GDC Weekly Test 2026 | Drugs acting on ANS- Medicinal Chemistry | Detailed Explanation |FREE Test

[00:00:48]Hello ma'am please start.

[00:00:55]Okay. Thank you.

[00:01:37]Hello everyone. Good evening.

[00:01:40]So how was your test before that? Welcome to GDC classes English channel. So has ev everyone given the weekly test? Yes. How was it?

[00:01:55]Is what is it easy or what? You find it difficult.

[00:02:01]How was the taste?

[00:02:05]Okay.

[00:02:10]Okay. So, today we are here to start with our or to we can say this is an explanation for our GDC weekly test for the medicinal chemistry. It was at 5:00 p.m. I guess everyone attempted. Hope so everyone attempted it. Okay. So basically on autonomic nervous system it was there. So let's start discussing it.

[00:02:33]How many questions you have done it right or what? Okay. So starting with our first question we have here that is the greatest adronergic activity of a direct sympathomiimetic agent is when aromatic ring is separated from an amino group by. Okay. So basically they are asking the question from the structure of our sympathomimetic agents. Okay. So basically they have four carbon atoms, three, two or five carbon atoms in that chain. Okay. So basically they are specifically asking about the number of the carbons. So as we know all the sympathomiatics they have similar kind of structure that is basically a ring we can say and then we have what we have two carbon chain and amino group is attached there like this. Okay. So basically that two carbon atoms between that is very important. Okay. Yes. The answer is the C1. Okay. So basically those carbon atoms if they are more than two or less than two that will result in the uh decrease in its we can say binding to its receptor. Okay. So basically that two carbon atom distance is very important. Okay. So that's why C is the answer. Like here also you can see in the phenile ethylamin. Okay.

[00:03:58]Okay, we have two carbons in between and that distance we uh consider it as it gives the better binding and better potency. Okay, next question. How does phenile aphrine different differ from adrenaline?

[00:04:15]Okay, so adrenaline adinergic drugs they are talking about all the catakolamins we can say. Okay, so that adin adrenaline is a type of catacolamin only. So if we talk about how this adrenaline is different from phenilein.

[00:04:30]So if I say adrenaline adrenaline has a kacol ring. Okay kakolum mean we can say kakol ring in its structure and along with that it has a uh n h and then here we have the chestries attached. Okay. And if we talk about phenilephrine. So in phenilephrine what we have? We do not have a kakol ring but we have a phenol ring here. Okay. So this is the ring that is present in phenile. Rest of the structure is almost similar. So lacking a four CO and H2 on a benzene ring. I guess no. On benzene ring we do not have a CO and H2 here. Lacking a 4 CO group on benzene ring. No, that is also not true. Not a not. We are not talking about a benzene ring here not CO lacking a four O group on the benzene ring. Yeah C1 I told you now that in phi in adrenaline we have two hydroxy groups and then phenile one hydroxy group is removed. Basically a phenol ring is present there. Okay. So that is our phenile aphrine. So C1 lac and four O group on the benzene ring. Okay. The C1 is the correct one. Here you can see and structure of an adrenaline. Okay. And here is a phenile. So basically do not have an four O group. Okay. So 1 2 3 4 at fourth position the O group has been removed in phenile. So that is the difference. Okay. Very easy question.

[00:06:06]Next. Dopamine hydrochloride. Dopamine again a type of kakolamin. Basically in the sympathetic system it works. The BP British Pharmacopia they're talking about a crystalline powder and a freely soluble in water is in which color basically they're asking about the color or the crystal what colored crystal we have of the dopamine hydrochloride they are they orange red green or white yes anyone who's going to answer me that one so basically if we talk about dopamine hydrochloride so they have what they basically have white crystals. Okay, the structure has what? White crystal. It is a white crystalline powder only.

[00:06:52]Dopamine hydrochloride. Okay, so it is white in color. If we talk about appearance and it is very freely soluble in water also. Okay, coming to the next.

[00:07:03]Which form of enumer of proprinol is active? Okay. So basically proprinol what is that? LOL drug that is it is a type of beta blocker and if we talk about its structure it has a napylene ring in its structure and they are saying that which form S form cis form R form or transform which form of this proprinol is most active okay so basically if we talk about its activeness so one form is more active than other basically the optical isomeism also says so s isomer Okay, that is a naturally occurring enanumer of proprenolon. Okay, and it is very much more active about 60 times more active as compared to its R form. Okay, so the S form is more active as compared to the R form. Okay, it is about 60 to 100 times more affinity than R in animer means much much much more affinity as compared to the other form. So the S in anumer is more active. Okay.

[00:08:13]Yes. Next. Dash ions. Okay. This one is simple one. Okay. At least everyone should try to answer. Now whatever you answered in the weekly test, you just give me that answers. These are the simple questions. They're not very difficult. So dash ions are responsible for the release of acetile colon from neuron. Okay. If you are wrong that doesn't matter. I'll not scold you or something just I want to know that where you are getting or where you are lacking so that I can help you better. Now if you give me wrong answers I will I will explain you more better. Okay. So that's why I'm saying at least participate whatever questions are there there.

[00:08:53]Okay. So ions dash ions. So basically if you remember acetylon is synthesized from the amino acid serin. Okay. Serin is an amino acid by which the acetyl colon is synthesized and after its synthesis the acetile colon is stored in the vesicle inside a neuron. Okay. So inside this vesicle we have acetyl colon. Why? Because in the cytoplasm that is outside there in the cytoplasm acetile colon eststerase enzyme is present and that acetylon eststerase can degrade this acetyl colon that's why we store it under the vesicle inside the vesicle the acetylon is stored and whenever required they are released okay and they are released by uh action of one ion there and that ion they are asking so those ions are basically What those are are the calcium ions. Okay, calcium ions are there. Those are responsible for the release of those acetile colon from the neuron so that they can get there bind to the receptor and can show their action. Okay. So calcium ions and this is the process I told you know that basically from the help of the seren amino acid the acetyl colon is synthesized and acetyl colon is stored under the vesicle okay like this vesicle why inside the vesicle because acetyl colon eststerase enzyme can degrade the acetyl colon that's why inside the vesicle they are stored and after the calcium enters there and due to the calcium influx What happens the vesicle it releases the acetile colon outside the sinap so that it can bind to its receptor and can show its action.

[00:10:46]Okay. Next pilocarbon is an alkyoid containing.

[00:10:53]Pilocarpin is an alkyoid containing.

[00:10:55]Pilocarpin is a very common drug that we have studied in autonomic nervous system. I guess everyone know about that. Even I told you a trick for that.

[00:11:04]P I L L O W It sounds same. Pillow.

[00:11:09]Pillow. And what alkyoid? What particular ring is contained in that? It is pyrozole, oxazol, emidazol, pyzole.

[00:11:17]Okay. The pyrol that is a simple one.

[00:11:20]Pillow carbon. I told you pillow. A soft cushiony pillow. Okay. Uh looks very cute. Okay. So cute shape. It has like a pentane cute shape. It is five cornered.

[00:11:33]And as pillow has O just like that this ring also has an O.

[00:11:40]Okay. So like this it has a pillow. So one ring is this that is present. So it is a furon kind of ring. Okay. So in options furan is not there. Other ring that is present in pillocarbon is the imidazolei. Yes. Now you are right.

[00:11:56]Okay. So yes you have to participate.

[00:11:58]Okay. Yes the C1 is the correct one. So basically two rings are there. One is the furan kind of ring. Other one is the emidazole ring. Okay. So that is our pilocarbon.

[00:12:10]Again the simple question is leop, dextroof, optically inactive or misoir.

[00:12:18]Okay that's a very simple question.

[00:12:20]Atropen is what form is atropen? Do you remember what is an atropen?

[00:12:26]Yes. Atropen is an Atropen is an Yes. Atropen is anic mixture.

[00:12:39]Remic mixture of what? Remic mixture of hyiosamin.

[00:12:44]Atropen is a reimmic mixture of hyiosamin means 50% L form and 50% R form of hyiosyamin. D form of hyiosyamine is present in the atropen and hyioyamine is formed by the asterification reaction of tropin and tropic acid. Yes, the answer here is yes we can say it is an yes it is an antiolinergic drug. You are right Samriti it is an anticolinergic drug and it is optically inactive. Why optically inactive? because it is a reimmic mixture so the net charge here is zero okay so that's why it is optically inactive so C1 is the correct one here okay I guess everyone got it right yes the C1 is the right answer okay so which of the following is a kakolamin okay I guess everyone knows what is kakolamin the compound that contains a catacol ring thing in its structure is called as catakolamine.

[00:13:57]So this kind is called as catakolamine. Okay, we can call it a catakolamine. Now it contains a catacol ring. So thyroxine, melanin, thyroamine or dopamine. Which one is a kacolamine?

[00:14:17]I guess everyone knows sympathomiimetic drugs are kakolamines. Okay. So that kakol ring is very important for the activity and there are three neurotransmitters of that particular system ephedrin nor ephedrin and dopamine. So yes the dopamine this structure belongs to dopamine only. So that is a type of catacolamine. Yes you all are right the D1 is the correct answer. Okay dopamine here you can see the structure of dopamine. Okay coming to the next question. Okay, I already answered this question. Okay, in the last to last question. Atropen is an aster of I told you atropin is a recemic mixture. Remic mixture of hyiosyamine and hyioamine is formed by the asterification of what? Tropin and menalic acid. Tropin and tropic acid.

[00:15:10]Tropic acid and salicylic acid or tropen and salicylic acid. Which one? Yes.

[00:15:15]Atropen is an aster of what? It's a easy one. I right now I told you about that it's a reimmic mixture of hioyamin formed by the asterification of yes the B1 is the correct one that is tropen and tropic acid very good okay so the yes the tropen and the tropic acid here you can see eststerified together the H2O is removed to form a tropen okay next question in which drug is the pharmacological activity activity associated with a specific optical isomer. Again the simple question the pharmacological activity is specifically associated with the optical isomer adrenaline aspirin pheninoarbone or acetylcoline. I guess this is the simplest one. Everyone knows the answer of this question according to me. The optically optical isomer we have done it so many times.

[00:16:17]specific optical isomer. Which one?

[00:16:23]Uh do you remember the form of epinephrine, R epinephrine, Sepinephrine? Yes. Do you remember that?

[00:16:32]So in that we study that basically the O that is present either it is present in the dash form, either it is present in the veg form. Based on that we have two forms of epinephrine that is R and S and R form. Which form is more active? Yes, the answer is the epinephrine or we can say adrenaline both are same. If we call it epinephrine or epi adrenaline both are same. And if we talk about which form is more active. So R form is 10 to 12 time more active than the S form.

[00:17:07]Okay. So adrenaline is a pharmacological activity associated with the specific optical isomer where the R form of adrenaline is more active than its S form. Okay, here you can see the R isomer that is a naturally occurring isomer and it is more active. Here you can see the D w ve wedge form and here you can see the dash wedge. Okay, based on that the optical activity is only changed. Okay, so that is the answer.

[00:17:35]Next, the chemical name of ephedrin is Okay. Ephedrin. Yes. Okay. Do you know the structure of ephedrin? Ephedrin is a very simple structure. Uh let me draw it for you. So, ephedrin is not a catacle ring. Doesn't contain a catacol ring.

[00:17:56]This is the ring we have here in ephedrin. And we have this kind. Okay.

[00:18:02]And what we have? We have an O group here.

[00:18:07]We have a CH3 group here. And then we have what? We have N H attached here. Okay. So this kind of structure is present in what? What? It is the ephedrin structure. And if we say the UPSC name. So UPSC name it has a what? It has a propane chain and that propane chain has O. So it has propenol.

[00:18:38]Okay. So propane one all that is present. Okay. So propane to all, propane to all, propane to all. Okay. So the we have a propenol here. Okay. So basically one at first position we have a propenol and at first position only we have a phenile ring attached also. Okay.

[00:18:56]So at first position only we have phenile ring and then at second position one this is one this is two. So at second position we have what? We have a methile amino attached. So at second position we have a methile amino attached. So the A1 is the correct UPSC name. Yes. Yes. The A1 is the correct UPSC name. Here you can see the structure also of our ephedrin. Okay.

[00:19:26]Next question. Which of the following isomer is more potent as a musceranic agonist? The question is from S. The structure activity relationship. Not a difficult question but it sounds difficult. Okay, it's very easy. Which of the following isomer is more potent as a muscerinic agonist? So in the basically the parasympathetic drugs they act either on the nicotenic receptor or muscerinic receptor. Okay they have a special kind of uh structure we can say like if we say a ring we have okay then we have a twocarbon bridge and then we have an amine group. Okay like this we have. So when we add a methile group according to its S when we add a methile group to the alpha position. When we add a methile group to the alpha position now that results in what? That results in increasing its nicotinic receptor activity. And if we add the what? If we add the methile group at the beta position. This is the alpha position.

[00:20:40]This is the beta position. If we add a methile group at alpha position that results in increase in its nicotinic activity and if we add a methile group to the beta position that results in increasing its activity for the muscerinic agonist or muscarinic activity is increased. So basically if we add the uh we can say any group to the beta position now that will result in making it a transisomer. Now basically trans isomer will be formed. Okay. So transform we can say it is more potent as the musceranic activity. Okay. The trans isomer. Okay. Next the drug which contain kacol nucleus. Okay. Again the simple question. The drug that contain a kacol nucleus. Again the kacol nucleus two hydroxy one and second position we have an hydroxy group that is a kacol nor adrenaline epidin phenilephrine or the pseudo ephedrin.

[00:21:47]Okay, if you look onto the options carefully, the B, the C and the D, all three are basically having the similar kind of structure that is ephedrin.

[00:21:58]Ephedrin and we already discussed the structure of ephedrin, it do not contain the kakol ring. It contains a simple phenile ring attached. Okay. Yes, remember. So basically if we talk about the kakola means they are basically sympathetics.

[00:22:16]They contain a kakol ring in its structure like epedrin epinephrine norepinephrine and the adrenaline nor adrenaline and dopamine.

[00:22:28]So here in the options you can see nor adrenaline.

[00:22:32]So the noradrenaline is a drug that contain a catacol ring in its structure.

[00:22:38]Nor means removal of an CH3 group. So it has NH2. So this kind of structure this is nor adrenaline.

[00:22:48]Okay. So nor adrenaline. Okay. That is the containing a kacol ring in its structure. Yes. The A1 is the correct answer. Okay. Everyone is not participating. Everyone should participate. At least if you say give me the wrong answers that doesn't matter.

[00:23:09]Okay, you will learn something if you give the wrong answer maybe next time you will be right. Okay. So which among the following do not have a active metabolite. Is it morphine? Is it neostigman? It is diazipam or is it deox digtoxin? So which do not have an active metabolite. So if you see the options carefully morphine, dazipam, digtoxin they all are metabolized to form an active metabolite. But if we talk about neostigmine it is active itself and further if it will be metabolized it will be metabolized to an inactive metabolite.

[00:23:52]Okay. So neostigmine is the only drug that do not have an active metabolite.

[00:23:59]It has an inactive metabolite. Okay, so the B1 is the correct one here. That is the neostigman.

[00:24:07]Okay, got it or not?

[00:24:12]Next question. Again the simple question that is the IUPAC name for proprinol. So before knowing the UPSAC name, we should know the structure of a propanol. So basically proprinol is a beta blocker.

[00:24:28]So it will contain a nucleus a oxyropenol amin. So it contains an aril oxy methropenol and then we have an amine group. Okay.

[00:24:44]And that amine group also like this. So basically like this kind of here we have and proprenal. So propanol means isopropile group we have here. Okay. And from prop we can say we have an isopropile group. From NO we can remember that it has a napylene ring. So a napylene ring we have here.

[00:25:07]Okay. So this is the structure of an propranol. And if you see we have what we have one 1 2 3. Okay. So basically what we have at second position we have an all group.

[00:25:24]Okay. So at second position we have all group that says nothing. So basically all the groups have second position O group. So propanol that is common. Next we have a neptil oxy group. So napylene ooxy group. Okay. So that is first position it is attached to here. So that's why one neepth. Okay. That is also common in all. Okay. Now this neptilog it is attached to what? So if we see it is it is it contains what?

[00:25:55]Here we have one position. It here we have two position. It is attached to a what? It is attached to a methile amino.

[00:26:04]Yes. A methile ethile amino group. So we have a methile ethile amino group here.

[00:26:12]But the position here is the one position. Okay. So one position we have.

[00:26:16]So one me mythile amino and this neptilox oxy is attached at the third position not the second one. So here the third position. So the C1 is the correct one here. Yes the C1 goes with that now one one methyl I mean no then third position we have an epileoxy group attached and propane 2 all is the common one. So the C1 is the correct one. Yes the C1 goes with the answer now here.

[00:26:45]Okay. So numbering starts from 1 2 and three. So at third position we have a neptily group and we have a propane to all and here also we can add basically the numbering. So 1 2 numbering we can say. Okay. So basically so that is the ring. Okay. So that is the basically the nucleus or we can say the UPSC name that is the simple one of the proprenal.

[00:27:11]Okay. Coming to the next question, which form of adrenaline is having the maximum biological activity? Adrenaline, we discussed it right now that adrenaline is optically active. One form of adrenaline is more active than the other. It is about 10 one form is more 10 to 12 times more active than the other form. So which form has the maximum biological activity? They are asking is it S form, R form, cis form or transform. Okay, this answer I need from you people. I told you it right. I told you about this right now. Okay, just two questions before I discussed this one.

[00:27:53]Which form of adrenaline is more active?

[00:27:56]Is it S form, R form, cis form or tri transform? I also told you 10 to 12 time more active it is is the that form is the other than that. Yes. Which one?

[00:28:09]Who's going to answer me? This one?

[00:28:12]Yes, the Which one? The R form. I told you now R form of adrenaline is 10 to 12 times more active than its S form. Okay, that is the optical form. Optical activity basically changes the potency.

[00:28:31]So 12 times more ductive than the S form. Okay, coming to the next question.

[00:28:41]Okay, which salt of hyiosin is most stable and popularly accepted? So, hyiosin is a drug and basically it's one form is more active. Okay, but it is more accepted with one salt. Okay, so methon nitrate, brutile bromide, hydro bromide and methyl bromide. So they have given the different salt names and we have to tell that which salt form it is active more. Hyiosin is active more with. So basically hyiosin to make it stable and to make it uh less degradable or to increase its stability we combine it with a salt and which salt is more favorable to be get combined with? So basically it is an hydro bromide. Okay.

[00:29:28]So hyiosin hydro bromide is uh basically taken because this form or this salt along with this salt hyiosin is more stable. Okay. It improves the stability of the hyiosin whenever it is taken with that. Okay. So basically decomposition is uh we can say less it less prone to decomposition. That word we can use.

[00:29:52]Okay. Next the IUPAC name for salvamol.

[00:29:58]Okay. Salvetamol. Uh again a drug that is used in the treatment of COPD and asthma. Okay. If we talk about its structure. So salutamol. So it looks like salicylic acid ring. Okay. What is salicylic acid? O CH.

[00:30:19]This is salicylic acid. So what they have done? They have done a minute change in the structure of salicylic acid. Instead of CO what they have said it is CH2O.

[00:30:31]Okay. What it is? It is CH2O.

[00:30:36]So this is the ring we have. And then it says the chain that is present and that amine group. And then from but we can say it has a butile group also. So butile isobbutile group also we have. So this is the structure that we have in what that we have in the salutamol and if we talk about its upupac name. So it has what? It has a tertiary butile amino. Okay. So 1 2 at first position what we we have what? We have a tertiary butile group. Okay. We have a tertiary butile group here.

[00:31:18]here and then we have a hydroxile group and that hydroxile group is attached to the ethile group. Okay, so we have one hydroxy ethile group and we have a hydroxythile phenol. So hydroxythile phenol we have. Okay, so basically A1 we can remove it as it says 1 comma 3 diol C1 1 comma 2 diol. So we don't have that B1 that is hydroxythile phenol that can go with the structure. So B1 is the correct one. Okay. So the B1 is the correct UPSC name of the salvalol. Here you can see the structure clearly. Okay.

[00:31:59]If you were facing some confusion in the what I have drawn. Okay. Yes. The answer is the B1. Okay. One of the following statement with respect to carbaol is not correct. Identify carbol. The structure is very similar to acetyl colon only of the carbakol also just the minute difference is there. And they are saying which statement is not correct for the carbol. It is administered orally. It can be administered orally. Yes. It possess both muscarinic and nicotenic action. Yes. It is non- selective drug.

[00:32:36]So it can act on both it is more susceptible to hydraysis under two catakolam acetylcholine. So basically the drug the carbocol is made only because because it has low hydraysis rate as compared to acetyloline. But here they are saying opposite. Okay. So this statement is not true about kakol sorry carbocol. Okay.

[00:33:01]So basically acetile colon is more gets hydrarolysis fast as compared to the carbakol. Okay. So C1 is the wrong one here. So C1 the C statement is correct.

[00:33:13]Yes the C1 Samrit is right. Okay. So carbacol the C statement is correct for that. Wrong for that or we can say the answer is correct.

[00:33:23]Then next question. a synthetic anticolineergic drug which is an asteraring zenthine hetroycle.

[00:33:33]Okay, antiolineergic drug they are talking about it. It it has an asteraring zenthine hetroycle. What is a zenthine ring?

[00:33:44]Okay. So a ring like this is called as a zenthine ring. So they are asking out of all these four anticolinergic drug that is homotropin, benropin, propenthelenin and biperidin which one has the zenthine hetroscycle nucleus in its structure. So yes, can anyone answer me that? Yes, the answer is the C1 propenthelin. Propenththelen is a drug that contains zenthine nucleus. All other has a tropin tropin structure we can say. Okay, tropen like structure in its structure. If we say homotropin, if we say benropin even from in their structure it's written tropin, tropin. So they have a tropin ring. But if we talk about the propenthelene, it contain a zenthine nucleus. Okay. So the C1 is the correct one that is the propenthalene. It contains a zenthin nucleus in its structure. Okay. Next find the given structure. So they have given us the structure and we have to tell that what structure is it.

[00:34:56]Pilocarpin carbacol acetylcoline or tacin. Okay. Pilocarbon I guess we have already discussed what is pilocarpine.

[00:35:04]Pilocarbon it contains a furonlike ring and it also has one more ring in its structure that is that is what that was an imidasole ring so the structure doesn't belong to that so pilocarbon we remove it from the list okay next we have tacin okay we come to D1 tacin so tacin contain a acridin ring in its structure what is an acridin ring basically a benzene ring that is attached to the pyodine ring and it has a This this kind of ring we have in tacin if you remember. Yes.

[00:35:41]This kind of ring we have in tacin.

[00:35:43]Okay. So again it doesn't look like this at all. So yeah tacrine is also not there. Acetile colon. Acetile colon contains an acetile group in its structure. Okay. So acetile group is CH2 not CH not NH2. So yes again it is not an acetile group also. So it contains a carbonate group here. So this group here is called as a carbamate group. Okay. So carbol is similar like cate like acetylcholine it structure is almost similar like acetylcholine just the difference is that instead of an acetile group we have a carbonate group. So this structure belongs to carbacol. Okay. So this is the carbakol. Yes, the 21 is B.

[00:36:33]Yes, Rajes, you are right. The B one is the correct one.

[00:36:37]Okay, carbol. This is acetyloline. Here you can see acetile group. Rest of the structure is same. This is carbakol. We have a carbamade group. Okay. Next question. The drug used in organo phosphate toxicity. I guess this is the simple one just directly from the class the classification they have asked. Now whatever the drug that do not that belongs to the opposite of the organo phosphate or we can say anticolin colon eststerase drug against that whichever drug works that works under uh against the organo phosphate toxicity is it neostigman it is palidoxyme is it carbacol or is it pilocarpin yes the answer is very simple the palidoxim it is an what it is used for the organo phosphate toxicity because it is a drug that is the class opposite to the organ of hospital we can say anticolinsterase it works opposite to that so that's why prid dom is the drug okay yes the B1 next find out the structure name given structure name so the structure is given is it clenium bromide okay it is tropicamide it is cyclopenolate hydro hydrochloride or is it dicyclloman?

[00:38:00]So if you see it carefully clinium so em is present at the end. So whenever there is em at the end I already told you that it means that nitrogen with a positive charge should be present in its structure. And if we see we do not have that. So yes it is not a clinium then tropicamite. Tropicamite it should contain a tropic ring. Okay that's why you are calling it tropomite. So we do not have that. So B is also not there.

[00:38:28]Cyclopentolate. Cyclopentilate contain a cyclopentane ring in its structure. So we do not have a cyclopentane ring here.

[00:38:36]So again it is not a cyclopentilate hydrochloride. Other one is dicyclomine.

[00:38:41]So we have two cyclic rings here. You can see two cyclic rings. So dylo means two cyclic ring. So yes the answer is the D1. Yes Rajes, you're right again the D1 is the correct one. This structure belongs to the dicyclomin only. Okay, I guess the questions are very simple. You just have to read them carefully. Then you get your answer there and there. Muscarin is generally again the question from classification.

[00:39:09]Muscarine is generally muscarin ericoline pillocarpin or echothophate.

[00:39:17]So they have given the structure. So uh Erica kachu you must have heard about.

[00:39:23]So muscarin is generally aricolin that we get it from the arica ketachu okay it is isolated from the seed era ketachu and this structure belongs to that only so this is aricolin muscarine is an aricolin okay next the ring system present in fiso stigman okay the question again the structure based question the ring system present in piso stigmanin I guess everyone remembers the structure of piso stigman it is an what it is an anticolinergic drug piso stigman okay it is a colonergic drug sorry so indole pyodin uh pyramidine or thofopin which particular ring is present in the structure of fiso stigman yes the stigman drugs which r which structure it has so basically if we talk about its structure. It has a indole ring. Indole ring attached to one more pyrol ring. So this ring is present there and indole ring here and that indole ring is further attached to the one more pyrol ring. So that is the structure of our what the fiso stigman.

[00:40:42]It is an anticolinsterase drug reversible one. Okay. So that is our fiso stigman. The indole ring is present in its structure. Okay.

[00:40:54]Next is carbomo B methile colon is okay. Methile colon carbam moil. So what is the structure of methile colon? So basically in the structure of an acetile colon when we add a methile group at beta position that is methile colon. So let's draw it. So like this kind of structure we have sorry ring we'll draw again. So basically what we have C double bond O CH3 O okay so here we have a chain here.

[00:41:36]Here we have an CH3 group. Okay. Here we have N positive. So this is the structure.

[00:41:46]Okay, sorry I have drawn three rings three only two carbon chain is there.

[00:41:56]Okay, so like this n and then we have a positive charge with n positive. So this is an methileen structure. Okay, now in this methileen they are saying carbamo.

[00:42:10]Carbomo means we add a carbonate group.

[00:42:13]So instead of a CH3 if we make it what if we make it an NH2.

[00:42:22]So they are asking what is the name of this structure. So this structure is the this structure is bethnocall.

[00:42:31]Okay. So what is bethino? when we add in the carol structure when we add in one CH3 group here at this position that becomes a bethno coil structure. Okay.

[00:42:45]So they all belongs to the derivatives of the acetylcholine only just a small small differences there. So carbam be methile at beta position they are saying that we have a methile group attached.

[00:42:57]Okay. So that is the bethninoall structure. Okay. Here you can see bethnyl. So this is the group here. So this is the carbomile group. Here we do not have any CH3. But when we add a CH3 here, that becomes okay. So the answer is the bethno. Okay, I guess that's a simple one. Again the structure the question from beta blockers the L drugs beta blockers. So Timol the ring present in tealol is so there is a very very simple uh trick to remember the rings that are present in timol. So t i m o l o l we just break down that word in three words. Okay. So t i. So from TI we can say we have a thazole ring. From M O we can remember that we have a morpholine ring. From lol we can remember it is a beta blocker. So it should have a oxy propenol amin nucleus. Okay. So that is about teemol. So Timol has two rings ring and a morpholine ring. So ring present in teeol morpholine diioazole pyodin and imidazole. So we have what the A1 that is the morpholine ring present. So this is the structure of timol. Here you can see a oxy propenol amine nucleus is present that's why it is beta blocker. Then we have a morpholine ring and thazole ring present. So that is a timol a type of beta blocker. Yes, the answer is the A1 that is morpholine. Yes Rajes you are right again. Okay. So the answer is morpholine.

[00:45:08]IUPAC name of Nefazolin.

[00:45:12]Nefazolin structure. I guess everyone is familiar with Nefazolin. What is the structure of nephasoline?

[00:45:19]A napylene ring. We have napylene ring between we have a CH3 and that is further attached with the imidazole ring. So nefa is for napylene.

[00:45:36]Zolin is for imidazole. Between we have CH3. So that is the basic structure of our nepholin. And if we talk about the what if we talk about the uppac name in nefa zolin. So yes the UPSC name that basically belongs. So one neptile at first position we have nepthile. Yes one neepth one neptilethile.

[00:46:03]Yes one neptilethile we have. Okay that is the correct one. So C or D can be correct. A and B we can remove. Okay.

[00:46:11]Next we have two emazole. Okay. So at second position neptil second position imidone. So this one is common. So at second position we have neptile. Yes. 1 2. So at second position have we have not the third one. So D1 is the correct one. Yes. The D1 goes with our answer.

[00:46:32]So yes this is the structure of a nefolin. You can see here and the right answer is the D1. Okay. So yes. So D1 is the correct one that is nefa solen.

[00:46:47]Now the next question we have here.

[00:46:58]Yes. Which of the following imidazolin derivative is selective alpha 2 agonist?

[00:47:05]Selective alpha 2. Yes. Which one can you name? Selective alpha to agonist. I guess that's a simple one. Selective alpha 2 agonist. The name of that drug is nefazolin cloned in xyometazolin or tolazolin. Which one? The selective alpha 2 agonist is our is our clonin. Yes, the B1 is the correct one. Clonidin is our selective alpha 2 blocker. What is the structure of clonidin? So basically a benzene ring with two chloro groups here like chloro clonidin and this here we attach it with what? We attach it with imidazoline ring. So that is the structure of our clonidin. Okay. So yes it is an alpha 2.

[00:48:00]Okay it is an alpha 2 agonist. Okay.

[00:48:06]Next.

[00:48:12]Yeah, this question was left. I was searching this one. Ring present in Prazosin. Prazosin is an alpha 1 selective blocker drug. Okay, it is an alpha 1 selective blocker. If we talk about yes the 30B then yeah if we talk about its structure what ring is present in prazosin. So prain has a specific line to remember its whole structure.

[00:48:39]Queen piper feels pressureless.

[00:48:51]Just remember this line.

[00:48:54]Okay. Queen piper feels pressureless. So what does this mean? So queen means it has a quina. It is a quina derivative. It contains a quinazolin ring. Okay. Next piper is for pipin ring.

[00:49:16]Okay. Fields. Okay. It has a fural ring.

[00:49:21]Fur royal ring or we can even say furon with an O group. Oh ketone group attached. So okay so that is fur. So basically how it is that so this ring is present and then this so this is the fural ring. Furan ring with an double ketone. Okay. So furil is present.

[00:49:41]Pressureless is for it is used as an anti-hypertensive drug. So that's why pressureless. So we have in praosin we have all three rings. We have furan ring also we have pipin ring also we have quinazolan ring also. Yes the answer is the D1. Okay. SR I don't know what's the name is and Rajes both are correct okay so the yes the answer is the D1 that is all of the above so that is the structure of prazosin let me show you this is the structure of prazosin quinazolin ring pipresin ring furan ring along with the this so this along together is also called as furile ring okay yes so coming to the next question in in the chemical synthesis of propran ol okay the synthesis of proprinol what are the two primary starting material reactants used in the first step so basically the reactants that we need for the synthesis of proprinol is it one nepthol isopropylamin parah hydroxyphenile acetamide and epic epiclorohydyrin one one nepthol and epicllohydrren three for dchloroepidate and morpholine so yes For the synthesis of our proprinol, we need a napylene containing ring like a napthol because we have a napthol in its structure and an epiclo hydrron. Okay. So these two together combines to form a napylene ring or we can say a propanol ring.

[00:51:17]Okay. So the C1 the A not a the C1 one nepthol epiclody. Okay. We can't see just nepthol. We have to see the other one also. So it is not isopropylamin, it is epiclorohydrin. Okay.

[00:51:33]Okay. So C1. Okay. Now you will not do it wrong. I know.

[00:51:40]So the C1 is the correct one. Next two acettoxy N trimethile ethanmium can be termed as. Okay. Second at second position we have an acettoxy group. Then at N we have three methile group attached. Em it says so it has a nitrogen with a positive charge. Then it says ethile means we have two carbon containing. Okay. And it says actoxy. So this is first position. This is second position. And at second position we have an acettoxy group.

[00:52:20]So this structure belongs to what? Yes, you are right. The structure belongs to acetile colon. Okay, just a simple question. Acetile colon. Okay.

[00:52:33]Yes, you're right. Next. The ring present in donapasil. Donapazil is again the reversible type of anticolinist and it is water insoluble. So it is used in the treatment of an Alzheimer also Alzheimer disease. Remember I told you that the water insoluble drugs and anti-olinistase drugs that are reversible in nature are used in the treatment of an Alzheimer. Okay, because they can easily cross the bloodb brain barrier and can show its action. Next, what is the ring present in its structure? Is it pyodin? It is pipridin.

[00:53:12]Is it pyramidin or pyrol? So in the donasil, the pipidin ring is present.

[00:53:17]Okay, it contains an Indian ring along with the pipid ring in its structure. So that is the donasel. Okay, the B1 is the correct one.

[00:53:31]Next question.

[00:53:34]The starting material used in the synthesis of anzosin.

[00:53:40]Okay, so the prazosin, what is the starting material that we used? We already discussed the structure of prazosin. So what is the starting material? 2 amino 45 dimethoxyenzoic acid. 2 amino 34 dimethoxyenzoic acid.

[00:53:58]Anthonolic acid or four amino 26 dimethoxyenzoic acid. So if we look onto the synthesis of a prazosin. So we have here two amino 5 dimethoxy benzoic acid along with sodium cyanide to further results in the formation of a brazosin.

[00:54:19]So what we need? We need two amino 5 dimethoxy benzoic acid at a starting material for the synthesis of our prazosin. Okay, easy one. Next question.

[00:54:34]One of the following drug is obtained from natural product. Okay, that's a simple one. Naturally occurring product.

[00:54:42]The natural product drug. Is it hedrophonium chloride? It is pilocarpin.

[00:54:47]Is it neostigman or is it echothophate iodide? Pilocarpin. Yes. Neostigman.

[00:54:54]Echioate iodide. Which one is a naturally occurring one? Okay. I guess this one is known to everyone. The naturally occurring.

[00:55:04]It is the plant called pilocarpus.

[00:55:09]That is the plant from where we get Yes.

[00:55:12]The answer is the B1. Yes. Yes. Yes. The pilocarpin is the answer. Okay. The B1.

[00:55:18]It is obtained from the leaves of a pilocarpus that is found in South America shs. Okay. It is a type of a South topical South American sh. Okay.

[00:55:29]So that is pilocarpin. Basically ey drops of pilocarbons are used. Okay.

[00:55:36]Next question. One of the following is a tertiary antis-spasmotic agent. Tertiary antis-pspasmotic agent.

[00:55:45]Yes. Which one? Basically, it has a tertiary amine. We can say that's why it is called as tertiary antis-pspasmotic.

[00:55:52]Which one? Yes, it was easy.

[00:55:56]The antis-pspasmotic. Tertiary antispasmotic.

[00:56:02]It is dicyyclomine. Okay. It contains a dicycyclomine. You can see em I told you nitrogen with a positive charge. It is not tertiary. Iatropium it is not tertiary. Dicycyclomine it is also not tertiary.

[00:56:22]Okay. The dicyclloman is tertiary.

[00:56:23]Sorry. The glyopyrolate it is also not tertiary. Dicycyclomine is a drug that is the tertiary one. Okay, it do not contain a quartinary nitrogen. So yes, the C1 is the correct one. Okay, next.

[00:56:40]Okay, this one is the good question. I many times it has been seen in different forms in your exams. So this one is the good one. Antiolinistase agent. Okay, it is an anticolinsterase drug. Quartinary ammonium compound. Quartinary ammonium compound means N with a positive charge it has as it is quartinary so it will be water soluble. Okay. So water soluble also we can say intermediate duration.

[00:57:04]Duration is intermediate. Next carbonilating agent. Okay. Carbonalating agent used in cobra bite and in mthenia gravis. It is a drug of choice in mastthenia gravis.

[00:57:18]Yes you are very very right. It is actually B1. Okay. Neostigine. Yes, you can lock it. So B1 the neostigman. Okay, cobra bite it is used and it is used in mthenia gravis. Neostigment none other than neostigman. By closing your eyes you can take it neostigman. Okay. Yes.

[00:57:39]The common N substituent present inol, metoprolol, bisprolol and bettoxol.

[00:57:47]Okay. All the beta blockers they are they are lol drugs and what is common in them? methile, ethile, tertiary butile, isopropile. So I guess everyone knows that almost all the beta blockers have similar kind of structure and to the end they have attached some group. Okay, some kind of group. So a oxy propenol amin. So this kind of structure is there. So here we have an ary group.

[00:58:19]Okay. And here we have something attached. So what is that attachment in all the all these groups they have given ethinol, protopol, bisoprolol or bittoxol? Yes the yes your answer is right again the isopropile group. Okay like this we have an isopropile group that is the common one here.

[00:58:43]Okay, you can see it in lot of structures. In all these structure you can see a isopropile group as common.

[00:58:50]Let me show you isopropile. Isopropile isopropile isopropile. Okay.

[00:58:56]Next carvidol. Again carvidol it is a generation third drug. It has an extra vessily property. The carvidol, leetitalol, okay, these comes under the same class and they are the beta blocker drugs possess both alpha and beta blocking activity. Okay, due to the alpha blocking activity, it has a velocilatory effect. Okay, what structural feature contributes to its alpha blocking activity? Is it a carbazole ring system, multiple aromatic rings, absence of a beta O group or tertiary amine group? Yes. Which one?

[00:59:33]The carvidol possess what? That it has an alpha blocking property.

[00:59:42]Yes.

[00:59:44]So basically carvidalol has a what? Has a special carbazole ring. Not C1. No no not C. The car carvidol has a special carbazole ring. From the car we can remember it has a carbazole ring and due to that carbazole ring due to the tricyclic ring it has a very good alpha blocking activity also. So it acts on both on beta receptor also and alpha receptor also. It is actually studied under beta blockers. It is a third generation beta blocker. And all the third generation beta blockers along with carvidol even the letolol they have a special feature that they can have a vzodilatory property and that vzodilatory property is due to the alpha blocking. It blocks the alpha 1 receptor and due to that vzodilation is seen.

[01:00:43]Okay. So that is our what that is our carvidol the A1 the carbazole ring that is present that is responsible for that.

[01:00:52]Okay.

[01:00:54]Next the two inter the key intermediates we have discussed the structure of timol. I told you the trick. Okay. So with the help of that trick you can easily identify which one is the correct one. The key intermediates for the synthesis of teemol are 34 dicchloro 125 thiodolin morphylin 34 dicchloro 125 thioin piproin dy bromo 1 to5 thiodolin pipresin or 125 thiodolin morpholine which one which one it's a simple one very simple I already told you the structure the even the trick I told you to remember the structure of timol t i m o l o TI for what? M O for what? Lol for what?

[01:01:39]Yes. Yes, it is. No, no, no, no, no, no.

[01:01:43]You are doing one mistake here. It is.

[01:01:45]You are almost correct. But one mistake you are doing that here you see it has three four chloro but three four two chloro groups are there. So you should say it dloro. So you come to the one. Okay. So yes 125 thiodol that is correct morpholine that is also correct but we need yes the A1 is the correct one okay that is 3 4 dchloro rest is similar A and D almost they are similar they have given that option to confuse you only so that you see these two and you just take them okay so basically it is dchloro so 34 dchloro A1 we will take okay not C D1 because it is 34 chloro So that is wrong. So A1. Okay. Yeah. A is locked.

[01:02:40]Next cyclopropile containing side chain present in cyclopropile. So this kind of side chain is present in only one drug.

[01:02:50]Okay. And this is also asked many time.

[01:02:52]Btoxol, bisoprolol, esol, timol all are beta blockers. Which beta blocker has a cyclopropile ring? Yes, it is. Yes, it is. What? It is again the answer is again the A1 bittoxol. Here you can see a cyclopropile ring. Okay. Bisprolol has two ether groups. Here ether groups are present in bisprolol. Two bis bis means two. So two ather group. So that is bisprolol. Esol it has an acetamide group. Esmol E S we can say acetamide.

[01:03:33]Okay. Esol it is a small acting ultrashort acting esol and it has an acetamide group also in its structure.

[01:03:42]Okay. Then we have what we have and Timol. Timol we have already done. So basically bettoxol is the drug. Next the amine present on an aliphatic side chain is a beta blocker. Pindol is okay it's also a simple one a amine present on the alifhatic side chain in the beta blocker pindol so they are talking about pendol pendol has an special property that it act as an partial it act as an partial agonist P for partial agonist indol indole ring it has an indole ring also and lol means it is a beta blocker. Yes. And the ring that is present there is no the most common ring that is present, the most common amine that is present in all the beta blockers, it is isopropile. Okay.

[01:04:42]So it has an isopropile group only. Not cyclopropile. No, no, no, no, not cyclopropile. It has an isopropile group only. So here you can see isopropile.

[01:04:50]Here you can see indo ring. Yes. And it is a partial agonist also. It has a partial agonist property also. It acts as fully antagonist. It has a partial agonist property. It has an indole ring and it has an isopropile group. That is our pindolon. Okay.

[01:05:10]Next, what heteroscyclic ring system is uniquely present in the structure of pendol? They have asked the question.

[01:05:17]So, what is the heteroscyclic ring that is present in pendolal? I right now told you P for partial agonist and indo means it has an indole ring and lol means it is a beta blocker. So yes we can lock the A1 that is the indole. Okay let me show you the structure of that also. So this is pindoleon. Okay. Indole aroxipropinolamin nucleus and here a is isopropile group.

[01:05:57]So that is our bindol. Yes, the a1 is the correct one. Next, which of these is a derivative of betaoalkky amin?

[01:06:07]Betaoyamin.

[01:06:09]Which drug comes under betaamin?

[01:06:12]I guess this one is very simple one. We have studied that one before.

[01:06:17]Phenoxyenzamin isopranine or phenileaphrin or pseudo ephedrin which one is a derivative of beta hellalkky amins. Beta helloky amines are irreversible in nature that we have already studied irreversible that's why they are not used too much but it is a very it has a very special kind of property that they are used in focytoma in the treatment of focytoma like condition yes it's an easy phoxyenzamin yeah it's easy it is very widely used in the treatment of fiochromes cyto. What is focytoma? When it is an rare type of we can say tumor. Okay. In that what happens there is an excessive release or excessive uh population of a kakolamins.

[01:07:13]Okay. So basically to reduce that excessive release or excessive level of kakolamins we use phoxyenzamin. It is an it is having an irreversible action and basically specially used for piochromtoma only and it is comes under the class beta haloalkky amamin. Okay. So that is our phoxybenzamin.

[01:07:37]Okay. This is the structure. Okay. Comes under betaamin.

[01:07:43]Next clinically used letolol. Lebtolol again it is a third generation beta blocker. Okay. third generation beta blocker. It has an extra vessodily property.

[01:07:55]Okay. And it it's most widely used basically naturally occurring letolol is is it SSR rs or SR? So the which form the RR form that is the more yes the RR form. Yes. The B1 is the correct one.

[01:08:14]The RR form of the labol that is very widely used. Okay. that has a very good activity. So it is used more so yes the B1 is the correct one the RR form okay yes bisprolol contains which additional structural feature is it bisprolol I told you the structure uh we discussed it with btoxol if you remember bisprolol contain what it contains an ether linkage it is an additional structural feature it is not present in others but bisprolol has a special feature a linkage amide linkage multiple ether linkage or caroxy acids which one? Yes, caroxilic acid which one I told you bis means two.

[01:09:03]So how many ether linkages were there?

[01:09:07]If you remember we had two in the side chain and one above. So almost three ether linkages were there in the structure of bisprolol. No no no no no not D1 it's C multiple ether linkages I told you know this means two so basically two ether linkages are here and one ether linkage is here as it is an beta blocker so it it will contain a nucleus called a oxy propenol amin so one ether linkage is here next in two are in side chain so it has multiple ether linkage okay so C1 we'll go with that multiple ether linkage Okay.

[01:09:47]Next. Which functional group is present in ethanol but absent in metoprolol?

[01:09:54]Ethinol. A for ethanol. A for ether secondary alcohol amide or isopropyl.

[01:10:06]So basically ethinol has a very special uh structure. We can say it has a special uh structure that is not present in metoprolol that is it contains an acetamide group. What it contains? Etinol has an acetamide group attached. So at means einol, at means acetamide, lol means it contain it is a beta blocker. So it is a type of beta blocker that contains an acetamide group in its structure. Okay.

[01:10:41]So that is our ethinol and metoprolol has a mythoxy group. So metoprolol metomythoxy.

[01:10:49]Okay. So mythoxy group is present here and einol has acetamide group. Okay. So the C1 that is an amide group that is present there. Okay.

[01:11:01]Coming to the next question.

[01:11:04]Phentoolamine the methile substituted phenile ring is known as benzile group paratlo group phoxy group or anil group.

[01:11:18]So which group is present in our fenolamin?

[01:11:22]The methile substituted phenile ring is known as what? Yes. In fentolamin which ring we have fentolamin. Yes, pentoolamin has a phenol fenl. So it has what? It has a phenile ring with CH3 that is also called as a toluine ring. So it has a toli group. Okay, let me show you. Yes, here we have a paratli group in its structure along with the phenol group.

[01:12:05]Okay. And we have an imidazole ring also. So that is fentolamin.

[01:12:11]Okay. So that is we have what? We have a paratile group. Yes, the B1. Yes, you are right Vive. Okay. Very good. The B1 is the correct one. That is the pento. Next, which starting material is utilized to and initiate the synthesis of ethinol. So remember the structure of ethinol. I told you the trick right now.

[01:12:33]A ect actinol acetamide. Okay. So einol contain an acetamide group in its structure. And if we talk about the synthesis of ethinol one nepol we do not have a neptile neepth ring in the structure of ethanol also. No not one nepthol. parahhydroxyphenile acetamide, thodiasol, hydrochloride or morpholine.

[01:12:55]So what is present? Yes, the answer is very obvious. So yes, the vic B1 B1 not nepthol. We do not need nepthol. We need what? We need an at 80 means ethanol. AT means acetamide. Okay. So we need an acetamide group. So second one you see parahhydroxyphenile acetamide. Okay. So B1 is used. Okay.

[01:13:19]acetamide group is used. Parah hydroxyphenile acetamide group is used in the synthesis of our ethinol. Okay, the B1 is the correct one.

[01:13:31]Now here we come to our last question for today's session. Okay, so basically this was the last question in your uh test I guess. So catacolamin metabolism involves the methilation on one of the catacol O group to give. Okay. So they are saying kakolamine. So kakolamine what is the structure of kakolamine? It contains a phenile ethanol amin nucleus. So phenile ethanol amin nucleus and it contains a kacol ring O. So they are saying catacolamin metabolism. So they are undergoing a catacolam in metabolism and involves methilation means adding of a methile group. So what group it will give? What particular group we will get after the methilation in the kakolamine. So let's say here the methilation occurs. So OC3.

[01:14:33]So it gives us the mythoxy group. So we can call yes the answer is the A1. Yes.

[01:14:40]We wait. It is A. Okay. Okay. So yes, the mythoxy derivative very good. Okay.

[01:14:45]So the mythoxy derivative we will get whenever we will metabolize the kacol ring. Yes. Yes you are very right. The A1 is the correct answer for the for the 50th question. Okay. So these were the questions of your today's we can say the weekly test. I hope your test was good. If it was not then I hope now you got all your answers. Okay, now you understood all your answers.

[01:15:14]Whatever you have done wrong in that test so that you don't do that wrong in your exam when you will give that. Okay.

[01:15:23]So I hope the lecture was um helpful for you today and you understood many things and your thing and your very doubts were clear here. Okay. If you have any questions, you can ask me. Otherwise, we will end the session here only. Okay?

[01:15:41]And if you have any questions, you can even ask me in the comment section. I'll uh I'll definitely answer them afterwards also. Okay? So, for today, I guess we should end the session here.

[01:15:53]Thank you so much. Bye-bye. Good night everyone. Okay.