J2026 ZIMSEC A LEVEL CHEMISTRY SECTION C

[00:00:00]All right, it's Nyaki. That's Niyaki Osteo Zone and today we are going to revise the June 2026 A level chemistry from the Zimsec exam board as you can see on the screen. Right. So if you're doing Zimse, [music] you need to pay attention to these instruction the candidates. If you're not doing just pay attention to data, right? So today we are simply going to uh discuss uh section C which is organic chemistry and in our previous tutorials we did section A, we did section B. So always remember always remember to subscribe so that you'll be notified [music] whenever whenever we post. So today we simply going to have number number seven right in the previous tutorial we did number we did number six. So today we are focusing with number [music] seven. So number seven says define the term the mechanism the reaction mechanism we required to define the term the reaction mechanism. So the reaction mechanism simply defined as the step this is the step by [music] step sequence right step by step sequence of elementary of elementary reactions or stages showing how the reactant bonds are simply going to break how the reactant reactant bonds are simply going are simply going to to break and how the new product bonds are simply going to form how the product product bonds are simply going to to form also including the movement of electrons using Kelly arrows. Right? The movement of electrons using Kelly arrows are together. So we're going to have the step-by-step sequence whereby we are having the reactant bonds being broken down and then the product bonds being formed and then we also having the movement of electrons by arrows. Are we together? So this is the definition of reaction mechanism. So you should know the reaction mechanism free radical substitution. You should know that one of nucleophilic substitution. You should know all the reaction mechanism. Are you together? So let us move on to the next the next part. So the next part says define what is meant by the substitution the substitution reaction. So substitution reaction is easily defined is a reaction whereby an atom or a group of atoms replaces another in an organic molecule. So we're having an atom or a group of atoms replacing another atoms or group of atoms in a molecule together. So here we're going to have this one is a substitution reaction. So we have electrophilic substitution, nucleophilic substitution, right? So you should be able to define also what is meant by the term nucleophil and also what is meant by the term electro all together. So you should be able to define these these two terms, right? And then we also have what you call addition reaction. So addition reaction, you should know that we're simply going to form one product, one single mole together. So it's a plus b to give us c. So this one is what you call the addition reaction. So where two or more molecules react together to form one one molecule. And we also have what you call the elimination elimination reaction. We are simply going to remove small molecules. Are we together? So here we can remove water. We can remove hydrogen chloride for example in concentration polymerization. Are we together? And then we also have what the hydraulysis reaction. We're simply going to break a molecule using using water.

[00:02:48]Are we together? And then we also have redu reaction whereby we are simply going to have oxidation and reduction using oxidizing and reducing agent.

[00:02:55]Using the oxidizing and reducing [music] reducing agent. Are we together? That's what you call the redux reaction. Right?

[00:03:00]The oxygen is simply going to come from the oxidizing agent and then oxygen from the oxidizing from the oxidizing agent and then the hydrogen from the reducing agent. Are we together? So this is what you call the redox reaction. Are we together? The rearrangement reaction. So this one we're still going to have a reaction whereby atoms or group of atoms shift to occupy different position in a molecule. Right? So this is what you call the rearrangement rearrangement reaction. Right? We're having the atoms or group of atoms shifting to occupy different position from the initial positions. Are we together? This is what you call the rearrangement reaction. Right? And we are and we are done. We are now moving on to the next we are now moving on to the next the next part. Right. So the next part says describe the reaction mechanism of the reaction between propan 2 and what hydrochloric hydrochloric acid. So required to have the reaction mechanism. So let us insert another page so that we can easily have the reaction the reaction mechanism together. So we're having propano to write. So here is prop. So the suffix prop simply mean to say we're having three carbon atom right. So we having three carbon atoms right the alco group on position number two mean say this one is a secondary alcohol the carbon alcohol contains one hydrogen atom like we have alluded to in our previous discussion. So we said primary alcohol we're simply going to have the carbon alcohol the carbon alcohol the one containing the hydroxy group right attached to two hydrogen atoms right so having two hydrogen atoms and then the secondary we're simply going to have the carbon alcohol attached to one hydrogen. So this one we're still going to have the other group and then we have the other group and then we have this hydrogen atom.

[00:04:29]Right? So this one is the alcoarbon. So this one is the secondary. And then in terms of the stationary we are simply going to have the carbon alcohol attached to zero hydrogen atom. So it is not attached [music] to hydrogen atom.

[00:04:39]We're simply going to have these other groups. So this one doesn't under go oxidation. This one is simply oxidized to alihide. This one is simply going to be oxidized to to ketone together. So ketone doesn't form any further oxidation. These ones they under go further oxidation to give the caroxilic the caroxilic acid are together. They're still going to form the weak acid which is the caroxilic acid. A weak acid defined as an acid which partially dissociate in solution to give the protons right. So we're still going to have the formation of acids from this our primary alcohol. So together and then the alcohol doesn't under go doesn't under go any oxidation together.

[00:05:12]So here we're simply going to have this one is our is our secondary our secondary alcohol. Right. So let me clear here so that we can easily have the structure of the secondary the secondary alkal right. So this one right we're having three carbon atoms right and then we're having this hydroxy group on position number two right [music] so here we're having the hydroxy the hydroxy group right so here we're having our hydroxy hydroxy group and then it is attached to one carbon atom right like we've alluded to before right so here we're having this these hydrogen atoms and we are done so this one is the display formula of the propon2 right away together and then we're simply going to have the reaction with hydrochloric hydrochloric acid Are we together? So this one, let me draw it [music] in this way. Simply going to have our hydrochloric acid here. Are we together? So since this one, it is a secondary alcohol. We are simply going to have the the favoring of the first order nucleophilic substitution. Right?

[00:06:05]So simply going to have the first order nucleophilic substitution. Are we together? So you must understand this one. You must know this one by heart. So for the second order, for the second alcohol in the tertiary, they under go the first the first order reaction.

[00:06:17]First order nucleophilic substitution because they've got a carbon cation. to form a stable carboation all together because of these because of these alcohol groups which you've got a positive inductive effect that pushes the electrons towards this carbon atom to stabilize the the carbon cation together and then for the primary alcohol these are the ones which under go the second order nucleophilic substitution the primary primary alcohol are together and we are and we are done you should not this one the first order it is unlecular unlecular in terms of the rection kinetics and then this one the second order it is bi molecular we are simply going to have the concentration of the nucleophile and also the concentration of the alcohol affecting the rate of the reaction.

[00:06:54]Right? So this one it is the rate law or the latest equation of the the second order. Right? So we simply going to have the rate constant and then we have the the alcohol here. Right? The alcohol here and then we also have the the nucleophil here. Right? This one is for the second order which is ble. Right?

[00:07:08]And then in terms of the first order we're simply going to only to have the concentration of the alcohol affecting the rate of the reaction. So here we're having the concentration of the of the alcohol affecting the rate of the reaction. So this one is uni molecular.

[00:07:20]So you should know this one in terms of your understanding from reaction kinetics. Are we together? So here we're going to have the reaction mechanism of this one. Right. So we're simply going to have the reaction mechanism right. So let me highlight the bond here. So simply going to have this one. So here let me clear again so that you can have the the [music] bond right. So here we know that chlorine is more electrogative than than this hydrogen. So I'm simply going to have a chlorine withdrawing all the the bonding electrons towards itself to attain a partial. So let me use another color. A partial negative charge is simply going to have all the electrons concentrated on the chlorine atom. So here partial negative this one partial positive all together. And then here we're simply going to have let us have let us have this one this side having this hydrogen atom here. And then we're having these two lone pairs.

[00:08:05]Right? So here we're simply going to have the electron cloud here on this oxygen atom. Right? So we're simply going to have this positive hydrogen atom attacked by this electron cloud.

[00:08:14]Right? So here we're simply going to have the fusion of this bond to form the chloride. the chloride ion, right? And then we're simply going to have positively charged hydrogen atom attached here, right? So, we're simply [music] going to have this one and then we're simply going to have oxygen here and then attached to two hydrogen atoms bearing a positive charge and then we are simply going to have this one H here and then we still going to have we're still going to have the fusion of this bond. We're simply going to have the fusion of this bond to stabilize this carbon atom and then we're simply going to have the next stage where to have the production of water molecule. Right? So, here we're simply going to have the stable carbon cation. We're simply going to have our carbon cation. So here we're having these um metal groups right these methyl the methyl groups right the methyl methyl groups and then here having this positive charged carbon atoms and then here we have production of water from this from this group all together. So here we have this nucleophile right which is the chloride.

[00:09:06]We're still going to have this one attacking this carbon cation together and then here we are simply going to have the production of this molecule together. [music] We're simply going to have this metal group these metal groups here together and then here we simply going to have this one and and this one right [music] and we and we done. So this is how we simply going to have the mechanism of the of the reaction. Right?

[00:09:28]And then [snorts] we now move on to the next uh the next part. So [music] the next part says explain why in the mechanism in this reaction we favor the use of propon2 and not propan one. So this one it is a a primary alcohol.

[00:09:40]We're simply going to have SN2 and then here we're having first nucleophilic substitution because of the positive inductive effect offered by the methyl groups together. And then this one we're not having the formation of a stable carbon cation. Right? So here formed is is unstable [music] and then here we simply going to have the stable carbon cation due to the positive inductive effect offered by these methyl groups.

[00:10:00]Right? [music] So here we're having the stable stable carbon cation all together. So you must know these principles by by heart. Right?

[00:10:07]>> [music] >> And then let us go on to the next part.

[00:10:09]So the next part we're having so let me clear here. So it's fix 7.1 shows the reaction scheme for the production of phenol. Right? So having the production of phenol. [music] So the first part we're having the nitration of benzene.

[00:10:21]Right? So here we're having the nitration of benzene to form nitro nitrobenzene. Right? So we're simply going to have to production of nitro nitroenzine. So we're simply going to have a nitrating mixture where we going to have sulfuric acid concentrated and also nitric nitric [music] acid. So we're simply going to form nitrous acid in situ. And then the conditions we're still going to have the reaction carried out at 50°. Right? So we must avoid the use of high temperatures because it will result in multiple nitrations are together. So here we're simply going to have this one as the first reaction in the production of M to give nitro nitroenzene. Right? So in terms of the mechanism of this one, we're simply going to have the benzene ring here.

[00:10:54]Right? So we're having the benzene ring and then here we're having the nitronium ion formed from this nitration mixture.

[00:11:00]Right? And then you're simply going to have this pi system of electrons in the benzin ring attacking this nitronium ion. And then we're simply going to have the formation of this arinium arinium ion. Right? So here we having this arinium ion this positive sign. And then we simply going to have this. And then we are going to have this hydrogen atom.

[00:11:17]We're simply going to have the fusion of these bonds to stabilize the the normal benzene the pi system of electrons to stabilize the the pi system of electrons in the in the benzene ring. Are we together? And then here we're going to have the formation of this the nitro the nitro benzene. Right. So this one the first step and then you go to this one the second one and then lastly we have this one which is the the third step [music] right this one these are the this one is the mechanism of the nitration of benzene right and then we go on to the next the next one so the next one we're having step two where we have n and then n going to react with sodium nitrate and acid and then we maintain low temperatures below 10°C [music] and then we're simply going to have this benzene chloride all together so here we're simply going to have the formation of phenom right so here we having the production of NSphenol amine which will then under diation reaction to give this benzene donium chloride. Are we together? So here this is what you call the diioidation reaction. Are you together? Which is carried out carried out at this low temperature. So you must always try to understand why we are to use these low temperatures because this ion this salt is very very unstable. So it above temperatures above 10°C.

[00:12:20][music] This salt is simply going to decompose to give pheno nitrogen and nitrogen hydrogen chloride. Right? So these are the products. If we to increase this temperature above 10°C [music] are we together? So here we have this salt and then here we're having phenol reacting the salt benzene dasonium chloride. So this one is what you call the coupling at the clapping reaction right so in terms of the coupling reaction we need to react this benzene donium ion with an activated activated benzene ring so that you can have the formation of the aso dies are together. So here we're having this one the first so we have highlighted the first reaction here. So for the second reaction we are simply going to make use of use of tin and concentrated hydrochloric acid. Right?

[00:12:58]So here tin and concentrated hydrochloric hydrochloric acid and then we reflex the the mixture. Right? So these are the conditions for step number two and then the step the conditions of the steps we are given and then here we are also given the salt. So we want to give the conditions of the coupling reaction to form this aso the ao right.

[00:13:15]So the coupling reaction is an electrophilic aromatic substitution reaction. Right? The coupling reaction is also known as an electro electrofphilic electrofilic aromatic substitution substitution reaction.

[00:13:27]Right? So we're simply going to have the benzene donium ion. This one acting as as the electrofil and then it is reacting with an activated benzene ring which is the the phenol. So this one is acting as the electro together. So this one is acting as the electrofil with an activated benzin ring. So we're simply going to have this one is the coupling coupling reaction. Right? [music] Whereby we are simply going to form the ao the ao right? So the aso linkage this one is our azo azo linkage right so you must know the ao the azo linkage and also the uses of azo aodoise are we together so here since this one we say this one is in the electro field and then we're having this activated group we're simply going to have what you call the or par substitution on the benzene ring or par substitution right but when we are to have the the coupling reaction you must always highlight [music] underain that we are only going to consider the par substitution why so because of the bulkiness of the benzene tonium chloride we are simply going to have what you call the steric hinderance like we have in the tetra chloromine where we simply going to explain why this one is not cannot allow hydrarolysis because of the steric hinderance principle where going to have the carbon atom clustered by this chlorine the big chlorine atom so we're simply going to have no space for the infiltration of the water molecules from the understanding of inorganic chemistry so here we're simply going to have the same reaction we're simply going to have the same condition where we are to have this bulkiness of this donium [music] chloride benzene donium chloride being a bulky molecule so We are simply going to have hinderance whereby we are to force the molecule to have par substitution not substitution. Right? So we are simply so you should know the reason why we are to avoid the or substitution. Are we together? Right? So here we're going to have this donium the formation of the donium the as right and then here we're having step four and then we're having pheno. So this one it is the hydrarolysis of this one. So we said if we to have temperatures greater than greater [music] than 10° we're still going to have this one the formation of this one. So we having the formation of pheno and nitrogen gas and hydrogen hydrogen chloride. Are we together? And we are and we are done. Right? So let us go now to the questions. Let us now go to the to the questions. Right? So the first part says draw the structure of M.

[00:15:29]So we have already done that. We having nitro nitro benzene. Right? We are we have already done that. This one is our M. And then that one of N we have you know am right. Are we together? And then state the conditions and re reagents for nutrition. We said we need to have the nutrition [music] mixture. We need to have the nutrition the nutrition mixture right. So the nitration mixture we're simply going to have concentrated nitric acid this one [music] and then concentrated sulfuric acid this one which act as the catalyst or to regenerate the electrofil which is the nitronium and nitronium ion. Right? Are we together? And then the temperature we need temperatures below 50. The temperatures must not exceed at that one because we have multiple nitrations. Are we together for the step two? So which one is our step two? So we said here we're having tin and concentrated hydrochloric acid and then we're simply going to reflux right and then write the equation for step four right which one is our step [music] four. So this one is our step four. So we said we're simply going to have the decomposition reaction. So here let me clear this one so that we can easily have the equation for step step number four. So step number four we're having a benzene denium chloride right. So here the hydrolysis of benzin donium chloride. So this one right from the benzene ring and then we having two nitrogen atoms and then this one and then we're having the chloride ion right and then we're having hydrolysis mean you're having the presence of water and then you're having the product phenol right so here we're given phenol as the as the product plus hydrogen chloride and then also plus plus nitrogen right so this one is the is the equation which we are simply going to to have are we together and then we now go on to the next part state the observations made in step [music] four. So in step four we are simply going to have the production of nitrogen gas. So we're simply going to have the efficence of of the gas are together and describe the industrial use of compound G. So which one is our G? So the G we said G is the AO die. So the uses of AO dies we're simply going to use them in printing inks in the manufacturing of printing inks and then also in the manufacturing of dyes and also in the manufacturing of coloring coloring leather coloring leather and also coloring paper in the manufacturing of paints and then plastic and rubbers all together. So these are the uses of a.

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