrolysis is going to give us a salt because the acid reacts with the 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 in AS 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. So arrows are not required in this mechanism, but I saw it in the specimen 2022 paper for AS chemistry paper two. 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, 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 chloromethane and the free radical. If you see this, the chlorine free radical has been regenerated at the end, right? 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 the ozone layer. And in termination, so the two free radicals, 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. 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 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 electropositive bromine over here and heterolytic fission occurs so in the second step we actually get a carbocation and the bromide anion so the bromide anion 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 the alkene is reacting with HBr. It already has an inherited dipole. Okay, so heterodatic fission 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 carbocation. So why is it called electrophilic substitution? Because the electrophile, right, H plus with a partial positive charge is actually um 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 step reaction 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 heterodotic fission of the halogen alkane we're going to get a carbocation this is the slow step okay and in the second step the Hydroxide actually attacks the carbocation and we get the alcohol from the halogen alkane. This is called hydrolysis of halogen alkanes. So in A2 you're gonna get a question like this. If it was a chiral carbon, how many products would we get? Like would we get only one isomer or both stereosomers? We would actually get both because the hydroxide can attack from any direction. But in SN2, hear me out. the hydroxide can only attack from one direction okay so we are only going to get one stereosomer okay so in sn2 why is it called sn2 the 2 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 hydrolytic fission occurs. Then we get a hybrid compound which has not been, we have not been able to 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. It's for carbonyl compounds only. It requires heat. KCN is the catalyst. So in the first step, the CN from the catalyst, KCN or NACN, actually attacks the electropositive carbon. It's attached to a electronegative oxygen, so a dipole is created, okay? And heterolytic 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 and h plus where did we get this h plus from the hcn reagent that we added in the first place hydrolytic fission 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 hydroxyl 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 ketone. Okay, we are getting a hydroxy nitrile here as well. Okay, so 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.

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 going to give timestamps in the description below so you can skip to them if you want.

Right. So let's start. So an organic compound typically contains carbon, hydrogen and 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. Structural isomers are of 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 5-carbon alkane in pentane. An isomer is 2-methylbutane and another one is 2,2-dimethylpropane. So remember to give commas in between numbers and dashes have to be placed after, before and after numbers, alright? That's the basic.

These are the basics. So anyway, in positional isomers, it's kind of the same, but here we're juggling the functional group around. For example, the 5-carbon alcohol. pentane 1l then an isomer exists pentane 2l and pentane 3l 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 the um carboxylic acids and alcohols are functional group isomers of ethers.

Ethers are not in our syllabus by the way. So for example propanol is the functional group isomer of propanone. Ethyl ethanoid 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. They are of two types, cis-trans or geometrical isomers, and on the other hand, we have optical isomers. So for cis-trans or geometrical 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 right 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 you 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, cis-3-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 laterally inverted. Look at how I drew the COOH, it's laterally inverted. Make sure the wedge shaped diagram is fine as well.

it has to be laterally inverted or you won't be getting marks okay uh 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 in a2 they rotate like physically they are the same Physically, they're the same physically and capillary. 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 his physics plane polarized light only oscillates in one plane basically so each an enchemer basically rotates the plane rotates plane polarized light in different directions For example, the one on the right may rotate it clockwise, dextro-rotatory, and the one on the left may rotate it anti-clockwise, which is lever-rotatory. So if we have a mixture containing equal amounts of both enantiomers, it's called a racemic mixture. If we pass plane polarized light through it, it will not change its direction of oscillation.

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. Homolytic 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 right 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 heterodic fission. It typically occurs in a polar molecule.

So for example, AB, 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 pen, right?

So that's heterodic fission basically. A 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 electrons in the shared pair. Ions are formed after heterolytic fission. Classification of reagents. Reagents 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, epitrophile without being charged, okay? For example, water and ammonia, they are nucleophiles because they have lone pairs. Or an alkene can be a nucleophile because it has an electron-rich pi bond.

So the curly arrow must be from the nucleophile to electrophile. Remember, the electrophile can attack the nucleophile, as we've seen in A2 chemistry, where the nitronium ion attacks the benzene ring, for example. You're going to learn this next year.

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 formed so easily? Typically because carbon is on comparatively it's on the left hand side of the periodic table. So it has low electron negativity.

So the other compounds except hydrogen which are attached to it have a tendency to take away its electron. So carbocations are of 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.

So, on the other hand, a primary carbocation, it only has one alkyl group. So, the intensity of the positive charge is not reduced that much, right, comparatively. We say that electron-reducing 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 have 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 function groups, their 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, right? by an addition reaction which is called hydrogenation using a nickel or platinum catalyst plus we require we also require heat 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 an alkene and this is done to produce more economically favorable smaller alkenes and alkenes which are more useful for us right they have a few reactions they can undergo combustion 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 incomplete 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 soot 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 uh 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 So, alkenes. All alkenes 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 a trigonal planar. So to produce alkenes, we have multiple routes. 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, right? 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 routes.

So what are the reactions? Mainly they undergo electrophilic addition. You will classify all of these as addition reactions, okay?

In the Marx scheme, sometimes they don't accept hydrogenation. It's an informal term. So even though it's in the syllabus, right?

So the first fraction is hydrogenation. This is also addition. We add 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 H3PO4 catalyst, right? To get alcohols.

They also react with hydrogen halides, HX 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 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, propene, if we add HBr to the compound, we have two roots.

We have two roots. Basically, we can either get one bromopropane or two bromopropane. However, we see that right so in Markovnikov addition we have two possible routes Either from propene we can we can get one bromopropane or two bromopropane But two bromopropane 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 bromopropane 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 caminophore acidified cold dilute caminophore with alkanes to get um diols or we can use hot concentrated k-amino4 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 right 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 going to 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 there are zero alkyl groups? So, for example, in ethene, we basically get methanoic acid first, which is unstable in the presence of KM4, right? So, further oxidative cleavage occurs.

You're going to see this in A2 chemistry a lot though. not in ace 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 chemnophore methanic acid and ethanedioic acid these further oxidize 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 ethene can undergo polymerization to form polyethene 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 alkanes in the presence of uv light with chlorine or bromine we can add hydrogen halides to alkenes in room temperature at room temperature okay so there are multiple ways we can produce them using alcohols we can use pcl5 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 thionyl chloride to get the halogen alkene. This also occurs at room temperature. We can use a hydrogen halide. We need heat or heat under reflux.

Okay. In this preparation is in-situ, inside the reaction vessel basically. Or we can use 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 halide 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 under 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 halogen alkanes undergo sn1 while one degree halogen alkanes undergo sn2 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 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 equals 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 NaOH 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 alkyl.

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 going to 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 March-Feb 2022 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 KCN, ethanolic KCN, in the presence of heat to form nitriles.

Okay, so there is another reaction though. This was mentioned in the syllabus. Silver nitrate, ethanolic silver nitrate can be reacted with the halogen alkane to actually figure out which halogen was present in the alkane, right? So if we 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 C-X, the C-I bond is the longest, right? It's the weakest and there's minimum orbital overlapping. So what's the next reaction?

Halogen alkanes also undergo elimination in the presence of ethanolic NaOH. we get alkenes right this is very important we we get alkenes so the inorganic progra products are actually um so hcl is removed from the halogen right so it actually reacts with noh afterwards to form na cl right plus h2o these are the inorganic products be careful okay so this follows the sides of rule Which states that if there are two possible elimination products, we'll get the major product as the one which has more alkyl groups. So for example, we had two chloropropane, right? So the product, the major product will actually be but-2-in rather than but-1-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 h3po4 catalyst oxidation of alkenes using cold dilute k1o4 right diols um hydrolysis of halogen alkenes using naoh aqueous plus heat okay and also we can reduce aldehydes and ketones using both li alh4 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 nascent hydrogen to show this. These 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 LiLH4 only to get the primary alcohol.

Okay. So aldehydes give us primary alcohols. Ketones give us secondary alcohols. What about the hydrolysis of esters?

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 ethanoate to form ethanol. What reactions do 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 SOCl2 at room temperature to form the halogen alkane. For PCl5, it will produce phosphoryl chloride POCl3 and steamy fumes of HCl, Abicene. It will react with PCl3 in the presence of heat to form phosphorous acid.

It will also react with SOCl2 at room temperature to form SO2 and HCl. This is the best way to prepare the halogen alkane. You'll find this in MCQs often. because all the other products are gaseous 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 get the bromo alkane or the iodalkane, 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 these 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 halogen bromides or iodides okay i mean bromalkins basically They also react with HX in the presence of heat under reflux and you know the in-situ preparation KBr plus 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 is too high, right? It needs heat. So the gaseous product 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 K2Cr2O7, acidified K2Cr2O7, which is preferred, or KMnO4, acidified. 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 oxidized to aldehydes.

Basically, the route is like this. Primary alcohols are oxidized to aldehydes followed by carboxylic 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 nitrogen 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 routes for dehydration, alumina plus heat, concentrated H2SO4 plus heat, and concentrated H3PO4. You will see that this one does not require heat, okay, for dehydration. We will get one molecule of water afterwards. So, for esterification, alcohols react with carboxylic acids in the presence of a few drops of concentrated H2SO4 and heat under reflux to form the ester. So some secondary alcohols can be identified by the Iodoform test.

We're going to 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 aldehydes are produced by oxidizing primary alcohols followed by distillation using K2Cr2O7 acidified or KMnO4 acidified. I'm showing you the reaction here. Ketones are produced by oxidizing secondary alcohols. So here we don't need distillation because that's the only product we're going to get.

But to collect it, we do need to distill it, but we don't need to do it immediately. Where is the characteristic reaction of carbonyl? They undergo nucleophilic addition reactions. OK, so remember, the reagent is HCN and the catalyst is KCN or NACN.

What is the function of the catalyst? Basically, HCN cannot provide a strong enough nucleophile. So we use KCN or NACN to provide the Cyanide ion Cn-in great amounts, okay, which which gets regenerated at the end of the reaction So what do we get at the end of this reaction? Hydroxynitrile are produced okay hydroxynitrile That's the class of the compound produced so they also undergo reduction Aldehydes get reduced in the presence of you know either Lial H4 or NABH4, right?

to produce primary 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 marks scheme 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 four dnph this is a condensation reaction as you can see i have shown you here they react with Ethanol is reacting with 2,4-DNPH 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 tolerance reagent or failings reagent.

Tolerance reagent is a solution of ammoniacal 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 failings reagent, this is basically a Bendix solution. The copper 2 plus ion turns into Cu plus. and we get a brick red color of co2o as seen in biology if a 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 this 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 a2 methanoic 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 taught you beforehand, right? Like it is oxidized in the presence of K104 and K2CO2O7, in fact, to form.

with just k-104 right k-104 to form co2 and h2o okay so moving on we have a test for methyl ketones it's called the iodoform test using aqueous alkaline iodine the positive result is a yellow precipitate in the book they mention warming the test tube but in the question 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 test for the iodoform test? 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 iodoform test. We have butanone over here, 4-carbon ketone. It reacts with 3 molecules of iodine and 4 hydroxyl ions 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 iodine 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 one degree alcohols or aldehydes using k2cr2o7 acidified or k1o4 plus heat under reflux so very few reactions require heat under 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 i'm halogen alkanes from alcohols using hx right so you can also prepare them by hydrolyzing nitrals um acidic hydrolysis h plus equals plus heat this is what you need to write to get marks in the question paper to get carboxylic acid the inorganic product is not important you can also hydrolyze it using an alkali but you need to acidify it 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 and A2.

You can differentiate between phenol and stuff 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 h2so4 and heat under reflex to form esters they can be reduced using lil h4 only we cannot use nabh4 because it's not strong enough um this is the reaction we're going to get an alcohol primary alcohol okay so esters we can produce them by react alcohols and carboxylic acids and constant H2SO4 plus heat under reflux to give us the ester ethyl ethanoate.

They can be hydrolyzed in two ways, either in acidic conditions or alkaline. Acidic hydrolysis of esters gives us 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 nitriles in two ways, either nucleophilic substitution of halogen alkanes using KC and ethanolic KC and heat. and the halogen alkane to form this nitrile or using nucleic addition of carbonyl compounds so this actually requires heat according to the syllabus although you won't see it in questionary period times it actually requires heat okay nucleic addition requires heat so Ethan L for example reacts with HCN the reagent in the presence of casein catalyst and heat to produce hydroxyl nitrile why are these two reactions so important 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 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 nitriles using allyl H4 or hydrogen plus nickel or platinum catalyst to get amines. Okay. This isn't in the syllabus, but I'm still going to say this because you're going to find this in paper one at times. This is important for A2.

So, MRIs can be hydrolyzed in two ways. Acetic 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 going to get a carboxylic acid.

And alkaline hydrolysis is going to give us a salt because the acid reacts with the 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 in AS 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. So arrows are not required in this mechanism, but I saw it in the specimen 2022 paper for AS chemistry paper two. 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, 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 chloromethane and the free radical. If you see this, the chlorine free radical has been regenerated at the end, right?

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 the ozone layer. And in termination, so the two free radicals, 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. 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 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 electropositive bromine over here and heterolytic fission occurs so in the second step we actually get a carbocation and the bromide anion so the bromide anion 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 the alkene is reacting with HBr. It already has an inherited dipole. Okay, so heterodatic fission 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 carbocation. So why is it called electrophilic substitution? Because the electrophile, right, H plus with a partial positive charge is actually um 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 step reaction 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 heterodotic fission of the halogen alkane we're going to get a carbocation this is the slow step okay and in the second step the Hydroxide actually attacks the carbocation and we get the alcohol from the halogen alkane. This is called hydrolysis of halogen alkanes.

So in A2 you're gonna get a question like this. If it was a chiral carbon, how many products would we get? Like would we get only one isomer or both stereosomers? We would actually get both because the hydroxide can attack from any direction.

But in SN2, hear me out. the hydroxide can only attack from one direction okay so we are only going to get one stereosomer okay so in sn2 why is it called sn2 the 2 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 hydrolytic fission occurs. Then we get a hybrid compound which has not been, we have not been able to 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. It's for carbonyl compounds only.

It requires heat. KCN is the catalyst. So in the first step, the CN from the catalyst, KCN or NACN, actually attacks the electropositive carbon.

It's attached to a electronegative oxygen, so a dipole is created, okay? And heterolytic 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 and h plus where did we get this h plus from the hcn reagent that we added in the first place hydrolytic fission 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 hydroxyl 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 ketone. Okay, we are getting a hydroxy nitrile here as well.

Okay, so 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.

Transcript for:AS Organic Chemistry Principles Overview

Transcript for:
AS Organic Chemistry Principles Overview