hello everybody my name is Iman welcome back to my YouTube channel today we're going to do a practice problem set that relates to our carboxilic acid derivatives lecture let's go ahead and get started this first problem says which of the following would be the best method of producing methyl propano now methyl propano is drawn drawn right here it's an Esther so we need to recall how we can produce an Esther all right it can be synthesized by reacting a carboxilic acid with an alcohol in the presence of an acid now here if we look at methyl propanoate we notice that the parent chain has three carbons three carbons so if we're thinking about starting off with a carboxilic acid reacting with an alcohol to get methyl propanoate then that carboxilic acid is also going to have three carbons and so it's going to be called propanoic acid so we're starting with a carboxilic acid called propanoic acid and then we're going to be reacting it with an alcohol what kind of alcohol well if we look at methyl propanoate the estring group is a methyl group so it's going to have to be an alcohol that has a methyl group and that is called methanol all right and so if you react propanoic acid with methanol you're going to get methyl propanoid of course in the presence of an eye Aid and the answer choice that best aligns with that is going to be answer Choice a one is a reacting propenoic acid and meth and methanol in the presence of a mineral acid fantastic two says what would be the products of this reaction all right this question's asking for the products when ammonia reacts with acidic and hydride recall from the lecture that an amide and a Caro IC acid is going to be formed all right so we talked about how we can do an hydride cleavage under our third objective nucleophilic AAL substitution reactions all right and we talked about the fact that anhydrites can be cleaved by the addition of a nucleophile and what that nucleophile is determines what the products are specifically with the addition of ammonia or an amine all right what your anhydr is going to be cleaved into is an amide and a carboxilic acid so you're going to have an amide and a carboxilic acid those are the two things that are going to be formed all right now here's the thing the carboxilic acid an acid is in the same environment as ammonia which is a base and actually what's going to happen is that the two are going to react forming the ammonium carb carboxilate that we see in answer Choice D all right so if in in the case that you are wondering what this is all right that is carboxilic acid but because this acid is in the same environment as ammonia which is a base you're going to form ammonium carboxilate so you're looking for your answer in your answer choice to have an amide which is present right here and a carboxilic acid or at least car you know ammonium carboxilate since it's in the presence of a base and you see both of those answers present in answer Choice D so two is d three says which of the following undergoes a Fisher esterfication most rapidly so a Fisher esterfication involves reacting a carboxilic acid and an alcohol with an acid Catalyst all right so carboxilic acid plus an alcohol is how you get fish esterification all right under these conditions the carbonal carbon is open to attack by the nucleophilic alcohol all right now the rate of this reaction really depends on the amount of steric hindrance around that carbonal carbon because there has to be room for the alcohol to approach the carboxilic acid substrate so which of the following under goes Fisher EST esterification most rapidly you're going to be looking looking at the carbonal carbon in all of these and you're trying to find which one of these options has the least amount of steric hindrance around that electrophilic carbon site all right now if you if you look at these answer choices all right especially CN D re right off the bat it looks like there's a lot going on near the that carbon that Alpha carbon um to the carbonal carbon there's like three groups of three ethyl groups in an answer Choice d three methyl groups in answer Choice C and here even in answer Choice B at that Alpha carbon right here there's two methyls whereas if you look at a there is just one methyl group so you're looking all right for an answer choice that has the least amount of steric hindrance around the carbonal carbon that's going to be answer Choice a b c and d all have more crowding around the carbonal carbon which will de decreased reactivity all right in addition those additional alkal groups in in BC and D also donate electron density to the carbonal carbon making it less electrophilic so that's going to make the alcohol less likely to react with those sites in comparison to answer Choice a all right so three is a four says each of the ACL compounds listed below contain a six membered ring except blank all right a says Delta lactum so that's right here all right B says cyclohexane carboxilic acid I drew that out right here C says uh gamma but uh butol lactone that is drawn right here all right and then D says the anhydride formed from intr molecular ring closure of pentane diode acid I went ahead and Drew that out for us now the question requires knowledge of The Gnome and clature of cyclic molecules all right right off the bat because all four of these options are cyclic molecules now deltal lactum has a bond between the nitrogen and the fourth carbon away from the carbonal carbon this is what it looks like the ring will have six elements yes the nitrogen the carbonal carbon and the four carbon in between all right answer Choice B all right so um for a it is a six membered ring okay so we're trying to figure out which one of these four options is not all right the exception so a is a six-membered ring so four the answer for four is not a we can cancel that out now if we look at answer Choice B cyclohexane carboxilic acid has a cyclohexene which is a six membered cyclohexene so this has a six membered rink so B is also not the the answer here now the anhydride formed from the pentane dioic acid in answer Choice D will have the five carbons in the parent chain in addition to this carbon right here all right the one oxygen atom that's closing the ring which means it also has a six membered ring so it's not going to be answer Choice D all right if we look at C however we notice it only has five elements in its ring all right and that's because it contains a bond between the Esther oxygen and the third carbon away from the carbonal carbon which results in a five membered ring and so the all of the choices contain a six membered ring except for answer Choice C so four is C beautiful five says which of the following would be most reactive towards nucleophiles we have propy ethano propenoic acid propanamide propanoic and hydride with the same R Group as you notice steric influence is the same in each case so we can therefore rely solely on electronic effects and when this is all that is taken into account reactivity towards nucleophiles like we said in leure is highest for anhydrides then Esters and carboxilic acids then amides all right so we said it right there highest reactivity towards nucleophiles is highest for anhydrides and so the correct answer for five is going to be D propenoic and hydride beautiful five says how might succinic and hydride shown below be formed from succinic Acid AKA buttin butane dioic acid all right so and hydrides particularly CYCC anhydrides like we see here will form spontaneously from dicarboxylic acids when heated that's what we always said in our lecture whenever we were dealing with intr molecular reactions within a anhydride molecule to form cyclic anhydride it happens when heated so cyclic anhydrides will form spontaneously from dicarboxylic acids when heated and so what do you require answer Choice C six to C heat beautiful seven says which of the following would react most readily with a carboxilic acid to form an amide so you're going from carboxilic acid plus something to form an amide what do you need all right so we know that we're going to need an amine all right right we talked about amide synthesis carox acid with an amine ammonia or an amine gives you an amide all right what you notice is all the answer choices are a means so there's another caveat that we learned that we have to remember here for this specific problem all right the rule here is the less substituted the nucleophile the easier it will be for the nucleophile to attack the carbonal carbon and form the amide so we're looking for the least substituted nucleophile in other words the least substituted amine we have methyl amine triethyl amine three ethyl groups dieny amine two pheny groups or ethylmethylamine which one of these is going to be the least substituted nucleophile it is obviously going to be a methylamine would re react readily to form an amide all right and that's because the less substituted the nucleophile the easier it will be for the nucleophile to attack the car caral carbon and form the amide beautiful 8 says if propane amide were treated with water what products would be observed all right so we have an amide and you are treating it with water let's think about this so propane amide is an amide as such it's going to be the least reactive of the carboxilic acid derivatives that we discussed in in this chapter now with without a strong acid or base propane amide will not be able to actually undergo nucleophilic AAL substitution all right it has to be treated with some sort of strong acid or base without that no reaction would occur and so you stay with propanamide nothing happens so 8 is a nine says betal betal lactum draw right here are blink one says cyclic forms of the least reactive type of carboxilic acid derivatives two says more reactive than their straight chain counterparts and three says molecules with high levels of ring strain all right beta lactum are amides in the form of four membered rings like you see here and they are generally the least reactive type of carboxilic acid derivatives remember the hierarchy for highest to lowest reactivity for our carboxilic acid derivative goes and hydrides then Esters then amides now betal lactum if you just look at this you can easily guess that they do experience significant ring strain from both eclipsing interactions AKA torsional strain and also the angle angle strain and so they are therefore more susceptible to hydrolysis than the linear form of the same molecule so if we look back at 1 2 and three each of these statements is absolutely true and so the correct answer for nine is going to be D 10 says the acid catalyzed conversion of propyl ethanoate to Benzel ethanoate is likely blank all right we're going from Propel ethanoate to Benzel ethanoate let's talk about this as far as we can tell we're just converting one Esther to another in this reaction all right the fact that this reaction is acid catalyzed should confirm all right that this is transesterification all right just like we talked about in lecture we're just converting from one Esther to another all right and there's and it's acid catalyze so this is transesterification 11 says the reaction shown which is important to break down of polypeptides would be favored under what conditions this reaction if you look at it this is the hydrolysis of an amide amide all right it's favored in strong acid acid protonates the carbonal oxygen which increases the electrophilicity of the carbonal carbon and this allows water to serve as the nucleophile attacking the bond and then hydroling the molecule so the answer here is that this would be favored under acidic condition I or an acid environment so 11 is B 12 says a positive charge on the molecule shown below would have greater stability than a positive charge on a straight chain alkane version of the same molecule what property would explain this effect all right this molecule is going to be more stable with a positive charge than a straight chain alkane due to the conjugation that we have in this Benzene ring this permits delocalization of the charge through resonance and that helps stabilize this structure all right now that's going to be due to again the conjugation of this Benzene ring all right which means that the answer here for 12 is going to be D all right again this molecule is going to be more stable with a positive charge than straight chain alkane because of the conjug ug a of the Benzene ring that conjugation permits delocalization of the charge through resonance 12 is D 13 says the molecule shown is blank one says synthesizable from hydroxy carboxilic acid two says this is a lactone three says a form of an Esther all right let's talk about this the molecule shown here is actually gamma non Nona lactone it's a cyclic Esther also called a lactone this molecule would arise from intr molecular attack in a gamma hydroxycarboxylic acid which means that all three of these statements are absolutely true and the correct answer for 13 is D 14 says which reactant would be combined with butanol to form butal acetate now in order to prepare butal uh butal acetate from butanol we need to perform nucleophilic AAL substitution reaction If the product is an Esther we're going to need to start with a reactant that's more reactive than the es Esther itself or the reaction will not pred now anhydrides are more reactive than Esters but amides are less reactive and so with that if we look at C and D we can go ahead and eliminate those now reactions with propenoic and hydride like we see in answer Choice B would actually result in butal propanoid and that means the correct answer here all right which reactant would be combined with butanol to form butal acetate it's going to be answer Choice a 15 says why should esterfication reactions not be carried out in water a says carboxilic acid ID from which Esters are made are generally insoluble in water B says the polar nature of water overshadows the polar nature of the leaving group C says the extensive hydrogen bonding of water interferes with a nucleophilic addition re mechanism and D says water molecules would hydrolize the desired products back into the parent carboxilic acid all right now the presence of water in an esterfication reaction would likely revert some of the desired Esters back into carboxilic acids small carboxilic acids though like formic or acetic acid are easily dissolved in water so this answer with generally insoluble in water not correct um the polarity of water plays a little role in affecting the leaving group if anything though water can be used to increase the electrophilicity of the carbonal carbon by protonating the carbonal oxygen so answer Choice B is also not a very good answer um C well finally we also should remember that this is a nucleophilic substitution mechanism not a nucleophilic addition mechanism and so C isn't a good answer all right what is true though is the water molecules would hydroly the desired product back into the parent carboxilic acid so 15 is D with that we complete our problem set let me know if you have any questions comments concerns down below other than that good luck happy be studying and have a beautiful beautiful day future doctors