this will be an exam three roughly what the topics might be covered on an organic chemistry one exam I have five problems it's gonna span everything from as you can see right here a proton NMR question it's kind of hard to nail down exactly when or if a teacher might include proton NMR in organic chemistry one so we'll run through this as a first question then we'll do a naming question with stereochemistry Enz nomenclature with some functional groups and then we will dive head first into completing the reaction completing the reagent questions involving things like epoxides alcohols the grignard reaction ethers carbocation rearrangements then we have a very pointed carbocation rearrangement mechanism question and then we will round things out with a synthesis question let's get started if if you're saying something in the chat probably won't be able to multitask but away we go okay in question one we are going to deduce the structure from the following information this molecular formula and this proton NMR Spectrum right so this is everything we've gotten this is enough information to actually go about and deduce what the structure this you know this amount of information is describing so I was personally and I'm still not amazing but I definitely struggled with this when I first learned it I always felt like there was too much information and not enough process I couldn't find some strategy to go about and try to consume everything I was given and then actually put it into use and kind of you know have a planning of attack so hopefully if you're in that same situation what we kind of do here and what I show in the videos on Joe came around proton NMR will help you out the first thing I always always do if you're given the molecular formula is I like to calculate the degrees of unsaturation basically you there's a formula but in this particular case I won't use the formula because we only have carbons hydrogens and oxygens what we know for sure is that a regular alkane abides by CN h2n plus two okay now what that means for us is we can compare that to our molecular formula and then if we come up short of you know if we come up short in terms of the amount of hydrogens in our molecule what that means is we have double bonds or a ring and if you have one or two of the like if you have either a double bond or a ring that basically subtracts two hydrogens off of your account because you don't have a fully saturated hydrocarbon so we can ignore the oxygen and again if you want no more details on degrees of unsaturation go ahead and check out that video on Joe chem I have the NMR series Linked In the video description here what we can do is say okay you know if we did C 10 times 2 and then h uh two times um a 2N plus two so again the N is the number of carbons what we have here is oh my goodness what am I doing C10 sorry sorry write the amount of hydrogens we have um is that we should expect 22 for a fully saturated you know 10 carbon chain we're actually for short of being fully saturated which means we have four unexplained missing hydrogens now every degree of unsaturation subtracts two hydrogens so basically we have four divided by two because we are four hydrogen short and every degree of unsaturation gets rid of two hydrogens we have two degrees of unsaturation here that we need to work with which is to say that means we have the possibility of two double bonds two rings or one ring one double bond so let's just keep that tucked away okay now we'll shift away from the degrees of unsaturation and actually start looking at this spectrum here and kind of digesting what's going on there now here's what I like to do personally I like to First label all of the different unique signals that I have so what I'll go ahead and do is just go right down the line I'll call this signal a signal B and Signal C so a is the signal at around two ish that has an integration number of four which means four hydrogens because this is a proton NMR Spectrum four hydrogens correspond to this peak two hydrogens correspond to Peak B and 12 hydrogens correspond to Peak C now here's what I like to do and and it's no surprise we have 18 hydrogens in the molecule form the molecular formula and that adds up to 18 over here as well all should be accounted for in terms of the full sum of the integration numbers and what we have over here now here's what I like to do before kind of freaking myself out I like to build out a table and it's during this table building out process that we kind of do some heavy lifting heavy lifting thinking so what I like to do is I kind of I make a chart over here so I'll have rows for all your signals a B and C and I'll draw some lines right here and I'll have some column labels up top so the The Columns you'll have are chemical shift so that'll be the value we you know take from the chart chemical shift some of these might be redundant but I kind of like concluding them all because it makes make sure you're kind of on top of what's in there um your coupling which is like singlet doublet triplet Etc then your integration number than just n and and again there's no particular reason to order this exactly the way I'm doing um but and then I like to have a column called a you know a possible appearance so here's what we'll do chemical shift so for a this is coming around just around two so maybe you could put in like 2.1 here on an exam maybe your professor will have exact numbers but I'm just gonna 2.1 around 2. B and C are around one so from this maybe we could say B is like 1.1 and C is like 0.9 again doesn't have to be exact right you're going to be comparing this against a chemical shift table the neighborhood the ballpark is what's important and then you know um what else you have going on the molecule as well now in terms of coupling all three of these are singlets right and you can tell that they're singlets because this is what a doublet would look like you have the two peaks a triplet would look like this you know um a quartet looks like that we only have one Peak on all of these signals so these are all singlets which is again some interesting information here because as you can already see we only have three signals and they're all singlets so when you have a lot of singlets or when you have very few Peaks sorry and a lot of hydrogens that means you have a lot of symmetry going on your molecule because that means we leave three unique hydrogen positions in this molecule with quite a bit of carbons and hydrogens integration number right a has four B has two C is 12. so C obviously definitely corresponds to many positions now n n kind of goes in hands with coupling because you get your coupling from n plus one your neighbor you know your neighboring hydrogens plus one the built-in one um so uh wow sometimes I include this sometimes I don't really what this means is we have zero neighbors all around okay zero neighboring hydrogens that is the reason why that's going to be important because now we're going to kind of make a somewhat of a guess in terms of what we look like so let's kind of take this row by row actually in fact let's start with c and whenever you see something around one and it's a singlet and especially if it's a multiple of three what that means is you have a methyl group most likely that doesn't have any neighboring hydrogens you have like a methyl group and then that so its next door neighbor is a quaternary carbon now the reason why that makes sense and you see this pattern quite a bit is because what this really means is that you know we're around one which means our chemical shift is very upfilled it's not near anything very electronegative when you're next door to electronegative atoms you know like like uh when you're next door to a carbonyl or attached to an oxygen you're very downfield okay so we're very upfilled we're we've no one's splitting us next door we're singlets there's 12 for integration number so the fact that we're so far up field the fact that we have a multiple of three that is screaming to us that there's probably four unique methyl groups and their next door neighbor is some type of carbon with four bonds meaning it has no hydrogens so I bet you a nickel that we look something like this and that there's four of these because that would give us three times four twelve hydrogens now for B Again The Chemical shift here is very closely similar and we only have two you know two hydrogens here for our integration number and again we're not splitting with anyone next door so with hopefully what you you know again this is all like trial and error what I'm thinking is that this is a carbon that has two neighboring carbons I'm just using like R Prime and r and this carbon so it's a secondary carbon with two hydrogens and the fact that no one's splitting it next door means these are also quaternary carbons so I'm just going to put like a four and a degree symbol meaning that this carbon is wedged in between two carbons that don't that they themselves do not have hydrogens now last but not least this is the most interesting thing we have a chemical shift about 2.1 and you will definitely have a chemical shift table when you do these types of problems this shift is very indicative of being next door or Alpha to a carbonyl something like this so if we had something like R Prime this is the position we'd be interested in and this would be R it's this carbon and these hydrogens being next door to this C double bonded o that yields this chemical shift value a little bit more down field at 2.1 now here's interesting not only does that kind of fit our chemical shift narrative here but basically saying that we need a carbonyl here takes care of one of these degrees of unsaturation I put a single check mark there so because there's also an integration number of four that probably means we have something like this foreign these two carbons being on either both sides of the carbonyl and me dotting them means that they're kind of equivalent positions so that's what I'm thinking here Prime okay the fact that we have this type of situation pretty good in that we're starting to build out what this might look like right we have this table now let me suggest something like this we've taken care of one degree of unsaturation and we still have one left what if we tried some type of ring like a six-membered ring and again that's just a guess let me just build out a ring like this so that would be one two three four five six carbons right it's an ugly six-membered ring that would be six carbons which means we would have four left now let me say say this if we had this position what I can do here is just say these positions here would be carbon a both a we still need to take care of uh we still need to take care of the fact that we have four methyl groups potentially and the reason why I'm bringing that up is because we have six carbons here if we create if we somehow position these four ch3s that gets us to our 10 carbons okay and I think adding these methyl groups will help us achieve this the the fact that we need three uh signals we need all this symmetry so hopefully what you're seeing is right now these two groups are not singlets right right now they would have two neighbors to their one side they would be n plus one two neighbors they would be triplets so hear me out on this what if we put two methyl groups here and two methyl groups here so I've added it one two three four ch3s and by doing so I have made a effectively singlets because a has no neighboring protons below it or two it's one side the carbonyl so a is satisfied in this scenario now let's look at the methyl groups which we know to be C there's three carbons here three carbons here three carbon or three hydrogen three hydrogens three hydrogens three hydrogens 3 6 9 12. good and they're the the only neighbor that they have right these are not Neighbors these are neighbors and the neighbor here is a quaternary carbon C is a bunch of singlets we're good to go here and what we accidentally did which is good is that down here for B right again a is downfield because of the carbonyl next door to it see sorry I meant a downfield c is up field because it's not near anything electronegative B also downfield because it's not um sorry up field I'm getting these mixed up up field because it's not near anything electronegative and there's only two hydrogens here which fits our integration number but we also took care of the fact that b is a singlet because B has two quaternary carbons as neighbors so B works there as well so A B C all good and we have two degrees of unsaturation the carbonyl double bond and then the ring that takes care of that so we did abide by our goal here we have one two three four five six seven eight nine ten carbons two four six and then six hydrogens and then 12 from the methyl groups this is our structure right here if these questions are difficult don't sweat it because they are it's it's they're tough you kind of just gotta if you're struggling struggle through a little bit longer hopefully a process like this can help you out and then eventually you just start seeing typical things like multiple methyl groups you know having a certain splitting pattern and then a high integration number a lot of things with the carbonyl position linking your degrees of unsaturation to how the information in this table can get flushed out don't lose hope you can get good at these if you're struggling I was terrible at them trust me on this okay cool that was a very long problem one but hopefully if you're doing these or if you know you're going to do in the future you can reference this all right so let's move on to problem two all right all righty let's throw our two up here and write the problem down and then we will go at it so let's see kind of like one more boring problem until the fun stuff okay make sure I don't incorrectly draw this because then we're going to be naming something that is not what I intended to and things will get go off the rails okay I believe this looks good all right for number two we're going to be naming the following structure now sometimes it might be explicitly called out on an exam reviewer sometimes it might not be but if you take note of double bonds and stereochemistry probably at this point in your organic chemistry class you are going to be expected to take care of two things stereochemistry and with double bonds e and z terminology or if your situation so calls for it says trans but in this case we're going to be looking at e and z because we are not in CIS trans um situations here okay so the first thing we need to do and again let's I'll say this worry about this stuff at the end foreign do not worry about it this information comes at the beginning of your name anyways don't let that stuff phase you just do what you know how to do find your highest priority functional group get your parent chain number name substituents and then at the very end remember to find out e and z as well as your stereo centers you know labeling and then we'll shut it we'll go through how to put that at the front of your name okay but back to business if you look at this structure right here two things kind of you know we obviously have an alkene we have alkene presence with the double bonds we have the Ketone and the alcohol if you're going to be naming you know hopefully your teacher can give you a functional group priority list but if you had to take a guess here hopefully you know if you don't have a list like that hopefully you'd guess that the Ketone takes priority over it would go Ketone in terms of priority then the alcohol and then the double bond but we only care about what's highest and that is the ketone so with the Ketone we're going to find our we're going to give this the lowest number and we're going to get the longest chain possible Right those you know the goal is not necessarily to wrap all the functional groups up in our chain the goal is to get the longest chain possible with this functional group and then give it the lowest number so I think it's we have one two three four five one two three four five six seven one two three four five six seven eight nine and that's the chain that is clearly the longest so we will have some type of no name but we won't be using the a any ending and we will worry about that at the end we'll be naming this as a ketone okay let's name our substituents three we have an ethyl six we have a methyl at eight we have a hydroxy because remember we don't have we're not naming this as an alcohol so then we have to use kind of the terminology to name the alcohol as a substituent okay excellent so what we can do is we can actually construct this name and not have to worry about this yet if we're going to alphabetize it'll go e first H second M third so we can name this as three Ethel eight hydroxy I don't forget six methyl and then we run everything together right we'll do none hyphen double bond at three double bond at five so that's three five die in so what I did there was non I'm not going to do a e because this is not an alkane from there I put the numbers where the double bonds start so we start at three and then go to four start at five go to six then you have to do an N right because it's an alkene that we're naming but we don't put the final e because that would be the end of the name and then you have to use a prefix to talk about how many double bonds there are seems very redundant but puts all the information together right and then what you do is you can actually identify where the Ketone is because we're going to finish off the name the Ketone is at position two and then you give it an O N E ending because we have one Ketone and it's at position two so this is the full name minus the stereochem in the Enz terminology now we're going to do is clean up after ourselves and give everyone who would read this name the full information on how to draw this so when you do that you have something out front you put this all in parentheses I did this very large on purpose just to show you like where the information is going to go and it goes stereochemistry then your double bond information so let's look at this I know our lowest priority group a hydrogen is facing towards us so what we do is the alcohol is a first in priority position seven is higher position than higher priority than position nine so we look like we're turning the wheel to the left it looks like s but our lowest priority group is facing us so it's not s it's actually R so you have two options here if you drew this out you would see that position 8 is a stereocenter you don't so we could just put R here if you want to be extra extra explicit you could put this as 8r which I will leave and again the reason why the 8 is optional is because there's only one stereocenter so if someone drew this and they they would recognize that position eight is a stereocenter they would look out front they would say oh the only stereo set the chemical information I have is R so they would know to draw this as r so eight is optional now position three remember with e and z we're still playing the priority game but in the context of a double bond so if we're comparing this ethyl group in this Ketone it's clear that this position 2 is higher priority than the ch2 up here because of the oxygen and remember you do this by molecular weight so we have a tie a carbon this carbs attached to a carbon or this carbon has two a one carbon and two hydrogens this carbon has one carbon and two oxygens clearly we went over here and then at position four we're actually pitting this carbonate five against this hydrogen so clearly we went over here if you walk this double bond we never cross a double the dot a double the dotted black line so we're on the same side that means and we can separate this with a comma at three we are Z because we're on the same side Z is like CIS now at position we'll do this again but for position five and six what you can see is on position five we have a hydrogen versus this so clearly this side wins on this side of the dotted line and then at position six we can go to the methyl group or down to position seven it's definitely over here position seven beats out that methyl group so again if we walk this dotted line you can see we cross it so while three to four with Z five to six is e which means down here we can say five e and that is your full name the full everything down here and then the extra information where you go stereo chemistry first and then your Enz terminology who that's it for naming all right I'll erase this and moving on into problem three we have seven total questions so I'm gonna break this up into a part A and A Part B so we'll go with a first do a little flip on my paper okay and then I will unfortunately have to bore you a little bit longer with me writing so just give me a second while I toss this up on the board thank you foreign part A and three in Part B I'm gonna try to do the best I can with space oh oops what I meant to do here is kind of highlight we're going to put stuff in here and this will take two steps so you saw a little bit of a preview and the product looks like this Plus enantiomer and then final one we have foreign okay all right let me see if I can give myself a little bit of breathing room up here we go we can say two steps here a little bit of hand okay excellent all right gang for part a it's a complete the reaction complete you know predict the product predict the product predict the product and then provide the reagent a little bit of a spin so in part A we can see we have this lovely structure with two alcohols and we see our reagent that we're treating this you know reactant with is PCC now if you're if you're not if your class doesn't use PCC this is basically the equivalent of um any oxidation reagent that will oxidize an alcohol but it will take it you know it'll create ketones and if it's a primary alcohol it'll make it into an aldehyde okay so in this scenario we can see we have a secondary alcohol and we have a tertiary alcohol one two three now remember when you're oxidizing alcohols whether you're using Jones reagents PCC what have you in your in your relative class you need to make sure that you have at least one hydrogen right here you can see we have one hydrogen because it's it's those CH bonds that you replace and repurpose to be Co bonds so when we redraw this product when we draw this product rather just know that we're not changing stereochemistry here so this Dash is still present and I have one two three four carbons one two three four this secondary alcohol this CH Bond will become another Co Bond and then this hydrogen will disappear we'll create a ketone at this position that I'm dotting in Black and when we do that we lose the stereochemistry there's no more this is no longer a tetrahedral carbon this is a SP2 hybridized trigonal planar carbon now the way I've seen these oxidation questions get asked in like a I got your way is in this particular carbon up here we have no CH bonds this carbon has four bonds none to hydrogen so you cannot oxidize this no matter what reagent you use so this position remains unchanged you still have the wedged o h and whatever else was present and that happens to be a dashed methyl group here so oxidation question for this first one with a bit of a twist okay now for the second question that has four steps and I might look scary I promise you it's not take it one step at a time so we have the structure here that has three carbons a bromine and no h now in this first step this is a strategy um and again what you can see further down here is we're going to create a grignard reagent which is a great nucleophile and that nucleophile is going to attack this carbial carbon now the first step here is actually to do some work such that we can disguise this oh because if you tried to create a grignard reagent with just this molecule right here you would basically do this this would become an mgbr and remember while grignards are very good nucleophiles they're also incredibly good bases this thing would either try to grab a hydrogen you know this this bomb would try to do something like this on itself or it would find another one of it you know another like you're obviously doing this like a beaker of some sort in real life or some type of flask this thing would find another one of the original starting material and just do an acid-base reaction right here and that's called quote unquote spoiling your grignard so one way you can have elements of a molecule you know if you have some type of protic nature to your molecule when you're doing grignard preparation you can protect your alcohol and create it temporarily into an ether and that's exactly what we're going to do in Step One with step one before this thing even touches your reactant h2so4 will protonate t-butanol and then t-butanol like the out the protonated alcohol now water leaves so you end up having something like this a tertiary carbocation and that tertiary carbocation gets attacked by the alcohol right here so long story short in a very long step one you end up doing this you have your alcohol or your oxygen but you replace the the hydrogen here with the four carbons right there so now it's an ether it's not an alcohol at all it's far less nucleophilic right sorry it's not acidic anymore that's what it meant to say now what you can do is move on to step two where you throw in magnesium sometimes thf as a solvent is included but the important thing is you're throwing magnesium in here so what that does here I'll do this I'll just add in a step two that will change this bromine into an mgbr which means now this carbon right here I believe I've lost I've lost a carbon one two three four oh no hold on goodness what am I doing I think I was just losing one one two three one two three there we go we'll have this after step two and then with step three what you're doing here I'm gonna just redraw this to be smaller to save some space is you include cyclopentanone a ketone and this carbon is in control of these electrons right here this carbon takes these electrons attacks over here and as we go to make a bond between this dotted carbon and this asterisk carbon we'll break the octet rule unless we do something and to avoid doing that we will break one of the bonds in our c c double bonded oxygen or carbonyl the Ketone kick it up as a lone pair onto oxygen because oxygen is more than happy to take on extra negativity I'll go ahead and draw the result of that over here it's going to look pretty big so I'm going to start with this thing we have this going on our mgbr is basically gone now you don't have to worry about it anymore when I draw a line from this dot carbon that I've redotted over there I can draw that line and know confidently it's going to be attached to an asterisk carbon the Carbonite asterisk which is right here that carbon is a part of the five-membered ring so I can just draw a pentagon around that carbon and then also on that asterisk carbon I have an O minus from the attack and that my friends is step three and now step four is going to do a lot for us step four the acid h3o plus it could be h2so4 specifically that is going to not only quench the negative charge here right quench meaning o minus we'll grab a proton it'll become an oh it'll become an alcohol but that acid's also going to unravel this ether back into the oh that it used to be so we protect it in step one step four protonates our grignard product but it also deprotects The Ether and flips it back to the alcohol it originated from so the product looks like this three and let me just make sure I don't didn't lose any carbons one two three and then eight from five more from here so eight total one two three one two three four five perfect okay so this is our product so I know that was a long question but that kind of wraps in a lot of things principles about grignards and you know making sure you're doing your due diligence for protecting acidic parts of your molecule when you're creating grignards cool this one down here so what you can see is we have a oh and hold on I'm very sorry what I forgot to reflect up in the second problem is when we attack this ketone we can do so from above this molecule or below because this is an SP2 trigonal planar carbonyl so that resulting o minus could be either a wedge or a dash so what I forgot to reflect here is that if I draw the wedge or the dash we need to show one of them if stereochemistry matters which it's always safer to assume it does and since that's the only stereocenter in this molecule if I took the opposite product the dash it would be the enantiomer so I can just draw one with the dash and then write Plus enantiomer along my answer and it shows that I'm considering both you know of the um products here from the attack on this trigonal planar carbon from above and below okay cool now down below here that run because I saw nantium over there we have to in two steps show how we can go from this ketone to having a bromine here secondary now in this you know Gatherings of chapters right we we know how we can get bromine on a molecule right we can do that through sn2 we can do it like sometimes with nabr or one other way we can do that is with this Nifty reagent called pbr3 okay but pbr3 works on an alcohol it does some stuff to make the alcohol into a better leaving group and then the BR minus comes and attacks but this isn't an alcohol but it can be if we in a first step use nabh4 and ethanol or you could use lithium aluminum hydride it doesn't necessarily need to be two steps this schlub made up the problems but if we use nadh4 and ethanol we basically have if I'm going to shorthand this hydride H minus attack like this electrons kick up so because again trigonal planar carbon you'd get a wedged o minus and a dashed o minus and then the ethanol is where this thing gets the o h from and then you could throw in some pbr3 and that will perform the sn2 for you oh what I should say is to show the product I actually Drew we can get the dashed but again we get Plus enantiomer then pbr3 does sn2 comes from the back side bromine attacks as a wedge that's how you get this and like I said you have the enantiomer over here which would yield the enantiomer over here with the pbr3 cool and last but not least here on 3A we have this molecule and then we have H2O so4 heat right I wrote as a Delta but you know Heat and reflux this problem is aiming at the dehydration of um an E1 reaction that functions with the dehydration with you know facilitating that with sulfuric acid so I'm going to give myself some room so I can show you some mechanistic things so basically just remember right seeing the heat remember that favors elimination and because we have an alcohol the first thing that is safe to assume is that you'll have h2so4 and that's something that this alcohol will be interested in because alcohols right we know they are infoteric they can act as acids and bases here it will grab h plus it'll act as a base so we have this right here oxygen gets the plus charge and then remember this is an E1 mechanism because we're in a polar protic environment sorry about that a polar protic environment because of that and now that we have a good leaving group solvolysis occurs and why that's important let me draw my arrows in Black why that's important is because now that we are in a world where we understand carbocation rearrangements every time you form a carbocation make sure stop and look around can you make it better can you improve it through a hydride shift or methyl shift and this is a secondary carbocation and if we look no further than just next door we can see that there is a dashed hydrogen that we can rearrange through hydride shift so this moves the secondary carbocation to the tertiary position foreign now be careful because again this has implications on your stereochemistry this carbon is now a tertiary or a trigonal plane or trigonal plane or carbocation it only has three bonds SP2 hybridized meaning this wedge is no longer a thing it's a flat methyl group okay cool now we actually can you know we have our carbocation in the right place now we actually need to do the elimination and finish the problem off because this is an E1 mechanism so we need to form the most stable double bond we can the most substituted double bond we can so we can do there is look next so we can either do tertiary secondary or tertiary tertiary and we definitely want to do tertiary tertiary so just to show you that you can bring back the conjugate base of sulfuric acid grab the hydrogen have the double bond form and your final answer finally looks like this okay that was a lot four problems we have three more let me erase this take a deep breath we'll get through the rest of them and also just know right I'm trying to make these as you know I'm not I'm not in your class if we're from wherever you're watching but I'm just trying to put in as many things together so if you're watching this you're getting a lot of bang for your buck so they're meant to be involved they're meant to be long and if you're thinking they're easy then hey guess what that's awesome okay let's erase these all right 3B so three more we're almost almost done with this monster okay let me draw these up here quickly you got that foreign we have a lot of uh ether and epoxide stuff coming our way right now it's a funky cyclopentane there we go a little bit better I am the worst at drawing epoxides I'm just never good at like judging how long certain bonds should be it's like I feel like at some point I got somewhat decent as drawing like cyclohexanes that kind of look good but man give me an epoxide I'm gonna Draw Something okay Okay cool so what we have going on here is first problem we're just predicting the product for all three of these first problem we have propanol three carbon alcohol and in the first step we just have sodium and you might see sodium like this you might see a little thing up here like that that's meant to be Elemental sodium so this is you getting the signal to form a Williamson ether okay so what that means or sorry about that we're forming an ether with that approach but what this does is this the elemental sodium basically just rips off the proton so in step one here you're really just creating an alkoxide you're literally just ripping that proton off and creating o minus okay that gives you the good nucleophile to then in step two execute a Williamson ether synthesis which is creating an ether where you use sodium to prepare this nucleophile and then in step two that good nucleophile finds a good thing to do sn2 on it attacks leaving group leaves and then you have one two three and then the alcohol one two three and then the alcohol and then you this is a good way right here of making sure you don't lose any carbons we have three carbons after the fact so this is your product here nice sweet short and simple cool now we have two epoxide problems so this you know right in a and uh an epoxide I always forget exactly how to say this in Oxo um cyclopropane you just have a ring of three two carbons and an oxygen and there can be things coming off of it as well so here we see that one two three ring so we know we have an epoxide so what we can do here is we have a grignard we have a good nucleophile and you know epoxides are susceptible to attack because right these two carbons are bonded to an oxygen something that's more electronegative than itself so there definitely is a partial positive thing going on here right so the thing is with epoxides you can either attack them in basic conditions or acidic conditions and if you have an asymmetric epoxide like we do here the conditions you're in basic or acidic depend on where you end up attacking okay that's why I wanted to include one of each in this situation when you so when you see something like this right this is if I expand this out this is what we look like here grignards are good bases if you're seeing anything with negative charge in the step where you believe you're going to attack the epoxide even if it was something like ETO minus like ETO n a and etoh the fact that you have ethoxide ETO minus balancing with an NA Plus right obviously here we have the grignard that is your signal that you are in a basic environment you have negativity around you you don't have an excess of h plus around you when you have when that is the case so this is a basic environment when you're attacking an epoxide in basic conditions your mode of thought is let me attack where the steric hindrance is the least right you're going for the best sn2 conditions okay so what I can show you here is ET mgbr I'll draw this bond this these will be the electrons that the F the you know the greenery region will attack with I can either attack this carbon up here that is you know tertiary or I can attack this carbon here which is secondary you're definitely going to go with the secondary option so the ethyl group attacks right here and then our leaving group is this Co Bond so we're going to bust the epoxide open okay so after the fact we look like this I'm going to redraw the ring this position largely remains what it looked like before we have the methyl group that did not change nothing changed with this dashed Co Bond so I'm just going to redraw that has an o minus now here's the part where you need to do a little brain you know use some brain power the ethyl group attacked at this position and the leaving group it forced off was a dash so as a result the ethyl group attached at this position as a wedge because this is an sn2 style attack so you must attach from the back side of the leaving group leaving group is a dash back side is a wedge and then the second step here is just to throw a proton on this o minus to quench that negative charge so your product here looks like this bingo cool all right now hopefully this will illustrate this will illustrate the difference if we look down here we have another epoxide it's in a straight chain as opposed to ring on ring now when you look here you're seeing ch3oh and h2so4 when you see this this screams acidic conditions h2so4 is a strong acid so when you can assume in like first step of this mechanism when you're thinking through this there's h plus everywhere this oxygen will be grabbing h plus from h2so4 so you can basically think of your epoxide looking like this okay now I would encourage you to watch the Joe chem video on epoxides if you don't know what I'm about to say but because we have oxygen in this mode right here you can draw resonance back and forth like this taking turns where this Bond breaks and this position becomes a carbocation and breaking this Bond and then this position becomes a carbocation as a result what you basically see is through the resonance whatever carbon is more substituted this position is secondary this position is primary this position is more stable overall as a carbocation so through the resonance because this carbon is more comfortable as a carbocation its resonance hybrid contributes to the overall resonance hybrid more so this position is more positive than this position I say all that to say when you're out your nucleophile the methanol more specifically the oxygen in methanol when it goes to attack it is more drawn to the more substituted position it's kind of the opposite thinking from up here here we were thinking sterics here we're thinking what carbon's more positive and the resonance influences that okay so we're attacking the more substituted position here and then we kick off the leaving group which is one of the bonds in the epoxide so when we draw our product here we have just an oh over here the reason why there's I didn't even need to draw this wedge over here because this one isn't a stereocenter because it's not attached to four different things so when I draw it normally I'll drop the wedge just make it a straight line but with this position right here this is a stereocenter at this point in time so this methanol must attack as a dash and it looks like this at first just to show you how that looks and then like if the first step was the epoxide grabbing h2so4 like a proton from h2so4 like this you can imagine conjugate base of this coming back cleaning this up because the oxygen is not a big fan of the positive charge so your product looks like this och3 so you end up with an ether actually cool all right that does it for all the complete the reactions whoo we have two more questions four and five this is more of a mechanistic thinking like you know concept question and then five is a synthesis and then my friends we are calling it an exam three review session so if anybody is still watching this if there's anybody out there you are a real one okay all right so for this one four is one of these type styles of problems let me write it down first it'll involve a little bit of writing and I'm going to take the time to write it just for the recording and for video editing purposes afterwards so this is one of those problems where you're kind of giving something you're unfamiliar with but you can do it because of things you you do know so it's it appears like you're given something that you you shouldn't have the knowledge to accomplish or it's something that your teacher like didn't teach you but you you do know enough but they're always scary when you're reading them because you're like what what the hell is a pinnacle rearrangement um that hey corvo sorry I keep talking and writing and then I lose my train of thought while I'm writing is stabilized okay so this is like given information right here given info and then the problem is and this thing up looks crazy foreign okay so this looks Wild I totally understand that but whenever if if these types of problems give you grief if they are the bane of your existence on exams I feel like just actually trusting that and again I don't know your teacher if maybe you have a evil maniacal Professor I would hope not um but what I hope that I think one big thing that helps me or helped me overcome these problems is they can't get they would not give you something that you couldn't solve it always links back to what you've covered so on an exam you know when that information starts and when it ends and if you have something that looks unfamiliar just try to find individual pieces and Link it back to stories or themes of the chapters that are on your exam so a reaction called The Pinnacle Arrangement depends on the fact that a carbocation important is stabilized when next to an oxygen okay so this is like them giving you extra information that maybe wasn't even necessary but this is like if there was a missing link here it is so paired with what you've learned and this you should be able to then explain the following reaction now again I totally understand this thing looks crazy but this is definitely about carbocations right we know that for sure so we just need to figure out where it links in with what we've learned recently if you look here right we have two alcohols and we do see a familiar reaction h2so4 heat and reflux a dehydration reaction and usually what that means is we create a carbocation and then we do an E1 reaction we do elimination but we don't see a double well we see a double bond but it's not a carbon-carbon double bond it's a carbon oxygen double bond we created a ketone AKA carbonyl but what I'm hoping you can see and how this links in is that we have two carbons each with you know two bonds to methyl groups what we have over here is one carbon with one bond to a methyl group and then this carbon has three bonds so we did a it doesn't look like it but we did a methyl shift here and I think if you can spot that amongst the chaos and the madness clearly this h2so4 heat reflux helps us lose one of these Waters we know that and in the process hopefully we'll find some good opportune moment to use this information to yield this carbo this Ketone but also to do a methyl shift to get closer to what we need to we have a symmetrical structure so let's just start doing this this dehydration mechanism that we know so I'm going to save some space here I'm going to draw h2so4 and I'm going to have this oxygen grab this so I'll go this way so here's what we have now in this part of the reaction the next thing that usually happens is our leaving group leaves right so I'm going to have is that same line with two methyl groups and then we have the same thing on the right hand side okay now here's the key part we've created a carbocation and hopefully this makes sense if we do a methyl shift here right if we do a methyl shift we accomplish two things in one go if I were to move either one of these methyl groups like this I now achieve one carbon that has three methyl groups and I'm going to shift the carbocation from this position to this position which means I will basically be using the information that was given to me so it's relevant and that will help me create a c double bonded oxygen which is what is in our product as well so it's this key step like this is the Once these problems are tough because they usually they usually take you off guard and they scare you out of applying the things that you do know because we had to do everything we learned in you know in terms of carbocations and alcohols that powered us to get to this point to then apply the given information to keep going so trust me these problems are tough but I know you can do them so trust your arrows we now have three methyl groups right here and then we have a um on this carbon right here which I'm going to dot over here we just have one ch3 we have the oh still and then we have a plus charge now what we can do is use this information here in this situation all we've done is resonance to swing down electrons um filling the octet of this carbon like this and that shifts the positive charge up top now luckily the R Group here is a hydrogen such a long Bond that's a ch3 so to finish this off we could even bring this thing back to do something like this and that gets us to the finish line so again one of these problems where they're trying to knock you off your balance they're trying to kind of with wave with one hand and do something with the other but if you just stop and take a breath hopefully hopefully you can use information that you recognize to start the problem you'll get to that inflection point where you do need to kind of like put two sides of the equation together but when you do you get to the finish line okay excellent we have one more problem and we are going to call it a review synthesis all right so this problem would be foreign so in this problem we are going to with carbons with pieces of carbon four or fewer right so what that means is you can grab as many pieces of carbon as you want but they have to have four carbons or less we need to synthesize this molecule right here so my first rule of synthesis always always always is if you're given hard P like if you were if we were given like methyl you know methane and cyclohexane as our sources of carbon I always like to relate what we have available for us to start with what we need to get back to and our Target molecule because your goal is to use as few pieces of starting material as possible the more starting material you have event more likely not in a synthesis problem you are sticking pieces of carbons together so you won't have to do that as few times as possible so we have one two three four step one is to count your carbons one two three four five six seven so ideally what that means if it makes sense when we look at the structure and cut it up we could do one piece of four and one piece of three that would make our lives pretty easy if the path to creating this allows for that okay now and again hopefully this kind of assumes you've done um I can't talk hopefully this this assumes you've done the grignard reaction at this point but in this molecule right here I am like one thing to do next is to look at your structure and maybe think about what was what was the piece where you stitched two things together or pieces of carbon together when I look at this I see the presence of an oxygen now when we do a grignard reaction right if I want to just draw a really simple simple meaning like small during your reaction this involves doing attacking carbonyl and you get an O minus that eventually with you know if we had h3o plus here becomes an alcohol okay so seeing alcohols is a great indicator that you did agree in your reaction however we don't see the presence of an alcohol but it's just because maybe we did something after we did something to that alcohol after the grignard reaction so remember when you're doing synthesis going backwards surprisingly enough is your friend doing retrosynthetic analysis so if I were to take this one step back and again I'm going to kind of like go this way to maximize my space on the board if I was thinking about what came before this thing I see a ketone it could have been an alcohol beforehand and if I step it back with like PCC or even the Jones reagents na2 CR2 07 and H2S04 either one will help you go backwards in time and have this thing be an alcohol and I'm going to ignore stereochemistry for the sake of this we're just more concerned with getting the products like taking this back to the starting material cool so once I've done that now I have something that looks like a grignard product excellent so what do I do from there when you get to a place where you can basically think of what you're looking at to be like the product of kind of combining pieces together you want to take a look at it and you want to try to make a cut in your molecule you want to take your molecular scissors out shout out Dr Erica Houston from pitt molecular scissors and you want to make a cut into your molecule some ideally in a place where we're going to go from seven and we would love to get to a piece of four and three now if we look at one two three four one two three you kind of want to do it where the your functional group is around so in a grignard reaction the connection point is where your alcohol like where your alcohol is so basically I want to make a cut somewhere around this carbon and I see one two three carbons right here I'm going to try and make a cup right here because what that does is above the blue line I have three carbons one two three and below I have one two three four what that means is if I take this one step back and over the arrow I'm going to put h3o plus because in a grignard reaction your final step is always to protonate the O minus to an o h so that's that reagent takes care of that I'll have a grignard piece and I'll have a carbonyl piece the carbonyl piece is always where the oxygen is so I know for a fact I'll have this instead of an oh I'll have a c double bonded o now again if you stamped it back and Drew it like this there's nothing wrong with that I just like drawing my out hides either like up or down if possible so either one is correct it's not a big deal carbonyl and then the grignard this piece right here that I'm going to asterisk that was your grignard okay now we're going to have mg and you'll see why I'm going to put a br here for a second okay now some teachers because if if they specify carbon pieces of four or fewer this is technically carbon pieces for fewer you could be done at this point I'm going to show you how to take this all the way back to one two three propane and then butane okay so let's do this first with the grignard you'll see why I I'm actually gonna yeah I'm gonna I'm gonna go like this for that and you'll see why I chose bromine so when you prep a grignard the last thing you're gonna do is treat it with magnesium and you can include thf here if you're trying to be fancy that means if I set this back I have sec butyl bromide now to take this together I have to get bromine on here now the reason I chose bromine instead of like chlorine is I can go from plain butane and put a bromine on here if I do the free radical bromination remember that bromine is extra is extremely picky so the bromine will flock to this either secondary position and then we are done here we have shown how this became this which feeds into the production of our Target molecule now we need to show how this gets back to propane we have an aldehyde here so the only way we know how to make aldehydes today is that they come from alcohols so I could show this now I need to oxidize this alcohol to get this and we can do it with PCC or whatever reagent you use in your class to produce alcohols or aldehydes from primary alcohols cool now when you get this alcohol to get primary alcohols you can whoops we could have something like this is a primary position so just regular old sn2 off of something like this would get the job done and then last but not least we can get a chlorine on this primary position with some free radical chlorination and that my friends shows the synthesis problem okay we are calling this review a wrap so if you're still here if anyone is still here I would be wow if there's a few people watching you're Legends for staying on and if you're watching the end of this thanks for watching and using Joe chem I hope it's been super helpful I hope your foray into carbon based chemistry is one that is exciting and interesting okay as maybe interesting or if you're liking it that's great too but hopefully it's not a source of stress because that's the whole point of what we're trying to do here is to demystify carbon-based chemistry to help Propel you to your goals whatever those might be higher education just getting your major getting that degree whatever thank you for watching hopefully you're subscribed hopefully hit the like button all the good stuff and no matter what I hope to see you all in the next video