GDC Weekly Test 2026 | ORGANIC CHEMISTRY- Resonance | Detailed Explanation

[00:00:00]Hello everyone, welcome back to GDC classes. So the students as you all know we are having weekly test of organic chemistry in this today uh of the weekly test schedule and uh now we are going to have the explanation of each questions that we have in our weekly test of this organic chemistry. The topics for this test was what reaction intermediates inductive effect resonance and hyper conjugations. So let's get started like what are the questions that we are having from this topics and let's see each question one by one with the detailed explanation. So the first question we are having is consider the following carboatines. We are having some carboatines that is methile carocatine ethile carocution dimethile dimethile substituted kion and trimethyl substituted kion. The order of the stability.

[00:00:50]So order of stability. Now here we are having very easy question you can say we can see clearly methile kines is having no any substitutions while in the ethile we are having at least one methile substitution on the kion in which we are having and then the next we are having the kion which is having two methile substitutions and again the kion is having three methile substitutions. So either you can have the inductive effect or the hyper conjugation. Hyper conjugation means number of alpha hydrogen's and inductive means 3° kine more stable than 2° 2° is more stable than 1°ree least stable will be what?

[00:01:24]Methile. So the correct order is in this like we are having three methile substitutions. So we are having 3° kine.

[00:01:31]This is 3° kine. This one is 2° carocatine. This is 1°ree carocatine and this is just methile kine. So the correct answer is 3° most stable. So 3° that is what? Fourth is most stable. All right. So two options is having two option is having fourth the most stable one after that two degree uh so three third option okay so here after four you are having the most stable as second no that is wrong so four then three then two and then the one so the correct order was what B option option B is the correct answer for this given question either you apply hyper conjugation or the inductive both of them are having the correct answer for this question all right now the same that is written Here as well you can go for the stabilities of the kion by two methods that are present here. Either you go for the number of alpha hydrogen's or as well as we are having the inductive like 3° most stable one. Next is what which carocatine is most resonance stabilized which will be having the most resonance stabilization. So resonance stabilization this is all. Okay, alli is having resonance and we can have the next resonance structure of this will be like something this. Okay, so this pi bond can sift on these two carbons. So the positive arises here like this one.

[00:02:47]Okay, this is what okay? So this is benzel kion. We are having phenile we are having phenile ring and we are having a CH2 and here in this carbon we are having this positive charge. So this can have maximum resonance stabilization because this positive charge can move at or para and again at the ortho position. So we are having more number of regionating structure. So more resonance is stabilized will be this. This is having no resonance possible. No resonance possible. So these are wrong. Two options can have resonance stabilizations and among A and B most stability due to resonance will be seen in the option B. So yes the B that is benzil cation will be the next answer because of having more number of resonance structure in this given compound.

[00:03:30]Which carb annion is most stable? So which carbonion is going to be the most stable one? We are having methile as three three methile and the carbon where we are having the negative charge. one methile on the carbon where we are having the negative charge and this is having what resonance is stabilized and this is having two methile substitutions for the stability of carbonion carbonions are stabilized by what inversely proportional to minus I group oh sorry plus I groups if you are having alkal groups electron donating groups so the lesser number of electron donating groups are present on the carbon which is having negative charge more will be the stability of that particular compound and if you are having what minus I groups.

[00:04:13]So greater the number of minus I groups more will be stability of your carbon annion. While for the kions that is totally reversed for the stability of kions the stability of kions depends on more number of plus I groups electron donating groups and inversely proportional to what? Inversely proportional to what? Electron withdrawing that is minus I groups. And among hyper conjugation resonance and inductive most powerful is which one?

[00:04:38]Resonance is the most powerful resonance. So resonance because of complete transfer of charge from one portion to the other makes this to have the best stabilized specy intermediate isy. So resonance then you are having the hyper conjugation and after hyper conjugation you are having the inductive effect after that. So in the C case we are having resonance stabilization this negative charge can move here and this pi bond can sift over there. So we can have a resonance stabilization.

[00:05:11]Hm. And this is the structure that we are going to have after resonance. So the most stable you are asked which is most stable. So because of having resonance in the option C this becomes the most stable one. And among all these three you are having 2°ree annion, 1°ree anion, 3°ree nion. So the order will be C is most stable because of resonance.

[00:05:31]And among C, D and A we are having what?

[00:05:34]1°ree. 1°ree that is the B most stable.

[00:05:38]Then two degree because of having two alkal substitution two plus I groups two plus I groups. So after that you are having D and after this three alkal groups three electron donating group three plus I groups lesser the number of plus I more will be the stability of anion. Greater the number of plus I groups electron donating group lesser will be the stability of carbonion. So after that we'll be having the A. So this is going to be our order while the most stable is which one? C because of resonance stabilizations.

[00:06:04]because of resonance stabilizations.

[00:06:06]This will be the correct answer. Option C is the correct one. All right. Option C is the correct one. All kion that was present here is having resonance stabilization and the negative charge of CH2 can have resonance with the alternative just alternately present the pi bond. Which species is not a planer specy? We are having four species.

[00:06:28]So which is not planer? Which is not planer? CHS3 kion a carboatine is having what planer structure this will be having three hydrogens's and one electron deficient p orbital so it's a planer structure this is what CHS3 minus carbon are these carbon this this this carbon is having what this is having pyramidal shape and pyramidal structure is having what if this is pyramidal this will be nonplaner this is nonplaner not planer yes this b option will be not in planer because it is not in resonance. If the negative charge is present in the conjugation resonance that will be having what planerity that can be having the planer structure but here in this case this structure will be this structure will be hot this structure will be non-planer because of having sp3 hybridizations this will be pyramidal in symmetry sp3 and pyramidal shape pyramidal next is what CH3 radical it is also having sp2 hybridization so this will be again a planer structure and we F3 most important this is levy acid having vacant p orbital on the boron planer structure so all these are planer while this is not planer so3 minus that is what methile nion will be methile nion will be a non-planer structure so the b option is going to be the right answer for this given question which intermediate has the greatest number of hyper conjugative structure so the better number of alpha hydrogen's more number of alpha hydrogen's more will be the hyper conjugates and structures hyper conjugative structures you are having kion here this carbon and the carbon directly bonded with the kinic carbon is alpha carbon and the hydrogen's present on this alpha carbon is termed as alpha hydrogen's so here in this case we are having how many alpha hydrogen's three alpha hydrogen's okay three only three hydrogen's are present so this will be having three alpha hydrogen's this is having What this will be having what carbons with two methile substitutions means this kionic carbon is directly connected with two carbons and both the carbons is having three three hydrogen's. So we are having six alpha hydrogen's here in this case. This is directly connected with three carbons and all the carbons are having three three alpha 3 hydrogen's.

[00:08:54]So this will be having nine alpha hydrogen's and this is having no any carbon directly attached with the kionic carbon. So the most number greatest number of hyper conjugative structure will be present on the compound that is having greater number of what alpha hydrogen's. So in the C option you can see in the C nine alpha hydrogen. So this will be having the most number of hyper conjugative structure for the stability of any kion, alken or radical.

[00:09:20]Greater the number of alpha hydrogen's greater will be the hyper conjugative structures and greater the number of hyper conjugative structure means greater is going to be the stability of that particular compound. So it means this C is the C is going to be the most stable kion here as well. Arrange the following in decreasing stabilities. We are having ali radical, benzil radical, tertiary butil radical and the methile radical. So benzil radical most resonance is stabilized.

[00:09:50]Benzil radical is having what? Most resonance stability. Sorry radical. It is radical. So because of having the radical machine because of having radical this can conjugate in the ring as well. So most number of resonance structure will be seen on the benzel that is second.

[00:10:10]Second is going to be the most stable one. Only two cases are having where second is the most stable. After that because of having resonance here in this case as well the first this will be the next most stable structure. So this is again wrong. And after that we are having the third why because we are having what? Nine alpha hydrogen's. We are having nine alpha hydrogen's and in the radical no alpha hydrogen's are there. So first this then this one then this one then this one. So 2 1 3 4 2 1 3 4 The C option the C option C option is the most correct way of writing the stability orders for these given radicals.

[00:10:52]Which carocatine under go the fastest SN1 reaction? Okay. So the SN1 reaction here we are having the SN1 reaction and this SN1 reaction is having carbotan intermediate formation. So in fact you are asked which of the given compound can have the most stable carbotan formation. Most stable carotion formation. If you remove the CL from here you will be having formation of CH3 CH2 and positive kion here because we are having elimination of the Cl minus form. From here you will be having elimination of the CL to give you what?

[00:11:26]CH2 double bond CH single bond CH2 and sorry and and removal of the removal of the chlorine will leads to the formation of kine at this position.

[00:11:40]Then you are having what? C6 H5 CH2 CL.

[00:11:43]So after the removal of chlorine from here you are going to have formation of CH2 positive elimination of Cl minus Cl minus and Cl minus and this CH3 whole a tertiary kion formation of a tertiary kion. So CH3 CH3 and CH3 and positive charge. So the most stable kion will be formed by which compound by eliminating the Cl minus. So the most stable kion is seen in this C option. So the correct answer will be what? The C most stable kion. So which carbotion under goes the fastest SN1 means which is going to be the fastest conversion into kion. So that will be the most stable one. the stable kion will be having the fastest formation of that particular species because that is stable the most stable one among all the given formation. So yes the C option is going to be the best answer for this question. Which carotion is least stable? Which carboation is the least stable? Okay. Venile carotion. No venile carotion is having no stability.

[00:12:49]Why? Positive charge is present on the carbon which is already sp2 hybridized.

[00:12:54]So sp2 means more electro negative than the sp3 and you are having what and you are having what kion on it. So that makes this particular specy to be less stabilized. Ali kine can have resonance stability. Tertiary butil carbutan again more number of alpha hydrogen more plus i groups more stable. Benzil carutan. So among all these given the most stable is what? This one D is the most stable.

[00:13:20]After that because of resonance it will be B and after this you are having C and after that only you are going to have the A. This is the correct way. So the correct answer the most stable is what?

[00:13:31]Benzil carocatine. Benzil car. Okay. So you are asked about the least stable.

[00:13:35]Yes. The least stable is what? Leastable is what? A. Least stable. Least stable is going to be the A. That is correct.

[00:13:42]Most stable is what? Benzil carotion.

[00:13:44]That is correct. That is correct. Okay.

[00:13:46]That is correct.

[00:13:47]Which molecule shows maximum resonance energy? Maximum resonance energy.

[00:13:53]Maximum resonance energy. So here given all option cycllohexine is having no resonance. Not at all resonance possible. This is your cycllohexine.

[00:14:02]Okay. Only one pi bond. So how can it have resonance? Benzene. Okay. Benzene is having resonance. Three pi bonds are there. 1 three butadine we are having. Okay. And propene again no resonance not at all resonance at possible resonance because we here we are having just two pi bonds we are having just two resonating structures two forms of resonance but here in the benzene we are having greater number of resonance structures greater the resonance structure and the amount of energy in the resonance involved in the resonance will be highest for the benzene and this is going to be benzene as the correct answer. Benzene will be going to have the most resonance energy here in all the given four options. All given four options. All right.

[00:14:50]Which compound exhibit both minus I and plus M effect? O because of oxygen connected with the carbon. So carbon oxygen is having what polarization. So this O will be acting as minus I. And because of presence of lone pair this will be acting as what? Plus group. That is correct. NO2. NO2 is minus I. as well as minus M group. CH3 is a plus I group and it will be having plus hyper conjugation.

[00:15:19]It will be having plus hyper conjugation phenomena because of having alpha hydrogen's directly when it attached to benzene ring. CF3 will be having minus I as well as minus H minus hyper conjugation as well. So the only option in which we are having minus I and plus M that is what a option O group O NH2 O R O O CH3 that is what and O R O O C double bond O R all of them are having what minus I but but but but but but but they all are having plus M natures plus M natures. is inductively they are having minus I but due to presence of lone pairs they will be acting as what plus M group. So the A option is the correct answer for this given question.

[00:16:10]Which carboatine is not resonance stabilized? Which is not having resonance stabilization? Yes, this can have resonance stabilization. This is all kion. This is what benzil kine resonance stabilizations. Pi bond sigma bond positive char. We are having resonance at this position and this trimethylilecine having no resonance not at all resonance stabilized. So D option is having no resonance possible. So the D option is going to be your right answer for this given question. This trimethyl trimethile kine is having not at all any resonance structures. Now moving to the next question we are having which alken has the maximum hyper conjugative stabilizations. Which of the given alken is having maximum hyper conjugative stabilization means which alken will be having the maximum number of hyper conjugation structures? That simply means that is direct proportional to the number of alpha hydrogens.

[00:17:06]Greater the number of alpha hydrogen's means more number of hyper conjugative structures or you can also have majority or you can have more number of stabilization of that particular alken.

[00:17:16]We are having ethine which is having no any alpha hydrogens. This ethine is having no any alpha carbon. So no alpha hydrogens's are there. So okay delete this. Then we are having what? Propine that is having what? CH2 double bond CH and we are having one methile substitutions. So we will be having three alpha hydrogen for this two methile propene. So we are having what?

[00:17:40]Two methile propene. So we having CH3 CH double bond CH2 and two methile. So we'll be having a methile substitution at here as well. We are having a methile substitution. So we are having this is our alken and this alken is having two alpha carbons. So three and three six alpha hydrogen's are there. Three alpha hydrogen are there. Now two methile to butine. So we are having butine butine.

[00:18:08]So we are having one carbon carbon double bond carbon and we are having this one. So we are having this is our butine. We are having two methile. We are having one two. So let's suppose this is what two methile butine. So this is the case. This is the case. And how many alpha hydrogen's are there? This is our two methile butine. We are having one alpha carbon, two alpha carbon and three alphaarbon. And because of three these three alpha carbon we are having three + 3 + 3 that is nine alpha hydrogen's. So in the last we are having nine alpha hydrogen's. So more number of alpha is present on which D option it means the best or you can say the most stable alken most number of hyper conjugative structures will be seen in which the D the D option that is this one. So D option is the correct answer.

[00:18:58]Which compound under goes electrphilic aromatic substitution most rapidly?

[00:19:05]Which compound under goes electrofilic aromatic substitution most rapidly?

[00:19:11]nitroenzene, chlorobenzene, benzene and analene. So which will be having the fastest ESR, electrofphilic aromatic substitution reaction. Now the ring which is having the highest electron density that will be undergoing the fastest ESR reaction that will be having fastest electrophilic substitution reactions. Nitro group will be having what minus M nature. Chloro is having what? Plus M but it is having minus I dominant. Benzene having no any substitution. Analine is having plus NH plus M group because of this NH2 because of this NH2 this analine we are having NH2 group and this NH2 will be having acting only as the plus M group.

[00:19:52]So the ring which is having highest electron density will be seen in the enoline because of having the plus nature of this and it means the electrofile can have the fastest reaction on the most electronrich center most electronrich ring and that is what that is what the analine and that is what the analine so analine will be having the fastest ESR reaction ESR fastest arrange the following in a decreasing order of resonance stabilization benzene phoxide ion all kion and the acetate ion. Arrange the following in decreasing order of resonance stabilization.

[00:20:28]This is what benzene, this is what phoxide. This is aliation and this is having what? Acetate ion.

[00:20:35]In the phoxide we are having what? Negative on oxygen.

[00:20:42]In the ali kion we are having positive on what? The CH2. And acetate is also having what? CH3 C double bond O minus no doubt we are having resonance stabilities and resonance stabilities will be seen highest for what and this was benzene let me also write down the benzene benzene is having no charge benzene is a neutral structure benzene is a normal structure and yes resonance is again present on the benzene but among all the other options we can have benzene because of having no charge formation. This will be having the best resonance, most stabilized resonance because no charge was there. After that we are having phoxide all and the what acetate ion. So most of you must be having like like phoxide will be having the next most stable resonance structures but that is wrong. Negative of oxygen after resonance goes to hot goes to hot carbon atom. Okay. This moves here. So the pi bond that is present here moves to this carbon. That means negative from oxygen sifts to the carbon atom and that makes this phenox oxide to be less stable as compared to the acetate. Any of the caroxilate ion is having equivalent resonating structures formation because of both these structures are having nearly same structure. negative from oxygen goes again to the negative from in the oxygen again the next one. So because of this among phoxide and acetate acetate will be the next most stable ion. So the most stable is what? First. Okay. So this is wrong. This is wrong. Only two options are having one as the most stable one.

[00:22:33]And after that we are having what?

[00:22:34]Acetate. So no that is wrong. After one we are having the fourth and after that we are having phenoxide. Then only we are having the alien. Ali kine that is carbon with the positives. So that means the correct stability order is what the A1 1 4 that is acetate and then phoxide then all. So 1 4 32 1 4 2 3 sorry. So A option option A is the correct resonating structures resonance structures stabilizations order. All right. Now the next one which carocatine is formed during the nitration of toluine at the para position. So you are having this toluine you are having CH3 group this will be acting as plus hyper conjugation making ortho and pyro to have a bit more electron density. So which carbotan will be formed during the nitration. So yes we are having a formation of carbutan but that carbotan will be better termed as what arenium or a sigma complex. We are having formation of what?

[00:23:36]We will be having formation of what? So we are having formation of a sigma complex. Sigma complex. All right. And that will be that will be suppose we are having methile at this position and we are having at the para position. So we are having NO2 addition at this H. So we are having positive charge.

[00:23:56]Positive charge. Suppose this will be like having positive charge at suppose this position. Then this positive we are having uh first we are having negatives electron density at this position then higher electron density at this position and after that we are having higher electron density at this particular position. So this will be having formation of what? This will be having formation of a positive center because of addition of this NO2 on the ring.

[00:24:27]This generates a positive center. This generates a positive centers and that is known as what? Sigma complex. And that is known as what? That is known as what?

[00:24:36]Sigma complex formation. Arinium ion formation. Arinium or sigma complex.

[00:24:46]Sigma complex. So that will be having arenium ion formation. Arinium ion formation. That is also known as what?

[00:24:53]Sigma complex. That is also known as what? Sigma complex. Which of the following compound has the lowest pKa?

[00:25:01]Lowest PK? Acidic strength.

[00:25:06]Acidic strength is inversely proportional to PK value.

[00:25:12]More the PK less will be the acidic strength or you can say like lesser the value of PK highest will be the acidic strength. Lowest PK it means most acidic.

[00:25:27]most acidic. So which is the most acidic compound here? Ethanol, phenol, paranitenol or the cycllohexenol?

[00:25:35]Ethanol is a aliphatic alcohol.

[00:25:38]Cycllohexenol again alifhatic like alcohol it is just a cyclic alcohol not aromatic alcohol. While in the case of paranitenol and what phenol in the phenol that is what aromatic alcohol and the paranitenol is having what? Nitro substitution at the para position and because of this nitro substitution acidic strength is directly proportional to presence of minus m groups inverse to plus m groups. So it means this is a minus m because of present at orthoposition. So this paranitenol leads to have the most acidic strength of the phenol and that is the reason most acidic means the lowest pk. So the correct answer will be what? C option.

[00:26:19]Correct option is what? Option number C.

[00:26:22]Option number C. Which compound is least reactive towards electrofphilic aromatic substitution?

[00:26:30]Which compound is least reactive towards electrofphilic aromatic substitution?

[00:26:35]The ring which is having the least electron density or you can say the ring that is having what? Substitution of a group which is minus M strongest minus M that pulls the electron density outside of the ring that makes the electron deficiency inside the ring and that is the reason a electrofile will won't be able to attack on that particular ring.

[00:26:57]So least reactive towards ESR any soul is having what OCS3 group it is a plus M group. So because of this plus M we are having faster attack of the electrofile towine is having CH3 because of having plus hyper conjugation this will be also having electron deficiency high in the ring nitro group nitro group makes the ring to have electron deficiency at ortho and para because of this nitro group is having what minus m character as well as a strong minus I both so because of this this nitro makes the ring to have less electron density and phenol O group is there O again a plus M group makes the electron rich at ortho and par positions. So all the A, B and D are electron density activator inside the ring but the nitro will be electron density decreased inside the ring at orthopera making electrofile to have less reaction less reaction towards this ring. So the correct answer is what nitroenzene nitroen this ring will be having the lesser chances for attack of electrofile.

[00:28:04]Which species aromatic? Cyclloadine, cyclopentadinyl anion, cycllo octatetrine, cyclopentadinyl kine. First one is what? Cyclloadine. So this is having what? Cyclloadine. This is having what? This will be having cyclic structure planer and it is having conjugation pi bond two pi bond conjugation. Total electrons that are here in the conjugation is what? 2 and two. So two pi bonds are there in conjugation. 2 + 2 that is four pi electrons will be there in conjugation.

[00:28:35]If we equivalent this to 4n we are having value of n is equals to 1 that makes this ring to have anti-aromatic structure.

[00:28:45]This ring is a anti-aromatic and if the total conjugated electrons are equivalent to 4 n + 2. If it is equal to 4 n + 2 and now solving this we are having n as 0 1 2 3 and so on a whole number complete number it means that was what a aromatic ring cyclopenta nion cyclopenta denil nion so this is having what negative charge yes this ring is cyclic this is planer conjugated total electrons are what two in this pi bond two electron in this pi bond and two electron on this negative charge. So 2 + 2 + 2 that is what 6 electron. Now if we equivalent this to 4n is equals to 6 we won't be going to have n as what a complete number. n will be what? 1.5. It means no this is not antiomatic structure. But if we make this total conjugated electrons equivalent to 4n + 2. Now in this we are having 4 n is equ= to 2 goes here. 6 - 2 that is 4. N is equals to what? 1. Yes. So this ring is what aromatic in nature. So the correct answer is what?

[00:29:55]Correct answer is what? Cycllopentadinal anion. Cylo octatetrine. Yes. This ring is having exceptional cases. That is what this ring is a nonplaner.

[00:30:09]This ring is nonplaner because of tub-shaped geometry. So this is non-aromatic.

[00:30:16]This one is a nonaromatic one.

[00:30:19]And this was anti-aromatic one. And cyclopenta kion cyclopenta denil kion is having what?

[00:30:31]You can count total four electrons are there in conjugation. So this makes again this ring to have a antiaromatic structure. So this is antiaromatic. This is non-aromatic. This is antiomatic. The correct answer the correct aromatic compound is what? The B cyclopentadanyl anion. This ring will be a aromatic one.

[00:30:50]This is a aromatic one. Which compound is anti-aromatic? Antiomatic. Benzene is aromatic.

[00:30:57]Cyclloadine just we have seen this is what an antiromatic one. Pyodine is aromatic and napylene is again aromatic.

[00:31:04]Napylene, anthraine, fenthrine, pyodine, pyrol, furine, thyophene, pyodazine, pyramidine all of them are aromatic. So aromatic aromatic aromatic only the ring that is antiomatic is what this cyclloadine cycllout dyne this is what an anti-aromatic ring consider the following orders benzil carocatine ali carocutine tertiary butil carocutine and the secondary carocutine the major factor responsible for this order is what benzyl carbotine is having resonance stabilizations alion is having resonance stabilizations. Tertiary but is having more number of alpha hydrogens's. Secondary carabotile is having again a lesser number of alpha hydrogen's. So this major factor responsible for this orderic hindrance.

[00:31:54]No, this is not the correct way to check for the stability orders. Resonance stabilizations. Yes, that is correct. In these two options, we are having resonance. So that is incorrect. That is incorrect. So yes, resonance stabilization followed by hyper conjugation. two structures is having resonance and then the order is followed by what hyper conjugations. So the B option is what? The correct answer.

[00:32:17]B option is the correct one for this given question. Choose the correct option. Assertion phenol is more acidic than ethanol. Phenol is more acidic than ethanol. Yes, that is correct because after the dissociation of phenol we are having formation of phoxide ion. The phoxide ion is stabilized by resonance where the ethoxide ion is not. Yes, that is correct. This phoxide is having resonance stabilization and according to the rule more stable conjugate base of any acid will be more acidic one. So yes that is correct. Both of them are correctly written. So both a and r are true and r is the correct explanation for the assertion a.

[00:32:58]Next question is what? Moving to the next. Okay. Hyper conjugation is also called as no bond resonance. Hyper conjugation is called as no bond resonance. That is also known as Baker Nathan effect. That is also known as what? Baker Nathan effect. Okay. Baker Nathan effect. Yes, that is correct.

[00:33:17]That is correct. One of the contributing structure of hyper conjugation depicts the cleavage of a carbon hydrogen sigma bond with electron deoization. So yes, that is a a correct one.

[00:33:30]Suppose you are having this uh C H and you are having what? Positive charge at this position and you are having this one, this one, this one. So this is your kion carbon and you are having a carbon that is directly attached with this kanic carbon. So this is your alpha carbon and on the alpha carbon we are having alpha hydrogen's. So what happens here? The electron of this alpha hydrogen and carbon bonds moves towards the carbonarbon double bond to give you structure like this. You will be having a formation of this structure and you are having formation of H+. Both the electrons goes here like this. We are having a pi bond formation and we are having a slight second a very fine very minute time for the formation of this H+ and again that goes back and we are having formation of a carbon hydrogen bond and the same phenomena will be happening with this alpha hydrogen and then by this. Yes, that is correct. Both of these options are also correct in which we are having deoization of carbon hydrogen sigma bond and followed by electron deoization with the adjacent carbon. So that is correct.

[00:34:46]The NO2 group nitro group is a meta directing group. Yes, NO2 is what? A minus M group and this decreases electron density at ortho and para positions. And now the incoming electrofile can have only attack at the meta position. So it is a metadirecting group in electrophilic aromatic substitution. Yes, that is correct. The nitro group donates the electron. Oh oh oh oh oh oh oh oh oh oh oh oh oh oh oh oh oh oh oh oh oh just a second. Nitro group donates electron to the benzene ring by resonance. So nitro group is a one of the strongest minus m group that pulls the electron density from the benzene ring towards themselves. So no they do not donate they actually withdraw they withdraws electron okay so statement two is wrong but statement one is true. So one is true but the statement two is false. So the correct answer is what? B here. B option is the correct one here in this case.

[00:35:51]Option B is the right answer. Hyper conjugation requires the presence of at least one alpha hydrogen atom.

[00:36:00]For the hyper conjugation to be present, you must be having at least one alpha hydrogen. That is correct. The methile carboation that is CH3 positive cannot undergo hyper conjugation. Methile cannot undergo hyper conjugation. Yes.

[00:36:14]Again that is also correct. This is the methile kation. You are having positive charge on this and you are having three hydrogens's that are directly connected with this kanic carbon. How can you have the hyper conjugation possible here? No alpha hydrogen's are there. Okay. No alpha carbon. So no alpha hydrogen's.

[00:36:34]Yes, that is also correct. Both the statements are true. Both the statement one and two are true. Both of them are correct here. Okay. Next question.

[00:36:43]Tertiary butil kion you are having tertiary butil kation. Okay. So let me draw the tertiary butil kion as well.

[00:36:52]This methile methile and methile and the positive charge is more stable than the isopropile kion. Isopropile. Okay.

[00:37:03]And you are having a hydrogen here and a positive on this carbon. Yes, that is correct.

[00:37:09]So tertiary but is having 33 36 39 9 alpha hydrogen 3° it is a 3° carbon and this one is just a 2°ree kion six alpha hydrogen's are there only tertiary butilicide has more alpha hydrogen tertiary butilicion has more alpha hydrogen yes nine alpha hydrogen's available for hyper conjugation than the isoprofile which is having just six alpha hydrogen's for the hyper conjugation both of of them are true and they are correctly explaining the assertion. Reason is correct correctly explaining the assertion. Okay. So that is again right answer. Carbonion stability decreases with increasing alulus substitutions. If you are having carbon with the negative charge and you are having it stability of this carbon decreases with increasing alkal substituents. Yes.

[00:38:02]Because of alkal groups are having plus I that is electron donating characters.

[00:38:07]And this electron donating characters reduces or diminishes the stability of your kion carbonion. Because annions are already electronrich they are actually reactive because of presence of high electron density. And you are again making such substituents which are already donating electrons. It means that will be having lesser stability.

[00:38:30]They will be having reduction in the stability order. So yes that is correct.

[00:38:34]Assertion is correct. Alkal group exerts a plus I effect. Yes. Which intensifies negative charge on the carbon and carbon. That is correct. That is the reason like the alkalle substitutions.

[00:38:48]More the number of alkal substitution leads to destabilization of what your anionic charge. Anionic charge. So both of them are again true and reason explains the assertion correctly.

[00:39:01]Triricchloro acidic acid you are having C double bond O and you are having triricchloro okay one chlorine second chlorine and the third chlorine triricchloro acidic acid is a stronger acid than acetic acid and this is your acetic acid this is your acetic acid by removal of the H you'll be having formation of this acetate ion resonance stability is present here and resonance is also present in this minus with the C double bond but the substituents that are present. This is a plus I donates electron here and this is having three chlorine. So this pulls the electrons towards chlorine. It means the negative charge the acetate triricloroacetate ion will be more stabilized because of this three chlorines and this is having destabilizations because of methile because of this alkal group. Chlorine exerts a plus I effect. No that is wrong which stabilizes the caroxilate. Now reason is false but the assertion is true. Assertion is true but the reason is false. C option assertion is true but the reason is false.

[00:40:10]Tricchloric acid is stronger acid than acetic acid. That is correct. But the triricloro 3 chlorine substitution leads to stabilization of the triricchloro acetate conjugate base. That is why we are having more stability of this triricloroastic acid than the acetic acid.

[00:40:30]The carbon carbon double bond connecting the methile group of the double bond in a propene is shorter than a typical carbon sp3 carbon sp3 single bond.

[00:40:41]Carbon carbon double bond connecting methile double bond. Okay. So you are having carbon carbon single bond that are connecting the methile groups to the double bond in propene.

[00:40:57]This is having a propene structure and you are having CH3. So this is your single bond and this is shorter than carbon carbon single bond of what? Let's suppose we are having propane CH3 CH2 CH3 this one this is because of what hyper conjugation between the methile that is sigma bond of carbon hydrogen bond and the adjacent pi bond imparts a partial double bond character to this bond as well. So look this is your propane this is the propane again here we are having carbon methile sigma bond carbon mythile single bond sigma bond but still this is shorter than this one. Why? What is the reason? If you are having sigma bond here sigma bond here so this sigma bond the single bond has to be same bond length but still it is seen this is shorter than this one. The reason is what? Because here we can have hyper conjugation. we are having methile means this is alken this is alpha carbon and we are having at least three alpha hydrogen's so because of this alpha hydrogens's we will be having carbon carbon and let's suppose this is again the third carbon I'm writing this like this and you are having three hydrogen so 1 2 hydrogen is having plus form the electrons are shifted here to make a pi bond formation and as soon as we are having this pi bond formation so suppose Let me have it here. Let me have it here. Just a second. Just a second.

[00:42:48]This is the propene 1 2 3 and we are having H and H. How this alpha hydrogen's affects the bond length. So we will be going to have one of the carbon hydrogen electrons shifting here. As soon as these electron sifts here, we will be having formation of a carbon carbon pi bond. As soon as we are having pi bond formation between these two carbons, this electron of the pi bond sifts on this carbon. And now we are going to have formation of carbon 1 H 1 H and we are having carbon carbon sigma bond and we are having a negative charge here. And because of this we are having carbon we are having H+ because of both the electrons were here involved in the pi bond formation by this carbon carbon double bond and we are having two hydrogen's left like this and that is the reason we are having pi bond that is seems to be having only present between these two carbons is now actually present here as well that is because of what hyper conjugation and this is the reason We can say we are having a partial double bond character between these two carbons as well. And because of this partial double bond character the bond length of this carbon carbon that is seems to be a single bond but still that is having partial double bond character and that is why it is having shrinkage in the bond length. We are having reduced in the bond length and that is the reason it is having bond length lesser than what a normal carbon carbon single bond. This single bond that is present between these two carbons here and this this is shorter.

[00:44:41]This is because of what? This is because of this is because of hyper conjugation possibility with this adjacent double bond system. And yes that is written here hyper conjugation.

[00:44:56]Hyper conjugation between the methile carbon hydrogen bond and the adjacent pi bond imparts partial double bond character to this bond and that is the reason we are having reduced in the bond length. reduced in the bond length. All right. So a option is again the correct one. Methile kation cannot be stabilized by hyper conjugation because of having no alpha hydrogens. No alpha hydrogen is present in methile. Yes. So because of no alpha hydrogen we cannot have re hyper conjugation stabilization. Methile kion has no alpha hydrogen on adscent carbon.

[00:45:37]Yes correct correct correct correct correct correct correctly explained.

[00:45:43]Paran nitroenoline is a much weaker base than analine. This is your paran nitro NO2 and NH2. So paran nitroenoline is a weak base as compared to the enoline that is correct. Basic strength basic strength is directly proportional to plus M groups substituted plus M groups and inversely proportional to substituted minus M groups. Nitro is actually a minus M group. That means this must be having reduction in the electron density of the nitrogen nitrogen of the NH2. And that is why we are having paranitinylene to be a weaker base as compared to any because no substitution any present here that reduces the electron density of the nitrogen in N2. So yes that is correct.

[00:46:43]The nitro group exerts only an inductive effect. The nitro group exerts only an inductive effect on NH2 group with no resonance interaction. With no resonance interaction. Is it true? No. That is totally wrong. It is having a strong minus M group. It is having strong resonance effect on the N2 as well because it is present at the para position. So reason is falsely written here while assertion was true. So this is true but the reason is false. Assertion is true but the reason is false. Reason is false. Okay. Which of the following carb annion is the most stable? Most stable carb annion. So again because of having resonance stabilization in this this will be having the most stable one.

[00:47:32]So the C option is what? The correct answer. Option C is the correct one.

[00:47:36]Which compound shows the greatest resonance stabilizations?

[00:47:40]Resonance is not at all present here.

[00:47:43]Resonance is never present here. Not present here. But this is what yes here we are having the resonance. How? This is CH2 double bond CH and we are having carbon double bond O H. So we can have possible resonance that looks like something this positive carbon hydrogen double bond carbon minus and H. So this is the resonance in the case of what?

[00:48:15]This is what propenal propenal. Okay. So this can have the resonance stabilization. This is having condition for the resonance pi sigma pi bond. So yes this B option can have resonance.

[00:48:28]This can have resonance formation of resonance.

[00:48:32]Which carocatine cannot undergo hyper conjugation. The carodine which is having no alpha hydrogen's. This is having three alpha hydrogen, six alpha hydrogens, nine alpha hydrogens, zero alpha hydrogen's. So because of presence of zero alpha hydrogens this will be having no alpha al no alpha hydrogen. So no hyper conjugation structure hyper conjugation is not at all possible. The hybridization of carbon in a carboation is carboation. If we are having carboation this will be having a sp2 hybridization and the geometry will be what? Trional planer.

[00:49:10]Triional trional planer.

[00:49:16]All right. Trional planer sp2 sp2 for the carbonion it will be having the sp three sp3 stability order of carocatons what will be the correct order of stability order?

[00:49:30]So not at all any resonance is here. So direct you can go for the 3° 2° 1°ree then methile. So the B option. Why?

[00:49:39]Because we are asked about the stability. Suppose you are asked about the energy of the carocutions.

[00:49:44]Energy. So what will the order of energy? Do let me know in the chat section. Energy. What will be the order of energy of the kions? Energy and stability is inversely related. So if this is the order of stability, energy will be just opposite. This most energetic less lesser least. Okay. So the correct stability order is what? The B 3 2 1. than methile.

[00:50:09]The geometry of a simple carb annion that is CH3 minus is best described as what is the geometry of CH3 minus. It will be sp3 only one option is having sp3 and that will be what pyramidal in shape is structural with what ammonia. Ammonia is also having the same thing but in the ammonia we are having a what what lone pair but here we are having what negative charge. So carbonions and ammonia are having the iso structural pyramidal shapes and they are having sp3 hybridizations for both the cases stability order of carbon ions. So for the carbon methile carbonion the most stable then 1° then 2° then 3° that will be the order. So for the stability of carb annions we are having this was the order for kions but this is the order of what nion methile then 1°ree then 2° then 3° carbon are reactive because of having higher electron density on the carbon and because of having no substitution no plus I substitutions on the methile that will be having the basis stabilizations in the one degree we are having one alkal group means one group that can donate electron that intensifies the negative negative charge of the carbon and means okay that is bearable. 2° nion. Two groups are more intensifying the electron density on the carbon.

[00:51:35]Already we are having electron rich center because of this negative charge.

[00:51:39]But two groups making more and more unstabilization of them. Three degree three alkal groups three electron donating groups least stability on the carbon. The negative charge will having the least stability. So B option is the correct one. Now the next one free radicals are best described as species with h a complete octate and negative charge complete octate free radicals are having incomplete octate seven electrons unpaired electrons. Okay and unpaired electrons and are electrically neutral. Yes, they are electrically neutral but they are having a one unpaired electron. That is correct. That is correct. That is why they are having paramagnetic paramagnetic behavior.

[00:52:33]They shows paramagnetic behavior.

[00:52:36]That is correct. Empty p orbital. No empty p orbital and positive charge.

[00:52:40]This is regarding what? Carocetions. Two unpaired electrons with no charge. Two unpaired electron. This is regarding what? triplet carbine.

[00:52:50]This is the option for the triplet carbine. So that is incorrect. That is incorrect. That is incorrect.

[00:52:57]This was for carbonion. This was for triplet carbine.

[00:53:03]Triplet carbine. Two unpaired. No unpaired no charge but still electron deficient. That is what? Singlet carbine.

[00:53:16]Correct answer is what? B. Correct answer is what? B. Stability order of free radicals. Free radicals will be what?

[00:53:27]Methile 1°ree, 2°, 3°. Radicals are stabilized by number of alpha hydrogens.

[00:53:36]In the 3° you are having nine alpha hydrogen's, 2° six alpha hydrogen's, one degree three alpha hydrogen. Methile no alpha hydrogen's. So the correct answer for the stability of radical will be this one that is 3 2 1 then methile for the free radicals. For the free radicals a singlet carbine has which electronic configuration a singlet carbine singlet carbine will be looking like this. We are having carbon. We are having a hybrid orbital and we are having two substitutions and we are having one vacant p orbital. So two electrons are there in this hybrid that is sp2 and we are having groups substitutions here.

[00:54:24]So group and group something present here. So this is your singlet carbine in which two electrons are present in the paired form. Two unpaired electrons in two different orbital sp2 plus p. No, that is incorrect. Two unpaired. All the electrons are paired here. So that is wrong. Paired electrons in one sp2 orbital. Empty p orbital. So we are having an empty p orbital.

[00:54:52]This is what p orbital. That is correct.

[00:54:57]We are having a hybrid sp2 orbital where both the electrons are filled. So that is correct. That is the correct one. Two paired electrons in a p orbital only.

[00:55:07]No, that is also incorrect. No electrons. That is one. That is also incorrect. So B is the correct one for your single carbine. In the case of triplet, we are having sp hybridization and two p orbitals.

[00:55:22]Each of them is having one one electron and we are having two groups attached on the carbon and this will be having the sp hybridization. This will be having the sp hybridization. So this is the triplet one and the one was what singlet carbine. So this is correct nitrines in which nitrogen is having just one substitution one lone pair and two unpaired electrons or the paired electrons depending on the case like what we are having in the singlet and triplet carbines. Same here as well for the nitrines.

[00:55:52]Nitrines are nitrogen analog of carbines. Yes, that is correct. The nitrogen in a nitrine is so the nitrogen in nitrine is having one group substitution.

[00:56:05]We are having a lone pair and either we can have one p orbital empty and two electrons filled in a sp2 orbital or we can have both the electrons unpaired in a different p orbital. So sp3 hybridization not at all possible either sp2 or sp electron deficient that is correct dalent nitrogen dalent no no it is having monalency only one group is present in the here case of nitrogen fully octate satisfied like N3 that is again incorrect always negatively charged okay so that is also incorrect electron deficient that is correct divalent nitrogen with a lone pair and six electrons so This is also not exactly correct. It has to be the monoalent. Here the nitrogen is having monoalency. Only one of the bond is present on the nitrogen. So technically all the four options are incorrect. But I hope so that this is the right answer that is correct. Why? Why? Why? Look nitrogen like nitrine simply nitrine is like R N this one and either we are having the lone pair and now we are having left with the two p orbitals either they are hybridized or unhybridized. If you are having one hybridized If you are having one hybridized P sp 2 orbital and we are left with what a vacant p orbital this is what this is what singlet nitrate or we can have r n and we are having filling of electron one here one here. So this is what your triplet nitrine. So the correct answer will be this one. But you have to be very clear here. Carbine is divalent.

[00:57:53]Carbon is having two bond in the case of carbine. But for the nitrine, nitrogen has to be monoalent. It is monoalent.

[00:58:00]Only one of the bond is present from the nitrogen directly. Okay. It is again uncharged specy. No charge is there. It is electron deficient. Is it having six electrons? Okay. It is having six electrons. So all the other options are correct. And it is having only sp2 or the sp hybridization. So that all the options are incorrect only the correct answer only the most appropriate mostly uh correct answer is what the B that is this one that is this one. Okay.

[00:58:31]Uh now the inductive effect operates through inductive effect operates through pi bond only. No sigma bond via permanent partial polarization. That is correct. Unpaired electrons in p orbital. Incorrect. Only through space.

[00:58:45]No no no. Sigma bonds and we are having displacement of the sigma bond electrons because of having difference in electro negativity of the atoms that are having actually sigma bond formations and they will be having more electro negative atom pulling the electrons toward themselves and that is how we are having partial polarization. One of the more electronegative atom is gaining partial negative charge and the lesser electro negative is having what partial positive charge formation.

[00:59:12]And the next one which is an example of plus I that is electron donating effect.

[00:59:18]NO2 is minus I caroxilic is again minus I. This is what plus I group alkal group. C and the florine is again a minus I group. So the correct answer is what? C. Alkal group will be having the plus I for here phenomena. Arrange the following groups in a decreasing order of minus I effect. NH2, O, florine and NO2. NO2 will be having the strongest minus I. Incorrect, incorrect, incorrect. Only one option is having NO2 as the highest. Then we are having florine, oxygen and nitrogen. So then florine, then oxygen, then nitrogen. A option is the correct one. Option A is the correct one. Uh moving to the next question we are having. Which acid is strongest?

[01:00:03]This is what acidic acid 1 chloro 2 chloro 3 chloro. So acetic acid monocchloro acidic acid dchloro acetic acid triricchloro acidic acid. So because of having three minus I groups triricchloro acidic acid will be the most acidic compound. Most acidic compound D option will be the correct answer. D option is the correct one. As the distance between an electron withdrawing group and the reacting center increases, the inductive effect distance between electron withdrawing group and reacting center increases. If the distance increases, inductive effect decreases.

[01:00:43]Inductive is effect is a distance dependent effect. So that is if distance is increasing it decreases rapidly.

[01:00:54]It decreases. Inverse relation is regarding the uh what you can say distance with the inductive. It is inverse. It is inverse. Correct order of acid strength is acid strength correct order. Four nitroenzoic acid is less acidic than three nitroenzoic acid which is less acidic than benzoic acid. That is totally wrong. Why? We are having substitution of nitro groups. So nitro groupoups always make benzoic acid to have a better acid. better acid. Okay.

[01:01:25]Substitution of nitro groups make the particular group particular benzoic acid a better acidic compound. So we are having four nitro means nitro is present at the par position. So it will be having the minus m as well as the minus I. But nitro is present at third means meta position. So it can have only minus I. Benzoic is having no any substitution. So the most acidic because of having minus I minus m both effect.

[01:01:52]Then with the minus I only nothing is there. So the correct order is what? B.

[01:01:57]Correct order is what? B. Acidic strength is directly proportional to minus M minus I inverse to plus M or plus I groups. So the correct answer is B. Resonance that is meism involves actual physical oxillation of electrons between two real structures. between two real structures we are having actual oxilation of electrons. Okay. Deoization of pi electrons or lone pairs through multiple contributing structures having the same atomic positions. There is no any change in the atomic positions.

[01:02:35]Movement of atoms between positions.

[01:02:37]Atomic moment is seen. Permanent unequal sharing of sigma electrons.

[01:02:44]Now in the first actual physical oxillation it is a hypothetical it is a hypothetical concept. There is no any actual uh physiological actual physical oxillation of electrons. Movement of atom between position between positions.

[01:03:02]Is there any movement of atoms was seen?

[01:03:05]No. Only bonds only electrons were having movement if they were present in the conjugation system. So that is incorrect. permanent unequal sharing of sigma electrons. No, it is only and only limited to pi bonds, lone pairs, p orbitals. That is again incorrect. So we are having permanent or inequal sharing of sigma electron that was inductive.

[01:03:29]All the options are incorrect. Yes, it is a deoization of pi electrons or lone pairs through multiple contributing structures. We are having multiple resonating structures that is known as canonical structures as well. And we are having a real hybrid of all the given canonical structures that is thought to be having the real structure. So yes B option is what correct and no atomic positions that is all of them are having same atomic position. So no atomic moment is seen. Okay. All of them makes the B option as the very true answer of this question. In the analine the lone pair on nitrogen is localized deoization. It involved in deoization because of presence of the conditions for the resonance deoiz deoizes into the benzene ring making nitrogen less basic than a cycllohexylamine.

[01:04:21]Cylohexylamine the nitrogen is what localized. This is your cycllohexylamine this. So here in this case we are having localized electron pair on nitrogen. So yes that is correct. that is more basic compared to the anal because in the analine we are having movement of lone pair of nitrogen makes nitrogen more basic than ammonia. No ammonia is more basic than analine. Why? Because of having localized electrons on the ammonia has no effect in the ring's reactivity.

[01:04:54]What are you talking about? Because of this electron pair deoization inside the ring. This makes ring more electron rich making more faster electrofilic substitution reactions. So yes it is having direct relations direct so many reactivity orders for the ESR electrophilic aromatic substitution reactions is seen the number of hyper conjugative structures alpha hydrogen's for isopropile kion is how many number of alpha hydrogen's are there in this case so we are having two alphaarbons directly attached with the car kic and two alpha carbons is having three three hydrogen's it means we are having total six alpha hydrogen's so the total number of alpha hydrogen Total number of hyper conjugative structures for this given compound will be what? Equivalent to six equivalent to six. So the B option is the correct one. C option is the correct one. Okay. So this is all about your today's weekly test explanations. I hope you all are having this understood each question by question and I hope so you have marked the questions correct in your examination in your test as well.

[01:06:02]Okay. So that's all for today. See you again in some other videos with some new questions or some new information, newer topics. Thank you so much.