Organic Chemistry 2 Practice Tests: Multiple Choice Review

welcome to my organic chemistry 2 multiple choice practice tests so let&#39;s jump right into it number one what is the major product in the following reaction so what do you think the answer is to this problem so notice that we have an aldehyde functional group and we have silver oxide whenever you have the ag plus ion it could be in the form of silver oxide or it can be written like this it can be complex to ammonia these reagents is associated with the tolerance reagent and it can oxidize an aldehyde to a carboxylic acid it&#39;s very specific for aldehyde functional groups and even alpha hydroxy ketones as well but for the most part they react with aldehydes so a is not an aldehyde it&#39;s out c is a ketone so it won&#39;t work um if we had an o h here where an alcohol on the alpha carbon then it would work that&#39;s the alpha hydroxy ketone but since it&#39;s not there c is not the correct answer so now we&#39;re between b and d b is the protonated form of the carboxylic acid d is the deprotonated form of it so to distinguish between b and d you need to look at step two which is h3o plus under acidic conditions the carboxylic acid will be in its pronated form so the hydrogen should be here so the answer is b number two complete the reaction by the way for these questions feel free to try the problem yourself pause the video and then unpause it when you&#39;re ready to see what the answer is so let&#39;s begin we have one etho 4 isopropyl benzene and we&#39;re gonna mix it with bromine in the presence of ultraviolet light so this is going to be a radical bromination reaction now we can eliminate answer choice b and d because this reagent will not put a bromine atom on the benzene ring the only way to do that is to use br2 in the presence of a lewis acid catalyst like febr3 so since we don&#39;t have that answer choice b and d is out so what this reaction will do radical bromination it&#39;s going to replace the benzylic hydrogen which could be the hydrogens in purple or in green now we have both answers the answer choice a we replace the purple hydrogen and for answer choice c we replace the green hydrogen so the question is which one is better what type of carbon do we have here notice that that carbon is connected to three other carbon atoms the two methyls and this carbon so it&#39;s tertiary and this carbon here is a secondary carbon because it&#39;s attached to two other carbon atoms so which radical is more stable a secondary radical or tertiary radical because tertiary radicals are more stable the bromine will selectively abstract the tertiary hydrogen to create a tertiary radical so therefore answer choice c is the correct answer let&#39;s review some reactions associated with bromine so let&#39;s say if you have an alkane and you use br2 with hv by the way this is the same it has the same effect as using mbs it&#39;s a radical bromination reaction now what&#39;s going to happen is bromine will selectively replace the tertiary hydrogen so it&#39;s just going to add a br to the tertiary carbon now let&#39;s say if we use nbs on pentane so here we have a primary hydrogen and here we have a secondary hydrogen this is another radical bromination reaction it&#39;s going to replace the highest substituted hydrogen in this case the secondary one rather than the primary one so we&#39;re going to get this as an answer now i can also replace this one as well because both of these are secondary and they lead to different products so the bromine atom is going to go on the most substituted carbon for this example the most substituted carbon is a tertiary one and for this one it&#39;s the secondary carbon so that&#39;s what happens if you add br2 with hb hv or mbs to an alkane now let&#39;s say if you add it to an alkene the bromine under radical conditions will not react with the double bond what&#39;s going to happen is it&#39;s going to replace the allylic hydrogen the one that&#39;s one carbon away from the double bond so you&#39;re going to get this as an answer and if you have a benzene ring it&#39;s going to replace the benzylic hydrogen so basically the bromine atom is going to go one carbon away from the benzene ring which is kind of the same as this situation where it&#39;s one carbon away from a double bond so the bromine atom is going to go it&#39;s going to replace the hydrogen on the benzylic carbon which is one away from the benzene ring now let&#39;s say if you have an alkene and let&#39;s say if you add br2 in the presence of dichloromethane this is not a radical reaction in this situation the bromine reacts with the double bond and a double bond is going to disappear so it&#39;s going to add two bromine atoms anti-addition now you also get the enantiomer which looks like this so you get a mixture of two products now if you have a benzene ring um br2 by itself will not react with the benzene ring if you want to add the br directly to the benzene ring you need to add a catalyst like fvbr3 so what&#39;s going to happen is it&#39;s going to replace the hydrogen with a bromine atom and so you get this as a product now let&#39;s say if you have an alkene if you add hbr it&#39;s going to react with the double bond and it&#39;s going to put a bromine on the more substituted carbon which is the secondary carbon and the hydrogen will go on the less substituted carbon now let&#39;s say if you add hpr and peroxides it can be written as h2o2 or ror in either case there&#39;s two oxygen atoms that are single bonded to each other and that&#39;s a peroxide so what&#39;s going to happen is the bromine is actually going to go on the less substituted carbon so we get the anti more carbon carb reaction rather than the markovnikov reaction with just regular hbr so just make sure you&#39;re familiar with those reactions um so you might see bromine in those different forms and just keep that in mind for your next organic chemistry final exam number three which compound has a proton with the lowest pka so let&#39;s compare different compounds let&#39;s start with the alpha hydrogen for the compound and answer choice c and an answer choice b which of those two hydrogens is more acidic well whenever you have a alpha hydrogen between two carbonyl groups the pka is around nine for a hydrogen next to one ketone it&#39;s around roughly around i believe 20. now this hydrogen here between the ketone and the ester is 11. the reason for that is because of this electron donating group because the oet group the oxygen can donate a pair of electrons to one of the carbonyl groups it makes that this hydrogen slightly less acidic so the pka goes up a little anytime you add an electron donating group to an acid the pka goes up it becomes less acidic if you add an electron withdrawing group to an acid the pka will go down it becomes more acidic now the most acidic hydrogen for the carboxylic acid is this one here the pka for a typical carboxylic acid is around four to five so um the carboxylic acid is much more acidic than compound b or compound c so we can eliminate b and c now the reason for this is if you notice the trend on a periodic table let&#39;s say this is carbon nitrogen oxygen fluorine chlorine bromine and so forth acid strength increases as you go to the right and as you go down so when comparing a ch bond versus an o h bond usually 99 of the time the o h bond is more acidic so here this is a carbon hydrogen bond the pka is 11. here this is an oxygen hydrogen bond it&#39;s much more acidic four to five but keep in mind these are general trends though so you might see an exception or two now this is a protonated carboxylic acid so this is even more acidic the pka of a pronated carboxylic acid is somewhere in the negatives it can be like negative four negative five but it&#39;s much less than positive four to five so the correct answer is d whenever you see a hydrogen that&#39;s attached to an atom with a positive charge it&#39;s very acidic for example the difference between h2o and h3o plus h2o has a pka of 15.7 it&#39;s neutral but h2o plus the pka is around like negative three or negative four as you can see that extra hydrogen makes it a lot more acidic number four which structure is most consistent with the following ir spectrum now the first thing you want to look at is this huge bulky peak around 25 to 3300 you can&#39;t miss it when you see a massive very strong very broad signal and in this region you have a carboxylic acid this is the oh stretch of a carboxylic acid so we can eliminate answer choice c and answer choice d because they&#39;re not carboxylic acids the next important signal that we have here is the carbonyl group at 1700 that&#39;s also part of the carboxylic acid functional group now the ketone would also have that signal at 1700 but the ketone wouldn&#39;t have this very broad signal now granted we do have an alcohol and an alcohol would generate a broad signal but it wouldn&#39;t be around 25 to 3300 this signal would be more like 33 to 3500 and it&#39;s not as broad as a carboxylic acid so if i can draw a rough sketch an alcohol oh stretch would look something like this it would be in that region it wouldn&#39;t cross over past three thousand it wouldn&#39;t go to the right of three thousand so we can eliminate d another reason why we can eliminate answer choice d is um this signal here that signal which is roughly around 1200 it&#39;s a co stretch but not for an alcohol but for a carboxylic acid there&#39;s two types of co stretches there&#39;s the co stretch of an alcohol and ether where this is a an sp3 carbon that&#39;s around usually a thousand to eleven fifty so we&#39;ll say like eleven hundred and then there&#39;s a co stretch um that&#39;s next where the carbon is sp2 hybridized and that&#39;s this carbon the carbon from a carboxylic acid that&#39;s sp2 hybridized so that co bond it&#39;s a little bit more than eleven hundred it&#39;s like twelve hundred to thirteen hundred so this signal which is roughly around 1200 that correlates to a co stretch where it&#39;s an sp2 hybridized carbon rather than an sp3 hybridized carbon so that&#39;s another reason why it&#39;s not the alcohol now there&#39;s another signal that we can look at and that is the c triple bond c at 2200 this will distinguish the right answer from the rest so only answer choice b has a carboxylic acid with an alkyne so that&#39;s sigma 2200 it could be a c triple bond c or even a nitrile a c triple bond n but we don&#39;t have a nitrile so we can eliminate a because we don&#39;t have an alkyne we have an alkene and an alkene we should see a signal somewhere between around 1600 for a seed of 1c so the correct answer for this problem is answer choice b we have the oh stretch of a carboxylic acid the carbonyl stretch and the alkyne and also the co stretch for carboxylic acid but to the right of 1500 is the fingerprint region so this region is not always reliable so you can use that as a last resort number five which set of reagents will produce para nitro benzoic acid from benzene with the greatest yield so go ahead and take a minute and see if you can get the right answer so i&#39;m going to go through every choice but first let me show you how you can quickly get the right answer the first thing you want to keep in mind is that we need uh para we need the no2 and the carboxylic acid group to be para with respect to each other so we know we can eliminate antichoice b because once you put the no2 on the ring it&#39;s going to tell the methyl group to go in the meta position which we don&#39;t want so b is out now looking at antichoice c we could eliminate that as well once we add the methyl group it&#39;s an ortho para director but um notice that in step two when you oxidize it with potassium permanganate it&#39;s going to convert it to a carboxylic acid which is a meta director and that&#39;s going to tell the no2 group to go into meta position so c is out now the only issue with d is once you add the no2 group in the first step and once you reduce it with tin and hydrochloric acid it turns into an nh2 group now granted an nh2 group is an orthopedic director however once you add a lewis acid catalyst you&#39;re going to deactivate the nh2 group and it becomes a meta director and so step three won&#39;t even work in the presence of an anti-street group the the frito crafts alkylation doesn&#39;t work well in the presence of an amine but a is the best answer so let&#39;s go through each one i&#39;m going to go over a last but let&#39;s start with answer choice b so starting with benzene so once we add a nitric acid and sulfuric acid we&#39;re going to replace a hydrogen with an no2 group so this is called nitration and then we&#39;re going to add methylchloride with an aluminum chloride acid catalyst now the nitro group is a meta director so it&#39;s going to tell the methyl group to go here so we&#39;re going to put the ch3 on this side and then once you add potassium permanganate with h3o plus it&#39;s going to oxidize this group and it&#39;s going to convert it to a carboxylic acid so we end up getting meta nitrobenzoic acid rather than pera so that&#39;s why we can eliminate answer choice b so now let&#39;s see what we get for answer choice c so starting with benzene we&#39;re going to add methyl chloride first as soon as we add the methyl group um what we now have is called toluene that&#39;s the benzene ring combined with the ch3 and then once we oxidize it with potassium permanganate in atrial plus it&#39;s going to convert into a carboxylic acid and then once we add nitric acid and sulfuric acid um the carboxylic acid is going to direct the no2 group to go to the meta position so we actually get the same answer that is meta nitro benzoic acid so now let&#39;s consider answer choice d so step one is nitration and then following step one we have reduction so you can use any active metal so you can use tin with hydrochloric acid you can use uh zinc metal with hydrochloric acid you can even use um iron metal with hydrochloric acid and what&#39;s going to happen is it&#39;s going to convert the nitro group to an aiming group so when you have an nh2 and a benzene ring the name is called aniline so now if we add methylchloride with the aluminum chloride catalyst notice what happens aniline has a lone here it&#39;s actually quite nucleophilic so what&#39;s going to happen is it&#39;s going to combine with the lewis acid catalyst so the methyl won&#39;t even go on the ring so now the nitrogen is bonded to the aluminum chloride and it also has three hydrogen atoms i mean two hydrogen atoms so when nitrogen has four bonds it now has a plus charge when when you have a nitrogen for lone pair it is an ortho para activator but now when it has a plus charge it is a meta direct and deactivated so if we mix a one-to-one ratio with aniline and methyl chloride with this catalyst the ring is deactivated and the catalyst binds to the nitrogen so it won&#39;t help the ring to react with the methyl chloride so the reaction stops here so there&#39;s no point in atom potassium permanganate because there&#39;s no methyl groups to oxidize so just keep in mind that the frito-craps alkylation reaction and the acellation reaction does not work in the presence of an nh2 group on a benzene ring so now let&#39;s look at our correct answer which is answer choice a so the first step um is photocraft&#39;s alkylation the catalyst aluminum chloride helps to remove the chlorine from the methyl group and so that gives us a toluene now in the next step we have nitration once you add toluene the ring is slightly activated alkyl groups are weakly activating groups so the ring is a little bit more reactive than before so once we have the no2 group um the methyl group will direct it ortho or para now we&#39;re going to write the para answer because that&#39;s the one that&#39;s going to lead us to our product and then once we add potassium permanganate the methyl group will be oxidized to a carboxylic acid and that&#39;s the answer that we want so thus we have para nitro benzoic acid by the way if you want more practice problems click a link that is in the description section of this video and you can get access to more practice problems to prepare for your exam the video that you&#39;re currently watching only has 15 free practice problems but the paid version includes those 15 practice problems in addition to another set of 85 questions for a total of 100 uh practice problems so if you think you might need more questions to work on to prepare for your exam you can check out that video and let me know if you have any questions number six which of the following reagents will carry out the reaction shown below so we have an acid chloride and we wish to convert it to an ester so let&#39;s look at each of our choices a is a granted reagent if you add a grain of reagent to an acid chloride it&#39;s going to add two r groups and convert the acid chloride into an alcohol since we don&#39;t have an alcohol a is out now b is called the gilman reagent or the lithium dialkyl cuprate reagent and if you put that with an acid chloride it&#39;s going to add one r group and convert the acid chloride into a ketone which is not an ester so answer choice b is not the correct one c is an alcohol an alcohol plus an acid chloride will make an ester so uh c is the right answer and d d is an ester if you put an s with an acid chloride i&#39;ve never seen it make another ester so we&#39;re going to eliminate ant&#39;s choice d now if you understand how these reactions work already feel free to fast forward to question seven but if you want to see the mechanism uh feel free to continue watching so let&#39;s start with the granite reagent the carbon that&#39;s attached to the magnesium is partially negative and it&#39;s attracted to this partially positive carbon and so we&#39;re going to create a new carbon-carbon bond as it attacks the carbonyl carbon the pi bond breaks and we&#39;re going to have this a tetrahedral intermediate so what we need to do is add the ethyl group to this carbon now the magnesium which has a plus charge it forms an ionic bond with the the negatively charged oxygen so what&#39;s going to happen now is this tetrahedral intermediate is going to collapse and chlorine the leaving group will leave as chloride and so right now we have ketone but the grenade reaction does not stop there the grenade reagent will react with the ketone as well so another uh griden molecule will attack the carbonyl carbon and then the pi bonds gonna break so this time there are no leaving groups so this is going to be our answer because i really didn&#39;t add step two but step two usually you would add a h3o plus once he had h3o plus then it converts into an alcohol so it&#39;s going to look like this a tertiary alcohol this oxygen is going to grab a proton from h3o plus so now for the second reaction that is between the acid chloride and the gilman reagent so we have a copper atom attached to two ethyl groups and copper has a negative formal charge because it&#39;s in the plus one oxidation state and lithium has a plus charge so only one of the r groups is nucleophilic and so this r group attacks the carbonyl carbon and this part is going to be very similar to the first step of the the mechanism of the grinded reagent with the acid chloride so we&#39;re going to get a tetrahedral intermediate and we&#39;re just going to add only one r group so right now the lithium ion is with the oxygen with the negative charge the copper now has only one r group and so it looks like this and so it&#39;s not going to react anymore so we&#39;re not going to worry about it so this oxygen is going to reform a double bond and expel the leaving group so if you choose to use the gammon reagent it stops at the ketone level so it&#39;s very good for converting esters and acid chlorides into ketones by the way if you put the ester with the given reagent the mechanism is exactly the same the only difference is instead of a chlorine you&#39;ll have like in a methoxy group in och3 okay so now let&#39;s consider the mechanism with the acid chloride and the alcohol so the alcohol will attack the carbonyl carbon and just like before we&#39;re going to get a tetrahedral intermediate but this time instead of having an ethyl group we have an ethoxy group an och2ch3 group because antichoice b is very close to instachoice c the only difference is it differs by an oxygen and just like before the leaving group is going to leave so as you can see the mechanism is very similar and now we have our desired product which is an ester and so that&#39;s the mechanism for the conversion of an acid chloride into an ester number seven complete the reaction sequence so we have cyclohexanol reacting with pcc then we have some other reagents as well so what does pc do to an alcohol if it&#39;s a primary alcohol pcc can oxidize it to an aldehyde and it can oxidize secondary alcohols into a ketone so that&#39;s the end result of the first part now what happens if we take cyclohexanone and react it with triphenylphosphine and alkyl halide and butyllithium what do you think the major product will be if you recognize these reagents you&#39;ll know that it&#39;s part of the vitic reaction or some people say the width reaction the width reaction or vedic reaction it converts ketones into alkenes so knowing that fact you can eliminate answer choice a and answer choice b so we&#39;re down between c and d now if you want to quickly get the answer notice where notice the carbon that has the iodine atom that is the carbon that is going to connect to this carbon so it&#39;s the secondary carbon notice how we have relative to that carbon we have an etho on the right and a methyl on the left d is going to be the answer we have a methyl on one side and an ethyl group on the other side let me show you the mechanism for this reaction so the first thing that happens triphenylphosphine reacts with two iodo butane the phosphorus atom is a very good nucleophile and so it attacks the carbon from the back expelling the iodine group so now we have a phosphorus attached to a carbon that has a methyl on one side and an ethyl on the other side but i&#39;m going to redraw it like this this carbon has a methyl group a hydrogen and an ethyl group when phosphorus whenever a phosphorus atom has four bonds it has a plus charge now we also have butyl lithium as well butyl lithium is a very powerful base and butyllithium is used to remove the hydrogen atom on the carbon that&#39;s attached to the uh the phosphorus group and then this bond is going to break it&#39;s going to put a lone pair on the carbon atom so now we have our our illid which looks like this and it has a negative charge now now the illite is stabilized by resonance but i&#39;m going to use this form for the mechanism so once we pair up cyclohexanone with the illidan the carbon with the negative charge is attracted to the carbonyl carbon because it has a positive a partial positive charge so whenever you have a nucleophilic carbon and an electrophilic carbon those two will connect and form a carbon-carbon bond and then this bond is going to break that&#39;s going to open up so now we have this intermediate and then what&#39;s going to happen next is this negatively charged oxygen atom is going to attack the positively charged phosphorus group to form a four-membered ring known as a a beta-end group or betaine group so whenever a bond breaks let&#39;s say a bond between carbon and bromine because bromine is more electronegative than carbon when the bond breaks the electrons will go towards the more electronegative atom so you would get a carbocation and a bromide ion conversion let&#39;s say if you have a carbon hydrogen bond carbon is more electronegative than hydrogen so when that bond breaks you get a carbonyl so understanding that&#39;s going to help us to see how we can write the mechanisms for the next step oxygen is more electronegative than phosphorus so oxygen bears a negative charge phosphorus bears the positive charge this carbon atom is partially positive with respect to the oxygen atom and the carbon that&#39;s attached to phosphorus this carbon is partially negative because carbon is more electronegative than phosphorus carbon is closer to fluorine than phosphorous so when the carbon-oxygen bond breaks those electrons will go towards the more electronegative phosphorus atom i mean the more electronegative oxygen atom and when the carbon phosphorus bond breaks those electrons will go towards the more electronegative carbon atom so as you can see the atom that&#39;s more electronegative pulls the electrons toward itself when the bond breaks so the product of this reaction we get triphenylphosphine oxide and we get a carbon-carbon double bond so as you can see this is the the end product for this reaction so whenever you have a ketone and if you&#39;re using the wittig or the wittig reaction you&#39;re going to get an alkene as your end product so that&#39;s what you want to take from this and make sure you attach the carbonyl carbon to the carbon that had the leaving group the iodine so just for the sake of practice let&#39;s say if we have this ketone and let&#39;s say the reagents are triphenylphosphine three iodo hexane and butyllithium if you quickly want to draw the major product get rid of the oxygen right the double bond and connect it to the carbon that has the iodine so notice that on the left side of that carbon you have an ethyl group and on the right side you have a propyl group so that&#39;s what you want to draw here you want to draw the ethyl group and the propyl group and that is going to be the answer the alkene that forms as a product of the wittig reaction number eight which of the following diene and dienophile will produce the product shown below so this is the diels-alder reaction and the diene is the one that has two double bonds and a dienophile is the electrophile that has just one double bond so this is the dienophile so which answer choices can we eliminate well let&#39;s count the carbon atoms we have one two three four five six seven carbon atoms not including the two nitro groups so we can eliminate choice c because there are six carbon atoms so choice c is eliminated and we could eliminate choice d as well choice d has a total of eight carbons not seven if you don&#39;t count the nitro groups so d is out now if you count the carbons for answer choice a and b they&#39;re both seven so we need to look at something else the fact that we have a bicyclic compound as a product means that we need to start with a ring the diels-alder reaction creates a ring but if we start with a ring we&#39;ll have a total of two rings and so knowing that this will tell you that choice b is the correct answer let&#39;s go over the mechanism real quick for answer choice a and b so for choice a here&#39;s what&#39;s going to happen let&#39;s uh say that this is carbon one two three four and this is five and six the deals out of reaction um it makes six uh carbon rings or i mean to say six membered rings so this double bond is going to the pi electrons in the double bond is going to connect carbon one and six and this double bond will move here and the double bond between carbon stream four uh the pi electrons will be used to connect four and five but let me use a different color so you can see the connections so here are the pi electrons between in the double bond between carbons five and six and here are the pi electrons a double bond between carbons three and four so there&#39;s also a double bond between carbons two and three and we still have a methyl group and notice that the nitrile groups are trans with respect to each other so they&#39;re going to remain that way so as you can see if the diene doesn&#39;t start as a ring you don&#39;t get a bicyclic compound you simply get just a regular monocyclic compound just wondering but if we if the diene was in the ring already we would get a bicyclic compound this will be carbon one two three four five and six and let&#39;s call this other carbon carbon a so here&#39;s a six membered ring drawn differently so this is carbon one two three and four so there&#39;s going to be a double bond between two and three just like we had it here and on five and six we have the two nitro groups but because they&#39;re trans with respect to each other one is going to be in the exo position the other is going to be in the endo position now this carbon carbon a is on top it&#39;s over here so when you start with a ring you&#39;re going to get two rings a bicycle compound if you don&#39;t start with a ring you&#39;ll just get a one ring system you won&#39;t get a bicycle kappa so those are just some basics of the deals auto reaction so make sure you review that and that&#39;s it for this question number nine what is the product of the reaction shown below so we have a diketone sodium hydroxide heat and looking at our answer choices we see that we&#39;re going to form a six membered ring this reaction is the intramolecular aldol reaction sodium hydroxide will be used to remove an alpha hydrogen but there are some mass choices that we can eliminate if you know it&#39;s an aldol reaction there are two types of products the first product is a beta hydroxy aldehyde or ketone what this means is that the oh group should be on the beta carbon so if we look at answer choice a relative to the ketone the first carbon away from it is the alpha carbon and this is a beta carbon so that&#39;s a beta hydroxy ketone but if you look at b this is the alpha carbon that&#39;s the beta carbon this is the gamma carbon the hydroxy group shouldn&#39;t be there so we can eliminate b now the second product that you can get if you add heat heat leads to dehydration loss of water so you&#39;re going to lose a hydrogen and an o h group and you&#39;re going to form a double bond between the two so dehydration will remove the alpha hydrogen and the oh and it&#39;s going to put a double bond right in the middle so that double bond should be between the alpha and beta carbon so free anti-choice d the double bond is between the beta and the gamma carbon notice that it&#39;s not conjugated with this carbonyl group whenever you form a double bond by means of the aldol reaction it should always be conjugated with the ketone or the aldehyde so choice d is out looking at a and c um a would be the product if we didn&#39;t add heat but because we heated the reaction um it&#39;s going to facilitate dehydration so c is the correct answer but now let&#39;s actually write up a mechanism to see how it works so under basic conditions the first step is deprotonation we need to remove the alpha hydrogen and we&#39;re going to get a carbanion now this negatively charged carbanide is going to attack the other carbonyl carbon so this is going to be carbon one two three four five six so we&#39;re going to get a six membered ring so carbon two has a ketone so i&#39;m gonna make this carbon two this is going to be carbon one three four five and six carbon six has a single bonded oxygen atom and instead of having two lone pairs it now has three and a negative charge and there&#39;s a methyl group on carbon 6 as well so notice that hydroxide when it grabbed the hydrogen and turned into water so we&#39;re going to be generating our hydroxide catalyst when this alkoxide ion grabs the hydrogen from water so at this point what we have is known as the aldol addition product we have an alcohol and a ketone sometimes the instead of a ketone it could be an aldehyde so if we if this reaction wasn&#39;t heated this would be the product but since it&#39;s heated we need to take it to the next step or to the next level so we still have an alpha hydrogen to that ketone and in the presence of heat the hydroxide ion will remove this hydrogen via the e1cb reaction mechanism so we&#39;re going to get the enolate ion initially and we&#39;re going to use that enely ion to get rid of the o h group and so now we have our final answer this is the alpha beta unsaturated ketone so that&#39;s the answer of the intramolecular aldol reaction after dehydration number 10 which of the following reagents will produce the product shown below with the greatest yield so looking at the product and comparing it to the reactant we can see that we&#39;ve added a methyl group to the left side or basically we replaced an alpha hydrogen with a methyl group and looking at our answer choices we see that step two is the same methyl bromide the difference is in step one should we use lda as a base or nah nah is sodium hydride it&#39;s a small base and lda is a bulky base lithium diisopropyl amide and the second thing we need to consider is the temperature should we use a very low temperature negative 80 degrees celsius or very high temperature now to get the right answer we need to realize that there are two products that compete against each other the kinetic product and the thermodynamic product and there are two alpha hydrogens that lead to those products the blue hydrogen and the green hydrogen now if we remove the blue hydrogen we&#39;re going to get the kinetic enolate ion which looks like this and if we remove the green hydrogen we&#39;re going to get the thermodynamic enolate ion which looks like that now you also need to know that if you use sodium high sodium hydride it favors the formation of the thermodynamic product sodium hydride is a small base and because the hydride ion is small it can grab both the blue or the green hydrogen so it prefers to grab the green hydrogen because it leads to the more stable thermodynamic product if you see if you look at the double bond it&#39;s next to the methyl group and that means that double bond is more substituted than the blue double bond the more substituted a double bond is or the more r groups it has the more stable that particular alkene is or that just that double bond because i guess this is really not alkene of the enolate ion now lda is a bulky base and because it&#39;s so huge it&#39;s not going to go for the green hydrogen even though it leads to the more stable enolate ion rather the bulky base is going to go for the hydrogen that&#39;s more accessible which is the blue hydrogen this methyl group um blocks the lda from grabbing the green hydrogen so it&#39;s going to go for the more accessible blue hydrogen now the second thing is the temperature a low temperature favors the kinetic product and the high temperature favors the thermodynamic product so knowing that looking at our answer do we have the kinetic product or the thermodynamic product because the methyl group is the methyl group that we added was on the left side that means that it&#39;s the kinetic product because it came from this double bond here which is on the left side the thermodynamic product if we had that we would have another method group on this side but since we don&#39;t have it it&#39;s the kinetic one so we need to choose lda so we can eliminate answer choice c and answer choice b and we need to choose a low temperature to get the best yield so therefore answer choice a is the correct answer so now let&#39;s write up a mechanism for this process so the methyl group that was already existing on this cyclohexanone ring i&#39;m going to put it to the left side i&#39;m going to flip it to the left before it was on the right it just makes it easier to show the mechanism so lda it&#39;s a nitrogen atom with two isopropyl groups and the nitrogen atom has two lone pairs so it has a negative charge and it&#39;s paired up with a lithium ion so this bulky base is going to grab a hydrogen and create the enolate ion so that&#39;s the first step depronation now once we have the enolate ion the second step is pretty straightforward we just need to add the methyl bromide so the oxygen with the negative charge is going to reform a double bond it&#39;s going to use the lump here to do that and it&#39;s going to cause this double bond to become nucleophilic forcing it to attack the methyl group expelling the bromide leaving group and so therefore our answer looks like this so here we get the kinetic product rather than the thermodynamic product number 11 complete the sequence so for this problem we need to find out which ring is more activated the ring on the left it sees this partially positive carbonyl group and the carbonyl group is a metadirecting deactivator it can pull electron density toward itself thus deactivating the ring making it more electrophilic or partially positive now the ring on the right is activated it sees this partially negative oxygen atom which can donate electron density to the ring by means of the resonance effect put in a partially negative charge on that ring so the ring on the right is more reactive than the ring on the left so therefore any any answer choice that has groups or substituents on the left ring is eliminated we can eliminate answer choice a and c so now we&#39;re down between b and d so the first thing that happens once we add this acid chloride to the ring we&#39;re going to get this group a ketone just on the right side and then once we add hydrazine which is h2n nh2 that&#39;s the wolf kitchen reaction it&#39;s going to reduce the ketone into a propyl group so as we can see both answers contain the propyl group and this oxygen is an ortho para director so the propyl group went to the para position relative to the oxygen now the only difference is where the bromine is going to go so will the bromine go ortho or meta with respect to the oxygen now keep in mind even though this position is ortho with respect to the oxygen the green arrow the red arrow is um ortho with respect to the propyl group and both the oxygen and the propyl group are ortho para directors so in a situation like this you have to find out who&#39;s going to win so who which group is more activated is it the oxygen or the alkyl group the propyl group it turns out oxygen is a much stronger electron donating group the propyl group is a weak activating group so the oxygen is going to win so therefore the oxygen is going to tell the bromine to go ortho with respect to itself so b is the answer but now let&#39;s actually let&#39;s write out everything so the first reaction is a frito-crafts acylation reaction the catalyst takes away the chlorine group and the carbonyl group adds itself to the ring and it goes on the para position initially and then in the next step once we use hydrazine with potassium hydroxide this is going to reduce the ketone into an alkane pretty much it gets rid of the carbonyl oxygen and it replaces it with two hydrogen atoms but we don&#39;t have to show the hydrogen atom so we can simply write it as a propyl group so once we get to this part oxygen will direct the bromine ortho with respect to each other with respect to itself i should say and the propyl group wants the bromine to go on the ortho position with respect to itself but because oxygen is a more electron donating group it&#39;s more it&#39;s a more powerful activator relative to the propyl group the bromine is going to go in this position and so that&#39;s why we get that answer number 12 which of the following compounds will show two signals in an hmr spectrum so we got to find out how many different types of hydrogens are in each of these compounds so let&#39;s start with compound one we have a methyl group on top so that would show as one signal these two hydrogen atoms are identical to each other so they would show as a second signal we&#39;ll call it signal b and there&#39;s another two hydrogens here which are identical to each other they&#39;re equidistant from the methyl group they&#39;re two carbons away from it so they have the exact same chemical or electronic environment so we&#39;ll call it proton c and then there&#39;s another proton here which we&#39;ll call it which would be the fourth signal so compound one would show four signals in an hmr spectrum so we want a compound that shows two signals so anything with compound one as an answer can be eliminated so we&#39;re down between b and d looking at compound two the methyl groups are identical to each other so we&#39;ll call a proton a as we can see there is symmetry across this line so therefore um these two are identical to each other and there&#39;s also symmetry here so we can call this proton b as well all four of those protons are identical to each other they have the exact same chemical environment if you compare these two protons um they&#39;re both one carbon away from the methyl group so compound two would show up as two signals compound three also has symmetry these three methyl groups are identical to each other each method group is two carbons away from another method and the hydrogen atoms are all equally distant from each other so they&#39;re all identical so we&#39;ll call it signal b so this will generate two signals as well now keep in mind there are no hydrogen atoms on this these carbons that i&#39;ve indicated in green because those carbons have four bonds so we can see compound two and three generated two signals so chances are that&#39;s the answer but for the sake of practice let&#39;s see if we can find out how many signals are in compound four the three methyl groups in the tert-butyl group are identical because they&#39;re attached to the same carbon atom and uh there&#39;s symmetry in the middle so those two protons are identical to each other and these two as well um we have at least one hydrogen on this isopropyl group so we&#39;ll call it signal d and then the two methyls which are attached to that ch carbon are identical to each other so compound four shows up as five signals in a graph so the correct answer is d compounds two and three shows two signals in hmr spectrum so number 13 what is the major product so here we have a ketone cyclohexanone and this amide i believe it&#39;s a semi-carbosome group but what do you think the major product of this reaction is do you think it&#39;s a b c or d well just review if you have a ketone and you add a primary amine where it&#39;s it has an nh2 group you&#39;re going to get the i mean which is a carbon double bonded to a nitrogen and if you add let&#39;s say like a secondary amine where that nitrogen only has one hydrogen instead of two hydrogens you get the enamine or enamine group where you have a carbon single bonded to a nitrogen but adjacent to a carbon double bond carbon bond like a c double bond c so the enemy is basically an amine with an alkene so if you look at answer choice b and d those are an enemy you could see the amine group next to an alkene and answer choice a and c represents uh an i mean functional group so what we really need to decide is which nitrogen is most nucleophilic which nitrogen will attack the carbonyl carbon because each of these nitrogen atoms have lone pairs but one of them is more nucleophilic than the other so would you say it&#39;s we&#39;ll call this nitrogen a nitrogen b or nitrogen c nitrogen a and b are not amy nitrogens they&#39;re amides amid a nitrogen that&#39;s in an amide functional group are not nucleophilic but the amine nitrogen is looking at nitrogen a its lone pairs aren&#39;t available for a nucleophilic attack as you can see those lone pairs are busy um resonating with the carbonyl oxygen so nitrogen a is not very reactive nitrogen b is not reactive either it can delocalize into the carbonyl group so the nitrogen and that&#39;s part of an amide functional group will not attack the carbonyl carbon nitrogen c doesn&#39;t delocalize into the carbonyl group is too far away so this is the amine group that&#39;s going to react with the carbonyl group and because it&#39;s an nh2 we&#39;re meaning it&#39;s a primary amine we should get the i mean and not the enemy so we can eliminate answer choice b and d now that we know we&#39;re supposed to get a carbon double bond double bonded to a nitrogen atom um we need to distinguish between a and c if we look at antichoice c um we could see that the amide nitrogen attacked the carbonyl group which shouldn&#39;t be the case because that lump here delocalizes into the carbonyl group so ancestry c is out answer choice a is the correct one because this is the nitrogen that&#39;s two carbons away from the carbonyl group that&#39;s the amine nitrogen and that&#39;s the one that&#39;s going to attack the carbonyl um so let&#39;s review some reactions between amines and ketones so let&#39;s say if you have a ketone and you add a primary aiming to it what you need to do is get rid of water what is going to be removed and so replace the oxygen atom with the nitrogen atom and then the nitrogen lost two hydrogens so it only have the methyl group at this point so that&#39;s the i mean or amine functional group you need the primary amine to get the amine by the way the the amine group works um under mildly acidic conditions meaning the ph has to be around four to five to get the greatest yield if it&#39;s not at that range your yield for the amine will be low now let&#39;s say if we react a ketone with a secondary amine let&#39;s use this one because it&#39;s very common especially in the stork enemy reaction now what&#39;s gonna happen is you&#39;re going to get the the enamine or the enemy which is an alkane an alkene plus an amine so you&#39;re still going to lose water but you&#39;re going to lose this oxygen this hydrogen and the one of the alpha hydrogens and that&#39;s going to create water so this time instead of having a carbon double bonded to a nitrogen it&#39;s going to be a single bond and then you&#39;re going to have your nitrogen with its two r groups it&#39;s a lump here but the double bond is going to be between the carbonyl group and alpha hydrogen so it&#39;s going to be here so you get the enemy by the way let&#39;s say if we have cyclohexanone and there&#39;s a methyl group and let&#39;s say you wish to add this same secondary uh amine so we know we&#39;re going to get the the enemy as a product but there&#39;s two possible locations in which the double bond can go it can go on the right side or it can go on the left side so which of these two products is favored if you have to choose pick the one that is that generates the most stable alkene the most stable alkene is the one that&#39;s closer to the r group the methyl group so this form is more favored it&#39;s more stable here we have uh this alkene has three r groups so it&#39;s tri-substituted and this alkene has uh two r groups so it&#39;s disubstituted not counting the nitrogen atom number 14 which compound is not aromatic now just to review for a compound to be aromatic um it needs to have six pie electrons based on huckel&#39;s number oracle&#39;s rule four and plus two and it needs to be uh cyclic completely conjugated around the entire ring all of the atoms must be sp2 hybridized so the whole structure has to be flat or planar so let&#39;s look at each one let&#39;s start with uh compound a known as a pyrrole and every double bond represents two pi electrons so this is going to be two four and the lone pair is six now if you look at b um it&#39;s two four also six pi electrons but we don&#39;t count this slope here now if you&#39;re wondering when you should know if you should count the lone pair or not here&#39;s a review so let&#39;s say if you have some generic atom a only count the lope here if it makes the fourth group so let&#39;s say if this is number one group number two and then this is number three don&#39;t count it because the lone pair make the third group but let&#39;s say if you have this scenario where this is group one group two group three and then group four count the lone pair as towards the pi electrons or let&#39;s say if you have this situation group one group two group three don&#39;t count it group four count the lone pair so if the lone pair makes the fourth group count it that&#39;s a quick and simple way to tell um if the lone pair is going to be counted so this loop here is not counted so we still have six pi electrons and if you look at a and b all of the carbon atoms are sp2 hybridized so and it&#39;s completely conjugated so a and b are both aromatic looking at compound c we have 2 4 four six pi electrons not eight because we don&#39;t count both lone pairs it&#39;s pretty much this situation we count one of the two lone pairs so answer choice c is also aromatic now d may appear to be aromatic it has two four we do count this lump here six we have this situation we have a carbon atom that has that&#39;s attached to two other carbon atoms there&#39;s an invisible hydrogen that you don&#39;t see and a lump here so the lone pair makes the fourth group so we count it towards the pi electrons so even though this compound has six pi electrons notice that this carbon which i indicated in red it&#39;s an sp3 hybridized carbon it has two hydrogens and because it&#39;s an sp3 hybrid carbon this lone pair cannot move in this direction that lone pair can only resonate around these five carbon atoms it can&#39;t fully resonate around the entire uh six carbon cyclic structure so um this sp3 carbon pretty much doesn&#39;t allow conjugation around the entire ring plus it&#39;s an sp3 carbon is not planar so if it&#39;s not planar it&#39;s not going to be aromatic so d is the answer d is a non-aromatic compound a b and c are aromatic number 15 which compound is consistent with the h nmr spectrum shown below so if we look at the functional groups present we can see that we have an aldehyde an ester an ether and a benzene ring the first thing that we could distinguish is the aldehyde so we&#39;re focused on the hydrogens though the aldehyde hydrogen has a signal around nine to ten and the fact that we don&#39;t see that on the graph means that we can eliminate answer choice a now the next thing that we can take uh that we can focus on is the signals around 6.5 to 8.5 if you see signals in this region um that that is the um the hydrogens on the benzene ring so the fact that we have those signals means that there should be a benzene ring so we can eliminate answer choice b now for the other signals that are between that are less than four there&#39;s some things that you need to know if you have a regular ch3 next to let&#39;s say a ch2 a regular so basically an alkane ch3 if nothing special around it the signal is going to be around one now let&#39;s say if you have a carbon next to a special group like an electronegative atom we&#39;ll call it x it can be an oxygen it can be a nitrogen it can be a chlorine or bromine usually that signals around three to four and let&#39;s say if you have let&#39;s see where i can put this one a ch3 next to a carbonyl group that signals around two to two point five so we&#39;ll say like 2.3 and also if you have let&#39;s say a ch3 next to a benzene ring that signal will also be around 2.3 so now let&#39;s analyze c and d let&#39;s start with compound d here we have a ch3 next to a carbonyl group so that&#39;s going to be around 2.3 and here we have a ch2 next to an oxygen so we&#39;re going to say 3.5 it&#39;s supposed to be between 3 to four and here we have a ch3 so we&#39;ll say that&#39;s around one the ch3 is next to the ch2 so we have those signals in the graph so now let&#39;s look at um institute c here we have a ch3 as well um that&#39;s 2.3 actually no this ch3 is not 2.3 this ch3 is actually next to an oxygen not to a carbonyl group it&#39;s a small little difference but it&#39;s very important so because that ch3 is next to an oxygen we have this signal 3.5 here we have a ch2 next to an oxygen so that&#39;s 3.5 and a ch3 next to a ch2 and that&#39;s around one so answer choice c we should have two signals around 3.5 not one so we could eliminate answer choice c which means the correct answer is d now just to review we can also look at the spin splitting pattern this ch3 doesn&#39;t have any adjacent carbons that have hydrogen so it&#39;s a singlet and as we can see we do have a singlet around two looking at the ch2 it&#39;s adjacent to a carbon that has three hydrogens so using the n plus one rule it should have a splitting pattern of four so it should appear as a corten and this we do see the quartzite here it splits in four the signal splits in into four smaller peaks and the ch3 is adjacent to a ch2 so two plus one is three so that&#39;s a triplet and we do see the triplet around one so compound d is the correct answer

welcome to my organic chemistry 2 multiple choice practice tests so let&amp;#39;s jump right into it number one what is the major product in the following reaction so what do you think the answer is to this problem so notice that we have an aldehyde functional group and we have silver oxide whenever you have the ag plus ion it could be in the form of silver oxide or it can be written like this it can be complex to ammonia these reagents is associated with the tolerance reagent and it can oxidize an aldehyde to a carboxylic acid it&amp;#39;s very specific for aldehyde functional groups and even alpha hydroxy ketones as well but for the most part they react with aldehydes so a is not an aldehyde it&amp;#39;s out c is a ketone so it won&amp;#39;t work um if we had an o h here where an alcohol on the alpha carbon then it would work that&amp;#39;s the alpha hydroxy ketone but since it&amp;#39;s not there c is not the correct answer so now we&amp;#39;re between b and d b is the protonated form of the carboxylic acid d is the deprotonated form of it so to distinguish between b and d you need to look at step two which is h3o plus under acidic conditions the carboxylic acid will be in its pronated form so the hydrogen should be here so the answer is b number two complete the reaction by the way for these questions feel free to try the problem yourself pause the video and then unpause it when you&amp;#39;re ready to see what the answer is so let&amp;#39;s begin we have one etho 4 isopropyl benzene and we&amp;#39;re gonna mix it with bromine in the presence of ultraviolet light so this is going to be a radical bromination reaction now we can eliminate answer choice b and d because this reagent will not put a bromine atom on the benzene ring the only way to do that is to use br2 in the presence of a lewis acid catalyst like febr3 so since we don&amp;#39;t have that answer choice b and d is out so what this reaction will do radical bromination it&amp;#39;s going to replace the benzylic hydrogen which could be the hydrogens in purple or in green now we have both answers the answer choice a we replace the purple hydrogen and for answer choice c we replace the green hydrogen so the question is which one is better what type of carbon do we have here notice that that carbon is connected to three other carbon atoms the two methyls and this carbon so it&amp;#39;s tertiary and this carbon here is a secondary carbon because it&amp;#39;s attached to two other carbon atoms so which radical is more stable a secondary radical or tertiary radical because tertiary radicals are more stable the bromine will selectively abstract the tertiary hydrogen to create a tertiary radical so therefore answer choice c is the correct answer let&amp;#39;s review some reactions associated with bromine so let&amp;#39;s say if you have an alkane and you use br2 with hv by the way this is the same it has the same effect as using mbs it&amp;#39;s a radical bromination reaction now what&amp;#39;s going to happen is bromine will selectively replace the tertiary hydrogen so it&amp;#39;s just going to add a br to the tertiary carbon now let&amp;#39;s say if we use nbs on pentane so here we have a primary hydrogen and here we have a secondary hydrogen this is another radical bromination reaction it&amp;#39;s going to replace the highest substituted hydrogen in this case the secondary one rather than the primary one so we&amp;#39;re going to get this as an answer now i can also replace this one as well because both of these are secondary and they lead to different products so the bromine atom is going to go on the most substituted carbon for this example the most substituted carbon is a tertiary one and for this one it&amp;#39;s the secondary carbon so that&amp;#39;s what happens if you add br2 with hb hv or mbs to an alkane now let&amp;#39;s say if you add it to an alkene the bromine under radical conditions will not react with the double bond what&amp;#39;s going to happen is it&amp;#39;s going to replace the allylic hydrogen the one that&amp;#39;s one carbon away from the double bond so you&amp;#39;re going to get this as an answer and if you have a benzene ring it&amp;#39;s going to replace the benzylic hydrogen so basically the bromine atom is going to go one carbon away from the benzene ring which is kind of the same as this situation where it&amp;#39;s one carbon away from a double bond so the bromine atom is going to go it&amp;#39;s going to replace the hydrogen on the benzylic carbon which is one away from the benzene ring now let&amp;#39;s say if you have an alkene and let&amp;#39;s say if you add br2 in the presence of dichloromethane this is not a radical reaction in this situation the bromine reacts with the double bond and a double bond is going to disappear so it&amp;#39;s going to add two bromine atoms anti-addition now you also get the enantiomer which looks like this so you get a mixture of two products now if you have a benzene ring um br2 by itself will not react with the benzene ring if you want to add the br directly to the benzene ring you need to add a catalyst like fvbr3 so what&amp;#39;s going to happen is it&amp;#39;s going to replace the hydrogen with a bromine atom and so you get this as a product now let&amp;#39;s say if you have an alkene if you add hbr it&amp;#39;s going to react with the double bond and it&amp;#39;s going to put a bromine on the more substituted carbon which is the secondary carbon and the hydrogen will go on the less substituted carbon now let&amp;#39;s say if you add hpr and peroxides it can be written as h2o2 or ror in either case there&amp;#39;s two oxygen atoms that are single bonded to each other and that&amp;#39;s a peroxide so what&amp;#39;s going to happen is the bromine is actually going to go on the less substituted carbon so we get the anti more carbon carb reaction rather than the markovnikov reaction with just regular hbr so just make sure you&amp;#39;re familiar with those reactions um so you might see bromine in those different forms and just keep that in mind for your next organic chemistry final exam number three which compound has a proton with the lowest pka so let&amp;#39;s compare different compounds let&amp;#39;s start with the alpha hydrogen for the compound and answer choice c and an answer choice b which of those two hydrogens is more acidic well whenever you have a alpha hydrogen between two carbonyl groups the pka is around nine for a hydrogen next to one ketone it&amp;#39;s around roughly around i believe 20. now this hydrogen here between the ketone and the ester is 11. the reason for that is because of this electron donating group because the oet group the oxygen can donate a pair of electrons to one of the carbonyl groups it makes that this hydrogen slightly less acidic so the pka goes up a little anytime you add an electron donating group to an acid the pka goes up it becomes less acidic if you add an electron withdrawing group to an acid the pka will go down it becomes more acidic now the most acidic hydrogen for the carboxylic acid is this one here the pka for a typical carboxylic acid is around four to five so um the carboxylic acid is much more acidic than compound b or compound c so we can eliminate b and c now the reason for this is if you notice the trend on a periodic table let&amp;#39;s say this is carbon nitrogen oxygen fluorine chlorine bromine and so forth acid strength increases as you go to the right and as you go down so when comparing a ch bond versus an o h bond usually 99 of the time the o h bond is more acidic so here this is a carbon hydrogen bond the pka is 11. here this is an oxygen hydrogen bond it&amp;#39;s much more acidic four to five but keep in mind these are general trends though so you might see an exception or two now this is a protonated carboxylic acid so this is even more acidic the pka of a pronated carboxylic acid is somewhere in the negatives it can be like negative four negative five but it&amp;#39;s much less than positive four to five so the correct answer is d whenever you see a hydrogen that&amp;#39;s attached to an atom with a positive charge it&amp;#39;s very acidic for example the difference between h2o and h3o plus h2o has a pka of 15.7 it&amp;#39;s neutral but h2o plus the pka is around like negative three or negative four as you can see that extra hydrogen makes it a lot more acidic number four which structure is most consistent with the following ir spectrum now the first thing you want to look at is this huge bulky peak around 25 to 3300 you can&amp;#39;t miss it when you see a massive very strong very broad signal and in this region you have a carboxylic acid this is the oh stretch of a carboxylic acid so we can eliminate answer choice c and answer choice d because they&amp;#39;re not carboxylic acids the next important signal that we have here is the carbonyl group at 1700 that&amp;#39;s also part of the carboxylic acid functional group now the ketone would also have that signal at 1700 but the ketone wouldn&amp;#39;t have this very broad signal now granted we do have an alcohol and an alcohol would generate a broad signal but it wouldn&amp;#39;t be around 25 to 3300 this signal would be more like 33 to 3500 and it&amp;#39;s not as broad as a carboxylic acid so if i can draw a rough sketch an alcohol oh stretch would look something like this it would be in that region it wouldn&amp;#39;t cross over past three thousand it wouldn&amp;#39;t go to the right of three thousand so we can eliminate d another reason why we can eliminate answer choice d is um this signal here that signal which is roughly around 1200 it&amp;#39;s a co stretch but not for an alcohol but for a carboxylic acid there&amp;#39;s two types of co stretches there&amp;#39;s the co stretch of an alcohol and ether where this is a an sp3 carbon that&amp;#39;s around usually a thousand to eleven fifty so we&amp;#39;ll say like eleven hundred and then there&amp;#39;s a co stretch um that&amp;#39;s next where the carbon is sp2 hybridized and that&amp;#39;s this carbon the carbon from a carboxylic acid that&amp;#39;s sp2 hybridized so that co bond it&amp;#39;s a little bit more than eleven hundred it&amp;#39;s like twelve hundred to thirteen hundred so this signal which is roughly around 1200 that correlates to a co stretch where it&amp;#39;s an sp2 hybridized carbon rather than an sp3 hybridized carbon so that&amp;#39;s another reason why it&amp;#39;s not the alcohol now there&amp;#39;s another signal that we can look at and that is the c triple bond c at 2200 this will distinguish the right answer from the rest so only answer choice b has a carboxylic acid with an alkyne so that&amp;#39;s sigma 2200 it could be a c triple bond c or even a nitrile a c triple bond n but we don&amp;#39;t have a nitrile so we can eliminate a because we don&amp;#39;t have an alkyne we have an alkene and an alkene we should see a signal somewhere between around 1600 for a seed of 1c so the correct answer for this problem is answer choice b we have the oh stretch of a carboxylic acid the carbonyl stretch and the alkyne and also the co stretch for carboxylic acid but to the right of 1500 is the fingerprint region so this region is not always reliable so you can use that as a last resort number five which set of reagents will produce para nitro benzoic acid from benzene with the greatest yield so go ahead and take a minute and see if you can get the right answer so i&amp;#39;m going to go through every choice but first let me show you how you can quickly get the right answer the first thing you want to keep in mind is that we need uh para we need the no2 and the carboxylic acid group to be para with respect to each other so we know we can eliminate antichoice b because once you put the no2 on the ring it&amp;#39;s going to tell the methyl group to go in the meta position which we don&amp;#39;t want so b is out now looking at antichoice c we could eliminate that as well once we add the methyl group it&amp;#39;s an ortho para director but um notice that in step two when you oxidize it with potassium permanganate it&amp;#39;s going to convert it to a carboxylic acid which is a meta director and that&amp;#39;s going to tell the no2 group to go into meta position so c is out now the only issue with d is once you add the no2 group in the first step and once you reduce it with tin and hydrochloric acid it turns into an nh2 group now granted an nh2 group is an orthopedic director however once you add a lewis acid catalyst you&amp;#39;re going to deactivate the nh2 group and it becomes a meta director and so step three won&amp;#39;t even work in the presence of an anti-street group the the frito crafts alkylation doesn&amp;#39;t work well in the presence of an amine but a is the best answer so let&amp;#39;s go through each one i&amp;#39;m going to go over a last but let&amp;#39;s start with answer choice b so starting with benzene so once we add a nitric acid and sulfuric acid we&amp;#39;re going to replace a hydrogen with an no2 group so this is called nitration and then we&amp;#39;re going to add methylchloride with an aluminum chloride acid catalyst now the nitro group is a meta director so it&amp;#39;s going to tell the methyl group to go here so we&amp;#39;re going to put the ch3 on this side and then once you add potassium permanganate with h3o plus it&amp;#39;s going to oxidize this group and it&amp;#39;s going to convert it to a carboxylic acid so we end up getting meta nitrobenzoic acid rather than pera so that&amp;#39;s why we can eliminate answer choice b so now let&amp;#39;s see what we get for answer choice c so starting with benzene we&amp;#39;re going to add methyl chloride first as soon as we add the methyl group um what we now have is called toluene that&amp;#39;s the benzene ring combined with the ch3 and then once we oxidize it with potassium permanganate in atrial plus it&amp;#39;s going to convert into a carboxylic acid and then once we add nitric acid and sulfuric acid um the carboxylic acid is going to direct the no2 group to go to the meta position so we actually get the same answer that is meta nitro benzoic acid so now let&amp;#39;s consider answer choice d so step one is nitration and then following step one we have reduction so you can use any active metal so you can use tin with hydrochloric acid you can use uh zinc metal with hydrochloric acid you can even use um iron metal with hydrochloric acid and what&amp;#39;s going to happen is it&amp;#39;s going to convert the nitro group to an aiming group so when you have an nh2 and a benzene ring the name is called aniline so now if we add methylchloride with the aluminum chloride catalyst notice what happens aniline has a lone here it&amp;#39;s actually quite nucleophilic so what&amp;#39;s going to happen is it&amp;#39;s going to combine with the lewis acid catalyst so the methyl won&amp;#39;t even go on the ring so now the nitrogen is bonded to the aluminum chloride and it also has three hydrogen atoms i mean two hydrogen atoms so when nitrogen has four bonds it now has a plus charge when when you have a nitrogen for lone pair it is an ortho para activator but now when it has a plus charge it is a meta direct and deactivated so if we mix a one-to-one ratio with aniline and methyl chloride with this catalyst the ring is deactivated and the catalyst binds to the nitrogen so it won&amp;#39;t help the ring to react with the methyl chloride so the reaction stops here so there&amp;#39;s no point in atom potassium permanganate because there&amp;#39;s no methyl groups to oxidize so just keep in mind that the frito-craps alkylation reaction and the acellation reaction does not work in the presence of an nh2 group on a benzene ring so now let&amp;#39;s look at our correct answer which is answer choice a so the first step um is photocraft&amp;#39;s alkylation the catalyst aluminum chloride helps to remove the chlorine from the methyl group and so that gives us a toluene now in the next step we have nitration once you add toluene the ring is slightly activated alkyl groups are weakly activating groups so the ring is a little bit more reactive than before so once we have the no2 group um the methyl group will direct it ortho or para now we&amp;#39;re going to write the para answer because that&amp;#39;s the one that&amp;#39;s going to lead us to our product and then once we add potassium permanganate the methyl group will be oxidized to a carboxylic acid and that&amp;#39;s the answer that we want so thus we have para nitro benzoic acid by the way if you want more practice problems click a link that is in the description section of this video and you can get access to more practice problems to prepare for your exam the video that you&amp;#39;re currently watching only has 15 free practice problems but the paid version includes those 15 practice problems in addition to another set of 85 questions for a total of 100 uh practice problems so if you think you might need more questions to work on to prepare for your exam you can check out that video and let me know if you have any questions number six which of the following reagents will carry out the reaction shown below so we have an acid chloride and we wish to convert it to an ester so let&amp;#39;s look at each of our choices a is a granted reagent if you add a grain of reagent to an acid chloride it&amp;#39;s going to add two r groups and convert the acid chloride into an alcohol since we don&amp;#39;t have an alcohol a is out now b is called the gilman reagent or the lithium dialkyl cuprate reagent and if you put that with an acid chloride it&amp;#39;s going to add one r group and convert the acid chloride into a ketone which is not an ester so answer choice b is not the correct one c is an alcohol an alcohol plus an acid chloride will make an ester so uh c is the right answer and d d is an ester if you put an s with an acid chloride i&amp;#39;ve never seen it make another ester so we&amp;#39;re going to eliminate ant&amp;#39;s choice d now if you understand how these reactions work already feel free to fast forward to question seven but if you want to see the mechanism uh feel free to continue watching so let&amp;#39;s start with the granite reagent the carbon that&amp;#39;s attached to the magnesium is partially negative and it&amp;#39;s attracted to this partially positive carbon and so we&amp;#39;re going to create a new carbon-carbon bond as it attacks the carbonyl carbon the pi bond breaks and we&amp;#39;re going to have this a tetrahedral intermediate so what we need to do is add the ethyl group to this carbon now the magnesium which has a plus charge it forms an ionic bond with the the negatively charged oxygen so what&amp;#39;s going to happen now is this tetrahedral intermediate is going to collapse and chlorine the leaving group will leave as chloride and so right now we have ketone but the grenade reaction does not stop there the grenade reagent will react with the ketone as well so another uh griden molecule will attack the carbonyl carbon and then the pi bonds gonna break so this time there are no leaving groups so this is going to be our answer because i really didn&amp;#39;t add step two but step two usually you would add a h3o plus once he had h3o plus then it converts into an alcohol so it&amp;#39;s going to look like this a tertiary alcohol this oxygen is going to grab a proton from h3o plus so now for the second reaction that is between the acid chloride and the gilman reagent so we have a copper atom attached to two ethyl groups and copper has a negative formal charge because it&amp;#39;s in the plus one oxidation state and lithium has a plus charge so only one of the r groups is nucleophilic and so this r group attacks the carbonyl carbon and this part is going to be very similar to the first step of the the mechanism of the grinded reagent with the acid chloride so we&amp;#39;re going to get a tetrahedral intermediate and we&amp;#39;re just going to add only one r group so right now the lithium ion is with the oxygen with the negative charge the copper now has only one r group and so it looks like this and so it&amp;#39;s not going to react anymore so we&amp;#39;re not going to worry about it so this oxygen is going to reform a double bond and expel the leaving group so if you choose to use the gammon reagent it stops at the ketone level so it&amp;#39;s very good for converting esters and acid chlorides into ketones by the way if you put the ester with the given reagent the mechanism is exactly the same the only difference is instead of a chlorine you&amp;#39;ll have like in a methoxy group in och3 okay so now let&amp;#39;s consider the mechanism with the acid chloride and the alcohol so the alcohol will attack the carbonyl carbon and just like before we&amp;#39;re going to get a tetrahedral intermediate but this time instead of having an ethyl group we have an ethoxy group an och2ch3 group because antichoice b is very close to instachoice c the only difference is it differs by an oxygen and just like before the leaving group is going to leave so as you can see the mechanism is very similar and now we have our desired product which is an ester and so that&amp;#39;s the mechanism for the conversion of an acid chloride into an ester number seven complete the reaction sequence so we have cyclohexanol reacting with pcc then we have some other reagents as well so what does pc do to an alcohol if it&amp;#39;s a primary alcohol pcc can oxidize it to an aldehyde and it can oxidize secondary alcohols into a ketone so that&amp;#39;s the end result of the first part now what happens if we take cyclohexanone and react it with triphenylphosphine and alkyl halide and butyllithium what do you think the major product will be if you recognize these reagents you&amp;#39;ll know that it&amp;#39;s part of the vitic reaction or some people say the width reaction the width reaction or vedic reaction it converts ketones into alkenes so knowing that fact you can eliminate answer choice a and answer choice b so we&amp;#39;re down between c and d now if you want to quickly get the answer notice where notice the carbon that has the iodine atom that is the carbon that is going to connect to this carbon so it&amp;#39;s the secondary carbon notice how we have relative to that carbon we have an etho on the right and a methyl on the left d is going to be the answer we have a methyl on one side and an ethyl group on the other side let me show you the mechanism for this reaction so the first thing that happens triphenylphosphine reacts with two iodo butane the phosphorus atom is a very good nucleophile and so it attacks the carbon from the back expelling the iodine group so now we have a phosphorus attached to a carbon that has a methyl on one side and an ethyl on the other side but i&amp;#39;m going to redraw it like this this carbon has a methyl group a hydrogen and an ethyl group when phosphorus whenever a phosphorus atom has four bonds it has a plus charge now we also have butyl lithium as well butyl lithium is a very powerful base and butyllithium is used to remove the hydrogen atom on the carbon that&amp;#39;s attached to the uh the phosphorus group and then this bond is going to break it&amp;#39;s going to put a lone pair on the carbon atom so now we have our our illid which looks like this and it has a negative charge now now the illite is stabilized by resonance but i&amp;#39;m going to use this form for the mechanism so once we pair up cyclohexanone with the illidan the carbon with the negative charge is attracted to the carbonyl carbon because it has a positive a partial positive charge so whenever you have a nucleophilic carbon and an electrophilic carbon those two will connect and form a carbon-carbon bond and then this bond is going to break that&amp;#39;s going to open up so now we have this intermediate and then what&amp;#39;s going to happen next is this negatively charged oxygen atom is going to attack the positively charged phosphorus group to form a four-membered ring known as a a beta-end group or betaine group so whenever a bond breaks let&amp;#39;s say a bond between carbon and bromine because bromine is more electronegative than carbon when the bond breaks the electrons will go towards the more electronegative atom so you would get a carbocation and a bromide ion conversion let&amp;#39;s say if you have a carbon hydrogen bond carbon is more electronegative than hydrogen so when that bond breaks you get a carbonyl so understanding that&amp;#39;s going to help us to see how we can write the mechanisms for the next step oxygen is more electronegative than phosphorus so oxygen bears a negative charge phosphorus bears the positive charge this carbon atom is partially positive with respect to the oxygen atom and the carbon that&amp;#39;s attached to phosphorus this carbon is partially negative because carbon is more electronegative than phosphorus carbon is closer to fluorine than phosphorous so when the carbon-oxygen bond breaks those electrons will go towards the more electronegative phosphorus atom i mean the more electronegative oxygen atom and when the carbon phosphorus bond breaks those electrons will go towards the more electronegative carbon atom so as you can see the atom that&amp;#39;s more electronegative pulls the electrons toward itself when the bond breaks so the product of this reaction we get triphenylphosphine oxide and we get a carbon-carbon double bond so as you can see this is the the end product for this reaction so whenever you have a ketone and if you&amp;#39;re using the wittig or the wittig reaction you&amp;#39;re going to get an alkene as your end product so that&amp;#39;s what you want to take from this and make sure you attach the carbonyl carbon to the carbon that had the leaving group the iodine so just for the sake of practice let&amp;#39;s say if we have this ketone and let&amp;#39;s say the reagents are triphenylphosphine three iodo hexane and butyllithium if you quickly want to draw the major product get rid of the oxygen right the double bond and connect it to the carbon that has the iodine so notice that on the left side of that carbon you have an ethyl group and on the right side you have a propyl group so that&amp;#39;s what you want to draw here you want to draw the ethyl group and the propyl group and that is going to be the answer the alkene that forms as a product of the wittig reaction number eight which of the following diene and dienophile will produce the product shown below so this is the diels-alder reaction and the diene is the one that has two double bonds and a dienophile is the electrophile that has just one double bond so this is the dienophile so which answer choices can we eliminate well let&amp;#39;s count the carbon atoms we have one two three four five six seven carbon atoms not including the two nitro groups so we can eliminate choice c because there are six carbon atoms so choice c is eliminated and we could eliminate choice d as well choice d has a total of eight carbons not seven if you don&amp;#39;t count the nitro groups so d is out now if you count the carbons for answer choice a and b they&amp;#39;re both seven so we need to look at something else the fact that we have a bicyclic compound as a product means that we need to start with a ring the diels-alder reaction creates a ring but if we start with a ring we&amp;#39;ll have a total of two rings and so knowing that this will tell you that choice b is the correct answer let&amp;#39;s go over the mechanism real quick for answer choice a and b so for choice a here&amp;#39;s what&amp;#39;s going to happen let&amp;#39;s uh say that this is carbon one two three four and this is five and six the deals out of reaction um it makes six uh carbon rings or i mean to say six membered rings so this double bond is going to the pi electrons in the double bond is going to connect carbon one and six and this double bond will move here and the double bond between carbon stream four uh the pi electrons will be used to connect four and five but let me use a different color so you can see the connections so here are the pi electrons between in the double bond between carbons five and six and here are the pi electrons a double bond between carbons three and four so there&amp;#39;s also a double bond between carbons two and three and we still have a methyl group and notice that the nitrile groups are trans with respect to each other so they&amp;#39;re going to remain that way so as you can see if the diene doesn&amp;#39;t start as a ring you don&amp;#39;t get a bicyclic compound you simply get just a regular monocyclic compound just wondering but if we if the diene was in the ring already we would get a bicyclic compound this will be carbon one two three four five and six and let&amp;#39;s call this other carbon carbon a so here&amp;#39;s a six membered ring drawn differently so this is carbon one two three and four so there&amp;#39;s going to be a double bond between two and three just like we had it here and on five and six we have the two nitro groups but because they&amp;#39;re trans with respect to each other one is going to be in the exo position the other is going to be in the endo position now this carbon carbon a is on top it&amp;#39;s over here so when you start with a ring you&amp;#39;re going to get two rings a bicycle compound if you don&amp;#39;t start with a ring you&amp;#39;ll just get a one ring system you won&amp;#39;t get a bicycle kappa so those are just some basics of the deals auto reaction so make sure you review that and that&amp;#39;s it for this question number nine what is the product of the reaction shown below so we have a diketone sodium hydroxide heat and looking at our answer choices we see that we&amp;#39;re going to form a six membered ring this reaction is the intramolecular aldol reaction sodium hydroxide will be used to remove an alpha hydrogen but there are some mass choices that we can eliminate if you know it&amp;#39;s an aldol reaction there are two types of products the first product is a beta hydroxy aldehyde or ketone what this means is that the oh group should be on the beta carbon so if we look at answer choice a relative to the ketone the first carbon away from it is the alpha carbon and this is a beta carbon so that&amp;#39;s a beta hydroxy ketone but if you look at b this is the alpha carbon that&amp;#39;s the beta carbon this is the gamma carbon the hydroxy group shouldn&amp;#39;t be there so we can eliminate b now the second product that you can get if you add heat heat leads to dehydration loss of water so you&amp;#39;re going to lose a hydrogen and an o h group and you&amp;#39;re going to form a double bond between the two so dehydration will remove the alpha hydrogen and the oh and it&amp;#39;s going to put a double bond right in the middle so that double bond should be between the alpha and beta carbon so free anti-choice d the double bond is between the beta and the gamma carbon notice that it&amp;#39;s not conjugated with this carbonyl group whenever you form a double bond by means of the aldol reaction it should always be conjugated with the ketone or the aldehyde so choice d is out looking at a and c um a would be the product if we didn&amp;#39;t add heat but because we heated the reaction um it&amp;#39;s going to facilitate dehydration so c is the correct answer but now let&amp;#39;s actually write up a mechanism to see how it works so under basic conditions the first step is deprotonation we need to remove the alpha hydrogen and we&amp;#39;re going to get a carbanion now this negatively charged carbanide is going to attack the other carbonyl carbon so this is going to be carbon one two three four five six so we&amp;#39;re going to get a six membered ring so carbon two has a ketone so i&amp;#39;m gonna make this carbon two this is going to be carbon one three four five and six carbon six has a single bonded oxygen atom and instead of having two lone pairs it now has three and a negative charge and there&amp;#39;s a methyl group on carbon 6 as well so notice that hydroxide when it grabbed the hydrogen and turned into water so we&amp;#39;re going to be generating our hydroxide catalyst when this alkoxide ion grabs the hydrogen from water so at this point what we have is known as the aldol addition product we have an alcohol and a ketone sometimes the instead of a ketone it could be an aldehyde so if we if this reaction wasn&amp;#39;t heated this would be the product but since it&amp;#39;s heated we need to take it to the next step or to the next level so we still have an alpha hydrogen to that ketone and in the presence of heat the hydroxide ion will remove this hydrogen via the e1cb reaction mechanism so we&amp;#39;re going to get the enolate ion initially and we&amp;#39;re going to use that enely ion to get rid of the o h group and so now we have our final answer this is the alpha beta unsaturated ketone so that&amp;#39;s the answer of the intramolecular aldol reaction after dehydration number 10 which of the following reagents will produce the product shown below with the greatest yield so looking at the product and comparing it to the reactant we can see that we&amp;#39;ve added a methyl group to the left side or basically we replaced an alpha hydrogen with a methyl group and looking at our answer choices we see that step two is the same methyl bromide the difference is in step one should we use lda as a base or nah nah is sodium hydride it&amp;#39;s a small base and lda is a bulky base lithium diisopropyl amide and the second thing we need to consider is the temperature should we use a very low temperature negative 80 degrees celsius or very high temperature now to get the right answer we need to realize that there are two products that compete against each other the kinetic product and the thermodynamic product and there are two alpha hydrogens that lead to those products the blue hydrogen and the green hydrogen now if we remove the blue hydrogen we&amp;#39;re going to get the kinetic enolate ion which looks like this and if we remove the green hydrogen we&amp;#39;re going to get the thermodynamic enolate ion which looks like that now you also need to know that if you use sodium high sodium hydride it favors the formation of the thermodynamic product sodium hydride is a small base and because the hydride ion is small it can grab both the blue or the green hydrogen so it prefers to grab the green hydrogen because it leads to the more stable thermodynamic product if you see if you look at the double bond it&amp;#39;s next to the methyl group and that means that double bond is more substituted than the blue double bond the more substituted a double bond is or the more r groups it has the more stable that particular alkene is or that just that double bond because i guess this is really not alkene of the enolate ion now lda is a bulky base and because it&amp;#39;s so huge it&amp;#39;s not going to go for the green hydrogen even though it leads to the more stable enolate ion rather the bulky base is going to go for the hydrogen that&amp;#39;s more accessible which is the blue hydrogen this methyl group um blocks the lda from grabbing the green hydrogen so it&amp;#39;s going to go for the more accessible blue hydrogen now the second thing is the temperature a low temperature favors the kinetic product and the high temperature favors the thermodynamic product so knowing that looking at our answer do we have the kinetic product or the thermodynamic product because the methyl group is the methyl group that we added was on the left side that means that it&amp;#39;s the kinetic product because it came from this double bond here which is on the left side the thermodynamic product if we had that we would have another method group on this side but since we don&amp;#39;t have it it&amp;#39;s the kinetic one so we need to choose lda so we can eliminate answer choice c and answer choice b and we need to choose a low temperature to get the best yield so therefore answer choice a is the correct answer so now let&amp;#39;s write up a mechanism for this process so the methyl group that was already existing on this cyclohexanone ring i&amp;#39;m going to put it to the left side i&amp;#39;m going to flip it to the left before it was on the right it just makes it easier to show the mechanism so lda it&amp;#39;s a nitrogen atom with two isopropyl groups and the nitrogen atom has two lone pairs so it has a negative charge and it&amp;#39;s paired up with a lithium ion so this bulky base is going to grab a hydrogen and create the enolate ion so that&amp;#39;s the first step depronation now once we have the enolate ion the second step is pretty straightforward we just need to add the methyl bromide so the oxygen with the negative charge is going to reform a double bond it&amp;#39;s going to use the lump here to do that and it&amp;#39;s going to cause this double bond to become nucleophilic forcing it to attack the methyl group expelling the bromide leaving group and so therefore our answer looks like this so here we get the kinetic product rather than the thermodynamic product number 11 complete the sequence so for this problem we need to find out which ring is more activated the ring on the left it sees this partially positive carbonyl group and the carbonyl group is a metadirecting deactivator it can pull electron density toward itself thus deactivating the ring making it more electrophilic or partially positive now the ring on the right is activated it sees this partially negative oxygen atom which can donate electron density to the ring by means of the resonance effect put in a partially negative charge on that ring so the ring on the right is more reactive than the ring on the left so therefore any any answer choice that has groups or substituents on the left ring is eliminated we can eliminate answer choice a and c so now we&amp;#39;re down between b and d so the first thing that happens once we add this acid chloride to the ring we&amp;#39;re going to get this group a ketone just on the right side and then once we add hydrazine which is h2n nh2 that&amp;#39;s the wolf kitchen reaction it&amp;#39;s going to reduce the ketone into a propyl group so as we can see both answers contain the propyl group and this oxygen is an ortho para director so the propyl group went to the para position relative to the oxygen now the only difference is where the bromine is going to go so will the bromine go ortho or meta with respect to the oxygen now keep in mind even though this position is ortho with respect to the oxygen the green arrow the red arrow is um ortho with respect to the propyl group and both the oxygen and the propyl group are ortho para directors so in a situation like this you have to find out who&amp;#39;s going to win so who which group is more activated is it the oxygen or the alkyl group the propyl group it turns out oxygen is a much stronger electron donating group the propyl group is a weak activating group so the oxygen is going to win so therefore the oxygen is going to tell the bromine to go ortho with respect to itself so b is the answer but now let&amp;#39;s actually let&amp;#39;s write out everything so the first reaction is a frito-crafts acylation reaction the catalyst takes away the chlorine group and the carbonyl group adds itself to the ring and it goes on the para position initially and then in the next step once we use hydrazine with potassium hydroxide this is going to reduce the ketone into an alkane pretty much it gets rid of the carbonyl oxygen and it replaces it with two hydrogen atoms but we don&amp;#39;t have to show the hydrogen atom so we can simply write it as a propyl group so once we get to this part oxygen will direct the bromine ortho with respect to each other with respect to itself i should say and the propyl group wants the bromine to go on the ortho position with respect to itself but because oxygen is a more electron donating group it&amp;#39;s more it&amp;#39;s a more powerful activator relative to the propyl group the bromine is going to go in this position and so that&amp;#39;s why we get that answer number 12 which of the following compounds will show two signals in an hmr spectrum so we got to find out how many different types of hydrogens are in each of these compounds so let&amp;#39;s start with compound one we have a methyl group on top so that would show as one signal these two hydrogen atoms are identical to each other so they would show as a second signal we&amp;#39;ll call it signal b and there&amp;#39;s another two hydrogens here which are identical to each other they&amp;#39;re equidistant from the methyl group they&amp;#39;re two carbons away from it so they have the exact same chemical or electronic environment so we&amp;#39;ll call it proton c and then there&amp;#39;s another proton here which we&amp;#39;ll call it which would be the fourth signal so compound one would show four signals in an hmr spectrum so we want a compound that shows two signals so anything with compound one as an answer can be eliminated so we&amp;#39;re down between b and d looking at compound two the methyl groups are identical to each other so we&amp;#39;ll call a proton a as we can see there is symmetry across this line so therefore um these two are identical to each other and there&amp;#39;s also symmetry here so we can call this proton b as well all four of those protons are identical to each other they have the exact same chemical environment if you compare these two protons um they&amp;#39;re both one carbon away from the methyl group so compound two would show up as two signals compound three also has symmetry these three methyl groups are identical to each other each method group is two carbons away from another method and the hydrogen atoms are all equally distant from each other so they&amp;#39;re all identical so we&amp;#39;ll call it signal b so this will generate two signals as well now keep in mind there are no hydrogen atoms on this these carbons that i&amp;#39;ve indicated in green because those carbons have four bonds so we can see compound two and three generated two signals so chances are that&amp;#39;s the answer but for the sake of practice let&amp;#39;s see if we can find out how many signals are in compound four the three methyl groups in the tert-butyl group are identical because they&amp;#39;re attached to the same carbon atom and uh there&amp;#39;s symmetry in the middle so those two protons are identical to each other and these two as well um we have at least one hydrogen on this isopropyl group so we&amp;#39;ll call it signal d and then the two methyls which are attached to that ch carbon are identical to each other so compound four shows up as five signals in a graph so the correct answer is d compounds two and three shows two signals in hmr spectrum so number 13 what is the major product so here we have a ketone cyclohexanone and this amide i believe it&amp;#39;s a semi-carbosome group but what do you think the major product of this reaction is do you think it&amp;#39;s a b c or d well just review if you have a ketone and you add a primary amine where it&amp;#39;s it has an nh2 group you&amp;#39;re going to get the i mean which is a carbon double bonded to a nitrogen and if you add let&amp;#39;s say like a secondary amine where that nitrogen only has one hydrogen instead of two hydrogens you get the enamine or enamine group where you have a carbon single bonded to a nitrogen but adjacent to a carbon double bond carbon bond like a c double bond c so the enemy is basically an amine with an alkene so if you look at answer choice b and d those are an enemy you could see the amine group next to an alkene and answer choice a and c represents uh an i mean functional group so what we really need to decide is which nitrogen is most nucleophilic which nitrogen will attack the carbonyl carbon because each of these nitrogen atoms have lone pairs but one of them is more nucleophilic than the other so would you say it&amp;#39;s we&amp;#39;ll call this nitrogen a nitrogen b or nitrogen c nitrogen a and b are not amy nitrogens they&amp;#39;re amides amid a nitrogen that&amp;#39;s in an amide functional group are not nucleophilic but the amine nitrogen is looking at nitrogen a its lone pairs aren&amp;#39;t available for a nucleophilic attack as you can see those lone pairs are busy um resonating with the carbonyl oxygen so nitrogen a is not very reactive nitrogen b is not reactive either it can delocalize into the carbonyl group so the nitrogen and that&amp;#39;s part of an amide functional group will not attack the carbonyl carbon nitrogen c doesn&amp;#39;t delocalize into the carbonyl group is too far away so this is the amine group that&amp;#39;s going to react with the carbonyl group and because it&amp;#39;s an nh2 we&amp;#39;re meaning it&amp;#39;s a primary amine we should get the i mean and not the enemy so we can eliminate answer choice b and d now that we know we&amp;#39;re supposed to get a carbon double bond double bonded to a nitrogen atom um we need to distinguish between a and c if we look at antichoice c um we could see that the amide nitrogen attacked the carbonyl group which shouldn&amp;#39;t be the case because that lump here delocalizes into the carbonyl group so ancestry c is out answer choice a is the correct one because this is the nitrogen that&amp;#39;s two carbons away from the carbonyl group that&amp;#39;s the amine nitrogen and that&amp;#39;s the one that&amp;#39;s going to attack the carbonyl um so let&amp;#39;s review some reactions between amines and ketones so let&amp;#39;s say if you have a ketone and you add a primary aiming to it what you need to do is get rid of water what is going to be removed and so replace the oxygen atom with the nitrogen atom and then the nitrogen lost two hydrogens so it only have the methyl group at this point so that&amp;#39;s the i mean or amine functional group you need the primary amine to get the amine by the way the the amine group works um under mildly acidic conditions meaning the ph has to be around four to five to get the greatest yield if it&amp;#39;s not at that range your yield for the amine will be low now let&amp;#39;s say if we react a ketone with a secondary amine let&amp;#39;s use this one because it&amp;#39;s very common especially in the stork enemy reaction now what&amp;#39;s gonna happen is you&amp;#39;re going to get the the enamine or the enemy which is an alkane an alkene plus an amine so you&amp;#39;re still going to lose water but you&amp;#39;re going to lose this oxygen this hydrogen and the one of the alpha hydrogens and that&amp;#39;s going to create water so this time instead of having a carbon double bonded to a nitrogen it&amp;#39;s going to be a single bond and then you&amp;#39;re going to have your nitrogen with its two r groups it&amp;#39;s a lump here but the double bond is going to be between the carbonyl group and alpha hydrogen so it&amp;#39;s going to be here so you get the enemy by the way let&amp;#39;s say if we have cyclohexanone and there&amp;#39;s a methyl group and let&amp;#39;s say you wish to add this same secondary uh amine so we know we&amp;#39;re going to get the the enemy as a product but there&amp;#39;s two possible locations in which the double bond can go it can go on the right side or it can go on the left side so which of these two products is favored if you have to choose pick the one that is that generates the most stable alkene the most stable alkene is the one that&amp;#39;s closer to the r group the methyl group so this form is more favored it&amp;#39;s more stable here we have uh this alkene has three r groups so it&amp;#39;s tri-substituted and this alkene has uh two r groups so it&amp;#39;s disubstituted not counting the nitrogen atom number 14 which compound is not aromatic now just to review for a compound to be aromatic um it needs to have six pie electrons based on huckel&amp;#39;s number oracle&amp;#39;s rule four and plus two and it needs to be uh cyclic completely conjugated around the entire ring all of the atoms must be sp2 hybridized so the whole structure has to be flat or planar so let&amp;#39;s look at each one let&amp;#39;s start with uh compound a known as a pyrrole and every double bond represents two pi electrons so this is going to be two four and the lone pair is six now if you look at b um it&amp;#39;s two four also six pi electrons but we don&amp;#39;t count this slope here now if you&amp;#39;re wondering when you should know if you should count the lone pair or not here&amp;#39;s a review so let&amp;#39;s say if you have some generic atom a only count the lope here if it makes the fourth group so let&amp;#39;s say if this is number one group number two and then this is number three don&amp;#39;t count it because the lone pair make the third group but let&amp;#39;s say if you have this scenario where this is group one group two group three and then group four count the lone pair as towards the pi electrons or let&amp;#39;s say if you have this situation group one group two group three don&amp;#39;t count it group four count the lone pair so if the lone pair makes the fourth group count it that&amp;#39;s a quick and simple way to tell um if the lone pair is going to be counted so this loop here is not counted so we still have six pi electrons and if you look at a and b all of the carbon atoms are sp2 hybridized so and it&amp;#39;s completely conjugated so a and b are both aromatic looking at compound c we have 2 4 four six pi electrons not eight because we don&amp;#39;t count both lone pairs it&amp;#39;s pretty much this situation we count one of the two lone pairs so answer choice c is also aromatic now d may appear to be aromatic it has two four we do count this lump here six we have this situation we have a carbon atom that has that&amp;#39;s attached to two other carbon atoms there&amp;#39;s an invisible hydrogen that you don&amp;#39;t see and a lump here so the lone pair makes the fourth group so we count it towards the pi electrons so even though this compound has six pi electrons notice that this carbon which i indicated in red it&amp;#39;s an sp3 hybridized carbon it has two hydrogens and because it&amp;#39;s an sp3 hybrid carbon this lone pair cannot move in this direction that lone pair can only resonate around these five carbon atoms it can&amp;#39;t fully resonate around the entire uh six carbon cyclic structure so um this sp3 carbon pretty much doesn&amp;#39;t allow conjugation around the entire ring plus it&amp;#39;s an sp3 carbon is not planar so if it&amp;#39;s not planar it&amp;#39;s not going to be aromatic so d is the answer d is a non-aromatic compound a b and c are aromatic number 15 which compound is consistent with the h nmr spectrum shown below so if we look at the functional groups present we can see that we have an aldehyde an ester an ether and a benzene ring the first thing that we could distinguish is the aldehyde so we&amp;#39;re focused on the hydrogens though the aldehyde hydrogen has a signal around nine to ten and the fact that we don&amp;#39;t see that on the graph means that we can eliminate answer choice a now the next thing that we can take uh that we can focus on is the signals around 6.5 to 8.5 if you see signals in this region um that that is the um the hydrogens on the benzene ring so the fact that we have those signals means that there should be a benzene ring so we can eliminate answer choice b now for the other signals that are between that are less than four there&amp;#39;s some things that you need to know if you have a regular ch3 next to let&amp;#39;s say a ch2 a regular so basically an alkane ch3 if nothing special around it the signal is going to be around one now let&amp;#39;s say if you have a carbon next to a special group like an electronegative atom we&amp;#39;ll call it x it can be an oxygen it can be a nitrogen it can be a chlorine or bromine usually that signals around three to four and let&amp;#39;s say if you have let&amp;#39;s see where i can put this one a ch3 next to a carbonyl group that signals around two to two point five so we&amp;#39;ll say like 2.3 and also if you have let&amp;#39;s say a ch3 next to a benzene ring that signal will also be around 2.3 so now let&amp;#39;s analyze c and d let&amp;#39;s start with compound d here we have a ch3 next to a carbonyl group so that&amp;#39;s going to be around 2.3 and here we have a ch2 next to an oxygen so we&amp;#39;re going to say 3.5 it&amp;#39;s supposed to be between 3 to four and here we have a ch3 so we&amp;#39;ll say that&amp;#39;s around one the ch3 is next to the ch2 so we have those signals in the graph so now let&amp;#39;s look at um institute c here we have a ch3 as well um that&amp;#39;s 2.3 actually no this ch3 is not 2.3 this ch3 is actually next to an oxygen not to a carbonyl group it&amp;#39;s a small little difference but it&amp;#39;s very important so because that ch3 is next to an oxygen we have this signal 3.5 here we have a ch2 next to an oxygen so that&amp;#39;s 3.5 and a ch3 next to a ch2 and that&amp;#39;s around one so answer choice c we should have two signals around 3.5 not one so we could eliminate answer choice c which means the correct answer is d now just to review we can also look at the spin splitting pattern this ch3 doesn&amp;#39;t have any adjacent carbons that have hydrogen so it&amp;#39;s a singlet and as we can see we do have a singlet around two looking at the ch2 it&amp;#39;s adjacent to a carbon that has three hydrogens so using the n plus one rule it should have a splitting pattern of four so it should appear as a corten and this we do see the quartzite here it splits in four the signal splits in into four smaller peaks and the ch3 is adjacent to a ch2 so two plus one is three so that&amp;#39;s a triplet and we do see the triplet around one so compound d is the correct answer

Transcript for:Organic Chemistry 2 Practice Tests: Multiple Choice Review

Transcript for:
Organic Chemistry 2 Practice Tests: Multiple Choice Review