um okay hello everybody um today we're going to be going over the or organic chemistry 2 exam 1 for spring of 2022 unfortunately um the recording that I had made on Monday did not record correctly so I'm going to be re-recording this and uploading this tonight the night of the 14th which is Valentine's Day wow okay um so to begin we're going to start with the nomenclature section so um the first thing to do with the first nomenclature is that we have an epoxide so um there are two ways to name this uh technically the correct way is that the O group is the uh parent chain but we will I'm also going to be naming it as if the the epoxide was the parent chain so the first one I'm going to do is the um alcohol parent chain so the alcohol parent chain uh we want the longest chain of carbons first starting with the alcohol so we're going to start numbering from 1 two 3 and four so now that we have this numbered we can give the um give the parent name so parent uh the parent chain is going to be uh four carbons long so butane but it's going to be an alcohol and so that would call it uh it would be an alcohol on carbon one specific specifically so we're going to call it 1- butenol and then uh for this s substituents uh Subs we know on carbon 4 we have bromine so we're going to put 4- bromo and uh we have to label this epoxide ring as a substituent as well so to do that we're going to be referring to this as 2 comma 3- epoxy e p x y right and then lastly to do stereo chemistry um we see that we have two hydrogens here and these are both the same groups so because of that we don't actually have to do RNs configuration we can actually just call this CIS but if we were going to do RNs then uh we would still be allowed to um if you wanted to do RNs you would look at carbons 2 and carbon 3 because they are both the um the carbons that have the four groups on them so um two three four okay so Carbon 2 is connected to a carbon right here a hydrogen so hydrogen is going to be group four it's connected to an oxygen this way it's connected to another carbon this way so between the wedge and the uh bottom of the oxirene it has or the epoxide it has um two it has two carbons so uh obviously the oxygen is going to be labeled as the biggest group so we're going to put one for the oxygen and then for across the this bottom side we have a carbon that's attached to an oxygen a hydrogen and another carbon and this carbon right here is attached to two hydrogens and an oxygen so oxygen and two hydrogens so because this one is attached to an oxygen hydrogen and a carbon this is going to be uh carbon number three is going to be our number two and then this last one is going to be our number three so if we were to do the arrows this would be a clockwise Arrow so our stereo chemistry would begin to R sorry not S I don't know why I wrote that 2 R for clockwise and as for the other side I'm going to choose a different color as for the other side we [Music] have uh carbon 3 which is connected to a carbon with a two hydrogens and a bromine and oxygen which is going to be one the hydrogen which is going to be four and another carbon going this way that is attached to an oxygen a carbon and a hydrogen so from this we have the oxygen uh the carbon with the oxygen carbon and hydrogen is going to be labeled as number two and then the carbon with two hydrogen and the bromine is going to be number three so whenever we do an arrow this way this is going to be 3s s considered because it's counterclockwise so 2 R comma 3 S so to put it all together um now we can put the CIS or 2 R comma 3s Dash um we're going to start with bromo and then epoxy so 4- bromo uh-2 comma 3 Das epoxy D1 D butenol and one thing you can also put is uh epoxy butan D1 and lastly I'm going to be showing you guys the um the epoxide naming so I'm going to redraw this actually so just so that I have like a clear thing I don't want to erase all everything I've done we are H it is an oh off of this and another H so whenever we label the whenever we label the epoxide as the parent chain we're going to start with laboring the oxygen is number one and then these are going to be either two or three so they can be either two or three and we're going to find out how we can determine that so uh for the parent chain the parent chain whenever we label the epoxide as the parent is going to be called the um sorry called the epoxy oxir sorry not epoxy it's going to be called an oxir so and then now we can for Subs we know we have the four uh this four bromo is not no longer Rec call that so these are now going to be two different complex Subs so in this one we have a one carbon chain with the bromine on it so our first sub is going to be a 1 D bromo I'm not putting the actual carbon the labeling carbons yet this is just the um this is just the parenthesis name one- bromomethyl and then another sub is going to be the 1- hydroxy meal and then Lastly lastly what we're going to want to do is to um get the stereo chemistry excuse me one second okay for the stereochemistry um we can do the same thing we can call this 2 r3s or CIS so I'm going to choose to call it CIS for Simplicity sake so now that we have that um we can finally name the compound uh well not now but um we would have to first figure out the B and H so because B comes before H this is going to be a two and this is going to be a three um and then finally we can put it all together so we have a CIS Dash uh two bromomethyl so because these are on carbon WS we don't actually have to put these in parenthesis I just did that for I don't a good practice that's just something I do I suppose um two Das bromo methyl dash3 Das hydroxymethyl oxirene and moving on this uh in this structure we have an ether uh remember for ethers that they actually don't have priority over other um over alkenes so um while it may be uh like maybe tempting to like start from like one two right here uh just remember that um it's not going to be like an alcohol it's going to be a little bit different so um these uh each don't have the same uh priority as alcohols do they actually have a similar priority to uh I believe haights they're a little bit on the lower end so we're going to be considering the alken as the alken as the priority group so whenever we start with alenes we have to make sure that we have as many alkenes in the chain as possible and not just the longest chain so we want want the longest chain of alenes with the most number of alkanes possible so if that is a three chain uh an a chain of three alkanes that's four carbons long is going to be more important so if we have a chain of alkenes that is let's say like this that is one two 3 4 5 six so this is a six carbon long chain of alkenes and then we have one that's going just say like we have like 15 carbons s way and then we just have an extra Aline right here so even if um this one had like as many carbons as it had and it let's say it had one extra right so if we went from this way up we would only get four alkenes so even though it'd be like a like a 12 carbon chain this alen this way with five alkenes but less carbons is actually going to be more uh beneficial or it's going to be more it's going to be higher priority so this is going to be our current chain this one so to expand on that now whenever we do this we have to make sure that we do the same thing so we're going to start by numbering one 2 3 4 five 6 and 7even and always make sure that you number backwards and forwards 1 2 3 4 five 6 7 so we see that our double bonds are both on on both the blue and red numbering are on one and six so that means we have to go to Subs to break the tie so with the red numbering we have our subs on three and and four and with the blue numbering we have our substituents on number four and five so because of that we're going to be going with the red numbering so in order to start naming let's erase the blue numbering and let's start so our first uh our first goal is to name the parent so parent would be a 1A 6- hepta and remember because we have two alkenes we're going to start with the prefix dine so we have a 16 heptad and then as for Subs we can begin by naming the sub one carbon 3 um so whenever we have an ether as a sub an an ether yes an ether has a sub then we're going to be U referring to it by the chain that it's that it is and then putting oxy at the end so in this case we have an ethyl group on this so we're going to start with 3- Ethyl uh and instead of putting ethyl oxy what we're actually going to do is we're going to erase the y l of ethyl so and we're going to replace that with oxy so it's going to be and E oxy whoa how I put a t there e oxy and as for the subon carbon 4 we have the 1 2 3 4 this is going to be a but group b u t y l and lastly we have to do stereo chemistry so we don't don't have stereo chemistry on any of these alken bonds because they don't have um these alken bonds both have hydrogens on both ends meaning that it wouldn't be our identical Hy not identical hydrogens but they're both um these hydrogens are both on the same group and so that's why we're not able to um figure out which one is higher or lower so for that reason the alkanes are not going to have any stereo chemistry but we are going to have stereo chemistry on carbon 3 so to uh figure out the stereochemistry on carbon 3 what we're going to do is four five six what we're going to do we're going to label this and then we're going to go every which way that we can so we have a hydrogen two carbons and an oxygen that is B bound to so hydrogen four and then we have two carbons here and then oxygen so the oxygen is going to be number one and now whenever we go to this carbon carbon 2 Carbon 2 is bonded to two is bonded twice to a carbon so C comma C and it has one hydrogen hanging off so H and then this carbon carbon 4 is bond to two carbons as well and one hydrogen so C comma C comma h so because of that both of these car uh carbons are tied so we're going to go to the next one so as for this one we're going to be going to carbon one right here so this carbon is bound to two hydrogens and uh for this case it wouldn't matter which carbon we went to if we went to this carbon or this carbon um but it's important to stay on the parent chain so we would go to this carbon and then now this carbon is attached to two hydrogens and then a carbon the carbon so because of that this is going to take priority on the left group so this is going to be number two and this right group is going to be number three so whenever we draw our Arrow we have it going in the clockwise Direction so our stereo well for clockwise is R but remember that we have our smallest group on a wedge and our biggest group on a dash so our R is going to be flipped to an S so now finally we have everything that we need to name this chain so we can put start with s or stereochemistry Dash we're going to make sure to put the but before the aoxy so four dasb das3 d ethoxy and then the parent to finish it off D1 comma 6 heying and one thing you can still do uh aoxy hepa-1 comma 6 ding so that is question two and then moving on we have another ether uh this time the ether is four carbons long on the left side four carbons and if you just counted from The Ether which is tempting 1 two three four you would see it's four Carbon on on this side too but it's important that we make sure that we have the longest chain so we will start numbering from this carbon right here one 2 3 four and five and so on this side on the right side we have a five carbon chain so that's going to take priority over the four meaning that this entire ether is going to be a pretty long I feel like complex substituent but we'll see so we're going to have two substituents one substituent on Carbon 2 and one substituent on carbon 3 if we numbered it backwards 1 two 3 4 and five it would be on three and four so we're going to be going with the first numbering okay and now uh what we can do is we can just name the parent chain so remember um that we don't treat uh ethers much differently as as part of the parent there wouldn't be part of the parent so this is just going to be a regular penten and now to go to Subs we have have on carbon 3 we have a methyl group so 3-methyl and on Carbon 2 we have our ether group so on The Ether group remember we have to name the chain that the ether is hanging off of and then put the uh put the oxy at the end so we're going to start by numbering 1 two 3 4 so we have a 33 Doro so the parentheses we can put the two sub that are on the substituent Chain get 3 comma 3 Doro and then uh to uh name the chain is going to be but so b u t y l but once again we're going to drop the Y L and add the oxy so but oxy and that is all we need because there's no um there's no stereochemistry to be had in this so we can just put it all together so we're going to start with the 2-3 comma 3 Doro butoxy D3 d methal pentan okay moving on uh we have rank the following carbocations in order of increasing stability so uh carbocation stability um actually goes from uh tertiary stability is going to be the highest is greater than secondary stability and that is greater than primary stability so whenever we look at this we're going to make sure that we determine if it's primary secondary or tertiary so on the first one we have a primary carbon this carbon is only connected to one carbon right here so it's a primary this uh this bond is only connected to two carbons so we have a secondary carbon but we do have some resonance right here so we'll take that into account and then our third one we have a tertiary um carbo on this carbon is attached to three carbons so now what we can do is that we can look at which one is primary so we have primaries and we have tertiary and we have secondary so we know that the primary is going to be the least stable so we're going to rank this number one and at first glance uh we would rank the secondary as secondly stable right but remember that resonance is a very good stabilizing Factor U stabilizes it a lot and so because of that this uh this one with resonance is actually going to be more stable than the tertiary carbocation so this is going to be number three and our tertiary carbocation will be number two um moving on the second one is asking us to put us put the halides in order of increasing reactivity in sodium iodide and acetone so uh this question may look a little bit confusing but remember sodium iodide and acetone ni AI acetone this is a reaction in orgo one that um might not look familiar to you guys but it um it is actually a strong nucle strong nucleophile with a a product solvent which means this is an sn2 reaction so one thing to note about sn2 reactions is that they have increasing stability have increasing reaction rate so its reaction rate increases with a methyl halide ch3br which is has the greatest um which has the greatest reaction rate and then primary and then secondary so we'll look at it through that lens and remember that there is no sn2 on SP2 carbons no sn2 on SP2 at all that is a rule so whenever we look at number two the first one we have this bromine on a vinyl position and uh if you remember vinyls are incredibly unreactive and just to put the to put the nail in the coffin this carbon that is on is an SP2 carbon so it has SP P so it's an SP2 carbon and we know that SP2 carbons do not undergo sn2 so immediately we can just put that as the least uh the least reactive now moving on we have two secondary halides and uh we have to figure out which one is going to be more reactive so whenever we're uh doing this we have to remember that uh these bromines are not the same so one of these bromines is two away from a double bond two carbons away and another one is on right across so it's one carbon away and that one carbon away position is called the LC position so this bromine is secondary alic so one thing to note about alyc is that it increases the reaction rate by quite a bit so because of that the secondary alic is going to be it's going to have a higher reaction rate than the secondary regular bromine so this is going to be the most reactive and this is going to be the middle reaction uh next number three place the following compounds in order of increasing reactivity with uh methyl mgbr and ether so this is a greenard reagent so one thing to note about greenard reagents is that they are incredibly basic so this is basic and this will come into uh come in later in the question so we have three um we have three structures we have two epoxides epoxide epoxide and an ether um so whenever we have an ether we know that graned reagents do not react with ethers so immediately this is going to be the least reactive and now we have to determine which epoxide is going to be more reactive with the greenyard so one thing uh that we do need to remember is that it is basic so that's going to tell us which one which epoxide is going to react with so whenever we have a basic solution with an epoxide the basic we have an acronym that we uh that we use called BL M blam well my think my pen is dying B o a m that means that basic attacks the least subbed and acidic attacks the most subbed so for this one because we have a greenard reagent we're going to be attacking the least subbed so uh whenever we have a grenard reagent uh it's going to attack the epoxide from one of these two partially positive carbons so as for the first one we have two carbons that have the that have near identical groups or they have the that have identical groups and so because of that um both of these carbons have two groups on them there's not really any preference for the greenyard reagent to attack which one whereas whenever we look at this one we have one of them with only one group on them and one of them with two groups so this grenard reagent is going to have the preference to attack this carbon with only one group on it because it's basic so because of that um we're going to be labeling this one as the most reactive this one as the second most reactive and one thing other that you can also to like that you can also think about it like this is that um this is going to have a lot more steric krance because this because this one has the extra methyl group whereas this one just has a hydrogen attached to it so it's going to be a little bit easier for the grard reagent to come to this less subside all right getting a little bit bored of red so let's go back to Blue so so okay on to the next one place the following resonance contributors in order of increasing stability so in this one we're going to look at all the charges and see which one where the charges are going to be the most stable so in this one we have the charges on an oxygen a nitrogen and a carbon so first let's see which ones are going to have a full octet so whenever we look at an oxygen we can draw the um can draw our lone pairs on that the nitrogen as well one two three four and then our carbon is going to have two lone pairs or one lone pair sorry and a hydrogen attached to it so now whenever we count we know the oxygen is going to have a full octet 1 2 3 4 5 6 7 8 so two in the bond and six in the alone pairs our nitrogen is going to have two bonds and four carbons or four four electrons in the L pairs so it's going to have the total of eight fills the octet rule and our carbon has two bonds three bonds sorry and the uh electron pair so it does fill the octed rule so now we notice that all of them fill the octed rule and now we have to go by a different way to solve the problem so we want to see we should see that which um which atoms take the negative charge the best so if you remember if we had the periodic table right here um we have a periodic trend of electr negativity which increases as we go up the group and increases as we go further right down the columns so now uh if we look at the periodic table and we see we compare carbon to nitrogen to oxygen we see that carbon comes at carbon number six nitrogen at number seven and oxygen at number eight so if we employ this same method we know that oxygen is going to be the most electronegative and carbon is going to be the least electronegative so from that we can uh put it in order of increasing stability because we want the most electronegative at to have the negative charge we're going to put oxygen at number three and carbon at number one and lastly we're going to put nitrogen at number two the second part of this question asks us for the hybridization of the carbon atom so if we were just to look at this carbon atom just where it's pointed you would see it has 1 two three and four groups it looks like a smiley face nice so it has four so we would consider this an sp3 just off the bat but uh we have to make sure that we look at the other resonance contributors so if we look to the left we have this one has a hydrogen on it so it would be one two and three this is an SP2 carbon and once again we have another SP2 carbon so SP2 so both of these carbons are SP2 and this carbon is sp3 so one thing to know is that the is that the um the hybrid or the hybrid is the one that exists in the real world so all of these structures this wouldn't um these wouldn't exist the hybrid is the only one that exists so because of that that carbon although sp3 in this um in this resonance structure it's the hybrid moves along along like all of these forms right the hybrid can move between like all of these just as it as it exists so because of that um this SP we know that the hybrid likes to have the lowest energy structure possible so because of that we have the SP2 structure which is going to be our lowest energy and so our Reon um our answer for Part B will be S2 next uh place the following compounds in order of increasing reactivity in a Deal's Alder reaction so in a Deal's Alder reaction we have a dine and a Diop file so d dile so what happens in the reaction is that one of these bonds is going to attack this B and these bonds this uh is going to attack this and it's going to form a cyclohexene ring so in that case because this one is going to be the aggressor let's just say that we have this top Bond being the aggressor right that means that this top bond is going to be part partially negative correct so we know that partially negatives attack attack partially positive bonds so partially negative so this must be partially positive and so because of that we want so to have increasing reactivity with a Deal's aler reaction we want the dine to be more negative and we want the d dile to be more positive so that means for the ding we want more electron donating groups and for the dile we want more electron withdrawing groups because electron withdrawing groups if they were like pulled over here they would withdraw electron density from the uh from the alken double bond and they would make this partially positive compared to its already partially negative state so now whenever we do this uh we see that we have Dina files so in this case we want more electron withd drawing groups so let's take a look so our first one we have one electron withd Drawing Group this oxygen is going to take some electron density from the a double bond and we have an electron donating group so the carbons are considered electron donating groups in this one we have two uh we have two um carboxy groups so we have two electron with drawing groups [Music] AWG AWG and lastly we have one carboxy group and this ether is a electron donating group because it has lone pairs that I can donate to the onto this carbon so this those pairs it is a electron donating group so now we have labeled all the groups that we have across our ding of Fes and now we can uh see which one is the most reactive so we know we want more electron withdrawing groups for Diop file so immediately this one has one electron withdrawing group this one has two electron withd drawing groups and this one only has one so the one with two electron withd drawing groups is going to be our most reactive then we have to consider which one is going to be a better electron donating group the carbon or the oxygen so the oxygen has lone pairs to give right so the O so the oxygen would be considered a better electron donating group so because of that because we don't like electron donating groups and Diop files our carbon is going to be the second most reactive and our ether is going to be our least reactive okay finally we're on to the last fax question um this one it is ask asking us about uh the spectroscopy about NMR so let's see so our first question is asking us for how many signals that they are going to be having so for that we're going to be counting all the all of the um unique carbons so in this case we have one and we have another methyl group right here right beside it this is also going to be called one so both of these methyl groups are considered the same groups 2 3 4 five 6 7 and eight and remember for alkenes because these cannot move around these are like stuck in place this hydrogen right here will never be in the spot of this hydrogen hydrogen 7 will never be in the spot of hydrogen 8 which means a h hydrogen 7 will always be across from a huge group while hydrogen 6 or hyd sorry hydrogen 8 will only be beside hydrogen 6 so that's why whenever we have just single bonds like CH the c h h h two an R all of these hydrogens are the same because this Bond can move around but these Al um these alkine bonds are not able to do that they're stuck in place so that is the reason so we have eight different signals so for our first one we can do signal number eight now we want to find the multiplicity of hydrogen a so on hydrogen a we want to look at every single hydrogen that's attached the every single carbon that that's attached to the carbon the hydrogen a is on so we're going to look at this carbon and this carbon as well as this carbon so we have three different carbons to look at so one of carbon uh carbon 3 has has two hydrogens on it so we can count that as A2 + one and then next uh we have two of the same groups and to calculate this we wouldn't have uh two different 3 + 1es would because they're the same groups what we're going to instead do is 3 + 3 + 1 so because they're the same groups they're going to go under the same parentheses and whenever we calculate this out this is going to be 21 so our multiplicity of ha is going to be 21 our multiplicity of hydrogen B which is is on carbon 5 this hydrogen is only looking at this carbon which is bonded to one hydrogen because of that we're just going to do a 1 + 1 and that gives us the multiplicity of two and finally we have the multiplicity of carbon C so if we look at Carbon C uh this carbon is bonded to three hydrogens so whenever we find the multiplicity of a carbon we only going to look at the hydrogens that are explicitly attached to that carbon so now we can just find that and now we have uh 3 + 1 and that is going to be four let me make sure that my mic is unmuted real quick um give me second yes it's unmuted okay perfect okay and now we have the multiplicity of carbon C which is four okay on to the reaction section so on the reaction section we're going to start with this first reaction br2 H2 this is a mark addition of O and and BR so the O is going to be the mark addition and the BR is going to go in place of where the least subbed area is so the first one we're going to have the O go onto this Bond and the bromine onto this so what we have is O and BR like this in this configuration now we have NaOH this is intr molecular Williamson and what this does is that whenever we have a whenever we have a structure like this where the um where the hydroxy group is one away from the haly is going to form sorry it is going to form an epoxide so for our next step we have an epoxide this and the Broman leaves so my battery dying okay now we have our step HCL ether so HCL ether this is an acid and what we know about acid is that whenever we have epoxide they attack the most subside so now what we can do is that we have the acid attacking this carbon which breaks off this Bond and then this oxygen is going to become an O minus and attack that hydrogen and become proteinated so now for our third step our product will look a little bit like this see [Music] l o just like that right 1 two 3 4 1 two 3 4 okay perfect and our last step we have PCC and ch2 cl2 this forms um alcohols into ketones or alahh it's important to remember that PCC only makes ketones and alahh highes and it is unable to make carboxylic acids so for our last step we have our aldah because this is a uh ter primary sorry primary alcohol so it's going to become an alahh next um we have two double bonds that can attack this h3o+ and we have a major and minor so for this one we need to figure out which one is going to have the um which one is going to be more reactive this double bond or this double bond so in order to find out let's react them with both and see how it turns out so I'll label this a if this double bond attacks and I'll label the other one B so for a a this double bond was to attack the h3o+ then we would have a structure that looked like this so the hydrogen would get bonded here and I'm choosing the structure like this even though these are both secondary even though these are both secondary positions I'm choosing the structure like this this because uh either way we would like some resonance so if we were to put the positive charge here it would obviously not uh have any resonance and so it wouldn't it wouldn't make sense to put it there so because we can have resonance with both then we would try to at least have resonance we would try to make a product that would have resonance so we have our structure like this so this is a secondary uh with resonance and then we have our for for if we had B this carbon attacked the this carbon attacked then we would have a oh sorry I misre so the hydrogen would go onto this carbon and positive charge would go on the tertiary so this one is going to be tertiary with resonance and because of that um we know that a tertiary carbocation is more stable and so even though they both have resonance the tertiary is going to be the winner so now what we can do is we can draw this tertiary structure so we have the tertiary structure but we do have resonance so let's draw the resonance contributors so now we have two different alkanes and we see that we have a 50° C so this is going to be in a hot temperature so in this case whenever we have hot we want a more substituted Aline will be the major product and if we have cold it's going to be the less subed Aline and usually if you had cold uh doctor Bean would put like 0° Cel so there would be like no confusion so if we had 0° Cel we would have the least subd Aline be the major product and if we had a hot we would have the more subd aline so whenever we look at this we have two subs on this alken and we have three on this alken so because we want more subs we want this one to be our major product so on this plus charge we can put an O group so oh and on this on this first one we can put um the O group onto the that tertiary carbon for the minor structure and we are on to the third the third question so our first step is hi EXs Heat this is an sn2 like reaction and remember uh sn2 does not react with SP2 so no SP2 so whenever we look at this um we have two potential things that we can do so we know that this ethyl is not ethyl sorry ether is a good leaving group so this is going to be where we react so we have the choice to either react here or here to cleave one of those bonds but remember since we have an sn2 leg reaction what we're going to have is we're going to have an iodine at one of these places so let's remember that because there's no SP2 on sn2 this iodine will not be able to form here it's not going to be able to form because this is an SP2 carbon this carbon is sp SP2 right whereas this carbon is an sp3 hybridized carbon SP p p p so that is an sp3 carbon so because of that we're going to be CLE this Bond and leaving this oxygen attached to the Benzene ring so the product from reaction one is going to be going to have our benzine ring the oxygen is going to be protonated from this hydrogen right here that the iodine is attached to and we're going to have our iodine on carbon 3 of four one 3 4 it's going to be on carbon 3 so we have our iodine right there next we have ch3oh and Heat this is an sn1 reaction so now what we can do with sn1 is that we can just react it with the iodine and that will replace it with the oc3 group so now for step two we have our Benzene ring oh our four carbon chain and now this is going to become an O3 uh finally we have NaOH so this is another intr molecular Williamson but this one is going to work a little bit differently so remember prior we had an inmolecular Williamson where the there the leaving group was almost was a visal to the um o in this case it is one two three four five and six carbons away so because of that um inmolecular Williamson does not make ethers so if we were to make a react over here it would not create an ether it would create an ether and so we can't have that with intermolecular Williamson so with Williamson we can only make AOC Ides so because of that what this does instead is that it de perinates the oxygen so it's going to become an O minus so our secondary product our third or the product from the third reaction is going to be an O minus and then the O3 is going to stay intact and lastly ch3br that is just an sn2 reaction U using this as the base and then ch3br as the um as the recipient I suppose or as the halide sorry yeah the methal halide and so our step four we're going to have our Benzene ring r o and then this is going to be having a ch3 on it from the ch3br and then our ether moving on uh we have for our fourth reaction n2150 and then and then H+ uh this creates a terminal Aline alkine sorry not alken so we have the terminal alkine so what that means is that the alkine will be on one of these two ends it wouldn't matter which because they're both the same but uh let's draw it out so our first product is going to be our terminal alkine so this is a five carbon chain so we're going to have five carbons in this chain as well one two 3 four five so we have a five carbon chain we have a five carbon chain Right Here and Now secondly hso4 h204 H2O this forms uh this creates ketones from alkine so it's Al alkine to Ketone so remember ketones have one R Group on either side of them so this is what a ketone looks like so uh because of that we need at least one one R Group on either side so we need a carbon on either side if we were to put the alkine uh for example or not the alkine if we were to put the carboxy group on this carbon this carbon instead we would have an aldah and this reaction does not make alahh so because of that we're going to be making it on this carbon wow we're going to be making it on this carbon right here so for our second step we're going to have it on the we're going to have the carboxy group on the second carbon so carbonal group sorry not carboxy so o 1 2 3 4 5 so it's on the second carbon and we have created a ketone and now finally uh this is an Al calide then h3o+ uh this works similar to grard in that it attacks The partially positive carbon that the oxygen is attached to and then breaks the bond off the oxygen becomes negatively charged and is then protonated by the h3o+ so the reaction so the product of this reaction is going to be the same five carbons this oxygen is going to become an alcohol because it was protonated and now we have our aide that is Bond bound to that same carbon so we have our C Tri Bond C- C H3 next um we have an Esther this is the the structure of an Esther and we have for our first reaction a strong reducing agent so what this reducing agent is going to do is it's going to reduce both of these oxygens and cleave this Bond because it is an Esther so whenever we do that our first um the product from our first reaction is going to cleave both cleave this Bond and make both of these into alcohols so it's going to bind in h to them so next um we're going to have two different molecules well not two different molecules I guess I'm spoiling it so we have a 1 2 3 so three carbons and then the o for on this side right here then on this side we have three carbons again and then an O so we have two of the same thing so two three carbon o so uh for this reaction U whenever you are putting the the product it's not necessary for you to put both uh you only need to show one um so only one is enough because they are the same thing and moving forward we're only going to be using one of these um three carbon uh alcohols so we're only going to be using one of these for step two and we're going to continue that throughout the rest of the um drawings I suppose so next we have a strong oxidizing agent this is sodium d chromate so remember sodium dichromate when with a primary alcohol is going to form a carboxylic acid so compared to if PCC was attached to this PCC would make this into an alahh um sodium dichromate is going to take it all the way to carboxilic acid so what that's going to look like is going to be like this uh finally we have right here uh ch3oh H+ this is sorry Fisher esterfication so with fish or esterification you have an alcohol and a carboxilic acid that form an Esther so we're going to go back to an Esther so our our product for step three is going to be the same wait one two three it's on the third carbon and then we're going to have our oxygen once again and then this is going to be bound to a ch3 o C3 so we have an Esther once again and lastly for our fourth step we have grenard and so this grenard is going to be a little bit special so whenever it attacks this way this oxygen is going to gain the negative charge and you might think that it is going to immediately get reprotonated but that is not the case this negative charge is actually um going to bounce back because this um this oc3 is a good leaving group so it's going to bounce back and then the oc3 is going to leave and then because we have another Ketone so whenever it bounces back we're going to have a structure that looks like this because we're going to have another Ketone we're going to have this attack once again and this time whenever it gets the minus charge it won't be able to attack again so so it will be protonated by the h3o+ so our final product is going to be a um an alcohol with two benzines on it so whoa o and then we can draw our two Benzene rings and that will be our final product and now for our last reaction we have a Deal's Alder reaction so in deals Alder what we have what we want is a Dio file that looks like this so to do that we're going to make this axis of symmetry right here and we're going to flip these over so now what we're going to do we're just going to draw this down here we're going to have our double bond have our oc3 and our methal group right here ch3 and this is going to be added to uh a Diop file remember the Diop file is going to attack one of these two so in this case um what we're going to do we're just going to rotate it around so that it's easier to look at and now finally we can do our reaction whenever this reacts right here we're going to have is the uh cyclohexene so our product is going to look like this with a little bit of thinking that will be going into the product so whenever we make the product we have to also put our groups on it so our methyl group is going to be on the top part of where the um where the alkine was so our methyl group can go right here then we have our oc3 this o 3 um whenever we have a dine we can have two positions we can have the oc3 right here or the O3 right here so this position is called the out position and this position is called the in position so whenever we have out we're going to put the oc3 on a dash and whenever we have in we're going to put the O3 on a wedge so because of that we're going to put our o 3 on a dash so right here put r o C3 and now um we we're almost done but we do have to figure out where this electron withd Drawing Group is going to go so when we have an electron with Drawing Group we want it one two away from the um from the electron donating group so we have this electron donating group and we want it one two away we want one two or 14 so what that means is that whenever we have something we want it to be on one carbon two carbons away or 1 2 3 and four so that means that it would either have to be on this carbon or this carbon so because of that the only place that the um that the uh car carbonal group can go is going to be on this bottom carbon right here so we're going to put our car right there and one last thing uh because we have to follow the Endo rule for uh deals Alder uh to show that we can what we can do is we can put dashes right here but that just sometimes makes it a little bit more difficult to see so what we can do instead is just put hydrogens on wedges right here in the middle so now we have our final product moving on to the mechanism so on this mechanism we have a cyclohexane that's formed and then an and then an ether ring looks a little bit strange but uh we'll see how we can do it so whenever we have uh this we have this oxygen uh this oxygen just seems like it's looking to be protein ated so let's make it attack in a hydrogen of the h3o+ so so now our first uh our first drawing is going to be like this like this uh now that we have a positive uh charge we can make a reaction so this is a partially negative uh double bond and this is a partially positive carbon so this partially negative double bond can actually attack this carbon but it can also attack this carbon so we have two choices uh we know that since it's acidic it's going to attack the more subside so it would attack the carbon on the left because that is a more subside but another thing that we need to remember is that we are making a cyclohexane so we need to make the product so if we look at this um this would create a bond with this carbon right here so 1 2 3 four five and six so that perfectly lines up so that will be the bond that is created this one um this um this Aline is going to react with that um that carbon and then this Bond on this oxygen is going to break off so we have a r we have a ring closing reaction so our next step is going to be just drawing the cyclohexane ring and we can just number it one two three four five and six and we can put the groups so one has no groups two has a methyl group on it three has no groups four has no groups five has no groups six has a methyl group and then a here uh six has a methyl group and then because this Bond was cleaved it has this ch2 oh so we we have ch2 o and now for our charge uh carbon 1 gained lost a bond with carbon 2 but it gained a bond with carbon six so carbon one is net neutral Carbon 2 only lost a bond with carbon 1 so Carbon 2 is going to be positive so this is going to be a positive charge now what we can do is that we can have this oxygen the lone pair ATT this positive charge and what that's going to do is that it's going to form an ether similar to what we see right here so if we did that then we would have our cyclohexane with the methyl group actually I'm going to put the methyl group facing a little bit away with our methyl group right here and then our bond to the oxygen that's bonded to the hydrogen which now has positive charge because it is bound to two things it is bound to carbon six right here which has another methyl group onto it so we are very close to the final product our one last step that we need to do is deprotonate the oxygen so how the way we can do that H2O this oxygen really likes that hydrogen and then the hydrogen breaks its bond with the oxygen and then we have our final product so pretty simple mechanism our next section synthesis so we need to make this using alcohols alkenes alkin of three carbons or less so always remember just make sure that you read the top so from this uh from this one thing we can do is cleave off this and have a grard reagent mg VR right here but if you think about it whenever we go through here we're going to have a br and a vinyl carbon and it's just it's going to get really messy from there so it's not a good idea to do that so one thing we can do is we can make this into an alkine and so the way to do that is that we can use H2 and the enlarge catalyst so H2 PD and lenars so that's also called quinoline so you can put either of those on your exam linar Catalyst or koline and they'll both be accepted now we have a structure that looks like this is that it so this is the triple bond and then it has one two three one two perfect okay so oh right here so this is our structure and now what we can do with this structure is that we can do what we did before we can cleave that Bond and we can put then h3o+ and now we'll have have our grenard reagents so what we can do here is we can call this oxygen this um this alcohol is going to become a ketone and that's going to be bound to an aide so not exactly greenyard but so the aide with the um triple bond so one thing we can do with the aide so to form an aide we need a reaction n2150 and we have seen this reaction before but we saw it with hydrogen and what that does um what that does is that it actually protonates the aide so if we didn't put the hydrogen then it would just just be de just a aite and ion so now what we can do here is just have a regular alkine so we have our regular alkine and now one thing that the next step that we can do here sorry I'm getting call give me two seconds okay so the next thing to do here is to um cleave this Bond right here and we can use this as a sn2 reaction so what we can have now is oh AR was pointing the wrong way sorry about that you can have the [Music] CH trond C or not CH I suppose it has to be a cide so c h triple bond C minus n A+ oh and also we would have to put an A+ over here too n A+ and that is going to be bound to this wrong structure so it be um it would react with that and that would be a sn1 reaction sn2 reaction and then to get a cide remember we just need n2150 and and nh250 leaves our triple bond ch ch and we like Alin with um three carbon so that is perfect and now next um what we can do is we can make this into an alcohol using PCC this Ketone uh remember uh alcohols can become ketones through PCC and honestly you can use PCC or you can use sodium dichromate r27 you can use either of those you can also use Jones Jones reagent and that will all create um from a secondary from a secondary alcohol which is important to note secondary alcohol into a ketone so all that can be used to form a secondary alcohol into a ketone these are both very strong so if you used it with a primary alcohol it would actually form a carboxilic acid so make sure only to use that whenever you have a ketone like this if you want to otherwise just PCC for ketones and alahh is just a good idea um so now we have an alcohol of three carbons or less and that's exactly what we like so we can Circle that that is one of our starting materials and our last step this BR what we can do for this BR we can just use the reaction pbr3 and that is going to turn that BR into an O for us and you know what we can do with o greenard so then h3o+ that Cleaves it and now we have a form alahh bonded to the greenard reagent mgbr this uh Al form alahh for this one you need to use PCC if you use Jones or if you use Jones or um sodium dichromate you uh this will be counted wrong so you need to use PCC whenever you have an alahh so now this is just going to become ch3 methanol which is a valid starting material now we have mgbr we can use mg/ ether that'll add our mg so we can now have BR only and to get that brr off what we can do this time is pbr3 again and what that's going to do is make our o and now we have a now we have a three carbon uh alcohol which is a valid starting material and that is our synthesis section moving on on to spectroscopy so our first step with spectroscopy is to look at this so we have C9 h122 so we have to calculate our unsaturation number so unset is going to be calculated by 2 C + 2 - H + nitrogens all divid by 2 so we don't have any nitrogens right now but in the future you guys will have nitrogens so or you guys probably will have nitrogen so it's important to remember that nitrogen's add to the unsaturation number so whenever we have whenever we have this to calculate the unsaturation number we can just plug everything in so 18 + 2 - 12 all / 2 that gives us an un saturation number of four so whenever we have an unsaturation number of four or more remember that that means that we have a Benzene ring so four means Benzene ring and now U moving to the spectroscopy you see this like very long Peak right here uh just around this this is going to be our o stretch so we have two oxygens to take note of and one of them is going to be an alcohol so we know we'll just put what we know right here so we know it's a Benzene we know we have an alcohol that's pretty much it so far so now what we can do is whenever we look at this structure this is called the doublet doublet structure and what that means is that we have a d Paras substituted Benzene so we have a Benzene that is substitut add both ends one and four and now if we keep moving we have remnants of the actual exam review um we have uh three hydrogen looking at uh Three Peaks which means it is looking at two hydrogens so we have ch3 ch2 this gives us uh ch3 bonded to ch2 two and it's this carbon and then over here we have one H we're going to come back to that we have two h's right here and this is bound to one two three so this has Four Peaks so it's bed three so this is going to be a ch2 b to a ch3 and it is somewhat downfield so this could be ether so let's just say o question mark and now finally we have ch2 and this ch2 is very downfill as well so we have a ch2 bound to oxygen that has or a carbon ch2 bound to a carbon but the thing is this is so downfield that we would assume that it is actually just oxygen so this one is going to be considered an oxygen and it has no carbons on this side either so no hydrogens I mean so it has nothing on this end so this side has no hydrogens well this side could have hydrogens actually it's not um Ultra Pure so it won't give any signal from this side but this side for sure it also won't give any signal so this ch2 is all alone with no hydrogen adjacent to it on either end well yet so now we have all of our hints pretty much more remnants um now we have all of our hints so now what we can do is we can start figuring this out so one last thing that we have right here is we have something slight downfield 1 H so we know that we have an alcohol so this is going to be our o Peak that is looking at this hydrogen is looking at this oxygen which is why it has this one Peak so we have our o so what we can do to draw this out is that we can just have our Benzene ring and we can start by using what we know so we know we have a ch2 ch3 and we know that ch2 ch3 is bound to an O because of how downfield it is so what we can do on one end we know it has two subs so we can do o ch2 ch3 and now on the other end we know we have uh nine Hydro nine carbons 12 hydrogens six uh two 12 hydrogen and two oxygen so if we count the number of carbons that we've used out of nine we have six in the Benzene ring and two in this ethoxy group so that means we only have one carbon to play around with and perfectly we have one carbon right here that doesn't have any Bonds on either that doesn't have any carbons bonded to it and it has an oxygen bonded to it which is exactly what we need so we can put our ch2 right here and that is going to be our final structure so if we had to guess and not guess and check but if we had to check our work then we could look at this ch2 this ch2 is only going to give a signal on this side and this carbon has no hydrogens that's exactly what we see and it's downfield perfectly where this o is so that is that makes sense our dou dou structure is fine right here then we have our ch2 that's bound to ch3 so we have a ch2 bound to a ch3 that is also quite down that is downfield which is why it's bound to this uh oxygen right here so it's an ether then moving on we have our o um o Peak right here and is somewhat downfield and then we have a ch3 ch2 usually these are going to come at around 1.2 to like8 PPM so it being around over here is somewhat like not a not the best indicator but it's somewhat of an indicator that it's going to be de D shielded a little bit so this ch3 ch2 slightly deshielded it makes sense that it would be right here so that would [Music] be our final uh answer for the spectroscopy question D cuz I can't have that circle around it just put it [Music] off but uh yeah that was our spring 2022 exam review um I wish you guys guys the best best of luck on the test I know you guys are going to do great so yeah um I'm sorry that this came out a little bit late uh I'm going to try to make sure that it doesn't happen again so yeah uh I appreciate appreciate uh you guys you know watching the video and everything so have a have a good day and good luck on your exam yep