Free Chemistry NEET Video Lectures for Alcohols, Phenols and Ethers

00:>00:17 [Music] Hello everyone. Uh today we are going to learn about an important class of organic compounds which are called alcohols. Alcohols are represented by the general formula R O H where R is the alkal group. Okay. So it is the hydrocarbon part of the molecule and this is the hydroxil functionality which imparts special properties to these hydrocarbons. Okay. So this is the hydroxile group and we can say it's a unique uh composite of a hydrocarbon and a water molecule. So these uh compounds

00:01:02 have several important uh roles in our day-to-day life. They are part of many important compounds and you know things which we use such as varnishes. They are present in perfumes. They are also present in the hand sanitizers which we use these days quite popular and the alcoholic beverages. So they are a part of several important compounds which we have been using. However, uh probably we need to learn something uh about their chemical structure and properties and that is what we are going to do exactly in

00:01:39 today's lecture. So when we refer to an alcohol in general uh the most common alcohol is an ethanol right. So ethanol is given by the structural formula CH3 CH2 O this is a twocarbon alifhatic alcohol and in general it is prepared by ferment by the metabolic process of yeast. it is prepared by the fermentation of sugars. So the fermentation of sugars yields ethanol and this is in fact one of the industrial ways of making and producing ethanol. So you start with the sugar and in the presence of yeast.

00:02:28 This is a metabolic process of yeast. we get ethanol as the major component along with the production of carbon dioxide gas. So these alcohols are uh important as you as we are saying and ethanol particularly is important because it it is used these days as a fuel in cars. So it can be a replacement of gasoline. In fact, ethanol, gasoline mixtures can be used in varying amounts as a fuel in automobile industry. So, uh with this background of the importance of alcohols and their applications in everyday life, let us look at from the

00:03:12 chemical perspective why they are so important class of compounds. So uh alcohols if I would say they can undergo various functional group transformations and they are the raw material to a variety of different organic compounds which we can obtain starting from the alcohols as the basic structural unit. So they are the raw material for several uh interesting class of compounds. So starting from alcohol we can do an elimination reaction. So via an elimination reaction the alcohols can furnish us what we know as

00:03:55 alkenes. So elimination reaction on alcohols can yield us different types of alkenes. If they can also act as a substrate for a nucleophilic substitution reaction. So if we carry out a nucleophilic substitution reaction on alcohols what we get can be a variety of substituted alkal derivatives like alkalh helides or other compounds. Likewise they can also undergo oxidation. Oxidation of alcohols by virtue of the presence of the hydroxile group. They can generate variety of alihides. We can have ketones. They can also produce

00:04:43 acids as well as the acid derivatives like esters. So all these are the possibilities when we are starting with alcohols as substrates. We can even use them as a nucleophile. So when they used as a nucleophile in a nucleophilic substitution reaction they even yield what we call as ethers. So we can get ethers also from alcohol. So these are the different functional group transformations which are possible on alcohols and that is why we say they are such important class of organic molecules. So what are some of the

00:05:19 physical properties? Let's look at the physical properties of alcohols which make them so important compounds compared to the counterpart hydrocarbons. All right. So the physical property is by virtue of the hydroxile group. Okay. So it is by the virtue of hydroxile group that there are many differences you would find in a hydrocarbon compared to any alcohol. So if you look at the structure of an alcohol, they have this unique ability to form what we call as the hydrogen bonds. So this lone pair of electrons on oxygen

00:06:04 which is electro negative it and the hydrogen which is rendered electron electropositive they have this unique ability to form hydrogen bonding between two molecules or more of alcohols and because of this ability to form hydrogen bonding which could be intermolecular or intramolecular depending upon the situation of the hydroxy group. So it forms essentially intermolecular hydrogen bonding between different molecules of alcohols and that is why they tend to exist as associated liquids. So alcohols usually exist in form of

00:06:49 associated liquids. In fact they can also be used as proteic solvents in many organic reactions. So some of the implications of this hydrogen bonding is it's manifested in two of the important properties. The boiling point of alcohols and the solubility properties of alcohol. So the solubility and the boiling point shows distinction uh compared to the corresponding hydrocarbon because of the probable possible hydrogen bonding in these molecules. So let us look at uh the boiling point scenario. What happens to

00:07:22 the boiling point of alcohols? If we compare the boiling point of an alcohol to that of a hydrocarbon or a corresponding ether for that matter, we say that the boiling point of the alcohols is much higher. It is much higher than the corresponding hydrocarbons. or ethers of the same molecular weight. So we can conveniently uh say that if we compare the boiling point of alcohols with that of a hydrocarbon or an ether of the same molecular weight, it's higher than than them. And this can be easily understood

00:08:16 in terms of the additional hydrogen bonding interactions. If we compare ethanol to ethane, then ethanol as we all know. So this is ethane. So ethanol is a liquid at room temperature. And the boiling point of ethanol is + 78° C. Compare this to ethane which is a gas at room temperature with a boiling point of 89° C. So this uh clearly shows such an increase in the boiling point because of the additional hydrogen bonding interactions. Also um we see that alcohols which are 12 carbon or more they exist as solids at room

00:09:03 temperature. So not only can they exist as a liquid they the additional interactions make them exist as solids at room temperature. The other properties in terms of the boiling point of alcohols is similar to as we see in hydrocarbons that as the carbon chain length increases the boiling point of alcohols tends to increase. So this is quite uh obvious because this is supported by the vender walls interactions which the longer alkal chains are going to have thereby raising the boiling points. Also as the

00:09:43 branching in the carbon chain increases as so as the branching in the carbon chain increases again as anticipated the boiling point decreases because of the reduced Vanderval's interaction. So these are characteristics similar to any other hydrocarbon except for the fact that the hydrogen bonding makes uh the boiling point higher than the corresponding hydrocarbons or the ethers of same molecular weight for that matter. Let us look at the solubility properties of these molecules. So by virtue of the hydroxil groups

00:10:22 these compounds are polar in nature. They are polar in nature and as we know that if there's a polar compound it'll be soluble in a polar solvent. So in the original form if these alcohols exist as hydrogen bonded associated liquids. And if you compare it to the structure of water which also has an intermolecular hydrogen bonding in which it exists. So both are polar molecules and if alcohol has to dissolve in water then it has to form new hydrogen bonds with water molecule. for this dissolution to take place. So

00:11:19 overall it means that for this solvation or the the ethanol or any alcohol to be missible in water the delta H solvation should be greater than the delta H required for dissociation of the alcohols. the dissociation energy required for breaking these hydrogen bonds and the dissociation energy required here for making breaking the hydrogen bonding in the water molecules. So if this exceeds then the substrate alcohol will be soluble in the solvent and also uh we witness that the alcohols with longer alkal chain. So

00:12:01 we are talking about so it has two parts the hydrophobic alkal part as well as the hydrophilic uh hydroxy group. So alcohols which have a longer alkal chain are dominating in the hydrophobic part and they are immissible with water. So this missibility and solubility in water is only by virtue of the hydrogen bonding interaction between the polar group and the uh O of water. All right. So if you have understood this physical properties, let us move on to classify the alcohols. Let us understand how can

00:12:39 we classify the alcohols based upon the number of hydroxal groups present in them. So the classification which we are going to discuss now is based on the number of hydroxal groups present in the alcohol. So based upon that we can have an alcohol with one hydroxile group which we call as the monohydric alcohol. So it is an alcohol with one hydroxile group given by a general formula CN H2N +1 O. It could be an alcohol with two hydroxile groups which we say is a dihydric alcohol and given by the general formula CN H2N

00:13:32 two hydroxils or it could be a trihydric alcohol with three hydroxile groups or it could be more than three. So we can always have a polyhydric alcohol. So if multiple hydroxy groupoups are present can be a polyhydric alcohol. So based upon that the properties the physical properties their solubilities the boiling points change as well. So we say that as the number of hydroxile groups in the alky chain increase okay from 1 to 2 to three. One, the properties the boiling point. What is the implication? The boiling point

00:14:18 increases and the missibility with water as expected also increases because of many more hydrogen bonds it can the multiple hydroxy groups can form. So the misibility with water increases as well. So a typical example would be that of ethylene glycol which is a dihydric alcohol. It's a twocarbon dihydric alcohol. It boils at 197° centigra. So if you compare it to ethanol, there's a substantial increase in the boiling point of this molecule by virtue of an additional hydroxil group. Okay. So in

00:15:02 the monohydric alcohols we have a further subclassification. So all those hydrocarbons bearing one hydroxy they can further be classified as a onederee or a primary alcohol which means that the hydroxy group is attached to a carbon which is monos substituted. So we have a monos substituted carbon to which the hydroxil is attached. It's called a primary alcohol. we can have a secondary or a twoderee alcohol in which the hydroxile group is attached to a carbon which is disubstituted. So there's one hydrogen and there is

00:15:40 there are two alkal substituents or it could be a three°ree or a tertiary alcohol where the hydroxil is attached to a carbon which is tri substituted. So based upon that we have uh different classifications of alcohols which we'll be we'll be using uh subsequently in the later part. Okay. So let us look at the structure of these molecules. What is the structure of alcohols in general? So alcohols are shown by the molecular formula RH. The carbon is sp3 hybridized carbon. The oxygen is again sp3 hybridized. So if

00:16:27 you just look at the oxygen electronic configuration 08 is 1 s2 2 s 2 px 2 2 p y 2pz. So with the sp3 hybridization on oxygen we get a tetrahedral arrangement in which there are two lone pair of electrons on the oxygen and there is one electron each in one of these hybridized orbitals which then bonds to the r the carbon and the hydrogen. to give this alcohol and the theoretically the angle the bond angle Roh theoretical is expected by virtue of the tetrahedral structure to be 109.5 but experimentally this angle has been

00:17:30 found out to be close to 105.5° so this decrease in the bond angle from from the theoretical to to the experimental values can be u seen in terms of the loan pairs the two lone pairs which are available on oxygen and by virtue of this loone pair loan pair repulsion this bond angle the ro bond angle squeezes and we see a lower experimental angle. All right. So going ahead uh to some important uh part the property of these molecules is the acidity and the basicity behavior of alcohols. So let us see the acidity and

00:18:16 basicity behavior of these molecules. So as we know that acidity is nothing but the ability of the molecule to lose a proton and basicity is its ability to accept a proton. So let us look at the proton donor and acceptor abilities. The pKa values of the the alcohols roughly fall in the range 16 to 19 which tells us that they are weaker acids than water. So they are weaker acids than water. And if we compare their acidity with the known acid like a caroxilic acid then you can see and we all know

00:18:55 that any alifhatic caroxilic acid aqua solution if you treat this with a base like sodium bicarbonate which is a weak base. So when we treat any caroxilic acid with a weak base like a sodium carbonate we get the sodium salt of that acid. we get the corresponding caroxilate annion along with the formation of carbonic acid. So we form this weak acid which dissociates to give us carbon dioxide and water. So whenever we have to confirm the presence of an acid we add sodium bicarbonate and we see the

00:19:35 effervesence we see for the presence of effervesence which is uh an indicator that the compound organic compound in question is a caroxilic acid. But if you run a similar experiment on an alcohol, I'm afraid we do not see a similar result which which is quite uh explainable because alcohols are much weaker acids as compared to the caroxilic acid. So if we have to test for alcohols, one of the tests is that you treat it with a reactive metal. You treat it with a reactive metal. The metal could be a sodium,

00:20:09 potassium, magnesium, aluminium. So any of these reactive metals when you treat the alcohol you get the corresponding metal aloxxide. So we tend to get these metal aloxxides along with generation of an evolution of hydrogen gas. So this uh shows that they do have acidic properties but they are much less acidic compared to uh the regular caroxilic acid. And this property can be used for making all these metal oxides which are used as bases. Some of the uh most used ones are we probably have come across different

00:20:52 places is like something like a sodium ethoxide which is used as a base in many organic synthetic reactions. Aluminium isopropoxide. So this is trialent aluminium. And we also have very popular which is uh potassium tertiary buttoxide. So all these are used as bases in in different reactions. Something about the basicity of these molecules. So this is the acidity. There are much weaker acids than water we have claimed. And if you have to compare the relative order of acidity of the alcohols with amines and water. So

00:21:36 we would uh we could rate it that in terms of their relative order of acidity water being the strongest followed by alcohol followed by ammonia followed by a hydrocarbon. So this this could be the relative order in which the acidity could be rated and uh the basicity of these molecules The ability to abstract a proton to get protonated is because of the presence of the lone pair of electrons on oxygen which can take up the proton get protonated and form this kind of a kionic intermediate which will come across in

00:22:28 many chemical reactions. the acid catalyzed uh reactions of alcohols all of them go through formation of this uh this kind of intermediates and if you have to compare the relative basicity of the counterions of the acids which we just discussed so between the hydroxide the aloxxide the amide and the carbonion the order of the relative order of basicity if we can predict would follow the order this being the strongest base and hydroxide being the weakest base. We we will discuss this in more details with

00:23:05 examples in subsequent sections. Okay. So with this brief introduction to the physical aspects of alcohol, let us move on to some of the chemical properties uh dealing with their preparation. So the preparation or the synthesis of alcohols is important as I told you they're important raw materials for so many reactions uh involved in our day-to-day life to get important molecules. The preparation essentially is from three important starting precursors. One is starting from alkenes. So we'll discuss the preparation of

00:23:45 alcohol starting from alkenes through the reaction on alkenes is the electrofphilic addition reaction. Electrphilic addition on alkenes gives alcohols. And this could either follow a maronov's pathway or it could be an antimaronic addition. So we have the maronic of and antimaronic of addition and conversion of the alkenes to alcohols through electrofphilic addition. Okay. The other important substrates are the alkyhelites and alkyhelites can again be converted to the corresponding alcohols through

00:24:38 substitution reactions. So substitution of alkaalh helites can yield us alcohols and this again can go via SN1 or SN2 pathway. All right. And the third in the list is the carbonal compounds. Very important. They are the backbone of the organic synthesis. The carbonal compounds they can also yield alcohols. And now in this case it is through a nucleophilic addition reaction. So the nucleophilic addition onto the carbonal double bond C double bond O the nucleophilic addition on the carbonal compound either again through

00:25:23 different reagents uh could be lithium aluminium hydide strong reducing agent could be the milder one sodium borohydide or even the addition of the grignard's reagents can give us a variety of different alcohol. So this is the layout the important uh starting materials for alcohols and now we'll take up each of them one by one. So let us see when we start from alkenes what are the different options and possibilities we have for making alcohol. So with alkenes as the starting substrates the most convenient way. So if you just

00:26:01 look at it carefully the alken has to be converted to an alcohol. So it is nothing but it is addition of a water molecule across the carbonarbon double bond. So there are different ways of doing this. A direct way would be which is as you can see you directly add water molecule to it. So what we say it's we can do direct hydration of alkenes. Okay. So hydration of alkenes is nothing but addition of water. So you take the alken any alken. Let's say we are working with one propene as an example

00:26:39 and you treat it with water and this is usually an acid catalyzed hydration. So you do an acid catalyzed hydration. So the first step would be addition of from the sulfuric acid unit the proton and HSO4 onto the carbonarbon double bond and this is followed by replacement of this HSO4 by the the the O unit from the water and we get this corresponding alcohol which you can now see for yourself is a secondary alcohol. alcohol because it is attached to a carbon which is a D substituted. So what we call here

00:27:23 is that the overall effect is addition of water molecule and hydrogen from this acid it has gone to add onto the carbon which already had more hydrogens attached to it. So that is why we call that this product is a maronov's product. It's a maronov's addition. The product is a maronic product where in general all this hydroxy group always goes and adds on to the carbon which forms a more stable carocatine. This is a general principle. So we usually gets the maronic of addition as the dominant pathway for all these

00:28:01 addition reactions on alkenes giving us alcohols. For example, in this case, the other uh important process which is industrially relevant as well is the oxo process for making alcohols. All right. So uh in the oxo process we start with the alken treat it with carbon monoxide hydrogen gas mixture in the presence of a cobalt catalyst which is octaarbonile dooobbalt. There is an intermediate formation of an aldihide which is subjected to reduction with hydrogen gas and nickel as the catalyst and this furnishes

00:28:57 what we say is an alcohol and the alcohol is a primary alcohol. So this auxo process is in fact an industrial uh process of generating these alcohols and the important thing about this process is as you would see here is that it helps in stepping up of the series. So if we are starting with a twocarbon system, let us say we starting with an ethine and we treat it with CO NH2 in the presence of the cobalt catalyst, the dicobbalt octaarbonil we get an aldihide which is now a 3carbon alihide. So there's a stepping

00:29:41 up of the series and this is followed by reduction to give you the 3carbon alcohol that is the one propenol. So we get a onederee alcohol through this because an aldihide reduction would yield up a primary alcohol. So this is u a very important uh industrial process for making these alcohols. Okay. So let us move on to the third important pathway starting from polyphen. It's what we call popularly as oxy mercuration demmercuration and as the name says you can see that it involves somewhere mercury in this

00:30:32 reaction. So in this reaction we start again with the alken. We are talking about all the methods from alkenes for generation of alcohols. You start with an alken treat it with a mercury salt which is a mercuric acetate in this case and this is in THF water. In between we get an intermediate which is a mercurinium salt and this is in the second step it is subjected to reduction so sodium borohhydide alkaline medium when we get the corresponding alcohol. So it's a two-step process. The first step is what is the oxy

00:31:22 mercuration and the second step which is the reduction is the demmercuration which is the removal of the mercury from the from the intermediate salt which is generated. So uh the reaction has certain important features. One this reaction gives you alcohols through an addition which follows maronikov's addition. So it gives you maronov's product. It does not involve any carocatine. So there is no rearrangement which is observed if you're making alcohols through oxy mercuration demarcuration.

00:32:00 There's no rearrangement because there is no formation of carocatine. So there's no carocatine intermediate which is because the carocatine has this unique ability to always rearrange to a more stable carocatine whenever it gets an opportunity. But in this particular case since there's no carocotion generated we do not see any rearrangement in the product and this addition the the oxy mercuration step it follows an anti- addition it is an anti-addition protocol. So uh let us look at the sequence the manner

00:32:36 which it follows in. So you start with an olifine, you treat it with the mercury salt, mercuric acetate in THF water. So this step involves loss of an acetic acid unit and what you get in the middle is an addition product in which there is a hydroxil added and there is HG A which is the acettoxy mercury which is which is uh present on the other carbon and this now has to be reduced with sodium borohydide as the reducing agent to give the final desired alcohol which overall involves addition of a

00:33:34 water molecule on the carbonarbon double bond. So this is the overall oxy mercuration demmercuration reaction in which there is addition of this water molecule as as we mentioned that from in order to get an alcohol from an alken the overall process involved is the addition of a water molecule. Okay. So let us uh look at uh some of the examples and the products which we expect to arise from these molecules. So if you start with this cyclic olifhen which is one methyl cyclopentine and you subject it to this oxy

00:34:13 mercuration demmercuration reaction. There are two possibilities. We can either get the product in which there is a hydroxy substituted on this side or we can get a product in which so this is corresponding to a methyl. So in which there is a hydrogen and the hyd hydroxil goes onto the other side. So of the two products uh which is the product which is formed through this reaction. If you look here in this case the hydroxil is adding to the carbon which has which is more substituted or in other words uh the hydrogen is added to

00:34:58 the carbon which is already having more hydrogens which is less substituted. So we say that this is the maronicovs product and we know that this reaction favors a maronic addition. So this will be the only product. So this is in terms of the rigiochemistry of this uh reaction it follows and allows you to get a maronikov's addition product. Okay. So if look at another olifhen which is it's a tertiary butil. Okay. So if you have to number this this will become a 33 uh dimethyl 1 2 3 4 and one butine. So this is a 33

00:35:45 dimethyl one butine. And if you again subject it to hydro oxy mercuration and demarcuration, what we get in the process is an alcohol uh which will be this one with hydrogen goes and add to the terminal carbon and you get this product which is 33 dimethyl 2 butinol. So you can see that there was a possibility of a rearrangement in this case to give you a more stable carocatine at this center but no there is no rearrangement there is uh no rearranged product there is no carocatine so you get this exclusively

00:36:26 as the only product which is obtained so this is interesting uh let us understand since uh a lot of this specificity is observed in this reaction what is the mechanism that this oxy mercuration demmercuration is going through. So let's peep into the mechanism and try to understand some of the aspects associated with the reaction. Okay. So we have an olifen to begin with. Okay. So we begin with an olifen and there is a mercury salt available in the form of mercuric acetate. So the first step

00:37:12 is believed to involve this attack of this pi electrons onto the mercury salt HG and with the elimination of the acetate annion. The first step we believe is the formation of uh this kind of an intermediate which is a three-member intermediate. You have the mercury which is forming a cyclic three-membered cyclic intermediate with the two carbons of the double bond. This positive charge sitting with the mercury. So this is u this is a cyclic mercurium ion which is generated as the reaction intermediate. So you can

00:38:07 see that there's a bridge formed on this side. Okay. So the next step which is the attack of water molecule onto this cyclic mercurium ion intermediate now has to be from the side other to the side from where the bridge is formed. So one side is blocked by formation of this bridged intermediate. So the water molecule attacks from the other side and this results in opening up of the bridge and formation of. So if I have to show this the O2 it adds from the side opposite to where the bridge was and we get

00:38:56 this as the open chain analog of this cyclic analog. So the next step now is the if you see this is a trans addition. So that is why this is an anti-addition. And this is very selective. It's quite stereospecific. This follows an anti-addition pathway. Okay. So the opening up we have the mercury and the water residing on opposite sides. Okay. And if we see that the water molecule it attacks onto the carbon which is more substituted. Okay. So there so it's it's attacking onto the carbon which is more substituted. So

00:39:35 again there are two reasons that this carocatine will be more stable one and that is why it is attacking here and the second is that the mercury would go onto sterically less hindered carbon. So it being a bulky uh group there it'll go and and sit on the less sterically hindered carbon of the two carbons of the alken. All right. So the next step now would be to take up this proton which is done with the help of the acetate annion which acts as a base and it deprotonates this intermediate to yield

00:40:23 to yield this intermediate in which you have a hydroxy. on the carbon which is more substituted and mercury is on the carbon which was less substituted and now is the last and the final step where we have the sodium borohydide reduction. Okay. So in this step is where it loses the stereospecificity. So the the borohhydride acting as a reducing agent it reduces this mercury right it breaks the CHG bond. So there's a breaking of the carbon mercury bond. And there is a creation a new CH bond takes its place. So we have

00:41:05 eventually the last step hydride from the sodium borohydride which replaces the HO A and we get uh finally our alcohol which is uh Maronikov's product. So uh this is overall what is expected out of the oxy mercuration demmercuration mechanism. It is also sometimes called oxy mercuration reduction because there's a reduction involved in the final step. A very interesting reaction the fourth one uh of what we have discussed from from alkenes which complements this oxy mercuration demmercuration is a very

00:41:43 popular reaction called hydroboration oxidation. So hydroboration oxidation it was for this reaction that Etsy Brown got the Nobel prize for discovering this unique reaction called hydroboration oxidation. So when I say that it is complimentary to oxycuration demarcuration why because you start from alkenes and you get alcohols which are antimaronic. So we get antimaronic product. So this is interesting because this complements your oxycuration demarcation alken treat it with boron hydride in a solvent like THF

00:42:34 or ether and what we get is an alkal borine. So there is addition of the boron onto the carbonarbon double bond of the alken. So there's addition of the boron and hydrogen. We get an alkal borine and then this is subsequently followed by again it's a two-step process. It is subsequently followed by the oxidation with hydrogen peroxide as the oxidant under alkaline conditions to give us finally the desired product the alcohol and the boric acid. So it has two components again. The addition of the borohydide over to the

00:43:15 alken is called the hydroboration and conversion of this alkal borine into the alcohol is the step which we call as the oxidation and overall it involves addition of a water molecule across a carbonarbon double bond to give us an alcohol which is an antimaronic product. That means we tend to get alcohols which are less substituted. Okay, looking at how this reaction works. Let us see and figure out. So we have an olifen treat it with borine. So a borane usually exist in the dic form as a diorane.

00:44:00 These reactions are executed in dry ether or THF. So the first step is the hydroboration that is addition of the boron hydrogen across the carbonarbon double bond and in the process what we get is something like this which we call as an alkal borine. So there is addition of one of the olifhen across this BH adds across one of these olifhen we get the ethile borine and this step repeats because still we have two of these boron hydrogen bonds available for addition over an olifhen and this again repeats twice it adds on

00:44:41 to another molecule of ethine we get di alkal borine which is a dethyl in this case and still we have a BH bond. So the first step is that you know these borines are leis acidic we have a empty p orbital and this pi bond they push the electrons into the mtp orbital of boron which is why the first step is initiated. So this adds on now to the last another molecule of olifhen since we have still a boron hydrogen available there right and finally what we get is a triethyl borine. So this is the hydrooration sequence

00:45:30 which goes one after the other and there's replacement of all the three boron hydrogen bonds by the boron alkal bonds and that's why it's called a hydro borration reaction. All right. So the next step is once we get the trial borine it under goes oxidation with hydrogen peroxide under alkaline conditions to furnish three molecules of ethanol since we started with ethine as the starting olifhen and boric acid. So this is eventually what happens in the during the oxidation. So if we start with um

00:46:11 let's say I'll take the same examples which I took for oxy mercuration demmercuration. I start with the same olifhen and we carry out a hydrogoration and oxidation on this molecule. So the product which we expect to see from this is complmentary to what we got for the other case. So in this case again you will see the hydroxy group. Okay. So this is a methyl this is a hydrogen. So you can see that in this case the hydrogen goes and adds to the carbon which is more substituted and we get the antimaronic

00:46:49 product. Another interesting feature here is that we get addition through a sin pathway. We get a sin addition product. So if I have to write the stereo uh chemistry of this product then the hydroxy as well as the hydrogen they add from the same side. So I've shown them by the bold the wedges and this by the dashed. So this methyl and this hydrogen are away from us. So this follows a sin addition pathway. And if we do the other substrate as well which we took for the previous case. In this case again the product will be

00:47:32 the other way to what we saw with the oxy mercuration demmercuration and we get in this case again a primary alcohol. So in this case we are getting a primary alcohol. So this is how this hydroarration oxidation complements the oxy mercuration demarculation and so important a reaction for obtaining the alcohols which are less substituted at the carbon. So what is the mechanism? it it's quite interesting to understand uh what is going on during the reaction that we are getting so uh interesting

00:48:02 products. So uh we believe let us start with any olifhen and this is uh let's say I start with this olifhen in which this carbon is substituted with two methyl groups and this carbon is non-substituted. So we have two hydrogens's shown with the dashed and the wedge and this is treated with diorane or the boron hydide. So the first step is the transfer of this uh electron density from the pi bond onto the electron uh deficient boron the mtp orbital. Right? So the first step would be the attack here. And

00:48:55 what we see is the formation of a four membered cyclic transition state. So we have a methyl. There's another methyl here. Okay. I will not write these methals on because it's uh kind of understood. So only for hydrogen's we put the hydrogens's. So the first step is that this formation of this carbon boron bond here we form this carbon boron bond and this boron it's delta positive and the carbon gets delta negative and it it pushes uh this electron density. So this carbon becomes delta positive and now to

00:49:38 satisfy this boron hydrogen the the other boron hydrogen it transfers this electron density from hydrogen and works as a hydride and forms an additional formal bond with this carbon. So we get this kind of a four membered uh cyclic transition state. So this formation of this four-membered cyclic transition state and this can explain the sin addition and eventually please remember that it is this boron which would get replaced by the hydroxil group. So there's formation of this cyclic four member

00:50:19 transition state. there's a sin addition. Eventually, this gives you this product which is the monoalkkal borine. So, there's a new CH bond here and this hydroboration. So, you have two hydrogens and you have added a boron onto this carbon. So, this is the first step of hydroarration where one of the BH bonds has has been added across a carbonarbon double bond. So this step now is repeated twice with two more molecules of the olifhen. It is repeated and what we get here is a trial borin.

00:50:55 So if we just have to represent so you have this hydrogen added up here and this gives you the trial borine. This unit is attached to boron and if I just write starting from the alken which we have the trialkken boron will look something like this. So you have boron now it is tri substituted. So you have one alkal group here then you have another alkal group and you have a third alkal group. So if you just have to uh give the general representation what we get here is a R3B where R is equivalent

00:51:38 to this alkal group. Okay. And this is connected to boron. Okay. So this is what happens during the hydroarration when you end up with the trial borine. Now the next step is the oxidation in the presence of alkaline hydrogen peroxide. when this trial borine now reacts with hydrogen peroxide under alkaline conditions to yield the corresponding alcohol and there's this formation of hydrooric acid. So what is happening here in the alkaline conditions the first step is that your hydrogen peroxide with

00:52:17 hydroxide annion to give you the hydro peroxide annion. So this is actually uh the active nucleophile which is generated in the reaction which initiates the oxidation process. So you have the trial borin now with the boron having the empty p orbital and the hydroparoxide annion which is generated. All right. It transfers its lone pair of electrons into the empty p orbital of boron by acting as a nucleophile and this is what triggers the reaction and we get a boron which is a tetraalent boron with the hydroparoxide unit attached

00:53:01 here. Boron bears a negative charge and you have the three alkal groups. This intermediate which is formed is quite unstable. So it's quite unstable and then this OO bond which is here it's unstable and this is stabilized by the transfer of this alkal group from boron to oxygen. So this carbon attached to the boron it migrates from boron to oxygen with the elimination of this anion as the hydroxide annion. So this is what happens uh during the oxidation process. So we get a boron with two of

00:53:38 these alkal groups and there is an O R. So we got we get this aloxxy borine and this aloxxy borine now the last step is left that we have to get the alcohol out of it. So again this reaction is uh initiated with the hydroxide annion which is available there and it again attacks boron and we get again a tetraalent boron a boron with one hydroxy one eloxy and two alkal groups and a negative charge. Okay. So now this is uh the last step when this O is lost as the aloxxide annion. Okay. And it gives us back this

00:54:30 B R and R. And the same uh is repeated. So we get the aloxxide annion. Okay. With the two lone pair of electrons here on oxygen and another. And then the water which is available we are using a THF water. This acts as the proton source and the aloxxide annion here acting as picks up this proton and get the corresponding alcohol and the hydroxide annion. So here uh this step this boron which is formed this can further in the same manner just the way we started with this it can again react with the hydroparoxide twice more and uh

00:55:14 give us two more units of the corresponding alcohol along with the hydroxide. The mechanism of a hydroarration oxidation very useful reaction follows sin addition it follows antimaronic's pathway and gives us alcohols which are primary in nature difficult to obtain from other methods. All right that's what we do for this time and next time we start with the other starting materials the alkalh helides and the aldihides and ketones for producing alcohols. All right. Thank you.