 okay can we start now tell me the order of acidity in the first one order of acidity will be what this is ortho substituted benzoic acid here we have what effect SIR steady inhibition of resonance okay this what effect will have here only plus i metaposition and here we have plus h plus h at ortho position right so you see this one is most stable sorry most acidic since it is ortho substituted right plus h effect will decrease the acidity more than plus i so this will be 2 and this will be order of acidity will be this if you compare here here we have what effect possible in the first one SIR steric inhibition of resonance what effect we have here what effect we have here minus h do we have minus h here see it is at metaposition vashnavi so we'll have here minus i metaposition only inductive effect at para position will have what minus h possible so this this you must know at metaposition will always have i effect why because when you draw the resonating or hyper conjugative structure then only ortho and para positions are affected correct so ortho substituted most acidic is this one this is first minus i minus h both will increase the acidity but minus h dominates over minus i this will be more acidic than this one will have okay one more question we'll discuss here c o o h c h 3 c h 3 c and then here we have n o 2 c o o h c n triple bond n c h 3 c h c in this question what happens it is not ortho substituted so what happens here because of this repulsion n o 2 will change its plane i have discussed one example here when n o 2 will change its plane so its electron with drawing nature is not affected over here because it is not in the same plane of this ring but here the cyanide group is linear right triple bond it's linear so here we do not have that much of repulsion so here minus m is possible here in comparison to this one and hence this one is more acidic got it okay okay so we have discussed so many questions see here for acidity order you may have o h also present here but in case of phenol when you have o h here then there is no ortho effect there okay so ortho effect we only apply in case of benzoic acid always you remember this okay if you have o h present here then only we'll see what we'll see various electronic effects present at different different position according to that we'll decide correct so you always remember one thing electric sorry ortho effect is only applicable in case of benzoic acid okay phenol also you can do the same way write down the effects possible whether it will destabilize or destabilize the conjugate base according to that you can say one order i am giving you this order sometimes helps you to you know write down the acidic nature okay sulfonic acid is maximum acidic s double bond o o h this acidic order you write down and try to memorize this keep in mind then we have carboxylic acid after this we have phenol then we have water r c h double bond c h 2 alkene then we have alkene this order no we always take this in carboxylic acid benzoic acid always in benzoic acid now like for one last thing we'll discuss here and then we'll see those questions okay five more minutes we'll discuss here and then we'll see the question or the assignment that i have given you okay you see for amines basicity order if you compare for any lane for any lane in case of benzoic acid we use s i r but here we are s i p and that this stands for hysteric inhibition inhibition of protonation of protonation now what is it you see when you have ortho substituted aniline suppose we have c h 3 and n h 2 ortho substituted aniline when you put this into water this forms this we have n h 3 plus c h 3 plus o h minus okay now because of this n h 3 plus here three hydrogen here right and here we have lone pair here we have three hydrogen instead of one again here we have the hysteric repulsion okay and to minimize this hysteria hindrance okay the reaction has more tendency to go into backward direction okay to minimize the hysteric repulsion the reaction has more tendency to go into the backward direction right when it goes to the backward direction the basicity decreases right basicity decreases why because n h 2 has lone pair when it loses its lone pair that is nothing but the basic nature since this hysteric hindrance is there because of this hindrance it has more tendency to go into backward direction and hence the basicity decreases right so what we can say ortho substituted aniline is less basic than para or meta substituted got it ortho substituted benzoic acid is more acidic but ortho substituted aniline is less basic okay that is why you see when h plus adds over here that is nothing but the protonation of this compound since this because of this hindrance the protonation is hindered right and that is why we are calling it as in hysteric inhibition of protonation is it clear hysteric inhibition of protonation got it let me guys quickly we'll see few problems on to this and then we'll discuss the assignment question now we'll see some questions so this too you must remember right ortho substituted benzoic acid more acidic ortho substituted aniline is less basic see the first question here n h 2 c cl 3 the city order you tell me tell me the answer the first one is meta para ortho right so ortho substituted aniline is least basic so this will be the third one right and this c cl 3 here it shows minus i this is very important okay minus i and here it is minus h so minus h will reduce the basicity correct so this will be one and this will be two so we'll have meta para and then ortho correct question to be the second one you see again this one is least because here we have s i p here also we have s i p here we have plus i plus h right because of plus the basicity increases so first then second and this one is okay what about the last one again you see we have hindrance here right because of hindrance the plane n o 2 will change its plane and for this case minus m is not possible because n o 2 change its plane minus m not possible but here we have minus m possible is the basicity of this will be more is it clear tell me okay so guys you must keep this in mind s i p and s i r okay you will definitely have questions on to this in your exam okay so don't forget this and then what all the effects possible according to that in the second okay so we'll you know we'll stop this session here only not the session actually now we'll discuss the assignment question you have some doubt right so i will not be able to see this screen now this youtube page the better what you do you write down your you give me your doubt on the whatsapp group i will see the questions there only is it fine send me your question numbers on that group i will discuss that here on this screen okay okay so text me on whatsapp whatever doubt you have okay so vashnavi has this doubt question number 22 all of you can see the question is this i will write down the question also this is a a next question is this and the last one is this see okay we have to find out that decreasing order of acidity okay now you'll see acidity means what if this hydrogen one of the hydrogen from here here it lose so we'll have a lone pair or negative charge here if it loses the hydrogen from here so we'll get a negative charge here right and here we'll get the negative charge of this carbon you see this ring is now the C it is aromatic because we have six pi electrons right it is aromatic this one is what this is resonance stabilized and there is no effect over here so stability is what stability of C is maximum then B and then A this is the order of acidity also third second and then one option D is correct here vashnavi got it vashnavi if you can question you see the next question number 25 which of the following has most acidic hydrogen there are two benzene ring attached question number 25 you see that have you finished it this question have you done vashnavi you understood this question number 22 i'll come to this question now first let me discuss this question okay question number 25 you see vashnavi in this question what happens option C you see option C right yeah i understood option C you see the hydrogen over there you see this there's a phenyl group present phenyl and here we have you know hydrogen and P CH3 positive charge you see this P has vacant D orbital right vacant D and when this hydrogen comes out so we'll get a lone pair over here right this lone pair is in resonance with this vacant D orbital and that is why this conjugate base is most stable when you see option D where we have nitrogen present here in nitrogen we do not have vacant D orbital right and that is why option C is the right answer here i hope you understood this option C is the right answer any other doubt vashnavi have you finished your assignment A is more stable than B can we say that A is more basic than B no we cannot say that stability has nothing to do with this base city or acidity order it depends on see base city means what when you lose is okay okay vashnavi see base city means what when you lose is lone pair of electron then you get conjugate acid if that conjugate acid is stable then we can say the compound with which this conjugate acid has formed is more basic correct acidity if you have to compare remove H plus iron the conjugate base that you get if that is stable then we can say the molecule or the compound from which this conjugate basis form that compound is acidic that's the point understood predict acidity you have to compare you remove H plus and then you compare the stability of conjugate base more stable conjugate base more will be the acidity that's the thing next class we'll see we'll take some more problems okay GOC I have there are a few things left in this but but we don't see theory much okay we'll solve we'll take the problem self solving session okay we'll solve only problems P block you want okay then we'll do a little bit not because we are so you cannot do P blocks complete theory but some important thing I'll discuss in P block okay and then we'll solve some questions also is it fine okay so so we'll wind up the class here only I will send you one more assignment on this questions GOC now okay you solve those questions okay because in this chapter you have to do more and more practice that only perfect that only make you perfect in this chapter understood so I am just like we'll wind up the class here only and I'm sending you the assignment okay okay thank you thank you for joining