 Hello, good morning, so all of you are prepared for the exam last I think last 10 to 10 days are left right? Okay, so before I start this before like we start this session, how many of you are doing inorganic chemistry, P-Blog and D-Blog, P-Blog, D-Blog, S-Blog? Have you finished NCRT and previous year questions? See you have last I think 10 days left so you must do at least half an hour or one hour inorganic chemistry because there is no learning into that maybe some point or fact that you are starting nowadays maybe they ask directly into the exam okay so whatever you are doing do that but don't leave inorganic chemistry these days okay at least half an hour 45 minutes or one hour maximum whatever chapter you like whatever chapter you feel you must go through okay and must see the suppose you are studying theory for like half an hour then must see the 10 15 minutes at least 20 30 problems every day okay it takes hardly for 20 problems it takes hardly around 10 minutes 10 15 minutes not more than that okay so you don't believe this thing okay these these days are very important for inorganic chemistry don't leave that okay and next thing okay next thing of physical chemistry you have done I think 12th class you have done recently what about 11th class have you done properly atomic structure various formula you have revised because in atomic structure you will get from generally you will get formula based questions so formulas you have revised radius velocity total energy all these okay good mathleys singed if you have finished it's fine okay atomic structure you must see because there are very there's so much formula there see more concept you don't put so much time on more concept basic concepts you have only you see the previous year questions like 10 15 good questions you see and that is it because more concept is a chapter like if you can do that particular questions from that particular chapter if you can do you will do it in a minute okay otherwise you keep on doing for five minutes you will not get the answer okay so you must must take care of these things don't waste your time in any of the questions in the exam and more concept don't solve 50 or 100 problems 10 15 selective problems you solve in more ones okay more concept when I say it includes both more concept to that equivalent weight concept in all and this thing right and you must remember with equivalent weight concept you will have questions in electrochemistry also that is Faraday's law of electrolysis to that also you cover okay don't leave that okay redox reaction only few specific example of like the redox oxidation state you see and then one or two questions to balance in acidic medium and basic medium that is it for it off so you will not get much much much question into that okay chemical bonding have you revised you must know how to calculate hybridization and all right bond order hybridization all these things you know hybridization there is one section in coordination compound also in coordination compound let me tell you don't leave that optical isomerism part at least twice you must revise optical isomerism part and facial and meridional isomerism isomers that I have done in the class that you must see that is very very important okay you must you're going to get one question from coordination compounds to do that particular chapter very properly okay in inorganic chemistry I'm talking about right thermodynamics have you finished thermo chemistry thermo chemistry is important there heat of formation heat of you know combustion and all thermo chemistry thermo chemistry you must do and then in thermodynamics this side the entropy and all we have there are very much formula basic understanding of like what formula you should apply when that you must know so these kind of like don't put so much time in one particular chapter okay little bit of theory don't read any other book try to when you solve 10 questions try to figure it out that what what knew that you get from these 10 questions what concept you understood from these particular questions okay try to do that only okay today also we'll solve some questions okay so last class we were doing just a second let me this one did you solve this question chemical kinetics these three questions I think this you have done let it be this this one you don't do I'll give you some questions of organic chemistry just a second okay this is a chapter of organic compound containing oxygen and then it means this may have aldehyde ketone carboxylic acid all these organic compound containing oxygen six question I've given you solve this one by one we'll discuss this in two minutes see like I'll give you 2025 general questions then few questions that has been asked in J exam okay like that we will practice one C poor big is getting first one most of you are getting D Ramjanan is getting a rhetoric is getting a mathly is getting a you see the question is arrange the following compounds in increasing order of boiling point okay so we know the boiling point of any molecule is directly proportional to its molecular mass more molecular mass more will be the boiling point molecular mass is directly proportional to the boiling point so first of all we'll check the molecular mass so obviously pentane has the maximum molecular mass right propane butane and pentane right so pentane has maximum so obviously cnd is not possible right cnd is not possible after pentane we have butane and then propane so a and b order is correct but in butane to all and butane one all which one will have more boiling point that's the question right these two are what if you write down butane to all which is nothing but C here we have OH and butane one all is this OH and then C one two three and four right this is butane one all and this is butane two all so these are the isomeric alcohol right what kind of isomers these are can you tell me what kind of isomers what kind of isomers these two position isomers because the position of OH is different the number of carbon atom in the parent chain is equal right so these two are positional isomers so among isomeric alcohol what can we say among isomeric alcohol the boiling point in decreases with branching as branching increases boiling point decreases because in branching increases surface area decreases and that's why the boiling point decreases because a decrease in surface area more branching less in the surface area and less will be the boiling point so boiling point of butane one all is more than to that of butane to all hence the order will be option question number two tell me question number two question number two all of you are getting see all of you are getting see it's correct why you see fourth one is dihydroxy sorry third one is one for dihydroxy benzene right one for dihydroxy benzene which is nothing but this and then again OH so in this intermolecular hydrogen bonding possible right so because of intermolecular hydrogen bonding and both side intermolecular hydrogen bonding is possible both side so that's why this has maximum boiling point right so third has maximum boiling point this one I'm talking about right one three dry hydroxy benzene one two dihydroxy benzene one information I would like to give you here right when we are talking about intermolecular hydrogen bonding this you must remember intermolecular hydrogen bonding the order of hydrogen bonding the order of hydrogen bonding for a compound which has this kind of bonding intermolecular hydrogen bonding the order is ortho we have minimum and then we have meta and then we have para right or to meta and para isomers if you have ortho meta and para compounds then the order of hydrogen bonding will be this means what if the hydrogen bonding is intermolecular hydrogen bonding is possible at para position then the then the tendency to forming this hydrogen bonding will be maximum meta will be lesser than the para position and ortho will be least okay so for intermolecular hydrogen bonding the order is this if intramolecular is possible intramolecular bonding the order of hydrogen bonding is at ortho position it will be maximum then we have meta and then we have para so these two order you must remember now the compound that has been given here in this intermolecular hydrogen bonding is possible right and that's why we'll go by this order the answer will be option see that all of you have got so I'm not discussing beyond this okay but this order you must remember in case of inter and intra okay third one what is the answer third one is a this one I have discussed so I think most of you have memorized it no need to discuss this right finoftiline right it's also given that the product gives pink color with alkali okay that is the indicator that we use finoftiline indicator into this okay so again you all know this that's good fourth one see this kind of direct question you will get in JMAs at least okay you will not don't try to solve tough questions these days okay only you solve basic basic questions basic understanding if you have you can easily get 170 plus marks in JMAs minimum I'm talking about okay so solve only basic questions fourth one sodium phenoxide reacts with CO2 at 400 Kelvin at this pressure what is the product we get wait wait wait I'll do that see this sodium phenoxide reacts with CO2 at 400 Kelvin 4.7 atmospheric pressure to give options are this catechol, cellicyl aldehyde, sodium cellicylate and benzoic acid clear okay I'll just increase that into the next slide did you get the questions Shweta this would be better okay fourth one see answer is correct okay so most of you are getting C only phenol sodium phenoxide reacts with CO2 sodium phenoxide is this we have benzene and on that benzene ONA group is attached which reacts with CO2 I have done the same reaction in the class and this gives you whatever the temperature and pressure we have here that is written there mechanism you know the C double bond O and all then this group attached onto this carbon atom and last this NA shift onto this CO group and here we'll get OH so mechanism I have done in the class I'm not discussing it now the answer will be OH here and COONA here and this we call it as sodium cellicylate option C is correct here tell me the fifth one phenol with NA OH write down this reaction phenol with NA OH CO2 fifth one is D again all of you are getting D aspirin what is the formula of aspirin I'll write down the structure weight so that reaction will be this phenol we have and this reacts with NA OH and CO2 NA OH CO2 which gives here NA and COO and two moles of NA OH actually will take you okay what's the efficient NA OH H plus and H2 when you put we have to find out this is a with H plus and H2O both sodium atom is replaced by hydrogen OH and COOH and then we'll use this and head ride and finally we get O as it is CO CS3 CS3 COH will go out and COOH will be here as it is this is aspirin okay you must know this acetyl salicylic acid it is this one is salicylic acid and this one is acetyl salicylic acid which is also called it as aspirin question number six option D is correct question number six what is the name of this reaction C is correct all of you are getting C C is correct what is the name of this reaction six one is COOH's reaction in which you'll get salicylic acid yeah right poor with scope reaction okay the name also you must keep in mind okay can we move on next question you see see all these questions that I'm giving you it is all about the organic compound containing oxygen so it contains alcohol phenol ether and dehydrate it on carbazolic acid all these things mixed question I'm giving you can is that a reaction question number seven hydrogen is getting B stress is getting A methyli is getting B kushal is getting B okay first of all option B is correct here okay you see here mechanism I have already discussed pH C double bond OH in this what happens the base that you have OH minus this OH minus will attack on to this carbonyl carbon this by bond goes here we'll get pH C O minus H and here we have OH now next what happens this when reacts with the another molecule of aldehyde which is pH C double bond OH transfer of hydride takes place right so this hydride ion the transfer of this hydride ion is difficult and slow transfer of this hydride ion means the base will attack on one of the aldehyde group right one of the one of the molecule from that molecule only in the next step the hydride ion will transfer on to the carbonyl carbon of the another molecule right and this step is the slowest step transfer of hydride ion is difficult and hence it is the slowest step hence you see the transfer of hydride ion to the carbonyl group is difficult option B is correct clear carboxylic group you can easily eliminate not possible deep protonation of this that is also not possible okay attack of OH minus the first step which is a fast step nucleophile attacks quickly on to this carbonyl carbon because of the electronegativity of or polarity of the C carbon oxygen bond option B is correct clear question number eight pram chidron is getting A Simeer B Sanjana B Pratik A Karnia A Suresh B so A B A B A B you are getting like that the structure of the compound that gives a tribromo derivative on treatment with bromine water okay so when it reacts with bromine water will get tribromo derivative and that bromine will attach at ortho and para position right so I think option A will correct okay see option option A is what this is metacrysol right here we have OH and here we have CS3 this is metacrysol and when this reacts with bromine water that is BR2 H2O then this bromine will attach at ortho and para position and the product we get here is this OH and since ortho and para position you see this ortho and para position these two are vacant here so bromine bromine and bromine and here we have CS3 as it is and CS3 sorry OH OH okay ortho chrysol if you have like I said the addition of bromine is always takes place I'm coming to that addition of bromine always takes place at ortho and para position if you have this compound which is option C you see this CS3 OH option C in this case again bromine will attach at ortho and para position but where it will attach those ortho and para position sorry CS3 those ortho and para position which is vacant which is available so here we are getting di bromide derivative just a second I'm coming someone's there yeah so bromine will attach at ortho and para position so this is the answer for the third one when you take this option B okay in this also we'll get di bromide derivative right in this also we'll get here and here we'll get bromine not at this position okay option B if you take when we have CS3 CS in this we'll get monobromo derivative we'll get ortho and para product right like this if you have CS2 OH attached over here with bromine water we'll get ortho and para substituted product CS2 OH bromine will attach here plus CS2 OH bromine will attach here this is a reaction given in the book I think you forgot but when we have OH present here at this position right then we'll get mono substitution I have done this also like why we get mono substituted and this thing here in this case you see when we have OH present here this position ortho position will become highly reactive right and at the same time either see any of these two position ortho and para position when you have OH present here then the effect of this group will be same at this position and this position right and it depends on the solvent also here you see the structure of the compound that gives a tribromode derivative on bromine water right so the point is when you have this it's not like we do not get you know like in organic chemistry we cannot nail it any of the answer in this the point is the major product we have to discuss right so here the major product will be bromine will attach at ortho and para position here also it will attach ortho and para positions by because OH is attached directly to the ring but when OH like I told you already this when OH is attached directly not attached on the ring then we'll get ortho and para substituted product like this and in one of these this para product will be slightly more dominant or major product we have here right so that's the reason in option we will get mono derivative or mono bromo product here C and D will get dibromo product and A will get tribromo product or tri derivative tribromo derivative whatever you is it clear yes so you must remember there are many reason we have here but what key point you have to keep in mind that this is the reaction of when we have OH group attached directly to the ring here it is not attached directly okay whenever OH attached directly to the ring then the bromine or the halogen group mainly where we take bromine only will directly attach to the ortho and para position of the ring right if it is no available right so here you see the this position is not available so accordingly the bromine will attach another position okay so that's the thing you have to keep in mind okay question number nine ninth one is D C some of you are getting D some of you are getting C Amchalan is getting C Rithik D Aditya D Sai Mahir C Matali C Kuchal C Vaishnavi is getting D Sai Yuzha is getting D okay now you see when we have para chrisol tell me what is the name of this reaction para chrisol is this we have OH CH3 plus it reacts with chloroform which is CHCl3 in alkylene medium so for that we'll write OH minus gives a compound A so in this case this reaction is rhymer time and reaction if you remember in rhymer time and reaction CHO group will attach with the ring CS3 will be as it is since it is there okay this reaction is rhymer time and reaction so the product will be this now the next thing is what that in this A this is A we add what HCN into it so where this HCN will go this will obviously react onto this carbonyl carbon most reactive right so CN minus and H plus will attach simply onto this okay so the product here it will be OH CH3 and C double bond O that I will not write this CN minus will come over here and this H plus will come onto in this oxygen so we'll get OH here fine now when you do the acidic hydrolysis of this H plus H2O that is what the case we have acidic hydrolysis then what happens in acidic hydrolysis of cyanide group it converts into what carboxylic acid when you do the acidic hydrolysis of any cyanide group that will convert into carboxylic acid which is given in the question itself the structure of the carboxylic acid we have to find out so OH will be as it is H will be as it is COH so I think option C is correct adjacent to OH ortho position option C is correct understood some of you were saying D you may not concentrate on this we have OH here and CS3 here so ortho position of OH will be this not this will this product form like major product will be this but what about this this product will form or not we'll get this product will not get this product because this is the meta position of OH group so reaction will not take place in meta position okay if it is at ortho then little bit of this amount also will get okay yeah correct for me so that's the another no question but okay can we move on next third one question number 10 I'll just zoom in a liquid was mixed with ethanol and a drop of concentrated H2SO4 was added see this 10th one you are getting D why it is D all of you are getting D so D is correct but why it is D like you have done this or you know or you have seen this question before which compound gives this fruity smell the acid gives ester okay so that's the thing ester is formed yes correct correct see one more thing you must keep in mind since this logic you have applied that acid may give ester that you have that yeah that's why you selected D that is correct but fruity smell is given by esters only fruity smell is the characteristics property of ester right so you must get ester as a compound which is only possible when acid reacts with reacts with alcohol in in in what in acid condition right so that's why the answer is option D right so fruity smell is the characteristics for acid plus alcohol gives you ester and ester gives fruity smells that's why acid is the answer okay 11th one option D you see I don't have a space option D is this okay CS2 CS2 CS2 OH now you see this question question number 11 C9 H12O the compound is if you are if it is not visible the compound is sorry just a second the compound is C9 H12O the question is a compound is given which is C9 H12O have four isomeric structures okay four different structure you can draw for C9 H12O all these compounds all the compounds react with sodium to give order less and color less gas except one right means what the four different structure that you have here out of the four all reacts with reacts with sodium and gives order less and color less gas except one means one molecule does not react next information is what the three out of the four are pre oxidized by hot KMNO4 right means out of the four compounds three is getting oxidized by KMNO4 to produce to produce benzoic acid one another isomer does not respond to IDO form test right one of the four does not respond to IDO form test but one isomer out of these four fulfills all condition isomer is okay the question is for a given compound there are four isomeric structure possible and for all those isomeric structures different different three four conditions are given right on the basis of these condition you have to identify the isomeric structures okay okay now you see I'll just solve this one you see the formula is C9 H12O right this is the formula of the compound and for this the possible structure if you draw that will be H2N OH general formula is this unknown compound this have four isomeric structure can you tell me what is the DOU of this degree of unsaturation what is the DOU of this compound yeah it's correct it's four because the formula of DOU is what it's not one way formula of DOU is this C plus 1 minus H plus X minus N divided by 2 so C plus 1 is 10 minus H is 12 X is 0 N is 0 so DOU is 4 it means four unsaturation is there that's why this formula represents it's a benzyl alcohol and the general formula of benzyl alcohol is this C9 H2N OH okay benzyl alcohol you see one ring and three double bonds so three plus one four degree of unsaturation that's what we are getting here so this formula actually represents that it is a benzyl alcohol now since the formula of this is given so the possible isomers of these of this molecular formula is C9 H12O the possible isomers of this is I'm drawing here CH CH2 OH CH3 one we can write CH2O CH2 CH3 next one is this CH CH3 CH3 OH and the last one is CH2 CH CH3 OH okay you see this particular compound this does not give IDO form test does not give IDO form test this one has CH OH CH3 it gives IDO form test and this actually this one also that does this gives IDO form test this one does not liberate H2 liberate H2 only this can liberate H2 and maybe oxidize also so this molecule fulfills all the condition that is given in the question okay that's why this one where is this CH2 CH CH3 OH CH2 CH CH3 option A is correct into this one understood so for this kind of question you should know the property of alcohol ether why does not give H2 H2 because this does not oxygen it's the ether right so this oxygen does not contain hydrogen that's why H2 is not possible into this one okay so option A is correct is it clear see only one thing if you keep in mind like this does not give IDO form test right this also does not give IDO form test and this also won't give IDO form test so only one of the molecule gives IDO form test and that's why this molecule has to fulfill all the condition given in the question that's why the answer is this one if you have the information of IDO form test only you can do this question okay next one question number 12 question number 12 pmq is given we'll do the acidic hydrolysis of this and the product that we get can be distinguished by first of all you check what is the product you are getting into this food week is getting B Simeer D Mathle C Rithvik Khushal D question number 12 Trash Ramcharan is getting A okay now you see the answer in this is option C filling solution how you see this first of all like I said you have to first form the product right so p the compound P we have is this H2C double bond COCO CH3 and here we have CH3 this is P when you do the acidic hydrolysis of this so obviously this CH3 CO group that we have here that will form CS3 COOH right and OH group will attach to this H2C double bond C OH we have CS3 plus we'll get CS3 COOH now in this this is the enol form and keto enol tautomerism we can easily product we can write down okay so with keto enol tautomerism what happens this double bond will go here and this H plus will attach onto this carbon atom so we'll get CS3 C double bond OH CS3 so we are getting ketone into this fine now Q we have which is nothing but CH3 CH double bond CH OC double bond O CS3 again when you do the acidic hydrolysis of this same kind of reaction takes place CS3 COOH will go out and here we get CH3 CH double bond CH OH again keto enol tautomerism will take place this pipe on will go here H plus will attach here so CH3 CH2 C double bond O and hydrogen will be as it is so here we are getting ketone here we are getting aldehyde which can be easily differentiated by failing solution option C is correct failing solution lucas reagent we use for alcohol right but here we are getting aldehyde and ketone NaH SO3 when we use NaH SO3 see NaH SO3 we use for the compounds which have larger group present around the oxygen atom larger means larger than CS3 group like we have acetophenone acetophenone we use NaH SO3 one more thing you write down here about NaS most of you are saying D so write down NaH SO3 reacts both with aldehyde and ketone where we have like one reaction or two reaction I'll write down probably these two reactions you see suppose I'm taking aldehyde CS3 CHO so we have CS3 C double bond OH reacts with NaH SO3 so Na plus H SO3 minus the product we get here CH3 C single bond OH and this H is coming from here this hydrogen is there and here we have hydrogen of carbon this hydrogen is here and we have SO3 minus Na plus like this we have the salt same kind of reaction we have with ketone also like suppose if I write down in this only suppose instead of this hydrogen if you have CS3 then here also instead of this hydrogen we'll get CS3 and the product will be same so you see the aldehyde and ketone does not both reacts with NaH SO3 so we cannot distinguish with these two this reagent so you must remember this whenever in the option you see failing solution and all try to get aldehyde and ketone you must think about the question is related with aldehyde and ketone okay okay next question you see question number 13 just a second 13 is C yeah series and resonance reaction RCOCl plus H2 gives this BA SO4 here for this kind of question you should know the mechanism the mechanism helps you so property of BA SO4 is what see this palladium is highly reactive okay and with this nature of palladium when you're using palladium into this reaction the reaction is very difficult to control okay 24 DNP I'll tell you I'm just a second let me finish this okay so when we use this PDB SO4 so palladium is highly reactive in nature right so without and this reactive nature of palladium with because of that only the reaction is very difficult to control right so we need to deactivate palladium little bit okay and for that purpose only we are using we are using BA SO4 so answer in this question will be it deactivates the nature or palladium it deactivates palladium simply see so answer will be option C in the previous question I'll just give you the you know what is 24 DNP just to you write down this 24 DNP is 2 comma 4 dinitrophenyl nitrophenyl hydrazine the structure can you write okay I'll write down you won't be able to write it the structure of this will be 2 comma 4 dinitrophenyl hydrazine so hydrazine is what NH2 NH2 with NH2 NH2 this phenyl group is attached right so NH2 this is NH and this is NH2 in N2 so you see 2 comma 4 dinitrophenyl the structure is this okay if you see the test of this just a second the dilation prepares okay this is 24 di phenyl hydrazine when you do the test of it we'll use a little bit of amount of mythenol that is the you know CS3 OH and sulfuric acid also H2SO4 so all these mixture we call it as brady's reagent b r a d y s brady's reagent this name you also remember see the two nitrogubes that we have here so in reaction what happens this reaction is slightly dependent on the nature of aldehyde or ketone that we are getting right so when we have any aldehyde let's write down the reaction here suppose if you have like both reacts with aldehyde and ketone first of all so again this won't distinguish but how it reacts that you see when this reacts with 24 dinitrophenol into this so what it's the reaction is very simple you see I write down this NH2 NH and then we have the ring which contains N02 group at 2 and 4 position so this H2 and this O comes out as H2O right and this two joins this O and H2O comes out and these two joins this N and H and then we have ring over here with N02 group sorry N02 group and here it will be R now this R and R can be two different alkyl group also can be same alkyl group also one can be alkyl group other one can be hydrogen also in the case of aldehyde it is right now in this this kind of reaction again it is a condensation reaction because H2O is coming out right so since one molecule of H2O is coming out and the two molecule joins together so reaction is condensation reaction right and this actually involves the nucleophilic addition and elimination reaction right elimination of water molecule takes place and carbon and nitrogen is joined together right so it is addition example of addition elimination reaction okay now the point is by the use of this reaction we can just understand or find out the presence of carbon oxygen double bond okay where the carbon oxygen double bond is present okay one important point in this you write down note down this point I am dictating you using this reaction using this reaction we can understand we can understand the presence of carbon oxygen double bond and we get this is important and we get orange or yellow precipitate and we get an orange or yellow precipitate a carbon oxygen double bond a carbon oxygen double bond in an aldehyde or ketone so we'll get orange or yellow precipitate into this okay from that we can identify specific aldehyde or ketone okay with that precipitate but that is not required the point is since both aldehyde and ketone reacts with these two for DNP that's why we cannot use this particular reagent to distinguish the two compound that is aldehyde and ketone clear so this property you must keep in mind to identify carbonyl group yes yes simply you can say it is there to identify carbonyl group C double bond O we cannot distinguish aldehyde or ketone but yes we can understand whether double bond C or carbon oxygen double bond is present or not and how do we get this because H2O is eliminating condensation reaction is there okay next one you see question number 13 we have done just now right question number 14 can you see this which of the following carbonyl compounds i'll just check carbonyl compounds on condensation gives an aromatic compounds which of the following carbonyl compounds on condensation gives aromatic compounds C is correct two moles of the compounds will take C is correct simul 2 C S 3 CO C S 3 H C H O is not possible for aromatic compounds you must have at least no six carbon atom right and with that only it is possible if you take two molecules of this or two molecules of this aldehyde won't give you this okay this may give you because of the symmetry also you can understand okay so 14 C is correct question number 15 just the name reaction question number 15 and we use Hvz helvolat zealinski stiffens reaction is correct stiffens reaction is correct alkyl cyanide gives aldehyde the reaction is known as stiffens reaction that's the name reaction it is given so stiffens reaction is the answer 15 16 sorry 15 a is correct next one yeah correct pulvic so all these reaction you see all these whatever see you're solving question number 15 but there are four name reaction is given so it's not like only actually stiffens is the answer so it's fine you should know all the four things four options what is this 14th see actually we are getting aromatic compounds okay for aromatic compounds we must have six carbon atom right six carbon atom and all these molecules actually that it is understood that we are getting only we cannot take only one molecule of all these abc the option that is given we can take at least two molecules you have to take right so when you take the two molecule of cs3 c double bond o mechanism i'm not explaining so what you can understand when you take two molecules of this and both molecule clubs together right and h2o molecule goes out okay so you see this h2o goes out from here and then another molecule we have cs3 c double bond o cs3 right so they're taking two molecules of this fine so when these two combines you can see there is a possibility of forming what a ring here when these two combines these two combines we'll get a ring over here and this h2o and this h2o oxygen will come out okay so h2o molecules comes out we have c h3 here cs3 here double bond single bond double bond okay so okay one more thing you see see we have cs3 i'll just write down this and cs3 in this the question is on condensation so obviously condensation is given so h2o molecules will go out right so when you have this with another molecule right so one two three and one two three so if you take simply one more molecule here so this is h2o this h2o goes out we'll get a bond here if you take one more here cs3c double bond o cs3 so one two three this two will join so you see here to understand this i think i have to discuss the mechanism only fine so what happens condensation reaction is condensation is taking place right so h2o will go out and we can will heat this simply and h2o will go out right so one possibility is what that this oxygen will take this hydrogen atom from here right and we'll get a double bond here correct and what happens we are actually taking here three molecules of this why because you see one cs3 is here and one cs3 is here so one of these cs3 molecule will take part into the condensation reaction and h2o molecule will go out and here also will have the same thing so either suppose these two are this is reacting with this this is reacting with this this will be left behind which will not take part in the take part in the reaction correct and from this again when you heat this from this h2o molecule goes out from this h2o molecule goes out from this h2o molecule goes out so we are getting three h2o molecule will go out from this right and since this carbon and this carbon again since we have double bond here and single bond here it's only three bonds with this carbon atom right so from this h2o molecule that goes out this carbon atom will join as a bridge between these two carbon atom right and the structure that we get here i'm not giving you the exact mechanism here right but little bit of idea you will get this the third molecule will act as a bridge between this molecule and this molecule right right and we get a six membered ring compound like this and when on which these two joints like one two three four five six here one cs3 is left one cs3 is here so this cs3 will be attached here here and here correct so if i take this as suppose if i write down the number then probably you can understand one two and three okay one two and three so this cs3 will have here and here we have five and six so one two three five even six and this h2o goes out so this is this seven will join with this six seven eight and nine correct so seven this seven eight and nine so like this it joins you see this carbon two and three this three joins with five and this carbon joins with this carbon here eight one right so like this the ring forms okay and this cs3 will be there at alternate position okay you have to actually memorize this the name of this why it is important because the name of this sometimes they have asked this mesutiline the name of this is mesutiline and they have also asked the degree of unsaturation of mesutiline that's why this one is important that's why this one is important so degree what is the value of degree of unsaturation of mesutiline degree of unsaturation of mesutiline what is the answer? Four. One ring and three double bond. Four. So this question they have asked, you must memorize this structure of mesodium. Okay, this kind of condensation reaction, chain reaction, which is one molecule is connecting with two and two is connecting with third. It is not possible when the unsaturation or this carbonyl group present in the, on the terminal carbon, like in the case of this one, D option, aldehyde. It is not possible over there. Okay, that's why this acetone gives this kind of reaction. I'm not acetone, this gives this kind of reaction. Yeah, I don't give this kind of reaction. So you have to memorize the structure of this mesotylene and degree of unsaturation into this. Anyways, so 16th one, we have done. Next one you see, one, two, three, fourth one. Question number 16. If it is possible, like in display, it's not covering the whole question. This is because in one pick, I have taken four or five questions. That's why the font size is small. 16 is a simple one. Why is alcohol? D is duty ADM. X, Y will have the structure. 16. Ramchandran is getting A. Sanjana is getting B. B, all of you are getting B except Shweta and Ramchandran. What happened in this? This OH- will attack onto this carbon, right? And this pi electron goes onto this oxygen. One molecule we are taking of this. And then this D- will go out, will get C double bond O again here. And finally H- also will go out. So we'll get carboxylet iron, which is X. I think A is correct. C double bond O- and alcohol is this. I think A is correct. We'll do this. Wait. You see the question, the compound we have here, D, C single bond D double bond O. And we are taking OH- here. So this OH- will attack onto this carbon atom. Pi electron goes here. And we get C single bond D O- OH. And then this D will go out as D plus, minus D plus. And this bond pair, sorry, it's not D minus. No. Then only we'll get the double bond. D minus. Okay. I'll just correct it as D minus. And we'll get double bond here. So we'll get D, C double bond O, OH. And finally, when you heat this, H plus will go out. And we'll get D, C double bond O, O minus carboxylet iron. Okay. So X is this. Now, when this reacts with Y is an alcohol. So how do we get alcohol? How do we get alcohol? You see the structure is this D, C double bond O. And D we have here. Okay. D plus OH- if we are taking. Or H plus OH- H2O. D is just isotope of it. So when this pi electron goes here, right? So D plus will attach. We have to form alcohol, right? So OH we have to get, right? Oh, for OH, for OH we required because we have already H plus there. D, C, D double bond O. So this will come here. Here we have OH, O minus. So for that we required H plus and D minus will take. Right? So this D minus will attack on to this carbon atom. So we'll get DC single bond D. D, O minus will take this H plus will get OH. And we are getting option A is correct. See, I'm taking this H plus and D minus according to the option. Okay. Because nothing is given in this question. If you are getting OH here, it means this O minus should take that H plus. That's why I've taken H plus. For the here also you see, you must take H plus so that OH you'll get. Here also the same thing, right? And in these two option you see, we have D. I think option B is also possible if you take H2O. Oh no, one thing, one thing, let's just do one thing. Oh, fine, fine, fine. No, it is correct. Yeah, it's correct. See, what happens here? This one molecule we have used over here. And in this molecule, D minus is going out. This D minus we are using here. And you see H plus is also going out, which we are using here. So this H plus is coming over here. D minus is coming over here. And that's why option A is correct. Understood. Next, question number 17. Do this. 17, you are getting B. How it is B? How it is B? Okay. Only tertiary alcohol. And alcohol A, when heated with copper gives a product B, not having oxygen atom, B on ozone, Ozonelysis gives two isomeric products, C and D. C on oxidation gives monobasic acid E, silver salt of which contains this percent of A, G, the structure of A is. Okay, so Ozonelysis possible in alkene, right? If you have any unsaturation. And that is possible if you have tertiary alcohol, because that only from that only will get the elimination reaction takes place and we'll get the alkene into that dehydration. Correct. So tertiary alcohol gives you alkene first and then Ozonelysis takes place. Fine. This is what you're done. Yeah, it's correct. D option is correct. 17, D is correct. Only tertiary alcohol gives you alkene and then Ozonelysis takes place. B is correct. Now you do this 18, 19 and 20. It's solved 18 and 19 first, then we'll do the 20. A is getting D, Ozonelysis is getting A, Simehir, D, Rittik, D, Mathli, A, Pratik, A, Kondinia, A, Tamchalin, C. This one is a bit easier. You see, answer D is correct. Option D is correct. You see this acid gives you H plus, right? So this H plus will come onto this oxygen. So we'll get OH positive charge onto it, right? Next we are heating it also. So because of this heating, what happens? This bond pair, this will dissociate because oxygen will have the positive charge here. This bond pair will shift over here and this sigma bond comes over here. The sigma electron comes over here and this bond will dissociate, right? And since this pi electron shift over here, so we'll get positive charge here, right? Where the chlorine will attach onto the next step, right? So that's why phenol and 3-chloro-1-butene will be the answer. You see this, I'll write down the reaction. See the question is this lone pair we have here and when we use H plus CL minus, so this H plus will attach here. So we have phenyl OH positive charge. Now this bond pair goes here and this bond pair goes here. So we'll get pH OH which is phenol plus we'll get this double bond and this where we have the positive charge here. This carbon will have the positive charge and with this the CL minus will attach over here and the product will be this 3-chloro-butene. 18, question number 18. What is the following true regarding why it is produced as per single, as pure single enantiomer? Why is the mixture of pair of geometrical isomers? Why is the mixture and why is chiral? What is the answer? 18. How do we get single enantiomer? 19th one. See, first of all, this is Y and this is X. So Y is forming from this carbocation, which is planar, sp2 hybridized. So attack of chlorine can be from any of the side, from this side also and that side also. So we'll get two different configuration and that's why it is a rest big mixture. The probability is equal of CL minus to attack from both side from the top or from the bottom. That's why we'll get a rest big mixture. We have a chiral carbon in this. In Y, we have a chiral carbon, but it is saying Y is a chiral. Question number 20. I don't think there is an option to crop this, that I can do initially only when I was adding the question. I don't think it's possible here, but I'll take care of it next time. I'll not take four or five questions in one pick. Okay, two, three questions we'll take so that the font size will be bigger. Okay, I'll check that. First you do this. I'll figure out that option. 20 you're getting A. How? So I mean, anyone question number 20. NaCl will go out. According to the option you try to do this. Tell me quickly. You're getting C. Okay. See, Y is this double bond. C2H5ONA. This Na plus will take this CL minus and we'll get what positive charge here and NaCl will form. You see the option here. 1, 3-beta-n is one of the options we have and 1, 2-beta-n. So 1, 2-beta-n is also not possible because removal of H plus is possible from this carbon atom. So when the H plus removal from this, so this plus and this negative will form a double bond here and this double bond will be as it is. So option one is possible like this or if another way if you see, we have double bond here and OC2H5 will join here. This is not there in the option. CH3, CH, double bond CHOC2H5 we have. This is even not there in the option. Okay. So my point is you can get confused into these two because when H plus will come out so that C2H5O minus we have there, that will take that H plus and form C2H5OH which is already there in the solution. In the option it is given C2H5OH we are getting here or we are taking C2H5ONA or C2H5ONA. So both what happens either in this case it will form NaCl and in this case it will form HCl. So if OC2H5 will join here, see they are asking major product. So major product will be what? In these two, which one is the major product? Depending upon the stability, two possible product will be this one. One is 13-betadine and other one is this. So whatever the product here, the decision for making this, like which one will be the major product is difficult over here. We have to have some factual information here which is not given in the question. That's why the only one option is given in the question, which is nothing but A. Is it clear? Okay. So it's 11.15 now we'll take a break. Okay. I will start at 11.30. Okay. Half an hour we are left with. So we'll start at 11.30 sharp. Okay. And tomorrow do you guys have any other class tomorrow? Physics or maths? All of you I'm asking. So tomorrow I'm planning to take these classes till 28. Okay. Continuously I'll take the class. Okay. Anyways, I'll let you know in the evening about tomorrow's class. Okay. So we'll start the class at 11.30 again and then we'll take till 12 o'clock. Okay. Fine. Fine. So HSR guys, accordingly I'll send you the schedule. Okay. So don't worry with that. Okay. So we'll take a break now. We'll start at 11.30. Okay. Can we start? Okay. These two. Tell me what is the answer? What happened? I'll give you a hint here. You see this reagent is for Perkin reaction. This anhydride and this salt we use for Perkin reaction. But property here you have to, you should know or the information that you should have here that F of essence with NaFCO3, the compound which gives F of essence with NaFCO3 is carboxylic acid. This you should know. Compound which gives F of essence with NaFCO3 is carboxylic acid. Okay. And positive bare test means what? Bare test is the test of unsaturation. Right. It means in this compound we must have any double bond present. Then only test of unsaturation will be there. Right. Ramchandra is getting D. Right. Okay. Not D. Okay. So the product here J will be a carboxylic acid with at least one double bond. That's the information we should have from this data, given data. Or if you see that DOU of this, if you calculate that also probably you'll get two or three. You can check. What is the DOU of this you tell me? Degree of unsaturation of C9H8O2. Yeah. C is correct Ramchandra. C is correct. See the degree of unsaturation of C9H8O2 for J. Which is C9H8O2. It is 6 if you calculate. 9 plus 1 is 10. 8 by 2 is 4. 10 minus 4 is 6. Okay. And what is Perkin reaction? In Perkin reaction, Benzyldehyde CHO reacts with this reagent. CS3CO whole twice O. CS3CONA. And it gives C6H5 CH double bond CHCOOH. Which is probably Z. Sorry Z. Correct. You see the unsaturation here. C6H5 phenyl groups. One ring, three double bond, four, five and one double bond here. Six. Okay. So this is the carboxylic acid we have, gives FFSNs with NAHCO3 and double bond gives positive Bayer's test. Okay. Now when this is allowed to react with H2PD. Right. Then reduction takes place and we'll get C6H5 CH2CH2 COOH. Hydrogen atom will attach. Now after this it is easy. SOCl2. Chlorine will attach with this. C6H5 CH2CH2 COCl. Correct. And in the last we have anhydrous anhydrous ALCl3. This will take this Cl minus from here. So if I write down the structure of this, CH2C double bond OCl. Okay. This ALCl3 will take this Cl minus forms ALCl4 minus here. And this carbon will attack onto this carbon and joints. Okay. So we'll get 123 and this will attach with this ring. So the product will be this here. Like this we get. Clear. Option C is correct. What about 35? I also I have taken. This is it. This is I. So this is option A. So this is nothing but K, J, I. Okay. Next question. Now see these are the questions has been asked in J. Question number 17. Wait, wait. All of you are getting C. What is this reaction? Answer is correct. C is correct. What is this reaction? Only canizaro. This is an example of aldol condensation. Repeated aldol condensation followed by canizaro reaction. Okay. Your answer is correct. Three is the answer. All of you have done or do I have to do this? Let me know. You want me to do this or what? I know reaction that occurs in the given transformation is. Okay. So, okay. I'll just explain this in the next one. It's a big mechanism we have. So you see the step one in this reaction. We have CH3, CHO and the reaction is going in basic medium. And if you remember in this what happens alpha hydrogen takes part in the reaction. So this hydrogen will take this OH minus H2O forms and we'll get CH2, CHO with one lone pair on it and H2O will get here. Now the another molecule which is HCHO, four molecules of this we have. So for the first one is this plus CH2CHO lone pair on this carbon or negative charge also you can write. That gives which is a reversible reaction here. HCO minus HCH2CHO. So what happens here? You see this lone pair will attack on to this carbon atom and this pi bond goes on to the oxygen. Okay. And in presence of H2O next this gives you this H plus will come over here OH minus will be as it is. So we'll have excuse me CH2OH CH2CHO. This is the step one reaction. Now again you see what happens since we have four molecules of HCHO. Only one I have used now. Okay. So one I have used and we have got one OH here. Similarly you have to keep on writing down the steps involved here. You see the second step. What happens into this? Step two. Now one molecule is this and another one is HCHO. So again what happens this one CH2OH CH2CHO which is this molecule plus the base we have OH minus and again this will take the alpha hydrogen which is this. Right. And we get what? HO CH2CH minus CHO plus H2O. So basically you can do this question in one single step. I am giving you the mechanism how do we get the product here. But since the question is how many reactions are there. So it depends on the number of alpha hydrogen that is present because you see this OH minus will take this alpha hydrogen one step. Again this OH minus will take this alpha hydrogen another step. And the third one we have we have one more left. So one more step we have after this. Okay. So we'll get this again in this one. The second step is what? This this compound which is HCHO we have plus CHCHO with lone pair here and CH2OH is attached with this. So again what happens? This lone pair this lone pair will attack on to this exactly same step. And this by electron goes here. You see we have HCO minus H which is nothing but this molecule is attached with CH CH2OH CHO. And again with H2O it gives COH H and then CH CH2OH CHO. Did you understand this? Now in the step three what happens? This is the molecule we have. And this alpha hydrogen will take part in the reaction. So again you see we are getting one OH here another OH here and then again in the next station we'll get OH here. Okay. And then next you'll get the product which is given in the question. Can you write down the step three or you understood or should I write down again? Step three. Fine. So you can write down you will get the product. Okay. So it's actually depends on the number of alpha hydrogen that we have. Okay. Next question. Question number 71. Finish it fast. I have another class also after this after 12. The compound that undergoes the carboxylation most readily under mild condition. 71 is B. Correct. See I'll give you this. This is just information you should have. B is beta keto acid. Okay. This is the this is alpha position. And this is beta position. So when you have the beta keto acid undergoes decarboxylation in very mild condition means what? If you simply heat this slightly if you heat decarboxylation takes place. Usually decarboxylation of carboxylic acid requires catalyst. Okay. For decarboxylation. Right. For any carboxylic acid if you have so on simple heating it won't decarboxylate. So for that purpose you have to add some catalyst into it. But if the carboxylic acid is beta keto carboxylic acid then it's decarboxylation is very easy under mild condition like on simple heating. So that's why the answer is option B. This point you write down in your notes. 72. I said the decarboxylation of beta keto acid. Beta keto acid means what? At beta operation we have ketone group present. Okay. The decarboxylation of beta keto acid is very easy is very easy for ordinary carboxylic acid or ordinary carboxylic acid beta sorry or ordinary carboxylic acid decarboxylation requires ordinary carboxylic acid decarbal decarboxylation requires catalyst in bracket you write down soda line catalyst we use soda line. One more thing I forgot to tell you decarboxylation of beta keto acid which contains alpha hydrogen also you write down alpha hydrogen must be present in beta keto acid here you must have at alpha position you must have hydrogen present alpha hydrogen question number 72 question number 72 A is correct option A in option B you will have H plus and C L minus will attach here. See when you have H C L Z and H C L then H plus and C L minus will attach onto these two carbon atom. So H C L actually H C L attacks on two degree alcohol as well as double bonded carbon atom also right H C L attacks on two degree alcohol as well as double bonded carbon atom also. So in this case what happens when you take H C L here only write down here will instead of OH you will get C L and here will get C L will attach over here and hydrogen will attach over here. The product we get here is this C L C H C L C H 2 C H 2 C H 3 C H 2 CO C H 3 sorry C H 2 CO C H 3 this is the product you will get in case of Z L H C L answer will be option A okay NA with NA BH4 the product will be very less okay NA with LH3 will replace this H from here oh NA will get over here if double bond is not there where where double bond if double bond is not there then the product will be different no see in this question yeah then it is possible if if alcohol is not there then it is possible yes then that's what in organic chemistry one group or one reagent if you change the whole answer will change okay but yes if alcohol is not there then it is fine 73 Idoform test I think D is correct 73 yeah D is correct Idoform can be prepared from all except isobutyl alcohol 3 alcohol be required okay 2 degree alcohol forms Idoform you must keep this in mind 2 degree alcohol forms Idoform isobutyl alcohol is 1 degree 73 is D 74 246 tribromofinol I have done this in the class right 246 tribromofinol 74 D is correct understood all of you done okay so we'll wind up the class here only I have another class at 1 o'clock okay so tomorrow I'll send you the timing okay when we have in the class okay tomorrow we'll do amines we'll solve the question of amines plus we'll also try to cover up atomic structure and chemical bonding right so you must revise amines atomic structure and chemical bonding okay we'll do that tomorrow