 Can you see the screen now? Why black you can't see the screen now Now it's fine. So we'll start chemical reactions of amines Chemical reaction you write down. Okay. So the first reaction we have that is formation of salt formation of salt In this you see write down See it is right on this reaction first. Suppose you have amines like I'm taking one degree amine Rns2 plus HCl one thing in the reaction of amines You must keep in mind that this hydrogen which is attached to nitrogen over here. This hydrogen is always Most of the time it takes part in the reaction. Okay, so when you keep this in mind, it helps you in memorizing the reaction Okay, so what happens here? You see this our ns2 right our ns2 Since it is a formation of salt we are so here the same kind of reaction is not that I'll show you this thing that I told you just now here you see this our ns2 will have one lone pair here Right this lone pair will take this H plus from this Acid right and it forms a salt and that we call it as Formation of salt so that forms like this our NH3 Plus and We have Cl minus. This is the formation of salt we have from one degree of amines Okay, which we also write is our NH3 CL simply Okay, now what happens this if you write or if I write down like this also our NH3 CL This our ns3 CL from this salt we can easily obtain the amines from which the salt has been formed and for that purpose We use an A OH any Alkaline solution, okay, so this what happens in this this any CL Forms H2O also forms and our ns2 will get into this So finally we'll get one degree amines plus we'll get salt and a CL and plus water Releases in this so that one degree of mines gives you a salt which on Reaction with alkaline solution right alkali solution it forms the Amine with which it has been formed right But the same thing is not true when we have two degree of mines for two degree of mines you see the reaction Suppose we have R2NH. This is two degree of mine when it's react with react with HCl So this forms the same kind of salt which is R2NH2 Plus Cl minus or we can write it as R2NH2Cl. Okay. Now for this when you heat this it dissociates into the Into the amines which is the primary amines we have here RNH2 will get and plus we'll get alkyl halide RCl Okay, the salt that you get from second-degree amine that is two degree amine on Heating that salt we don't get the amine itself But we'll get one degree amine and alkyl halide This is also the preparation method of one degree amine from The from secondary amine from first we'll get the salt of it and then we'll heat this we'll get one degree of mines into this Okay Now the same kind of reaction we have when the mine is three degree tertiary R3N. R3N reacts with HCl Okay, so it forms again salt of it R3NHCl but when you heat this since it is Tertiary amines on heating it gives primary. Sorry secondary amines R2NH plus again RCl Okay, so these are the three different types of reaction We have in the formation of salt for amines one more important reaction here You write down one note you write down here three degree amine three degree amine forms Borein This is important. Okay, three degree amine is suppose we have CH3 whole thrice N CH3 whole thrice N when it reacts with Borein BH3 this Borein is electron deficient molecule, right? It is electron deficient Boron has in this state boron has six electron hence we call it as electron deficient and this lone pair of nitrogen It can donate to the vacant P orbital of boron here because this boron has one vacant P orbital So this forms and And what we say Dative bonding here BNH3 and this we call it as adept AWD lucity This is the reaction with Borein. You must remember Then then see first of all if you have RR dash two different group we have here, right? So in this what happens one of the alkyl group will take this Cl Right and forms our NH2. So suppose we have CH3 Right and C2H5 Cht and C2H5. So in this the See because this chlorine will be Cl minus electronegative, right? So this will take the lesser plus i nature the alkyl group which shows lesser plus i nature So if you have CS3 and C2H5 it forms CH3 Cl here and C2H5 NH2 Understood? One other thing you see if Yeah, correct. CS3 Cl and C2H5 NH2 will get so lesser plus i will go with halide and Cl Okay. Okay. Next you write down reaction with alkyl halide reaction with alkyl halide I'm just directly giving you the reaction here Okay, because we have 2R 30 unit 2 and half hour class only today. So I wanted to finish this Okay. So in this you see we have our NH2 it simply primary amines gives you secondary and secondary gives you tertiary And then tertiary gives you salt quaternary ammonia salt R dash X So it forms what Hx comes out like I said this hydrogen takes part in the reaction So this hydrogen and X forms Hx and we get our NH R dash plus Hx Again when this reacts with R dash X again and again it forms Rn R dash 2 plus Hx Okay, again when this reacts with R dash X. So we'll get quaternary ammonia salt into this R n R dash 3 plus X minus this is quaternary ammonia salt Again, whenever Amines reacts with alkyl halide will get primary Converts into secondaries then tertiary and then quaternary ammonia salt. Okay So this again it is again the preparation method of secondary amine and tertiary amine But it is not useful because we are getting mixture of amines into it. Okay Now third reaction you write down third reaction you write down reaction with acid chloride Reaction acid chloride. What is acid chloride? We also call it as acylation acid chloride is this C double bond O Cl or acid halide is this what is the suffix for this in nomenclature we use What is suffix of acid halide? Suppose I'll give you this compound What is the name of this compound CH3? C double bond O Cl What is the name of this compound suffix is oil halide If halogen is there generally, I'm giving you it is oil halide Okay, but since chloride is there and one remember one thing we start numbing from this carbon here From the carbon of the functional groups. So two carbon is there. So we call it as what? It's an oil chloride I upsc name Fine. Yes or no Okay, so reaction with Acid halide you see One degree of mine like this RNH2 This RNH2 when reacts with supposed cl C double bond O CH3 Again, like I said that this hydrogen in this cl comes out and we get RNH C double bond O CH3 Plus at cl Okay, and the name of this compound is what what is the name of this compound? N with N alkyl group is attached. So we call it as N alkyl Acetamide Okay, co you see co ns2 is amide, right? co ns2 is Amide With one hydrogen is replaced by alkyl group. So we call it as N alkyl acetamide Or if you write down the name here, it will be N alkyl ethanamide Understood 2 degree of mine R2NH When reacts with cl C double bond O CS3 What we get into this again at cl goes out and we'll get R2N C double bond O CS3. What is the name of this compound? Yeah, N comma N di alkyl acetamide Okay, write it on the name N comma N di alkyl acetamide in all these reaction You see the hydrogen which is attached with nitrogen here takes part in the reaction And that is why tertiary amine does not react with acid chloride write down tertiary amide does not react with acid chloride third one Sorry next one you see reaction with nitrous acid nitrous acid Nitrous acid is what? HNO2 and how do we get this HNO2? If this NaNO2 reacts with HCl Then NaCl forms and we'll get HNO2 plus NaCl So in the reaction they may give you this react this Thing also NaNO2 plus HCl or they can give you directly HNO2 also Okay, any one of these they can give you so both you have to keep in mind the reaction is same Okay So what happens when this RNH2? reacts with HNO2 This gives you alcohol N2 gas evolves And H2O forms into this Okay, N2 gas evolves H2O forms into this you see this HNO2 in this It is it exists like NaO plus and OH minus NaO plus and OH minus, okay Understood So here you see one thing just which is not important what I'll tell you here when you have this RNH2 how this reaction proceeds That reacts with HNO2 So with this HNO2 what happens are I'll write here NaNO2 And HCl first this thing So what happens in this it forms RN double bond N Cl right And this we call it as Dysonium salt Okay, this we call it as Dysonium salt You must have seen this Dysonium salt in phenyl in instead of this R if you add phenyl ring benzen ring You will get that particular compound only okay, and then after this what happens This Cl comes out from this and this bond pair goes here and we get RN triple bond N positive charge here Right now from this positive charge when you When OH minus attacks onto this On to this alkyl group This bond pair is taken up by this electron deficient nitrogen atom and we get ROH Plus N2 Right, this is how the reaction proceeds So basically this is not important. This is not important in this the only thing you have to keep in mind Whatever R group you are taking here with that only the alcohol forms Okay, and in this reaction hydride and methyl shifting also possible into this Okay, because you see what happens when this This I have written in one single step, but actually what happens this take this bond pair here And then we'll get positive charge here Where this OH minus attacks, okay, so I have written it one single step But actual thing is first this N2 goes out we'll get positive charge here and then OH minus attacks Okay, so this reaction follows actually SN1 type reaction in which the carbocation forms first and then the product we get Okay, so in this you write down hydride and methyl shifting is possible note this point Hydride and methyl shift is possible into this Okay, so I'll just write down one reaction here and tell me the product CS3 CH2 CH2 NH2 now when this reacts with NaNO2 Plus HCl Or we can also write write down H NaNO2 What is the product we get here? Propen to all because here this is our group right R is this only And we are getting R OH But when you see when this reaction proceeds we'll get positive charge here on this carbon atom here you see We'll get positive charge here right so this reaction process here It is not we should not write down here, but the reaction goes like this CS3 CH2 CH2 N double bond NCl This is what here we have Now in this Cl minus goes out and N2 also goes out. So we'll get CS3 CH2 CH2 positive charge here Plus N2 right now in this what happens hydride shift And when hydride shift takes place and CS3 CH CS3 and here we have the positive sign which on reaction with OH minus it forms CS3 CH OH CH3 Okay, this is the product you must remember in this thing that Shifting is possible into this Okay, one exception you write down this reaction is important. Okay, one exception you write down into this exception you write down Methyl amine Does not give does not give alcohol In this reaction But forms but forms methyl nitrite methyl nitrite Or dimethyl ether But forms methyl nitrite or diethyl dimethyl ether. So this is important exception. We have you must keep in mind So if the reaction is this CH3 NH2 Plus HNO2 which is nothing but HONO And on reaction it gives CH3 O N double bond O Plus N2 Plus H2O or If you take two moles of this CS3 and NH2 Plus two moles of HONO It gives CS3 O CS3 Plus two N2 Plus three H2 here. We have two moles of this. Okay, this is the exception you must keep in mind experimental thing we have one more point to write down secondary amines secondary amines forms nitrosoamines Which are yellow liquid And insoluble in water. This is actually a test of amines. Okay when the amines reacts with HNO2 So primary amine if you have then in that case nitrogen gas evolves Okay, that you that is the test of amines. We have okay nitrogen gas evolves secondary amines. What happens? They forms nitrosoamines Okay, which are yellow in liquid and insoluble in water. This is the test of secondary amines yellow liquid and insoluble in water Next write down tertiary amines tertiary amines react with nitrous acid to form to form nitrite salts Which are soluble in water. So these three Fact you must keep in mind Okay, secondary amines forms yellow liquid insoluble tertiary amines forms Nitrite salt which are soluble in water. Okay soluble insoluble. You must remember Next write down next reaction is alkylation write down Aniline reacts with alkyl halide Aniline reacts with alkyl halide to form secondary and tertiary amines to form secondary and tertiary amines and quaternary salt Okay, so the reaction you write down C6H5NH2 Plus CS3I What happens? HI forms And we'll get C6H5NH CS3 This is what? 2-degree amine When this further reacts with CS3I So this forms C6H5NH2 CS3I So this forms C6H5 N CS3 whole twice This is tertiary amine 3-degree and when this again reacts with CS3I Then it forms C6H5N CS3 whole thrice plus I minus This we call it as quaternary Anilineum salt Anilineum One important reaction we have which we'll see later on also in the reaction of aniline will see a few reactions This is aniline NH2 When this reacts with aryl halide which is Cl In presence of CuCl 200 degrees Celsius we are heating Now in this the product will be This at Cl goes out and we get Di-phenyl amine Understood? Okay, in this only you write down the next one That is acylation. This is the same reaction. Okay, so That's why we are moving fast over here This is the same type of reaction acylation in which acyl halide reacts Okay, with aniline So in this only the second part of this you write down acylation acylation means if you have again aniline like C6H5 NH2 Plus ClC double bond O CH3 Now this acylation reaction takes place in presence of pyridine. Generally will take pyridine Or we can take NaOH also Pyridine or NaOH again HCl goes out into this same reaction here You see HCl goes out and we'll get C6H5 and H C double bond O CH3 N-phenyl acetamide correct N-phenyl acetamide Just a second Okay, next I write down test of amines One test of nitrous acid test that we have already done Okay, test of amines This you have to again memorize First one is Hinsberg method Hinsberg method write down it involves the reaction of amines it involves the reaction of amines With Hinsberg reagent Hinsberg reagent Which is nothing words C6H5 SO2Cl C6H5 SO2Cl Okay, this is a Hinsberg reagent this you must remember. Okay Sometimes they ask what is Hinsberg reagent not in Je other exams write on primary amines on reaction with primary amines on reaction with Benzene sulphonyl chloride this C6H5 SO2Cl Is benzene sulphonyl chloride Hinsberg reagent sulphonyl chloride primary amines on reaction with benzene sulphonyl chloride forms forms an alkyl Benzene sulphonamide forms an alkyl Benzene sulphonamide Which on treatment with alkali Which on treatment with alkali solution forms soluble alkali salt forms soluble alkali salt Now you see suppose i'm taking one degree amine. This is the test of this one degree amine is rnH2 When it reacts with benzene sulphonyl chloride C6H5 SO2Cl Can you tell me what is the product we get here? guess HCl goes out no this H and Cl This HCl goes out and it forms what? C6H5 SO2 NHR This we call it as n alkyl benzene sulphonamide Okay, it's the n alkyl benzene sulphonamide Now when this salt you put into any alkali solution suppose koH right So what happens rH2O forms into this again this H and OH minus forms H2O and we'll get C6H5 SO2 n R and K potassium This salt potassium salt this is this is soluble And this is what the test of one degree amine you'll get soluble salt it confirms It is one degree amine Okay, this is also soluble in water This salt is insoluble right So soluble amines reacts with this gets insoluble product and again when you react with koH it gets soluble this is for tertiary amine remember one thing here this is for primary amine Second reaction you see this reaction takes place because of what this hydrogen yes or no Yes Yeah, so when you take secondary amines here then what happens? Test of secondary amines is what you take secondary amine that is R2NH reacts with C6H5 SO2Cl So what it forms it forms R2N R2N O2S C6H5 right And and dialkyl benzene sulfonamide correct So this is again soluble this becomes insoluble And when you put into any alkali solutions suppose koH Since this end does not contain any hydrogen here. So this does not give any Reactions no reaction will have here conforms the presence of secondary amines Understood all these reactions you see whether it is first reaction or second reaction All these reactions takes place because of this hydrogen Okay, this hydrogen or this hydrogen tertiary amine if you see tertiary amine what happens in tertiary amine tertiary amine does not give this reaction does not give this test write down tertiary amine does not react at all Okay, next you write down the second test Yes, see See one thing is there That you cannot take the mixture of see you cannot differentiate Suppose if you take a mixture of all three kind of amines right and when the reaction takes place Then you will get then you cannot actually differentiate That is not at all. You know, you know that is that is pointless When you take a mixture of alkene in mixture of amines because one degree amines will react completely, right Soluble insoluble insoluble secondary amines forms only insoluble Salt tertiary amine does not react at all. So in that mixture all three kinds of reaction will be there Will have soluble salt will have insoluble salt and there are few amines which does not react at all So the point eventually comes at how do you differentiate that which one is reacting which one is not reacting Understood So so this kind of reaction goes like this I'll give you one particular compound and we know that this is amines Okay, and we do not know whether it is one degree is two degree or three degree amines and then we'll perform the reaction Okay, make sure of amines if you take you won't get anything Like you cannot Actually say that whether we have one degree two degree or three degree amines Because all three kinds of reaction possible. Okay. So eventually you cannot differentiate Okay, that's a thing. That's why we do not take any Make sure of amines for this kind of reaction. We'll take one by one Okay, and then we try to do this reaction and according to the observation We say the compound is one degree two degree or three degree Understood Yes, okay carbide amine reaction Next one Write down this reaction is used For the detection of primary amines this question they have asked, okay Carbide amine reaction is used only for primary amines secondary and tertiary amine Does not give this reaction secondary and tertiary amine does not give this reaction Next line primary amines on reaction with chloroform in presence of primary amines on reaction with chloroform In presence of alcoholic koH in presence of alcoholic koH Forms isocyanides isocyanides, which is nothing but carbide amine carbide amine and isocyanides are same thing Okay forms isocyanide So now you see the reaction here We have one degree amine that is r n h two reacts with chloroform the formula is c h cl three plus koH And this is alcoholic koH three moles of this This gives isocyanide that is r n c Plus three kcl And plus we'll get three h two This is alkyl isocyanides isocyanides Which has a very bad smell That only confirms the presence of primary amines this bad smell Okay, this isocyanides. We also call it as carbide amine. That's why it is carbide amine reaction Okay, third one that is just you write down the name we have done this Lieberman L i e b e r Lieberman Nitroso reaction Nitroso reaction This involves the reaction with the same reagent that I told you na n o two plus h cl Which gives h n o two Plus na cl Okay, and this we have already done. So that is the only reaction we have here Also, we are not doing the thing again. Okay Next write down the last test we have That is Hoffman mustard oil test mustard oil test Suppose you're taking one degree of mines in this Suppose we have one degree of mine r n h two And in this test it is allowed to react with carbon disulfide which is c double bond s double bond s C s two carbon disulfide And in this we get first R n h positive sign here C s minus double bond s Actually, this lone pairs attacks onto this carbon atom and this pi electron goes here Okay, now what happens in this in this proton exchange takes place. So write down here Proton exchange One of this hydrogen comes over here So we'll get r n h C double bond s double bond s double bond s h also Just a second wait, uh, or we should not have double bond here This is single bond only. Okay s h fine now now When this this actually bond pair goes here so that utilize the positive charge The final step will allow this to react with hg cl2 and the product we get here is r n double bond c double bond s plus hg s plus 2 h cl This we call it as iso thiocyanate Iso thiocyanate And the smell of this iso thiocyanate is pungent smell Pungent or we also call it as mustard oil smell Mustard oil and that's why we are calling it as mustard oil test Okay Now here you see this hydrogen will take this cl So secondary amines this is for one degree of mine you write down For secondary amine you see if you have r2 n h reacts with cs2 in short i'm writing down now It will be r n c s h Double bond s and here we have r after this when you put hg cl2 Then there will be no reaction since nitrogen does not contain any hydrogen here Okay Now for tertiary amines the first reaction does not take place at all Plus cs2 this has no reaction Tertiary amine doesn't really does not react okay So these are the test of amines Finally see we have done like most of the reaction But there are few reactions which i give you directly you should know the product into that Okay, the various reactions of aniline and what product we get There is no mechanism mechanism at all it is not required, but it is again based on the Electrophilic aromatic substitution reaction, okay So that mechanism will not do few reactions, which you need to Memorize is this write down reaction with grignard reagent write down aniline forms alkane with grignard reagent aniline forms alkane with grignard reagent reaction you see c6 h5 n h2 plus ch3 mg i That gives ch4 You see the alkene forms from the alkyl group of group of grignard reagent that you must remember ch4 plus mg i And then we have n h c6 h5 n h So this is the alkene we get that we form from the alkyl group of grignard reagent okay Oxidation of aniline you write down as it is. I am writing down. Okay. I'll just give you this table in this the reaction. I'll give you So we have n h2 here Aniline So for oxidation we can use a different oxidizing agent Different oxidizing agent and we'll get different different product also in all these cases Okay, aniline when reacts with acidified kman of 4 acidified kman of 4 or with n a2 cr2o7 n a2 cr2o7 plus c us of 4 plus dilute acid dilute acid When these reagent you use you will get you will get aniline only in this reaction but it will be Aniline black means the color will be black aniline black. We also call it as simply aniline black any lean li any Aniline black. Okay, when you use alkaline kman of 4 This you have to memorize all this reaction alkaline kman of 4 you'll get azo benzene This one is important Azo benzene is this we have a benzene ring that is c6 h5 n double bond n c6 h5 Two phenyl group attached with this n double bond n This is azo benzene two moles of this we are using because the reaction is not balanced product. You must remember When we use chromic acid What is chromic acid h2 cr o4? Right, or we also write it as na2 cr2 o7 plus concentrated h2 so4 Concentrated as to so4 In this we'll get para benzo Q non para benzo q non the structure is this Oh, sorry Just a second. Okay, non is this here. We have double bond here. We have double bond And your double bond o and double bond o this is para benzo q non Next we have sodium hypo chloride sodium This one is not important. Just you write down here. This one's Sodium hypo chloride or h ocn here. We get para amino phenol the amino phenol Okay, and that is structure we can draw easily we have phenol And at para position we have NH2 group Keros acid. What is the formula of keros acid that is h2 so5 keros acid When it reacts with keros acid we'll get nitro benzene. This one is important nitro benzene and nitroso benzene mixture of this Nitroso benzene is c6 h5 NO and nitro benzene is c6 h5 NO too With bleaching powder It gives It gives deep violet color not important Deep violet color that is it These are the few oxidation reaction of enylene in which we are using different different reagent and we'll get different different product I'm channel write on next benzene disonium chloride Benzene disonium chloride It's reaction This reaction again you have to memorize Right on the reagent we use this actually benzene right on first line here. Benzene disonium chloride We can obtain or we can prepare With the help of enylene Benzene disonium chloride we can prepare with the help of enylene and then the reagent You write down in this the reagent we use for For the purplation of this from enylene is NaNO2 plus HCl NaNO2 is sodium nitrite sodium nitrite plus HCl Or we also simply write HNO2 because when these two reacts it forms nitrous acid and an NaCl Goes out into this. Okay. So reaction here. You see We have enylene That is 6th H5 NH2 reacts with NaNO2 Just a second. Okay NaNO2 plus HCl in excess And this reaction usually takes place at very low temperature around zero to five degrees Celsius the product we get here is Benzene disonium chloride Which is ring N double bond NCl plus H2O goes out And NaCl forms Now to understand this reaction you see first of all this um Enylene That is R S6H5 NH2 reacts with HCl what it forms. Can you tell me we have done this reaction? It forms salt, right? Yes or no We'll get a salt here first, which is NH3 plus and Cl minus Now this salt you see I'll just write down the reaction like this with three hydrogen and one chlorine plus This reacts with O double bond N N OH Just a second. What is that? Hello Yeah So and these two react you see easily in this reaction you can understand that this Three hydrogen we have here this hydrogen and this OH and O Combines together so this forms H2O And this forms H2O, right? So we'll get two moles of H2O from this and the product we get here is Dysonium chloride And double bond N Cl This particular compound we also write it as like this Both has actually same thing N triple bond N plus Cl like this Cl minus Dysonium chloride salt it is Okay Done now a few reactions of Dysonium chloride, which is again Important, okay, they directly ask this question. What is a product we get into this particular reaction? Okay, so a few reactions we have which are important and that will see Reaction of Dysonium chloride. So I'll write down same as we have done just now So I'll write down here Dysonium chloride that is plus Cl minus First reaction you see when this Dysonium chloride reacts with Hx hydrogen halide Okay, and this x can be chlorine or bromine Generally, we'll take chlorine or bromine Right and the product we get here is aryl halide. That is the ring and x aryl halide here When you take this as CuCn here Right, or we can also write it as copper in presence of potassium cyanide Right both are same thing actually you'll get this I'll write down the ring now like this Here and Cn here Now these two reactions we call it as sandmere's reaction Okay, sandmere's reaction When this reacts with Ki When you heat with this Ki solution will get again aryl iodide I here Okay, very simple straightforward reactions we have hbf4 For hbf4 we use and we'll get This see these four are important here for chlorine for aryl chloride and aryl bromide We can directly use HCl and HBr For iodide will get Ki for Fluorine will get hbf4 for this also we have one reason. Okay. Fluorine is most electronegative That's why we cannot use this here in Hx or with this Ki in kf, right? So we use with this Fluoride of boron, okay boron hydrofluoride. We also call it as Hbf4 is a reagent we use for the preparation of aryl fluoride. Okay, that is again important when you use copper with Hx And when you heat this where this X is again X is again chlorine or bromine So we'll again get aryl halide X Okay, this reaction we call it as Gatterman reaction Gatterman reaction When you use this H2 with SnCl2 Sorry SnCl2 and HCl So here we get phenyl hydrazine NH NH2 This is phenyl hydrazine With benzene itself C6H6 and NaOH will get This as the product C6H5 Okay, this reaction we call it as Gomberg reaction This one is not that important just to write down with NaNO2 We heat this and we also use CuO2 here Cu2O, sorry We'll get nitrobenzene Okay, and the last one more reaction with this I'll write down here Any Acid hydrolysis if you do H plus H2O Then you will get phenol and in all these reactions N2 Comes out nitrogen gas evolves in this reaction Okay So these are the few reactions of disonium chloride which is important Just a second We have one coupling reaction is there Next coupling reaction of this Coupling reaction of disonium chloride Yeah, we'll finish this in few what is that this is 550 no we'll finish this by 615 620 Okay, so last half an hour 40 minutes we'll have problems already Okay, next right on coupling reaction only two reaction we have into this We have disonium chloride N2 plus Cl minus And when this reacts with phenol right, so phenol In basic medium The pH generally for this medium will be 9 to 10 Okay, basic medium then coupling takes place. Okay, this Two joins what happens this hydrogen this Cl comes out. Okay, and the product we get here is this N double bond N OH this is the product we get coupling reaction plus HCl comes out into this What happens if we take here? any lean NH2 This reaction takes place in acidic medium We're here. We get what the same kind of reaction and double bond N HCl goes out And we'll get NH2 here next write down electrophilic substitution reaction for any lean electrophilic substitution reaction For any lean right on the first one That is halogenation chloride that are chlorine and bromine water chlorine and bromine water reacts readily with Sorry readily with chlorine and bromine water reacts readily with any lean and forms and forms trichloro and tribromo benzene trichloro and tribromo benzene So if you have any lean NH2 reacts with three br2 and this we use in glacial glacial lactic acid Which is nothing but dry carboxylic acid. We have glacial lactic acid Then we'll get 246 I bro what I told you 246 tribromo benzene I told you 246 tribromo benzene or what aniline Yeah, where were you what I told you 246 tribromo benzene or tribromo aniline You know, huh, correct 246 try What happened ask him text him Condonia, anyways So, um, okay, so this is the reaction we have and I see this is important reaction. That's why I'm uh giving you To one more reaction. I'll give up after this will get different product into that one note you write down into this Simply aniline if you are using Then we are getting 246 tribromo aniline. Okay after this one reaction. I'll give you first you write down one note in chlorination Right down in chlorination water free solvent such as in chlorination Water free solvent such as chloroform should be used Otherwise oxidation takes place Okay, aniline will get oxidized then if water will be present over there So in case of chlorination we use chloroform water free solvent Okay, that is one important thing that you must keep in mind Otherwise oxidation takes place Okay, next write down next line Note I have finished next time write down monochloro Or monobromo derivative monochloro or monobromo derivative Of aniline can be prepared aniline can be prepared If ns2 group is first preserved by Acetylation if ns2 group is first preserved by Acetylation on account of oxidation of ns2 So what happens you see a little bit of difference we have and we'll get different simply different product If you have aniline if you do direct bromination of this Then we'll get 246 tribromo aniline right But when you have to form monobromo derivative or monochloro derivative Then first first of all You have to preserve this ns2 group to get oxidized and for that purpose We use acetyl chloride here CS3 COCl we can use Or we can also use anhydride CS3CO Whole twice O So when you use this the product that you get here will be an H Then we have C double bond O CS3 right Correct Yeah If you use CS3 COCl then another product will be HCl If you use this then another product will be CS3 COOH fine Right now because of this C double bond O group here electron withdrawing group the oxidation of this part is difficult now That is what we use for to preserve the oxidation of ns2 Now this what we do this we allow to react with suppose bromo derivative product We have to form so this will allow to react with br2 with CS3 COOH babozoic acid Then we'll get what ortho and para substituted product so this Position we have the same product that is NH CO CS3 And here we'll get bromine Plus para substituted bromine will get here Which is nothing but NH CO CS3 and bromine here Now when this reaction this product allowed to acidic hydrolysis Right then H plus will attach onto this nitrogen and this group will come out as it is right. So we'll get here The product that we get here is NH2 we get here is NH2 br Plus again we get NH2 and br This is what the product we have so if you have to form mono derivative product Right mono bromo or monoculor derivative product Then first we have to preserve the oxidation of NH2 and that we can do by acetylation or by reaction with anhydride understood Para will be the major product into this This one is the major product one note. You write down halogenation of any lean halogenation of any lean is faster in polar solvent And slower in non-polar solvent faster in polar solvent And slower in non-polar solvent sulfonation second one write down the reaction NH2 Plus H2SO4 giving sulfuric acid conch also we can use And we'll get para substituted product as the major product here NH2 and SO3 H here this we call it as sulfanilic acid Is nitration. This is also the same reaction as we have done for halogenation NH2 first of all we use CH3 COCl Or we can use CH3 C whole twice O And the product we'll get is NHCOCH3 And when this reacts with conch HNO3 And conch H2SO4 Both are concentrated See if you use this directly Okay, then we'll get 246 trinitroaniline NO2 NO2 NO2 will have here, but when we preserve this NH here NH2 here then we'll get ortho and para substituted product That will be NH COCH3 NO2 Plus para substituted product Which is NH COCH3 NO2 this one is the major product we have Now when you do the acid hydrolysis of this H plus H2 You will get Para nitroaniline and ortho nitroaniline NH2 here NO2 Plus We'll get here NO2 And NH2 para will be the major one. Okay. These are a few important reactions of aniline Next write down basicity of aniline Okay, write down basicity of aniline See uh eline is Benzene ring with NH2 so basicity is what tendency to lose electron Right, so any electron withdrawing group if it is attached onto this ring Any electron withdrawing group that will Decrease the basicity right Yes, or no Because electron withdrawing group will withdraw electron and then tendency of tendency to lose lone pair of electron will be less correct And any electron releasing group will be directly proportional to The basicity fine or not if I write down few examples say we have NH2 then we have NH2 NO2 NH2 NO2 And the last one we have NH2 NO2 Tell me the Basity order of this second one NH2 and then we have NH2 CH3 NH2 CH3 CH3 What is the basicity order? Yes, first one is right. What about the second one? Okay, see the first one is correct. The order is this Because NH2 or what? NO2 are electron Withdrawing group EWG EWG right These are electron withdrawing group. So electron withdrawing group will decrease the basicity fine So obviously this aniline is the most basic Then we have meta then we have para and then we have ortho because the meta position will have the least effect onto this Because NH2 is what? NH2 is ortho-paradirecting group Yes or no? NH2 is ortho-paradirecting group. So any group that is attached to ortho or para position the effect of that group will be more Whether it is decreasing basicity effect or increasing in basicity effect, whatever it is But the effect will be more when it is at ortho and para position But that is also not exactly true because this is what Here when ortho-paration is there There will have ortho effect takes place Okay, so because of ortho effect the basicity of aniline Decreases regardless of the fact Whatever the nature of this compound we have Whatever the nature of this group we have whether it is electron releasing or electron withdrawing So one note you write down first. You must keep this in mind that ortho substituted aniline ortho substituted aniline are weaker base are weaker base then aniline itself Regardless of the nature In bracket you write down after nature Regardless of the nature in bracket you write down EWG or ERG Regardless of the nature EWG ERG of the group. This is known as ortho effect This is known as ortho effect understood So because of ortho effect Ortho substituted aniline is always weak base, right? So here in this case you see CS3 are electron releasing group So electron releasing group will obviously increase the basicity of aniline But this one is ortho substituted So this one is the least basic fourth one is this Right, obviously these two have electron releasing group attached with it and we know electron releasing group increases the basicity of Aniline so obviously these two will be at first and second position This will be third Yes, and this one will have what plus i effect we have to see so plus i effect is maximum for CS3 is electron releasing, right? So we have a two position meta and para position Okay, fine is your ortho para directing group No, and this too is the ortho para directing obviously. So this first one will be this Second one will be this third is this and fourth Right para substituted aniline if The substitution is electron releasing group substituted group is electron releasing group then that will be most basic understood Yes Okay, one one more point you write down here Aniline are also weaker base then Amines Aniline are also Weaker base then aliphatic amines due to resonance few more example you see There's two more Tell me the order Three greater than two than one Why the condonia? No, we are talking about the basic nature Okay, so this will be because these group are what these group are having plus i nature Because of plus i nature electron density here Increases and hence the density already electron will be more so order will be this Similarly for this one One greater than two how it is one First of all this ccs five or what electron releasing electron withdrawing group, right? These are electron withdrawing group electron withdrawing group. Okay So obviously the electron density on nitrogen decreases So obviously the first one will be this second one will be This because only one ccs five and third one will be this Understood Okay, now last two things we'll see the last two things of this chapter See, uh, there are few reactions of isocyanides and cyanides. Okay The very important reaction of isocyanides and cyanides that we have that I give you Okay other reactions you can go through in the book. Okay. Suppose you have isocyanides is r and c r and c This is isocyanides When you do the acidic hydrolysis of this r and c isocyanides Then it forms amines one degree of mine r n h 2 plus h c o o h When you take isocyanides if you take cyanides that is r c n And when you do The hydrolysis of this write down h2o in alkaline h2o2 It gives you r c o n h 2 first acid amide Amide you get first and then if you do the acidic hydrolysis of this You will get acid r c o h Plus n h 3 so you see two difference we have here ammonia gas evolves acid forms and here amines we get into this Isocyanides gives you amines cyanide gives you acid that is one important reaction We have that you must remember reduction you see isocyanide reduction r n double bond c plus hydrogen And for reduction we use na with c2 h5 o h Under reduction with nickel platinum catalyst we can use It gives r n h c h 3 2 degree amine we get here But if you have cyanides r c n And when you do the reduction of this With li al h4 You get one degree of mine r n h 2 So there's a few difference we have those are important and question has been asked on these two Okay, there are some more reactions of cyanide and isocyanides are given in the book that you can go through but those are not that much important Okay, so this is it for amine's chapter Now, what do you want to solve? Did you revise iNIC? Any 11 chapter you want? Just a second see this question You have to find out the final product in this We have phenol which reacts with c h c l 3 k o h Which gives a and then 50 percent k o h b What is a and b or if only b you tell me? Yeah, we have Final product these are the options option d What about you condonia? why See first of all phenol reacts with c h c l 3 and k o h okay When phenol reacts with chloroform with k o h it gives salicyl aldehyde Okay So the first product a that you get here will be this o h here and c h o This is a product a and when this aldehyde goes under 50 percent k o h solution Then this reaction we call it as what can is that our reaction? right For can is that our reaction oxidation and reduction takes place Right disproportionation reaction it is this c h o will get oxidized first and then we'll get reduced also So one product will be c o h here other will be salt of this c o k So product will be this annual condensation we use 50 percent any o h But for that the condition is we have alpha hydrogen required Alpha hydrogen condition is there. So this is can is our reaction. Okay, so this gives you oxidation and reduction of the same compound Okay, two moles of this will take if you remember if you don't don't remember then revise this reaction. Okay Two moles of this reagent we take one mole will get oxidized other one will get reduced So oxidize gives you salt of this and reduction gives you alcohol Fine Yes or no c 6 h 6 h 2 s o 4 350 kelvin a Then fusion with alkali Then bromine water vr h 2 c What is the final product c Three options are given c or to bromophenol para bromophenol meta bromophenol and last one is 246 tribromophenol ortho para meta And tribromo 246 tribromophenol Yeah, it's correct then first of all First first for what we get benzene sulphonic acid A will be what benzene when reacts with concentrated or h 2 s o 4 will get s o 3 h here benzene sulphonic acid and when it's reacts with alkali Solution right naOH means what fusion when takes place. It means we are taking naOH with h plus h2o and this gives you h 2 s o 4 comes out into this and will get phenol in this product as to s o 4 will go out And we'll take phenol and now when phenol this reacts with vr2 and h2o Then all these vr will attach here here and here 246 tribromophenol. These are direct questions understood acidity what happens ramsel of ortho para and meta nitro phenol you tell me acidity order of this ortho para and meta nitro phenol Okay, tell me first one thing just a second You are saying this Condonia, what is your answer ortho and then meta What about yours ramsel ortho Then para and then meta you see Nitro phenol is this We know Na2 group is what Na2 group is the electron with drawing group And electron with drawing group increases the acidity of phenol Yes So now here you see the effect of Na2 will be more when it is placed at ortho and para position fine Because this group is again what ortho para directing Yes effect of effect of Na2 will be more when it is placed in ortho and para position So meta position will be least acidity that is for sure. Yes or no. So this is correct Right, but when Na2 is placed at ortho position there will be intramolecular hydrogen bonding So because of intramolecular hydrogen bonding its acidity Decreases of it. Okay. So here the para para becomes More acidic than ortho. So correct order will be this Para is not always more stable In case of amine's reaction para is more stable Your other reaction you'll see ortho is also more stable according to the product What tell me The one more question you see strength of acid Bitsat this question has been asked in bitsat. First we have phenol and the next one we have OH here and CS3 NO2 OH acidity order, what about you? Condonia 341 to viso. So correct order will be 341 and then 2 Correct now next question. I am sending you on a whatsapp. Okay, because I have to write down so much then. Okay So I'm sending you on whatsapp. You just see that question and tell me the answer This question was asked in jay advance No structural change is made to x Read the question properly condonia why a First you tell me in which of these are You know How this reaction takes place? Can you tell me that? That compound which is given With acid with presence of acid how this reaction goes why so Which is not released? That is what See if you if you try to understand the mechanism of reaction in this how reaction may go Then probably you can answer this question if you have acid, right? What does this acid do it produce? What? H plus so that h plus is taken by taken up by The oxygen which is there in the ether Or that we have See suppose we have one phenyl group here Another one here another one here And this we have or right H plus from the acid will get attacked by the lone pair of this oxygen And we get what here we get this OH R and plus correct What happened next this bond will break and we'll get one carbocation here plus R OH This is what happened, right? And then some OH minus will attack onto this and we'll get the product So basically we have to see in which of these given options this carbocation will be stable That is what the question Yeah, it is stable, but if if there is any possibility you see what is the question you see It is a third degree carbon is stable, but what is the question? Acetic hydrolysis of either Sean below is fastest when Any of these given option will stabilize this carbocation or not that we have to find out The option C. Yes Because if this any of these phenyl group if you replace by methoxy phenyl group, which is nothing but this Sorry Etoxy phenyl group is this only know we have OCS three here, right? Now this OCS three is what electron releasing group So this electron releasing group is obviously stabilize this carbocation. Yes or no Fine So first option is not possible because if you place here CS three that is electron releasing group that will also stabilize right But we have CS three over there only Okay, second option you see we have a paramethoxy phenyl paramethoxy phenyl is this that will also stabilize this carbocation Fine third option two phenyl group are replaced means we have one more phenyl group here Which is replaced by paramethoxy phenyl group. So in this case, obviously this carbocation is most stable and the reaction will be fastest understood So this is how you see the question they ask Okay, j advance question Next one you see HBR Tell me the product phenol and CS three BR. Why so? So actually what happens first of all this x plus will come over here Right And we'll have positive charge with this Now to supply this oxygen first positive charge on this oxygen either this bond or this bond will break Right. So when this breaks, this is phenyl cation, which is highly unstable fine So this bond will break and we'll get phenol Plus CS three BR This is the answer understood So this question was asked in j means j means Tripoli, which is nothing but j means Another one you see bits that question It is given that ethanol C2H5 OH The boiling point of ethanol is 78 degrees Celsius. What is the boiling point of diethyl ether C2H5O and C2H5O options are given hundred 78 86 And 34 degrees Celsius These are the four options What is 34? Why? What is the molecular weight of diethyl ether and ethanol ethanol? Yeah, so when boiling point when molecular mass increases boiling point boiling point also increases Right When molecular mass increases boiling point also increases But here in this diethyl ether, what happens? There is Intermolecular hydrogen bonding takes place in diethyl ether intermolecular No, sorry uh We have hydrogen bonding in this. No, okay. Fine. We have hydrogen bonding in ethanol, right? Intermolecular hydrogen bonding present in ethanol and because of that its boiling point decreases Because of intermolecular hydrogen bonding boiling point decreases, right? Two things you keep in mind when even you have intermolecular hydrogen bonding boiling point decreases and melting point increases No hydrogen See when when hydrogen bonding we have two types intermolecular and intramolecular So because of intermolecular hydrogen bonding Because of intermolecular hydrogen bonding the boiling point and melting point will be more than the compounds having no hydrogen bonding So there is no hydrogen bonding in diethyl ether Okay, so due to intermolecular hydrogen bonding ethyl alcohol has Has obviously intermolecular hydrogen bonding and hence its boiling point Is more correct boiling point of ethyl alcohol will be more than diethyl ether So obviously the boiling point of diethyl ether should be less than this So answer will be 34. See all these things you can assume but if you see Obviously you see one thing you cannot memorize Yeah, ha logically also it's 34 Like logically also and you can easily look at the option and say because you see one thing you keep in mind boiling point you cannot memorize of any compound fine Because there are so many compounds So when they ask you like this question for one compound the boiling point is given and other one they are Calculating it. They're asking you to find it So obviously it should be either less than this or more than this Fine, so when you see the option there are two options which are more than 78 degrees Celsius So if this is true, this may also be true Why because they know that you cannot memorize boiling point of any molecule So they won't ask you this question Understanding this fact that that you know the boiling point of Few molecules. No, you cannot memorize the boiling point of any molecules So they ask you either it is more than or less than so there are two options Which are more than 78 degrees Celsius. So obviously these two are not correct So answer will be this only if you do not know anything then also Basic common sense you can apply You understand what I'm trying to make you understand Boiling point against you see boiling point you cannot memorize that the examiner also know So they won't ask you such questions If they ask you in this they will simply give you suppose if then they'll give you the boiling point of diethyl ether is suppose One question if I give you suppose in this only they give the boiling point of diethyl ether is 86 degrees Celsius Okay, then what is the boiling point of ethanol c2h5 OH and these are the four options given What will be your answer? Diethyl ether is given 86 degree Celsius Ethanol you have to find out and the options are given These are the four options 100 because there is only one option which is more than 86 Leave that fact that whether that the boiling point is more than or less than that you let it be You you just look at the option There are only one option which is more than 86 and there are two options which is less than 86 So obviously if the boiling point of c2h5 is less that is not possible because there are two options here So only one option is correct that is 100 degrees Celsius You understand my point that they won't ask you they don't they know this thing that they you cannot memorize boiling point of any compound So whatever value if it is given your answer will be either less than this or more than this depending on the option If only one option is there which is more than the given value then then then that will be the answer If only one option is given which is less than the given value then that will be the answer Understood Okay, another one you see this one we have done Here you see this j This one you see ch OH cs3 cs3 p plus br2 Gives you Oh, it's not given here then na Gives you x x you have to find out. Yeah You want option also options is this we have ch CH2 cs3 cs3 cs3 Then we have c cs3 cs3 cs3 cs3 All these are methi then we have cs3 cs3 ch cs3 cs3 ch2 ch2 cs3 Last question this is Yeah, why? First of all we'll get here we'll get here alkyl halide right bromine will get here And then na so these are wood's reactions will get two molecules of this will attach this answer Like you see like I told you know that Organic chemistry whether you know reaction or not with option they'll give you the idea that what could be the answer I told you this thing. That's why you see when I don't give you an option You feel like okay, there's you cannot understand what what is the product we should have or we could have into this But when I give you the option you can understand then okay This could be the product and then the reaction may go like this And then you actually understand that how the reaction is going on So that's why organic you see it is a bit difficult to memorize All the reactions again, but if you go through once properly With option you will have the idea that what is the reaction? What should be the product? Okay, that's why it is a bit easier also at the same time Okay, anyways, so like we'll wind up the class here only okay, so Sunday, uh, I will let you know tomorrow. What timing okay sunday. We will be doing a metallurgy right and Then next class biomolecules and then again half an hour 40 minutes will solve some questions. Okay Okay, so we have next letter 7 30. It's already seven almost seven. Okay. So take care. Bye. Bye