ALKYL HALIDE: Nucleophilic Substituitiion: PART-1 (SN2)

インデックス作成: 2026-05-15

[00:00:01]Hello students, stay happy, stay healthy.

[00:00:07]Today we are going to discuss about class 12th organic chemistry in that most important topic nucleophilic substitution.

[00:00:21]This is very very important topic students with this almost 30% of 12th organic chemistry will be covered.

[00:00:34]So first we'll discuss before discussing about nucleophilic substitution we need to know what is a substitution reaction.

[00:00:45]What is a substitution reaction?

[00:00:49]Substitution reaction is a chemical reaction during which an atom or one functional group in a chemical compound is replaced by another atom or other functional group. Simply we can say an atom or group of atoms are replaced by a another atom or another functional group. that is the substitution. But in the organic chemistry nucleophile substitution is a class of reaction in which in which a leaving group is replaced by an electronrich species.

[00:01:36]See students a leaving group new term has been introduced leaving group generally halides like Cl, Br, iodine and tossile groups they all are act as leaving groups right and electronri species they all are nucleophiles. What are nucleophile and nucleophilicity that we will discuss in upcoming lecture. So what is the reaction of nucleophilic substitution? I have written general form. So this is the alky hallide. In this alkyhalide ion is a leaving group and nucleophile is attacking on the alky group. Simply halide ion is replaced by a nucleophile.

[00:02:28]This is a nucleophilic substitution reaction.

[00:02:34]Clear students?

[00:02:36]Then this is the general representation of nucleophilic substitution. General representation of nucleophilic substitution. In this three terms you should remember three components mainly is used in this reaction. What are those? First one is the most important one is alky group. Alky group in the alky hallide. Always remember alkyle group must be an sp3 hybridized carbon.

[00:03:09]Sp2 and sparbon generally do not undergo any nucleophilic substitution. Alky group must be sp3 hydbridized carbon.

[00:03:21]Right? Another important thing is X in the alky hallide in the substrate X generally X is H hallogen not only hogen we can also take any leaving group X is any leaving group which is capable of accepting electron and it is leaving from the reactant substance that is the hallide group and third component is this is the most important one nucleophile file. A nucleophile is electronri species that tends to attack on substract at a position of low electron density. See this position of low electron density. It attacks from the position of a low electron density reactant. So do you understand the nucleophilic substitution we need mainly three components. One is alky group, another one is all. Another one is nucleophile. Now we'll see we'll discuss a little bit deeper into the concept.

[00:04:30]See this reaction general equation for nucleophilic substitution. This is the general form of nucleophilic substitution.

[00:04:40]In this general form, the bond between R and leaving group is a weaker sigma bond. This bond is a weaker sigma bond.

[00:04:55]That is the reason leaving group is trying to go away from the substrate and after this the formation of bond between alky group and nucleophile is a stronger sigma bond.

[00:05:09]This is stronger sigma bond. Simply what is happening in this reaction? Weaker sigma bond is replaced by a stronger sigma bond. Another interesting information we will get from this equation that is the alky group is present over there. Alky group never the bond between the alky group and leaving group is never a never a double bond.

[00:05:42]Never a double bond or a partial double bond or a partial double bond.

[00:05:56]Right?

[00:05:57]This is the most interesting point.

[00:06:03]Another and another thing is that the RN leaving group leaving group suppose if I consider leaving group is analides then in this case RF RC RB RA among all RF bond is not act as a leaving group because alkal group is having 2 p orbital.

[00:06:33]Florine is having 2p orbital. This 2p 2p overlapping is 2p 2p overlapping is very very very strong bond. Strong overlapping. So it is very difficult to break the bond between alkal group and florine. That's why RF do not undergo any nucleophilic substitution. Whereas in the case of RCL and RBR, RA RCL, RBr and RA, RA can undergo very fast because the bond length of iodines increases. So bond energy is decreases. R and I bond is very very weak. So the RA reacts very fast. This is another interesting point we can study from the general form of nucleophilic substitution. Now we'll see the types of nucleophilic substitution in that main main one is SN2 reaction.

[00:07:31]What is SN2? SN2 means biomolecular nucleophilic substitution. SN2 means this two represents biomolecular nucleophilic substitution reaction that is SN2 reaction. What is happening here?

[00:07:49]When an alkalhalide is treated with alkalhalide is treated with aqueous KO Br act as a leaving group and O minus act as a nucleophile. So the Br is replaced by O group that is the biomolecular nucleophilic substitution.

[00:08:09]So to study about this we need to focus on this mechanism.

[00:08:14]How the reaction is taking place sir?

[00:08:17]how the reaction is taking place. See carefully students in this reaction suppose I am taking alkal group is like this alky group right I'm trying to attack what is happening here trying to attack the nucleophile O minus from the front side I'll try to attack the nucleophile from the front side can it be possible no you cannot attack from the front side. Why sir? Because the bromine is having the lone pairs or nucleophile is having the negative charge. There exist a repulsion between negatively charged electrons and this lone pair electrons. That's why nucleophile cannot attack on the carbon from the front side. So what will happen sir? So nucleophile attacks. See here nucleophile attacks on the saturated carbon from the backside right. The moment nucleophile attacks see here nucleophile attacks from the backside 180° away from the leaving group because lone pair of electron being to form a new bond. See here. So when the nucleophile is attacking from the backside the bond between carbon and bromine begins to break and bond between carbon and O begins to form. In such case the bond between carbon and bromine is breaking. Bond between carbon and O is forming. At this stage if you notice the partially partially carbon is having how many bonds total how many bonds are having carbon partially five bonds are there carbon with five bonds is stable no carbon with five bonds are highly unstable that's the reason this is a transition state a transition state highly unstable state more energetic state so ultimately Being a unstable form of carbon what it will do the bromine will go away as a good leaving group bromine go away then O is formed.

[00:10:39]O is formed on the back side of the carbon that is what umbrella flipped inside out. Inversion umbrella umbrella inversion also we can say this umbrella inversion. So in the mechanism in SN2 reaction in SN2 reaction what we have understood mainly in SN2 mechanism it is not possible for the nucleophile to attack on the carbon from front side because there is a repulsion between a lone pair of electrons of leaving group and the negative charge of the nucleophile. So nucleophile attacks from the backside that is exactly 180° away from the leaving group. As a result the bond between carbon and bromine is breaking and the bond between carbon and incoming nucleophile is forming at this stage. At this stage carbon is having partially five bonds. So highly unstable. So it is called as transition state. So ultimately what will happen broine will go away from the reactant and it converted into the reactant molecule is converted into the product by the umbrella inversion product. Let's see some more interesting concept we'll discuss through the energy diagram how it is forming. See here students the very nicely energy diagram it is reactant molecule I have and reactant molecule the moment nucleophile is attacked the moment nucleophile is attacked it forms a transition state transition state is stable no s transition state is highly unstable because of this transition state is highly unstable so its energy you see as the energy increases stability decreases that's why then immediately at a higher energy state is unstable so immediately it converted into product it is immediately converted see carefully another important point I want to tell the bond between carbon and hallad is weaker sigma bond bond between carbon and nucleus is stronger sigma bond so what is happening the weaker sigma bond replaced a stronger sigma bond. That means that during the during the reaction during this SN2 reaction energy is released or absorbed. What is happening here? Energy of reactants are greater than energy of products.

[00:13:25]Therefore, what will be the delh? Delh will be negative and it is what?

[00:13:31]Exothermic reaction. We can also say one extra point SN2 reaction the formation of a stable bond is an exothermic reaction. Clear students?

[00:13:44]Next. Do you understand energy diagram?

[00:13:48]Just look into the diagram carefully.

[00:13:50]How reactants are converted into transition state then transition state to product. Overall nature of reaction is exothermic.

[00:13:59]Let's see the characteristics.

[00:14:03]Characteristics of SN2 reaction. The first one rate of reaction is directly proportional to rate of reaction is directly proportional to concentration of nucleophile as well as concentration of reactant alkyhalide.

[00:14:24]So according to the chemical kinetics the sum of the powers of concentration terms is nothing but order. What is order of reaction? The sum of the powers that is 1 + 1 that is two. And how many reactants are involved in the reaction?

[00:14:42]Two reactants are involved. One is alkalhalid another one is nucleophile.

[00:14:48]That's why the molecularity of reaction is molecularity of reaction is two.

[00:14:55]Clear students? This is the first characteristic first information we are getting. And next is the most important one. Already we discussed weaker sigma bond is replaced by a stronger sigma bond.

[00:15:11]Next one is the following compounds won't give SN2 reaction. See one example we discussed already. The following compounds won't give ascent to reaction.

[00:15:22]First one is RF. Sir why sir? Why RF won't give SN2 reaction? Because in the RF that is nothing but uh we can say carbon and florine. What is the carbon orbital is present? 2p. Florine is also having 2p. The interaction between carbon and florine is what? The bond between those two. to be more stronger bond. So it is not possible the bond won't break. So no SN2 in this case. No SN2 in this case. Second example a highlights.

[00:15:59]Second example is a halides. Ah Hallid is nothing but H hallogen with H hallogen with a benzene ring. This is the arin. This is the H hallogen.

[00:16:10]Already we discussed alkyle group never be a double bond.

[00:16:16]The bond between the bond between halogen and alkal group alkal r and x the bond between R and X never be a double bond or partial double bond. But when alkal highlights are present due to the resonance what will happen due to the resonance? The partial double bond character is developed between developed between carbon and halogen.

[00:16:51]Clear students? Carbon and halogen partial double bond is developed due to resonance. So if partial double bond is forming is it stable? Is it possible essential reaction? No. That is the reason ah high helides do not undergo do not undergo nucleophilic substitution of SN2.

[00:17:13]Next another one you see venile halides.

[00:17:17]What are venilehides?

[00:17:19]Venile group. Venile group is nothing but CH2 double bond CHX. This is the venile group. This is the venile group. Again on the double bonded carbon which carbon it another explanation also we can give alky group must be alky group must be sp3.

[00:17:41]The among the three components we discussed the first component what they said alky group must have sp3 hybridized carbon but what it is here we are having sp2 carbon when sp2 carbon is attached to the h hallogen what's the main problem occurs due to the resonance partial double bond character developed so multiple reasons we can discuss why they are not not undergoing nucleophilic substitution Is that clear students?

[00:18:13]Excuse me. Then we'll see next one.

[00:18:17]Halogen at bridge head carbon. Halogen at bridge head carbon. Suppose I'm taking one of the bridge head carbon.

[00:18:29]This is the bicycllo bridg carbon and it is having some broine I'm having here.

[00:18:35]Now I'll try to attack on this particular carbon with O minus. Which carbon? On this carbon with O minus. Can I attack from the front side?

[00:18:47]No sir. From the front side it is not possible sir. Then can I attack from the back side? No. From the back side. How I can attack? Because on the back side I have a complete ring. No, we have a backside entire ring. bulky ring is present. A bridge is present. So it is not at all possible to attack from the back side in the bridge head carbon. So bridge head carbon is blocked completely. From the front side the leaving group loan pairs are not allowing. From the back side bulkiness of the bridge group is not allowing.

[00:19:26]That is the reason bridge head carbons also do not undergo a cento reaction.

[00:19:32]Let's see what is another example for this tertiary alkalhalide.

[00:19:38]Tertiary alkalhalide.

[00:19:40]What is the tertiary alkalhide? What happened with that? Let's see this is the CH3. See CH3 tertiary alkalhalide.

[00:19:53]As usual from the front side I cannot attack due to the lone pair. Now sir what happened from the backside? See backside of this carbon backside of this carbon is surrounded by what? Three bulky groups.

[00:20:12]How many are there? There are total three bulky groups are present. Three bulky groups. In this in these three bulky groups what it causes? It causes steeric hindrance.

[00:20:30]It causes steeric hindrance. Because of this nature, because of steeric hindrance, tertiary alkalhalide do not undergo SN2 reaction. Right?

[00:20:44]Simple way students easily we can say which of the compounds undergo SN2 reaction. Simple whichever having not possible the backside attack all are the best example for SN2 reaction. Another one you see neopentile.

[00:21:01]Neopentile hallide. I want to give a clarity about neopentile. Sir in which case we will use the term neopentile.

[00:21:11]See the second carbon other than the alpha carbon the attached carbon. The second carbon suppose the next carbon is having quartonary degree 4° carbon is presented. If the adjacent carbon is 4° then we can use the term neopentile.

[00:21:32]Same reason students here you may think sir this is a primary alkalhalide. It is a primary alkalhalide one degree. So in primary alkalhalide can we can do the SN2 reaction but yes of course it is a primary alkalhalide only. But what is happening? The presence of this four degree carbon on the back side the presence of this four degree carbon on backside it is totally totally creating more and more steeric hindrance.

[00:22:07]That's the reason neopentile also do not undergo asto reaction. Clear? Then next one is see here betaquarbon is quartinary.

[00:22:21]Another important one cyclopenta denile chloride. Cyclopenta dionile chloride front side not possible because of this repulsion. Back side to perform the backside attack. See this pi bond how the pi bonds are there. Happily they have arranged this lateral overlap right and the electron cloud is electron cloud is moving.

[00:22:50]So this charge completely the back side of the back side of the carbon is completely surrounded by what?

[00:22:58]Completely surrounded by what? Pi bonded electrons. Pi electrons. Front side also electron repulsion. Back side also electron repulsion. So it is not possible any SN2 reaction. It is not possible SN2 reaction in this example also. Right? Next one. If you go to the another example this is very easy cyclopropyl chloride you know whenever the three member ring you see whenever the three member ring you see only one important point we can in our mind only one point is in our mind that is it is highly unstable. What is present sir?

[00:23:37]Due to angle strain suppose if you nucleophile attacks from backside due to the angle strain the ring will collapse the ring may break that's the reason here SN2 reaction is not possible in the case of cyclropile all also is that clear students so these are the main important examples that we can see just recolct whichever the which compounds do not Conjugation. This is the question. This is the area. We are going to get the question. Which of the following or else integer based question they may raised? How many of the following compounds do not undergo SN2?

[00:24:22]How many of the following undergo SN2 reaction? That is the question integer based. Let's see what you have written.

[00:24:29]First one, sir. Alky fluoride is the first example. Second one is what? Aile Hallide.

[00:24:37]Third one is what? Venile hallide.

[00:24:41]Right. Fourth one is what? Bridgehead carbon.

[00:24:47]Third one is sorry. Fourth one is having bridge head carbon. Fifth one is having tertiary alkalhalide.

[00:24:55]Tertiary alkalhalide.

[00:24:58]Next sixth one is what? Neopentile.

[00:25:02]Neopentile.

[00:25:04]Next one is what? Cyclopentaile.

[00:25:09]Clear.

[00:25:11]Cyclopentadenile helide. Next cyclopropile.

[00:25:16]So these are the examples definitely do not give SN2 reactions. There are some more other examples are there but these all are main examples we are considering. Let's see another characteristics.

[00:25:29]Another characteristics we'll study.

[00:25:31]Next one is inversion at site of nucleophilic and is a beautiful point very very very very beautiful and important point it is inversion at site of nucleophilic substitution takes place at SN2 is called wen inversion what is this sir not understand anything can you explain clearly yes we'll discuss see this example students see This example it is R2 brooane. The reactant is R2 bromutin.

[00:26:07]Sir I forgot RS configuration. Can you elaborate clearly once? Simple students.

[00:26:13]What is RS configuration? This is the alkal group. Uh highest priority one.

[00:26:19]Then ethile is two. Then this is three.

[00:26:22]How it is rotating? Clockwise direction it is rotating least priority group on dotted line. So which when it is rotating clockwise direction it is R configuration and see this one this is highest one 2 3 how it is rotating counterclockwise direction. So it is S configuration this is the concept RS configuration. Now see we are not discussing any optical just simply because of backside attack what is happening due to the backside effect the configuration reactant is having R configuration and product converted into S configuration. The R configuration product is converted into S configuration because of the back side detect that is what as a result what has happened what happened in the configuration. What happened in the configuration? Inversion takes place.

[00:27:22]Inversion takes place. So this is another crucial point. If they want to make a difficult question paper definitely question will ask from the optical isomem of SN2. What is that?

[00:27:36]Because of SN2 reaction inversion of configuration takes place. Why sir? due to the backside attack. Now we'll discuss another characteristic another characteristics that is rate of reaction is directly proportional to directly proportional to the nature of leaving group. Nature of leaving group among RF, RCL, RBR, RA we discussed the bond length of iodine. Iodine is larger size. So bond length is very very high.

[00:28:14]So easily you can break it. In the case of bromine it is less and that's in the case of chlorine it is much more less bond length. So as the bond bond length is increases bond energy decreases. So iodine can leave the substrate very very easily. Iodine is a very good leaving group. So order of leaving tendency is here. I minus greater than Br minus greater than CL minus right F minus I don't write also next reactivity order of alkalhalide we discussed in tertiary alkalhalide there is a steeric hindrance back is not possible backside attack is very very easy in the case of primary alkalhalide so primary alkalhid is the highly reactive towards SN2 reaction ction.

[00:29:08]This is the concept students. So now we'll discuss some questions related to this. Some questions related to SN2 reaction. Let's see the first question.

[00:29:20]Compare rate of SN2. One chlorobutane and one iodane.

[00:29:27]One chlorobutane and one ibutane. Let's see what is this one chlorobutane.

[00:29:33]The leaving group is 1 2 3 4 chlorine and here leaving group is iodine. Tell me one chlorobutane and one iodine.

[00:29:43]Which is what is the correct answer here? The good living group is what among chlorine and iodine. Good living group is iodine. So one iodin is very good. Next one.

[00:29:57]One iodin is very good one. Next we'll see second example chlorobenene and benzile chloride. Chloroben and benzile chloride name seems to be familiar.

[00:30:11]Majority students confuse with the structure of chloroben and benzile chloride. So this is the chloroben and benzile chloride is benzene with CH2.

[00:30:26]CH2. Now you see this is what a halide.

[00:30:30]This is what a in a hallide SN2 reaction takes place. No. And this is what alky because hogen attached to the sp3 carbon alkalhalide with primary alky primary alkalhalide. So benzile chloride is faster. Here one is faster. Next one.

[00:30:57]1 chloro 3methile pentane. One chloro 3methile pentane. This is the pentane.

[00:31:05]First chloro and three methile group is there. Another one is two chloro 3methile pentane. This is the two chloro 3 methile pentane. You know very easily you can say 1 chloro 3 methile. This is what which carbon it is primary alkalhalide and it is what? Secondary alkalhalide. You know the reactivity what is the reactivity order we discussed primary greater than secondary greater than tertiary. So primary alkalh highlights are much clear. Next one students when question I gave you just make a pause and try to solve after that you check the answers.

[00:31:47]Now in the fourth question see both the reactants are same only. Then where the question is changed? Question is changed with the nucleophile.

[00:31:57]Nucleophile which is the stronger nucleophile. Neutral spaces are charged spaces. Neutral component or charged one which is the strong nucleophile. Which one can attack very easily? Definitely which is having more electron density that is charged species. So charged one is very faster. And this information we'll discuss in the upcoming lecture that is about nucleophilicity.

[00:32:23]There we'll get more clarity about the nucleophile strength. Another one. Next one is so minus and S minus among those two S minus is better nucleophile than the O minus. S minus is better nucleophile than the O minus. This question also we will discuss in upcoming video because you should know which is a better nucleophile or we can discuss even among oxygen and sulfur sulfur electro negativity electro negativity of electro negativity of sulfur is less than oxygen. So therefore donating tendency donating tendency or attacking tendency of sulfur is greater than oxygen. Not only this example for clarity we can also discuss CH2 minus CH3 minus is a stronger nucleophile than NH2 minus then O minus than F minus because carbon is a carbon is a less electro negative element and it is having a very high electron density so it can donate very easily. So here nucleophilicity strength also depends on the electro negativity of an atom. Right? Next question.

[00:33:56]See here among nitrogen and phosphorus which is less electrogative which can do attack easily which can attack easily the phosphorus can attack very easily.

[00:34:08]Next O CH3 minus and O C3 both everything is same but here there is given different concentrations they have mentioned they have mentioned the different concentration sir here what is this how to approach this question. So we know rate of reaction is directly proportional to concentration of alkalhalide as well as concentration of nucleophile also. So if the nucleophile concentration is more rate of reaction is more so 2 m is having very faster rate.

[00:34:46]Next question.

[00:34:50]Find the products of the following. Find the products of the following. So see the reactant both the cases the bromine is present.

[00:35:04]Bromine is present here. Here the bromin is tertiary. Here the brin is secondary.

[00:35:12]So which one will react? Which one will react? Only faster reacting takes place with less hindered position. So here SH will form remaining broine as it is said. Remaining is broine as it is said.

[00:35:29]Here in the bracket we must give one equivalent one equivalent of SH minus.

[00:35:34]We are using one equivalent of SH minus.

[00:35:39]Next second one I'll treat this compound with O minus with O minus. Which broine can react? We discussed already. If broine attached to the sp2 carbon double bonded carbon no substitution.

[00:35:55]So only this broine can replace. So answer will be O CH2 CH double bond CH Br the same component will be there clear. So this broine is replaced.

[00:36:15]Another one. Another one is there among bromine and chlorine. Bromine is a good leaving group. But but see students small confusion here. But broine attached to whom? Broine attached to sp2 carbonarhalide. So there is no nucleophilic substitution here. So but here what we can do alkyhalide sp3 carbon. Even though less leaving tendency of chlorine also but it attached to the alky carbon that's why here nucleophilic substitution is taking place not on the not on the ben ring right these are the important and nice examples next we'll see this is all about the information related to SN2 reaction. So there are combined questions will come different type of nucleophilic substitution all the things we will study we'll discuss in the upcoming lectures just uh keep follow the channel keep follow this videos and you will get benefit that's it thank you so upcoming concept is nucleophilicity the most important because two to three question I had given what is nuclear philicity to give introduction and thank you. Keep learning, keep going, follow the channel the chemistry serve for IITJ need classes. Thank you guys.