 Okay Merry Christmas to all of you. Merry Christmas. There's any plan for the Eve today Going somewhere Okay, can we start? Tell me no plans Yes Yes see from now onwards Whatever chapter you have in organic chemistry that Is no mostly it is memory based because reactions you have to memorize Okay another thing there are Little bit of mechanism there are mechanism those mechanism If you understand it is fine, but at the end of the day eventually you have to memorize the reactions only that This particular reactions in this reagent will get this product Okay, and you can simply by read out the reaction. You cannot memorize you have to practice. Okay, like I always says Organic chemistry for je point of view You have to do at least twice at least two times We have to finish organic chemistry from hydrocarbon onwards like whatever chapter we have after this hydrocarbon Alkyl halide alyl halide ldehyde ketone carboxylic acid amines all these chapters what all reactions we have into this To memorize those reactions at least twice you have to solve before J means okay next 2020 january Okay, so that particular thing you keep in mind and it is not very straightforward that you put some logic and you know formula and you get the answer like we have We have done in physical chemistry, right? so there are exceptions also there are You know a little bit concepts also all those things you have to keep in mind So with questions only you will understand. Okay, how and when what product we get? okay, so Hydrocarbon you see as the name suggests In this particular chapter we'll discuss about the compound Which are made up of hydrogen and carbon mainly Okay, there are other compounds although but mean constituents of one of any Compounds organic compounds is hydrogen and carbon Okay, so these are the compounds Which are made up of Hydrogen right if you write down ch4 it is a hydrocarbon c2s this is a hydrocarbon But obviously there are many derivatives of hydrocarbon also present, right? So organic organic chemistry Yeah, yeah, yeah, it is still four Yeah, so organic chemistry is mainly deals with hydrocarbon and its derivative correct so if you see the classification of hydrocarbon this is classified into Mainly two categories mainly two categories The first one is open chain hydrocarbon. I'll write down here only open chain and The second one is closed chain Close chain or nothing but cyclic compound ring like a structure open chain is the Straight chain compound. We also call it as acyclic acyclic is open chain cyclic is closed chain Right this open chain Classified again into two categories and that is saturated and unsaturated unsaturated, okay Saturated compounds are nothing but alkane Unsaturated are again two types that you already know alkene and alkyne alkene and alkyne Okay, general formula of alkene. We already know cn H2n plus 2 This is cn h2n And this one is cn h2n minus 2 One more term that we that you probably see in the question that is homolog And homologous series, okay So for example, if I write down the various member of alkene family, we'll start with methane ethane Sorry propane Butane and so on right so all these methane ethane propane butane belongs to alkene family So what we say they are of Same homologous series Okay, if you see the molecular formula of methane I have written there ch4 and ethane it is c2 h6 So if you see the difference of these two is difference of ch2 molecule Right if you take ethane c2 h6 and propane That is c3 h8 In this also difference if you see difference of ch2 molecule so all these methane ethane propane butane pentane hexane and so on Methane ethane propane butane pentane hexane and so on they all belongs to the same Family alkene family and we call it as they belongs to same homologous series Okay, so this is nothing but a homologous series This is also a homologous series and this is also a homologous series Right and the compounds belongs to same homologous series And having difference of ch2 molecule are called homolog Okay, so these are Methane and ethane are homolog of each other ethane and propane are homolog of each other ethane and methane are Homolog of each other right butane and propane are homolog of each other So all those molecule compounds belongs to the same homologous series And the difference in the molecular formula in those compounds is if it is ch2 molecule Then they are said to be the homolog of each other right So two terms I have given you homologous series and the second one is homolog So homologs are like Methane and ethane And like this if you can write down the example many examples you can write down So homolog is a term defined for a pair of compounds in which the difference of In the molecular formula is ch2 In all these organic compounds all is like Chapters hydrocarbon and the coming chapters The whole chapter is classified is basically into four different segments. You can say One is you will see a little bit of introduction Then We'll see the preparation method of one particular compound And then we'll see the properties In properties, we have two different points chemical and physical properties And the last one is uses Uses are not that much important for a competition point of view But for any compounds a little bit of information you should have Methane and propane are not homologous of each other Successive carbon atom, yes, correct The difference should be ch2 Okay So in all these chapters the chapters are divided into these four segments introduction preparation properties uses Order can be anything, but the content will be this only Right. So now coming back to this chapter hydrocarbon and the first thing we are discussing here is alkane Okay, so write down The heading here that is alkane Okay, so first of all in alkane Like all these information you have but I'll just write down So that it will be there in your notes. Okay, you can also write it down some general information General formula is cn h2n plus 2. I have already written this What is the value of dbe? or Do you any one of you? Can you answer this? Yeah, correct Formula is correct at the way. So obviously The general formula is this this Number of carbon atom a hydrogen atom you have And you can substitute this n2n plus 2 into the formula. You'll get the answer Okay, and answer you'll get is zero Okay formula is correct. Now the point is you can calculate this It's not two first of all The formula of degree of unsaturation is c plus 1 minus h Plus x minus n divided by 2 this is the formula we have Now you see the number of carbon atom is n so n plus 1 Number of hydrogen atom is 2n plus 2s 2 into n plus 1 x and n value is zero so obviously when you solve this you'll get it is zero correct So this is one thing that the formula you can calculate and this Informs you that how many double bond triple bond or ring Is possible for a given compound right and since it is a saturated hydrocarbon There's no double bond Present into this. So that's why its degree of unsaturation is zero Okay, alkene's you can also say sir. Why not this one? This is also Cyclo alkene right cyclohexene For this the degree of unsaturation should be one because there is one ring right so point is No do you is Yeah, it's correct Yeah, correct. It's right. That's what I'm telling you You can also directly say there is no double bond or triple bond present or ring present. That's why do you is zero right So when I say alkene the definition of alkene is what it is the open chain saturated hydrocarbon That's why do you is zero in this we do not consider the cyclic compounds. Okay, so You must keep this in mind alkene means what it is it is An open chain Saturated hydrocarbon. This is the definition of alkene hydrocarbon That's why when I say alkene in alkene, we do not consider this kind of molecule cyclohexene cyclobutane and all Right, so open chain saturated hydrocarbon alkene. That's why dbe is zero if you have closed chain closed chain saturated hydrocarbon Which is nothing but this saturated hydrocarbon Then it is known as closed chain saturated hydrocarbons are Cyclo alkene. It's not alkene. It is cyclo alkene Like we have cyclohexene cyclo propane cyclobutane, etc All these are cyclo alkene not alkene next you write down physical properties physical properties First one and all these things you have to memorize from c1 to c4 carbon c1 to c4 compounds The compounds are colorless orderless gas Do you understand what is this c1 to c4? means methane ethane propane butane Or in all those compounds where the number of carbon atom belongs to 1 to 4 maximum Okay, like methane ethane propane butane All these molecules are colorless orderless gas c5 to c17 17 carbon it is colorless orderless If the number of carbon atom If the number of carbon atom Is more than or equal to 18 then the compound is Exist in the form of vex like solid room temperature In all these we have one exception in Organic chemistry you will get many exceptions And what is that exception? The exception is c5 pentane it is It exists in gaseous form room temperature Because of this structure neopentane this one i'm talking about This one is in gaseous form at room temperature Because of the branching okay, but this is an exception you can keep this in mind Color is not required and it is very uh, there are many compounds. Okay, so color of solid hydrocarbon See we are not talking about see actually When I say this c1 to c4 we are talking about the naturally occurring compounds okay color You can you know you can Put any color onto that That is a different thing Okay, but we are talking about the naturally occurring compound. It's not like When some compounds you see with different different color the color that you can put in according to your choice. Okay, that's a different thing Okay, you can paint any substance with any color, but that doesn't mean that the color of that compound is that particular color Okay, all these are for naturally occurring compounds So no general color is defined for them. Okay now in this the important One of the important property On which the question that they ask is boiling point Boiling point is inversely proportional to molar mass As the mass increases molar mass the boiling point decreases Okay, so boiling point of ch4 Is more than to that of c2s6 and c3h8 As molecular mass increases another thing is boiling point also decreases With branching in alkene branching in alkene. Yeah, it's right sriram. It's correct So branching in alkene means what suppose you have paint in like I have taken one exception of paint in So when you write down normal paint in it will be cccc and c Normal print in Yeah, yeah, we'll go we'll go faster strong. Just give me a few minutes And then when you write down this and then it is This one So boiling point is maximum for normal then this one and then this one as branching increases boiling point decreases So you see the boiling point of this is minimum in all these four all these three That's that's why at room temperature also it exists in gaseous form That's why it is an exception Next we'll write down properties quickly Alkanes these are non-polar compounds Why it is non-polar? Why these are non-polars because whatever compound you write All carbon atom are sp3 hybridized hybridized. So there is no electronegativity difference That's why there is no charge separation into this polarity is because of electronegativity difference and charge If it is there Okay, the charge accumulation is possible when we have electronegativity difference If there is no electronegativity difference no charge accumulation and compounds will be non-polar right one last property you write down as Branching increases Yeah, that's what that's what I'm writing down. Just a second as branching increases Surface area surface area decreases more surface area more will be the boiling point More surface area more will be the boiling point Okay, and that's why see actually when When a branching increases Surface area decreases because of the van der Waal forces. Okay, but all those things are not required Right molecules with the branched molecule like this one its surface area is lesser than this and then this Okay, because of branching the molecules the atoms are close Comparatively closer to each other and that's why the surface area is less This is because of the That a more van der Waal forces over here attraction forces Okay, these are the few physical properties of alkane we have now the important thing here in all these chapter is The preparation method preparation method of a compound and then the chemical properties is also important So chemical properties will discuss later or later on physical we have discussed now Chemical properties nothing, but they also we have chemical reactions Okay, so in all the chapters we have mainly preparation Where we have reactions and properties chemical properties where again we have reactions Okay, so these two things are The most important thing in organic chemistry in any chapter. Okay, so preparation method of alkane Preparation of alkane you write down the heading Okay, now the first method of preparation The first method of preparation we have by alkene and alkyne alkene and alkyne in this The first way by catalytic Hydrogenation hydrogenation Okay, so for this the catalyst that we use here catalytic hydrogenation we have and catalytic catalyst that we use for this purpose is nickel Or palladium also we can use Platinum also we can use all these catalysts we can use all these things you have to memorize right on the reaction Oh, Shreya is asking melting point. See melting point is Is actually also has the same order as boiling point of alkane the melting point and boiling point has same order But there is no definite You know order of melting point we have for alkane. That's why we are not discussing Okay, there is a large variation But usually we say what as boiling as boiling point increases melting point also increases or as molecular mass increases melting point also increases But again, there are little bit of variation we have in melting point That's why we are not discussing it over here and they also won't ask this question in the exam But if they ask you can write down that melting point also follows the same order as boiling point Okay, but there is no any regular trend. Okay, that is the one thing Okay, so the catalyst we use here is nickel platinum and palladium now, suppose the reaction I am writing down is I'm taking suppose ethene which is CH2 double bond CH2 and if this is allowed to react with H2O hydrogenation means what hydrogenation means addition of a reaction with hydrogen in presence of any of these catalyst nickel palladium Platinum around 200 degrees Celsius Around 200 degrees Celsius These hydrogen atom will attach at one of these Both of these double bonded carbon atom It's not ren see it's reny nickel r a and e y I think It's reny nickel. It's nothing but see reny nickel is nothing but it is the small The size is very small of the nickel for reaction purpose The size of the reactant molecule even the catalyst should be smaller so that there will be proper interaction Okay, so reny nickel is We actually take very small small size Of those things so you can write down reny nickel also r a and e y not c y Okay, and all these Um catalyst palladium or platinum so both these hydrogen atom attach on this carbon atom and will get CS3 CS3 as the product Okay, for this reaction it is Exothermic in nature Del H is less than zero Exothermic the catalyst that we use here are metal catalyst. All these are metal So these metals I just write down here These are metal catalyst and they have Very high adsorbing properties adsorbing properties, you know, what is adsorption? adsorption What is adsorption? adsorbing properties t i e s and hence and hence they provide Very high surface area very high surface area for the reaction to proceed Okay, understood It's very high adsorption properties and all so this thing we have now if you take ethyne Okay, alkyne We are taking suppose ch triple bond ch Plus we require two moles of h2 same catalyst nickel platinum palladium and all Right product we get here CS3 CS3 Okay, both two hydrogen atom attached over here two hydrogen atom attached over here Yeah, if the molecule accumulates at the outer surface only It is not going into the bulk of the Of the molecule, whatever you have Right So that is adsorption Okay, absorption is when the molecule goes into the bulk of it Okay Now in this like I said as to attach onto the carbon atom, but it is not as simple as it is Looking over here Right So here is what in catalytic hydrogenation Sin addition of hydrogen takes place Okay, sin addition means what? You have done stereo chemistry in the school Stereochemistry, have you finished note down this point first in catalytic hydrogenation hydrogenation sin addition sin addition Of hydrogen of hydrogen takes place. Okay, you haven't done stereo chemistry Okay, that's what that should be done before only anyways You see, what do you understand by sin addition? Sin Suppose we have alkene C double bond C like this Okay, when hydrogen attached onto this when you do I'm not writing down the catalyst and all here Suppose h2 is there So what happens in the transition phase? This double bond is about to break and one of this hydrogen will come over here and other one comes over here So in the transition phase, what happens the C single bond C will be as it is and the two hydrogen atom Joins like this One hydrogen here and one hydrogen here because we have bond Between this hydrogen atom We have a double bond between the carbon atom. So what happens this This double bond pi bond is about to break And hydrogen atom is about to attach find this is the transition phase. Okay Finally, what happens this hydrogen or this pi bond breaks completely and the carbon atom Takes one one hydrogen atom like this And hence the product will get Okay Now what do you mean by this sin addition? That's a question Right, if you see this carbon atom is what sp2 hybridized and this carbon atom is what? hybridization What is the hybridization? sp2 So both carbon atom are sp2 hybridized. It means it is a planar molecule Correct It is a planar molecule when planar molecule is there Then one possibility is what that one hydrogen will come from the top and other one is come from the bottom This is one possibility Another possibility is what both hydrogen comes from the same direction either from the top or from the bottom Right, so when the hydrogen atom comes from the same direction either from the top or from the bottom It is said to be sin addition When they are coming from the opposite side one is from the top and other one is from the bottom Then that is anti addition Right, so in this case you must remember catalytic hydrogenation Sin addition of hydrogen takes place. It means here you see both hydrogen is coming from the same direction Okay, this is the transition state we have and we call it as cyclic Transition cyclic transition state Because it forms a cyclic compound here like you say is it clear did you understand sin addition Now a few examples you see like if I give you these questions Here we have double bond when you heat this with hydrogen Or I'll take this one. So of course it is d2 In presence of nickel. What is d? d is d dadm. Okay, it is the isotope of hydrogen Okay, so whenever you see this d h You it has to it it will behave in the same manner as hydrogen behaves Okay, so what you can consider you can consider this d as a hydrogen only and you can write down the reaction But you have to write down d not h Okay, so the product here will get what either the 2d atom will join from this Up a top from the top or from the bottom, right? So one possibility is there that the 2d Will come from the will come out of the plane like this d and d and Another possibility is what The 2d will come from the Bottom of the plane so dash Okay, both both possibility here and these two molecules These two molecules are exactly same identical Understood a note you write down here write down this note hydrogenation of alkene in hydrogenation of alkene If we use only Nickel as metal catalyst if we use only Nickel as metal catalyst Then the reaction is known as Then the reaction is known as if only nickel we use Then the reaction is known as subatier-sendrance reaction If we use only nickel for example, you see CS3 CH Double bond CH2 with H2 nickel Then the product will be CH3 this reaction is subatier-sendrance reaction Next point you write down due to steric hindrance Steric hindrance, you know, what is steric hindrance? You know, what is steric hindrance? due to steric hindrance in alkene alkyne is Alkyne is more reactive reactive towards catalytic hydrogenation Yeah, obviously Trudeb, you see sp2 hybridized carbon is converting into sp3 That will change because the pi bond is breaking Okay, pi bond is breaking. I'll explain this steric hindrance just a second. Okay, so We can write the rate of hydrogenation is inversely proportional to the steric crowding pi bond So for example, you see if I ask you the order of rate of hydrogenation The first molecule is CH2 double bond CH2 The second one is CH2 double bond CHCH3 The third one is CH3CH double bond CHCH3 Now, you see the steric crowding is what? Suppose we have carbon carbon double bond And on these double bonded carbon atom Only hydrogen is present or suppose the another one if also I take CH3 H H This one Steric hindrance means what? Because the size of the atom or molecule there will be little bit of repulsion between them Size it's just like you have small space and in suppose on Like in the school the bench where you sit, right? there Maximum two people can sit comfortably, right? But what what happens if I? Ask three or four students to sit on the same bench, then what happens? Everyone will push each other, right? That is nothing but the hindrance or steric hindrance Did you understand if in the less space if you want to fit more items or more people into the small Bench that obviously everyone will push each other So that is nothing but the hindrance hindrance because of less available space or more size of the people right So that is what steric hindrance means. Okay. Now. What is the meaning of this here? What is the use of this steric hindrance here? Here you see We have hydrogen and hydrogen since the size of hydrogen atom is very small So we do not consider the steric hindrance here between these two hydrogen atom Correct, but when you compare the steric hindrance of cs3 and h here It is obviously more than the steric hindrance hydrogen and hydrogen here because the size of the methyl methyl Molecule is obviously more than to that of hydrogen Correct. So these two Will have more steric hindrance than this one So if I compare the rate of hydrogenation over here in these two molecule because of the less steric hindrance here The rate of hydrogen of this one is more than to that of this Fine clear Did you understand steric hindrance because this term you will see Very often in organic chemistry steric hindrance It affects a lot of property of the molecule or the reactions Now if I ask you the steric the rate of hydrogenation order of rate of hydrogenation of this molecule What will be the order if it is first? Second and third Where we have the maximum rate of hydrogenation Correct. First we'll have maximum Then we have second only one cs3 and then we have third Two cs3 group Now this molecule you see this molecule. I'll try to draw the structure of this We have C double bond C One cs3 I can write this side Hydrogen may be this side cs3 here And then hydrogen here Another possibility is what both cs3 group are on the opposite side Like this What is this and this are the same? These structures are same. Yes or no Tell me are they same tell me Okay. Now I got lost the connection actually so Yeah, they are They're assistant transfer it. Yeah, okay fine. Huh? Okay, so this one is cis form Right. This is see the molecular formula is same It is cis and this one is trans Two cs3 group are on the opposite side and here the two cs3 group are on the same side. Okay This is actually they are in stereo isomerism, which I will discuss Okay, but little bit of idea you can have when the same molecule Actually, we have to understand the priority first then we can do this But those things in detail I will discuss in the studio isomerism Here you see two cs3 group are on the opposite side trans And same side cis, but if you see the name of both the molecule it is It is what it is But two in only to the point is the name of the two compounds are same So if I ask you draw the structure of two butene Or but two in which one you will draw this one or this one? That's the confusion, right? So to clear this they have used this term cis and trans they have introduced this term cis and trans to clear the Structure of the molecule. So if I ask you draw cis to butane, you can draw this one trans to butane You can draw this one. Okay. Now here. Why I've why I've discussed it here because here you see both cs3 group are on the same side So comparatively they have more repulsion here because the distance between the two csp molecule is less comparatively here the distance is more comparatively Okay, so because of less distance here the steric hindrance is comparatively more over here That's why the rate of hydrogenation of cis form is less than to that of Trans form. Is it clear? understood this Now you see this first reaction we have seen the preparation of alkene from alkene and alkyne The second one you write down second method alkyl halide From alkyl halide. What is alkyl halide? You know already Rx right in alkene when one hydrogen atom is Substituted with halogen group Okay, so you see there are various reaction different different reaction we have For for the preparation of alkene from alkyl right, so I'll just write down here in one A simple page one page only and you also write down the same way when you use li alh4 nabh4 First you write down. I'll explain this then we have na with dry ether r1 zn And then we have r1 cu li this li alh4 at nabh4. You will see this reactant again very often in organic chemistry These are reducing agent both reducing Agent this is the Preparation of alkyne from a reduction method the name of li alh4 is lithium aluminium hydride lithium aluminium hydride the name of nabh4 is sodium borohydride sodium Borohydride these two are reducing agent it provides hydrogen to this molecule It has nature to provide hydrogen to this molecule. Okay, so in this what happens this x halogen group gets substituted From hydrogen and we'll get alkene as rh So whatever alkyl group you have here Yeah, sodium borohydride whatever alkyl group you have here that will Accept one hydrogen from this agent li alh4 and all right Na and dry ether this one is very important na ether and we get here Na x this method We call it as bourge reaction w ur Tz bourge reaction is used for the preparation of alkene See i'm writing down here general reaction, but when you take this bourge reaction We'll take two molecules of alkyl halide two molecules of alkyl halide means we'll take two rx reacts with na i'll just write down this later on okay R1 z n this also gives you r r1 plus z n x This reaction we call it as Franklin reaction Franklin reaction when you use this r2 c u li the product we get here is r r1 plus c u x plus li x Okay, and this we call it as c o r e y cori house synthesis cori house synthesis Okay, so these three reactions the third the three last reaction that i have written These are actually coupling reaction okay Why coupling reaction? because you see This here we use two molecules of this so this r and r gets coupled here This r and r dash joins together r and r1 joins together. That's why these are coupling reaction. Okay Our Grignard reagent we'll discuss later on through the view rated b Rmgx that is okay Rmgx this reagent We call it as Gilman's reagent not important, but you write down No need to memorize this just you write down Gilman's reagent r1 c u li All these name reactions are important. You have to memorize this one is very very important Burj reaction you is used for the preparation of alkane Franklin preparation of alkane Cori house synthesis preparation of alkane Okay, now another thing is what from all these three methods Can we prepare methane or the alkane which has only one carbon atom? R1 is any alkyl group. You see Shreya What happens? We have written here r right r is the alkyl group And r1 is also an alkyl group, but these two are different alkyl because one is methyl so other one is ethyl One is Propyls other one is butyl like that Okay, it's just different from this one That's the point. So we cannot prepare methane from words Franklin and Cori house synthesis because since these are coupling reaction You see r and r so at least two carbon atom must be there We cannot have any compound which has only one carbon atom present Okay, so these three reactions you write down into this This note you write down write on these coupling reaction These coupling reaction Can not form cannot form Alkane with one carbon atom can not form Alkane with One carbon atom Important this point is important Okay, next we'll discuss little bit woods reaction in detail That you see So you write down the heading woods reaction write down This is the preparation method Of symmetrical alkane is the You see the previous slide here this one Here we have the symmetry right Here we have symmetry But since this r and r1 is different r and r1 is different. So we don't have symmetry over here right, that's why What we say woods reaction is useful for the preparation of symmetrical alkane Understood this. This is the preparation method of symmetrical alkane. It follows free radical Mechanism have you studied free radical mechanism radical mechanism? Have you studied? So this particular reaction discussion also they ask that woods reaction follows what mechanism So free radical mechanism now you'll see mechanism. We'll discuss quickly here the metal that we are using n a This metal has free electron So that free electron it can release and this is the properties of the metal metal we use in the reaction So that it can provide free electrons for the reaction okay, so This you write down here. This is the metal purpose is to provide free electrons Okay, free electrons. So next is what the alkyl halide we have rx And we take two molecules must you write down here For woods reaction. I have told you already We take two molecules of alkyl halide the alkyl halide that you have here this electron the free electron We'll take this halogen atom And we have here two electrons two Bond pair of electrons One electron is taken up by this hello by this halogen atom and other one is taken up by this alkyl group. So we'll get r dash plus x With one pair of electron and negative charge on it This is nothing but the free radical one dot means what? free radical okay, so and this is True for both the molecules this happens with both the molecules means one alkyl free radical we get from one molecule another one also gives one molecule, right? So what happens then this free radical is highly reactive in nature So this two alkyl free radical combines together and form r r that is the alkyl group and na plus Which is here takes this x minus and forms na x Right, this is how the reaction proceeds in woods reaction understood Mechanism is not that important if you know the product here You don't need to write down this because You have multiple choice question. So you have to choose one of the option So you should know what product we are getting how the product is forming Right now, you'll see the alkyl halide. We have ch3 cl Example i'm giving you cs3 cl allowed to react with na in dry ether And two molecules of this we are taking. So what is the product we get here? Can you tell me? Yes, the product will be ethane Which is ch3 ch3 plus na cl major product is this Okay, we also write down this reaction in this manner also in the book. You must have seen They have written like this cs3 cl Plus na plus cl cs3 And they write down ether over here And this is the coupling we have Right two molecules of na also we take To na so we'll get here ch3 ch3 These alkyl group alkyl group joins plus we'll get two molecules of NaCl Some limitations we have forward reaction. You write down. There's two points. We have in this First point you write down. I don't have any book now Any book you see hope it and then and all it will be given like this. They are written Okay Any book you can see So limitations are what? Methane cannot be obtained I've already told you methane Can't be obtained from this method Since it has only one carbon and one important limitation we have here Fails with this reaction is not possible or it fails with Tertiary alkyl halide. What is tertiary alkyl halide? It is suitable only for It is suitable only for symmetrical alkane next one you see the reaction Which is franklin reaction in this reaction Write down in this reaction Kyle halide is heated with with zinc in inert solvent So you see again one alkyl that is rx Zn Xr dash When you heat with zinc then the product will get is rr dash And zn x it is actually the same kind of reaction Like originally But the difference here is what we are getting two different alkyl halide here Right, so we are getting what unsymmetrical alkane Unsymmetrical alkane Okay, this is see in organic chemistry. It's not like only two product will get into this Okay, I'll give you tertiary. Just give me a second tertiary. I'll give you see It's not like only one compound will get here We'll get little bit of this also. I'll write down here r dash r dash symmetrical alkane also We'll get little bit of r r also here Right, but the major amount will be this This one unsymmetrical one Okay, so mind this if they ask you how many total product we'll get then one two three will be the answer If they ask you what is the product we get here Then the answer will be this if they say only product it means major product If they say total number of product then major minor all you have to count right so in this in this Question in this reaction If they ask you what is the total product we get then the product will be symmetrical unsymmetrical all product whether it is major or minor correct tertiary alkane reduce this today Carbon with which the halogen group is attached That carbon atom must attached with three different carbon atom This this is tertiary alkane. No that and x2 we don't ask we don't consider Because this is the preparation of alkane. So we consider only alkane whether it is symmetrical or unsymmetrical next one you see The another reaction we have body house synthesis write down in this reaction in this reaction alkyl halide is treated with in this reaction alkyl halide is treated with lithium dialkyl cuprate lithium dialkyl cuprate following up following up the reaction with following up the reaction with alkyl halide the reaction with alkyl halide again to form an alkane to form an alkane alkane, okay Here you make one change. I missed here. This is r2 The same alkyl group with r2 here This one is right Now you will understand this. How do we get this here? Okay so you see this reaction First of all, suppose the alkyl halide we take is cs3 br and this is first reacts with li dialkyl cuprate is this r2 cu 2 alkyl groups dialkyl cuprate, okay, lithium dialkyl cuprate It reacts with 2 li in presence of dry ether and we get here ch3 li plus li br this ch3 li 2 moles of it allowed to react with cui and it forms li ch3 whole twice cu plus lithium iodide This is dialkyl cuprate lithium dialkyl or dimethyl we can say No Franklin reaction mechanism is not required. Whatever it is required. I'll give you Okay mechanism for all reactions are not important So all those things which is required. I'll give you don't worry with that And eventually if you know the product you can do the Questions easily it is not exactly same more or less same Okay, but that is not a mechanism is not required So li ch3 whole twice cu plus li this is dimethyl cuprate So these two reactions are the preparation method of this particular reagent That is Gilman's reagent dimethyl lithium dimethyl cuprate Now when this is allowed to react with The another alkyl halide. I am taking c2 h5 br here anyone you can take cs3 br here Here I have taken the different one c2 h5 br plus li ch3 whole twice cu This gives This alkyl group and this alkyl group joins So we'll get c2 h5 ch3 Plus li br lithium bromide This reaction follows The last one This is the major reaction of corey house synthesis. These two are not Major reactions the reaction of corey house synthesis is this These two reactions. I have given you for the preparation of gillman's reagent, which is nothing but this one Okay, so if they ask you to write down the reaction of corey house synthesis write down this one Okay, this reaction actually follows SN2 mechanism What is SN2 mechanism? I will discuss that just you write down now SN2 So this one you see this is unsymmetrical alkane Part number of carbon atom unsymmetrical One condition we have here that this alkyl halide that we are taking here this one This must not be tertiary halide tertiary halide If it is tertiary halide Then in this case elimination reaction takes place So this also I will discuss in mechanism Okay reaction mechanism Then you can connect all the understood So these are the few conditions. That's why this thing is a little bit Difficult because there are many many things you have to keep in mind and it's not like only 10 15 reactions you have one chapter has One chapter has 100 200 reactions Okay, so that's why the organic is a bit difficult Once you are doing this Once you are starting if you are starting you will feel okay It is easy we'll get like this this will get the product But all the reactions you have to keep in mind that is a bit difficult part Okay, next preparation method you write down that is From gregorad reagent When you have semester end examination of 11th class So feb will be off right complete holiday. Okay Fine See what is gregorad reagent? gregorad reagent is alkyl magnesium halide The formula of this is r mg x alkyl magnesium halide and in this r mg x The charge you must Keep in mind because it is important The alkyl group has positive charge r plus and mg x will have minus charge r plus mg x minus Just a second This is negative and this is positive r minus mg x plus Okay, this you must keep in mind Now In this what happens alkyl group and magnesium bond breaks Whenever the reaction involves gregorad reagent Because magnesium is a metal and here we have carbon So c mg bond if you see carbon and magnesium bond This bond is quite polar This is because of high electronegativity of Of carbon C mg bond is quite polar. That's why this bond breaks and the reaction Proceeds. Okay, so these are the few information you should have this reagent You will get in organic chemistry very often. So you must remember all these things now in this you write down gregorad reagent undergo gregorad reagent undergo decomposition reactions decomposition reactions with water with water Ammonia alcohol mines or amines having active hydrogen atom of amines having Active hydrogen atom. What is active hydrogen atom? I'll tell you first you write down having active hydrogen atom to give alkanes to give alkanes corresponding to to give alkanes corresponding to The alkyl group of gregorad reagent corresponding to an alkyl group of gregorad reagent active hydrogen atom means what? all those hydrogen atom hydrogen atom Attached with attached with highly electronegative elements the electronegative elements such as oxygen nitrogen sulfur fluorine or triple bond also all those hydrogen attached to Oxygen nitrogen sulfur fluorine triple bond are called active hydrogen. Okay So now you see the reactions the preparation method of gregorad reagent is this This also you must remember when you have alkyl halide It reacts with magnesium In presence of ether and we get Rmgx that's just this magnesium introduced in between this R and x We'll get Rmgx. This is the preparation of gregorad reagent. Sometimes this is also required. So you must remember Now You see the reaction suppose. I am taking the gregorad reagent as c2h5 mgbr plus H2o h plus acidic medium Okay like I said The alkane forms here corresponding to the alkyl group of gregorad reagent So the alkyl group the gregorad the alkyl group of gregorad reagent is nothing but this c2h5 And this will form alkyl when it takes h plus So r minus obviously here we have minus And mgbr we have plus So this h plus will taken up by this c2h5 and forms c2h6 h5 and h Plus Here mg plus we have will take OH minus from this water molecule and we get what mg OH Br will be as it is So this is what the alkane we get Lithium and sodium I have given you the mechanism of sodium We are not taking two molecules of this here Right and magnesium if it releases electron it will release two electron mg plus to second group element Right. So here the form is first of all we are not taking two molecules of this formation of r r is difficult here Second thing is what this requires one electron only in the outer most shell to complete the rocked it But this will give two electron because mg plus is not stable that much mg plus is not that stable So whenever it loses electron it will lose two electron and it requires only one electron So that's why the reaction in that way is not possible. Is it clear to them? Yes or no clear There may be any doubt Can you proceed further? Okay So this is the reaction with water like I said water With water ammonia alcohol anything so suppose I'll take the example of alcohol now now with this only If it reacts with alcohol I am taking HO C2H5 Okay, so what happens here This H plus is taken up by this C2H5 We'll get C2H5 H Plus mg Br will be as it is What is left? Oh C2H5 right or C2H5 will attach with this If you take ammonia ammonia is NH3. So I'll write down like this H NH2 So this H will taken up again by this alkyl group C2H5 H Plus mg Br will be as it is With NH2 attached with this main issue if you have any alkyne at C triple bond C CH3 With this the product will be again C2H5 H Plus mg C triple bond CH3 C triple bond C single bond CH3 and Br One last one you write down If you have acid like C double bond O CH3 Again, this hydrogen is taken up by this C2H5 So C2H5 H Plus mg Br And here we have O O C CH3 Is it clear? So wherever you see when the hydrogen atom is attached with oxygen nitrogen triple bond The hydrogen is taken up by this alkyl group of Grignard reagent and alkane forms This is the preparation method of alkane from Grignard reagent next can you move on next write down clay mention reduction See last product See all these reaction what happens you see in all this reaction Whatever is left after hydrogen since This hydrogen is taken up by this C2H5 This hydrogen this hydrogen and this hydrogen is taken up by the C2H5 And that's why we get C2H5 H C2H5 H H and H and whatever left here. What is left? O C2H5 So this O C2H5 is attached with mg here Similarly, what is left here NH2 attached with mg? You see here NH2 What is left here C triple bond CCS3 attached with mg? What is left here? CS3CO O O CCS3 attached with mg must take care of one thing This magnesium attached with the atom here With which sorry atom here from which the hydrogen has been eliminated So hydrogen is eliminated from oxygen. So mg is attached with O C2H5 not C C2H5O Hydrogen is eliminated from Nitrogen so mg attached with nitrogen Similarly here with carbon here with this oxygen Next one you see clay mention reduction write down in this It involves reduction of carbonyl groups It involves the reduction of carbonyl group of aldehyde and ketone This is also important clay mention reduction It involves the reduction of carbonyl groups of aldehyde and ketone To methylene group to methylene To methylene group In bracket you write down methylene group is this Group is CH2 carbonyl group of aldehyde and ketone is this carbonyl group is C double bond O So in this reaction carbonyl group converts into methylene group. It means C double bond O simply converts into CH2 Okay, the reagent we use here reagent is important ZnHg Concentrated HCl This is the reagent of clay mention reduction must remember this ZnHg With concentrated HCl So if you see here if the reaction is this see 6H5 C double bond O CH3 This when it is allowed to react with ZnHg with Concentrated HCl Then this C double bond O converted into CH2 simply Everything will be same C6H5 will be as it is instead of C double bond O Write down CH2 and then CH3 will be as it is understood Mechanism if you see Suppose I am taking this R C double bond O and R So the product if this is allowed to react with this will be R CH2 R Correct Now you see the mechanism how it goes H plus This H plus comes from this acid conch HCl Now this H plus Will this Lone pair Will take this H plus And C double bond O H positive charge on it And R will get Further what happens here Since this oxygen is positively charged electron deficient So this will drag the bond pair of pi electron here So it forms R COH R with positive charge on this carbon atom Right now this Zn that you have this Zn can easily Give two electrons Right So this will give two electrons and forms Zn plus two plus two electrons This two electron is taken up by this carbon atom positively charged carbon atom RCOH With two electron here R and one negative charge Again from the acid will get two H plus Which attached with The carbon and the OH group one H plus will come over here and one H plus will attach on the carbon since it has one negative charge So we'll take OH2 Positive charge here Rn hydrogen will be here Right now in this this H2O molecules comes out And we get R CR H Positive charge on it plus H2O Right Again that zinc Has two electrons it gives this electron to this carbon atom And it forms R C with two electron R H one negative charge And in the last Will add H plus into it The H plus will take this carbon atom And the product we get here is R CH2 R this is the answer So basically this mechanism you don't have to memorize. Okay. You have to just keep in mind Whenever this reagent we have Zn Hg With conch at cl This one Zn Hg with conch at cl C double bond O you have to replace with CH2 that is it. Is it clear? Tell me if you understood this next write down write down by decarboxylation By decarboxylation of carboxylic acid next method Decarboxylation of carboxylic acid means removal of CO2 from carboxylic acid But this decarboxylation takes place in the form of Na2 CO3 Okay, so write down into this first when sodium salt is heated with Soda lime When sodium salt is heated with soda lime What is the formula of soda lime? What is the formula of soda lime? When sodium salt is heated with soda lime see soda lime is is NaOH But in this process we add some amount of CaO also into this quick line So we'll take the mixture of NaOH and CaO the purpose of addition of this CaO is See this CaO is hygroscopic in nature hygroscopic in nature it Keeps NaOH dry. That's why we add CaO into it. Okay So decarboxylation when sodium salt is heated with soda lime soda lime is NaOH But in this reaction we'll take the mixture of NaOH and CaO reason is this Alcane forms, cane forms and CO2 eliminated CO2 eliminated Alcane forms and CO2 eliminated in the form of carbonate CO2 eliminated in the form of carbonate Hence we call it as decarboxylation Hence we call it as decarboxylation So sodium salt Okay, sodium salt of carboxylic acid will be what? Sodium salt is this are C-double bond O O Na plus when it is heated with NaOH and we have CaO also So what happens this Na2 CO3? Comes out and we get R H How this Na2 CO3 comes out you see this NaO Na2 CO3 comes out and forms R H Okay So one example you write down C2 H5 COO Na plus NaOH Heated with CaO So it forms C2 H5 H plus Na2 CO3 Right, so basically this hydrogen and this C2 H5 joins and into CO3 forms Here you see one important point here It is what the number of carbon atom here it is what? three carbon atom Okay, and the number of carbon atom we get here into this alkane is two only Right, so in this one note you write down alkane formed by this process alkane forms By this process as always as always one carbon Less than has one carbon less than the number of carbon atom less than the number of carbon atom Present the number of carbon atom Present in the salt present in the salt Hence this method is useful Hence this method is useful Hence this method is useful for the preparation of lower alkanes This method is useful for the preparation of lower alkanes Is it clear any doubt you have till here? Tell me any doubt Okay, fine. So we'll take a break. Okay Uh, I'll take my lunch 15 minutes break will take and then we'll start again at 335 fine Yeah, we can get Correctly there we can get methane clear. We'll start at 335 fine. Yeah, it's still four So no see you again at 335 So Okay, can you start so next Preparation method you see next preparation method is by electrolysis of electrolysis of sodium or potassium salt By electrolysis of sodium or potassium salt. Okay Write down into this in this process in this process electrolysis of electrolysis of a concentrated concentrated aqueous solution of aqueous solution of sodium or potassium salt a saturated saturated monocarboxylic acid monocarboxylic acid higher alkane higher alkane higher alkane at anode This reaction follows free radical mechanism Overall a reaction if I write down in one step the reaction is this 2r na plus 2h2o electrolysis it gives r r 2co2 plus 2 NaOH plus H2 So NaOH plus H2 this is the reaction we have mechanism you see the salt we have of saturated monocarboxylic acid We can take here potassium salt also cs3 cook also This gives you cs3 coo minus plus Na plus two different one cation one anion will get Now this cation and anion will react at cathode and anode So you see the reaction of anode two molecules of this we are taking so two cs3 coo minus at anode oxidation takes place will get two cs3 coo radical plus two electrons How do we get radical radical is nothing but only one electron this single electron this oxygen has So one electron is given by one ion of this since we have two ions total two electron will get out right Now this is a free radical and the free radical has property that it can dissociates into dissociate to form another Radical also it can also free radical has this property to eliminate Neutral molecule to get new free radical. Okay, so what happens into this cs3 coo radical we have it eliminates co2 molecule from itself to form methyl radical which is cs3 right All this reaction takes place at Anode this co2 Will go out now this two cs3 radical will combine together cs3 cs3 and forms ethane cs3 cs3 So this is what our r we have cs3 cs3 we'll get at anode so anode at anode will get the uh What will get the alkane now the reaction of cathode you see at cathode Oxide reduction takes place. So na Electron will get consumed. So we have na plus right to na plus So this two na plus will take this two electron and forms na metal which further combines with H2o because we are taking what we are taking here concentrated aqua solution right so h2o we have there Which forms two na OH and h2o Right this we call it as coal electrolysis this method is coal electrolysis the name of this uh method is k o l Be Electrolysis This is also important. Is it clear? So two three points we have to memorize here first of all in coal electrolysis method hydrogen gas evolves at cathode alkane forms at anode It follows free radical mechanism. This is also important. So these three points you must remember One note you write down into this the last point in this one note you write down When mixture of two salts Are taken suppose i'm taking this one r c o o na plus R2 c o o na when the mixture of two salts are taken then we get The mixture of hydrocarbons will get the mixture of hydrocarbons will get r1 r1 as one of the compound r2 r2 It's one of the compound. This is one important point next point It is used for the preparation of for the preparation of symmetrical alkene Which is the another method which is used for the preparation of symmetrical alkene What is the another method? Which is used for the preparation of symmetrical alkene Wurz reaction. Okay, that's We have done already. So can we prepare alkene methane from this? Can we prepare methane from this? Because again in this two alkyl group is joining. So we cannot prepare methane. So all these things you must keep in mind Okay Next one write down next alcohol comma aldehyde comma ketones alcohol aldehyde ketones or fatty acids Fatty acid or nothing but carboxylic acid alcohol aldehyde ketones or fatty acids and their derivatives Fatty acids and their derivatives are reduced to alkene reduced to alkene by the use of Okay And the derivatives are reduced to alkene by the use of red phosphorus Like this you write it is also an important reagent red phosphorus red p you write down red phosphorus with h i This is also a reducing agent in this process next line in this process The number of carbon atom The number of carbon atom in alkene will be same as will be same as the number of carbon atom the number of carbon atom present in the reactant molecule So very simple straightforward reaction we have in this Suppose we have alcohol like I said alcohol aldehyde and all alcohol. Suppose I have cs3 OH and if this is reduced with the help of h i red phosphorus Temperature we use around 150 degrees Celsius The product alkene will have the same number of carbon atom since we have one carbon here. So here we'll get ch4 Plus i2 also evolve and h2 This is the reaction we have in this another one you see If I take cs3 chO aldehyde now first one is alcohol second one is aldehyde red phosphorus Temperature we use again around 150 degrees Celsius two carbon we have here So c2 h6 plus h2o Plus two i2 Second one I'm taking ketone third one cs3 co cs3 Plus hi red phosphorus Again 150 degrees Celsius We'll get what three carbon we have so c3 h8 equal number of carbon atom Plus h2o Plus i2 ketone and then acid So acid will take fatty acid is the abusalic acid cs3 co OH plus hi Red phosphorus temperature around 200 degrees Celsius Again, we have two carbon atom here. So c2 h6 plus 2h2o Plus three i2 one example for the derivative of this cs3 co cl Plus hi Again red phosphorus 200 degrees Celsius two carbon here So again c2 h6 Plus h2o Plus hcl Plus three i2 This is what the reaction is. Okay These are the you know these are not important or these product The important thing is the alkane that you are getting and whatever the number of carbon atom here we have in the reactant molecule The same number of carbon atom will present In the alkane that we get in this process Okay, even this temperature is also not required temperature is also not Required for all these and in organic chemistry Balance reaction is also not important. You don't have to write down the balance reaction Like I have written this reaction all these reactions are balanced But if you don't know this also, it's fine. You don't need to balance the reaction The only concern in this organic chemistry. We have is the product. What product we get what is the major product we get Okay, so these are the reactions in all these reactions red phosphorus and h i we are using and we get The alkane in this Okay So these are the few Method we have seen for the preparation of alkanes Okay, after this we have already done physical properties. We are left with chemical properties Okay, so chemical properties We will not start today because we don't have time. Okay, so we'll finish chemical properties next class and then we start alkene right second thing is that You will have I will take one more class this week. Okay, when I let you know Okay, maybe friday or thursday. I will let you know So I'll finish this alkene and then alkanes is very small chapter. There are very few reactions we have into this See one more thing. Let me tell you this hydrocarbon chapter is very important because whatever chapter you are going to study in organic chemistry There will have questions like this. They'll ask I'll just Give you one example Suppose we have some compound here x Some reagent here r1 gives a Then another reagent r2 gives b Then another reagent r3 gives c and they'll ask you what is a b and c like this Suppose the chapter is aldehyde only aldehyde ketone only so in this these are the series reactions which is given So anywhere you will have the application of all these reactions that we are studied We have studied in this chapter so far and all those reactions which we are going to study in this so alkene alkene These reactions will be useful in All the chapters that we are going to study in coming future Okay, so this particular chapter you do it properly Maybe there are chances that this particular reaction suppose it requires some information of alkene Okay, here it is for Aldehyde suppose you're doing aldehyde chapter If you do not know this one if you know this also if you know this reagent also you cannot do this question So that's the point. That's why in organic chemistry all the chapters are related Whatever see we have done many we have done preparation of alkene From alkene right there we have seen the reaction of alkene and how they gives alkene So whatever the preparation method of alkene here we have In preparation of alkene here we have with alkene This particular reaction will become the chemical reactions of alkene Okay, so like this the same thing will repeat Okay, so you have to connect this that's why I said I always say That you have to do organic chemistry twice then only you will Start connecting things from one chapter to the another chapter Anyway, so I'll keep on discussing these things In classes don't worry with that Okay, so next class we'll start with chemical properties and then we'll do alkene alkene also Fine. Okay, so we'll see you soon Enjoy Merry Christmas Any plan you are going somewhere or not evening Or you are here only at home