️ Unit-2 | Chemistry-I | B.Sc Semester-I | Last Minute Revision | Osmania University

[00:00:00]So, let's check the important questions from the second chapter. Now for enolysis in the second chapter of organic chemistry, we have all the reactions as well as the mechanisms. So it's easy, I know it's a little complicated, but I've also prepared some simple tips for you to remember these mechanisms. So, let's look at the first question - ozenolysis. So, ozonelysis, from the name itself, we can understand that in the presence of ozone, it means that the bond breaks. So, with the help of ozone, the bond between carbon atoms is broken or the carbon chain is broken.

[00:00:54]So, what happens in this reaction, we'll see. So here's the enolysis of alkynes, alkanes, you know, it's a double carbon chain.

[00:01:06]So here, ozone breaks the carbon- carbon double bond. The double bind is broken here. So, for the formation of which compound, the first intermediate product formed is ozonide. Ozenide is an intermediate product that is formed in the presence of zinc in water as a reducing agent.

[00:01:35]So now let's see what happens in the reaction. So, here ozenolysis can occur for alkanes. Ozenolysis can occur using two types of reagents. The first is zinc and water. The second is by using hydrogen peroxide, such as an oxidizing agent, such as hydrogen peroxide. So, hydrogen peroxide in the presence of hydrogen.

[00:02:11]So, depending on the reactant, the products also change. So let's see if we use zinc and water as a reducing agent, we get aldehydes or ketones, and carbonyl compounds are formed. If we use peroxide, we will get carbonyl acids.

[00:02:35]So here we have two types of reagents. In the presence of a reducing agent, we obtain carbonyl compounds. In the presence of oxidizing agents, we obtain acids from alkenes through the process of enolysis. So, this is about enolysis. Now let's understand the mechanism of what is happening here. So in the first step... So here we first get an ozone molecule, O3, which attaches to the double bond, forming a molonide intermediate- molonide intermediate. Then, after the formation of mazenide, it rearranges to become a stable ozenide. First, mosenide is formed, then as a result of rearrangement we obtain osenide, and the third, third step is reduction or oxidation. Depending on the reagent used, we obtain end products from ozone.

[00:03:43]So, as we've already discussed, if we take a reducing agent, we'll get carbonyl compounds, if we take an oxidizing agent, we'll get carbonyl acids.

[00:03:54]So this last step forms the products. Now let's look at the equation. So the equation here that we have shows that when we add ozone, we get mazenide, which is unstable.

[00:04:21]As a result of the rearrangement, we get this molazinide, which is converted into ozonite, and in the presence of zinc oxide or DMSO, that is, dimethyl sulfoxide, we get a ketone.

[00:04:35]We have two ketones here.

[00:04:38]For example, if we take ethyne, we get two molecules of formaldehyde. Here is the mechanism. To remember this, I have drawn the mechanism in one line so that you can remember all three steps in just one line. So, first ozone attaches to the double bond, we get molozenide, which turns into ozenide. The second step is cleavage, which gives either carbonyl products or a carboxylic acid, depending on the reagent used. Here we used zinc and zinc oxide. So, we only get carbon products.

[00:05:24]Right? So easy to remember. The first step is ozone joining the double bond, we get a mole.

[00:05:34]The second step is the rearrangement of the moleide into the ozonide.

[00:05:37]Then the third step is ozone breakdown, which yields carbonyl products. Done.

[00:05:43]Let's look at the next question.

[00:05:50]Maronnikov's rule, sorry, anti-Maronnikov's. Therefore, Antimaronnikov's rule can also be stated as the reaction of an alkene with hydrogen bromide in the presence of peroxide. It can be set like this or like this.

[00:06:09]It is sometimes also called the peroxide effect or the Karash effect. So, these are the common names given to the anti-anti-Maroon rule. This question was repeated twice in previous exams. So let's see what happens in the reaction. So when hydrogen bromide HBr is added to a tropic or any alkene in the presence of organic peroxides, not hydrogen peroxide, here we get an organic peroxide, the reaction will follow the anti-Maron rule, the Maron rule, if you understand, then we will get, we will also come back to this question, also the Marconikov rule.

[00:06:55]Now let's see what's going on here. So what's happening here is that the bromine in the reactant that we added hydrogen bromide to is attaching to the carbon with more hydrogen. This mechanism is called the anti-Maron rule or peroxide effect.

[00:07:20]This mechanism occurs through the free radical pathway. This means that free radicals are formed in this reaction. So let's look at peroxide first. The first step - peroxide - is a homolytic cleavage, forming alkoxy radicals.

[00:07:45]Aloxy radicals are RO, denoted as a single dot. These alkoxy radicals will react with HBr to form free bromine radicals. We will get free bromine radicals.

[00:08:00]Look here, R gives two alkoxy radicals.

[00:08:05]Here it is organic peroxide or alkaline peroxide. Here we get free alko-alkyl alkoxy radicals, and this alkoxy radical reacts with hydrogen bromide to form 0 and 0.5 bromine free radicals. This is the first step. Then, in the second step, a free bromine radical is formed. It attacks the prop and attaches to the less substituted carbon, forming a secondary carbon radical.

[00:08:43]So here the bromine radical will attack the less substituted carbon. The less substituted carbon is usually the terminal carbon because the terminal carbon has only hydrogen.

[00:08:54]There are no other groups. Therefore, there are no other substituents except hydrogen. Therefore, the free bromine radical will attack the terminal carbon, forming a secondary free carbon radical. A secondary carbon free radical is formed. This is the second step. Then, in the third step, the equation here, you can see that this prop forms a terminal carbon with fewer substituents. Here we have this one. Among these two, this one is terminal. This one is also terminal.

[00:09:29]But there are fewer substitutes here, right? So, with fewer substituents, this carbon is attacked by the free bromine radical, forming bromide and a secondary free radical. Right?

[00:09:47]So this carbon then takes off a hydrogen atom from another HBr molecule. So, from the H of another HBr molecule, this free radical will attract a hydrogen atom and HBr bromide, and regeneration of the bromine free radical occurs. So here we get this free radical reacting with HBr again, we get propyl bromide and a bromine free radical. The reaction proceeds again and the major product formed is one propane bromide due to the anti-maroon orientation favoring a free radical mechanism. Right?

[00:10:34]This is the Karash effect or the anti-Maron effect.

[00:10:42]One mechanism is that peroxide forms alkoxy radicals and bromine radicals.

[00:10:51]Bromine radicals attach to an alkane, and upon formation, they form free carbon free radicals. Then they attract hydrogen and then form the anti-maroon product, which is one bromide. Right? Do you all understand? This is the easiest way to remember the reaction.

[00:11:15]The first step is the formation of peroxide radicals and bromine radicals. The second step is the addition of a bromine radical to the alkene, forming a secondary free carbon/carbon. Then we will get the attraction or extraction of hydrogen from HBr to get the final product, one bromoprotein.

[00:11:42]Right. Next question. So now we are back to the Maronic rule.

[00:11:53]Previously, we considered the anti-Maroon rule. Now that's Maronic's rule. Now, while this rule is completely opposite to the anti-Maronic one, what is happening here is what is the Maronic rule. What does it say? That is, the rule of asymmetric reagents says that when an asymmetric reagent, such as a hydrogen halide, is present, the hydrogen is asymmetric because the electrons are not shared equally between the two atoms present, so the bond pair is more attracted to the halogen than to the hydrogen.

[00:12:30]Right? That's why it's called asymmetric. When an unsymmetrical reagent is added to an unsymmetrical alkene ( unsymmetrical alk), a hydrogen atom attaches to a carbon atom that already has more hydrogen atoms. This means that here, in Maronic's rule, a halogen atom, bromide ion, bromide ion, or chloride ion will attack the carbon with the greater number of hydrogen atoms. So, this is the opposite of anti-Maroon. In the anti-Maron rule, fewer substituents on carbon will be attacked, but here there are more hydrogens. Now let's see.

[00:13:19]Therefore, the gaugen atom bonds to the carbon with fewer atoms.

[00:13:25]Hydrogen will attach to the carbon with more hydrogen, and bromine will attach to the carbon with less hydrogen. So, this rule is based on the formation of the most stable carbocation. Here, an intermediate product is formed - carbocation. In the anti-Maronite rule, this is a free radical. Here it is a carboat. Now let's see.

[00:13:49]So, the first step of carbocation formation. The first step is the formation of carbocation. Therefore, the pi electrons of the double bond are broken and it joins the double bond due to the attack of hydrogen. Thus, carbocation is formed. Then the proton in the proton that was formed here will attach to the carbon with more hydrogen atoms. So, this is the most important thing in this rule. The proton will attach to the carbon with more hydrogen atoms. This means less carbon is replaced. So, look at the equation for propyne, the proton is attacked from the propyne side. Of these three carbon atoms, the middle one is the least substituted, so the number of hydrogens here is less, we have CH⁺.

[00:14:55]Currently, secondary carbation is stable compared to primary carbation.

[00:15:00]Secondary carbation is more stable.

[00:15:04]Next, a nucleophilic attack occurs.

[00:15:09]Here, halide ion-ion attack occurs. We get the addition of carbation, which leads to the formation of the final product. The carbaation is directly attached to this second carbon. So, from here to the right. So, we get a secondary substituted alkyl. This means two chloropropanes or two bromopropanes. So, in the anti-maroon rule, we saw how the halide ion attacks the carbon atom present on the first carbon in the final product, but here it is on the middle carbon.

[00:15:58]So, this is the difference between the anti-Maron rule and the Maron rule. So now Markonov's rule is applicable to electrophilic addition reactions involving alkanes and reagents such as hx, hx stands for hydrogen halides, water or h o x h, similar to acids, chlorides, hydrogen, hypochlorous acids, hypochlorous acids, similar to these acids, all of these reagents will follow Markonov's rule when they react with alkanes, which is a common thing to remember.

[00:16:40]Then let's look at the mechanism in one line. What happens first is that the hydrogen attaches first, forming a more stable carbo, then the halide attacks, forming a maroon product. So, first the hydrogen attaches to the less substituted carbon.

[00:17:08]Carbon with less hydrogen, and the Galil ion attacks the carbon with a positively charged proton.

[00:17:22]This is usually a secondary carbo. This is the maroon part.

[00:17:34]Next comes rule Z. This question is also repeated twice.

[00:17:39]So, let's look at this rule first. During an elimination reaction. Previously we were dealing with accession, the accession reaction.

[00:17:47]Now this is an elimination reaction. So here's an alkene that has a higher number of basic groups attached to the double bond to the carbon. So the major product formed here, by this rule we will discuss the major and minor products which are formed in larger quantities compared to other products. Now let's see what happens. Here, the main product formed by the elimination of an alkene contains a larger number of basic groups attached to the carbon atoms with the double bond.

[00:18:23]Now we'll see what it is. So, according to this rule of elimination reactions, the hydrogen atom is preferentially removed from the beta carbon. Beta carbon means that, let's say, if the carbon has an O functional group, yes, an alcohol; This is an alpha carbon, and if another carbon is attached here, that carbon is called a beta carbon. This is beta, this is alpha. So this beta carbon removed from the beta carbon contains fewer hydrogen atoms.

[00:19:01]Here, this carbon will have fewer hydrogen atoms. Therefore, in an elimination reaction, hydrogen atoms are removed from the beta carbon present in this molecule. So, let's take a look. This results in more substituted alkene being formed. So it is stable due to hyperconjugation and the positive inductive effect of the alkaline groups. So, this is formed as the main product.

[00:19:35]Let's see how it goes.

[00:19:39]For example, we have two chloropentides, when treated with alcoholic potassium hydroxide, they dehydrohalogenate, forming one less substituted pentyne, and two more substituted pentynes. What is going on here?

[00:19:55]We'll see. That's it. So this is the reaction we have here: these are two chloropentides, two chloropentides, now this is this carbon, I don't know, I'll write it here. So, these are two chloropentides 1 2 3 4 5 2 chloropentides.

[00:20:48]So this carbon is alpha, and this is beta. Here it is also beta, but according to the rule, the beta carbon with the fewer hydrogen atoms is attacked.

[00:21:01]So, that means he is eliminated.

[00:21:05]Hydrogen is eliminated from here, forming a double bond. That's why we get two pentines, two pentines compared to one pentine. So if one hydrogen from this compound is eliminated along with this molecule, this double bond will be attacked here.

[00:21:25]So, we will get two pectins. So, two bends are formed. This is salvation.

[00:21:49]I hope you understand. Let's look at the single line mechanism. Like it's not a mechanism, but just a compound. How can we remember this? So, here hydrogen acts first, forming the most stable carbocation.

[00:22:05]Hydroxyethylene glycol then attacks, forming a brown product compound. It also occurs as a maroon product compound. But what happens next in organic chemistry?

[00:22:15]The following organic chemistry question concerns rule s.

[00:22:24]So, in rule s, this question has been asked for four points twice or maybe more, and this question is important because it tells us about how elimination occurs in alkenes. Let's see what happens in this reaction. Therefore, in an elimination reaction, the main product formed is an alkene, which has a higher number of alkyl groups attached to the doubly bonded carbon atoms. This means that when elimination occurs, it means the removal of hydrogen. Therefore, the alkene formed from the residue after the removal of hydrogen will have a greater number of alkyl groups on the doubly bonded carbon atom. So let's look at the equation so you can understand it easily. So now this rule applies to which molecules, and which molecules can we see this type of rule for. So here it's a beta elimination, meaning the functional group on the carbon is called the alpha carbon, the carbon that the functional group is attached to is called the alpha carbon, and the carbon next to the alpha carbon is called the beta carbon. So, when beta elimination occurs, it means that a product or an atom is removed from the beta carbon, this is called a beta elimination reaction.

[00:24:14]Therefore, beta- elimination reactions, especially the dehydrohalogenation of alkali halides and the dehydration of alcohols, are applicable to beta-elimination dehydrogenation dehydrohalogenation of alkyl compounds and the dehydration of alcohols.

[00:24:32]Dehydrohalogenation is the removal of one hydrogen atom and a halogen atom. And dehydration of alcohols is the removal of water from alcohols. These types of reactions follow Serre's rule.

[00:24:49]Let's see what happens. So, according to this rule, the hydrogen atom is preferentially removed from the beta carbon, which has fewer hydrogen atoms.

[00:25:07]This means fewer hydrogen atoms. The beta carbon with fewer hydrogen atoms is preferentially substituted, meaning it is useful for removing a hydrogen atom from the beta carbon with fewer hydrogen atoms. So, due to this, an alkene is formed due to stability effects such as positive inductive effect as well as hyperconjugation.

[00:25:44]Thus, the formation of an alkene by removing a hydrogen from the beta carbon results in the formation of an alkene with a higher number of substituted groups, which is more substituted and stable due to the hyperconjugation effect and the positive inductive effect.

[00:26:02]This stability increases the amount of product. Therefore, it is formed as the main product. So let's look at the equation here. The example they gave was the treatment of two chloropentynes ​​with alcoholic KOH. It undergoes preliminary hydrogenation with dehydrogen. What does this give? Two types of products. One is one pentine, and the other is two pentines. One pentyne is less substituted, hence minor, and two pentynes ​​are more substituted, hence major, according to the same root. Now, if you see this rule, I'll show you how, since it's not visible here, I write the equation of how the elimination happens. So here we have two chlorine dependent dangling atoms three.

[00:27:14]Another one is CH, here we have chlorine, and this is CH3. These are two chlorine-dependent dangling atoms. So, after adding alcoholic KOH. So this is an alpha carbon, and this is a beta carbon.

[00:27:39]Right? So what's happening here is the removal of a hydrogen from this beta carbon, which is also beta here, but it has three more hydrogens. In the rule, they assumed that elimination occurs from the beta carbon, which has fewer hydrogens. So compared to that, this beta carbon has a smaller amount. So, there are only two hydrogens here.

[00:28:09]So one hydrogen here and this chlorine atom will be removed or eliminated, forming a double bond. So the product will be two chloropentins, which is written as: CH3 CH2 CH double bond CH CH3. So these are two chloropentins. Even though this elimination is happening, there's also a small amount of one chloropentyne formed here with the elimination of this hydrogen atom, one hydrogen atom from here, and this chlorine atom, one chloropentyne is also formed, but it's a minor product.

[00:29:02]This is our savings rule. So, you have to remember that hydrogen first attaches to the more stable carbon. So to remember this, you can see that you can remember this as a beta carbon with fewer hydrogen atoms being eliminated. So one proton removed from here, and the neighboring molecule will be removed, will be eliminated along with the halogen atom. This is how two chloropentins are formed.

[00:29:40]Right? So, the following is about the acidic nature of acetylene.

[00:29:51]So, this question can be asked in different forms.

[00:29:54]The wording may be different, but the answer is the same. Explain the acidic nature of acetylene or explain the acidity of acetylenic hydrogen, or sometimes describe the acidity of a single alkyl. So, this question can be asked in any way, and this question is also repeated in the four- point answers. Now let's check if you all know that acetylene is C2H2. So, this C2H2 is classified or called a terminal alkyne, because the triple bond is missing between it and the end.

[00:30:29]For example, here in acetylene C, there are only two carbon atoms, right? So there are no particular problems here.

[00:30:39]But in some alkynes, what happens is, for example, if we take this one. So this one has one or two carbon atoms, but here the triple bond is present between them.

[00:30:57]So, it is not terminal. If the triple bond is present here, at this position, then it is called a terminal alkyne. So, here we took acetylene as an example, so it's terminal. So you don't need to worry. If you take for example molecules like butyne or pentene, sorry, butyne or pentyne, then we need to take examples of placing the triple bond on this end, either on the left side, or on the right side, on the right. So here, because of this triple bond, the protons attached to this carbon are acidic in nature, or weakly acidic in nature.

[00:31:52]Why is the carbon that has the triple bond in acetylene or any alkyne sp -hybridized? It is sp-hybridized.

[00:32:03]Due to sp hybridization, the s orbital has 50%s character. Therefore, due to this 50%s-character, the electronegativity of a particular carbon with sp hybridization will increase. This causes electrons to move away from the hydrogen atom. For example, I can show you this clearly here. So, because of the higher electronegativity of this carbon, the bond pair will be closer to the carbon atom than to the hydrogen. Right? This will make this proton more acidic. The hydrogen will become more acidic and can be removed as a proton. This property is called the acidity of alkyls, especially terminal alkyls. So it can be measured. We can also measure this, and some special reactions occur due to this property, such as the reaction with sodium metal.

[00:33:16]Acetylene or terminal alkyls react with sodium metal to form acetylide and release hydrogen gas.

[00:33:28]So, this is roughly the acidic nature of terminal alkyls. Now look at the equation sodium acetylene, sodium acetylide, and hydrogen is released. This reaction confirms that acetylene acts as an acid. So, if you remember this reaction, you can write a complete answer.

[00:34:00]Remember that the triple bond between carbon atoms makes them more electronegative, and because of this greater electronegativity, the hydrogen attached to these carbon atoms will be weakly acidic and can easily be removed as a proton. This is the acidic nature of acetylene or any terminal alkyne. So let's look at the acronym to remember this answer. This is spenH + spenH +. SP is a hydrogen attached to an sp- hybridized carbon, is acidic, and EN represents electronegativity; The electronegativity of sp-carbon is high, and H+ removes a proton, meaning that hydrogen is easily removed as a proton. So, if you remember this answer, you can use this abbreviation, develop it into an answer, and write the answer correctly. Right.

[00:35:07]Next question: aromaticity, what is aromaticity, and Hackl's rule. So, this is the question: aromaticity is a special type of stability that is especially observed in cyclic organic compounds and is associated with deionization or easy movement of pi electrons.

[00:35:35]Dezoization of pi- electrons. This deoization occurs especially in a conjugated planar ring system. So, conjugated means alternating double bonds. This means that there is one single bond, one carbon double bond, another carbon single bond, another double bond.

[00:35:58]So, there is an alternating space, one space. Such complex compounds are called conjugated and planar rings.

[00:36:08]This means that the entire ring is present in a single plane, not a double plane. This means that if one ring is like this and one other ring is like this, then it is not planar non-planar.

[00:36:25]So, this aromaticity is only present in conjugated planar ring systems, and due to deionization, the energy state is low and this becomes more stable, which makes the compounds more stable due to the lower bond energy. Therefore, these compounds are called aromatic compounds, and these aromatic compounds usually undergo substitution reactions.

[00:36:55]Right? Although double bonds are usually present, so far alkenes are different types of alkanes, we have seen addition reactions, but due to the delocalization of electrons in conjugated double bonds, these aromatic compounds undergo substitution reactions instead of addition reactions to maintain the stabilization caused by resonance.

[00:37:27]Right? So, this is about aromaticity. Then what is Hackle's rule? Hackle's rule determines whether a compound is aromatic or not. Now let's see what happens. If a compound is to be aromatic, it must have four conditions. Hackle's rule implies four conditions. And for a compound to be aromatic, it must satisfy all four conditions. Let's take a look at them. The first is a cyclic closed ring. Not an open ring, but a closed ring. The molecule must have a cyclic structure as well as closed rings. The next one is planar. So, the ring must be planar. This means that all the rings, all the atoms in the ring, must be present in the same plane. All atoms in the ring must be present in the same plane and fully conjugated. This means that each atom in the ring has a p orbital that participates in deionization.

[00:38:41]All the carbon atoms present in the ring must have p orbitals, and these p orbitals together form another shared orbital or electron cloud so that electrons can easily deionize.

[00:38:56]This means that the system must be fully coupled. The first condition is cyclic closure. The second is planar, and the third is fully conjugated.

[00:39:08]Then it must contain 4n + 2 pi electrons. The number of pi electrons must satisfy this condition 4n + 2. So what is n? Why is it equal to n here? Here we have n, right? So, n is, where n is a non-negative integer. N means a non-negative integer from 0, 1, 2, 3...not a negative integer, 1, -2, no, just 0, 1, 2, 3 and so on, right? Therefore, this rule helps to predict the aromatic behavior of compounds. So, the best example of an aromatic compound, as we all know, is benzene, right? This is a classic aromatic compound, it's cyclic, planar, fully conjugated, because all six carbon atoms and benzene have p orbitals, and they're both desolated, and here the number of pi electrons is 6 pi, n is 1, so it satisfies the 4n + 2 rule. So this is about aromaticity and the rule, right? To remember everything, let's look at it like this : uh, this can be written as CPC 4N + 2. C-C stands for cyclic, which means the molecule has to be a ring.

[00:40:37]T means planar and C means conjugated, which means continuous overlap of P orbitals, and 4N+2 is the total number of pi electrons. N = 012.

[00:40:48]If all these four conditions are met, then the compound is called an aromatic compound. Right? This is an easy way to remember the answer. Right.

[00:41:01]Let's move on to the next question.

[00:41:09]Groups that activate and deactivate the ring of aromatic compounds. This question is also repeated twice. So, let's look at aromatic compounds.

[00:41:20]As we discussed earlier, they undergo substitution reactions, namely electrophilic substitution. Therefore, electrophilic substitution is influenced by the nature of the substituents already present in the benzene ring. A regular benzene ring without any substituents would look like this. You all group, right? It will be like this. But if there is some substance, for example, a halogen atom H or a methyl group, or a nitro group O. So, there are many such molecules.

[00:42:01]They all act as substances because they are already present in the ring. So, based on the nature of these substances, these molecules cause an electrophilic substitution reaction on benzene.

[00:42:18]So what happens here is these groups are classified into activating groups and deactivating groups.

[00:42:27]Activating and deactivating groups depend on how they affect the reactivity of the ring to substitution.

[00:42:35]So let's look at what activating and deactivating groups are.

[00:42:39]Activating groups are those molecules that increase the reactivity of aromatic compounds.

[00:42:48]Activation increases reactivity.

[00:42:51]Obviously, the opposite would be deactivation. This means that they reduce the aromatic activity of the drink.

[00:43:03]So, this is due to this increase in reactivity due to activating groups because they donate electron density to the ring through positive resonance effect plus R effect or through positive inductive effect or positive resonance effect, they donate electrons to the aromatic ring and make it active so that it easily participates in substitution. So these activating groups make the ordo and para positions of the ring more electron-rich, so you know what the para and para positions are.

[00:43:48]Let's assume that chlorobenzene is present here. Now the positions adjacent to the chlorine, this, this, this, this, are ortho, and this, this, is meta, and this, is para, the para position is opposite the substituent, and the meta position is third from the substituent that is already present. So, this is the first, let's say this is the first carbon, the second, and the third, this is four, five. So, the third and fifth positions become the goal.

[00:44:32]The fourth position becomes para, the second, and sixth positions become ortho. These are positions... Because of the reactivating groups, these are the ortho and para positions 1, 2, 6 and four, these three positions are electron- rich, which means that upon deionization they will become positive, they have more electrons. These three positions will have more electrons, so the electrophile will attack these positions 2, 4, and 6. Right? So, which ones? We will see here O, NH2, O-CH 3 and CH3 hydroxyl group, amino group, methoxy group and methyl group. All of them are activating groups of the ring. They form electron-rich positions and electron-rich nature, as well as ortho- and para-, para-positions of the benzene ring.

[00:45:30]Now let's go. Next, let's go. Next, let's go. Next, let's go. Next, let's go. Next, let's go. Next, let's go. Next, let's go. Next, let's go. Next, let's go. Next, let's go. Next, let's go. Next, let's see what's up. So, another point: we see that these groups increase the electron density in the ring, promoting a faster substitution reaction. To understand this, let's consider the reaction of phenol. Phenol means this one O-group, it bends with the O-group, the hydroxy group, so it's called phenol. Now the phenol group donates electrons to the ring through resonance. So what's happening here is that in phenol, this hydrogen is lost due to resonance, and here we're going to have a negative charge. So, this negative charge is transferred to the ring.

[00:46:24]Right? So, this positive charge will deodorize again. It will move continuously in the ring, making the aro- and para- positions electron-rich, and substitution reactions occur easily.

[00:46:41]These are activating groups.

[00:46:46]Then the groups are deactivated, which means they take away electron density from the ring, reducing reactivity, thereby reducing reactivity.

[00:47:00]So, this effect arises due to minus induction minus inductive effect or minus resonant effect.

[00:47:08]Negative inductive effect or negative resonant effect.

[00:47:14]So let's see what types of groups will take electrons from the ring.

[00:47:25]Thus, due to electron withdrawal, the ring becomes less reactive, and therefore this also occurs in positions such as aro and para. So, because of this low electron density in the apo and para positions, substitution becomes slower, right. So we see the groups here: nitro group, carboxyl group, aldehyde, sulfonic acid, and cyano group. All of them will withdraw electrons from the ring, making the ortho and para positions less electron dense, and electrophilic substitution becomes slow. Right?

[00:48:11]Now another example : the nitro group of nitrobenzene, which withdraws electrons through negative inductive and negative resonance effects, making the ring less reactive, and directs the incoming groups to the meta position. So what happens in activating groups is because of the electron-rich ortho and para positions, it's an activating group, let's say an activating group. Therefore, these positions are highly reactive, meaning substitution will only occur at these three positions.

[00:48:58]Right? But if it's an acceptor group, a deactivating group, or a deactivating group, then those positions are less reactive, but because they're less reactive, there are fewer electrons present in those positions, and those two positions are substituted.

[00:49:28]So, here the products formed will be or, provided, here the products will be meta- positions. So, here's how activating and deactivating groups determine the reactivity of electrophilic substitution of benzene in electrophilic substitution. Now let's see in one line if we want to formulate this answer.

[00:49:50]Electron-donating groups activate the ring.

[00:49:53]Electron-withdrawing groups deactivate the ring. Again, to remember, you need to write down which groups are activated, activated, and which are deactivated.

[00:50:05]There are abbreviations for this: O, N, H, CH3, CH3 – they are all activated. M, COH, SO, CL–they all deactivate.

[00:50:19]Instead of SO, you can also write SO3H sulfonic acid.

[00:50:25]Right? It's about activating and deactivating groups.

[00:50:30]Let's take a look at the following. Long answer from this section. Inductive effect and acidity of carboxylic acids using the inductive effect.

[00:50:41]What is the inductive effect and how to explain the acidity of carboxylic acid?

[00:50:45]So let's see this question repeated twice. So, what is the inductive effect? The inductive effect occurs between sigma bonds, and it is a constant displacement of electrons from one to another throughout the chain.

[00:51:03]Right? So here it arises because of the electronegativity difference.

[00:51:12]Electronegativity difference.

[00:51:14]So, look here, the electronic effect is transmitted through sigma bonds along the carbon chain and plays an important role in the chemical properties of these molecules. So, this is a permanent shift that occurs across sigma bonds due to the difference in electronegativity in organic compounds, i.e., inductive effect.

[00:51:44]So, here let's see what some of the features are: it only occurs through sigma bonds and decreases with distance. This means that it decreases with distance. If the chain of the molecule is long, then the carbon chain decreases.

[00:52:03]It is reduced mainly to four carbon atoms, where the influence of the inductive effect is possible. After that, it will be insignificant.

[00:52:13]Then the inductive effect is constant and the electrons are not deionized.

[00:52:20]There is no electron desorption.

[00:52:22]These are two important characteristics of the inductive effect.

[00:52:25]Now let's look at the types. The negative inductive effect is the first. The negative inductive effect is electron-withdrawing groups. They attract electron density to themselves like Gauguin atoms. Right?

[00:52:46]Halogens attract electron density.

[00:52:49]That is, these groups are called negative or electron-withdrawing groups, and they cause a negative inductive effect, while a positive inductive effect is caused by electron-donating groups. They give away electrons, but they don't take them away, they give them away. So this is a positive inductive effect, and this inductive effect can be applied to understand the acidity of carboxylic acids.

[00:53:23]Now we all know that acidity depends on stability.

[00:53:27]Acidity in this case is a proton donor. To easily donate a proton, a carboxylate ion is formed. If, for example, it's a carboxylic acid, to give up this proton, if it has to give up a proton to become acidic, then it forms a proton and a carboxylate ion, a carboxylate ion.

[00:54:02]So this carboxylate- carboxylate ion should be stable. If it is stable, then the acid is strong. If it is not, then the acid is weak. Now let's look at the effect.

[00:54:18]This is for the normal state without substitutes.

[00:54:20]What happens if there are substituents, such as electron-withdrawing or electron-donating groups? What will happen?

[00:54:27]We'll see. Therefore, the effect of electron-withdrawing groups or negative groups such as chloride, nitro, and cyano groups is to take away electrons. This reduces the electron density on the carboxylate ion and stabilizes the ion.

[00:54:55]Stabilizes the negative charge because RC is negative, if there is a substituent like a chlorine atom here, then this charge will be like this, the electron density will be more directed towards the chlorine atom rather than towards the carboxyl; therefore, it becomes stable. It is distributed, the negative charge is distributed. That's why these acids are strong, because they can easily remove a proton.

[00:55:31]For example, trichlorous acid is more acidic than ordinary acid because it has three chlorine atoms, which exhibit a strong negative inductive effect, since the three chlorine halogens they remove distribute the negative charge among themselves, and thus this leads to easy removal of protons, and therefore to a more acidic state. Let's look at the order of acidity. So, here, acetic acid is the least acidic, followed by monochlorous acid, dichlorous acid, and trichloroacetic acid. Right?

[00:56:12]If we compare this monochloric acid to hydrofluoric acid, if there is fluorine instead of CL, then it will be more acidic, because compared to chlorine, fluorine is more electron-rich, therefore it is more acidic. Right?

[00:56:30]Then the effect of positive groups is electron-donating groups. Donor groups mean groups such as alkyl, ethyl or methoxy groups. These groups donate electrons, they donate electrons to the carboxylate ion.

[00:56:47]Because of this, there will be more electrons around the carboxylate. This destabilizes the carboxylate ion, and therefore the acid becomes weak. Let's see how this becomes weak in propionic acid. Propionic acid is less acidic than acetic acid because there is only one CH3 or one methyl group in the acidic acid. There is an ethyl group here. So, through this ethyl group, more electrons will be donated to the carboxylate ion, and therefore the electron density will be greater, the number of electrons will be greater. Therefore, the proton will not be easily removed. That's why this propionic acid is less acidic than acetic acid.

[00:57:35]Right? Then the distance from the substance to the carboxylate ion affects one way, but if the substance is far from the carboxylic acid, what happens? The closer the electron-withdrawing group is to the CO group, the stronger its effect.

[00:58:03]For example, this is a CO group, and here we have one alkyl group, and there is a fluorine. So it's just one bond, right, just two bonds, sorry, two bonds, that's two bonds, the distance is present between the carboxylic acid group and the fluoride atom. So, it must be close. If it is closed, then the nature of the electron association will be greater. Let's say R3 R2 are different basic groups, this is CO₂H, and here, not here, we have fluorine.

[00:58:47]So there's a distance between this fluorine atom, the gauge atom is on one end, and the carboxylic acid is on the other, so we have less of an inductive effect. So, distance also matters.

[00:59:03]Further distance will affect electronegativity.

[00:59:16]So, for this example, hydrochloric acid is given.

[00:59:21]Chloropropionic acid is given. So you see, there are two methyl groups between them here. There is only one methyl group here. So, this is weaker, and this is stronger. So to remember the answer, we see that you all know sigma bonds, it happens in electronegativity. The inductive effect is due to sigma bonds that form on sigma bonds due to the difference in electronegativity, and electrons are displaced. This is called the inductive effect.

[00:59:57]Now, in the acidity of carboxylic acids, negative inductive groups increase the acidity, similar to electron-withdrawing groups increasing the acidity of acids, and electron-donating groups decreasing the acidity of acids.

[01:00:12]This is about the answer. If you remember these abbreviations, you will be able to expand them in your answer and get maximum points. So, this is about the second chapter, yes, mostly all the issues are covered, these are important issues, and you can get...