Transcript for:
Understanding Hydrocarbons and Their Properties

you are braver than you believe you are braver than you believe stronger than you seem and smarter than you think so hello my shimmering stars i surya is going to start one shot video lecture on chapter hydrocarbons students in our victory batch we have already covered iupac isomerism and goc now we are going to start the most important chapter that is hydrocarbons so get ready with your books and let us start this chapter first of all students the one thing that arises in our mind is what are hydrocarbons to understand hydrocarbons let us recall about organic chemistry if i go to define what is an organic chemistry so you would say that organic chemistry is a branch it is a branch of science it is a branch of chemistry in which we deal or which we study with the hydrocarbons and its derivatives so what i am saying it is the branch in which we study about hydrocarbons and its derivatives now students question arises what are hydrocarbons hydrocarbons are those organic compounds that are made up of carbon and hydrogen okay now we got to know what are hydrocarbons but what do you mean by derivatives derivatives are the one that we derive from something if i am saying hydrocarbons and derivative that refers to that we are making organic compounds from carbon and hydrogen as well as with the help of them we are also deriving other compounds for example if i talk about the simplest organic compound simplest hydrocarbon i would say it would be methane that is ch4 i am talking about the simplest hydrocarbon that is methane which is ch4 why it is the simplest hydrocarbon because it contains only one carbon in it right now students what i will say i will say that kindly remove one hydrogen from here so if you are going to remove one hydrogen and you are going to add another group here for example you are adding alcohol you are adding halogen group or you are adding an alkyl group also so what you are going to obtain you are going to obtain some more organic compounds so that means from any hydrocarbon you can further derive another hydrocarbons or you can further derive organic compounds they are referred to as the derivatives right so hydrocarbons are basically what if you want to mention you can kindly write here they are made up of carbon plus hydrogen right the compounds which have both of these carbon as well as hydrogen like example is methane in front of you now if i take one more example let us see c2h6 now how is c2h6 obtained c2h6 is basically ethane it contains two carbons in it so it is ethane we have already started in the iupac nomenclature whenever we are having one carbon the word root is known as myth whenever we are having two carbons the word root is known as it likewise we move on it prop b butte pent hex in these ways like putting the value of n as one two three four five putting the values of number of carbons and we find out word route right so here what i'm going to tell you i'm giving you some of the examples of hydrocarbons now students coming to our topic today we know that hydrocarbons are classified into basically two sub categories one are saturated one another one are unsaturated one now what do we need to study in this chapter we need to study saturated hydrocarbons in detail and unsaturated hydrocarbons in detail likewise if i say if i'm going to classify hydrocarbons into two basic subcategories number first is saturated hydrocarbons and if i talk about second category it is unsaturated hydrocarbons now students what are saturated one and unsaturated one the question arises saturated hydrocarbons are the one which contain single bond between the carbon atoms if i say carbon carbon contains single bond they are referred to as saturated hydrocarbons but if i talk about if i talk about unsaturated hydrocarbons these are the ones which contain either double bond between the carbon atom or triple bond between the carbon atom right so these are what these are unsaturated one so when i talk about single bond i would say that these are referred to as alkanes these are referred to as alkanes and whenever students i am talking about unsaturated one whenever i am talking about unsaturated one we have double bond as well as triple bond with us so the double bond ones are known as alkenes and the triple ones are known as alkynes they are known as alkynes so students now moving to what do we observe in alkanes alkenes and alkynes if i make a compound of alkanes the simplest compound of alkane that is the methane molecule in which there is only one carbon atom so i would say this is ch4 this is methane if i am going to write it just like this so what are you observing here you are observing that there is a single bond between carbon and hydrogen and whenever there is a single bond it is referred to as it is referred to as sigma bond it is referred to as sigma bond right all these bonds are what all these bones are sigma bond right now if i talk about alkenes and alkynes if i talk about alkenes and alkynes let us see one example over here in this example students what do you observe we observe that basically this bond is sigma bond because it is single bond okay this bond is sigma bond because it is single bond this bond is sigma bond because it is single bond but but students look here the second bond is pi bond the second bond is pi bond every first bond is sigma bond second or third bond are pi bonds okay in double bond case second bond is pi bond in triple bond case second and third both are pi bonds now if i talk about the alkyne structures alkyne is basically like this okay now students if i see in alkyne's case first bond is known as sigma bond while the other two remaining two are referred to as what the remaining two are referred to as pi bonds okay now students these are the ones these three examples are the one that are of aliphatic that means of chains i would say straight chain or open shade cup out there are some organic compounds that are aromatic in nature which have basically cyclic now what are the aromatic ones or i would say cyclic ones if we say these are further classified into open chain and closed chain so open chain i have already shown you in which the first carbon atom is not attached with the last carbon atom that is open chin compound in closed chain compound the first carbon atom is attached with the last carbon atom so if i take example of an alkane let us take an example of alkane over here so it is carbon carbon carbon carbon keeping the valency of carbon in mind i am making bonds with hydrogen now this compound is basically what this compound is basically the cyclic compound it is basically the cyclic alkane right students here you can see there are single bonds present between them in between the carbon atoms so these are referred to as what these are referred to as alkanes so alkanes are the one in which we are going to observe single bond right alkenes are the one in which we are going to observe double bond and alkynes are the one where we are going to observe triple bond if i take one example of this uh cyclic over here let us take one example of alkynes okay now this is the cyclic one over here you can see there is a for that first carbon atom is attached first carbon atom is attached with the last carbon atom if this is first this is second this is third this is fourth first is attached with force then it forms a ring like structure or a cyclic structure so this can comes under the category of what cyclic chain like structure right now students this was the basic that we started in the iupac or we started in the isomerism during the classification of hydrocarbons but now we are going to start our chapter that is that is about hydrocarbons in which we are going to start with what alkanes now what are alkanes first of all we have started started this very carefully but i want you all you all to write some of the features okay to write here some of the features of alkanes some general characteristics of alkanes so alkanes are the one students in which we observe single bonds so they are referred to as saturated hydrocarbons they are saturated hydro carbons okay now if i talk if i talk over here that during the formation now we are going to see the methods of preparation of alkanes in the coming section what we need to know here is whenever you are making some of the alkanes like you are making with anything in propane so do they have a general formula yes every homologous series have a general formula so if i talk about alkanes they have a general formula they have a general formula the general formula students for alkanes is c n h two n plus two this is the general formula for the case of alkanes that you need to keep in your mind students other than this students what are the points here if you keep the value of n the value of n as 1 what you are going to get c1 h2 into 1 plus 2 that comes out to be ch4 which is referred to as what methane why methane because here is one carbon atom and all of them contain single bonds so ain will come at the end for alkane we write suffix as n right so what we are going to get over here we are going to get over here as methane okay now if i talk about n is equals to two students what i'm going to get c 2 h 2 into 2 plus 2 kindly write the value of n over here and start making all these right so here you are going to get what here you're going to get c 2 h6 that is going to be what ethane now coming towards n is equals to 3 what we are going to get c3 h3 into 2 plus 2 that is going to be c3h8 that is going to be propane that is going to be propane now let us take one more case n value has a 4 so c4 h4 into 2 plus 2 that is c4h10 that comes out to be what butane and so on you can write n is equals to 5 and is equals to 6 you can do do that structures right now students what i want you all to understand over here that these all are basically what these all are basically a homologous series homo logos series now what do you mean by homologous series this i have already told you in the iupac section that homologous series basically defer by the consecutive members of the homologous series defer by ch2 group let us see over here if i talk about methane and ethane both of the formulas are written in front of you molecular formulas are given to you what is the difference between both of these there is a difference of one carbon atom and two hydrogen that is the ch2 group so yes if you see any of the you know consecutive members of the homologous series you are going to get the gap of ch2 that means it lies in the category of homologous series other than this homologous series contains the members of the same family i would say all of these contain single bond between them so all of these are alkanes hence they are a homologous series right now students other than this what are alkenes known as alkanes are called as paraffins what are alkanes called alkanes are called as alkanes are also called para fins now what do you mean by paraffin students if i divide paraffins into two categories para refers to as little and if i talk about effins that means affinity affinity refers to the capability or to make a bond right so these are also known as paraffins that you need to keep in your mind students now students now students other than this what we need to know is that alkanes are basically inert at room temperature and normal conditions and normal conditions now if i talk about the simplest alkene that is the methane right if i talk about methane how does it looks like it is c it has hydrogen over here it has one hydrogen over here right and one hydrogen over here now what what is this let me tell you let me tell you in detail here students what you need to know is that two hydrogens are in plane one hydrogen is towards the observer and one hydrogen is away from the observer the hydrogen that is towards the observer that means you who are watching this video right so this hydrogen is towards you which is represented by the wedge bond and if i talk about this hydrogen that is away from you that is at the back side of mind so that is away from the observer that means the person who is looking at the methane molecule it is away from that and these two are in the same plane now what do you want to know here what you have to learn over here is the hybridization of carbon first of all carbon is making four bonds hence the hybridization over here is sp3 hybridized right students and the angle here angle basically here is 109.5 degree if i talk about bond angle over here it is 1.9.5 degree students these are the basic things that you need to keep in your mind other than this what is the shape here tetrahedral shape is there which shape tetrahedral so this was the basic that you need to keep in your mind about alkane students now moving forward towards our main topic that is methods of preparation of alkanes kindly put a star mark over here because it is going to be most important topic for you okay this is going to be most important topic for you methods of preparation of alkane now whenever i am talking about methods of preparation the very first methods that come in our mind is hydrogenation of unsaturated hydrocarbons what number first is hydrogenation of unsaturated unsaturated hydrocarbon hydrocarbons now students as the name suggests look at the name very carefully okay look at the name very carefully here you have hydrogenation if you are going to split hydrogenation into two uh parts you have hydroangenation where hydro refers to as hydrogen and generation refers to as addition that means that means addition of addition of hydrogen now where to add hydrogen where to add to unsaturated hydrocarbon addition of hydrogen in unsaturated hydrocarbon okay now students you got to know about hydrogenation you know what is unsaturated unsaturated refers to as the double bond ones or the triple bond ones right and hydrocarbon refers to as the uh organic compounds that are made of carbon and hydrogen right now students to start with the hydrogenation first of all you need to keep in your mind is that during hydrogenation i'm going to take both of the steps first of all i'm going to take about alkanes then i'm going to talk about alkynes if i talk about alkenes okay i'm going to write over there i'm going to write over here okay if i talk about alkenes this is my alkene here you can put hydrogen over here the bonds that are left are uh basically the hydrogen bonds right now what you need to do is you need to do this reaction in the presence of what in the presence of catalyst like nickel platinum palladium right and here you have to add hydrogen that means you have to add h2 over here as the name suggests and this will occur in the presence of nickel palladium or platinum any of these now these are what these are the positive catalyst that basically increases the rate of reaction whenever they are used now students when you are even you will be studying about you know surface chemistry you will get to know that there are positive as well as negative catalyst one that increases the rate of reaction and the another one that decreases the rate of reaction but here i am talking about what positive catalyst which are going to increase the rate of reaction right because they are going to provide the surface area for hydrogen to get adsorbed now what are these terms i am talking about adsorb surface area don't worry in the mechanism of hydrogenation i'm going to explain you in detail first of all learn to understand the basic general reaction over here i have taken this alkene with me this is what this is alkene now here as the name suggests we are going to add hydrogen over here so i'm adding hydrogen over here in the presence of this catalyst now what is going to happen see it is a very short kind of trick you would say or anything you have to break this bond you have to break this bond why you have to break this sigma sorry pi bond here because when you will break the pi bond all the bonds will become what sigma 1 because there will be only single bond present between the carbon carbon and you want to obtain alkanes only and alkanes are the one in which carbon and carbon contains single bond so you know the product that you want to obtain are alkanes but how to obtain it i'm going to tell you here you have to break this pi bond and you have to add this h2 one hydrogen over here and another hydrogen over here so what you are going to get over here see the carbon and carbon initially had one you know the sigma bond that is represented as it is and these two bonds are also represented as it is now we have broken this uh pi bond and one hydrogen is added to this carbon and another hydrogen is added to the other another carbon so what you are going to get one hydrogen over here and another hydrogen over here right so kindly identify is it an alkane molecule yes ma'am because we have seen that i have already told you that kindly consider these as hydrogens hydrogens you can take the for the case of another you know instead of hydrogen you can put an alkyl group also it should be alkene molecule that's it right but here i am taking the general case so i am writing it like this see here carbon carbon contains single bond that means here we are obtaining what here we are obtaining alkane here we are obtaining alkene now if i talk about alkyne molecule this was alkenes right now let us take off what alkyne molecules over here one second okay now let us take one more case over here that is alkynes if i talk about alkynes alkynes refers to as triple bond between the carbon atoms here let us take the simplest alkyne this is alkyne now here also we are going to add hydrogen in the presence of what nickel platinum or palladium any of these catalyst now what is going to happen here kindly of the observe this very very carefully first of all you are going to break number one this pi bond the very first one right and these hydrogens are going to attack over here on both of these carbon so what are you going to get see carbon carbon contain first of all the single bond then you have one more bond here but the third bond is being broken over here and two hydrogens are already present over here like this and two have attacked over here like this right now here we are getting alkene molecule first first we are getting what we are getting an alkene molecule now students now see here very carefully what we are going to do another step here you need to remember whenever you are preparing alkane from alkene then you have to take one mole of h2 but whenever you are preparing alkane from alkyne molecule you need to take two moles of h2 now if you take two moles of h2 it will again react with h2 in the presence of what nickel palladium or platinum now what you're going to get next time also this bond is going to break the pi bond and these hydrogens are going to attack over both of these carbons right so what we are going to get over here see carbon and carbon initially contained a single bond that was not broken but the pi bond was broken and hence hydrogens were attached over here so what you are getting from this you are getting an alkane molecule from this you are getting what you are getting an alkane molecule so students kindly observe that what is happening over here if i talk about alkene's case i am getting alkanes from alkenes by one mole of h2 but if i talk about alkyne case then i'm first of all converting alkyne into alkene and then alkene into alkane right so this is what this is the method of preparation from hydrogenation of unsaturated hydrocarbons now students we need to understand the basic mechanism of this also so that you don't get confused in your examination this i have marked for you so that you don't get confused right now let us move towards the mechanism of hydrogenation we are going to do now the mechanism if i talk about mechanism students see first of all we are going to take a nickel surface over here what we are going to take we are going to take this nickel surface this is the nickel surface or either take any of the catalyst surface okay this is the surface now see students what happens first of all first of all we know that there is addition of hydrogen right so when there is addition of hydrogen over the surface over the surface this is my h2 it will show interaction with the surface it will show interaction with the surface right so these are the bonds that means they are showing interaction with the surface of the catalyst right now students what you need to observe over here that we are basically taking an alkene case first so we want that in the unsaturated hydrocarbon there should be hydrogenation so whatever alkene molecule will be doing it will be approaching towards the hydrogen to make a bond with hydrogen right now whenever this alkene is approaching towards hydrogen to make a bond what is going to happen over here kindly see very carefully again this is my surface right and over here what i observe that hydrogen and hydrogen bonds will break so that it can combine with carbon because c carbon has basically a valencia four okay so it will make only four bonds to come and interact with it it should break its one bond otherwise it will be five valency for carbon that is not possible so what it is going to do is first for these uh this hydrogen is also going to break from here it is also going to break from here so what is going to happen i am going to write this with the dotted with the dash bond over here that it is going to break right this bond is going to break and now this carbon will be attacking over this hydrogen this will be reaching towards this hydrogen to show the interaction now what is going to happen this is the carbon this is the carbon right here we have two bonds over here like this these two bonds i have shown here it had a single bond yes it had a single bond but now the pi bond is going to break in order to interact with hydrogen atom in order to interact with hydrogen atom the pi bond is going to break so we are going to write it with again dash right now what is it is going to happen the carbons are going to interact with the hydrogen hence there will be bond formation between carbon and hydrogen so all i would say let us represent this also with the dotted ones now first of all let us write what is happening over here here this bond is going to break so this is bond breaking that i'm going to write with bb here this bond is going to form this bond is going to form and this bond is going to break so this is what this is bond formation clear students so now this state whenever we are seeing that old bonds are breaking and new bonds are being formed this state is basically known as transition state kindly understand this state is called what this is called transition state students if you remember if you remember in the goc portion in the starting i told you what are chemical reactions we started from this topic and i told you chemical reactions are the one in which substrate is there in which attacking reagent is there and whenever they react with each other in the presence of some solvent we obtain either transition state or we obtain either intermediates now intermediates we have started in detail about carbocations carbon ions carbon free radical but what what were transition state during that i told you old bonds are going to break and new bonds are going to form so this is what transition state looks like here you can see the uh the bond breakage is also occurring and bond formation is also occurring simultaneously right so this is what this is transition state kindly keep in your mind here they can ask you in your mcqs that is there any carbocation being formed no is there any carbon ion being formed no is there any carbon free radical being formed no there is a big no for the car for the intermediates portion there is no intermediate formation in ga creek but if they ask you is there any transition state so you will say yes and it is going to be four membered students how come it is going to be four membered because see one two three it is making a cycle you know whenever there is a ring formation so with that that what i am going to tell you there is a four membered ring being formed over here now what it is going to happen after this man we got to know that yes bond breakage is occurring bond formation is occurring but does this whole you know structure is going to land over the surface of this catalyst no the catalyst here acts as a medium for the reaction to complete that's it it will get separate as it is without being consumed it will get separate it is just providing a medium for the reactant to be converted into product that's it so whatever what we are getting what we are observing over here that catalyst is separated like this okay and other than this what we are getting over here see this carbon contains single bond yes here there are two bonds other than this yes this bond is broken so we are not going to show down and this these two bonds are being formed over here so here these bonds are being formed and these are formed with which hydrogen atoms now you can either take hydrogen hydrogen hydrogen hydrogen over here or you can either take any of the alkyl group but what you are observing over here that the carbon carbon contains which bond single bond and single bond is sigma 1 and sigma 1 is what alkanes molecule alkane molecule over here so what we are obtaining from here we are obtaining alkane from here so this was the basic mechanism that you need to keep in your mind about hydrogenation of unsaturated hydrocarbon students now there are some of the important points that you need to keep in your mind write important key points transition state is formed first of all other than the students no intermediates are formed no intermediates are formed now intermediates are basically here carbocations carbo annoyance carbon free radical okay all of these are not formed over here right other than this students four membered transition state is obtained okay now one different point that you need to keep in your mind the rate of reaction if i talk about rate of reaction in the case of alkenes and alkynes which will have more rate of reaction kindly consider this question very seriously which will have more rate of reaction i'm talking about alkenes and alkynes if i'm going to take a reactant as my alkene alkyne i obtain that alkynes are the basically ones which have more rate of reaction why because it has a linear structure when it has a linear structure it will interact with the surface of nickel or platinum or palladium very easily but alkenes due to its structure will not interact so much as compared to alkynes right so if i talk about the rate of reaction in alkanes and alkynes alkynes basically have more rate of reaction as compared to alkenes other than this if i talk about rate of reaction that rate of reaction is inversely proportional to steric hindrance now what do you mean by steric hindrance over here steric hindrance basically refers to as the bulky group what are you talking about this ma'am kindly explain in detail if i talk in detail i would say let us take one example of alkenes over here okay for example you have the simplest alkene molecule first of all here then you are going to get second basically here mb refers to as methyl group that is ch3 rupee you have these four structures if i'm going to name this this is your a it is going to be your b this is c while this is d now i'm asking you what is the order of rate of reaction in abcd what you're going to say is that my bulky groups are the one see hydrogen is not a bulky group but if i say i will replace one hydrogen with one methyl group it is a bulky group this is a bulky group over here if i talk in this case these two are the bulky groups that are present over here and if i see this case there are basically four bulky groups that are present over here now what do bulky groups do bulky groups basically what they are going to they are going to repel these of these structures right so whenever there are more bulky groups either you see directly there are four bulky groups over here there are two over here there is one over here so more the bulky groups more is the steric hindrance and more is the steric hindrance the rate of reaction decreases so i would say that a will have rate of reaction more than b than c and at the last d so d has the least you know rate of reaction why why d has the least rate of reaction students because because d has more bulky groups in it right so this was all about hydrogenation of unsaturated hydrocarbon students now comes the next method of preparation that is basically from alkyl halides if i talk about from alkyl halides number first that comes in our mind is direct combination if i talk about first of all what are alkyl halides so alkyl halides are basically rx where r is the alkyl group and x is the halide one halogen one fluorine chlorine bromine iodine but in this case what you are going to observe first of all let us take one case one one example of the alkyl halide the simplest one that is methylchloride ch3cl okay this is your alkyl halide now what you need to do is you need to react this in the presence of zedin and h-positive you need to react this in the presence of zinc and h-positive that is acid you need to react it what you're going to obtain over here is that one of the hydrogen is going to react with ch3 and one hydrogen is going to react with cl this bond will break from this this rx bond is going to break so when it is going to break one one hydrogen is going to attach with these sides and we are going to get ch3 h plus hcl see this bond is going to break the rx bond is going to break in both of them are going to get one one hydrogen over here so this is direct combination which occurs in the presence of zn and h positive okay now let us take one more case over here ch3 ch2cl so here also h2 is going to react one hydrogen is coming here and one hydrogen is coming over here so what we are going to get ch3ch3 plus hcl right students now one basic thing that you need to keep in your mind what are you obtaining from here you are obtaining an alkane molecule from here this alkane molecule is what it is methane while if i talk about this alkane molecule it is what it is ethane this is ethane this is methane right students so in this way you can clearly do these kinds of questions very easily now if i talk in alkyl halides only there is one exception there is one exception if i talk about the exception that is alkyl fluorides for the case of alkyl floralites direct combination won't occur why it won't occur because students alkyl fluorides are highly polar in nature why alkyl fluorides are highly polar because you know alkyl fluorides that means rf here we know that this r group contains basically the carbon right and fluorine is the most electronegative atom it is the most electronegative atom due to which there is highly polar bond between these due to which it is strong in nature and hence to break this bond because we are breaking this bond no so to break this one more amount of energy is required that's why there is no direct combination that occurs for the case of alkyl fluoride but it will occur for the case of alkyl chloride bromide and iodide clear students so this was from the direct combination method now comes the most important topic that is going to be the woods reaction it also lies in the category of alkyl halides but but students it is the most important you need to keep in your mind about woods reaction what is woods reaction if i talk about woods reaction so wood reactions basically occur in the presence of sodium and dry ether what is happening over here see for example i take again methyl chloride i am reacting this with sodium and i am taking two moles of this right so here i am writing this ch3cl twice and it occurs in the presence of dry ether and sodium also you have to take twice right so what is happening over here kindly listen very very carefully see this two and they are going to combine with the two cl which are present in both of these molecules right so first of all we are going to get two nacl other than this what we are going to get c when two nacl are combined two nacl are separated what is left over here one is ch3 and one is ch3 so they both will combine and will form what what it will form an ethane these ch3 will combine and will form ethane right so what is the basic thing that we observe over here that the reactant which we are using contains a single carbon but the product that we are obtaining contain twice the number of carbons as that of reactant so here we have taken one carbon yes we have taken two moles but we have taken one carbon but here if i talk about the structure we are getting two carbons so whenever you are using woods reaction you will see very different cases in one of the cases you are going to get the double of it in some of the cases you are not going to get the double of it there are many cases i am going to tell you one by one let us take one more case like this only first of all let us take this general reactions ch3 ch2 cl plus 2 na now take this twice right directly you can write this what you're going to get two nas are going to combine with 2 cl so yes 2 nacl is going to come out other than this what is left we are left with 2 ch3 ch2 now what is going to happen ch2 ch2 ch3 now what do you observe over here two carbons were taken but we are getting how many carbons four carbons we are getting butane from here now students if i ask you kindly prepare propane from this reactant only how will you obtain propane from this reaction so there are basically different cases as i already told you the very first case is the simple case that we have already seen right now comes the next case students that we need to know over here okay this was the very simplest case that we have seen over here now if i talk about the next case students we can see that there is a mechanism that we need to understand first of all the mechanism is free radical mechanism and if i talk about free radical mechanism it will help us to know that which of the following product is being obtained over here so first of all write mechanism here okay if i talk about mechanism it is free radical mechanism it is what it is a free radical mechanism now what is uh happening over here students first of all the sodium it splits into an a positive and an electron this is going to happen then what happens we know that we are taking alkyl halide for example we took methyl chloride over here so what is going to be happen over here that these contain two electrons one electron this will take another it will take so what did we are going to get we are going to get one r radical and one x radical now this x radical students combines with electron and forms x negative okay and this x negative further reacts with this n a positive to form nx like we form nacl similarly this is the basic thing that we need to keep in our mind now what is left over here this electron is being consumed this x radical is combining with electron this x negative is joining with n a positive the one thing that is not being used over here is this r radical now what we are going to observe from this that our radical now what happens this r radical either combines with itself and from rr bond t okay now this is basically what this is the combination this is the combination reaction this is what this is combination let us take one example to understand combination in detail if i talk about combination i am taking here ch3ch2 if i took ch3ch2x then ch3 ch2 radical will be obtained so i'm directly writing it now both of the times both of the times these radical will form a bond and will form ch3 ch2 ch2 ch3 which you have already seen butane is formed but now comes the next case which is disproportionation now in the case of disproportionation what happens students you will write these are same but in a different way how come see ch3ch2 radical plus kindly open one hydrogen from this ch3 and write this as hch2 ch2 radical now what will happen over here kindly see what will happen this bond will break and this one electron will react with this one electron of ch2 and this one electron will react with this ch2 radical right one electron so what we are going to see we are going to see this bond is going to break from here and we are going to get ch3 ch3 plus now see this hydrogen is reacting with this forming ch3ch3 but this here one radical one radical is making one bond hence here ch2 double bond ch2 is obtained that means an alkene can also be prepared from the woods reaction see we have seen in the case of woods reaction we can obtain alkanes also double alkanes that is butane we can obtain alkenes also in the case of disproportionation now for different alkyl halides which one is going to be valid if i talk about primary and secondary alkyl halides for one degree alkyl halide and two degree alkyl halide you know this is valid this one combination one if i talk about disproportionation one three degree alkyl halide is valid see all of these can make both of these structures like these can also make this these can also make this but most favorable are these one the most favorable for one degree alkyl halide and two degree alkyl halide is the combination one and most favorable for our disproportionation for three degree alkyl hair is disproportionation now students comes the next topic basically that is the types of voltage reaction okay if i talk about the types of wood reaction basically there are intramolecular and intermolecular now first of all i am talking about intermolecular in inter molecule we have already seen the case so i am directly writing the case for you that is simple if i take r r radical plus r radical it will form rr bond it is the basic that we have already seen but now what i do if i take another case if i take another case if i am taking 2 alkylides for example r1x plus r2x and i am taking again 2na what i am getting from here i am going to obtain 2 nx from here and i know that these two are going to combine like we did in the previous structure rnr is going to combine so here r1 r2 can combine moreover r1 can combine with cell itself also r1 can combine with itself also forming r1 r1 bond and r2 can combine with itself also forming r2 r2 bond so here in this what do we obtain we can see that basically three more structures are made with nx clear so this was our first kind of reaction that is intermolecular now comes the next one in which we are going to see two cases one is one two dihalide another one is one three dihalide now if i take the next case okay if i take the next case that is one two dihalide now for the case of one two dialyte what happens what is one two dialyte for example this is the carbons these are the halogen that are attached over first and second carbon right so this is one two dihalide now for example this is first carbon this is second so these are one two dihalide die refers to two halide refers to halogen groups now for this case students what is going to happen we are going to observe that this bond is going to break this one electron it is going to take with itself this is going to take electron with itself this is going to take electron with itself this is going to take electron with itself what we are observing over here that here here students these one one electrons are going to make a bond in between these and making a double bond over here like this these one one combine this also contain one radical this also contain one radical they both will combine with each other now what happens if i take one three radical one three sorry one three not one three radical one three dihalide for the case of one three dialite for example now on the first carbon and on the third carbon we are going to see the halogen groups like this right now what it is going to happen in the same way this one electron is moving here this one electron is moving here this one electron is moving here this one electron is moving here so what we are observing here there is a c radical over here then ch2 then c radical now this 1c radical will combine with this uh c radical forming forming what forming a ring like structure like this carbon carbon carbon forming hydrogen hydrogen hydrogen you can either make this you know in another way but you will form a ring like structure over here you can see that these two will combine and the first carbon is when it is attached with the last carbon it makes a ring like structure students right so these are the types of wood reaction now students we will proceed towards our next kind of reaction so students now comes over next topic that is decarboxylation if i talk about decarboxylation first of all we are taking here carboxylic acid that is ch3cooh now if we are going to name this ch3cooh what it is name it is ethanoic acid why because here carboxylic group that is coh is also included in the naming of parent chain right so during the naming this is included this is the first carbon this is the second carbon hence two carbon atoms so the word root is it there are single bond between the carbon atoms so n will come and we know that it is carboxylic acid so the suffix that we are going to use over here is oec acid and ethane ends with e and oak acid starts with o hence e will not come ethanoic acid will come right so here as a reactant we are taking ethanoic acid now it is going to react with naoh it is going to react with naoh students now what is going to happen over here water is going to come out from this how come water is going to come out see one from this hydrogen and if you see here if you see over here this oh group and this etch is going to come out that is h2o will come out what you are going to obtain over here is ch3 c double bond o o n a now students this is very important for you to understand this is sodium ethonate it is known as sodium ethonate okay what we have done is we have taken ethanoic acid reacted with any which water is being taken out and we obtain ch3coona where students you know that it further reacts with soda lime that is cao plus naoh it further reacts with soda lime cao plus naoh students now when it reacts with soda lime students we get here methane what is the product obtained here is methane that is ch4 okay other than the students we obtain na2co3 here na2co3 here right so what happens over here you can directly see what we have done we have taken ethanoic acid reacted with naoh water being removed we obtained sodium ethynate which further reacted with soda lime which is cao calcium oxide plus naoh and gave us the product methane right plus we obtained nh2co3 from this side so this was all about decarboxylation students now comes now comes our next reaction that is coal based electrolysis what do you mean by coal based electrolysis here we are going to take salt of carboxylic acid what we are going to take salt off carboxylic acid okay we are going to take the salt of carboxylic acid we in the presence of electrolysis now what do you mean by electrolysis in the uh chapter if you will see in your 12th class when you are going to study electrochemistry you will observe that basically electrolysis is the conversion of electrical energy into chemical to perform the chemical reactions right so what we are going to see over here is that when we take salt of carboxylic acid and electrolysis is done we obtain co2 and here the byproduct now if i take the carboxylic acid as ch3cooh this is going to be my reactant students i am going to take two moles of this okay now when it is going to react it is going to react with water okay it is going to react with water and electrolysis occurs over here and we obtain co2 okay we obtain co2 from this side and we obtain ch3ch3 that means ethane here ethane is produced here ethane is produced students so this is basically the coal based electrolysis process other than this we can also take instead of water we can take naoh and when we take naoh like decarboxylation the reaction occurs when we take ch3coh and react it with naoh what do we obtain we obtain that water is being removed from this side we obtain sodium ethylate that is ch3c double bond o o n a like we did in the decarboxylation process now the next step is similar to this what we are going to see we are going to see that two water molecules are taken out and hence we obtain naoh plus co2 students and main product that is ethane which is being obtained here okay so these are the two reactions that you need to keep in your mind while doing coal based electrolysis students hopefully it is clear to you both of these reactions it is quite simple i'm going to repeat it again for you in the very first reaction we took water and electrolysis happened and we obtained a thin molecule plus co2 in the next reaction we took naoh and the reaction occur and we got sodium ethanol just like the previous case and we obtained naoh plus co2 plus ethane right other than this students what we need to keep in our mind we need to keep in our mind it's it's mechanism what is the mechanism for this if i talk about mechanism students first of all see in some of the cases what you're going to see in some of the cases you are going to get rcok there there may be some of the cases in that also you need to first make just like structure just like ethernet you're going to you know free radical mechanism and you will separate our radical from that and that our radical is going to combine and form the product for example if i talk about this case also here what do we see here when i'm going to explain you the mechanism see here co2 will separate out and this ch3 will you know combine with its own radical ch3 and will form ethane so in the when you are going to study the mechanism you are going to get the the reaction in depth so in the previous cases i told you about general cases first then i told you about mechanism so this was the way of doing a reaction now in this case also i've given you general but these general might be sound confusing for you but don't worry when you will get to know the mechanism it will be very easy for you to understand for mechanism here we obtain that there are two cases one is anode one another is cathode one so sorry if i see here c h three o we have this with us right sodium ethonate now here oxygen has negative while sodium has a positive sign on it what happens over here this sodium gets separated when it gets separated it becomes n a positive and what do we obtain over here we obtain ch3 c double bond o o negative and yes n a positive is separated from this side now students what happens as you know this is an anion so an ion will move towards the anode because it is electrolysis so here an odd and cathode both will be considered now if i talk about anode at at an odd as i told you which will go and iron will go because from a a and odd a nine you can start in this way now this c field take this is what this is your anion molecule now what happens over here kindly take out two electrons from here if you'll take out two electrons from here i have taken two moles of this right what do you observe here there is a bond breakage here this bond will break over here now this uh you know electron will move here and you know we have taken two moles why too much because ch3 radical will combine with ch3 radical right and this will be separated out here this will be separated out now for the next case what you are going to do what you are going to observe over here you are going to observe over here that this bond is separated and this bond is separated right now next students what next you are going to get ch3 single bond ch3 hence it is clear to you so this was what at enod we obtained ethane as well as co2 is being produced here co2 is being produced as well as what ethane is being uh produced how it in this ch3 radical is obtained and this ch3 will combine with ch3 forming ethane molecule now this was at an odd now what will occur at k thought the question arises what will occur what will occur at cathode at an odd kindly remember you have taken the anion it will going to react at cathode you had water with you okay so what will happen the two water molecules that you took will lose electrons and split into h positive and oh right this thing you clearly know but here students this is the basic reaction that we know but i am telling you it will lose two electrons leaving when it will lose two electrons we will get over here two h radicals and o h negative okay when you are taking out two electrons from here you are getting 2h radical i can write in the radical form why i have written the radical form for you to understand it very carefully now this h radical you can directly also write it as like this h radical combines with this radical forming h2 either you can directly write it as like this or you can write it at uh write it as in the form of radical now the next stem is cup that two o h negative will combine with two n a positive forming two any which now if i talk about the important points in this reaction what you have seen kindly see here you have seen at anode we obtain ethane and co2 that means at an odd we are obtaining what alkane and co2 so the most important point is alkane and co2 are obtained at anode while while naoh is obtained at any which also also h2 is obtained at cathode these were the two points that you need to keep in your mind while doing this reaction other than the students here what you observe that you can never obtain methane you can never obtain why see here see here why you can never obtain methane because at least here if you have one ch3 ch3 will combine with ch3 forming a two carbon compound that means you can never form methane you will always form the simplest cell gain that can be formed here is it not methane so methane can never be formed methane can never be formed ethane is the simplest alkane obtained ethane will be your sink simplest alkane that can be obtained right students now you need to keep in your mind at an odd oxidation occurs while that cathode reduction occurs now this was all about cold waste electrolysis students now we are going to see our next reaction so students now comes our next reaction next method of preparation that is going to be franklin reaction in franklin reaction we are going to take two alkyl halides and we are going to react it with zn see i have written this opposite it is the same thing now what happens over here this combines that means z and x 2 is liberated over here and the other product that we get over here is the bond between r and r right so it is also method of preparation for alkanes now let us see one of the cases for it to understand more clearly if i take ch3 ch2cl reacted with ch3 ch2cl and with zn so if i ask you what is the product first of all you will rearrange them you will write ch3 ch2cl plus zn plus cl ch2ch3 we will see that here zn's cl2 will be liberated out z and cl2 will be liberated out leaving ch3 ch2 ch2 also as a product c now this ch3 ch2 will combine with this ch2 ch3 forming ch3 ch2 ch2 ch3 which is an alkane so this is also our method of preparation now students comes our next method of preparation that is corey how synthesis if i talk about corey how synthesis here also we are going to take alkyl halide in three steps it is going to occur that is li then the next step is cucl the next step is rx if i see over here what is going to happen in a direct method whenever you are going to get in the question paper you are not going to do with the mechanism step you are going to do it directly so what you need to keep in your mind is that just connect both of these if it is r dash then it will be r r dash if it is same r then it is connected how come c r radical will combine with our radical forming our bond r that is alkane it is a direct method no need to see what is the byproducts obtained we just want to obtain alkane that's it right just combine both of these if they are different then also combine both of these there is no worry now comes the mechanism portion students if i talk about mechanism first of all this rx will react with li forming rli this is the very first step now comes the second step that is with the cucl forming gilman's reagent students which is r2 cucl see you sorry li because here is li here is cu it will form r2 cu li which is gilman's reagent which on further reaction students with which on further reaction with rx is going to form two kinds of product one is ionic another one is covalent ionic one is going to be rli that means here r is negative l i is positive plus r cu this is our covalent while this is the ionic product that is being obtained clear now in the very next step what is going to happen students this r will combine with this r directly or you can see here kindly forget this kindly forget the students what you're going to observe over here is that you are going to see that here x negative was liberated first of all see this x negative was liberated that x negative will combine with this or i would say it will combine with this leaving l i x or either you can say here we have taken rx it will combine with this it is the same thing we took here alkyl halide again we took here alkyl halide so this alkyl halide this halogen will react with the positive portion and this r will be reacting with this because here see what happens this is bond is broken this bond is broken right so here this r is going to combine with this r forming r r bond forming rr bond other than this what is left over here this covalent bond rcu so these are the products that are obtained over here that is lix rr and another one is rcu now students here the attack of this this one carbon 9 the attack of this carbon 9. this step is basically the rate determining step that you need to keep in your mind and this is basically useful for preparing unsymmetrical you know unsymmetrical alkanes clear now students how for unsymmetrical because here r dash can be different and here r can be different so you can obtain for unsymmetrical also now students this is what this was our methods of preparation of alkanes which we have covered now i am going to move forward towards what towards physical properties and chemical properties once i want you all to revise it because these are the reactions that you can't forget okay now see these were these were all about alkanes that was the general thing here what you need to keep in mind i'm just putting a star mark over here on all those topics that are important why i'm revising it because you know first of all we will cover alkanes dell and kales then alkenes and then alkynes so what we are going to observe first we will do method of preparation physical property chemical properties so i don't want in one short video lecture for you to get mixed up that's why i'm re-revising it see firstly we're going we are going to do is this is the molecular formula that we need to remember other than this they are known as paraffins right because uh the ability for the bond formation they have the little ability for the bond formation little para is too little affinity refers to the tendency to make the bonds other than this the structure right here carbon is sp3 hybridized and it is tetrahedral while they are inert at the room temperature and the normal conditions this was about alkanes moving to the methods of preparation from where the important segment starts here you need to keep in your mind whenever you are doing hydrogenation my dear students you have to break the pi bonds and you have to add the hydrogen for the alkenes only you require one h2 molecule it will occur in the presence of water nickel palladium and platinum clear and for the alkynes case first it will convert into alkenes and then into alkanes portion clear it will also occur in the presence of water nickel platinum and palladium now its mechanism portion i have told you for the mechanism portion you require a catalyst and this is a surface phenomena this is a surface phenomena adsorption is occurring over here don't get confused between absorption and adsorption ads option refers to the surface phenomena the interaction with the surface molecules while absorption refers to the interaction with the bulk one here the interaction of h2 molecule with the catalyst is occurring on the surface only right so do not forget this point also here adsorption is occurring what happens over here their bond is being broken between hydrogen hydrogen and the carbon is going to attack here so here bond breakage and bond formation occurs which is known as transition state which is known as transition state and at last we are obtaining what alkanes clear okay now comes the important key points transition status formed no intermediates are formed these two are very very important other than this rate of reaction of alkynes is more than alkenes why because alkynes are linear in shape now you can revise with me good okay here rate of reaction is inversely proportional to what steric hindrance clear this was the question which i've already told you okay other than this we have the method of preparation for the alkyl halide now for the hairlight this is the general case and i know you people can do this very easily so i won't mark this as most important because it is not most important is it it is okay this topic is very easy because here one hydrogen is going to attack on the alkyl group and another on the halide group right so we are going to obtain the product now comes the woods reaction the most important one we have already given the star marks over here so i would give another here i would give this no we i would give a flag over here yes because it is important wood's reaction is important inwards reaction do remember you have to take dry ether do remember you have to take dry ether right now students here you are taking two moles two moles of the reactant plus two moles of ni so that two nacl can be obtained and this ch3 ch3 will combine forming ethane if we take here two n a two c uh two cl will combine with two na forming nacl and we will be left with ch3ch2 they both will combine with each other leading to the formation of alkane that is the butane now here one more question arised why do we obtain a product that contains twice the number of carbons can we obtain another one also yes so there is one more example for you this one if i talk about the combination reaction it is quite simple which we have seen and it occurs for the primary alkyl halide in secondary halide but if i talk about disproportionation one here what happens here alkene is formed this is very important here alkene is formed in this case clear okay now coming to the types of wood reaction this is also important where you are taking different kinds of alkyl halides for example here r one x is taken here r two x is taken so different alkyl halides are taken so first cases they both will combine with each other second it will combine with itself it will combine with itself so here three products are obtained rather other than nx right now students come one two dial iodine one three dialite in one two dihalide you obtain alkene iron one three you obtain cyclo cycloalkane right okay now comes decarboxylation for decarboxylation it is quite easy for you to understand that you have to do in the in the presence of any which water will be liberated when water will be vibrated sodium ethernet will be formed and with the help of sodium lime the product that is obtained is methane and na2co3 okay now students this is quite important coal based electrolysis it is quite important here what happens here is we are going to obtain we are going to obtain this anion now what happens here two electrons are liberated this bond is broken up and here this ch3 will combine with ch3 forming a thane molecule and co2 is liberated what we need to keep in our mind at an odd alkanes are produced and co2 are produced while while add cathode h2 and nuh is produced other than this you have to uh just observe the mechanism for anode and cathode sake right okay students now seeing the franklin reduction it is also easy i won't mark it as an important because it is quite easy whenever you are going to see in the reaction you just have to take z and cl2 out is that an x2 out that's it and the rest over is combined with each other so it is quite easy okay now comes the corey house synthesis for corey house synthesis i would say there is a three-step reaction that we have already seen the best trick to remember is kindly attach both of these because you there in examination there is no need for you to do the mechanism work right so students this was all all about our methods of preparation now we will move forward towards the physical properties of alkanes so students now looking towards the physical properties the very first physical property if i talk about alkanes is that alkanes basically are non-polar in nature they have very less polarity so the very first physical property is that alkanes are non-polar in nature in bracket you can also write very very less polarity why is it so students because carbon and hydrogen they almost have same electronegativity that's why if i talk about this case right ethane or any case you can take off any alkenes i am saying that the carbon and hydrogen the bond that they make both of them have almost same electronegativity so there is no electronegativity difference see whenever we see the electronegativity difference their arises poles that means polarity difference arises but if there is no electronegativity difference then it is referred to as non-polar so hence it is non-polar in nature students next is that we are going to see the nature for example if i am saying at room temperature you know carbon first to carbon fourth are gas at room temperature then carbon fifth to carbon 17 are liquid at room temperature you should remember this that carbon number first i'm talking about alkanes from carbon 1 to carbon fourth okay sorry okay carbon one two carbon four that means from methane to butane all of these are what all of these are gas at room temperature and if i talk about carbon 5 to carbon 17 students they are liquid at room temperature you know these are the physical states if we talk about and from you know carbon 18 they are waxy solid like structure like for example you know vaseline they have actually solid like structure so i would say from carbon 18 to onwards they are waxy solid again at room temperature so the very first property was quite easy the second one also was quite easy now if i talk about being non-polar in nature in which of the following they will be soluble or insoluble if they are nonpolar nonpolar in nature then they are easily soluble in non-polar solvents okay so the third point is due to their non-polar nature they are easily soluble in non-polar solvents non-polar solvents clear students now students moving on to the next property that is the boiling point if i talk about boiling point boiling point is directly proportional to the molecular mass it is directly proportional to the molecular mass students as well as if i talk about boiling point it is inversely proportional to branching also so whenever you observe branching their boiling point comes out to be low clear now if i talk about melting point case if i talk about melting point case then melting point is also directly proportional to molecular mass hence if number of carbon increases the molecular sorry the melting point also increases with the students but alkanes with even number of carbon atoms have more melting point okay do remember that but alkanes with even number of carbon have more melting point more melting point so these are the points that you need to remember during the physical properties of alkanes i would repeat it and the important one are these ones these both why because uh the question arises in your examination regarding the branching portion okay so you should remember where there is branching there is less boiling point okay do remember that and for the more melting point also do remember it is directly proportional to the number of carbons that means molecular mass as well as the even number of carbons will have more melting point clear students now we will see the chemical properties of alkanes so students looking forward towards the chemical properties chemical properties of alkanes now it is also an important topic students this is also an important topic okay now if i talk about chemical properties the very first chemical property that i am going to tell you is a substitution reaction if i talk in chemical properties number first is substitution reaction now what do you mean by substitution reaction see we have already covered this in the goc section at the very end when i told you about the types of reaction basically substitution the reactions are those reaction in which one of the reactant is being substituted by another group you know it takes place of another group that means is to substitute to take place of another that is substitution reaction so alkanes basically is either replaced by halogen nitro or sulfonic group okay so you need to remember alkane is replaced by either halogen either halogen nitro group or sulfur group okay so whenever it is replaced by halogen it is known as halogenation do remember now here one thing more here alkane hydrogen will come because because see here because see here if i talk about ch4 now how to substitute from ch4 1 hydrogen is being substituted by another group you know from here i can also write it as ch3h so there will be removal of this hydrogen and something will come in place of it so alkanes hydrogen is replaced by either halogen nitro group or sulfur group kindly corrected okay now when it is replaced by halogens halogens are your what chlorine chlorine bromine iodine by them okay when they are replaced by halogens it is called as halogenation it is known as halogenation students and when i talk about nitration hno3 is used and when i talk about sulfonation h2so4 is used okay now one by one we are going to see so the very first that we need to understand is hello generation the very first we will understand is halogenation now you have got that hydrogen will be substituted by halogen that is halogenation that is quite easy for you to understand students other than this it always occurs in the presence of sunlight or at higher temperature do remember that always occurs in the presence in the presence of sunlight and at higher temperatures or high temperature clear now students in halogenation we see various cases like you know bromination iodination chlorination we see a lot of cases so if i talk about the order order of reactivity the three degree alkanes order will be more than the 2 degree and then the 1 degree kindly remember this in your mind keep this in your mind now i'm going to do chlorination process in the chlorination process let us take one example if i take one example that is of ch4 and it is reacted with cl2 the right halogenation means the addition of halogen and the replacement of hydrogen so from ch4 one hydrogen will be replaced by and will be substituted by what by halogen by which halogen chlorine so what we are getting over here we will get ch3cl plus here we will get what hcl right see here this hydrogen that is being removed from ch4 is now attached with this cl now once cl is left that will be uh reacting with the ch3 right now it occurs in the presence of sunlight it occurs in the presence of sunlight now in the next step now see when we are taking excess of chlorine then it keeps on reacting again and again now here ch3cl is produced it will again react with cl2 in the presence of sunlight leading to the formation of ch2cl2 again one hydrogen will be removed and one cl will substitute and it takes this plates so here hcl will be taken out now in the very next step ch2cl2 will react with cl2 leading to the formation of chcl3 plus hcl this ch chcl3 is known as chloroform now this chcl3 students will react with cl2 leading to the formation of ccl4 ccl4 clear so basically this is what this is the chlorination process here we have t taken cl2 in excess students right now students i2 reacts slowly this is chlorination if i talk about iodination iodination reacts slowly therefore it must be used with the reagent known as hio3 right so during halogenation this occurs so this was about chlorination if i talk about iodination you shall write it because in the case of iodination we are using hio3 for this reaction to takes place do remember this iodination will not occur if it is done just like chlorination because it is a slow process because it is a slow process so we use a reagent hio3 and if i show you the reaction it is going to be ch4 plus i2 basically what happens over here when ch4 plus i2 is done it is done in the presence of 770 kelvins sorry 773 kelvins my dear students and here what do we observe according to chlorination process one hydrogen was taken out and iodine would come here so here ch3i would be formed plus hi but yes it doesn't occur so why because it is a reversible process when ch3 in hiv produced it will revert back to the reactants so we are using hio3 so that reversible reaction doesn't takes place so hi03 is used so to stop this reversible process right now how does it react hi03 students will react with this hi now hi03 will react with hi when this will react with hi it will lead to the formation of iodine students iodine will be formed plus water will be liberated out so hence now it is not in reversible reaction the reaction stops here so basically iodination is not possible because it is a reversible reaction to make it possible we require hio3 but when we make hio3 the product formed is not an alkane molecule it is i2 it is not an alkane molecule i2 is formed now comes students about nitration as we have seen in halogenation halogen was substituted at the place of hydrogen and it took its place right now in the case of nitration we are going to use h3 now what happens over here we are going to take our alkane molecule we are going to react with it at no3 here there is a coordinate bond between nitrogen and oxygen now what happens over here students kindly see this very carefully see here a bond breakage occurred this bond is broken up so here we get o h radical similarly here the bond breakage occur we get h radical now o h radical and h radical combines to form what water and what is left over here is our radical and this one so what we get over here is r o2 what we are getting over here is our no2 is obtained so this is very simple process for nitrogen it is just like that okay now in the examination only r group will be changed like ch3 c2h uh ch3ch2 okay so you just have to do the direct reaction that's it okay kindly take the water out and rest combined both of them now comes the next case similar to this is sulphur nation students now if i talk about sulfonation only here also we are going to take our alkane molecule students as i told you we are going to take h2so4 right now okay so here okay students so here h2 is a force h2so4 right this occurs now again similar to nitrogen here also bond cleavage will occur and so this bond will be broken r radical h radical now what happens over here h radical will combine with whom o h radical right so here this bond cleavage occurs what we are getting over here is water because oh h radical plus h radical forms water now this r radical is going to combine with this sulfur forming r s double bond o double bond o o h clear students this is r rso3h this is the process for sulfonation this is the easiest step i i would say for nitration and sulfuric so right students now come students the next chemical property that is combustion that is combustion if i talk about combustion what do you refer to as combustion when there is a reaction in the presence of oxygen that is referred to as combustion combustion so here see ch4 plus o2 gives co2 plus water so during combustion we obtain co2 plus water it occurs in the presence of oxygen do write this occurs in presence of oxygen that is o2 now this was the reaction for the methane case you can kindly balance it out accordingly and here heat is also released students now what happens over here you can also write heat released if i talk about the general reaction if i take general reaction of alkane what it would be cn h2n plus 2 okay this is going to be my alkane molecule where i have to put the value of ns123 according to the carbon atoms required then here i have to use 3 n plus 1 by 2 in front of o2 to balance it out students now what comes it comes out to be n co2 plus n plus 1 water this is the general reaction just you need to write the number of atoms like for n values right now combustion is basically when used in limited supply what is the reaction when combustion is used in limited supply when it is used in limited supply what do we observe see when methane is reacting with o2 students now co2 is not formed here carbon is formed here carbon is formed plus water is released now this is basically used in the inks this carbon is basically used in things so this is the reaction when combustion is used in limited supply here you can also write limited over here so this was all about combustion chemical reaction now moving forward is our next reaction that is the reaction with the steam it is quite easy see when ch4 reacts with the steam now this is steam form this is what this is your steam form well alkane molecule reacts with steam it forms carbon monoxide co is formed and h2 is released and h2 is released now this is what this is syn gas this is syn gas kindly put a star mark over here because it is going to be the most important reaction it is going to be very most important reaction right okay students now proceeding towards our next reaction that is aromatization now what is aromatization the formation of aromatic ring from your open chain you need to form a closed chain that means an aromatic chain and for this you at least require carbon atoms to be at least six n values should be at least six or more than six not below six because then aromatic ring will not be formed so if i take the case of n hexane if i take the case of n hexane that is ch3 ch2 ch2 ch2 ch2 ch3 let us count the number of atoms one two three four five six this is n hexane my dear students this is what n hexane it is now it occurs in the presence of various reagents you know one of the reagents can be v2o5 cr2o3 or al2o3 you can take two in the pairs okay take either of these two in pairs and it occurs basically in the presence of again 773 kelvin near about and you obtain what an aromatic ring you obtain an aromatic ring students benzene is formed over here benzene is formed over here clear okay now come students our next reaction this next reaction is basically isomerization isomerization so you know during uh the start of organic chemistry we did iupac then we did iso um basically isomerization i have already explained you what are isomer students isomers are those organic compounds which have same molecular formula but at least one different property now what we are going to do over here is that we are going to take any of the alkanes and we are going to obtain different alkanes but we are going to obtain isomers of that how let us take off hexane only ch3ch2 ch2 ch2 ch2ch now i guess all of the students would be able to answer this question and in a only few seconds just the thing they need to remember over here is the reagent that you need to hear uses anhydrous alcl3 and hcl students so if i take anhydrous okay anhydrous alcl3 if i'm going to take anhydrous alcl3 and hcl over here and hcl over here so what i'm going to observe is i'm going to make its isomers so what will it it's iso mercy i can make this of 5 membered ch3 ch2 ch2 ch2 here ch will come then here's ch3 is it an isomer of this yes for sure one more isomer that we can obtain over here is ch3 ch2 c8 ch3 ch2 ch3 we can also obtain this isomer students so this is what this is isomerization case clear it is very easy students now comes our next reaction so students as we were doing pyrolysis and cracking let us take one example to understand it a bit more if we take ch3 ch2 ch3 and now we are going to do basically two steps over here one is going to be dehydrogenation and another one that is going to be is cracking when i'm going to take the very first step what i'm going to obtain over here is that i have to break the bond between carbon and hydrogen so what i'm getting over here is ch3 single bond ch double bond ch2 right plus h2 because i am going to break the bond between carbon and hydrogen i'm going to break between these carbon and hydrogen between this carbon and hydrogen and another one is cracking in cracking what we do we break the bond between carbon and hydrogen and also in between what carbon and carbon so here what i am going to get i am going to get ch2 double bond ch2 students plus what i am going to get i am going to get ch4 now students it was all about pyrolysis and cracking now we will move forward towards our next segment that is alkenes if i talk about alkenes alkenes are basically the unsaturated the unsaturated hydrocarbons right what do we observe over here we observe that carbon carbon contain a double bond over here in alkenes and other than this students alkenes are basically referred as orphans why they are referred to as all fin students because the very first member of the alkene family you know um it the very first member on the reaction with the chlorine produced and oily substance the very first member of this family okay now students if i talk about the general formula it is cnh2n students the general formula for alkenes is cnh2n in alkanes what do we observed cn h2n plus 2 here what are we getting cnh2n clear now students we are going to see the methods of preparation of alkene and if i talk about the very first method it is going to be hydrogenation what is the very first method the very first method is hydrogenation now what happens over hydrogenation we are going to use a reagent over here now what are these reagent students what are these reagents we are going to use here lindlard's catalyst what we are going to use over here lindlards reagent or catalyst you can say now what happens over here how it is made what is the combination we are going to take a poison catalyst over here which is going to do what now there are basically various kind of catalyst one which increases the rate of reaction another which decreases the rate of reaction what are the reagents first of all let me tell you the reagents are h2 okay plus this we are going to get over here paladon paladinized carbon now this basically paladinoized carbon is poisoned by sulfur or cunylon it is it is poisoned by sulfur or cuneinone why so why um this happens students why does this happen it happens because whenever we are doing hydrogenation reaction what we are going to observe we are going to get our product alkene and we have already seen in the reactions what we have seen that basically if we have this alkyne with us then in the presence of nickel platinum palladium what do we do we take hydrogen atoms and we do the attack of hydrogen atom over the carbon atoms right so what do we get over here we get an alkene molecule then further this reaction proceeds and again one hydrogen is attacking the another one and we are getting what we are getting our alkanes we are getting what we are getting our alkanes so students we know that we want to produce alkene but we are getting what we are getting our alkanes to stop this reaction at this point we are using the poison catalyst which going which is going to stop this reaction over here and when it is poisoned by sulphur or cumulon it is basically the large reagent it is basically that now students okay as i have told you we need to stop it at what alkene so for that we have taken this catalyst now in the given reactions what you're going to get this is your alkyne this is your alkyne and h2 is going to react over here now you have taken this to poison it you have taken sulphur as well as with you or cunelon now what you are getting here you will get over here the cis product you will obtain the cis product over here do remember this do remember this you are going to get this product i am also taking one of the example if you have this as your alkyne molecule okay now now this will not proceed further because we have poisoned use the poison catalyst now this is with you or other than this you can also see baso4 is also present here now you are going to get what is this product in this product see you are going to get an alkene you know that you know that now both the hydrogens will be present on the same side to obtain to obtain what a cis product now what is happening over here this h2 this h2 students this hydrogen will attack over here another hydrogen will attack over here and this bond will break this is the basic process that we did during the formation of alkanes it is the same process that we are doing over here what is the same process we need to break the double bond and when then we need to attack the hydrogen over there but during the formation of product do remember we are going to get us this product and in this product both the hydrogens are going to be in the same direction students this point is the most important point and what have we used over here we have used over here h2 and pd carbon paladinized carbon and it is poisoned by sulphur cumulon why because we need to stop this reaction over here otherwise this h2 is going to attack again and will produce an alkane molecule but we didn't want that now students this was the formation of cysts now to obtain a trans uh you know product which of the following reaction we are going to do we are going to do bridge reduction in bridge reduction we are going to get the trans product we are going to get the trans product again we are having an alkene molecule let us say we are having r dash c triple 1 c r over here again this reaction is happening now this reaction should happen the presence of sodium and nh3 do remember this to obtain a trans trans product you need this reaction to occur in the presence of sodium and nh3 right now what is going to be happening over here what is going to happen over here you know that h2 is going to attack over here from this side or from this side and this double bond is going to break now to obtain the trans project uh product you are going to see that r dash is present over here r is present over here hydrogen is present over here and another hydrogen is present over here clear so this is what this is our trans product that we obtained clear for this product it was very simple you used pd carbon and it was the large reagent and for you know the transfer product you did birch reduction now comes the next method of preparation students for next method of preparation that is dehydration of alcohol if i talk about a dehydration of alcohol if i talk about dehydration of alcohol what we are going to do we are going to take an alcohol that is roh and we know that we have to obtain our product that is alkene it will be done in the presence of concentrated h2so4 h3po4 or h positive right and we are going to obtain our alkene from the side we are going to get our alkene now what happens over here students what happens over here see this roh basically when it reacts there is an electrophile that is h-positive that is working from h2so4 or h3po4 then this oh basically has lone pair of electrons which attack over the h positive and what do we get over here we get over here roh roh positive we are getting this substance now now what happens over here here water molecule liberates out here this water molecule liberates out and we get our carbocation that is r positive now this carbocation students this carbocation can either do rearrangement it can either work on rearrangement and if it's not doing any rearrangement then directly students hydrogen will come out directly h-positive will come out and will lead to the production of alkene so the main part over here is the rearrangement in some of the cases you are going to see rearrangement is occurring in some of the cases you are going to get where rearrangement is not occurring so what i am going to tell over here is that dehydration of alcohol contains what we are going to dehydration refers to what see we are using concentrated h2so4 and h3po4 that's it and with the help of alcohol we are obtaining alkene so now let us see the rearrangement one task if i take hydride shift it is an example of rearrangement if i talk about hydride shift students let us take one example where i am going to tell you hydride shift refers to as the h positive shift now what is going to happen over here i told you that r positive is being generated from the roh right we have seen this case now let us take this example ch3c chch3 here students we want r positive so let me say that there is no ch3 over here let us say there is a hydrogen over here and there is a positive sign over here we have this we have this with us now what happens you know that this carbon has basically four valency that means it has a hydrogen with it and you know that this is a one degree carbocation but you know that three degrees more stable than two degree than one degree so what will happen over here this hydrogen will take its electron and moves towards the ch2 positive and will break this bond between carbon and hydrogen leading to the formation of ch3 c positive then comes h over here and then comes ch3 over here by ch3 because this hydrogen is attached to the ch2 forming ch3 and this bond is broken over here electrons are moved from the carbon away from the carbon so carbon doesn't have any electrons so hence it is positive now this is what this is the rearranged product and the next step is what next step is just you have to remove one hydrogen if you remove this hydrogen there will be a double bond between this and alkene will be formed clear okay now comes from vicinal dihalide what do you mean by vicinal dihalide dihalide refers to two halogen groups and vicinal dihalide refers to like this when the adjacent carbon atoms have the halide groups with it that is the vicinal dihalide it this reaction is now going to occur in the presence of zed in students in the presence of zinc and with 300 degrees celsius you know with 300 degree celsius what you are going to observe over here is that now this zone will react with this x2 and these two bonds will break over and there will be a double bond between these two leading to the formation of alkene molecule plus zn x2 is liberated over here now to complete the valency of carbon other hydrogens are attached with it so it is very easily see this is the vicinal dialite if i talk about geminal dihalide what do you mean by germinal dihalide see if i talk about geminal dihalide then we will see that only a single carbon atom contains two halogen groups that is geminal dihalide but in vicinal dihalide we see that the adjacent carbon atoms that the adjacent one basically contains adjacent one contain two halogens if halogens are present on the single carbon then they are referred to as geminal but if it is present on you know the single carbon atom here s will not come then it is known as vicinal dihalide clear it is clear here you can see brbr also then z and br2 will be released this is the formation of alkene now comes next method of preparation which is the most important that is dehydrohalogenation dehydrohalogenation what do you observe over here it is basically beta elimination here you will see that halogen will come out from the reactant how it will come out see if i see this is my reactant if i see this is my reactant and halogen is attached over this carbon atom so this carbon atom is alpha carbon and the carbon next to the carbon is known as beta carbon if it is beta carbon then it is referred to what it is referred to beta hydrogen so here what you are going to see beta elimination that means beta hydrogen will be eliminated also what will be eliminated alpha alpha halogen alpha halogen so two things that you are going to observe over here are going to eliminate that is beta hydrogen and alpha halogen now let me tell you students what is happening over here first of all identify alpha carbon how you will identify the halogen that is attached to the carbon is alpha carbon and the carbon next to the alpha carbon is beta carbon the carbon uh which contains hydrogen with it the beta carbon that will be refers to as beta hydrogen now when this will be removed and this will be removed there will be a double bond between both of these there will be a double bond and we are going to get this alkene as a product we are going to get this alkene as our product clear students and this now hydrogen will react with this halogen and will be forming hx over here okay this is the basic reaction now the rate of reactivity is the order this is the order for rate of reactivity what that you know the rate for here our if i talk about uh alkyl fluoride it is very less why it is very less because you know fluorine is most electronegative it has electronegativity so high that this bond is so strong that it is unable to break that's why here we need to break this bond and whose bond will be broken who have less electronegativity so that it can be easily broken so ri has basically more rate of reaction why more rate of reactivity why because this bond can easily be broken while uh rf bro rf bond takes a lot of time to break because it is a strong bond because there is a lot of electronegativity due to fluorine atom now students here no intermediate formation occurs over here other than the students only transition state will be formed so do remember this it is the most important reaction that will come in your examination if you see examination point of view it is beta elimination reaction clear students okay if i talk about the properties of alkene now comes the properties of elkin we have seen what we have seen the preparation if i talk about the properties the very first property is that first three members are gaseous in nature then the next 14 members basically are liquid and higher ones are solid just like in the alkanes case right other than this they are slightly polar than alkanes why they are slightly polar than alkane because here you see there is a double bond between the carbon atoms and if i see the hybridization over here it is sp2 hybridized this carbon is also sp2 hybridized due to the sp2 hybridization we know that electronegativity directly depends upon the s character more the s character more is the electronegativity so here what we are observing that these are if there is any difference in the electronegativity poles arises which lead to the formation of polarity so what i'm telling over here that due to in alkanes what we observed there was carbon and hydrogen they almost have electronegativity same so they were non-polar in nature but if i talk about alkenes due to the increase in s character due to the increase in s character electronegativity got increased and due to the increase in electronegativity poles arised which lead to the formation of polarity so it is more polar than the alkanes now if i talk about um you know the color and the adore of them the ethereal is colorless but it have a faint sweet smell but rest of the you know molecules are colorless and orderless if i talk about boiling point they have low boiling point but if the number of carbon atoms increases the boiling point will increase so yes due to the non-polar nature there are insoluble in water now students comes the first chemical property that is addition of hydrogen this you have already done chemical property chemical property is addition of hydrogen now if i talk about addition of hydrogen let us take this case and we are going to add hydrogen in the presence of nickel platinum or palladium we have also done in the previous case what we are going to do we are going to break this bond and one hydrogen will attack over here and one will attack over here so this you will get as your byproduct clear students other than this let us take one more example if i take this example over here now again what you are going to do over here this is occurring in the presence of nickel platinum and palladium here also you are going to break this double bond and when this double bond is going to break one hydrogen is going to attack over here and another here so you are going to get this as your product you are going to get this as your product so this is basically what this is basically chemical property of alkene in which from alkene we can obtain what we can obtain the alkane molecules if you remember in the alkanes we have started the preparation of alkanes from the alkynes you know from the alkynes then we have also seen the preparation of alkane from the alkene molecule if you remember there that i told you why it is happening in the presence of nickel platinum or palladium because here nickel platinum prelidium is basically a catalyst and here there is a adsorption process which process occurs over here adsorption occurs and what happened in adsorption students in adsorption we saw that it is a surface phenomena when it is a surface phenomena the h2 molecule comes towards the surface of the nickel now i am not making the mechanism again and again because we have already done in the alkene case right so when the h2 molecule comes over the surface of the nickel then it does the interaction with the alkene molecule and the bond gets break and we obtain the alkane as a product and the catalyst comes out right so this was the first chemical property that is the addition of hydrogen students if i talk about the next chemical property it is addition of halogen it is addition of halogen addition of halogen now if i talk about addition of halogen let us take this case r c h double bond ch2 and we are going to add any of the halogen for example i am taking x as our halogen and this x can be fluorine chlorine bromine iodine and it occurs in the presence of what ccl4 so what we are getting over here we are getting a product that this double bond will break over and yes one x is going to attack at this carbon and one x is going to attack at this carbon so this leads to the formation of this right this leads to this formation now it is also referred to as a test for unsaturation why we can also calculate the test for unsaturation if you will if you are going to use br2 over here instead of x2 x2 can be fluorine chlorine bromine iodine so if you are going to take it is our ch double bond ch2 plus br2 in the presence of ccl4 so what you are getting over here the double bond is going to break over and yes you are going to get this as a product right now students what happened this br2 basically is reddish brown in color but when we are getting this uh as a product this product is basically what it is an alkane now if it is an alkane the color if the color becomes colorless first it was reddish brown when we add br2 in it but when the color become colorless then this is a test for unsaturation that the reactant which we used to check it is unsaturated or not it is unsaturated because the color has changed so from reddish brown color students from reddish brown color what do we got we obtained a colorless substance we obtain a colorless substance so this is a test for and saturation that will be coming in your examination it is asked many times right students now other than this chemical property is the addition of a hydrogen halide now what do you mean by hydrogen halide edition of now see if you have done alkanes know then alkenes and alkynes are very simple for you now you will feel that these reactions are very much easy if i talk about hydrogen halide it is hx it is what it is hx hf hcl hbr hi and if i talk about their order basically what happens again hf then we have h cl then we have hbr then we have hi this is the order of reactivity why again because here fluorine is most electronegative due to the more electronegativity of fluorine atom this bond is very strong and it will not break because it has more electronegativity and we want the addition of hydrogen allied so we will break the bond if i talk about the symmetrical alkene if i talk about this symmetrical alkene and i'm going to do a reaction of hbr over here so what i want i want that this bond is going to break and also this double bond is going to break so to break this bond either we are going to take hcl hb or hi but we are not going to use hf by why because it is you know um very strong bond right now what we are getting over here if it is symmetrical then uh you can either attach hydrogen over here or over here it is the same thing so i am attacking hydrogen over here and br over here what i am obtaining over here i am obtaining ch and yes there are two more bonds over here and single bond over here then we are obtaining b r over here and there is a hydrogen over here like this right clear now but if we get an unsymmetrical if we obtain any unsymmetrical but before this let us take one more example over here if i talk about ch3 ch double bond ch ch3 okay now in this case again i'm going to take hbr and i want you all to write the answers in the comment section i want you all to write the answer in the comment section it is quite easy just similar to this you can do any of the things either get hydrogen attack to this carbon or to this carbon or attack br to this carbon or to this carbon it is the same thing students why it is the same thing because it is symmetrical how you can check the symmetrical see this from the double bond it is same on both the sides right i want the answer in the comment section with your name students okay now we take the unsymmetrical one now if we take the unsymmetrical one in unsymmetrical one rises marconikov's rule now what is the unsymmetrical one let us take this case ch3 single bond ch double bond ch2 and i am taking hbr over here i want my product so you know that this hydrogen halide is going to break yes and both of these are going to attack now which will attack on which see if hydrogen is attacking on this carbon atom then what we are going to get ch3ch2ch2br how come c if hydrogen is attacking if hydrogen is attacking on this carbon then the then br is attacking over this one we are getting this as of a product but if hydrogen is attacking one second if hydrogen is attacking now on this carbon and br on this then what is the another product that we get over here ch3 ch3ch br ch3 so we are getting two products over here but the question arises which is in the more quantity so this is in the more quantity it is referred to as a major product because it is more stable while this is the minor product now how come i got to know that this is the major product or this is the minor product it is because of the markovnikov rule it is because of the mark onikov rule now what is mark konic off rules the question arises that the negative part of the hydrogen halide or the negative part of the attacking reagent is going to attack on the reactant on which carbon atom the carbon atom which have least number of hydrogen atom negative part will attack will attack on that carbon atom on that carbon atom of alkene of alkene which has which has least number of least number of hydrogen atoms it is most important do remember in mariconico we are going to see the least number of hydrogen atom but in anti-markovnikov we are going to see more number of hydrogen atom so this is the process that occurred over here let us see once again see why it is major because the negative part that is the br negative attached to that carbon atom of the doubly bonded which has less number it has one hydrogen it has two hydrogen so it has less number of hydrogen so br is going to be our negative is going to attack on this carbon atom hence this is the major product right students okay now students okay now moving forward towards the next chemical property that is the addition of sulfuric acid now comes what addition of sulfuric acid now these are the basic you know chemical reactions you just need to practice once or twice right if i'm taking again ch3ch double bond ch2 now i want the addition of sulfuric acid that means h2so4 edition will be done over here now if h2so4 addition is happening over here now do remember here you have h positive ion with you here what you have h positive with you and hso4 negative with you so you are going to see the negative part is going to attack to which carbon atom which has less number of hydrogen atoms this is the markovnikov rule so this will attack over this because it has less number of hydrogen so what we are getting over here ch3c here now see this ch3c now double bond is broken over here we are going to get hso4 over here now again single bond because double bond is gone now also h is also present with this carbon atom and now this edge positive will attack on this carbon atom so it will get ch3 so this is the product that we are getting over here the mechanism that is working over here is markonikov's rule again i am repeating it is the one that is markovnikov's rule which states us that the negative part will attack on that carbon atom which has the least number of hydrogen atoms do not get confused in this whenever you are getting an unsymmetrical alkene kindly attack the negative part over it here negative part is hso4 negative so it will attack on that carbon atom which is less number of hydrogen if i talk about only yes do remember here we are not going to observe this carbon atom why we are not going to observe this carbon atom because this is the chemical property of alkene what we are doing here we are observing on the alkene that means we are going to observe on the double bond portion clear so we are going to observe these carbon atoms now comes the next important chemical property student that is ozonolysis now this is the most important in ozonolysis you know there will be addition of a zone that is o3 again i'm going to take an unsymmetrical one that is ch3 ch double bond ch2 and there will be an addition of o3 now what happens over here do listen very carefully see there will be three oxygen that they are going to attack and they are going to attack on this alkene first of all so what will happen over here this is ch3 this is the carbon atom and this double bond will break over and one oxygen will come in between both of these this ch2 will remain as it is now one oxygen has came in between two oxygens are still pending what will happen over here one oxygen two oxygen and like this see it is a quite a different thing that you are observing over here now if you will complete the valency of this carbon it has a hydrogen over here also what i have done i have done the attack of o3 or three on this doubly bonded now this doubly bonded is going to break and in both of these carbon atoms one oxygen is coming on the top and other two are coming together right students now what is the very next step look very carefully what will be the very next step in the very next step students this is going to break what you're going to get over here ch3c over here this ch3 is written as it is this ch is written as it is and there will be a double bond between this oxygen because it is going to break other than this what you are going to get over here other than this this will break other than this this will break and what we are going to get over here is if you see it is a ch2 again it is ch2 and this with a bond with oxygen so what we will see c h h double bond oxygen clear so this is the most important reaction students it can be asked in your examination ozonolysis and yes in 12th standard also you are going to study this so what i have done it is very simple there is attack of o3 and o3 attacks in such a situation that there will be one oxygen between the doubly bonded carbons atom this double bond will break and one oxygen will come in between these and the other two are going to come together clear now comes the last chemical property that is polymerization that is polymerization students in polymerization what do we see we are obtaining a polymer with the help of a monomer basically the small unit monomers join together to form a polymer now if i talk about this alkene that is ch2 double bond ch2 this is my monomer here i'm writing n here refers to as number of monomers now when the polymerization process occur it leads to the formation of you know um a chain like structure this double bond will break it will come like ch2 single bond ch2 and will move like this so this is what it is a polymer it is what it is a polymer now let us take one more case over here if i see ch double bond ch2 and i have ch3 with me this is my monomer now n times monomer so what i'm going to get are a straight chain i'm going to get ch now this double bond is going to break ch2 it is going to make a bond further here this ch3 is at attached and this is the polymeric chain this is the polymeric chain so basically what do we do from with the help of monomers these n number of monomers will join together to form a straight chain like structure that is the polymers and polymers in the everyday retail life has the most important usage students so i took two of the examples i took very first example of two carbon atoms the very first simplest alkene i would say okay now this is again the monomer end times of this alkene molecule okay formed this polyethylene for us right where all the double bonds will break out and will lead to the formation of the single bonds and if i talk about the next case students in this case we have seen this is the propane right three carbon atoms propane now this propane monomer is taken and which lead to the formation of what again a polymer again a polymer so students this were the chemical properties of alkene clear these were interesting and i want you all to practice it more and more okay now students comes over next segment that is the alkynes one if i talk about the alkynes one basically alkynes are what they are unsaturated unsaturated hydrocarbons they are unsaturated hydrocarbons right students and you know they are highly reactive they are highly reactive they are highly reactive ones and if i talk about more about alkynes they have a molecular formula of cnh2n minus 2 students if i talk about alkane they had cn h2n plus 2. if i talk about alkene it had c and h2n if i talk about alkyne they have c and h2 and minus 2. and if i take the very first case of the alkyne that is for second number of carbon atom the very first member is when n value is two so we are going to get what c2h2 c2h2 this is we are going to get and it is ethylene which is gaseous in nature it is ethane ethane which is gaseous in nature students okay this one is gaseous in nature also known as acetylene gas also known as acetylene gas so you can also write the n values 3 4 5 to obtain the other products if you put the n value as 3 what you are going to get c 3 you know 3 into 2 is 6 6 minus 2 is 4 so c c3h4 then is the next member then you are going to put the n value as a 4 c 4 h6 clear students so likewise you can form all the homologous series for the alkynes now students we will move forward towards our next topic so yes students now comes our methods of preparation of alkyne how do we prepare alkyne so our first method is from calcium carbide see first of all what happens calcium carbonate is taken which is caco3 and it is heated and we obtain calcium oxide plus co2 now the very next step is cao that is calcium oxide it reacts with three carbons forming calcium monoxide and calcium carbide now with the help of this calcium carbide students we are going to get our alkyne as our product now how come this calcium carbide which is ca c2 it reacts with water molecule leading to the formation of calcium hydroxide plus our ethane now from here ethane is obtained how come see there are basically what it is h2o so i can write h2sh which now what happens over here this ca combines with this twice of oh forming caoh2 and what is left now we are left with two carbons and two hydrogens so this forms what ethane which is basically our what alkyne so this is the formation of alkyne very first method the next method is vicinal dihalide again i have told you what is vicinal when the two adjacent carbon atoms contain the halogen groups right so it has brbr with it the two different halogen groups now what happens here alcoholic koh is used do remember students do remember aqueous koh will not be used over here alcoholic koh will be used over here and when alcoholic koh basically reacts what happens over here what happens now this br from here from here again the process you remember this carbon atom attached to the halogen is alpha the carbon next to the it is beta it has hydrogen over here so there what will happen this halogen of alpha and this hydrogen of beta is going to eliminate it is going to eliminate and there will be bond formation between both of these and now in the next step when we are obtaining an alkene molecule by beta elimination by the elimination of alpha halogen and beta hydrogen we obtain an alkene molecule but now what will happen now in the presence of sodamite students nh2 nh2 this again again this br is going to remove and this hydrogen is going to remove now again hbr removal will be there see sodomized is used for the formation of removal of our halide hydrogen halide so hbr is removed and again a triple bond arises which leads to the formation of what it leads to the formation of alkyne it is the most important the most important personal one right okay now students moving to the physical properties of alkyne if i talk about the physical property the very first property is that first three members of alkyne groups are gases okay the next eight members are liquid and the next one after the eight one first three then the eight and then from i guess it's like um from 13th or 14th member they are going to be the solids one right so you need to remember first three gaseous then liquid then solid just similar to the alkenes and the alkane ones right now students if i talk about the next physical properties that it is colorless do you remember this other than this all are adornless okay and if i see it's weakly polar we have seen the polarity the polarity arises due to the s corrector over here there is a 50 percent s corrector in the alkynes case due to which electronegativity difference arised and the more the s character more is the electronegativity more is the electronegativity that mean more the poles will be arised so if we talk about the alkanes case they were non-polar why they were non-polar because there was no s corrector over here there was very less s character and if i talk about alkanes only the electronegativity of alkanes the carbon and hydrogen was always similar so there were no poles but if i talk about alkenes or alkynes we see that the poles are rised because the nature of the double bond or the triple bond due to which the s character got increased and when the s character got increased the polarity came right now students other than this they are immiscible in water but yes they are uh soluble in benzene uh like solvents right so these were the physical properties students that you need to keep in your mind okay now moving forward to the chemical properties i want you all to remember the preparation methods again so i'm going to you know repeat the preparation method because these uh if we have seen alkanes there are a lot of preparation methods in alkenes there are preparation method but in alkaline there are only two or three preparation method that you need to keep remember in mind the very first is calcium carbide in many of the books directly this reaction is given to you so i have given you all of the steps from the calcium carbonate caco3 you are obtaining this calcium oxide plus co2 then you are obtaining this that calcium oxide plus three carbon you are getting calcium monoxide plus calcium over carbide now this is very important how calcium carbide is formed this you should know okay then it will react with water leading to the formation of alkyne other than the students what is happening over here vicinal diahilide more important very important because we are not taking germinal dihalide in geminal dihalide it won't take place because we want um alpha halogen and beta hydrogen do remember this why we are not taking germinal one because in that because in the germinal one what we observed you observed that the same carbon atom is having the two halogen groups but during the beta elimination we want beta hydrogen that should be eliminated so that's why we took the vicinal one and the visible medicinal one again i'm going to write over here beta elimination from here if i see over here alpha halogen and beta hydrogen is eliminated do not forget this because it is going to help you in alkynes case also and in alkenes case also in both of these cases it is going to help you out right students now we have seen the physical properties we have compared the properties with the alkanes and another one also now we will see our what we will see our chemical properties students if i talk about the chemical properties see alkynes is very small topic it's not a big deal if i talk about the chemical properties students in the chemical properties they are acidic they are acidic how they are acidic see they are acidic if i talk about alkanes and alkenes they are not acidic as alkynes why they are acidic students because when there was a reaction of alkyne with sodium metal or when there was a reaction of alkyne with na na nh2 when alkyne reacted with nanh2 the reaction occurred the reaction occurred and also when alkyne reacted with sodium metal or any another metal again this reaction occurred again this reaction occurred students when it reacted there was a salt formation over here and there was h2 that was released which explained us these two reactions explained us through experiments that yes alkynes are acidic in nature that's why they are making here a salt when it reacted with metal or when it reacted with any nh to the reaction occurred but if i talk about alkenes and if i talk about alkanes they are not acidic in nature do remember this and due to this now if they are acidic what are alkynes going to do observe this if they are acidic students if they are acidic then what will happen see we know that we know that yes there is a s corrector in these carbon atoms more arise divide due to the triple bond here now what do we observe the hybridization if i talk about hybridization there are two sigma bonds see whenever whenever you are having four sigma bonds here sigma bonds plus lone pair sum so if there are four sigma bonds then the hybridization is sp3 if it is three then the hybridization is sp2 when it is two it is hybridization is sp so over here there are two sigma bonds other are pi bonds this one is sigma this one is sigma right other are pi ones so two sigma bonds that means hybridization is sp here also hybridization is sp now due to sp uh hybridization they have the electronegativity difference because more the s character more the electronegativity and when the carbon have more electronegativity what it will do it will break this bond it will break this bond will contain negative charges over here and yes these protons are going to release these protons this s h positive are going to release which states us what that they are acidic in nature now what happens let us take one example let us take one example we have this as a reactant and i tell you that kindly reacted with sodium so what you are going to do it is going to be a substitution kind of reaction this hydrogen will come out because it is acidic in nature this h positive will break out and this sodium will takes its place so what we are going to get over here is we are going to get this as our structure and yes h2 will be released and to balance it out half h2 will come because you know h2 when released will be released in the form of gaseous only so now again i'm going to react to this molecule which is obtained as a product with na again with n a now what will happen now this bond will break and this now n a will attack at this carbon atom so we are going to get n a carbon triple bond carbon n a and again h2 will release okay so if i write the overall reaction students all over here if i write the overall reaction what i'm getting my alkyne i have reacted it with 2na leading to the formation of what a single bond carbon triple bond carbon n a clear clear students and also also now complete h2 will release half h2 plus half h2 gives one h2 is it clear to you is it is this chemical property clear to you now how we got to know that this is the chemical property due to its acidic nature acidic nature they ask you these types of reactions in the examination now if you know both of these steps it is quite simple for you you know that h h positive that is the proton is going to release out when proton is going to release out sodium is going to take its position now sodium metal when reacted would take this position and will lead to the formation of this first of all this bond is going to break or either you can break that bond it's the same thing but the bond between and carbon and hydrogen is going to break why because they're acidic in nature right students other than this they do additional reaction other than this they do additional reaction and if i talk about additional reaction let us take one case generally i am talking about addition reaction in the chemical property when i say carbon triple bond carbon is present and yes it reacts with hz let us say that z here is more electronegative now when it is more electronegative then what will happen over here it will arise a negative charge it will arise a positive charge when it is going to break right students what is going to happen now this h-positive is going to attack on the carbon atom and yes its bond is going to break its bond is going to break when its bond is going to break what is happening over here what is happening over here it will this bond is broken kindly correct this arrow okay this h-positive is attacking now when this h-positive is attacking the electrons are moving then the electrons are moving from one direction what happens over here this hydrogen is attached over here this bond is broken and it is electron deficient so arises with a positive charge now what happens over here z negative is also produced z negative is also produced now this z negative will attack on what c positive leading to the formation of this as our product it was quite simple it is a general reaction that addition occurs how first of all the double the triple bond is going to break and the electronegative part comes out separates out and yes the h positive attacks and when the h positive attacks yes the carbocation is formed and then the z negative will be attacking on the carbocation leading to the formation of our product right students in addition reaction there are various cases that we will study uh just right now so let us take the very first case for the addition reaction so yes students now we are going to see the additional reaction properties only that is the addition of die halogen addition of dihalogen now what happens in the case of dihalogen two halogens are going to attack and they're going to add over the alkyne molecule now if i take the simplest alkyne that is ethane and i want here dihalogen let us take br2 okay now if i'm going to take br2 over here i want its addition now what is the basic step it is going to occur in the presence of ccl4 first of all do remember this now we know that when br2 is going to attack over here we know that it is symmetrical so this double bond is going to break one bear is going to attack and another one is going to attack on this carbon it is quite a easy process so what will happen over here hydrogen carbon double bond will be still present because triple bond is being broken one br is added over here another br is added over here now in the next step what is going to happen again in the presence of br2 ccl4 it is going to happen again this br is going to attack on these carbon atoms on these carbon atom students and this bond is going to break so carbon single bond carbon hydrogen br br br br hydrogen yes this is the basic reaction of you know dihalogen similarly you can do the addition of dihydrogen this also you have done addition of dihydrogen if i talk about the addition of dihydrogen what is occurring over here let us see for the case of dihydrogen this is my simplest molecule it will occur in the presence of again walked nickel platinum palladium this you have already done students many a times and this bond is going to break it is going to attack over here what we are getting we are getting this as a structure now again what in the presence of nickel platinum or palladium it is going to occur and this h2 is going to attack on both of these carbons leading to the formation of water leading to the formation of wart alkane so these are kinds of you know additional reaction that you should know first i told you about addition of die halogen that means two halogen groups are going to attack and then i told you addition of dihydrogen then two hydrogens are going to attack over here see in the very first case and in the second case they both have the similar steps they both have the similar steps now these are symmetrical if it comes out to be unsymmetrical students then what you need to do you need to follow the markovnikov's rule what is the markonikov's rule the negative part will move to that carbon atom which has less number of hydrogen atom but do remember students when does um the negative part moves to that carbon atom which is more number of hydrogen atom when it is anti-marconic of rule and it will occur in the presence of peroxide so whenever you are given the uh you know the reagent on the arrow the en the peroxide that means you need to remember that here anti-markonic of rule is going to happen and the negative part will move to that carbon atom which is going to have more number of hydrogen atoms is that clear to your students okay now this ends our section for the alkynes now students we are going to move further to the aromatic hydrocarbons now what are aromatic hydrocarbons if i talk about aromatic hydrocarbon they are also referred to as also referred to as aryans students okay now in the aromatic one we have already done in the goc section also what are the aromatic compounds they are further classified into benzenoid and non-benzonite if you remember aromatic hydrocarbons aromatic hydrocarbons are further classified into what into benzenoid and the non-benzenoid ones but in this lecture in this chapter we are going to focus more on benzenoid one benzeroid one are the ones which contain they they have benzene ring with them they have benzene ring in them so mostly we are going to focus on this only the benzeroid one other than this if i talk about benzene students benzene basically has a molecular formula of c6h6 which basically looks like this students you clearly know this you have done in the geos section also it is highly unsaturated it is what it is highly unsaturated and yes for a compound to be aromatic you should know that it should follow what it should follow huckel's rule these all things we have already done in the goc section that is four n plus two pi electrons it should follow this four n plus 2 pi electrons okay plus 2 pi electrons to be an aromatic compound to be an aromatic compound other than this it should be planar it should be cyclic these things we have already discussed over there okay students now we will see the preparation of aromatic hydrocarbons so yes my dear students now let us see towards the methods of preparation of aromatic hydrocarbons see aromatic hydrocarbons can be prepared from alkynes but not from alkenes and alkanes so let us see if i talk from alkynes if i talk from alkynes let us take the simplest one this is my alkyne i am going to take three alkynes over here now what happens students they are going to connect with each other in such a way in such a way that the triple bond between them is going to break one one triple bond is going to break and so here what will happen a single bond will come which will lead to the formation of what which will lead to the formation of what this structure this structure over here so this is what this is our benzene this is what this is our benzene so this is the basic reaction that we can do from the alkynes which will be done in the presence of red hot iron tunings right this is going to occur in red hot fp tuning do remember my point that from alkanes and alkenes you cannot obtain aromatic hydrocarbon i will write over here from alkanes and alkenes we cannot obtain this product directly obtain aromatic hydrocarbon clear is it clear to your students so this was the i guess very simplest method for you for the preparation of a benzene now students comes our next method of preparation that is from phenol if i talk about phenol phenol is the one in which the benzene ring contains oh group this is phenol this is what this is phenol now we need to prepare a benzene and we have phenol so what we are going to do the very basic reaction which we used to do again and again is with a zedin that is zinc is with zinc we are going to heat it up in the presence of zinc what we are going to get over here is that zinc will combine with oxygen will form zinc oxide it will form zinc oxide that is zno and our product that is benzene will be formed you always say my video does hydrogen went hydrogen is still here it is still here these are still here is it clear to you is it clear to you in the previous also students here i have shown carbon carbon carbon if you want to show in the terms of carbon this is the bond line rotation you can either show carbon over here then you can show all the hydrogens also with them okay it is the same thing so don't get confused you can either write in the form of representing each and every carbon atom or either you can show with the bond line notation of the benzene ring it is the same thing here i have written here i've directly used phenol and with the help of phenol we have obtained benzene you can also open it and write in such a way you can write in such a way also you can write in such a way also students it is the same thing it is the same thing don't get confused in this small small things right okay so this was students method of preparation from final students now we will see the next method of preparation that how can the aromatic hydrocarbons be obtained now students the next reaction can be decarboxylic reaction in this you are going to take sodium benzonate if i talk about sodium benzonate what it actually looks like this is a sodium benzonate this is what sodium benzonates students now it is going to react in the presence of naoh okay also in the presence of calcium oxide now what is going to be happening over here na2co3 will separate out and the product benzene will be obtained na2 co3 will separate out and we will obtain our product that is benzene that is the formation of aromatic hydrocarbon so that is d carboxylic acid the carboxylic nut acid decarboxylic reaction claire students it was quite easy very simple now you can easily learn the formation of method of preparations of aromatic one because these are basically three or four that you need to keep in your mind the carboxylic one then the phenol one and the previous one that simplest one is the alkynes one is it clear to you now students we will move forward towards the physical properties towards what to us the physical properties of the aromatic one okay now if i talk about the physical properties if i talk about the physical properties students first of all what we need to understand they follow huckel's rule you clearly know other than this they are non-polar in nature these are non-polar nature and hence you can see in benzene also resonance occur okay these are colorless okay you know what you can compare all the physical properties and learn together that will be easy for you now students they have a characteristic or door for example the nephilim balls right so they have a characteristic or door characteristic or door example the naphthalene walls clear example the nephilim ones now if i talk about the physical state basically benzene is liquid in state but if i talk about the nephilim ones they are solid right okay for this it is liquid and for this it is solid either it can be liquid or it can be solid next is they are immiscible in water okay but yes they are soluble in what they are soluble in benzene you know now comes the next chemical properties if i talk about the chemical properties the very first chemical property that comes in our mind is electrophilic substitution of benzene see what happens over here electrophilic that means referred to as electron loving philip refers to as loving electro is refers to as electron now who will love electron the one which has a positive charge that means electrophile that means electrophile so now we will see the substitution reaction of the electrophile that is the electron loving species now what will be substituted yes the hydrogen from the benzene ring will be substituted by the electrophile now we will look forward towards the mechanism students first of all we will look through the mechanism and with the help of mechanism only we will learn the reactions the very first is the generation of electrophile generation of electrophile how electrophile is going to be generated electrophile is the one which is electron loving which is going to attract electron towards itself and to do this we are going to use a certain catalyst we are used going to generate electrophile with the help of with the help of catalyst now this catalyst students is lewis base sorry this catalyst is lewis acid sorry this catalyst is lewis acid not lewis space now there are certain steps for the electrophilic substitution reaction the very first is generation of electrophile that means electron loving species will be generated right now this can be generated with the help of a catalyst which is lewis acid now basically generation of electrophiles we are going to see in four or five cases students it can be nitration it can be you know sulfonation one by one we are going to observe each one of them the very first in which we are going to observe is chlorination in if i talk about chlorination that means addition of chlorine and if i talk about chlorine it is going to be cl2 that is clcl now as i told you we are going to generate an electrophile and we are going to use a catalyst here the catalyst which i am going to take is alcl3 and now if i take alcl3 what happens this is aluminium this is chlorine this is chlorine this is chlorine now what happens over here this bond is being broken and this chlorine basically attacks on the aluminum when it attacks on the aluminum now this bond is broken the electrons are moved with this chlorine and this chlorine is now electron deficient hence what happens over here cl positive is arised and yes this chlorine that comes to the aluminum what happens it basically already had three chlorine with it and one chlorine came over here which contained electrons with it which contained students electron with it which when electrons were present with it it had a negative charge so yes this hole will have a negative charge over this now what is the generation of electrophile over here the electrophile is the one which is already having a positive charge and you know is electron loving which loves electron so what is electrophile over here cl is the electrophile over here which itself itself has a positive charge this is the electrophile this is the electrophile in this case is it clear to you electrophile is the one which itself has a positive charge and attracts electron towards it next comes friday craft alkylation if i talk about a friedel-craft alkylation again i'm going to use my catalyst alcl3 again i'm going to do now what is here reactant the reactant which we are using is methylchloride that is ch3cl again the same step is going to happen over here students now this bond will be broken and it will attack on aluminum the electrons chlorine takes with itself and hence this carbon is now electron deficient making it as ch3 positive and here what we are going to get al is now making bonds with four chlorine atoms and the overall charge students is negative over here again negative now this is what this is the generation of electrophile over here this is the generation of electrophile this is the first steps that we are doing in each of these reactions in each of these reaction this is the very first step if i talk about the next one that is friedel craft acylation in acylation students what we are going to take over here see very carefully it is also the same okay now check over here it is also the same what is going to happen over here again alc this bond is going to break this chlorine is going to attack over here what we are going to get generation of electrophile and it is the same thing negative charge over this surface is it clear to you is it clear to you it is quite simple students see here it can also be written in this form ch3 c double bond o positive is it clear to you now this is what this is friday craft acylation this is the previous one was friday craft alkylation this is friday craft acilation right now let us see the next case okay the next case is nitration students likewise it we did in the previous section we uh took our reactant and vertical catalyst and with the help of catalyst we got our uh generation of electrophile now in the case of nitration and sulfonation we have already done this the same process will occur one single bond is going to break and then there will be electrophile and yet and yes you are going to also get one negative charge species students if i talk about nitration we are going to take hno3 now this is the structure and yes here we are not going to take alcl3 but we are going to take h2so4 when i am taking h2so4 what happens over here the bond between this brokens up and yes this oxygen attacks over this hydrogen the lone pairs attacks over here what we are going to get what we are going to get over here is that we are going to remove hso4 because this bond is going to break this portion is removed now what we have got we got this one hydrogen is now attached this hydrogen moves to this oxygen now this is formed now what happens over here what happens water molecule gets removed this water molecules break the bond and gets removed and yes there is a generation of no2 positive which is a generation of electrophile this is the generation for electrophile in the case of nitration if i talk about the case of sulfonation here i'm going to take twice of h2so4 in nitration i took hno3 and h2so4 here i'm going to take twice of h2so4 and again the same step is going to happen this bond is going to break and there is a removal of hso4 this is going to remove it this on this oxygen this hydrogen is going to attack now as it is electrophile you know this h positive basically this h positive will move towards the electron and will form this and again here this water will come out and we are getting this generation of electrophile this generation of electrophile quite easy now you got to know how to generate electrophile students okay now these were the first step that was generation of electrophile the next step that comes out to be formation of carbocation see we know that kindly you know you have now making this you know how to generate this now after generating what you need to do is you need to attack this on the benzene ring on the aromatic hydrocarbon when you are attacking this double bond is going to break and this electrophile is going to attack on this carbon and when this double point is going to break see if it is going to break it is having a negative charge when it is a negative charge positive will come over it it is having negative so it will have positive because negative attracts positive so this attacks over here these double bonds are like this and here is hydrogen here is electrophile and yes it has a positive charge over here this is the formation of carbocation students now the next step is removal of proton to remove the proton now we require the byproduct that we obtained in the generation of electrophile that was alcl4 negative now for this what is going to happen now for this cl now cl will come cl negative will come it will take hydrogen with itself this bond is going to break and this will break over here making a double bond over here making a double bond over here hcl is gone because from here from here once cl negative is going to come out it will take hydrogen with it forming a hcl now it is left with alcl3 left with alcl3 when it becomes it goes from here then this bond is broken over here making a doubly bond and this is now neutralized and hence we are getting this as our product now if you know what is electrophile kindly attach it to the benzene ring that is your answer now let us see let us see for the nitration this is the process what you need to do first you will generate electrophile what was the electrophile generated in nitration and no2 so kindly directly attac attack uh attach over here no2 because you know the product that you are obtaining in each of the cases is this only so if you know what is electrophile you can directly write the answer so for this case sulfonation you know that so3 h was the electrophile so directly right so 3h right now for the case of halogenation you know cl will come so directly cl positive was the electrophile so cl will attack over here now for friedel craft acylation alkylation you know that you know that what was the uh electrophile ch3 positive so ch3 is directly attached over here now for friedel craft acylation what was the electrophile what was the electrophile ch3c double bond oh positive was the electrophile so yes ch3c double bond o positive was there so it is directly attached over here is it clear to you yes so it is quite simple if you know all the mechanism in depth so yes my dear students this ends our chapter of hydrocarbons i want you all to practice dpp questions why because we have done a lot of reactions today in this uh lecture one short lecture but without practice i would say that would be incomplete but you need to do if you are not doing students then you will be confined to a particular state of notes only but if you will practice you know if you'll put your concepts into the question then you are going to get that yes i have understood the concept so if you want to make a concept strong listen to the video once then read your notes and the next step you want to do is if you are forgetting any of the thing if you don't know how to remember a reaction apply it on the question section whenever you are doing questions you know you are learning do not say that no first i will learn all the reactions and then i will apply questions no that learning will take a lot of time what you have to do is you have to watch the lecture you have to make the notes watch the lecture once read the notes once and then apply the questions and if you're not getting the questions come back to the video lecture listen to the lecture again listen to uh read the notes again then apply on the question and i am telling you i am assuring you you will be able to do questions and this is the best way to do any of the questions it is the best way to learn the chapter right so students i want you all to practice because practice makes a man perfect and yes you can do everything you can achieve everything if you believe on yourself so with this note i end this video lecture over here and we will see you in the last chapter of organic chemistry thank you so much have a good day