9701/12/F/M/26 - PART 4 - (Qns 31 - 40) | AS LEVEL CHEMISTRY PAPER 12 FEB/MAR 2026

Added: 2026-05-28

[00:00:00]But for this is for questions that one two for the question that one three alkenes are listed. We have hex-1-ene.

[00:00:09]This one is CH3 CH2 CH2 CH2 then CH then the double bond CH2. This is 1 2 3 4 5 6 hex-1-ene.

[00:00:26]Then cis-hex-2-ene So we can have carbon double bond to this one hex-2-ene That means I can have this one here and then the CH3 here.

[00:00:43]Uh-huh. Then if they said cis, I can put the two hydrogens on the same what?

[00:00:53]Uh on the same side. Then the rest is going to be This is 1 2 3. So we are remaining with three.

[00:01:01]So this will be CH2 CH2 then CH3.

[00:01:07]The propyl group here.

[00:01:09]Uh-huh.

[00:01:10]>> [clears throat] >> Then trans- hex-3-ene.

[00:01:13]So we have one I mean that that carbon on this one here.

[00:01:18]This is on the third one. So meaning I can have C H2 and then CH3 so that we have one two then three. You see?

[00:01:32]Then I can have a hydrogen here since they have said trans. Then I can put this hydrogen this other side. Then this way we have um C H2 and so then C H3 If you can count this is 2 plus the 2 4 then 6.

[00:01:55]So we have something like that. Mhm.

[00:01:59]So, this is one, this is two, this is three. In fact, let me put them here.

[00:02:09]Put this is my one.

[00:02:13]This is my two.

[00:02:15]And then three here.

[00:02:20]Then they are saying, "Which alkenes can be formed by dehydration of hexan-2-ol?"

[00:02:26]So, hexan-2-ol Let me duplicate this.

[00:02:32]And I'll put the OH here.

[00:02:38]Mhm.

[00:02:40]Then after that, I can put a CH3 here.

[00:02:45]1 2 3 4 5 6. Yes. Hexan-2-ol is that one there.

[00:02:50]Now, when we dehydrate this, we can actually put conc. sulfuric acid, the one that can dehydrate an alcohol.

[00:03:01]And it gives you an alkene. Or we can also use hot aluminum what? Oxide. So, we heat it uh by passing it on top of that aluminum oxide. So, it can still give us the alkene. Now, what happens is that this carbon here holding the OH will lose the OH. And then the adjacent carbons here can lose the hydrogen. So, we can have these two here.

[00:03:38]Uh so, the first one I can consider is this one here, CH2 CH2 then CH2 then this carbon has lost the OH.

[00:03:52]Mhm. but it has its hydrogen here. Then this other one loses the hydrogen, so we can have CH2.

[00:04:00]Or we can have CH3, CH2, CH2.

[00:04:05]Then uh this one loses a hydrogen.

[00:04:09]So there were two, now it becomes one now a double bond.

[00:04:13]Now this one will be CH, it has lost the OH.

[00:04:17]And then the CH3.

[00:04:19]You see?

[00:04:20]Now this one can have the cis and the trans.

[00:04:26]Okay?

[00:04:27]Hex what?

[00:04:28]This is hex-2-ene.

[00:04:30]Okay?

[00:04:31]We can have the cis, then the trans.

[00:04:34]Hex two ene.

[00:04:38]Now there is hex-2-ene, so this one is gotten.

[00:04:43]You see?

[00:04:44]>> [gasps] >> Aha, then uh how about this one here? It is hex-1-ene.

[00:04:51]So hex-1-ene is this one here.

[00:04:54]Aha.

[00:04:54]>> [clears throat] >> But we have not gotten this one here. So it is one and two.

[00:05:00]One and two only. The answer is therefore what?

[00:05:03]B. Question 32, which compound would give positive tests with alkaline uh iodine and with Fehling's reagent? So Fehling's reagent it is an aldehyde that gives uh a positive test. Okay?

[00:05:24]There must be an aldehyde. And then here we must have a CH3, C O.

[00:05:32]Okay? Group.

[00:05:35]For ketones uh we call it the methyl ketone or this CH3, C then OH.

[00:05:45]Aha, with the [snorts] H here.

[00:05:47]So, that group there.

[00:05:49]So, this one here can be oxidized to this one, and then the hydrolysis takes place, okay? So, they first oxidize this, then they iodines are going to replace all of these hydrogens on the this methyl group here. Then hydrolysis takes place, and then you get the yellow precipitated, which is the triiodo what?

[00:06:10]Methane. Now, we are supposed to look for either this group or this one here.

[00:06:15]When you look at this one here, this is an aldehyde. So, it can give a positive test with the Fehling's solution. Are we there?

[00:06:23]Uh-huh. [clears throat] But, this is a ketone.

[00:06:27]Yes, it's a ketone here, but we don't have a CH3 group here to give us the group here at the positive test with alkaline iodine. So, A has to be off.

[00:06:39]Then we go to this one here. We don't have a CH3 group directly connected to this carbonyl here.

[00:06:46]So, even this one will be off.

[00:06:49]You see that? We come to this one here.

[00:06:51]It has a CH3 group connected to this, so this one can give a positive test with alkaline iodine. And then, this one is a ketone.

[00:07:01]Okay? Because this one is a ketone, and also this one here is a ketone, they cannot give a positive test with Fehling's reagent.

[00:07:11]Okay? So, uh this one is also off.

[00:07:16]Then we go to D. An aldehyde here gives a positive test, and then [clears throat] this one here, methyl ketone, gives a positive test with now alkaline iodine.

[00:07:26]So, our answer is therefore going to be D here.

[00:07:31]Then question 33, compound Y is heated with mild oxidizing agent. One of the products of the reaction reacts with hydrogen cyanide forming 2-hydroxy butanenitrile. Now, this hydrogen cyanide reacts with uh it reacts with carbonyl compounds to give us the 2-hydroxy uh those nitrile compounds. So, let us look at 2-hydroxybutanenitrile.

[00:08:09]Um 2-hydroxy, the OH is there.

[00:08:13]Uh-huh.

[00:08:14]The nitrile is here.

[00:08:16]Then, we must have these are two carbons. We must have more two. So, we have CH3 CH2.

[00:08:26]And of course, now we must have a hydrogen here to make four bonds here.

[00:08:31]So, we have something like that.

[00:08:38]That is a 2-hydroxybutanenitrile.

[00:08:43]Now, this one here, you're having hydrogen cyanide and you put trace amounts of, let's say, potassium cyanide or sodium cyanide, okay?

[00:08:58]Just to catalyze the reaction.

[00:09:01]So, here we must have something like this.

[00:09:06]Uh-huh.

[00:09:07]>> [clears throat] >> Then, CH3 CH2. This one is going to remain.

[00:09:13]Now, hydrogen cyanide I think this one has to be hydrogen because if we put this, we are just going to add this one here.

[00:09:25]It's an addition reaction, okay?

[00:09:27]Uh we are going to have this one added here.

[00:09:34]And the hydrogen remains put for this one here.

[00:09:38]Now, what are we going to have? So, this one now breaks off.

[00:09:43]Then you are going to have CH3 C H2. Of course, this one is partially positive. This one is partially negative. That's why this one can't donate the electrons to this, okay?

[00:09:54]Uh-huh. Then C then O minus.

[00:09:59]So, this one breaks first because there is limited rotation around the pi system. So, when electrons are coming in here, this one has to be broken first.

[00:10:09]You see? Then the cyanide is added. And these are reactions where we extend or we grow the carbon chain. Because do you see the carbon that is being added here?

[00:10:21]Uh-huh.

[00:10:22]>> [clears throat] >> Now, what happens is that this one because this is hydrogen cyanide, it can get the proton. Because this one in solution is going to be having those hydrogen ions in there.

[00:10:38]And then you get the CH3 CH2 CH CN and then the OH. You see?

[00:10:46]Uh-huh. Now, for us to get this one here, it is coming from a mild oxidation.

[00:10:55]And also, mild oxidation. So, since the oxidation is mild, that means it is coming from uh a primary what? Alcohol.

[00:11:08]Because it is a primary alcohol you are going to oxidize to give you an aldehyde.

[00:11:15]Okay?

[00:11:16]Then the aldehyde can be oxidized further to give you a carboxylic what? Acid. So, primary alcohol.

[00:11:25]And of course, this one has to be CH3 because you can see this is propanal.

[00:11:30]And so, then the propanal is grown uh by one carbon to give you the butanenitrile. You see that? So, this one should come from uh CH3CH2CH2OH, which is propan-1-ol.

[00:11:51]Now, it can come from the two, I mean from three carbons from here because we are also seeing three carbons here.

[00:11:58]But then, which of them? Is it the primary or the secondary? So, they put for you the propan-1-ol and then the propan-2-ol.

[00:12:07]So, you have to be very careful. This one has to be Alcohol. Because it is the one that is going to be mildly oxidized to the aldehyde. Okay?

[00:12:24]A Aha, then what else?

[00:12:28]Suppose it is propan-2-ol.

[00:12:31]You are going to form this one here, CH3C then this one here. Suppose we had this CH3COH.

[00:12:42]Let me show you also this one how it comes about.

[00:12:45]So, we have this one here. And so, now, the reaction here under hydrogen cyanide uh we are going to have CH3C.

[00:12:55]Let me put maybe the CH3 here, then the CN here and then the OH and so, uh I mean the OH. This one here it is a 2-hydroxynitrile, but what is its name?

[00:13:12]This is 2-hydroxy Okay, 2-hydroxy then there is 2- methyl.

[00:13:23]You see? 2-methyl.

[00:13:26]Then here it is propane nitrile.

[00:13:30]Propane nitrile.

[00:13:33]So, you are not getting this one here.

[00:13:36]So, that's is why we are taking for primary what? Alcohol. This is secondary alcohol.

[00:13:43]You see that?

[00:13:44]Uh-huh, and still the alcohol cannot be butan-1-ol or butan-2-ol because remember we are adding on the cyanide.

[00:13:53]So, the moment you add on the cyanide right here, the carbon chain is going to be grown by one or it is going to be extended further by one carbon and then they will go to five. So, it will be a pentane nitrile.

[00:14:07]You see that?

[00:14:08]Okay, so that's why our answer is what?

[00:14:11]Is C propan-1-ol.

[00:14:13]Question 34, an ester of structure formula this one is heated with aqueous sodium hydroxide. This is what we call alkaline hydrolysis.

[00:14:24]Alkaline hydrolysis.

[00:14:27]So, you are going to break this one down.

[00:14:31]CH3COOCH3 You cut this one into two.

[00:14:37]So, you are going to get CH3COO H and then you are going to have CH3OH.

[00:14:50]But, because this is an alkaline uh solution here, the sodium hydroxide is going to react with this acid and then you form a salt.

[00:15:02]So, instead of this one here, the ethanoic acid, you are going to have uh sodium ethanoate. So, you get a salt.

[00:15:14]And then an alcohol. This alcohol here does not react with the sodium hydroxide here, okay? Because of the positive These ones we explain them in a level. There's a positive inductive effect of this alkyl group, so it donates the electrons towards this oxygen here, strengthening this OH bond in between the two. So that this one cannot release the hydrogen ions, okay?

[00:15:41]To act as an acid that it can react with the base. So that's why alcohols do not react with sodium hydroxide or those alkali. You see that? So we are supposed to get now if you rush, you would say that the answer is A.

[00:15:59]Ethanoic acid and methanol. No.

[00:16:02]So it has to be the salt and then the methanol sodium ethanoate and methanol. So the answer has to be C.

[00:16:11]Question 35, a student suggests two uses of lithium aluminum tetrahydride. We don't use this one here for alkenes.

[00:16:20]Lithium aluminum tetrahydride, then you get an alkane. Here we use we we can't have nickel.

[00:16:28]So we put hydrogen in presence of nickel as a catalyst or platinum, but in most cases we are using what? Nickel. Then this one here is a carboxylic acid can be reduced by lithium aluminum tetrahydride to give you the primary alcohol. So it first takes it to the aldehyde, then to the primary alcohol.

[00:16:50]And we also have sodium hydride borohydride.

[00:16:54]Sodium borohydride, this one here. But this one does not reduce carboxylic acids. It can reduce ketones or aldehydes to their respective alcohols. You see? But not carboxylic acids. So this one is okay.

[00:17:12]Which reactions would give the product shown? Reaction two.

[00:17:16]Okay?

[00:17:17]So it is two only. Me out take C. This one is not. Okay?

[00:17:23]Uh-huh. [clears throat] So, that is that. Then, question 36. The diagram shows a section of an addition polymer. This one was formed when the alkene opens up. Okay?

[00:17:37]So, when we are to find the monomer, what we do, we close it. Uh we close it in. Okay? That's what I wanted to say.

[00:17:47]Now, you close, you also do this.

[00:17:51]You do this.

[00:17:53]You do that.

[00:17:54]You do that.

[00:17:56]You do this.

[00:17:57]Now, there is a double bond established here. So, we shall have a double bond there.

[00:18:03]Then, there is a chlorine.

[00:18:05]There is a hydrogen.

[00:18:08]Then, there's also a hydrogen and a chlorine.

[00:18:11]There's a hydrogen and then a chlorine.

[00:18:14]So, this one is 1 2 1 2 chloro.

[00:18:19]Mhm?

[00:18:21]Not chloro, cuz there are two chlorines.

[00:18:24]Dichloro.

[00:18:26]Uh-huh.

[00:18:26]>> [clears throat] >> Uh these are two.

[00:18:30]So, two carbons, that is ethene.

[00:18:33]So, dichloroethene.

[00:18:35]Mhm?

[00:18:36]>> [clears throat] >> So, because this is a polymer that is coming from this, so it will be poly 1 2 dichloroethene.

[00:18:45]So, that means our answer is going to be B.

[00:18:48]Like that. Cuz all of these ones So, this one here is broken.

[00:18:55]Even this one.

[00:18:56]Okay? Even this other one is broken here. So, all of these ones are giving us this. So, this is the monomer.

[00:19:05]The monomer unit.

[00:19:07]And that is that.

[00:19:09]So, if they you us to draw just one repeat unit, then that one would be C.

[00:19:15]Uh-huh. There is the hydrogen.

[00:19:18]If they had said you circle a repeat unit for this, so you you would do [clears throat] something like this.

[00:19:27]That is the one.

[00:19:29]Question 37, 0.2 moles of P reacts to form 7.83 g of primary alcohol Q. The percentage yield of Q is 45%. Then they give us the AR values here.

[00:19:44]Which row is correct?

[00:19:46]Now, we are having P reacting with this. So, if you have a halogenoalkane and you're reacting it with a potassium cyanide in ethanol, uh what we are getting here is a substitution reaction. It's a nucleophilic substitution one.

[00:20:06]reaction.

[00:20:07]So, we shall remove the chlorine, then we'll put the cyanide.

[00:20:13]But, remember they said we are forming the primary amine Q. So, here we are forming a cyanide.

[00:20:21]Is that a cyanide? No, we are forming a nitrile. So, that means A is off.

[00:20:26]Even this one is going to be the same.

[00:20:30]Same conditions, not so?

[00:20:33]Uh-huh. So, you will have a a cyanide attached here. So, it's a nitrile. Even this one will be off. Then here, when we put ammonia in ethanol, heated under pressure, then we can get the amine.

[00:20:45]Okay? But, this one has to be in excess, okay? Uh to avoid the amine you get from attacking again this halogenoalkane. So, in that case, you can get the secondary amine.

[00:21:01]You can get the tertiary amine like that. So, here we are going to actually have its I think it's all the same here. So, just do this.

[00:21:15]And then I'll put an amine here.

[00:21:20]Mhm.

[00:21:21]Just like that. So, this is NH2.

[00:21:25]Then NH2.

[00:21:27]Ahem.

[00:21:27]>> [clears throat] >> Now, from here we can have So, 1 mole when we are reacting 1 mole of this, of course, produces 1 mole of this one here. Now, if this is 0.2 moles, then we'll produce this 0.2 moles. Okay? Now, let us look at the MR here.

[00:21:49]The MR for this one. This is How How many carbons do we see here? 1 2 3 4 5.

[00:21:55]Mhm.

[00:21:56]So, that will be 5 * 12, not so?

[00:22:01]Then plus Mhm.

[00:22:04]Hydrogens CH2. CH2. CH2. CH2 here. So, these are eight hydrogens. Then there is one hidden hydrogen here.

[00:22:13]So, those are nine.

[00:22:15]Then plus the two.

[00:22:17]9 + 2, what do you get? So, that's 11.

[00:22:20]So, 11. Then plus nitrogen is 14.

[00:22:26]So, what do we get all together? This is 60.

[00:22:29]Then plus 25, that will be 85 g, not so?

[00:22:35]How about this one here?

[00:22:37]What is its MR?

[00:22:40]Uh this is 1 2 3 4 5 carbons. So, 5 * 12 also.

[00:22:47]Mhm.

[00:22:48]Then there is a CH3.

[00:22:50]There's a CH here. There's a CH2 here.

[00:22:53]There's a CH2. There's a CH3 here, not so?

[00:22:57]Mhm.

[00:22:58]So, hydrogens, how many are they?

[00:23:02]Uh we have three, five, six, nine, 11.

[00:23:09]Then plus two.

[00:23:12]Which is 13.

[00:23:14]Then 13 plus uh nitrogen, [clears throat] 14.

[00:23:19]So this is 27, then plus the 60, which is 87 g.

[00:23:26]Uh-huh.

[00:23:27]Now, 1 mol of this or maybe if you get the MR of this, those grams produce these grams. So, 0.2 mols will also produce 0.2 mols of this. So what would be then the mass?

[00:23:46]I'll get this one times the 0.2.

[00:23:49]Remember MR times the mols, you get the mass, not so?

[00:23:54]Uh-huh.

[00:23:55]>> [clears throat] >> So, let me try to take this one a little bit down.

[00:24:00]Then we say mass is going to be 0.2 times the 85. What do we get?

[00:24:09]0.2 times 85.

[00:24:13]That one gives us 17 g.

[00:24:16]Now, instead of us getting 17 g, we got 17.83.

[00:24:24]Okay?

[00:24:25]So, what is the percentage here?

[00:24:28]Percentage yield.

[00:24:31]The percentage yield is going to be the of the 17. So, you get 7.83 divided by the 17 then times 100. What do we get here?

[00:24:44]7.83, we divide by 17.

[00:24:49]Then we multiply by 100.

[00:24:51]This one gives us 46.1.

[00:24:54]So I can write this one as 46.1% but remember they said here the percentage yield is 45.

[00:25:03]Now, I'm centering my answers on these two the C and D because they are the ones that are giving us the amides, okay? So, I'm not minding about these ones they they do not give us the amide. Is that clear?

[00:25:17]Okay. Now, here we shall have just like what we have done here, the mass is going to be 0.2 * the 87.

[00:25:31]What do we get? 0.2 * 87 this one is 17.4.

[00:25:39]17.4 g.

[00:25:41]Now, we also find the what?

[00:25:45]The percentage.

[00:25:47]Percentage yield.

[00:25:49]So, percentage yield here is going to be uh 7.83 that is the one which is the one which was gotten out of the then times 100.

[00:26:02]Uh-huh, what does this give us?

[00:26:04]7.83 / 17.4 then * 100.

[00:26:11]Uh-huh, this one is giving us 45%.

[00:26:15]45%.

[00:26:17]So, you see that here they said the percentage yield was 45%. That means uh this is the Q which was produced, okay?

[00:26:28]And our P our answer is supposed to come from here because it is this one that has given us this which gives us 45 what? Percent.

[00:26:38]So, our answer is therefore supposed to be D.

[00:26:42]Question 38, bromine reacts with ethene in the dark.

[00:26:47]Which description of the organic intermediate of this reaction is correct. Now, we have ethene here.

[00:26:59]It has a high electron density here.

[00:27:01]Now, bromine they are saying in the dark because if it is under UV light, uh >> [clears throat] >> uh something else is happening. So, there we have free radical substitution.

[00:27:14]They they hydrogens on all the alkanes on those alkanes are going to be substituted. So, in the dark, this is a test actually that shows the presence of what we call unsaturation.

[00:27:31]Okay? Yeah, because this one is broken down and it is decolorized. So, this one is not charged because the electronegativity of bromine here is the same. They are the same here. So, this one pulls this way and this one pulls this way. So, the dipoles cancel. However, but as this one draws near to this one here because of the high electron density here, the electrons here will repel the electrons around this bromine here, making it partially positive.

[00:28:05]It's like it has lost electrons and they are pushed further to this other bromine this side. So, it will attain a partial negative charge. Then, this one here, because it appears to have lost electrons, now, this one acts as a nucleophile.

[00:28:25]Okay? Nucleophile. So, it will donate the electrons to this one here. You see?

[00:28:32]Uh-huh. Now, actually, this one we call it this reaction we call it electrophilic addition because the bromine here is acting as an electrophile here. It is gaining electrons from here. Okay?

[00:28:47]All right. then this one will be broken.

[00:28:51]And then we are getting the intermediate is going to be so bromine can go on any of those, okay?

[00:28:58]A so we can have CH2 maybe uh BR then we can have CH2 this one with a plus. So this is the intermediate, okay? So the organic intermediate of this reaction here is having a positive charge. So here they are saying and because it is having a positive charge, it can receive electrons. So because it can receive electrons, it acts as an electrofile.

[00:29:29]You see that?

[00:29:31]Uh-huh. So we can have now the bromine here which is completely negatively charged that came from here. Then it donates the electrons to this.

[00:29:44]You see? And then you get this CH2 BRCH2 BR. So because the bond between these two is completely broken and we know that bromine is brown, it's a brown liquid, then it is going to be decolorized because of this molecule which has been broken down. So we are no longer having this bond in between the bromine and the bromine. So that is how the color disappears in bromine, okay?

[00:30:12]Now our answer is therefore what? D. It is an electrophile, the organic what?

[00:30:18]Intermediate here. It is receiving electrons, okay?

[00:30:23]From >> [clears throat] >> uh it is receiving a lone pair of electrons from a species. So that is an electrophile. Electro from electrons.

[00:30:34]Uh-huh. File love. So the one that loves electrons receives them, all right?

[00:30:41]Uh-huh.

[00:30:42]>> [clears throat] >> So the one that loves electrons, remember electrons are negatively charged, so it must be positively charged.

[00:30:48]Because those that are positively charged, uh they're the ones that can take in the electrons. Okay?

[00:30:55]So, that is that.

[00:30:58]Question 39. When 1-bromopropane is treated in succession with two reagents X and Y, it produces propanoic acid.

[00:31:08]Uh-huh.

[00:31:10]Then, 1-bromopropane. Okay. So, this is CH3, CH2, CH2, then Br.

[00:31:18]So, we have reagent X.

[00:31:21]Mhm. So, [clears throat] we are forming propanoic acid. Okay. So, I can put here sodium hydroxide.

[00:31:31]Let me remove this. Sodium hydroxide aqueous.

[00:31:36]And then you can heat. So, if it is sodium hydroxide aqueous, it is going to be a substitution reaction. So, we shall form an alcohol here. CH3CH2, CH2, OH. Now, if we want propanoic acid, then we can put a strong oxidizing agent here that can take it to it can oxidize it to an aldehyde.

[00:32:00]But, that is a mild oxidizing agent.

[00:32:04]But, if we put a strong oxidizing agent, then it will take it to the carboxylic what? Acid.

[00:32:11]So, we must have sodium hydroxide.

[00:32:15]Okay?

[00:32:16]Uh aqueous to get an alcohol. But, if you put sodium hydroxide in ethanol, okay? In ethanol, you are going to form an alkene. This one it's an elimination reaction. So, this one will lose the Br.

[00:32:32]And then, this carbon here will lose the what? The hydrogen. So, you have CH3, then CH, then double bond CH2. This is what you can form here.

[00:32:44]You see that?

[00:32:46]And then, how do you get the carboxylic acid here? You need to first convert this one to an alcohol. Then you can oxidize. You see that?

[00:32:56]And yet, uh we are only using these two reagents here. So, we can't take these two. We can take this one and this one here. However, uh for this one, it has to be a strong oxidizing agent, which is potassium dichromate here, acidified potassium dichromate. So, it is this one here. And then our answer becomes what?

[00:33:19]A. Question 40 is about spectroscopy.

[00:33:24]We are given this table here.

[00:33:27]And they are saying that an organic compound X has composition shown.

[00:33:32]54.54% carbon, 9.09% hydrogen, and then 36.37% oxygen. The infrared spectrum of X is shown here.

[00:33:45]Now, we can identify the functional groups here.

[00:33:48]But here they are saying butanoic acid and whatever. So, let's first write these ones here. This is CH3, CH2, CH2, COOH.

[00:34:01]If I can even find the MR, uh this is 1 2 3 4. That would be 48, not so? Four carbons, not so?

[00:34:13]Then times the 12.

[00:34:16]Then what else? We have hydrogens. How many are you seeing? Five, seven, eight.

[00:34:22]So, plus eight. And then oxygens are two. 16 time 16 and then the 16.

[00:34:28]That would be 32. So, this one gives me 88. Mhm, 88. [clears throat] Then ethanoic acid, CH3COOH.

[00:34:38]What is the MR here? This one is having two carbons, so two times the 12.

[00:34:45]Then plus hydrogens are going to be four and then the other 32 for oxygen. That is a 60.

[00:34:53]Then how about ethyl ethanoate? This is CH3CH2 CH3.

[00:35:02]The MR here is like as as if the same as this because these are four carbons, two uh oxygens, and then eight hydrogens, so it is the same. MR is also 88. How about methyl methanoate?

[00:35:19]HCOOCH3.

[00:35:23]Yeah, this one is also like the same as this.

[00:35:26]The MR is also going to be what? 60.

[00:35:30]Okay?

[00:35:31]Now, um what do we have here?

[00:35:38]Uh after finding that let's look at the what we are having in here.

[00:35:44]The functional groups. So, this one here is nearer to 3,000. So, if it is near 3,000, I can uh should we look at this one 2,500 No, this one is not so broad.

[00:36:01]It is just very near here. So, we are not we are not going to take that.

[00:36:05]But, there is this one here 2,800 uh 50 to 2,950. So, we having this one.

[00:36:12]It is for the alkane.

[00:36:14]Okay? So, each of these ones have gotten an alkane. I can see an alkane on all of them.

[00:36:22]So, they will all give this peak.

[00:36:24]Okay?

[00:36:25]Then uh when I look at this one here, this is 1 5 1 6 1 7.

[00:36:33]This is 1 8. Okay? So, between 1 7 and 1 8 here, Mhm?

[00:36:42]A So, I think it is going to be this one here. It is in 1 7 and in between 1 7 and 1 8. So, it is here. So, that is where there is an ester. So, because it is that one is for the ester, which esters do we have?

[00:37:01]I have this one here and this one. So, that means I'm going to put off these ones.

[00:37:07]Okay?

[00:37:08]Just that. I'm not even going to look at whatever is coming this side in the fingerprint region, but we are still having this one here is 1,000 uh something, whatever. But, I'm going to take these two.

[00:37:24]And then, there is this other information they gave us here. They said that carbon is 54.54%.

[00:37:33]Now, using these two here, let us look at the percentage of carbon.

[00:37:40]Percentage of carbon is going to be Now, this is 1 2 3 4. So, 4 * 12, not so? Then, out of the total is 88 * 100. What will be the percentage of carbon?

[00:37:56]4 * 12 then 88 then * 100.

[00:38:03]This one is 54.54 recurring.

[00:38:07]Uh 54.

[00:38:09]54.

[00:38:11]Mhm?

[00:38:12]These two are recurring percent. And they said here that the percentage of carbon is what?

[00:38:18]54.54. Don't you think that is the answer?

[00:38:22]So, I think it is this one here.

[00:38:25]However, let's look at this one here, the percentage of carbon.

[00:38:30]Percentage of carbon is going to be there, two carbons here times the 12.

[00:38:36]Then I will divide by 60, then times 100. What we get? 24 divide by the 60, then times 100. The percentage is 40%.

[00:38:47]So, if it is 40% and the other one is 54.4, that means we are taking this one here.

[00:38:53]As simple as that.