 Okay, can we start now? The chapter we are going to start today is Amidst. You know, I'm getting married next month You didn't know this? Yes, at that time he told you now, but that time it was not at all fixed Now it is done 12th of December. Thank you. Thank you. You'll also get married. You want what you don't want Okay, anyways, see this is ammonia, right and S3 is ammonia Right and when I replace one of this hydrogen atom by any alkyl group like this our NH2 Right then this we call it as amines or one degree of mines Right, if you remove two hydrogen by alkyl group, it becomes R2 NH this is two degree of mines and when you replace all the hydrogen by alkyl group R3 and It becomes tertiary amine or three degree of mine Okay, so these are the derivatives of amine derivative of amine Okay, so this is the one thing. I think you have to keep in mind one more thing here The most of the reaction of amine if you see this Nitrogen and hydrogen bond takes part in the reaction This hydrogen actually takes part in the reaction and it did it goes on, you know Step-wise reaction actually we have over here all these hydrogens because of the electronegativity difference Has tendency to react in the reaction Understood fine. So we'll see those reactions one by one. Okay, preparation and reactions first point in this write down due to presence of lone pair presence of lone pair Amines are Amines are basic in nature and this particular And this particular, you know relation is important This particular relation is important. Okay, so and there are two different order. We have for this Okay, and that is the basicity order you write down in gaseous phase Basicity order in gaseous phase. Okay, these two you must remember it is very important in gaseous phase the basicity order is found to be this tertiary amine tertiary amine three-degree amine is most basic then Secondary and then primary in gaseous phase. Why it is very clear from this particular example, you see tertiary amine is this right tertiary amine is this with one lone pair on the nitrogen atom and Tendency to lose lone pair is nothing but what? basic nature Correct, so when all these alkyl group has tendency of plus I effect The electron density on nitrogen increases as the number of alkyl group increases and hence the tendency to donate lone pair of electron is maximum Correct, and that's why the basicity order is found to be this Yes Now in aqueous phase in aqueous these two order is very important. Okay in aqueous phase The order is found to be going to be this two degree of mine is Most basic then we have one degree of mine and Then we have three degree of mine. Okay Now in this you write down the reason This is only true. This is only true. Why it is important. You see if The alkyl group if the alkyl group is Methyl ps3 Okay, in case of ethyl next right on explain in case of ethyl It's high the basicity is going to be two degree maximum Then we have three degree and then we have one degree. Okay, so in this example also you write down Alkyl group means what CS3 whole twice NH and then we have CS3 NH2 and then we have CS3 Whole thrice and this is the order in case of methyl in case of ethyl the order is C2 H5 whole twice NH C2 H5 And it's two and Then we have C2 H5 Whole thrice and sorry it is whole thrice for 231 and this too. This is the order we have Now the reason of this you write down Write down in aqua solution See in gaseous phase what happens the basicity depends only on the plus I effect of these groups Right, but in aqua space along with this plus R effect We have two more effects that takes part into this Process and that is hydration effect and a steady effect. Okay, write down this thing that in aqua's medium In aqua's medium the basic is strength of amines the basic is strength of amines depends Not only upon depends not only upon the electron releasing effect electron releasing effect but also not only on the electron releasing effect, but also on Static effect and hydration effect static effect and hydration effect What is steric and hydration effect you see? See first of all first of all if you have this compound that is CH3 and Edge and edge and then we have CH3 and edge CS3 and The last one we have CS3 and CS3 and CS3 Now in this when you put these molecules into water that is in aqua's medium So in aqua's medium, what happens all these hydrogen has tendency to form hydrogen bonding of with oxygen Like this and this hydrogen bonding provides what? stability Yes, or no These nitrogen has lone pair one lone pair in This case since there is no hydrogen attached to nitrogen So there is no hydrogen bonding here. No edge bonding, right Along with this at the same time this CS3 also increases the plus i nature or Electron density on this nitrogen atom Right, so when you only see the i effect, which is plus i effect over here According to that the best city of this should be maximum then this and then this right Right, but when you see the stability factor hydro this is the hydration effect hydrogen morning That takes this because of hydrogen bonding this one is more stable, right and this one is least stable Right, so according to this if you see if this is more stable the tendency to donate electron pair will be More here, right? Will be less here Again, what happens you see in case of any of the trial group present over here then around this nitrogen The most of the space are occupied by the ethyl group, right? If you have bulky group present over here, the most of the space are occupied by the Alkyl group present over here and hence there is no space over here So that this electron pair this nitrogen can donate and forms and show the basic nature If all the space around the nitrogen atom is covered by this alkyl group Which is possible when the alkyl group is bulky then tendency to lose lone pair is less and hence the basicity decreases Right the first point see these two are contra-dictory point Because of plus i nature the electron density is increasing here and hence the tendency to donate lone pair should be more But it so if you see only plus i nature of this the basicity should increase But at the same time when the plus i nature is increasing or alkyl group is increasing at the same time What happens the space around the nitrogen that is occupied by these alkyl groups? And then we don't have many space or more space here. So that this lone pair can be donated, right? So since and that is the static effect we have steric hindrance or steric effect So because of this is steric effect the tendency to lose this lone pair of electron is less and hence the basicity decreases Right so these two these two logic that I told you these two are contra-dictory to each other So the net effect of this molecule is the you know a resultant of these two effect Right in between these two effect so that is what it is found to be when CS3 is there as alkyl group Then the order is fine to be Maximum for two then one and then three but when C2H5 is there ethyl group suppose we have ethyl group here here and here Obviously larger alkyl group larger plus i nature Large will more will be the electron density here more should be the basicity of this compound But again what happens when you have larger alkyl group large will be the repulsion also steric hindrance also more space around Nitrogen atom will be occupied by the larger alkyl group that will decrease the electron releasing tendency even more Right that is why the net effect of these two the net you know The effect of these two effect that is plus i and steric effect is found to be two three and one two degree then three degree and then one degree for Any alkyl group like C2H5 in aqueous medium But for CS3 like species or like molecule when we have CS3 it is two one and three degree for CS3 so net effect is found to be this Right, this is the explanation Understood. Okay, so these three effect you have to keep in mind. Okay. You have to write down all this thing Okay, so that the same thing you write down In case of larger alkyl group in case of larger larger alkyl group in Bracket you write down bigger than CH3 Larger alkyl group bracket write down bigger than CS3 For example, ethyl profile, etc. There will be steric hindrance. There will be steric hindrance and Hence the stability due to plus i effect stability due to plus i effect predominates over the stability predominates over the stability due to hydrogen bonding and Three degree of mines are more basic than one degree of mines Understood. So whenever like this see whenever like this if you have two different effects Contradicting to each other. So we cannot say by looking at the molecule. Okay, this is this should be the order or this should be the order In that case whenever more than one effect Taking place in one particular molecule then all those, you know, the net Order that we write in these cases all those order becomes experimentally in that case experimentally we Determine the order and then we try to explain it Okay, so in both the cases Secondary of mine is most basic 3 and 1 we have in case of C2H5 when we have ethyl E 231 order you have to follow 231 and when we have methyl M it is 230 21 and this you have to memorize it is important Okay, I will see the next one more point to write down as percent is as character increases character increases Basisity Decreases why basically decreases? Why business degree decreases because as percentage as character increases electronegativity Electronegativity increases and when electronegativity increases tendency to lose electron tendency to lose electron decreases and That is why the basicity Is it clear tell me the basicity order of this CS 3 and this to don't bear CS 3 and double bond CH CS 3 CS 3 C triple bond and What is the basicity order in this one two and three? This is what this is SP hybridized this is SP and This is SP Percentage as character is maximum in SP hybridized 50% right so basically your third one is Minimum then second and then one Right so these two factor you must keep in mind. I see the various reactions You copied on I was just coming in a minute. Okay, just okay so See like there are Various intermediates we have Okay And like we know carpet and carbon and all these right similarly there are one more Carbines is also one of the intermediate we have Similarly, there is one more intermediate and that we call it as nitrines. You must have remembered this I told you this in Aldehyde ketone chapter that will discuss nitrines in Amai's chapter, right? so there are various reactions of this chapter that Takes place by the formation of nitrines. It is an intermediate right so in all like what all the cases we have where the formation of Nitrines takes place that first of all we have to understand so what is nitrines? Nitrines is the nitrogen atom which has two lone pair on it and one vacant P orbital This is vacant P orbital. Okay. I'm not drawing that P orbital because every time we have to you know, write down this I'm just using a box to represent the vacant P orbit, right? this box is the vacant P Orbital correct few points about this nitrines you write down It is you obviously you see two plus two four electron. We have so it is electron deficient electron deficient Next what we can write it is neutral. There is no charge on it It is neutral third one is Monovalent actually we have one bond also here. It is attached with with carbon. So that's why it is monovalent third one Total number of electron is six Total six electron how it is six two plus two four and then two electrons will be here in this bond, right? So two plus two plus four plus two six six electron electron deficient, okay? And obviously with six electron we have here. So octet is incomplete one more point you write down here Can never be the product in any reaction Since it is an intermediate Can never be the product in any reaction since it is an intermediate and we know already this thing that any intermediates can never Present in the product, right? It forms during the reaction intermediates are forms during the reaction and destroys in during that reaction only right finally will get the product formation of nitrines you see Which is possible in various cases and then you write down whenever we have this kind of molecule Right you write down Rn Double bond N double bond N Okay, where here we have the positive charge Right and here we have the negative charge. It is as I'd kind of compound. Okay, I'll go as I'd NN3 NN3 is what it is a sodium azide Right, this is alkyl kind of when whenever this kind of compound we have in three cases nitrines formations are possible and those three cases are what when you When you heat this compound right When you strike with light onto this compound or when you are using very high temperatures Very high temperature In all these three cases formation of nitrines possible. Okay How this nitrene formation takes place? You see in this this nitrogen also has two lone pair on it for negative charge Right, what happens this lone pair comes over here in this bond Right and this Five odd comes on to this nitrogen So what happens here this forms are N Which lone pair on it one lone pair was already present over here This lone pair as it is this one pair goes here. This becomes a lone pair then N The positive charge will be as it is triple bond N Right triple bond N like this we have and one lone pair on this This nitrogen says it takes this one pair of electrons. This will have the negative charge on it Now finally what happens this from this particular, you know intermediate this nitrogen electron deficient now It takes this bond pair of electron and goes out as N2 so N2 comes out plus See here This he will have one electron here one electron here when both this electron pair is taken up by this nitrogen So obviously the orbital in which this electrons are present becomes vacant now That orbital is the vacant P orbital and will show this vacant P orbital like this Okay, this is how the nitrines formation takes place So here the nitrines will be our lone pair lone pair vacant P orbital plus understood write down the various product into this when you heat this PH as double bond O double bond O and double bond double bond N double bond N Positive and negative and you are hitting this One more to write down CS3 C double bond O And double bond and double bond and negative positive Okay, what is the product we get here? Atch and To lone pair and this plus and to Yes Here what we get PH as double bond O double bond O and plus N2 and Here we get what? CS3 C double bond O and Plus N2 Okay, but this thing here it is again not stable. Okay, so here after this you write down here Only this one you write on in this product write down here but if nitrines forms adjacent to carbonyl group if nitrines forms adjacent to carbonyl group Then it is not stable Then it is not stable wolf rearrangement takes place wolf rearrangement takes place and It gets stabilized Now what is wolf rearrangement CS3? C double bond O and And lone pair on it and one vacant orbital wolf rearrangement means what? This CS3 this CS3 Take this bond pair from here and this will migrate on to this nitrogen atom Okay, because this nitrogen has one vacant P orbital it can accommodate these two electron easily over here and When it migrates over here This electron pair comes over here and this pi electron goes on to this oxygen Sorry, this is not this will not go because all already this bond has been no Broken no this bond has been broken this lone pair only comes over here So what we get here you see this CS3 combines with this nitrogen will get CS3 and Lone pair comes into the bond pair so double bond C double bond O This we call it as with one lone pair on the nitrogen atom left. We call it as isosynate isosynate correct So this kind of rearrangement of migration of this alkyl group to this nitrene Nitrogen to this nitrogen Takes place only when we have carbonyl group present into this when the nitrene formation Takes place adjacent to this carb carbonyl group. Then only this bullfrey arrangement takes place. Is it clear? Okay, so this is the one possibilities one type of compound Which is the as I type compound we have in which the formation of nitrene is possible another Possibility is this second possibility if this Nitrogen attached with one alkyl group One hydrogen and other living group we have here L any living group now when this reacts with Any base in presence of any base, right? This is negative sign here, right? This is negative sign presence of any base this base takes this H plus from here Right and this pie electron this base takes this H plus from here The spy electron comes over here and this L comes out as living group. This is the living group We have like protein bromine anything. Okay, so when you heat this in presence of a base here Then this forms Rn one lone pair was present already and when this living group comes out it creates a vacant P orbital here The edge will be as it is Living groups takes the lone pair and has one negative charge on it This is how the formation of nitrene takes place in this kind of compound in presence of base Right now what all possible bases are we can have OH minus Right, we can have our oh minus We can have NH two minus Or The H oh minus all these bases are possible in for this Understood few examples you see that is suppose CS3 NH Cn They suppose I'm taking OH minus right on the product CS3 Cdouble bond oh and H see In presence of base right on the product. Yeah, so there will be there will be wolf rearrangement in this, right? So here the product will be what CS3 and Lone pair one vacant P orbital H2O plus Cl minus and here it will be the first will get this CS3 Cdouble bond oh and With two lone pair one vacant P orbital plus H2O Plus Cl minus and in this what happens wolf rearrangement takes place This will migrate over here and we'll get CS3 double bond or a single bond N Double bond C double bond oh I society, right? Now you see there are few reactions which involve the formation of Nitrene and the reaction proceeds right the first light on the reactions Which involves the formation of night which involves nitrenes formation? Reaction which involves nitrene formation which involves nitrene formation Okay, the first reaction you write down that is Hoffman Hoffman bromide reaction of men bromide reaction this reaction. We have done courteous reaction skimmit reaction And the fourth one is Beckman rearrangement Beckman rearrangement. We also have done In aldehyde keto all these reactions follows formation of nitrenes and the reaction proceeds Right Then the first reaction you write down that is Hoffman bromide bromide reaction The reaction is Reaction the reaction is what when amide our C double bond O NH2 When amide reacts with bromine in Presence of any base can be KOH can be an OH Right and when we heat this what we'll get we'll get our NH2 Right that is one degree am I you are getting one degree of mine from this Okay, this is a preparation method of amines also. So now we are going towards the preparation method of amines Okay, one degree amine we are getting okay, and this amine has one carbon Less you see our CO NH2 and gives you our MS right? So now you see the mechanism of this Okay, like one or two reaction. I'll show you the mechanism, but all these mechanism are not important This question was asked in the test, right? Anya you asked me this NH2 and the two was there right our C double bond O NH2 I mean this sorry NH2 when this reacts with this OH minus from the base What this OH minus do it takes this H plus from this? right This H plus takes this OH minus. They'll get our C double bond O and It's two lone pair here and H Plus H2 okay now this lone pair Will react with this BR2 molecule this lone pair comes over here and This bromine will take this lone bond pair and goes out. So this forms our C double bond O and H Br and one and one lone pair on it Again, we'll have a negative charge also here. Okay So we'll get this now from this what you can do when you have a base again OH minus What happens then from this? Can you tell me? Yeah, what we'll get here? So this is the N hydrogen and one living group, right? So certainly it forms it forms what our C double bond O N and one wagon P orbital And one wagon P orbital plus H2O Plus BR minus right now in this what happens Wolf rearrangement, so it gives you our N double bond C double bond O. This is what we get understood Now in this use this is a first step reaction where we get this Nitrene forms and we get this iso sign it now when this iso sign is the last step reaction is what our N N double bond C double bond O with the base OH minus and This base will attack onto this carbon. Okay, and this pi electron goes here on to this night So we'll get this R N C OH double bond O and on this nitrogen we have Now in this what happens There is proton exchange takes place. Okay, so this Pi electron actually goes here and This H plus comes out from here minus H plus and this H plus will attach onto this nitrogen atom What do you get R N? H C double bond O minus This we call it as proton exchange. Okay, another if you Draw this drug draw the you know another form of this this O minus comes over here This nitrogen will take this bond pair and comes out as R and H With two lone pair on it plus CO2 goes out now in the last step H plus from the solvent H plus From the solvent or this base that you are using here KOH and OH from there This H plus will attack onto this lone pair of nitrogen and finally you get R and H2 right plus Plus OH minus Okay, so this H plus you write down here It is coming from the solvent which is the base over here, which is coming from the solvent understood this This wolf Rearrangement is the rate determining a step Right, this has been asked in J. Okay. This is step is RDS Asked in J Remember this is important Okay, just I'll give you a few examples on this. Okay, then we'll start the next reaction after the break. Okay Just see what you how do you write down the product? You have started from this right amide and Finally is what this R and H2 joints? R and H2 will get this as answer. This is again one degree of mine we have so this reaction is You know helpful to get one degree of mine Okay, and one thing you must keep in mind and that was the question in the test this hydrogen that you get here It is coming from the solvent because you see here this hydrogen One thing you see this BR comes over here hydrogen goes out from H2 Right here. You see this hydrogen attached over here. This H2 is coming from The solvent if you take an AOD here your answer will be R and D2 KOH here R and D2 like that. So that you must write down in your own words. Okay, so this Ha KOH will give NS2 Correct only only thing is what are NS2 you have to join and this hydrogen must be Coming from this hydrogen if it is D then he also will write on D Few examples you see tell me the product in all these quickly find C double bond O and H2 and we are using PR to NaOH we are to NaOH and We are heating this mixture TH C double bond O and H2 we are to KOD D2 H2 also you write on it R C double bond O and it's true and a OBR or K OBR Tell me the product. Yeah, since we have hydrogen here. So here the product will be ET and it's true Here we get what? PH and D2 and D2 Here we get what? You also get R See this if you write or if you write PR to NaOH both are same thing Right, actually you see this reaction goes like this when you take this NaOH plus BR2. So this fine this actually this OH minus Combines with BR2 as H OBR and BR- goes out Right now this HOBR With this OH minus from the base this hydrolyse H plus take out so it forms H2O and OBR minus this OBR minus and Na plus will join and it forms and a OBR So when you write NaOBR or you write BR2 plus NaOH both are same thing There's no difference into it both are off man bromide only understood Yes, all these reaction you see we are getting one degree of mind one degree of mind one degree of mind everywhere, okay What is the product here in this one? We'll get the last one CS3 C double bond O and H CS3 With BR2 OH minus and we are hitting What is the product here? We get CS3 NH2 CS3 CS3 NS2, right? CS3 CS3 CS3 hold to NH that is what we get this CS3 and this group will join up This is this is not a living group. This is not a living group. That's why that's why you know This reaction does not take place understood This nitrene does not form into this and anyways if you have this CS3 So if you write down you'll probably get right down two degree amines But all these reaction you see we are getting one degree of mind off man bromide only gives you one degree of mind It never gives you two degree of mind. So for Hoffman bromide reaction this nitrogen Must contains only hydrogen. We should not have any alkyl group present over here Understood my point. Oh, it's fine because CL may come out But we should not have any Alkyl group attached with this nitrogen because these methiels ethiels any alkyl groups are not a good living group Understood Okay, so we'll start the class after 10 15 minutes. Okay. Okay. You there See one question that has been asked on this in J exam that I'll give you now the last one for this then we'll move on to the other the reaction and The question is this we have a benzene ring and with this It is attached with Cn d2 and double bond for this side plus We have the another compounds here. We have CH3 and Cn is to Double bond this one reacts with we add to and And an average and Be the product in this Okay So the product we get into this see this and this will Independently reacts with this we are doing So it's just first of all you see assume this Reacting is not given How this will react with this the product will be what a Ring benzene ring and then here what we have NH2 When this reacts with this correct plus Now this will react. So this has nothing to do with this reaction. Okay, so the another product will be Here we have CS3 and then It is NH2 This is the answer This question was asked in J exam I'm telling you there Okay, they should write down the next reaction that is called the S reaction in this reaction You see a side chloride our C double bond O. Cn is heated in presence of sodium aside That's true product that we get here is Rn is to which is nothing but one degree am I? Okay, whatever alkyl group we have With that only the product forms How this reaction goes you see any n3? Obviously this any will take this living group and a cl forms and This entry attached with the C double bond O in any entry the structure is this and a and Plus and minus Okay, so now here. What happened and CL goes out and it forms our C double bond O Can you tell me next what happens? When you heat this So this is loan pair comes over here, so we'll get our C double bond O and With they can be your correct Let's end to after this what happens Work for the arrangement that forms our N Double bond C double bond oh ISO sign it again this ISO sign it reacts with OH minus from this H2O this OH minus will attack on to this carbon and this By electron goes on to the nitrogen It forms our N right negative charge on it C OH and double bond O Okay, next what happens, but this H plus an H2O this comes over on this side It forms our NH COH Then again what happens this edge will give this bond pair of like click down here And this bond breaks CO2 goes out and RMS to form See I'll write down here When this H plus comes over here, you'll get this our NH C OH And double bond O with one lone pair Now when this by one pair comes over here, right? This bond pair on This nitrogen and this H plus will attack on to this nitrogen. So we'll get finally what? R and H2 plus CO2 CO sorry See you this is for this Acetyl acyl chloride for this reaction takes right we use any of the three and H2O for ester also this reaction They see here also We are getting one degree of mine So courteous reaction also gives you one degree of mine. This you must remember, okay? The same reaction you see with the help of ester with ester how this reaction proceeds So with ester RC double bond O R- Now for ester we required Three different reagent first is hydrogen and this to NH2 Then we are using HNO3 HNO3. We also write it as NA NO2 Plus HCl HNO2, sorry, not NA NO2 third one is H2O So in this when the reaction proceeds first of all the reaction the react that the Product that you get here is our NH2 one degree of mine plus R-OH plus CO2 Okay, our NH2 you see the alkyl group which amines forms is The alkyl group which is attached to the carbonyl group not with oxygen take care of this thing fine Yes or no So you see how this reaction goes See this reaction we use first of all when you use NH2 and H and H This one of this hydrogen takes this R-O and forms see what happens this lone pair of this oxygen We'll take this H plus from here and finally R-OH goes out Okay, and the product that you get here in this is RC double bond O and H and H2 correct Plus what we get we get R-OH Yes, now in this when you use HNO2 here then we get our HNO2 RC double bond O and H triple bond N With positive channel in this when you heat this So which is this N2O goes out and it forms RC double bond O and NNH right RC just a second and it goes out. So we'll have that so we'll have one lone pair One H and the vacant In this again, what happens? H2O that you have from that H2O Stakes H plus forms H2O plus and we'll get here the lone pair again, so that will be RC double bond O and It's two lone pair and one vacant P orbital Then isosinate forms are N double bond C double bond O and from this again The reaction is same OH minus attack onto this carbon This bond goes here then H plus comes on to this nitrogen and all these things are same Okay, so like this you get with H plus H2O We'll get R and H2 again Next write down skimmit reaction the skimmit reaction Right on in this reaction carboxylic acid in this reaction carboxylic acid reacts with N3H Hydrazoic acid Hydrazoic acid or N3H reacts with N3H followed by followed by the hydrolysis in acidic or Basic medium any medium we can take so the reaction you see our C double bond O OH reacts with N3H Then we have H2O with H plus or OH minus anyone So in this again, we get our NH2 Plus H2O and CO2 goes out into this again You are getting 1 degree amine into it all these reactions you see we are getting 1 degree amine the mechanism I am not writing down. Okay, you can do this easily okay This H2O first of all OH minus OH take this H Comps H2O out and this N3 attached with this and then N2 comes out from that wolf rearrangement takes place It forms isosinate and then the reaction goes like those reactions we have already done Okay Understood. Eschemic reaction we have done in case of aldehyde and ketone, right? You have done it aldehyde and ketone Okay, so that will not do again and this is just one point you write down note in Eschemic reaction Carboxylic acid Carboxylic acid the reaction that we saw just now gives 1 degree amine If you have aldehyde Aldehyde it gives cyanide secondary amide secondary amide Okay If you have ketone, then it gives secondary amide always Mechanism I have given you I hope in aldehyde and ketone chapter Okay, now there are very few common reactions. We have the main Major reactions we have already done. Okay, there are few more reactions There are few more reactions which involves the preparation of amines and those reactions you write down the next reaction from nitro compounds from nitro compounds Write down quickly now these things things you have to get memorize. So write down write down nitro compounds on reduction Gives primary amines primary amines nitro compounds On reduction gives primary amines Okay, the reagent used for reduction. There are many reagents Those reagents you write down for reduction purpose. We can use various reagent like SNHCN We can also use ZNHCN H2 nickel we can use we can use LiAlH4 in dry ether We can use that is it. Okay, these are the reagents we can use If you have any nitro compounds say RNO2, the nascent hydrogen we get from the Reducing agent suppose LiAlH4 is the agent here So this will combines with this Nitro compound and forms RNH2 1 degree amine plus H2 Right six hydro just two we have here two H2 will get okay RNH2 will get 1 degree amine This you must remember what degree amine we get from this. Okay, next one you write down From nitriles from nitriles Okay, the reagent we use we use here H2 in Rennie nickel We can use Na in C2H5OH ethanol That is Rennie nickel we call it as it is just you know fine Nickel is we use the size is very small nickel also you can write down Rennie any by the spelling we have sorry Or we can also use LiAlH4 Okay, again, it is nothing but reduction only right where you have any Nitriles that is cyanide only R CN a reducing agent will provide nascent hydrogen here LiAlH4 and that gives R CH2NH2 Remember one thing here the number of carbon atom in the product and this reactant must be same Okay, so if you have one series of reactions, suppose they have ROH When this reaction reacts with SOCl2 We'll get some Product here and then it reacts with KCl Then if you have LiAlH4 Can you write down the product in all these? If you have ROH in SOCl2, this gives you alkyl halide RCl Correct RCl with KCl, KCl goes out and forms nitriles RCl RCl with LiAlH4 in forms R CH2 Okay, this is the reaction we get next write down By reduction of oxyms Reduction of oxyms You have another class now maths no reduction of Oxyms this actually we use with aldehyde aldehyde, we know whenever it reacts with NH2OH it forms what? Oxyms R CH double bond NOH This is oxyms Right and when this is reduced with LiAlH4 1-1 hydrogen this H2O goes out and hydrogen attached onto this carbon and nitrogen and we'll get RCH2 NH2 again 1 degree of mines These are the reactions we can write down the mechanism also, but that is again not important I'm just giving you the product here see number of carbon atom is again same in all these products right that you must remember next reaction you write down that is Gabriel thalamide reaction Gabriel thalamide reaction right now this only gives 1 degree amine again all the preparation method we have seen till now that is for 1 degree amine Okay, this is also important like sometimes you also ask this gives 1 degree amine 2 degree amine like this Okay, this also you keep in mind Now if you see the thalic acid is this we have a ring and C double bond O OH C double bond O OH This is thalic acid when this reacts with NH3 NH3 This two hydrogen from this nitrogen takes this OH and OH forms two moles of H2O and and the compound that you get here is this C double bond O and C double bond O will be as it is and these two carbons are attached with NH Understood, okay Now in this what happens we use here any base that is any OH or anything we can use Any base and then we are using any alkyl halide RCL Remember important thing here It is what the alkyl group that you are getting getting here with that alkyl group only you will get the Amines the final product that you get here That amines you'll get that NS2 you'll get from this are only we'll get this are NS2 as the product So like if you have to form ethyl amine, right then you write on your CS3 CS2 CN Methyl amine and CS3 CN like that You know you can use this alkyl group here and you will get the desired product, right? What happens this base will take this H plus from here and H2O goes out, right? And this gives This gives you our C double bond O C double bond O O O O and this is attached with nitrogen with two lone pair on this nitrogen atom, right? Now this lone pair of nitrogen RCL we have that we are using here this lone pair will attack on to this alkyl group And this Cl minus comes out as a living group. This attack is SN2 type right Cl minus comes out as living group and the product that you get here is This C double bond O C double bond O and this is attached with N and R One lone pair of nitrogen This we call it as alkyl substituted. I'll write down the name in the last. Okay, so first we can be right Alkyl substituted halamide Right now this H2O that two moles of H2O that have that we have used So we are using two H2O So what happens here? You see the product that we have here is This C double bond O C double bond O and R Now in this you have H2O molecule this H2O molecule this lone pair Attack on to this carbon Right and another H2O molecule with this lone pair attack on to this carbon Right and these two bond gets break Right and what happens finally? Here since this oxygen loses its lone pair so oxygen will have positive charge on it Here also oxygen will have the positive charge So one of this hydrogen atom comes out as H plus and here also our hydrogen comes out as S plus and both H plus Joins on to this nitrogen. Okay, so finally we'll get our NH2 Plus thalic acid will get in the end as it is C double bond O OH C double bond O OH This is the a mine we have which is one degree This we call it as gabriel thalamide synthesis or reaction. Okay, both are actually same thing So this reaction is also useful For the preparation of one degree amines Yes What what see is shemit reaction generally we use for mono carboxylic acid generally Okay, this is di carboxylic acid two COOH group are there That also maybe takes place, but we never know because I have never seen this reaction Maybe here also we'll get our ns2 and ns2 like this But again, we are not sure this is the you know experimental thing we do it on the lab in the lab So we cannot say generally shemit reaction we do when we have mono carboxylic acid One COOH group must be present Fine This compound we call it as alkyl substituted thalamide l-i-m-i-t-e Okay, you must remember This R group that you are using here That only forms the amide so that you can use Accordingly like Whatever it is required according to that you can use this Okay, one node you write down into this One node you write down Aniline cannot be prepared by this So you can use this One node you write down Aniline cannot be prepared by this method Aniline cannot be prepared by this method since since aryl halide Since aryl halide does not go under Does not go under nucleophilic substitution reaction aryl halide does not go under nucleophilic Substitution reaction Then the another probably the last method of preparation we have Okay Write down by direct ammoniation by direct ammoniation In this it is simple write on alkyl halide in this reaction alkyl halide reacts with ammonia reacts with ammonia And we'll get mixture of amines mixture of amines And quaternary ammonium salt mixture of amines And quaternary ammonium salt Okay, you see suppose we have rcl alkyl halide And when this reacts with NH3 So what happens this hydrogen one of this hydrogen this cl combines and comes out as hcl And forms rNH2 This is one-degree amine In this method we'll get mixture of amines again. You substitute here rcl Again hcl comes out and we'll get r2 NH It is two-degree amine Next rcl and you'll get r3 N Three-degree amine now when you put again rcl here You'll get r4 N plus And cl minus this is quaternary ammonium salt quaternary ammonium salt Okay, the direct ammoniation gives you mixture of Amine that's why it is not that useful Suppose you required one-degree amine But if you use this method to prepare one-degree amine you will not get only one degree you will get two-degree three-degree and salt also Right. So all these reactions which gives you mixture of amines Those reactions are not that useful Okay, because you'll get mixture of again what you have to do You have to separate this amine one-degree amine from this two-degree and then three-degree right So that will also again require much more cost, right? So this is not cost effective method Generally, we don't use this method to you prepare Amines right so another method we have that gives you mixture of amines is Ammonolysis of alcohol Ammonolysis of alcohol right on to this alcohol on reaction with ammonia in presence of reagent like al2o3 alumina or thoria tho2 In presence of reagent like al2o3 or thoria Gives mixture of amines Gives mixture of amines The reaction is c2h5 OH plus c2h5 OH plus NH3 with al2o3 This OH and H comes out at H2o And you will get c2h5 NH2 one-degree amine again, you put here c2h5 OH Then you get c2h5 whole twice NH two-degree amines One more time the same reaction proceeds it forms c2h5 whole thrice And so one degree two-degree and three-degree amines again Okay Next write down properties of amines We have almost done with this chapter. Okay, there are some chemical reactions of this amines we have But those reactions are based on the same mechanism that we have discussed today only Okay, so that you can understand easily. Okay, so we'll do it Next class, okay Just two three properties you write down then we'll start this in next class the reactions of this And that will finish probably in the next class. So we will do this amines chapter into the Separate classes. Okay, this is the extra class Regular classes will do whatever we are doing over there Okay, so what we are left with any other thing that is left Also d&f blog. I'll just give you the important thing. Okay, there is not much to teach right you have to memorize those things fine Okay, and then what is left Metallurgy is also there now I'll tell you metallurgy also I'll tell you because those are the complete memory based chapter we have Okay, so most of the thing we have finished anyways write down the properties of amines Write down the first point lower amines are gases Lower amines are gases or low boiling point liquid Right lower amines are gases or low boiling point liquid next point Boiling point Is directly proportional to the molar mass solubility in water solubility in water Due to H bonding Okay solubility Is inversely proportional to molar Okay These are the few reactions few properties we have okay So next class we'll start with the chemical reactions reactions of amines. Okay Okay, okay. Bye. Bye. Take care. Thank you