 Hello friends today. We are going to discuss the question which has been asked in recent J main examination Before starting the discussion here. Let me remind you these questions are Not those questions which has been disclosed officially by J. Okay. These are the memory based questions. Okay, so few You know few questions you may get you may get you know different over here since these are the Memory based question. We have correct. I have taken this question From shift one of date 91 to 129 2019, right? So the first question we have arranged the following in order of kb value right So we have to arrange these you know given molecule In according to the kb value. Now. What is this kb value means? Okay, what we can say if one molecule will have more kb value, right? So kb value is nothing but the Basisity of the molecule Correct more kb value means The molecule is more basic Okay, so kb is directly proportional to basicity And it is inversely proportional to pk b Okay, so for all these molecule pq and r we have to find out the basic order Right, so first of all you see the first compound. We have this p which is spirited in Okay so the What is the basic nature of these compounds? How do we define basic nature? So basicity is nothing but the tendency to lose lone pair of electron. Okay Tendency to lose Lone pair of electrons Right, so all these nitrogen atom will have one one lone pair In the given molecule here. Also. We have one lone pair here. Also. We have one lone pair Okay, so each of these molecule has maximum tendency to donate its lone pair of electron The molecule which has the maximum tendency to donate its lone pair of electron will have maximum basicity Correct. So first of all, you see q that you have option you see Uh in one of the option q has the minimum basic order Right minimum basicity minimum kb value we can say correct So if I decide this or if I No draw this conclusion that q will have maximum or minimum basic order then We can directly Solve this question. Okay to minimize the time. Okay. So since only one option has minimum q value Minimum q has the basic nature. So we'll first check q right So you see this molecule is what the q that is given over here This molecule is aromatic in nature. This lone pair is involved in resonance lone pair This lone pair is delocalized delocalized with the pi bond of the ring Of the ring So it has six pi electrons. So aromatic in nature since lone pair is involved in resonance So obviously this lone pair is not available here Right, so its basic nature will not be will not that strong. Okay. Now when you talk about this p This lone pair is there on the nitrogen atom, but it is not involved in resonance. Okay This lone pair Is not involved in resonance because that p orbital which is perpendicular to the plane of this ring. Okay. So this lone pair is not involved in the Resonance It is not delocalized. We can also say Right when it is not involved in resonance. So it is available on the nitrogen atom. So obviously this p Will have one lone pair this nitrogen will have one lone pair on it and hence the basicity of p will be more than to that of q Right, the basicity of p will be more than to that of q. So with this fact you can easily eliminate option 2 and option 4 correct Now you see the third one that is r the lone pair is there but Obviously, it is not a conjugated system. So resonance is not possible here. Right here. Also lone pair is available on the nitrogen atom Here also lone pair is available Now we compare the you know tendency to donate lone pair of p and r over here This high nitrogen This nitrogen Is sp2 hybridized Right and this nitrogen is sp3 hybridized We know more s-corrector will have more electronegativity And that's why the electronegativity order is this sp hybridization is maximum Then we have sp2 and then we have sp3. So when electronegativity is more tendency to lose lone pair of electron Will be less. Okay. So this nitrogen is More electronegative than this nitrogen because of its hybridization. Okay, hence Since it is more electronegative So tendency for this nitrogen to lose this lone pair of electron will be lesser than to this nitrogen Correct. That's why the r is more basic than p and q is the least basic compound We have right. That's why you see when you know, this q is the least basic acid only one option is given and hence option 3 is right Okay, answer will be option 3 next question number 2. Okay, the ring reacts with br2 and then et over to ethyl alcohol Okay, so first of all you see that with this br2 we have addition reaction Correct. And this addition reaction gives you what the anti addition of bromine atom on this double bonded carbon atom So if I write down the reaction here, so in this you see what happens Addition reaction takes place and what kind of addition anti addition of bromine atom Okay, so here we have one bromine atom from The round and other bromine atom is coming Out of the plane towards the absorber here. We have the another bromine atom This ring will be as it is This is the anti idea Now the second molecule the second reaction we have and that is et OH ethyl ethyl C2H5 OH So what happens one of this bromine atom will come out as a leaving group and forms positive charge Carbocation over here. Okay, so you see when this bromine atom comes out as a leaving group Here we have positive charge and here we'll get positive if this if this bromine atom comes out Which bromine atom will comes out that depends on that depends on which carbocation will be more stable, right? So this reaction is nothing, but it is the sn1 reaction we have here The nature of this reaction is first order nucleophilic substitution reaction. What is nucleophile here? C2H5 O minus and this is H plus right. So when this H plus This H plus combines with one of this br atom the question is which br atom comes out Okay, if this br atom comes out, we'll get OC2H5 here Right and the answer will be option two Otherwise if this bromine atom comes out, then the answer will be option one. You can easily eliminate Sorry option three then because we have here we have bromine only so you can easily eliminate option one Option one you can eliminate and you can also eliminate option four Okay, now coming back to this question here You see if this bromine atom comes out through sn1 reaction, which is followed up by the formation of carbocation so this carbocation is Stabilized through this molecule right resonance is stabilizing this hence this carbocation here It is more stable and this bromine atom comes out. So the final product here you will get is This bromine will be as it is and here we have OET Hence the answer will be option three right Next question are in the following in order of k value. Okay, so k value is what? Like kb k is also directly proportional To acidity and inversely proportional to pKa value, right? Now, what is this acidic acidity we have? What is acidic nature? It is the tendency to lose Tendency to lose H plus ion Okay, H plus ion and this thing is possible when we have any electron with drawing group attached Right, which is nothing but which which can show minus i or minus r Okay, any of these nature of group will facilitate the removal of H plus and hence the acidity will be more right So the following order of k value means we have to arrange these molecules in the increasing acidic order Fine So now you see we have first of all we have CH2 C double bond O OH And here we have one fluorine attached CH2 C double bond O OH fluorine attached CH2 C double bond O OH and O2 attached here CH2 C double bond OH C and attached Okay So first of all all these atom which is attached. What is the difference in all this molecule? We have difference over here. You see Difference in the atoms or group that has been attached with this Here we have fluorine fluorine bromine Sorry, NO2 and then CL So here you see all these atoms will show minus i effect because there is no conjugation Correct. Now when you talk about minus i effect So fluorine and fluorine easily you can decide the minus i of fluorine Is more than to that of fluorine because of more electro negativity. So obviously this is p and this is q Right. So p will be more acidic than q Fine. So first option you see p is more acidic than q p is more acidic than q p is more So this option we can eliminate option four Correct. p is more acidic than q now when you compare the You know the minus i effect of NO2 and CL Okay, so here this nitrogen is directly attached to this carbon atom Here this carbon is directly attached to this nitrogen atom But what happens here in the c and we have c triple bond n So since nitrogen is more electronegative So this will withdraw electron towards the side then this carbon becomes slightly positive This carbon will also withdraw electron from this bond pair, right? So since nitrogen here it is actually attached to the carbon atom here. We have carbon. So this one is more Uh, you know shows more minus i than cn Correct. This one will show more minus i than cn. So what we can write? This is r and this is s we have so r also will have more acidic than s Okay That's more than this foreign one is there third one is there for so we cannot differentiate this. Okay, so when I compare Now we have to compare what we have to compare this rodin and NO2 Okay, so electron with drawing nature of rodin and NO2 if you see Any two will have the maximum electron withdrawing, you know effect electron withdrawing nature Okay, so that's why this molecule is the most acidic molecule. We have so r will have the maximum one Correct. So when r is maximum so we can eliminate option one also 2 and 3 if you compare s and p Where is this s s is the last one and p is the first one, right? So s is obviously this cyanide group, which is again more electronegative or it has more electron withdrawing power Then fluorine, right? So s also will have more acidic nature than t and hence option Third is also not true. Second one is the answer we have Okay, so option two is right Next question is the presence of If will be unsuitable for drinking. Okay, this is actually a factual question Okay You cannot Put any logic into this it is a factual question and what Is the data you should have to solve this question? Okay, it is Given in ncrt actually the presence of manganese Right of concentration Right if manganese concentration Is greater than or equal to 0.05 ppm then The water is Water is unsuitable for drinking unsuitable for Drinking hence you see option third is right over here. So this data you should have it is Given in ncrt probably Okay, so option three three is right Next Which of the following is the strongest acid? Okay, so how do we compare the strongest acid? Suppose if I take any example of ha If you have to compare the you know the acidic nature of this What we do will form the this by removal of h plus ion So once you remove one h plus ion you'll get a minus and h plus So this is what this is a conjugate base of this acid So we can say a minus and ha a minus And ha or conjugate acid base pair Second thing is what if this conjugate base is stable, right? If this conjugate base is stable The tendency to lose h plus ion will be more and the stability of conjugate base is directly proportional to the acidity of that particular compound Correct. So now when you see The conjugate base of ch cl3 This is nothing but C cl3 minus Conjugate base of ch i3 ci3 minus Conjugate base of ch br3 C br3 minus Conjugate base of ch cn3 c cn3 minus Now This base will be stable when we have any electron withdrawing group electron releasing group if it is attached to this carbon atom That will increase the electron density here on to this carbon atom and hence the basicity or sorry And hence the stability will decrease Correct to stabilize this conjugate base. We must require some electron withdrawing group all these You know all these Group or atom that you have if you consider chlorine iodine bromine and cn cyanide All these atoms or groups are electron withdrawing in nature But in all these electron withdrawing group Cn or cyanide group has the maximum electron withdrawing power Right. So because of this maximum electron withdrawing power of cyanide Its basicity is sorry its stability is the maximum and hence the acidity of ch cn3 Will be maximum in all the four. That's why the answer will be option four, okay Next question match the following drugs with correct functional group test, okay chloro chloro xylenol, okay chloro xylenol if you See the structure of this chloro xylenol. I'll draw it here over here Then we have penicillin Penicillin the structure is this This is penicillin The next one is sulfur pyridine Sulfur pyridine the structure is this This is sulfur pyridine and the last one we have That is Not ethyl drone. This is structure is a bit, you know complex We have double bond here double bond O another ring And with this we have again another ring attached in this one Okay, here we have C triple bond ch Sorry We have OH here C triple bond ch Here we have CH3 and all other places we have hydrogen. So these are the structure which is given um In the question, okay, chloro xylenol penicillin and all, okay So the question is match the following drugs with correct functional group Test, okay, so carbyl amine test is what it is a test of one degree amine Okay, if you remember carbyl amine test is used for one degree amine Okay, and that is why One degree amine In this what we do we just mix this Amine with chloroform and kowet and we get alkyl isocyanide Okay, which follows alpha elimination reaction. Okay So all these you know facts Are given in the book the only thing we have to keep in mind is what carbyl amine test is used for One degree amine, right? So one degree amine where we have in this structure. Okay One degree amine we have We should have what The p we didn't write because this is the one degree amine. So p should be c This is the first thing we have so where p is c given all these options will have c sp It's the first option is not correct right Now the second one is what bears reagent What is bears reagent bears reagent is used For you know the you know the detection of double or triple bond any any kind of unsaturation. Okay So for that we use alkaline k mn of 4 cold alkaline k mn of 4 we use and the color fades from Pink or in some book it isn't pink directly or more specific if you want to say it is purplish pink So the color purplish pink changes to brown. Okay, and sin addition takes place into this So bear reagent the reagent is what cold alkaline k mn of 4 and it is used for the Reduction of any unsaturation. Correct. So Where we have unsaturation present in all this molecule? We have this here the presence of alkyne. We have correct, which is nothing but not a thin drone Correct. So d or the bears in q should be d. Correct. So this d should be q The second one Where d is q given. Okay. So this fourth option is not correct now Okay, so we are left with only two and three So next is what fe cl3 test fe cl3 test is what it is a test which we which we do To detection of the presence of phenol right for phenol. We do this Ph enol, okay If you know this if you have this information for phenol official the test has been done So it is only possible in case of chloro xylenol. Okay, so this a must be what? A must be r one we have a must be r. So where a is r You see this a option is right Second one is also not true. So option three is correct here Right option three is right. So these are the answers Question number seven the reaction we have to find out what is x and y Right, suppose we have this benzene ring and the reagent we are using here is what H C double bond OH Plus HCl So what would be the product here? Okay, so see this molecule here Here the product That we get is CH2 Cl with OH What happens this bond pair? Pi electron goes over here and this will take H plus from here Here we have the positive charge and that will get attacked by this Cl minus. So we'll get this Okay, in this question you see the benzene ring that we have this benzene ring this pi bond Will attack onto this carbon atom back onto this carbon atom And this OH group comes out as leaving group. This is OH minus. Okay The product we get here is this Double bond here. We have CH2 Cl Here we have positive charge and this H will be as it is Correct. Now to gain aromaticity again, this H will come out as H plus And this bond pair comes back into the ring. So we'll get a benzene ring With CH2 Cl here Right plus H2 will get H plus and OH minus combines will get H2 Now this reaction is simple now AgCN we have so what happens this AgCl comes out and Cl attached over here X will be what X will be ph phenyl with CH2 Cl attached with it Okay, the first product is this one. This is nothing but X And when this react with AgCN AgCl forms the precipitate and we'll get CH2 CN here the product will be option Here that catches what in we have actually two leaving group here one is chlorine and another one is OH Okay, so which of this group will go out that is the you know The you know main thing over here we have So since oxygen is more electronegative here, so this bond is comparatively weaker Okay, and negative charge on oxygen is more stable than chlorine That's why this OH minus will go out and we'll get CH2 CN correct answer will be option a first option will have the answer Question number eight RCN with isobutyl alcoholic we have here Okay, so it is a direct reaction the CN will convert into This CN will convert into something is missing here Because they have written the answer directly here. The question is this I'll just make the correction over here. The question is this Here we have X And we have to find out what is X? Okay, so this is a direct reaction NH3 will go out into this and we'll get aldehyde into this Okay, okay. So this CN converts into CHO. So the final answer will be our CHO. It is a direct reaction Okay, this is di isobutyl aluminium hydride. It's a direct reaction. Okay, so answer will be option a That is the following amino acids in order of their pka order. Okay lysine aspartic acid Arginine and glycine Okay, so the PI value or isoelectric point Of all these amino acid I'll give you right so the PI value of lysine PI value of lysine is 9.8 isoelectric point Arginine it is It is 10.8 Aspartic acid aspartic acid It is three And for glycine glycine It is 6.0 Right, so these are the PI value right isoelectric point if this value you should you know memorize Okay, these are also this is also the factual question. We have few volume You should have that glycine and aspartic acid you should know lysine also We have discussed the class it is given in ncrt also So with this if you know one or two values also you can you know compare the order here So we have to find out pka value pka value and pi value are directly proportional according to according to this we can Solve this answer right so more PI value more will be the pka value of that Okay, so we'll have it as maximum for arginine then lysine and then lysine and aspartic acid Okay, so the answer will be option for arginine lysine lysine and aspartic acid Correct fourth option is correct over here The next question write the product of given reaction correct so what happens if this molecule which is The structure given in the question Is this We have a ring one bromine attached over here And then This is a compound we have So when you when this molecule reacts with koH Okay, so when this molecule reacts with koH then what happens Okay So obviously we know this kbr comes out and OH will attach but which kbr will come out That's the question Okay, suppose when this br comes out then this positive charge is highly unstable Okay positive charge on benzen ring is not stable. Okay directly on the benzen ring is not stable So if these two you'll compare so this here the positive charge is comparably more stable And hence the br comes out from this carbon the product here we get is this kbr forms And here we have OH group Okay, now this is one degree alcohol Which on oxidation with this Which on oxidation with this forms what one degree alcohol converts into ld height first and then it converts into acid Okay, so this converts into acid with equal number of carbon atom Okay, it is not partial oxidation is not given if partial oxidation is there then we'll get what Then we'll get ld height into that So we'll get acid with equal number of carbon atom same number of carbon atom now with this acid is allowed to react with H2SO4 in acidic medium. We are hitting this acid H2SO4 we are hitting this so this H2SO4 will give H plus and that H plus From the acid this will this H plus will get attacked by the lone pair of this oxygen atom here So we'll get OH2 plus here and finally this OH2 plus this oxygen is electronegative element positive charge on it It is highly unstable. So this will draw this bond pair of electron This bond will dissociate H2O comes out and we'll get the carbocation here So the product one like the intermediate product we'll get here is bromine here will be as it is And then double bond O with positive charge on it Okay Now this positive charge what happens since H plus comes out from this. So this positive charge will attack on will get attacked by the Pi bond pi electron over here if I draw the structure here, you see This will get attacked by this pi bond So what happens you see We have a ring Double bond as it is double bond O here We have the positive charge and this H will be as it is This bromine will be there now finally what happens H plus attacks onto this HSO4 minus is there So from this to gain aromaticity H plus will also come out Okay, this H plus like this bond will lose this electron to gain the aromaticity again and the ring will be like this Br will be as it is plus This H plus will combines with this HSO4 minus and forms H2 SO4. Okay. So the answer here it will be option One right, this is the option will have bromine one two second carbon. So bromine one two second carbon first carbon Not possible. Okay. So first option is right over here correct So these are the first 10 question we have discussed For shift one of 9th of January 2019. Okay. We'll come soon with the next 10 question again. Thank you