Transcript for:
Organic Chemistry Reactions Overview

the following lecture is going to be a quick crash course on all es organic chemistry reactions and I'm going to try and cover all the es organic chemistry reactions as quickly as possible so the first thing that we're going to discuss is called the cracking of long-chain hydrocarbons the problem with long-chain hydrocarbons is which obtained from crude oil mostly they are large molecules and they have the bigger the molecule the lower is going to be the flammability of the molecule it's going to be very viscous which means it won't be able to flow very easily and it's going to have higher melting and boiling point so it's going to be in solid or semi-solid state so these properties are not very suitable for use as fuel instead what we do is we break these large molecules into smaller molecules which are going to be more flammable and they're going to have lesser viscosity they would flow very very easily and they're going to have lower melting and boiling points which means that there would be they would easily evaporate and they could easily be transported or they could pice they could pass through pipes etc so here I've written one cracking reaction where you have C 60 H 122 that's provide breaking down into smaller molecules and it's a random process could break down into any smaller molecules so I've broken it down into an alkene another alkene and an alkane the conductor the condition is that whatever is formed it should all the carbon atoms and all the hydrogen atoms they must add up to form this molecule this long chain hydrocarbon because all these pieces tiny pieces are obtained from this long chain hydrocarbon so another way I can crack because it's a random process I can another way I can crack this molecule is that in this next cracking what I've done is I've produced an alkene and alkyne so all of them are alkenes the smaller molecules that are produced and when all of them are alkenes just to complete all the atoms I would need a hydrogen gas would also be produced next to it so if you sum up all the atoms they were all combined to form this molecule over here so hydrogen gas would be produced if you have all of them our alkenes so remember whenever you're cracking you're going to get smaller molecules they could be alkanes alkenes hydrogen and there could be a mixture and a mixture of these molecules the conditions that are needed for catalytic cracking are zeolite catalyst is needed which is a combination of live anymore Seiden silicon oxide and 500 degrees centigrade temperature would be needed for catalytic cracking or you can use thermal correction which requires high temperatures and high pressures temperatures generally in the range of 450 to 700 degree centigrade and pressures are around 70 atmospheres the next reaction we go to study is the free-radical substitution of alkanes and/or alkyl chains now remember free-radical substitution reaction is not only possible in alkanes but any carbon chain any carbon chain in any molecule whether it's a carboxylic acid or an alcohol any CH bond or carbon or hydrogen chain that could undergo free-radical substitution so here's one example reaction where you have a propane molecule you have carbon with all these hydrogen it's a carbon and hydrogen chain it's reacting with chlorine and and UV light is or sunlight is needed for this reaction what happens is any of the hydrogens over here it's going to get substituted substituted by chlorine so over here i've it could be the hydrogen on the first carbon atom so that is substituted by chlorine or it could be the hydrogen on the second carbon atom or it could be multiple hydrogens all of the hydrants can also get substituted and HCl is also formed as a byproduct in this reaction here's another example reaction where you have a propanoic acid molecule so i told you that it could happen this reaction will happen with any alkyl chain so chlorine is reacting and the UV light is required so any of the hydrogen the carbon chain can get substituted by CL and again it could be a byproduct so this reaction not only happens with alkanes but any carbon chain any carbon chain in any molecule that could undergo free-radical substitution free-radical substitution reaction is usually done with chlorine or bromine the reaction fluorine is very very explosive because fluorine is very reactive so that reaction is usually not done and with iodine the reaction is extremely slow or non-existent because iodine is extremely unreactive so usually it's these two that would be involved in free-radical substitution another thing that you must know is that the detailed mechanism of this reaction although I have not described a detailed mechanism over here but a detailed mechanism of this particular reaction is required and knowledge of that mechanism is needed for 4es organic chemistry in this third reaction we're going to discuss the electrophilic addition of alkenes so all the reactions will apply to alkenes or unsaturated hydrocarbons that have carbon double bond carbons all these molecules having unsaturated that are unsaturated hydrocarbons they move undergoes this particular reaction so the first reaction is called bromination where you have an alkene a double bond is present but bromine is reacted with it what happens is the double bond the first thing is that the double bond breaks into a single bond and both carbon atoms to complete their fourth bond remember they if you look carefully they're making three bonds are they're gonna make a fourth bond with the bromine atom from this molecule and this carbon does exactly the same so this isn't it this is an addition reaction and two bromine they're going to get added to the double bond carbon atoms conditions that are needed for needed for this reaction are you need absence of light because if the light is present then you're going to get free radical substitution as in alkenes the the other conditions are are it's room temperature in precious or no special conditions bromine would be in a queston of bromine would be dissolved in an inert solvent called CCL 4 or a tetrachloromethane so either of the two bromine are going to be used but this one issue with bromine Aquos and that is that when water is present water remember is not inert so when water is present what's going to happen is then instead of two bromine is getting attached they are chances all the two brewings are going to get attached but they would be chances that which ions from the water molecules they might also join with the carbon atoms so you're going to get another molecule as well which is going to be this molecule over here so one Romijn and 108 from water might also get a task but this is only the case of bromine acquisition bromine opposites use you could get two molecules instead of just one molecule over here now the important significance of this reaction is that it's used as a test for identification of unsaturated compounds so all alkenes can be identified by the reaction with bromine the red brown bromine is going to react it's going to get decolorized so if if with an organic compound bromine gets decolorized that means that that particular organic compound must have an unsaturated it must be unsaturated must have a carbon double bond born in it so it must be an alkene so this next reaction is called hydrogenation of alkenes in which you have an alkene or in any unsaturated hydrocarbon which has carbon double bond carbon atoms in it and you're going to react it with hydrogen again a similar reaction would happen the carbon double bond changes into a single bond it becomes saturated and the carbon then bonds the two carbon are transferred were making double bonds previously they're now going to bond with the hydrogens that we added over here so the hydrants are going to get added to the position with double bonds were present the conditions that are needed for this reaction are nickel and platinum catalyst and the temperature that's required is 150 degrees centigrade the significance of this particular reaction is that unsaturated hydrocarbons or any molecule that's unsaturated that would become saturated if it had double bonds all the double bonds are going to get removed if this reaction is performed change in physical properties would be that the melting and boiling points would also increase slightly this would have a smaller but less a boiling point this would have a greater boiling point the reason being that the molecule size has increased if the molecule size increases by the addition of these hydrogen's the molecule is slightly bigger than this molecule over here so if the molecule size increases van der Waals forces are going to be created and there would be more attraction between molecules so the melting and boiling point would be creative one one application for this is making of marjorine so in this application unsaturated vegetable oil is changed into margarine by the addition of h2 what h2 does is that it increases the size of the molecule so vegetable oils arginine liquid state but when they change into margarine when all the double bonds are removed and they change into single bonds and additions hydrogen gets added the size of the molecule increases and the melting in boiling point is higher which is why marjorine is in solid state whereas a vegetable oil is in liquid state this reaction is called the hydration of alkenes again another electrophilic addition reaction in which you have an alkene unsaturated carbon double bond carbon containing carbon double bond carbons and it's reacted with water and water gets added the double bond breaks down into a single bond and the carbon atoms that were making double bonds they get they have hydrogen and which groups attached to them and you get a molecule of an alcohol alcohols are formed in this reaction from alkenes the conditions for this particular reaction are 3 are in degrees centigrade temperature 60 atmospheres the pressure and an acid catalyst is used usually phosphoric acid but it could also be sulfuric acid that's used as a catalyst makes electrophilic addition reaction for alkenes is with hydrogen halides it's here hbr or a try and again you're going to have a similar reaction you'll have an alkene unsaturated carbon double bond carbons hbr is going to react with it the double bond breaks and changes into a single bond and both carbon atoms that have making double bonds one each would attach to one carbon atom BR would attach to another carbon atom so this is a reaction that I've shown with HBR with HCl in a chai the reactions going to be very similar and every time the conditions are going to be that constant created HBR is going to be used or if you're using HCl in concentrated it's here or concentrate H I is going to be used in this reaction now the very important issue is the Marconi cause rule for electrophilic addition since we are dealing with electrophilic addition reactions this rule is very important especially for unsymmetric alkenes now in unsymmetric alkene is the one where the the there is symmetry absent around the double bond for example if you look at the right side of the double bond this carbon is bonded to two hydrogen's but if you look at the left side of this double bond this carbon is bonded to one hydrogen and on the other side there is a carbon chain so this particular molecule is unsymmetric around its double bond so for these alkenes if they undergo an electrophilic addition reaction for example they react with I'm reacting it with water two products are going to be formed because the double bond is going to break and when the hydration occurs our one carbon bonds with H and the other one bonds with Oh H but the position could be switched which carbon atom bonds with hydrogen which carbon atom bonds with which either this one once with H this one mantra the which or this one bonds with each and this one bonds with which so you were to get two different products yeah the two products drawn and you would notice that the position of the eighth and which groups that are attached they are different between the two Molly Hughes so the double bond breaks changes into a single bond each gets attached on one carbon atom which gets a task to the other carbon atom or over here which gets attached to the carbon in the middle and each gets attached to the carbon at the extreme right so this is the hydration reaction happening and you're getting two different products now the mikaze comes rule is that the rule is that the hydrogen atom prefers the carbon atom that is already bonded to more hydrogen's so it's would bond to the carbon in season one C which is bonded to more hydrogen so if you look at the two carbon atoms that are making double bond over here you'll notice this carbon atom is bonded to two hydrogens well as this carbon atom is bonded to a hydrogen and on one side it's more need to a carbon atom so this carbon atom is bonded to two hydrogen it's bonded to more hydrogen's so each would prefer to bond with the carbon atom which is bonded to more hydrogen so when you're attaching H in which the H is going to be it's going to prefer the carbon atom that is bonded to more hydrogen's and the O H wise versa is going to prefer the carbon atom that is bonded to a lesser hydrogen so this is going to be your major product this will be your major product there's a higher probability of this product forming and there's but this other product will form but it would be lesser than quantity it's it's it would be call your minor product over here H is bonded to the carbon atom that is bonded to fewer hydrogen's whereas always responding to a carbon atom that responded to more hydrogen's so remember this this is going to be amazing product whenever electrophilic addition occurs for unsymmetric alkenes extraction that we're going to study is called the milder oxidation of alkenes on this alkene which is propane or it could be any alkene or any unsaturated hydrocarbon that has carbon double bond carbons and mine oxidation occurs when you reacted with an oxidizing agent specifically came in a for potassium magnate 7 and the conditions that are used are cold dilute or alkaline conditions these are mild conditions so under these mild conditions mild oxidation would happen and what happens in my low oxidation is that the double bond breaks into a single bond so here I've redrawn propane but without the double bond instead in the place of the double bond there is a single bond now and two complete the bonds these two carbon atoms that were previously making double bonds they would have hydroxyl groups attached to them and alcohols would be found specifically dials diodes are formed during my low oxidation sedation is pretty simple whenever you see a double bond and you see coal diluted as nine conditions I just get rid of the double bond convert it into a single bond and the two carbon atoms add which groups to both carbon atoms so - Oh H groups are going to get it asked during my lock sedation extraction that we're gonna discuss is a very similar reaction and that is strong oxidation of alkenes but what happens in strong and oxidation of alkenes it's pick an alkene first picked a molecule of 2 methyl butene which is an alkene you can pick any molecule any molecule that has a double bond and strong oxidation will happen and the conditions that are needed for strong oxidation are in hot concentrated and acidified came to know for potassium magnet 7 is going to be off sizing agent and the conditions are going to be extreme it's going to be heated it's going to be concentrated and it's going to be as divided so so the oxidation is going to be strong oxidation and the very first thing that's going to happen in a strong oxidation is that a double bond is going to break completely so you see this double bond over here the alkene double bond it's going to break completely and the molecule is going to split into two parts so this part is going to separate from the other part so the double bond is going to completely break off and there would be three scenarios now when the double bond breaks how do you determine the product of the oxidation depends on three scenarios which I'm going to write down now the first scenario is that when the double bond breaks off focus on the carbon atom that's making that was making the double bond ignore all the other more other atoms the other atoms enough that nothing is going to happen to the other atoms focus on the carbon atoms that were making the double bond for example in this molecule fucus on these two carbon atoms that were previously making double bonds so coming back to our first scenario the double bond breaks off and if the carbon atom that was making that our bond is bonded to two hydrogen's if it's bonded to two hydrogen's then that particular carbon atom is free to get converted into carbon outside and a water molecule second scenario is that the carbon atom that was previously making double bonds is bonded on one side with a carbon chain this R represents a carbon chain alcohol chain or any group any carbon group so any chain attached over here and one side is attached to hydrogen so if you have this earlier than that the carbon atom only that carbon atom that was previously making the double bond that carbon atom is going to get converted into a carboxylic acid so you can see that I've attached a double bond o and in which group with that carbon atom third scenario is so this third scenario is that the double bond carbon atom is bonded on both sides with the carbon chain so if it's morning on both sides with a carbon chain then that carbon atom is going to get converted into a ketone only that nothing happens to the rest of the atoms or molecules or anything attached to it only that carbon atom will be converted into a ketone it's going to form acetyl mono instead of previously was bonded it was making a double bond but now it's bonded to an oxygen so let's get back to a molecule and let's try and figure out the strong oxidation product for this particular molecule over here the first thing that you should notice is I've broken the double bond this part this entire group is now separate focus of this carbon atom that was previously making double bond exponent on one side with a hydrogen and on the other side it's bonded to a carbon chain so which out of the three options does it look similar to it's it's option number two there's a carbon atom that's bonded on one side by a hydrogen atom and on the other side there's a carbon chain what does it form it forms eventually ends up forming the carboxylic acid so let's go back and it's focus on this let's redraw this now and here I've drawn the product I've changed this carbon atom into a carboxylic acid so so this carbon atom only changed this carbon atom nothing happens to the rest of the molecule this carbon atom changes into a carboxylic acid group so the CC is still there everything's still there this part has broken off so I remove that that carbon atom has now changed into a carboxylic acid it's now forming a double bond over and in it's a - twin H group let's focus on this other carbon atom now this other carbon atom was also part of double bond and this double bond has now come broken look focus on this carbon atom it's bonded on two sides with a carbon chain so which out of the three options does it resemble it's it's probably assembling this third option when a carbon atom that was previously making a double bond is bonded to two carbons chains then that carbon atom gets converted into a ketone nothing happens to the carbon chain only that carbon atom gets converted into a ketone and that is exactly how I'm going to do what I'm going to draw over here so this is going to be my product the carbon everything is the same everything is the same this carbon atom has changed its it has changed into a ketone I've made it I made it into a C double bond o so only this carbon atom that I'm underlying this is the only part that is changing very quickly I'm going to try and do two more examples of strong oxidation one is this molecule over here and one is this molecule over here the first thing that's going to happen is during strong opposition the double bond is going to completely break so I'm going to break the double bonds in both molecules completely and I'm going to focus on these carbon atoms the ones that were making double bonds are only those carbon atoms are going to change so first focus on this carbon atom over here if you look at this carbon atom over here this is a carbon art and that is bonded to two hydrogen so if you add the double bond carbon atom that is bonded to two hydrogen then that particular fragment is going to be converted into a carbon dioxide and a water molecule when it oxidizes then really if you focus on this other carbon atom over here you would notice that this particular carbon atom is bonded on both sides with carbon chains on both sides there is a carbon chain there is no hydrogen attached to it so I've told you that this carbon atom is going to be converted into a ketone so I've redrawn this particular fragment as it is and just focus on this carbon atom this carbon atom is going to be converted into a ketone so I'm going to add as see they'll 1:02 it so this is what's going to happen the rest of the molecule remains exactly the same only the carbon atom that was previously making double bonds is the one that's going to change these over here are my two oxidation strong oxidation products the double bond breaks and the molecule splits into two parts now let's focus on this other molecule again the double bond is going to completely break focus on this carbon item that was making double bonds first and if you focus on this carbon atom you would notice that this carbon atom is bonded on one side with a carbon chain and on one side there is a hydrogen so if you have this carbon atom on one side one side carbon chain then this carbon atom is going to be converted into a carboxylic acid group so as you can see everything remains exactly the same only this carbon atom over here change that only rest of the groups should be exactly the same only change that carbon atom it changes into a carboxylic acid group and it oxidizes let's focus on this other side now and focus on this carbon atom over here this carbon atom is if you focus on this carbon atom on both sides it's bonded to carbon chains so if it's bonded to carbon chains on both sides if this carbon atom will be converted into a ketone so I've redrawn this fragment over here and focus on this carbon atom I'm going to convert it into a ketone and that's going to be my product so again the molecule splits into two the double bond completely breaks one side gets converted into a carboxylic acid the other side gets converted into a ketone the next reaction that we could study is called the addition polymerization of alkenes and this what happens is that you have thousands of different monomers which are which are small simple alkenes so for example I've drawn the different I've drawn for eighteen molecules in an actual polymer there would be thousands of eating molecules and they were to join together what might happen is that and they're high pressure in high temperatures the double bond is going to be converted into a single bond and once those double bonds get converted into single bonds each carbon is short of one bond so they would start interlinking with each other so they would start bonding with each other to complete the fourth bond and so this interlinking would continue in thousands of different eating molecules they're going to join together to form a very very large macromolecule another way to represent this addition polymer is to draw one repeat unit for example if you look carefully this repeat unit of this repeat monomer of this monomer is repeating over and over again so so you draw one repeat unit and you put brackets around it and put a small in around it n represents the number of times this repeat unit is repeating over and over again in this polymer chain there are some examples of addition polymers that you must know the first one that we drew which ended up forming this molecule over here it's a polythene molecule remember that was the Miraval was a theme so when lots of heating molecules they got together the joint combined they formed polythene another molecule a polymer that he must remember is a PVC polymer in PVC you have a molecule that's similar to e team and except for the fact that it has one chlorine attached to it so it's chloro ethene and lots of n number of chlorine molecules they're going to join together similar similarly in exactly the same manner as the first example I did and when they join together they would form this polymer over here these are continuation bonds which means that this is the repeat unit that is repeating over and over again and this is your molecule of PVC which is used in pipes and in tiles etc it's a construction with you so it's used in pipes and tiles etc another example molecule is all for Teflon which is made from tetra fluoro 18 so lots of them they combine together to form this polymer which with the double bond finishes and you have single bonds with these continuing bonds so this this repeat unit is going to repeat over and over again this is Teflon poly tetra fluoro 18 and death on it's used as a as a nonstick in cooking utensils the next reaction that we're going to study is called nucleophilic substitution of halogen au alkanes and the first reaction in nucleophilic substitution that you will to study is with anyways Aquos and reflux and in which a quest plus reflux produces this shine and the carbon the CL gets substituted by the a china this gobble has a partial positive charge so it attracts this which -1 and the seal gets knocked out so what's happening over here is that any halogen in a molecule it's going to get knocked out by the nucleophile and it's going to the neutral file OS my this one is going to take its place so that's the first reaction let's move to the second reaction in this reaction the conditions that I use are n ACN and the solid that's used as ethanol EQ you're not going to use water as a solvent because water contains OS - benign so you can avoid Ohi and so use ethanol as a solvent or alcohol as a solvent so le chien is ethanol ik not a quiz and you use reflux when na CN is ethanol ik it contains the CN minus 1 iron and again what happens is that this chlorine all halogen any halogen is going to get substituted but it's going to get knocked out by this CN and the scene is going to take its place so you can look carefully this CL is going to get knocked out and it's going to be replaced by the scene 1 significance of this reaction is that the carbon chain length increases remember this molecule originally has two carbon atoms but or after C and has been attached the number of carbon atoms in the molecule are now three so this reaction has a particular significance which is that increases the length of the carbon chain let's move to the third reaction which is in this third reaction use a nucleophile n is three ethanolic and you get a reflux say o you what you can do is you can use any three gas and heat it in a sealed container so again the same thing happens the chlorine or any halogen is when we get it's going to be substituted by the amine group the lone pairs on n are going to be attracted to the positive carbon atom and they go to knock out the seal so the N is going to be born in with this carbon atom so this is a simple reaction this is what happens you'll get substituted by the amine group one thing that must be remembered by these reactions is that primary halogen Joaquin's and they go is in two mechanism where the nucleophile directly attacks the positive carbon atom and the CL gets knocked out so they have a transition state it's a one-step reaction which - one slowly approaches the carbon atom which was partial positive and then CL partial negative is going to be the belt it's electrons are going to be repelled and in the next step the electrons are going to leave an OS is going to take its place so remember that primary halogen of alkenes are going to undergo sn2 mechanism meanwhile T halogen of alkenes are going to add sn1 mechanism in that you have a dishy halogen or alkyne we the carbon has a positive charge and the CL has a partial negative charge in the first step the electrons are going to get knocked out the CL bond is going to break during collision so you're going to heat and the CL would get knocked out so this is a tertiary no alkenes the Oasis - one cannot directly attack this positive carbon atom because it's surrounded by very large groups so we go to wait for the seal to get knocked out once seal is removed see a minus 1 gets knocked out a carbo cation intermediate is formed and the OS minus one then approaches the carbo cation intermediate and quickly attaches to the positive carbon and resulting in the formation of the alcohol now this reaction is a two-step reaction it has a slow step where the CL gets knocked out and then it has a fast step way the OS minus one quickly attacks it through the positive carbo cation and an intermediate is formed over here you had a transition state where it ends as where as in this molecule in this reaction isn't 1 you have an intermediate that is going to be formed another thing about these nucleophilic substitution reactions are all these 3 nucleophilic substitution reactions with H minus 1 i n CN minus 1 ions and with the NH 3 for halogen Joaquin's you must know that the see any hair is an alkene containing iodine is going to have the fastest reaction because the CII bond is very weak it's very weak and the bond length is larger so since the CI bond is weak it's going to have a faster reaction so CI iodine or iro alkanes are going to have fast reactions when ICF bonds are very strong so they are hard to break so they're going to have relatively slower reactions the speed of the of the reaction for a Latino alkenes is going to vary according to this CF it's going to have the slowest reactions where where as I drove alkenes or halogen o alkenes containing I already not going to have the fastest reaction the next reaction of view we're going to study is called the elimination reaction for halogen Joaquin's and in this reaction what will happen is that the halogen in the halogen alkane is going to be eliminated so one bromine would be lost from one carbon atom and the other thing that's going to happen is that the halogen would be lost and hydrogen from the neighboring carbon atom is also going to be lost so the neighboring carbon atom could be this carbon atom over here or it could also be this carbon atom over here as well but since it's in it's a symmetric alkene alkene it doesn't matter so one hydrogen would be lost from the neighboring carbon atom as well and an alkene would be formed so as you can see over here the eight and the B are a lost ER and an alkene would be formed in its place a double bond would be formed in hbr would be produced as a result of the elimination reaction so this elimination reaction could happen with any halogen no alkene a halogen would be lost and hydrogen from the neighboring carbon atom would be lost as well and in their place there's going to be a double bond an alkene that would be formed the conditions that are used for this reaction are anyways ethanol ik and reflux remember these conditions are very similar to the nucleophilic substitution reactions where the conditions were anyways Equus and reflux so we'll remember try to differentiate between an elimination reaction and a substitution reaction for elimination you can use any way to ethanolic and reflux would be used the next reaction that we're going to study is the substitution reaction of alcohols where the OAH gets substituted and is substituted by a halogen this is the first reaction where you have an alcohol and the Oh H is going to be substituted by a chlorine the first condition for this reaction that could be used is you can use dry hydrogen chloride gas which can be produced during the reaction when you mix any CL with concentrated sulfuric acid that ends up producing its eel gas and Nadia till gas is going to knock out this Oh H and replace it with CL the final product which are which is our halogen no alkane that is formed is then distilled off right at the end to remove it from the reaction mixture another condition that is used to substitute chlorine is socl2 plus heat and this is this gives the best seal for the reaction the equation for the reaction is given you have an alcohol socl2 has reacted the antigen substitutes the Oh H group and each year and so2 is produced but in this reaction no distillation is required because HCl and so2 are gases and they escape the container so in this reaction no need for distillation is needed for to obtain the halogen or alkane the third condition to produce chloral alkanes is you can use pcl5 at room temperature or pcl3 plus heat can be used pcl5 is much more reactive so I've given you an equation for pcl5 reacts with an alcohol the H gets substituted by a CL atom and ETA and pocl3 are produced it so biggest reaction with pcl5 at room temperature and you can observe misty white fumes of HCl gas that are going to be produced and again no distillation would be needed to obtain the high general alkane because the rest of the of them are gases and they agree to escape the container so you need you don't need to distill to obtain the halogen you alkane the last halogen substitution reaction that you should know is with bromine or iodine remember the previous reactions were all substituting the Oh s group with chlorine you can use pbr3 or pi/3 and heat to obtain a bromine substituted product you were to use BB RC to obtain an iodine substituted product you're going to use pi/3 and pbr3 and pi/3 are produced the reaction in C in situ which means that red phosphorous and bromine reacted during the mixture to produce pbr3 or at phosphorus and I already energy acted during the mixture to produce this pi/3 this over here is the reaction then you're going to get if you reacted with pi/3 V o H gets substituted by inhc p o3 is produced you don't need to remember the whole entire equation because it's very difficult to remember the equation it's very clear that this equation comes in your es exams what you just need to know is the conditions that would be needed to substitute the X group in alcohols with a halogen so HCl socl2 PCL 5 or PCL 3 are used for substitution with CL and the other thing is pbr3 and pi/3 are used for substitution with the bromine or with iodine next we actually go to study is the reaction of alcohols that we are and they react with sodium metal or any group 1 very reactive metal one thing before discussing this you must know that alcohols they very very weakly dissociate they are slightly very slightly acid a oh you can call that you can say that the acid et is almost non-existent the u.s. group dissociates but it this will see it so very less and it's even weaker compared to water so you can't really call them acids but when it comes to reactions with sodium very reactive sodium metals then this reaction happens but at a very slow rate so this is the only acidic reaction that could be categorized as an acid agree acid reaction when acids react with metals and the reaction is extremely slow so I've written down this reaction over here and alcohol reacting with sodium the H gets substituted by sodium I am then the solid form and this salt is called sodium ethoxide and hydrogen gas is released just like a metal plus acid reaction which produces salt plus h2 and this particularly reaction can be used as a test for identifying alcohols because it's a slow reaction sodium usually doesn't react with anything in a slow way it's a very vigorous reaction sodium usually as by the witness reactions but with alcohols the reaction is extremely slow and you're gonna see very slight bubbles that would form on the surface of sodium next reaction a certification is in all of this reaction so application of alcohols with carboxylic acids it is a condensation reaction with an alcohol reacts with the carboxylic acid ends up producing ester and a water molecule so a small molecule is released in this condensation reaction conditions that are needed for this reaction are concentrated sulfuric acid which is used as a catalyst in this reaction and reflux is used so I've drawn an alcohol and a carboxylic acid and they and make sure that the functional groups are facing each other because it's the functional groups that are going to react with each other so one molecule must be drawn or the other way down so carboxylic acid and the O h broke the alcohol group they are facing each other what's going to happen is that the alcohol is going to lose its H and the carboxylic acids will lose its H and water is when you produce by them and the two molecules the two parts are now going to join together at this point over here so I have just brought them closer and I'm going to connect them together and this is going to be called my ester link this is my ester link it's C double bond o and single oh right in the middle of the chain the Nats you are a sterling and this is going to be your estates of fruity smelling substance that is used in perfumes and food flavorings and this reaction is a reversible reaction which means that the ester can be broken back again and the conditions are needed for breaking the ester the reaction is called hydrolysis because you're breaking the ester with the addition of water so the forward reaction called condensation the reverse reaction is called hydrolysis and the conditions that are needed are that you need to heat the ester with that you'd acid or dilute Ashlee so it's diluted acid eddying acid is to any mineral acid could work so each twist of for HCl and I do doubtfully like any which and reflux has needed so you need to heat it so the ester is going to be broken back and it's going to form a carboxylic acid and an alcohol molecule back again so this is what I'm going to do what's going to happen is that the ista link is going to break the wave formed and the two parts are going to separate now and they're going to form the carboxylic acid and the alcohol group back again so here I have broken them the o single oh parts would change into an alcohol group again the double bond o carbon it's going to change into a carboxylic acid back again and that's how the east of the reverse reaction would happen the one thing that you must remember is when you're reacting it with a dilute alpha li the problem would be that the carboxylic acid that would be formed is an acid so if the conditions when you're breaking in ester are dilute out Li the NaOH is alkaline and the carboxylic acid that is produced as acidic they're going to react with each other instead what you're going to get is going to be a salt of the carboxylic acid so you're not going to get this carboxylic acid because it's going to immensely react with the dilute actually that's being used to break it so the Alpha Li that has been used as any which are then you go to get this solved not the cup of the gases not siedel bondo in which but instead you're going to get the salt c del bond o and o minus 1 na plus 1 i suppose if we actually use this koh then you will get C double bond o and O minus 1 and k plus 1 so instead of getting this carboxylic acid under - conditions you're going to get this particular salt that would be produced the next reaction we go to study is the dehydration of alcohols to form alkenes now here's an example of an alcohol Molly you know what happens if this reaction is dehydration is that the witch is removed and in each from the neighboring atom from the next door atom next door carbon atom is going to be removed so I'm going to remove these two things so once these two groups are removed the two carbon atoms are going to end up forming a double bond they will end up forming an alkene and a water molecule would be produced as a byproduct the conditions that are used in this reaction are L 2 3 in the form of Pumas or broken pot plus heat is needed this over here is a diagram for the reaction how this reaction is going to be carried out you can see Pumas over here and alcohol a soaked wool is placed over here the vapors of the alcohol are going to pass over the few mist and when they pass over the PM is they're gonna change into alkanes and alkenes are usually gases compared to our cause and those gases can be collected over water you will bubble them through the water and they will be collected over here another condition that could be used is concentrated sulfuric acid as a dehydrating agent that would be used but the problem with this reaction is that silica said is not only a dehydrating hydrating agent it's going to produce an alkene it's going to do exactly the same thing as what L 2 O 3 is going to do it's going to produce an alkene and it's going to dehydrate the alcohol but the problem with concentrated sulfuric acid is that concentrate sulfuric acid is also a very good oxidizing agent so some of the alcohol instead of getting dehydrated might get oxidized and if it gets oxidized it might change into a carboxylic acid and you're going to get many different acetic byproducts those acetic bi products include carboxylic acid if the alcohol gets oxidized or if the sulfuric acid gets reduced in the process then so2 gas would be produced which is also acid it now these acidic bi products need to be removed so instead of collecting the gas over water like the case in al2o3 will collect the gas over anyways when you're using concentrated sulfuric acid as the dehydrating agent next reaction that we were to study is the oxidation of alcohols so the first reaction we do oxidation discusses the oxidation of primary alcohols that get converted to a Liat and then get converted into carboxylic acids now the conditions that are needed for oxidation of primary alcohols is acidified potassium dichromate 6 and reflux would be required so k2cr2o7 and reflux and aldehydes are going to be found and those aldehydes can be further oxidized to carboxylic acids for that the conditions are gained a stiff add potassium dichromate 6 and reflux or you can use tollens reagent or you can also use fellings reagent now the color changes must be remembered as well remember potassium dichromate 6 is goes from orange to green when it oxidizes primary alcohols to aldehydes and further oxidizes them to carboxylic acid so the color change is always to green and if you are oxidizing a leeet's to carboxylic acid you can also use tollens reagent for that silvermere it is produced silver ions get get reduced as Andy heads get oxidized so silver middle and a black precipitate is produced and with felling's reagent which is copper 2 plus ions contains copper 2 plus signs it's blue colored it's a blue correct solution anyway when aldehydes get off sighs two carboxylic acid the filling agent gets reduced to copper one oxide which is a brick red in color so brick red precipitate is formed in that case so you must remember the reagents for these paths and you must also remember the color changes as well another thing that needs to be remembered about this oxidation path is that if you want this reaction remi alcohols getting oxides to aldehyde and further getting oxides to carboxylic acid if you want the reactions to stop at aldehydes what you're going to do is that instead of reflux saying if what you're going to do is you're going to use distillation distillation apparatus would be used you're going to distill the mixture because as soon as a laser produced a d-h don't form hydrogen bonds carboxylic acids and alcohols form hydrogen bonds so they have a relatively higher melting and boiling points are but so aldehydes are going to quickly evaporate and you can distill them off and you can remove them from the reaction mixture so if you want the reaction to stop at aldehydes so you use distillation instead of reflux so I am now going to do some example questions take a primary alcohol remember a primary alcohol is an alcohol way the carbon the alpha carbon that is bonded to H groups is bonded to two hydrogen's and on one side there's only there's only a carbon chain so if you look at this carbon atom this is your alpha carbon its Bonnie to which it's bonded on two sides with hydrogen and on one side there is a carbon chain so there's only one carbon chain attached on one side of the carbon all the rest of the two bonds are going to be hydrogen they would be bonded with hydrogen so so make sure primary alcohols are the ones where you have an O which and which is attached to a carbon that is bonded to two hydrogen at least two hydrogen's and the rest would be the carbon chain so this is a primary alcohol and all the changes are going to happen with this carbon atom over here that I have underlined the rest of the molecule remains exactly the same this carbon atom changes it changes into an aldehyde so I'm going to change this carbon atom into an aldehyde so changes into C double bond o each and if you don't remove it if you don't distill it then it's going to further get oxidized into a carboxylic acid keeping the rest of the molecule as it is and just going to change this into a carboxylic acid C double bond o and H let's now discuss the second oxidation of alcohol reaction in which you have a secondary alcohol and that secondary alcohol gets oxidized into a ketone and again the reagent data required AB dash in dichromate sex k2cr2o7 as a fire and reflux and the concern is again going to be orange to green so secondary alcohols I will get oxides and under this condition so I'm going to do an example question now so this over here is a molecule of a secondary alcohol a secondary alcohol is you're going to focus on the alpha carbon atom which is this one over here the one that is attached to which that carbon atom would be bonded to two carbons chains and only one hydrogen would be attached to it so if weight is attached to a carbon atom that is attached only one hydrogen and on two sites they are carbon chains then that carbon is going to be a secondary alcohol and secondary alcohols are going to get converted into into ketones so don't change anything else apart from this carbon atom a carbon atom is going to change so I've redrawn the molecule over here and I'm going to change this carbon atom into a ketone I'm just going to attach a double bond and Oh to it so this secondary alcohol got converted into a ketone so the rest of the molecule remains exactly the same only the carbon atom the one alpha carbon that is attached to which that is the one that's going to change into a ketone the condition is you must you should be able to identify whether it's a secondary alcohol or a primary alcohol let's move through the third reaction which is potentially our cause the good thing about this ER cause is that they don't undergo any oxidation they are resistant to oxidation so Tracy I'll cause don't undergo oxidation you must be able to identify what is it a she alcohol so I've drawn over here in red Dilley alcohol again focus on the carbon atom that is attached or bonded to that carbon atom would be bonded to three carbon chains on all three sites over here there's a carbon chain bottom there's a carbon chain left side there's a carbon chain so if a carbon atom is not bonded to any hydrogen and it's pointing to three carbon chains than that that particular alpha carbon atom the one that is pointed to O is that would be a tricky alcohol and it's not going to undergo any oxidation so where the cables here 207 or or tollens or felling no oxidation reaction is going to happen just like those addition of sedation reactions you're going to have reduction reactions wake up most Acosta's aldehydes and ketones are going to be converted back into alcohols this is the oxidation process that we have studied you can do the reverse you can change the carboxylic acid back back into an aldehyde and add here can be converted back into a prime e alcohol and the conditions that are needed for the reverse reaction or the reduction reaction are so to convert carboxylic acid back into aliens you need nih pour in dry ether which is a powerful reducing agent and to convert and remember this reaction cannot be stopped at analyze if you add an NH 4 to carboxylic acids we know when she gets converted it will be converted into alcohol and to convert aldehyde back into primary alcohol you will need an alias for in dry ether or naba poor and pretty much the same conditions the same conditions would be needed for converting ketone back into secondary alcohols remember secondary alcohols car oxides to ketones so to reduce them back into secondary alcohols you would need you need the same conditions and I it is for dry ether or any bh4 in ethanol solvent these are the solids that would be used for the reactions we're gonna do some we're going to do some example questions so I've drawn this molecule over here and I'm adding an alias for to this molecule I'm adding a reducing agent focus only on the carbon atoms nothing happens to the rest of the molecule only focus of the carbon atoms that a carboxylic acid carboxylic acid is going to get reduced and this ketone focused on the ketone this ketone is going to be reduced as well and what you need to do is you just need to convert them into alcohols back again so I'm going to draw the molecule so here I've drawn the molecule and the only thing I need to do is I need to change these carbons into back into alcohols so this carbon is going to have an O H attached to it again so that means an HS well this carbon as well is going to have in which attached again and it's going to have two hydrogen so so all these carbon atoms carboxylic acid ketones aldehydes etc they will be converted so you need to remember this any carboxylic acid any ketone any aldehyde it's going to be converted back into an alcohol where the primary or secondary but you just need to do is in the questions what you just need to do is you just need to focus on those carbon atoms and make them attach Oh H proofs back again to those particular carbon atoms next reaction that we could study is the nucleophilic addition of a li Ed's and ketones remember this reaction applies to both aldehydes and ketones so the reaction is that a molecule of an aldehyde or a ketone focus on the siedel bond Oh it applies to any carbonyl group so C double bond o is going to react with each CN and what's going to happen is the dis carbon atom CN of HCN is going to get attached to this carbon atom over here so I've redrawn this molecule over here and what I'm going to do is I'm going to attach a seen do this particular carbon atom the one over here this carbon atom seen gets attached and the on top it is into a single bond and it change to an wage group so it's it ends up forming a cyano hydron that is formed in this reaction so it's a simple reaction C double bond o carbonyl group ah in the perk and getting converted into scene gets a task to that carbon atom and the double bond o gets converted into an Oh H group a single bond connects the carbon and the O H group the conditions that are required for this reaction is that you need to produce h ZN which is produced in situ which means that any CN and concentrated sulfuric acid are reacted which results in the formation of h ZN NAC n is also the source of CN minus one ions which act as a catalyst any CN over here is also acting as a catalyst you must also know the reaction conditions and the and the reaction mechanism for this reaction which is covered in detail in other video lectures basically happens in the reaction mechanism is that the C and - benign from any CN which is acting as a catalyst attacks is attracted it's acting as a nucleophile attracted to the positive carbon atom this is a polar bond over here so the positive carbon atom so CN minus one gets a task to this positive carbon atom and the electrons over here in the double bond they get repelled by the negative charge over here so in the next step this intermediate is formed we've seen gets attached to the carbon atom the electrons over here they get repelled and oxygen gets a negative charge and in the last step the second step what happens is that the H+ ion in the solution from sulfuric acid is now attracted to the lone pairs on this oxygen and they end up forming an h group over here this in the end of ACN's are producing this cyanohydrin so this is your entire reaction mixture which should be learned by heart as well and make sure to focus on the positive and negative charges and the lone pairs and the direction of the arrows over here in in the mechanism in this reaction we're going to learn how to test carbonyl compounds Alys and ketones so - for dye nitro phenol hydrazine or 240 mph is used for testing aldehydes and ketones here I've drawn a 2-14 pH molecule it's a benzene that has two nitro compounds attached to it and hydrazine molecule is also attached on one side so remember what 214 PHS because you will be asked to even write equations who using 240 in peds I'm going to draw an ally I'm going to tell you what happens when a 214 pH molecule reacts with an aldehyde or a ketone so here I have drawn a propanone molecule what's going to happen is that the hydrazine part is going to lose its hydrogen and the ketone part is going to lose its oxygen and water is going to be produced so it's going to be a condensation reaction that's going to happen so this middle part is going to disappear and this carbon is then going to form a double bond with this nitrogen over here so I'm going to just do that I'm going to cut this out and drag it and join it over here and this is going to be the product that's going to be formed with two for udn pH and on with a byproduct a small water molecule is going to be produced as well and the results of this test this product is an orange precipitate so whenever you treat a carbonyl compound with 240 in pH you're going to cater to get an orange precipitate the carbon double bond o is going to go it's going to the water would be removed and it's going to form a double bond with nitrogen in to for the in pH so this would be your orange precipitate that's going to be formed the next test that we will study is called the Ida foam test instead of being a test for a functional group it's a test for a particular structure and those structures are these two structures it's going to be a methyl group that would be attached to a c del baño or a carbonyl group are over here could be a carbon chain or it could be hydrogen over here the other structure that it's going to identify is going to be a methyl group about it worth CH which has an O it's a task to it and this R could be a carbon chain or it could be hydrogen so it has to be exact this exact structure in this exact structure this entire exact structure let me highlight it so if these two exact structures are present either of them then the I deform test is going to be positive what happens in the I deform test is that nuh and iodine or alkaline aqueous iodine is added what happens to both of these structures is that these structures break this see part breaks in both structures and for both structures the CST bud breaks apart and ends up forming ch3 which is a yellow precipitate it's a yellow solid that's going to be produced and this happens in both cases both these structures whether this structure is present or this structure is been the under alkaline conditions iodine is added the CST part in both of the methyl group is going to break away in form try I to methane yellow precipitate what happens to the rest of the molecule what happens to the rest of the molecule is focus on this carbon atom over here this carbon atom changes into a carboxylic acid in both cases so this carbon atom over here changes into a carboxylic acid but since the conditions are alkaline remember that the carboxylic acid formed would eventually end up forming a salt under alpha in conditions so if you if you're using any weights as an athlete at any wage is going to react with the carboxylic acid with the with whatever carboxylic acid that's being formed and it would end up forming a salt instead of oh it's over here you're going to get o minus 1 in na plus 1 so I'm going to do just one example question let's say I have this molecule over here this is a molecule and does it contain the structures that we've discussed over here Leigha fire to form tests for the for the test to be positive it must have a ch3 next to a siedel bond over if you look carefully it has that it has a ch3 next to C double bond o what happens if you do the I reform test the ch3 is going to break away so this part is going to be converted into CH I 3 which is going to be your yellow precipitate and what would happen to this carbon atom over here on the other side this carbon atom is going to be converted into a carboxylic acid so I've drawn the structure this carbon atom over here is going to be converted into a carboxylic acid and since the conditions are alkaline a salt of the carboxylic acid would be formed instead so these would be your two products of the hydrophone test another important reaction is the hydrolysis of nitriles so a nitrile is any molecule that has a CN in it so a carbon chain attached to a CNC and has C triple bond in and if you heat it with a dagger I said ordered annually what happens to this nitrile is that this nitride over here breaks into two and the carbon changes into a carboxylic acid so this particular carbon over here changes into a carboxylic acid so instead of having a night tragic way to have a R and C and that C changes into a carboxylic acid the end breaks away in forms ends up forming ammonium ions on ammonia gas depending whether the conditions are acidic or alkaline always remember Nitra is heating up the end up changing into carboxylic acid and you must also remember be aware of the fact that if you use the conditions alkaline then instead you're going to get a salt of the carboxylic acid not the carboxylic acid so instead of carrying our Co each you're going to get our c o minus 1 na plus 1 the alpha li would end up reacting with the carboxylic acid so be careful with that that you'd acid you will get a carboxylic acid with anthony you're going to get a salt of the carboxylic acid the last thing that you must know is that carboxylic acids are weak acids which means that they partially ionized so here I've drawn an example of ethanoic acid it breaks down and produces at the no 8 ions and it's very real easy it's plus one not a lot of them it's going to push the ionize it's a weak acid now acids have very typical reactions the first reaction is that s acids react with metals to form salt plus hydrogen and that is exactly what carboxylic acids are also going to do so here in this equation I have a phenolic acid to balance the equation that I have ethanoic acid which is reacting with magnesium what big medium would do is that it's going to displace the hydrogen in this carboxylic acid group so it's going to do that and once that H is displaced empty spirit take its place co minus one has a negative one charge mg has a two plus charge so there would be two ethanoate ions and hydrogen gas is going to be produced so this the name of this salt is going to be magnesium ethanoate another typical reaction is that acids are going to react with bases metal oxides metal hydroxides to end they would end up producing salt and water molecules so here I have a throwing acid reacting with the basin in which DNA goes and displaces the eight in the carboxylic acid group ends up forming a salt and water molecule is produced along with this salt with the salt in this case is going to be sodium ethanoate let's move to the next reaction typical reaction of acids with metal carbonates so with carbonates acid are going to react with carbonates to produce salt water and carbon dioxide so let's write down a reaction so this here is a reaction of ethanoic acid again reacting with magnesium carbonate the mg goes in this place is the H same G is 2 plus 2 the H is displaced the carboxylate anion is has a minus 1 charge so there are two carboxylate ions ethanoate ions to be specific the salt that is produced is being medium ethanoate and water and carbon dioxide are also produced in the reaction let's move to the third we are the fourth reaction of assets that reaction is an acid reacting with ammonia to produce an ammonium salt so ammonia is a weak base it's a basic gas it could be an aqua state as well so I have a thick acid over here I have a Moni over here and ammonium salt would be produced ethanoic acid would lose its H and that each is going to be displaced by or replaced by an ammonium ion so mo neum into no 8 would be produced in this reaction so so simple reaction carboxylic acid group ammonia comes in and it's 4 plus 1 gets displaced over here the H gets substituted by an export plus 1 and that would be your salt