this video is for those of you who are taking organic chemistry 1 but you're studying for exam number two this video covers topics on stereochemistry sn2 sn1 E1 E2 reactions alkene reactions and alkyne reactions so if your exam covers those topics you're in the right place so let's go ahead and begin number one which of the following represents the name of the compound shown below so let's go ahead and name it this is going to be carbon one and this is going to be carbon 2. we have a six-membered ring so this is cyclohexane which we can see that in the parent name for all of the answer choices we have a bromine on one and a methyl on carbon two so it's going to be one bromo two methyl cyclohexane but what we really need to focus on in this problem is the configuration at each chiral Center we need to determine if it's r or s so let's start with this one focusing on that chiral Center keep in mind at Carl's Center is basically typically a carbon with four different groups so bromine is going to have the highest priority it has a higher atomic number than carbon next we have hydrogen which is typically number four and then comparing these two carbons which one is going to win the carbon at the bottom is a secondary carbon because it's attached to two other carbon atoms the carbon at the top is a tertiary carbon it's connected to three other carbon atoms a tertiary carbon will have a higher priority than a secondary carbon so the tertiary carbon is going to be group number two the secondary carbon will be group number three and of course H is number four so around the chiral Center we're going to go from one two to three and notice we're going in the counterclockwise Direction so that gives us the S configuration but hydrogen group number four it's not in the back it's in the front and we want it to be in the back when we try to figure out what the absolute configuration of the chiral Center is so we're going to have to reverse our answer so we're going to go in the other direction that is in the clockwise Direction which will give us r so we have the r configuration for that chiral Center and the carbon with the bromine atom that's carbon number one so it's going to be one r which means we could eliminate answer choices C and D now let's move on to the other chiral Center carbon number two now let's look at the four different groups connected to that chiral Center so we have a hydrogen and that's going to be group number four and we have three carbon atoms to compare so this carbon atom let's put in a different color this is a primary carbon this carbon atom here that's going to be secondary and this one is also secondary but this carbon has the bromine atom which has a higher atomic number than carbon so this entire group will be group number one comparing these two we have a primary carbon versus a secondary carbon the secondary carbon is going to have a higher priority so this will be group number two and this is number three so going from one to two to three notice that we're going in the clockwise Direction so we get the r configuration and H is in the back which means we don't need to reverse it so we're going to keep it the way it is so for carbon number two the second chiral Center we have the r configuration so the correct answer is going to be answer Choice a one r two r one bromo two methyl cyclohexane number six which of the following reagents will convert three methyl one butane to 3-methyl tubutanol so let's draw a picture so that's butane this is one butene this will be carbon one two three and four and we have a methyl on carbon three so that's three methyl one butane here's butane here we have a methyl and on carbon two we're going to put an oh group so that's three methyl tubuanol so which of these reagents here we'll convert this alkene to that alcohol let's go through each one starting with answer Choice a we have bh3 thf with hydrogen peroxide hydroxide and water this reaction is called hydroboration oxidation and what you need to know is that it proceeds with Mark harmonicle of regiochemistry the nucleophile the oh group it's going to go on the less substituted carbon of the double bond so that is the primary carbon therefore we're not going to get the answer we're looking for so it's not answer Choice a this is the product that answers a will give us now let's consider answer Choice B Mercury acetate with water followed by sodium borohydride this is the oxymercorbation demerculation reaction it proceeds with markovnikovagiochemistry without rearrangement so the oh group is going to go on the secondary carbon so this is the answer that we're looking for now you need to be familiar with the other reactions because there's a very high chance you'll see it on the exam so looking at answer Choice C what we have is water and sulfuric acid when these two combine they will react and form h3o Plus now atrial plus will react with the alkene giving us an alcohol but the oh group is going to go on the most substituted carbon that it can possibly get on so this reaction can proceed with rearrangement the end result of this reaction is that the oh group is going to go on the tertiary carbon we'll talk about the mechanism for that later now for answer Choice d once we use MC PBA you need to know that this is a peroxy acid the end result for this reaction is you're going to get an epoxide now in the next step when you add h3o Plus the ring is going to open and you're going to get a diol is basically two alcohol groups but with anti-addition now this first carbon is not Carlos so we don't have to worry about the sterochemistry there the second carbon is chiral and so you're going to get a mixture of enantiomers you need to get both the r and the s stereoisomer but let's say if you have something like this and you were to add h3o Plus the result will be starting from the alkene you will call that anti-addition you're going to have one oh group in the front one in the back so you need to be familiar with the sterochemistry associated with that reaction but depending on the carbon structure you may or may not need to show it but for a ring you'll need to show it so just know that this overall reaction proceeds with anti-addition it's going to add two alcohols two oh groups across the double bond where they're opposite to each other now for answer Choice e osune tetroxide followed by sodium bisulfite with water this too will give you a dial the only difference is it's going to be synodition as opposed to anti-addition so for this reaction you don't need to show it but if you had let's say cyclohexene and you had osum tetroxide with the same stuff rather than getting anti addition you'll get sin Edition and I'm running out of space here either case these two will not give us the product that we're looking for now let's focus on this reaction but with HBO Plus you need to be familiar with a mechanism for this reaction because it's a common one so the nucleophile the alkene is going to grab a proton that proton is going to go on a primary carbon of the double bond so we can get a more stable secondary carbocation intermediate now because this secondary carbon is adjacent to a tertiary carbon we're going to get something known as a hydride shift so we're going to get a carbocation rearrangement the driving force for that rearrangement is going to be increase alkene I mean increase carbocation stability so now we're going to have a tertiary carbocation and then water is going to react with it now we don't need to worry about the stereochemistry here because this carbon is not chiral then we're going to use another water molecule to remove this hydrogen so the end result is that we're going to get a tertiary alcohol 2 methyl two butanol number 13. what is the major substitution product of the reaction shown below so we have an alkyl halide reactant with methanol what mechanism will this reaction most likely occur by now what we have is a secondary alkyl halide the carbon that has the bromine atom is attached to two of the carbon atoms methanol is a product solvent so the fact that we have a secondary alkyl halide and a product a polar product solvent this is going to favor the S01 reaction over the sn2 reaction now the sn1 reaction typically occurs with the E1 reaction however we're not focused on a major elimination product we want to find the major substitution product so we're not going to worry about the E1 mechanism we're going to focus on the sn1 mechanism but you should know that these two they occur together and temperature will favor E1 over S1 if you increase the temperature so if you raise the temperature the yield of the S1 product will go down and the yield of the E1 product will go up so you you'll get more of the substitution product at low temperatures but more of the E1 product at higher temperatures now for us to find the major product for this reaction we need to write out the mechanism in the first step for an S1 reaction the leaving group is going to leave and so we're going to get a secondary carbocation which is adjacent to a tertiary carbon so a rearrangement will happen here but notice that we have an unstable for carbon ring when you see that look out for ring expansions a full carbon ring can easily expand to a five carbon ring depending on a carbon structure and sometimes you'll see a five carbon ring expand into a six carbon ring six carbon rings have minimal ring strain or angle strain so they tend to be pretty stable if you have a six carbon ring it's rare that you'll get another ring expansion or ring contraction so what's going to happen here one of these bonds are going to break but let's number the carbon atoms let's call this carbon one two three four five so we can break the bomb between carbon two and five or carbons two and three it doesn't matter but let's break the bond between carbons two and five so the electrons that connects carbons two and five those are going to be used to connect carbons one and five together so we're going to get a five carbon ring let's call this carbon one two three four and five now carbon one has a methyl group carbon two and five they lost the bond but carbons one and five they gained a bond carbon five got the bond back carbon one gain a bond carbon two lost the bond it didn't get any back so carbon two will have the positive charge so right now we still have a secondary carbocata intermediate so we're going to get another carbocation rearrangement this secondary carbocation is next to a tertiary carbon so we're going to get a hydride shift once the hydrogen moves we're going to get a plus charge on this carbon so now we have a tertiary carbocation with a five carbon ring so this carbocation is much more stable than this one so now that we have our the most stable carbocation we can get for this structure the solvent is going to behave as a nucleophile reactant with the carbocation when the solvent is the nucleophile the reaction is known as a solvolysis reaction this reaction is typically associated with an sn1 reaction but sometimes you'll see it with an sn2 reaction but for the most part if you see solvolysis 80 to 90 percent of time it's an S1 reaction but that's when the solvent behaves as a nucleophile in the substitution reaction so now we have an oxygen with a hydrogen and the methyl group now the next step is going to be a proton transfer step or an acid-base reaction so notice the difference between how methanol is behaving in this step and how it will behave in this step in this step methanol is acting as a nucleophile it's reactant with the electrophilic carbocation but here it's going to act as a base in that it's going to abstract a proton because that's what bases do so that's the difference between a base and a nucleophile bases they go for protons they abstract hydrogen atoms nucleophiles in the context of organic chemistry they typically react with electrophilic carbon atoms so the end result is that we get an ether so this is one methoxy one methyl cyclohex I mean cyclopentane so that's the major product for this reaction 21 which of the following statements is not true the rate of an essential reaction will double if the concentration of the nucleophile doubles is that true or false we know the rate law expression for an sn2 reaction is that the rate depends on the concentration of the substrate and the concentration of the nucleophile and this is raised to the first power so if the concentration of the nucleophile doubles we'll put a 2 for that that means that the concentration of the substrate stays the same so we'll put a one K is a constant so we don't need to worry about that one times two is two the rate is going to double so answer Choice a is a true statement we're looking for the false statement so answer Choice a is not our answer now let's look at answer Choice B the rate of an E1 reaction will triple if the concentration of the alkyl halide triples is that true or false for an E1 reaction the rate only depends on the concentration of the substring it doesn't depend on the concentration of the base now if we triple the concentration of the substrate or the alkyl halide the rate is going to Triple so B is a true statement we can eliminate that into choice moving on to C the rate of an E2 reaction will increase by a factor of six if the concentration of the alcohol and light doubles and the concentration of the base triples we know that for an E2 reaction the rate depends on the concentration of the substrate and the concentration of the base so if we double the concentration of the substrate or the alkyl halide and we triple the concentration of the base 2 times 3 is 6 the rate will increase by a factor of six so C is a true statement looking at answer Choice d the rate of an essential reaction will quadruple if the concentration of the nucleophile quadruples true or false the rate for an S1 reaction doesn't depend on the concentration of the substrate it only depends on the let me say that again the rate of an S1 reaction doesn't depend on the concentration of the nucleophile it only depends on the concentration of the substrate so this is the rate law expression for an S1 reaction the nucleophile is not in it as you can see as a result it doesn't matter what we do with the concentration of the nucleophile if we double it if we triple it it will have no impact on the rate of an S1 reaction so this is the false statement which means D is the answer we're looking for now let's talk about why let's say we have terbutyl bromide and we're reacting it with water so we have a tertiary alkyl halide protic solvent that's going to favor an sm1 E1 reaction but we'll focus on the S1 reaction the leaving group is going to leave that's the first step this step is the slow step the second step is fast that's when the solvent which acts as a nucleophile reacts with the carbocation given us this intermediate now the rate for the overall S1 reaction doesn't depend on the fast step it depends on the slow step and notice the nucleophile is not part of the slow step only the substrate is that's why the rate for an S1 reaction and also an E1 reaction doesn't depend on the concentration of the nucleophile or the base is because they're not involved in the first step which is the slow step or the rate determine step that's why it only depends on the concentration of the substrate so that's it for 21. the correct answer is to answer Choice d now let's move on to 34. which of the following reagents will convert one butane to two butenone so let's draw a picture so here we have one but I we can see we have a total of four carbon atoms the alkyne is on carbon one or between carbons one and two and we need to convert it into buno how can we do this how can we get the carbonyl group on this carbon here well let's go through each reaction so first we have hydrogen gas with a Palladium catalyst this will convert the alkyne all the way to an alkane of course you'll need two equivalents but it's not going to stop at the alkene level for the next reaction we have solid lithium metal in liquid methylamine now that reaction is very similar to this reaction where you have solid sodium metal and liquid ammonia this reaction will convert in alkyne into a trans alkene but because we're dealing with a terminal alkyne we're going to get a terminal alkene and CIS entrance doesn't really apply for that now methylamine and ammonia they're very similar the only difference is instead of having three hydrogen atoms on the nitrogen one of them has been substituted with a methyl group but the end result is going to be the same we're going to get an alkene now for the next one we have hydroboration oxidation but with an alkyne we have R2 BH which works very similar to bh3 and in the second step we have oxidation hydrogen peroxide with sodium hydroxide now what's going to happen is this reaction is going to proceed with anti marcarbonyl car regiochemistry initially it's going to convert the alkyne to an alkene but this is also a hydration reaction we're going to get an oh group on the less substituted carbon atom of the triple bond so this carbon is primary this one is secondary the oh group is going to be on the primary carbon so this is called an enol we have an alcohol and an alkene now the enol it's not very stable relative to what it can convert to there's an equilibrium in this reaction here but it's going to quickly convert to an aldehyde so this is known as tautomerization where the enol can convert into an aldehyde or ketone now for answer Choice d we have mercury sulfate with sulfuric acid and water so think of this reaction AS h3o Plus where it's acid catalyzed hydration but Mercury acts as the Catalyst for this reaction this is kind of similar to the oxymerculation demerculation reaction but somewhat different this reaction proceeds with more common car vitrochemistry it's very similar to this one we're going to get an enol but the difference is the oh group is going to go on the more substituted carbon in this case the secondary carbon and that enol is going to tautomerize into a ketone and this is the product that we want to butanone so the correct answer for this problem is answer Choice d 43 what is the major product of the reaction shown below so we have a molecule that has two functional groups an alkene and an alcohol whenever you see that when there's multiple functional groups in a molecule pay attention for intermolecular reactions now we know that the alkene can react with bromine we've seen how those two react there can also be side products as well this oxygen could react with bromine but since we haven't seen that reaction in this type of content in organic chemistry we're not going to touch that but we do know that bromine will react with the alkene the double bond is going to attack bromine and bromine is going to attack the double bond simultaneously kicking out this bromide ion and so we're going to get a cyclic romanium ion now here we have a secondary carbon and here we have a tertiary carbon the tertiary carbon is going to have more partial positive charge than the secondary carbon and so this oxygen which has a partial negative charge is going to be attracted to The partially positive tertiary carbon so it's going to attack that carbon this bond is going to break those electrons will go to the bromine atom and so this is going to close and form a ring we'll call this atom one two three four five six so we're going to have a six-membered ring with an oxygen atom being part of it now on carbon 6 we have an ethyl group here and carbon 6 also has a methyl group so here is the methyl group and this is the ethyl group now on the Etho group this spawn is broken so carbon 6 is not directly attached to a bromine but this carbon the first carbon in the ethyl group still has a bond to a bromine atom so that's what we now have but we still have this hydrogen atom on the oxygen that we need to deal with now we could use another molecule like this with an oh group to remove the hydrogen we could use bromide or something else I'm just going to put a generic base so a generic base is going to remove this hydrogen and put a lone pair on that oxygen there's already one but it's going to have two so this is going to be the final product for this reaction so that's what we're going to get in this reaction by the way if you like this video and if you think it's helpful feel free to share it with your fellow classmates if you know of any other students who might be struggling with this topic feel free to text it or email it to them whichever is convenient and feel free to share it on Facebook Twitter Pinterest Instagram or any other social media platform that you might enjoy using but now let's get back to the video 51. what is the major product of the following reaction so here we have two bromo four floral pentane and we're going to react with potassium cyanide and the solver of choice is hexamethrow phosphoramide hmpa when you think of an amide an amod is basically a carbonyl group attached to a nitrogen a phosphoramide is very similar but instead of a carbon we have a phosphorus group so that's a phosphoramide now think of the word hexamethyl we got six methyls we can put two methyls on this nitrogen and if we add two more nitrogen atoms we can have a total of six methyls so this is hexamethyl phosphoramide we got six methyls and a phosphoring Mite group so this is a polar a protic solvent that's what you need to know when you see hmpa so we have a secondary alkyl halide with two halogens we have a good nucleophile Cyanide and we have a polar a protic solver these conditions favor an sn2 reaction now notice that we have two leaving groups fluoride and bromide which leaving group is better with the halogens we know that leaving group stability increases as you go down the group so iodide is the best leaving group out of the halogens but comparing bromide and fluoride bromide is a Better Living group fluoride is the poor leader group so cyanide is going to preferentially attack the carbon with the better leaving group so we're going to kick out the bromine and we're going to get this now because we're dealing with an sn2 reaction we're going to have inversion of stereochemistry Flo well first I got to get rid of the bromine let's put cyanide instead the fluorine is going to remain it's not easy to kick out a fluorine group because it's a poor leaving group but this is going to be the major product so we just got to replace bromine with cyanide and with inversion of configuration so that's the answer for 51. number 60 which of the following molecules is chiral all right let's go through each one so looking at answer Choice a notice that we have a line of symmetry we can see the bromine atoms are the same across that line and on the same side also there are no chiral centers here this carbon is not chiral because the left side is the same as the right side so for answer Choice a this is an a chiral molecule a car molecules are optically inactive now looking at answer Choice B we do have a plane of symmetry and we can see it right here the left side is the same as the right side so this is going to be an a chiral molecule for answer Choice C we have an allele not for allenes what you want to do is look on each side on the right side we have two different groups on the left side we have two of the same groups if one of the sides have two identical groups it's going to be a chiral in order for it to be chiral the right side has to have two different groups and the left side has to have two different groups in order for it to be chiral now for answer Choice d notice that we don't have a plane of symmetry the left side is not the same as the right side so this is going to be a this molecule is Cairo which means it's optically active so D is the correct answer for this problem 78 what is the major product of the reaction so we have one methyl cyclohexene with high duration oxidation now we know this reaction proceeds with anti-micomic regiochemistry in the first step the alkene is going to react with bd3 the double bond attacks boron and deuterium attacks a double bond as well so initially bd2 adds to the less substituted carbon deuterium which has taken the place of hydrogen is added to the more substituted carbon but these two they're added on the same side with syn addition now when you add step two hydrogen peroxide and hydroxide this gets replaced with an oh group from hydrogen peroxide this hydroxide simply removes the hydrogen from hydrogen peroxide but the oh group that this will be replaced by comes from the hydrogen peroxide it's important to know that in the event that you have d2o2 because then this group will be OD instead of oh so the o h group and deuterium they're going to be on the same side it's in addition which means methyl has to be on the other side so the first thing we want to look for is we want to make sure that the oh group is on the less substituted carbon we could eliminate answer Choice B actual note we can eliminate s Choice C the o h group is on the tertiary carbon and we eliminate acid Choice d now we're between A and B the oh group is on the less substituted secondary carbon now looking at deuterium and the oh group they have to be on the same side because this is sin addition here we see them on the same side here they're on opposite sides anti so we can eliminate answer Choice B therefore s Choice a is the correct answer we need to make sure that deuterium and hydroxide are on the same side and hydroxide is going to be or the the hydroxyl group is going to be anti with respect to the methyl group but sin with respect to deuterium 81 what is the major product of the reaction shown below so notice that this reaction is Stereo specific hydroxide has to grab the hydrogen that's one carbon away from the bromine now notice that we only have one hydrogen if there were two hydrogen atoms we would get a mixture of products we can get more than one product but because there's only one hydrogen that the hydroxide base can abstract we're going to get one particular stir isomer so if there were like two hydrogen atoms here we can get both the CIS and the trans isomer the e and z isomers but because we have one we're going to get one of the e or the Z isomers not both so be mindful of that so the base is going to grab the proton we're going to form a pi Bond and kick out the leaving group the good thing about this problem is that the hydrogen and the bromine atom they were already anti with respect to each other so we don't need to change anything we don't need to rotate the molecule so we're going to get a double bond notice that the methyl groups they're on the same side they're on different carbons but on the same side and after the E2 reaction they will continue to remain on the same side so we have a hydrogen on this side left over and The Ether group on the other carbon so this is going to be the major product so comparing ethyl and methyl Esso has a higher priority comparing methyl and hydrogen hydrogen has a higher priority so the two high priority groups they're on opposite sides of each other so therefore we have the e i smart so that's the major product for this particular reaction we get the e-i smart as opposed to the Z isomer