in this video we're going to go over some common organic chemistry reactions that you'll need if you're studying for your final exam so let's begin let's say if we have one butane now let's react it with hydrobromic acid so what's going to be the major product in this reaction the alkene is a nucleophile and it's going to react with the electrophile which is the acid so it's going to acquire the hydrogen and expel the bromide ion now the hydrogen is going to go on the primary carbon of the double bond and the reason why that happens is so that we can put a positive charge on a secondary carbon you don't want to put the positive charge on a primary carbon secondary carbons or rather secondary carbocations are more stable than primary carbocations and that's why it works this way if we were to put the hydrogen on the secondary carbon we would get an unstable primary carbocation and so that's not going to happen in the last step the bromide ion is going to attack the carbocation it could attack it from the front or from the back giving us a racemic mixture of products so we can draw the two products like this we could put the bromine group in the front that is on the the wedge or on the dash so on a dash it's going into the page on the wedge is coming out of the page so we get a racemic mixture of two products or two stereoisomers now let's say if we have this alkene what's going to be the major product if we react it with hydrobromic acid and let's compare that with hydrobromic acid and peroxides so this is an organic peroxide or if you see h2o2 it will have the same effect now hbr without peroxides occurs with markovnikov regiochemistry as such the bromine atom will go on the more substituted carbon atom of the double bond that is the tertiary carbon but when you use an hbr with peroxides the bromine atom will go on the least substituted carbon atom of the double bond in this case the primary carbon and so you can get those two different products if you use peroxides or not now for the next example we're going to start with cyclohexene what's going to happen if we react it with bromine using dichloromethane dichloromethane is an inert solvent so it really doesn't do anything but when you react an alkene with br2 the reaction proceeds with anti-addition one bromine atom will be on the wedge the other will be on the dash now you get this product plus you also get the enactment so you get a mixture of two products and so those are the products for that reaction now let's start with cyclohexene again what's going to happen if we react to this alkene with nbs we need to know is that this is a radical reaction and an allylic hydrogen that is a hydrogen that's one carbon away from the double bond will be replaced with a bromine atom so notice that the double bond is unaffected it will remain in the same position now granted you could get some resonance structures as well but for symmetrical alkene like this you're only going to get one product but the end product is going to have the bromine atom that's allylic or one carbon away from the double bond now what if we have an alkane what if we reacted with bromine but in the presence of ultraviolet light what's going to happen this is going to be another radical reaction it's a free radical substitution reaction so you're going to replace or substitute a hydrogen with the bromine atom now bromine is selective and so it's going to replace the most substituted hydrogen that is the tertiary hydrogen and the reason for that is because it's going to form a tertiary radical tertiary radicals are more stable than secondary or primary radicals so for radical reactions dealing with bromine bromine will selectively abstract the tertiary hydrogen if you use chlorine which is reactive and non-selective chlorine will give you a mixture of products it can abstract the primary secondary or tertiary hydrogen leading to a bunch of different products but bromine is selective and so the major product will be this one by the way if you want a more detailed review on the mechanisms of these reactions feel free to check the description section of this video and i'm going to post my playlist my organic chemistry playlist where you can find these other videos where i go into the mechanism and explain why these reactions occur the way they do so i'll have some mechanisms in this video but a lot of reactions you'll see i'm just going to summarize the end result but you can check out my playlist if you want a more detailed review on those reactions so the next example is going to be the acid catalyzed hydration of an alkene so let's say if we were to react this with water and hydrochloric acid which will eventually form the hydronium ion what's going to happen what's going to be the major product in this reaction now for this one i'm going to go over the mechanism when you have an alkene with an acid the alkene will abstract a proton the proton is the electrophile the alkene is the nucleophile and just like in the reaction with hbr and alkene the hydrogen will go on the least substituted carbon of the double bond in this case the primary carbon and this is going to put a positive charge on the secondary carbon now notice that we have a tertiary carbon next to the alkene when you see that a carbocation rearrangement will occur in this case a hydride shift and the reason why this happens is to produce a more stable carbocation so now that we have a more stable tertiary carbocation water is going to combine with it and so we're going to get an oxonium species now another water molecule is going to be used as a weak base to remove the hydrogen atom on the oxygen and so the end result is that we get an alcohol so whenever you react to water with an alkene in the presence of an acid it will produce an alcohol and typically the oh group will be found on usually the most substituted carbon in this case the tertiary carbon now let's go back to cyclohexene so starting with cyclohexene what's going to happen if we react it with hydrogen gas using a metal catalyst the hydrogen atoms will add across the double bond turning it into an alkane particularly a cycloalkane and so we don't have to show the hydrogen atoms in this case so this is going to be the answer now let's use an isotope of hydrogen called deuterium now let's use a palladium catalyst instead of a platinum catalyst now this time we do need to show the deuterium atoms and so the double bond will be reduced to a single bond and we're going to add the two determine atoms cis with respect to each other so this reaction occurs with sin stereochemistry or sin addition now you can also draw your product like this now it's important to understand that these two compounds are meso compounds notice that we have an internal plane of symmetry they're not enantiomers so therefore these are identical compounds which means we only get one product in this reaction these two compounds are the same now let's go over the hydroboration oxidation reaction of alkanes so we're going to react this with bh3 thf with hydrogen peroxide hydroxide and water so what's going to be the major product of this reaction the first step is called hydration and the second step is oxidation but combined the net effect is to convert an alkene to an alcohol without any carbocation rearrangements so this carbon is secondary and this carbon is tertiary you need to know that this reaction proceeds with anti-more carbon carbohydrate chemistry so what that means is that we're going to put the oh group on the least substituted carbon atom as opposed to the most substituted carbon atom so the major product will be an alcohol now this reaction occurs with syn stereochemistry and so on the tertiary carbon a hydrogen will be added and on the secondary carbon we're going to put the o h group so it's going to look like this which means the methyl group has to be in the back if the hydrogen is in the front now on a typical exam you may not see the hydrogen so all you'll see is this product or it's enachment or you might see both so these are the two products that we can get in this reaction but keep in mind it's syn addition when you compare the hydrogen and the oh group because those are the two groups that you're adding across the double bond but your answer may look like this on a multiple choice final exam by the way for those of you who might be studying for the final exam in organic chemistry i posted a video on my patreon page which you can access in the description section of this video and it's basically a multiple choice practice test with video solutions so if you need help with the organic chemistry final exam feel free to take a look at that but now let's move on so let's consider this alkene one methyl cyclohexene we're going to react it with mercury acetate in water followed by sodium borohydride so this reaction is called oxymercuration demercuration and it's different from the previous reaction this reaction proceeds with more carbon carbon vitrochemistry now it's similar to the last reaction in that it will convert an alkene into an alcohol but the oh group it's going to go on the more substituted carbon atom of the double bond so we're going to get a tertiary alcohol in this example now let's go back to cyclohexene so what's going to happen if we react this alkene with something called mcpba mcpba stands for metachloroperoxybenzoic acid now if you see this group rco3h the effect will be the same what's going to happen is that the alkene will be converted into something called an epoxide now starting with the epoxide what's going to happen if we react with h2o plus if you see this it's equivalent to water and hdl now the epoxide is going to open and we're going to get two diols anti-addition so we're going to have two hydroxyl groups one is going to be on the wedge and the other will be on the dash and so we're going to get a mixture of enantiomers or racemic mixture of products so basically this reaction these two steps and this one too they can convert an alkene into a trans-diol now starting with the same alkane how can we convert it into let's say a cis style one reaction in which we can do so is osmium tetroxide with sodium bisulfite and so the end result of this reaction will be a cystile so this reaction proceeds with syn stereochemistry now keep in mind this is a miso product and so there's only one product that we can get if we draw the other way it will be the same as this one now let's start with an alkyne two butane what's going to happen if we react with hydrogen gas and the platinum catalyst and let's compare that with sodium metal and liquid ammonia now if you see lithium metal it will have the same effect as sodium metal because they're both alkali metals and also hydrogen gas with something called the lindler's catalyst so using the same reactant what products will we get using these three different reagents so if we react it with hydrogen gas and the platinum catalyst the alkyne will be reduced all the way to an alkane the triple bond will be converted into a single bond and we need to add two equivalents of hydrogen gas to do so now if we react with sodium metal and liquid ammonia it's going to stop at the trans alkene level now if we use hydrogen gas and the lindler's catalyst it's going to stop at the cis alkene level and so make sure that you understand what each of these reagents will do to an alkyne so now let's continue so let's say if we have an alkyne but instead of an internal alkyne we have a terminal alkyne now let's react it with r2bh and thf followed by hydrogen peroxide and hydroxide so this is the hydroboration reaction but for alkynes initially the triple bond will be reduced to a double bond and we're still going to get anti more carbon to carve geochemistry the oh group will go on the primary carbon as opposed to the secondary carbon this is called an enol now the enol is going to tautomerize into the aldehyde form and so this is going to be the major product of this reaction so the hydroboration reaction of a terminal alkyne will give you an aldehyde now let's start with the same reagent i'm in the same reactant propine but this time let's react with mercury sulfate in water and sulfuric acid so this reaction will proceed with more carbon the carbohydrochemistry and so the o-h group of the enol will be on the secondary carbon as opposed to the primary carbon and then this is going to tautomerize into a ketone and so this is going to be the major product for this reaction now let's say if we have acetylene which looks like this it's basically a two carbon molecule with a triple bond in the middle what's going to be the major product if we react with let's say sodium amide followed by methyl bromide so go ahead and try this problem so what do you think the major product will be in this reaction so the first thing that's going to happen is the amide ion is going to act as a base and so it's going to it's strong enough to remove the hydrogen of the alkyne and so this is going to give us the acetylide ion and then in the second step it's going to react with methyl bromide so this is going to be an sn2 reaction this carbon with the negative charge will attack the methyl group expel in the leaving group and so this reaction is a great way to create carbon carbon bonds this is going to be the product for the reaction now let's work on another example so starting with acetylene and the first step we're going to react with nanh2 and then in the second step we're going to use this time ethel bromide instead of methyl bromide and then in the third step we're going to use na nh2 again and then in the fourth step we'll use propyl bromide and then finally in the fifth step we're going to react it with h2 and the lidler's catalyst which you might see it as a mixture of palladium and barium sulfate with quinoline so go ahead and try this reaction so we know what's going to happen in the first step the nh2 minus ion is going to act as a base and deprotonate the acetylene molecule giving us the acetylide ion and so this is going to react with ethyl bromide and so this carbon will attack from the back kicking out the leaving groups so we're going to have a total of four carbons so right now what we have is this if you want to draw the line structure but i'm going to write this out so drawing it backwards it's ch3 ch2 attached to this triple bond now we have another hydrogen atom that we can remove and so let's use this to take off the hydrogen so now this is going to react with the alkyl halide in step four propyl bromide and so now we should have a total of seven carbon atoms now the last step is to react this with hydrogen gas and the liners catalyst so we're going to get an alkene but the alkene is going to be cis instead of trans so the two hydrogen atoms that will be added across the triple bond will be cis with respect to each other so the propyl group will be here and here is the ethyl group and so this is going to be the final product of this reaction now let's say if we have 2-bromobutane what's going to happen if we react it with potassium iodide and acetone so here we have a secondary alkyl halide a polar aprotic solvent and iodide is a good nucleophile when you see these conditions these conditions favor the sn2 reaction so the iodide will attack from the back kicking out the leaving group because of that backside attack we're going to have inversion of configuration and so the iodide ion will be on the wedge as opposed to the dash i mean it's on the the dash as opposed to the wedge and so we get the inverted product for an sn2 reaction so keep in mind the sn2 reaction works well in polar aprotic solvents now let's use the same secondary alkyl halide but this time we're going to react with water now water is a protic solvent and even though we have a secondary alkyl halide protic solvents favor the sn1 and the e1 reaction so we can get two products here the s1 product and the e1 product but for now let's focus on the sm1 product so for the sm1 reaction the leaving group is going to leave and we're going to get a secondary carbocation and then water could attack from the front which will give us the retention product or it could attack from the back giving us the inverted product now these two products are not equal because we still have the leaving group in this vicinity and so we're going to get a little bit more of the inverted product less of the retention because as water approaches the carbocation the bromide ion will slightly repel the oxygen part of water as it approaches the carbonyl cation so we're going to get a mixture of products but for now let's continue with the mechanism so once water adds to the carbocation we're going to get something that looks like this and then we need to use another water molecule to take off the hydrogen so the end result is that we're going to get an alcohol but an unequal racemic mixture so these are the sm1 products of this reaction now let's try another problem actually let's finish the last problem we need to do the e1 reaction so to do the e1 reaction the leaving group have to leave and then instead of water coming in and acting as a nucleophile which will give us the sm1 reaction first order nucleophilic substitution reaction water is going to act as a base and so it's going to go for the adjacent hydrogen forming an alkene and so the e1 reaction the elimination reaction produces alkanes we can get the trans isomer we can get the cis isomer or water can also go for this hydrogen giving us the terminal alkene but this right here is going to be the major product this is the zeta product it's the most stable alkene the hofmann product which is this one that's the least stable alkene so this right here is going to be the major product trans 2 butane but we do get a mixture of all three so just keep that in mind now what if we were to use let's say a strong base like let's use methoxide in water or rather in methanol methoxide is a strong base and as a result this is going to favor the e2 reaction which doesn't produce any carbocation intermediates and so no rearrangements are possible the strong base is going to go for the hydrogen form a double bond kick out the leaving group all at once and so the end result is that we're going to get the most stable alkene the zeta product now if you have a bulky base like terpetoxide which looks like this and if the substrate is also bulky for example let's say if we have something that looks like that then the sterically hindered base will prefer to go for the most accessible hydrogen it's difficult for it to grab this hydrogen because this is a tertiary carbon and so that hydrogen is sterically hindered by the methyl groups so it's going to go for the most accessible hydrogen in this case given us the hofmann product so sterically hindered bases and substrates tend to lead to the hoffman product whereas if you have a strong base that is not sterically hindered or a substrate that is also not sterically hindered then you're going to get the zaitsev product consider this reaction let's say if we have an alcohol and we decide to react it with sulfuric acid and we decide to heat the solution what's going to happen this alcohol is a secondary alcohol and whenever you have secondary and tertiary alcohols if you react it with an acid and you heat it dehydration will be favored and so you're going to get the e1 reaction and the alcohol will be converted into an alkene so the first step is protonation by the acid so you can represent the acid using h-plus just to keep things simple so after protonation we're going to get a good leaving group the hydroxyl group is a poor leaving group but water is the better leaving group and so water is going to leave and so we're going to have a secondary carbocation that is adjacent to a coronary carbon and so a rearrangement will occur specifically a methyl shift and so the whole entire carbon structure will change this occurs because the driving force is stability now we have a more stable tertiary carbocation intermediate now we can use a base at this point the base could be water or it could be the bisulfate ion h so4 minus now the base is going to go for the hydrogen that will lead us to the most stable alkene and so it's going to go for this hydrogen generating this alkane by the way before we finish this notice that the plus charge is here we could form a double bond here here or here however the best place to form a double bond is in the middle so the base is going to go for this hydrogen forming the double bond in the middle this is a tetra substituted alkene and so it's the most stable alkane notice that the number of carbon atoms attached to those two double bonded carbon atoms highlighted in red is four so this alkene has four r groups it's a tetra substitute alkene here are the other alkenes that we can make we could put a double bond here or we can put one in this region so this is a disubstituted alkene because there's two carbon atoms attached to any one of these two double bonded carbon atoms a tetra substitute alkene with four r groups is more stable than a disubstituted alkene with two r groups so therefore this is the major product it's the most stable alkene that we can form now let's say if we have 2-butanol what's going to happen if we react with hbr and then if we react it with let's say pbr3 and finally socl2 when you mix hbr or hi with a secondary or tertiary alcohol it's going to proceed by the sm1 reaction mechanism if it's a primary alcohol then it's going to favor the sn2 mechanism so because this reaction is sn1 we're going to get a racemic mixture of products so the bromine atom could be on a dash or it could be on the wedge now with pbr3 it's going to be an sn2 reaction and so we're going to get the inverted product so it's going to be on the dash it's going to be opposite to what we see here for the last one it's going to be an sn2 reaction as well so we're also going to get the inverted product therefore the chlorine atom will be on the dash so make sure you know these reactions they're very common so let's go over the mechanism for the first reaction where we have the alcohol and we're going to react with hydrobromic acid so we know it's an s1 mechanism the first thing is protonation we need to turn the poor leaving group into a good leaving group and so we're going to have oh2 and then water is going to leave now that we have a better leaving group giving us a secondary carbocation and then at this point the bromide ion can either attack from the front or the back giving us a racemic mixture of products and so that's how we can convert an alcohol into an alkyl halide now let's say if we have one butanol and let's react it with hydroiodic acid show the mechanism that will lead to the major product for this reaction since we have a primary alcohol this is going to be an sn2 mechanism and like the other one the first step is to convert the oh group into a better leaving group now we're not going to get a carbocation because if we do form a primary carbocation is not good primary carbocations are very unstable so that's not going to happen this is going to be an sn2 reaction the iodide ion will come from the back attack the carbon and expel the water molecule so that we don't form a primary unstable carbocation intermediate so this is going to be the end result of this reaction and so whenever you're dealing with an sn2 reaction you should not form any carbocation intermediates now let's briefly review oxidation reactions what's going to happen if we mix a primary alcohol with pcc pcc will oxidize a primary alcohol to an aldehyde it's a mild oxidizing agent but let's say if you were to use a strong oxidizing agent like chromic acid with a primary alcohol it will oxidize it to a carboxylic acid now if you have a secondary alcohol it really doesn't matter if you use a mild oxidizing agent or a strong oxidizing agent the net effect is that it's going to go to a ketone if you use a secondary alcohol now if you're trying to oxidize a tertiary alcohol let's say with pcc or a chromium reagent tertiary alcohols are resistant to oxidation and so you're not going to get any reaction in this example now let's talk about reducing agents let's say if you have an aldehyde or ketone and you react it with sodium borohydride what do you think is going to happen sodium borohydride can reduce aldehydes and ketones and even acid chlorides but it can't reduce esters and carboxylic acids so in the first example where we have a ketone with nabh4 it's going to reduce it to an alcohol and the aldehyde will also be reduced to an alcohol and the way it does this is by adding a hydrogen atom to it particularly a hydride ion now the acid chloride will also be reduced to an alcohol but by the addition of two hydride ions instead and so sodium borohydride produces alcohols from ketones aldehydes and acid chlorides now another stronger reducing agent lithium aluminum hydride can do all that nabh4 can but also it could reduce esters and carboxylic acids so to predict the product for this reaction you need to realize that this bond is going to break and so this group is going to leave it's going to collect a hydrogen from the solution turning into methanol and the carbonyl group will be reduced to an oh group and so this will be a product by the addition of two hydride ions now same is true for the carboxylic acid this is going to grab a hydrogen so that's going to leave us water and this is going to be reduced to an alcohol by adding two hydride ions and so those are two main reducing agents that you need to know now let's talk about the mechanism for this process so let's use cyclopentanone as an example and i'm going to reduce it with sodium borohydride the boron atom has a negative formal charge and sodium has a positive formal charge the hydride ion we'll attack this carbon causing the carbonyl bond to break and so we're going to get an alkoxide ion and here's the hydrogen that we added and then the alkoxide ion will react with water it could be h2 or plus two but i'm going to use water and so that's where it picks up the second hydrogen atom turning it into an alcohol but this is the hydride ion that we added to it and so that's a simple mechanism for the reduction of a ketone into an alcohol using sodium borohydride now another reagent need to be familiar with is the greener reagent so let's react to this ketone with methyl magnesium bromide followed by h2o plus now the greener reagent can reduce a ketone into an alcohol but by the addition of a methyl group as opposed to a hydride group so carbon is more electronegative than a metal so therefore carbon will bear the partial negative charge so this is a nucleophilic carbon and it's going to attack the electrophilic carbon of the carbonyl group this carbon has a partial positive charge because oxygen is more electronegative than it and so we're going to add this methyl group to this carbon and we're going to get an alkoxide ion and then we're going to acidify the solution so this is going to pick up a hydrogen turning into a tertiary alcohol and so the grain of reagent can reduce aldehydes ketones and even esters and acid chlorides into alcohols so this is the final product so that's it for this video thanks for watching you