J2026 ZIMSEC A LEVEL CHEMISTRY SECTION C

[00:00:00]All right, it's Niyake. It's Niyake Austin was on it. Today we are simply going to revise the June 2026 A-level chemistry from the Zimsec exam board as you can see on the screen. So, we are simply going to dissect this paper uh step by step. And if you are doing Zimsec, you need to pay attention to these instructions the candidates. But if you are not doing Zimsec, just pay attention to the data because chemistry is still the same, right? So, we are simply going to dissect this paper. In you know, our previous tutorial, we did section A, which is the physical chemistry section. And uh the next section we did uh section B, which is the inorganic. And today we are simply going to focus with uh section C, which is the organic chemistry section. So, if you haven't checked on our playlist, you need to do so. And always remember, always remember to subscribe so that you'll be notified whenever whenever we post, right? So, we are simply going to have number number eight, right? So, number eight says that the structure of the compound A and the compound B are shown in fig 8.1. So, we are having the structure A, uh the compound A and compound B, right? So, [music] here we are simply going to have uh this amino group, this amino group, and then this benzene ring, this benzene ring. And then we are going to have CH2 here, and then this uh carbonyl group. And then here we are having this uh carbonyl group, right? So, this one is an aldehyde, and then this one it is a it is a ketone, right? So, these two uh we are simply going to have what we call the functional uh the functional group isomers. These uh two they've got the same molecular formula, but different uh structural formula, right?

[00:01:15]So, this uh So, in terms of the structural isomers, we need to know this mnemonic scheme, P F I. So, you have the chain, we have the position, and then we have this functional group isomers. So, these two, A and B, are what we call the functional group isomers. They've got the same molecular formula, but different functional functional group.

[00:01:30]So, this one is a ketone, this one is an aldehyde. We are going to say these two are functional group isomers. Are we together? So, here we are simply going to have this one, the amino group, as the activating group, right? So, we are simply going to have this one as an activating group. So, this benzene ring has been activated. We are simply going to have the lone pairs of electron on this nitrogen atom delocalizing into the benzene ring through what we call the reso- the resonance, right? So, we are simply going to have resonance, whereby the lone pairs of this nitrogen atom being delocalized into the benzene ring.

[00:01:56]We are simply going to have this benzene ring being activated by simply going to have the benzene ring increasing in electron electron density, right?

[00:02:03]Meaning to say it is more reactive towards electro electrophiles, right?

[00:02:06]Electro electrophiles. So again, you have what you call electrophilic substitution substitution reaction, right? So this one, the benzene ring activated, meaning to say it has got an increase in electron density, right? And then it is a stronger nucleophile and it's got a a larger ability to attract electrophiles. Are we together? So this is the effect of this amino group on this benzene ring. It is an activating group. We're simply going to have the delocalization of this lone pair of electron into the benzene ring, thereby increasing the electron density of the benzene ring, thereby making the benzene ring more or a stronger nucleophile and a more attractive towards electrophiles, right? So you should know the definition of a nucleophile and the definition of an electro electrophile. Are we together? So you should be able to define each and every term in organic chemistry. Are we together? So here we're simply going to have the benzene ring activated and then we're simply going to have ortho para substitution, right? So here, activating groups, they activate ortho para regions. Are we together? So this, we're simply going to have ortho para substitution. Are we together? And then we now move on to the the question, right? So the first part says, "Describe and explain effect of the amino group on the reactivity of the benzene ring." So we highlighted this one. We say that we're simply going to have the delocalization of the electrons from this nitrogen atom into the benzene ring, the pi system of the benzene ring through what you call the resonance, right? So we're simply going to have this benzene ring being activated, making it reactive towards electrophilic substitution, right? So this one, it is an activating ring. Are we together? So we explained the issue of the activated ring when we also did the coupling the coupling reaction. So a coupling reaction is also example of electrophilic substitution reaction. Are we together? So we're simply going to have the diazonium reaction and etc. So we evaluated this one before in our previous tutorial. Are we together? So here, let us go to the next part which says, "Describe how A and B are reacted." So [music] we said these two, so let me clear here.

[00:03:49]Let me clear here. We said these two, they've got the same molecular formula, right? They've got the same molecular formula, but different functional group.

[00:03:56]Here we having this one at the carbonyl, we having the the ketone, right? And then in terms of the carbonyl, we having the the aldehyde. Are we together? So, these two are different functional groups, right? So, these are what you call the structural isomers and the class which you call the functional functional group isomers. Are we together? So, also you should be able to define isomerism isomers from your all level chemistry. Are we together? And then let us go to the next part which says describe a simple test that can be used to distinguish A from B, right? So, we are basically going to distinguish a ketone from an aldehyde, right? So, here we are simply going to have the aldehyde undergoing oxidation. So, we can make use of oxidizing agent potassium permanganate. So, we are simply going to have this one undergoing oxidation and then here we are not having any oxidation. So, this the aldehyde can be oxidized to a carboxylic carboxylic acid, right? So, here we are simply going to have the decolorization of potassium permanganate from colorless from purple to colorless. Are we together? From purple to to colorless.

[00:04:47]Are we together? And then here no observable change, no reaction, right?

[00:04:49]Because ketones cannot go under oxidation. Are we together? And then we are also going to have what we call the Tollens Tollens reagent, right? Tollens Tollens reagent [music] which is also called the silver the silver mirror the silver mirror test. Like so, the silver mirror test we are simply going to add the Tollens reagent to to the aldehyde and then we are simply going to observe the presence of a silver a silver mirror in B, right? And then no observable change in in the ketone. Are we together? One, so we are also going to make use of another one which we call the Fehling's or the Benedict's the Benedict's solution, right? So, here [music] we are simply going to have the Fehling's or the Benedict's the Benedict's test, right? So, here we are simply going to have copper two the blue ions the copper two sulfate solution being reduced by this reducing sugar the aldehyde, right? So, this one it is a test which is being used in biology, right? So, we are simply going to have the aldehyde reducing this copper two into copper one the brick red precipitate. So, we are simply going to observe a brick red color, right? And then here no observable change with the with the ketone. Are we together? Or we can also make use of what we call the iodoform or the triiodomethane test, right? So, we can make use of uh the iodo- uh the iodoform test, right? So, here we're simply going to have uh the use of iodoform test. So, let me clear here so that you can easily see how we are to use the iodoform test. So, here we're simply going to have a methyl ketone, right? So, this methyl group attached to this carbonyl uh carbonyl carbon, which contains the the ketone functional group, right? So, here we're having what we call the methyl a methyl ketone, right? Ketone, right?

[00:06:09]Are we together? So, this one is positive for the trihalomethane test to give uh the yellow precipitate. Are we together? So, here we're having yellow precipitate, [music] and then here no observable change, right? So, we're simply going to have uh this one positive to the trihalomethane test because of this methyl the methyl ketone. Are we together? And then here we're simply going to have uh So, let us summarize. So, in terms of the uh oxidation oxidation with a potassium permanganate or potassium dichromate, we're simply going to have B positive, and then A. So, it's A here, and then here is B. So, B is positive, and then A is negative. And then uh with Fehling's or old Benedict's solution, we're simply going to have A negative, B positive.

[00:06:43]And then with uh Tollens' reagent, we're simply going to have um A negative, B positive. And then with the iodo- iodoform test, we're simply going to have A positive, and then B negative, right? So, these are the tests which we are simply going to have uh distinguish uh A from B. Are we together? And we're And we're done, right? And then we go on to the next part which says, "Identify with the reason the compound with the higher boiling point." Right? So, we're required to have the compound with the higher boiling point between A and and B. So, let us clear here so that we can easily compare, right? So, here in terms of these two, we're simply going to have difference only uh in the functional groups. So, here that's where the whole difference is. This one uh it is a uh it is an aldehyde, and then this one is is a a ketone. Are we together? So, that's where the whole difference is. Are we together? And then here we're simply going to have uh the lone pairs being delocalized into the benzene ring, right? To intensify the pi system, right? So, here we're simply going to have the reduction in the electron density on this nitrogen atom. Thereby, we're simply going to have a reduction in the ability to form uh hydrogen uh hydrogen bonding, right? So, reduction in the ability to form hydrogen bonding due to the delocalization effect where the lone pair is being delocalized into the benzene ring. Are we together? So, here we have this reduction in the hydrogen bonds effect is same to both both compound A and B, right? So, this one is negligible. Why? Because it is same effect is it is the same in both compounds A and B, right? So, when we did physical chemistry, that's where we explained the criteria in which we have to have the formation of the hydrogen bonding. We said we need to have hydrogen bonded directly to one of the most electronegative element which bears at least one lone pair of electrons. So, you need to have hydrogen directly bonded to one of the most electronegative element which bears at least one lone pair of electrons. So, here we having this one amino group where hydrogen is bonded directly to nitrogen which is one of the most electronegative element. So, in terms of the most electronegative element, we have nitrogen, oxygen, and fluorine. So, fluorine is the most electronegative element with a polling value of four.

[00:08:32]Are we together? So, you must know you must understand this one from your physical chemistry. So, here we have the reduction in the ability to form hydrogen bonds on both uh compound A and compound B, right? So, we are not going to explain the difference using the issue of hydrogen bonding. Are we together? So, the difference it is there present in the issue of these two functional group, right? Here we having this one the ketone and then here we having this one the aldehyde. So, here in terms of the ketone we having this other group, we having this other group.

[00:08:57]And then here we only having this one is the other group. Are we together? We having the compound compound A attached to two electron donating alkyl groups, right? It is attached to two electron donating alkyl alkyl groups. This one R1 and R2, right? But since we going to have these two alkyl groups having a positive inductive effect, whereby they're simply going to push the electrons towards this carbonyl functional group. So, they're simply going to have the intensification of the dipole moments on this carbonyl functional group. So, the dipole moments are simply going to become more more polar. So, the dipole are simply going to become more more polar, right?

[00:09:27]Because of these two alkyl alkyl groups. Are we together? And then here we only have this one alkyl group.

[00:09:33]So, positive inductive effect is simply going to be reduced on aldehyde compared to this this to this ketone. Are we together? So, we're simply going to have compound A having a stronger or a higher boiling point compared to how to this one to this one common B. Are we together? Those are the strong dipole dipole moments which are there present in A.

[00:09:49]>> [music] >> Right? So, you must understand the difference and then we now proceed at the next part. We now proceed to the next part. So, number B says uh fig 8.2 shows the structure of an organic compound X. Right? So, here we're having the structure of X. So, this one X uh we're simply going to have this one the hydroxyl group attached to this uh benzene ring to having this carbon to carbon double bond. Then we're simply going to have this carbonium carbonium functional group. So, these are the functional groups which we're simply going to have in the in the compound X.

[00:10:13]Right? Uh hydroxyl group attached to directly to the benzene. This is what you call the the phenol. Right? And then phenol [music] it is uh inactivated benzene ring because we're simply going to have this one as an activating group.

[00:10:23]Right? So, the lone pairs of electron on the oxygen atom are simply going to be delocalized again like we have alluded to here when we have the amino group.

[00:10:30]Right? So, we're simply going to have the same effect. Right? And then here we're having this carbon to carbon double bond which is the the alkene functional group. Right? So, this one you're simply going to have a direction hydrogenation reaction where we're simply going to add uh hydrogen across the carbon to carbon double bond. We're simply also going to have uh the oxidation reaction using a potassium permanganate. So, we're simply going to use what is it called?

[00:10:50]Potassium permanganate. Right? So, the cold one we're simply going to have the manufacturing of the production of diols. Are we together? Right? And then here we're simply going to have the uh the carbonium functional group. So, we have fully highlighted uh all the reactions of these uh carbonium functional groups in our previous tutorial. Right? So, you should go and have uh the full video the full series in our topical in our topical playlist.

[00:11:08]Right? The first part says name functional groups in X. So, we have highlighted that we're simply going to have the phenol. Going to have the alkene the carbon to carbon double bond.

[00:11:15]And you're also going to have the uh the keto. Right? So, functional group is simply defined as the atoms atoms or group of atoms or arrangement of bonds which determines the chemical nature of an organic compound. So, we're having atom or group of atoms or arrangement of bonds in the issue of keto of carbon to carbon double bond, right? So, you must know the definition of functional group, homologous family. You must know this these definitions by heart, right? So, here the first part says draw the structure of the organic compound formed when X reacts with 2,4-dinitrophenylhydrazine, right? So, we're simply going to [music] react this one with the the carbonyl functional group, right? So, we're simply going So, let me clear here so that we're simply going to have this one as our phenol, right?

[00:11:59]And then we're simply going to have this one attached to this one.

[00:12:03]We are simply going to draw our our benzene our benzene ring, right? And then here, that's where we're simply going to have this one, right?

[00:12:11]And then we're simply going to have And then we're simply going to have CH3, right?

[00:12:16]So, here we're simply going to remove the the oxygen and then we [music] add 2,4-dinitrophenylhydrazine, right? So, we're simply going to have a nitrogen here. And then you're having nitrogen there and then you're having hydrogen. Then we're simply going to have this one again, the benzene the benzene ring, [music] right?

[00:12:32]And then we're simply going to have this one.

[00:12:37]This one And we are we are done, right?

[00:12:39]And we are we are done. So, this one we're simply going to observe the orange the orange precipitate of 2,4-dinitrophenylhydrazone, right? So, here is hydrazone. Are we together? And then we go on to the next one which says hydrogen cyanide in traces of sodium hydroxide. So, we're simply going to have the production of what you call the cyanohydrin, right?

[00:12:56]So, we're simply going to have the production of the cyanohydrin.

[00:12:59]hydrin, right? So, here we're simply going to have nucleophilic addition to this keto functional group. So, we're simply going to clear here. Let us clear this one so that you can easily use this one, right? So, here we're simply going to have this one is the alcohol group and then this one is the cyanide group. So, this one is what you call the cyanohydrin, right? So, you can also be examined on the issue the mechanism of this nucleophilic the nucleophilic addition reaction, right?

[00:13:24]So, you can also be you on this one. So, you should know this one by heart, right? So, let me clear here so that I can illustrate the the mechanism, right?

[00:13:31]So, here let us say we're having this one R, and then we're simply going to have this one as the carbonyl group, so this one is R2, right? And then we're simply going to have this one, oxygen being more electronegative than this carbon atom, and then we're simply going to have this one attaining partial negative charge, this one attaining a partial positive charge. And then here we're having this one at the cyanide group, we're having this one as the the cyanide group, right? So, we're simply going to have this carbonyl carbon being attached by this cyanide cyanide group.

[00:13:57]So, we're simply going to have this one here. And then we're simply going to have the next step where we're simply going to have this carbon atom again, and then here R1, and then here R2, and then we're having the cyanide group being attached, so we're having carbon and then N, and then we're simply going to have this one, the oxygen with a negative charge, right? And then here we're having hydrogen cyanide, right?

[00:14:16]So, we're simply going to have hydrogen being bonded to the cyanide [music] group, and then here we're simply going to have this one uh drawing all the electrons towards itself to attain a partial negative charge, this one attaining a partial positive charge. So, this one is simply going to be polarized uh by this oxygen, the negatively charged oxygen atom, right? So, we're simply going to have this one attacking this positively charged >> [music] >> hydrogen atom to give the uh to give what you call the the alcohol group, right? So, here we're simply going to have this one, carbon atom, and then we have R1, and then here we have R2, and then here we have the cyanide group, and then here we have uh the hydroxyl group. So, this one is what you call the cyano- cyanohydrin. Are we together? And we have regenerated the So, here from the fission of this bond, uh we're simply going to uh regenerate the cyanide group, right? We have regenerated group, which is the catalyst of the of the reaction. Are we together?

[00:14:59]And we are And we are done, right? And then uh we now move on to the next part where we're going to have the reaction with a cold uh potassium permanganate, right?

[00:15:07]So, we said with cold potassium permanganate, we're simply going to have what you call the production of of diols, right? So, we're simply going to oxidize this carbon-to-carbon double bond to form a a diol. So, here we're having O, and then here we're having O, and we are And we are done, right?

[00:15:20][music] So, we remove this double bond, oxidize it to 280 all together and we are and we are done, right? And then let us move on to the next part.

[00:15:27]So, we are now on section section D. So, always remember to subscribe so that you'll be notified whenever whenever we post, right?

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