hi everyone mrs hansen once again uh it's a nice sunday morning and i'm going to talk to you further in our chapter on uh qualifying acidity so kind of picking up with our acid-base lessons of chapter 3 in organic chemistry you know the last time we were together we were using a pka chart to compare strengths of acids and bases looking at a quantitative value you know a number provided to us in a published chart here in our next section 3.4 qualifying acidity we begin to compare acids and bases by analyzing and comparing their structures really without using the published chart of pka values we're really able to determine what side of an equilibrium would be favored which one is going to be the stronger acids just by comparing two structures and their ability to stabilize the negative charge so let's begin really through just thinking through what stabilizes a negative charge so to determine the relative strengths of two acids and we're not being given a pka value we are able to compare the stability of their conjugate base and understand that the stronger the acid the more stabilized its conjugate base becomes so an important statement there the conjugate base must be able to stabilize the lone pair of electrons that resulted from the loss of the proton so to determine the stability of a conjugate base we're actually looking at the stability of the lone pair for example let's say for example we have oh just an acid such as acetic acid uh ch3cooh you hear that at the end this is acetic acid its conjugate base of course is the loss of its proton and you would see this then as o negative we're going to ask ourselves for the conjugate base and i have to put all those dots in there there's this conjugate base how well is this negative charge stabilized by the structure when i say the negative charge remember that that's really a lone pair of electrons in other words remember when this whatever this base was when it reached out to abstract the proton this shared pair of electrons in the covalent bond collapsed back onto the oxygen the bond became a lone pair of electrons on the oxygen how well does this resulting species this conjugate base stabilize that negative pair we want the most stabilized structure so that this negative charge isn't being placed on an atom that can't handle it so remember the strength of an acid how easy is it to remove the proton the easier it is the more stabilized that negative charge is the stronger the acid becomes so that's what our game is here in this little section is deciding is this going to be a strong acid and if so it's because the resulting conjugate base does a fine job of stabilizing the negative charge and in order to stabilize the negative charge the conjugate base we look at four main factors right so the effectiveness of a conjugate base's ability to stabilize its negative charge that's the lone pair of electrons remember just looking at that acetic acid example what used to be a bond between the oxygen and hydrogen if the proton is removed this bond collapses and we form a lone pair that lone pair creates a negative formal charge on the oxygen how good of a job does that oxygen do at stabilizing the negative charge so the more stable that lone pair of electrons becomes on the conjugate base the stronger the original acid was there are four major factors that influence the ability to stabilize the negative charge and we're going to go through these one at a time notice the first says the type of atom that is carried on the charge here the type of atom was an oxygen so identify the element that's carrying the lone pair of electrons through the loss of a proton in a proton transfer is that negative charge going to be stabilized by resonance is there a high electron withdrawing group that will influence the ability to draw electron charge towards itself is there a highly electronegative element that process we'll look at is called induction and what type of orbital resides on the charge is it a tetrahedral three domain sp3 is it a trigonal planar two a total of three domains this is four domains three domains trigonal planar sp2 or is there just two domains in a linear molecule of an sp hybridized orbital all four of those considerations help us determine the the uh stability of the conjugate base and from there we're able to determine the strength of acids in comparison of one structure to another we remember this order of factors as the term areo a-r-i-o what type of atom is on the negative charge is the negative charge remember the negative charge is a pair of electrons is the pair of electrons being stabilized by resonance do we have a nearby electron withdrawing group producing the induction effect and what type of orbital is that negative charge reside in is it a tetrahedral sp3 a trigonal planar sp2 or a linear sp hybridized orbital as we go through each one of these examples we begin to examine the structure of the actual conjugate base in its ability based on its molecular structure to stabilize negative charges that's the topic of our fourth section erio the first letter in our little acronym area stands for the term atom what atom is carrying the negative charge so in order to compare acidity of two compounds let's take a straight chain carbon remember this is a saturated hydrocarbon and at the terminal end of the straight chain we have a hydrogen four carbons in a chain we'll learn to call butane here we have one two three carbons in a row and at the end we have a hydroxyl group the o h we've learned to recognize this as an alcohol functional group so what we're comparing is the hydrogen that's attached to an oxygen or the hydrogen that's attached to a carbon to analyze the acidity of the terminal proton of butane and remember there's again like protons because we know every carbon must have four bonds there are one two three four five six hydrogens that are equivalent hydrogens in that structure we're just looking at one of the six possible hydrogens doesn't matter which of the six we look at they're all equivalent hydrogens at the end of a carbon chain when we look to see the removal of the proton we draw what's known as their conjugate base remember that an acid and its conjugate base only differ by the presence or absence of the proton when i remove the proton remember removal of a proton causes a bond to collapse we get a lone pair of electrons this is now a carb and ion it has a set of lone pair electrons on that terminal carbon here we have an oxide ion the lone pair that collapsed from the bond when the proton was abstracted now gets placed on the oxygen so when we think about stability we have to envision where that negative charge is landed and to do so just sketch out the conjugate base of the acids provided and now you have a visual where we see the negative ion here is on the carbon the extra lone pair electrons the negative charge here is on the oxygen all right so now we have an envisioned where the negative charges where the extra lone pair electrons have been placed which one is more stable and so really we have two periodic trends to consider when we want to determine whether oxygen or carbon will be a better atom at stabilization ask yourself two questions and it depends on if the periodic trend is in the same row or is the periodic trend in the same column so number one says when i want to look at two atoms this is carbon versus oxygen the larger the atom the more stable a negative charge will become that's the most important factor the more you know the more electron shells there are the larger more polarizable electron cloud we have it has a greater ability to stabilize the negative charge however if you think about the location of carbon and oxygen they're the same period period two straight across the row carbon and oxygen really are not that different in size they have the same number of valence shells that's in period two they're both occupying their outermost electron energy in valence shell two so since they're in the same period they're very similar sizes so in that case we defer to a second way of determining who stabilizes the better the electric charge the negative charge a little bit better than the other and that goes by its electronegativity the more electronegative atom will be better at stabilizing the negative charge now just visually on our periodic table we had mentioned that carbon and oxygen were in the same row we said that that's period two so really the size is not the biggest influence because they're very similar in size because the size from the same row gives us the same number of energy levels really the size difference is not significant and so then we have to determine using electronegativity remember fluorine is the most electronegative element it has a published value of 4.0 but just ask yourself in that same row who's closer to fluorine and clearly you see oxygen is the more electronegative element it is because it has more protons packed in its nucleus as compared to carbon and therefore more protons will attract electrons more so than an atom with fewer protons so between carbon and oxygen oxygen is more electronegative therefore you can go back to the original question and then determining which one is the stronger acid answering that says well the stronger acid is the one that stabilized the negative charge better as a conjugate base and therefore the more acidic proton will be attached to the oxygen and not the carbon remember i just wanted to emphasize if the trend is comparing columns such as o versus s who would better stabilize a proton such as this example this is the same column not the same row these guys both live in group 6a on the periodic table we want to pick the larger atom it has a more polarizable more think of it as a squishier cloud a larger electron cloud this is uh n equal three in period three for sulfur and equal to only two electron shells in the carbon atom the larger carbo the larger orbitals outside of the nucleus are able to stabilize the charge there's just a greater distance to do so so since this is a larger electron cloud a greater surface area to distribute the pair of electrons in will help to stabilize more so than a smaller atom which has a smaller surface area to pack in some more electrons to try to stabilize the electron charge so in the same same row the more electronegative element will be able to stabilize the negative charge better in the same group same column you want to pick the larger atom to help stabilize the electron charge and so putting that together ultimately we were asked which one was the stronger acid because the oxygen was more electronegative it was able to stabilize the negative charge and therefore when you think back to the original acid this acid that proton on the alcohol functional group is a stronger acid than the butane which did not do such a good job as stabilizing the negative charge when i say stronger acid which one of those two acids is more likely to give up its proton so more likely to give up its proton and in answering that question it's more likely to be able to give up its proton because when that bond collapses and places a lone pair onto the oxygen the oxygen can stabilize it it's a more electronegative element and it has more protons in its nucleus to help stabilize the extra electrons so let's take a look at some practices i just pulled from your homework we're going to compare two protons in this structure and remember we have a series of things to consider the atom does it exhibit resonance resonance is there an inductive effect and what about the orbital but so far i've only said one of those four criteria so that's going to be our focus i'd like to do all four of those and then at the end say okay let's consider everything we've learned and try some more practice but one at a time then we'll put it all together so when i'm asking you to compare the two protons in this structure to find the more acidic proton i'm really only asking you consider one of the four criteria at this point what about the atom so here's our protons kind of highlighted in red and highlighted in blue the first highlighted in red is attached to a nitrogen the next highlighted in blue is attached to an oxygen since these are in the same row [Music] and that row is number two same period size is not really an influence because they're in the same row we have to go by electronegativity which one of these is best able to stabilize the negative charge is the more electronegative element so who's closer to fluorine is it nitrogen or is it oxygen so kind of thinking that through you clearly see oxygen is the more electronegative element and so here's kind of the out loud thinking visually do you you know look at the conjugate base when hydrogen is removed from the nitrogen remember how this bond collapses and it forms the second lone pair on the nitrogen this is the conjugate base of the red proton up above here if this hydrogen is removed the bond collapses and puts an extra pair of electrons on the oxygen this is the conjugate base of the blue proton which one is doing a better job at stabilizing that negative charge and we agreed that it would be the oxygen the more electronegative charge so even though they're in the same period we see that the more acidic proton comes from the more electronegative element this proton colored in blue was more acidic due to the fact that its conjugate base was able to stabilize the negative lone pair electrons better here's a couple more from homework you're going to see quite a few of these just trying to decide which one would be a better uh better at donating its proton to be a stronger acid it's a good proton donor and the answer is which one is best able to stabilize the negative charge the proton that's highlighted in red is attached to a carbon the proton highlighted in blue is attached to an oxygen now these two atoms are in the same row they're both found in period two so size really is not the influence here so therefore since it's in the same row we have to go by electronegativity who is the more electronegative element between carbon and oxygen and clearly you see oxygen as it's closer to fluorine because oxygen is more electronegative it's able to stabilize the negative charge think about what i've just said this oxygen when it loses a proton has a better job at stabilizing the charge because it's a more electronegative element therefore the blue proton is indeed more acidic here's an example where the proton is attached to a sulfur and the blue is attached to an oxygen i'm going to write that above because remember now oxygen and sulfur well they're both in the same group number group 6a and we know that as you go down the same column on the periodic table atoms get bigger so the bigger atoms are adding valence shells so remember here we have sulfur and period two i'm sorry oxygen and period two so it's putting its electrons in the outermost energy level of the second shell since sulfur lives in period three it has three valence shells which is a larger surface area which means a greater surface area can distribute the negative charge better you know it has more area to distribute the negative charge and therefore it will be able to stabilize that negative charge better that means the red proton is more acidic than the blue proton sulfur's larger surface area distributes the negative charge stabilizing it better than the smaller atom of oxygen let's try the next letter together the r area stands for resonance the letter r the ability to stabilize a negative charge which is that lone pair of electrons really increases with the ability to spread it out across multiple bonds so if you notice a proton adjacent to a pi bond we have the ability to form resonance structures and those resonance structures highly increase the stability because we can take that negative charge and distribute it through multiple places so for example in an ethanol this is the functional group we are learning that is an alcohol the proton when it collapses like so this a hydrogen gets abstracted by some base and it collapses the bond and so you have a negative bond here a negative charge you have this oxide anion o negative but here in acetic acid when this proton gets abstracted and the oxygen collapse you know this oxygen receives a pair of electrons when this hydrogen is removed what that does is form a negative charge here now they're both on the oxygen so both pair of electrons you know both negative charges let's just draw that both negative charges are on the same atom they're both on an oxygen so the first area the atom doesn't play a role here because it's the same atom the atom that's got the negative charge is oxygen in both both case so then i go down to the next letter r r stands for resonance if the atom didn't help me determine which is the better acid and the better acid is one that stabilizes its conjugate base i go to r which is resonance can this negative charge be resonance stabilized well the answer in the alcohol is no there's no pi bond in this particular molecule at all these electrons are what we refer to as localized electrons they have no ability to move through a process of resonance however the bond when it collapsed on the carboxylic acid carboxylic acid can be delocalized in other words they can be resonance stabilized the extra set of electrons can form a pi bond here this pi bond would then come up onto the oxygen and we form a resonance stabilized structure where now the oxygen is carrying the negative charge up here remember there were two dots i didn't draw them in and we could have a pi bond forming in that particular spot that's a resonance stabilized structure it can go back and forth and delocalize the pair of electrons delocalizing a pair of electrons increases stability resonance stabilized structures increasing stability through the delocalization of the pair of electrons so which one is more acidic well this guy right here because his lone pair of electrons will be resonance stabilized and you can see the structure that we drew for resonance structure we called the oxide anion from the alcohol having very localized electrons there is no resonance available no pi bonding at all in that structure versus the carboxylate and ion that does exhibit delocalized electrons resonant structures are delocalized electrons and therefore that produces a more stable conjugate base and therefore acetic acid is a stronger acid than an alcohol such as ethanol so in our area we look at the atom if we can't make a judgment by the atom we go to the next letter r resonance is any of the electrons the pair that's forming from the abstraction of a proton does that have any ability to be stabilized by resonance and if so that's the more stable structure therefore its parent acid is stronger for example which proton is more acidic the red proton or the blue proton notice the a for an area the atom well it's the same in both cases so that's a non factor this proton is attached to an n this proton is attached to an n so the atom will not play a role in determining our answer so we go to the next letter that next letter is resonance r resonance does either of those a lone pair electrons have the ability to be stabilized through resonance and clearly you see a nearby pi bond and the blue proton the resonance structure if i were to abstract the proton in red we would get a lone pair and a negative charge on the red proton if this proton is abstracted it's left that bond collapses i see two sets of dots and a formal charge of negative one now on that nitrogen well these are localized electrons there is no nearby pi bond to distribute the negative charge however if i redraw and i take off the nitrogen i take off the hydrogen on the nitrogen that's uh this this highlighted blue now all of a sudden i have a negative charge on a nitrogen that has a pi bond nearby in just the right position the adjacent carbon is involved in this carbonyl group and that possesses a pi bond remember that vocabulary carbonyl is a c double bond o that means we could have a resonance structure form where the set of electrons on the nitrogen that came from the original covalent bond here that set of electrons can be delocalized and we start structures that move that double bond around forming for example that particular structure i've moved the electrons from up here and the oxygen is now carrying the negative charge and the nitrogen developed no formal charge three bonds and one lone pair so this is what we refer to as a delocalized set of electrons and here notice that when we draw sometimes i like to have my figures drawn as well um just in case my drawings end up a little sloppy we get this first we had called localized we get the second conjugate base the conjugate base for each of those protons is shown here the proton was abstracted the bond collapsed the first nitrogen that was in the left side is called localized the second hydrogen on the right when it's removed forms a delocalized charge the adjacent carbon has a pi bond and that's the criteria then for making sure that resonance can occur the delocalization occurs because this extra set of electrons can collapse form a double bond this double bond then forms another lone pair and you've moved the negative charge all the way up to the oxygen that helps to stabilize when you spread out the negative charge between bonds you're stabilizing the the negative charge remember it's not one structure or the other structure it's not flipping back and forth it's a hybrid of both structures so keep in mind you could think of that as a dashed line a delocalization of those electrons involved in that bond the more acidic proton is the one that will be resonance stabilized how about this one in ariel the first question we ask ourselves is to consider the atom attached to the proton will that will be removed well it's the same atom in this case they're both attached to carbons all right so the atom will not play a factor so no influence here because it's the same we go to the next letter in our arial acronym and that's the term resonance are either of these resonance stabilized so let's suppose i remove the first hydrogen i'll say is colored in red that puts an anion here we still have this hydrogen i didn't take it off this is the conjugate base for the red hydrogen remember there's a lone pair of electrons this is now called a carb and ion look to the left look to the right is it attached to an atom that's involved in a pi bond the answer is no therefore these are very localized electrons they have no ability to be distributed across multiple bonds let's compare that to the blue hydrogen and draw its conjugate base that hydrogen is still on but now that hydrogen has had its elect uh its proton removed and therefore i'm starting to draw a little face there's an extra pair of electrons and therefore a negative charge still a carb and ion the lone pair of electrons when the hydrogen was abstracted by some base the base pulls off this this hydrogen and that bond collapses and there's the the dots now that are on that carbon so now does this have the ability to stabilize its electron charge by distributing it through multiple bonds and clearly you see adjacent is a carbon that's involved in a pi bond and that's a good thing for resonance that makes these delocalized so the resonance structure that can set up these electrons can move to form a double bond these electrons can place the terminal carbon as the extra electrons and so what happened one two three four five six one two three four five six there we go the original double bond like so just follow the arrows here now is the double bond here now is the negative electron so this is resonance stabilized we are moving the double bond across a series of atoms and remember it's not one structure or the other it's not flipping back and forth it truly is a delocalized where you can think of that being evenly distributed between all of the atoms there in a further distribution of electron charge through multiple atoms helps to create stability so this guy is the more acidic proton how about this one well look here the red hydrogen is attached to a carbon and remember our a r i o the first criteria is the term atom well i always consider that first we go in order and if atom doesn't help us then we go to the next as well here the hydrogen is attached to a carbon here the hydrogen is attached to an oxygen the atoms are different they're in the same row and so therefore size is not going to be an influence size would dominate over electronegativity so remember if they're in the same column we compare size if they're in the same row we compare electronegativity oxygen is more electronegative it is going to be able to distribute the charge better on the oxygen to stabilize the charge and so right now i'm already having a hunch that this blue hydrogen will be more acidic just because it's attached to the more electronegative oxygen as compared to carbon now notice this negative charge you know when i form a conjugate base i want you to notice that when i abstract that proton you know this guy gets plucked off this bond collapses this has a negative charge and i can see that it could be resonance stabilized this guy here could be resin oops i forgot this guy um coming back here we have bum bum bum bum o h here's that double bond o when i think about what could happen in terms of resonance this negative charge could absolutely be stabilized through resonance this particular set of electrons can form a pi bond forcing this to come up and make the oxygen carry the negative charge and the double bond would be placed here and when i abstract the proton that was blue up there let's do that one two three four up to an h uh here we have a carbonyl and this hydrogen i'm leaving on just so you can see that's the one i'm leaving on and i'm going to focus on the blue colored hydrogen when it abstracts when it's gone when it's taken the extra lone pair of electrons that collapse get placed on that particular oxygen and so is this have the ability to be resonance stabilized the answer is yes because the set of lone electrons can form a pi bond there forcing this pi bond onto the oxygen and so that hydrogen is still there the oxygen now has two dots here and we have a double bond this double bond that was present heading down and that carboxylic acid group now has a negative charge here so both are both have the ability to be resident stabilized right so both the blue that's adjacent to a pi bond and the red is adjacent to a pi bond there so they both have the ability to form resonance so what would be the dominating factor and it's the first letter in that whole story problem is a the atom itself is going to create a stable conjugate base because both exhibit residents residence really wasn't the factor here it was the first letter the first letter stands for atom an atom of oxygen is more electronegative than the atom of carbon and therefore the proton attached to the oxygen is the more acidic proton let's move to the third letter in ariel and this is called induction induction is what's used to stabilize a formal negative charge by helping to spread it out and let's compare the term induction from the term resonance induction is when we have an electron withdrawing atom or groups of atoms that inductively withdraw electrons from their surroundings and i just think of this as based on polarity of the bond in the polarity of the bond we can see electron withdrawing groups not delocalizing not spreading out via pi bonds but simply by an induction effect is based on the polarity of the bond for example here is a proton attached to an oxygen here is a proton attached to an oxygen these a's stand for atoms and that is not an influence because they're attached to the exact same atom oxygen the next letter is r r stands for resonance well aren't these both going to have the exact same ability to exhibit resonance because both oxygens are adjacent to the pi bond in the exact same position so this means that there is no difference they're the same so the r cannot be used to determine the acidity it's the same for both structures the a atom cannot be used to determine more acidity because it's the same for both structures but what about i finally the inductive effect do you notice on the carbon there are very rich electrons around the chlorine these are known as electron withdrawing groups based on polarity of the bond you can see that the electron distribution is being pulled by a massive amount out towards the chlorine which help to stabilize negative charge the inductive effect these highly polar bonds are pulling electron density away oops let me catch up electron density away from the oxygen that's trying to stabilize it let me say these polar bonds are pulling electron density about towards the chlorine the more electronegative element and that's a place where you want it to go because chlorine has a great ability to stabilize negative charge so in other words when i think about these these charges if oxygen here has a negative charge it's going to be resident stabilized and it's going to be stabilized because the negative charge is also being pulled in this direction so this oxygen has all kinds of help stabilizing that negative charge it can do so through resonance and it can do so through the ability of the inductive effect where the chlorine help to pull electron density towards itself making this a more electron rich area whereas the acetic acid with this boring little hydrogen tail here has no ability to stabilize negative charge and therefore because the a is the same for atom r is the same for resonance i look to see if there's an inductive effect are the highly electronegative elements present in the molecule such as chlorine to help stabilize negative charge and that's certainly true with trichloroacetic acid it is a more acidic proton than plain old acetic acid when we have more than one electron withdrawing group such as monochlorine then dichlorine then trichlorine you can see an even greater inductive effect so here notice that acetic acid with three hydrogens here is a pretty strong acid 4.75 if i remove one hydrogen the three that we started with remove one hydrogen and add a chlorine this becomes an even stronger acid this proton is even easier to remove because this oxygen can stabilize the negative charge even better because of the inductive effect chlorine places on the electron polarity all of the electrons are being pulled towards the more electronegative element in addition to being pulled up as a resonance structure and now look two chlorines when this oxygen is trying to stabilize its negative charge it can do so not only through resonance but it can do so in two different directions towards the chlorine and the strongest acid here is the oxygen that can have not only resonance stabilization but can we kind of stabilize in three different directions onto those electronegative electron withdrawing groups known as chlorine so look to see if there's any nearby interesting atoms that can help to stabilize through the process of induction here we have an idea of a little molecule where i have a carbon attached to an alcohol group here is a carbon attached to an alcohol group this in red this in blue which would be the more acidic proton well a r i o ario they both are attached to the same atom it's carbon so that's a non-factor resonance well there are no pi bonds here at all so none of them are going to exhibit resonance so a and r did not help me at all in determining the more acidic proton so i'm left with the next letter the inductive effect do you see any electron withdrawing groups near the proton well i certainly do see all these fluorines those fluorines have the ability to withdraw electrons and when i say that here's the electrons on the oxygen these are leading out to carbons that have three fluorines on them all kinds of electronegativity in that region of the molecule so this negative charge on the oxygen can be can be stabilized by these electron withdrawing groups fluorine and i have a direction that's being pulled that these negative very highly polar region on the fluorines can help to stabilize the negative charge on the oxide notice that it's not delocalized these are localized electrons this is not resonance they're localized but the distribution of charge can be brought into greater stability through the inductive effect the ability of a polar bond to pull electron density towards itself not move the bond but help to stabilize the bomb and so clearly the red is more acidic than the blue and you can see just based on the structures we drew the higher electron withdrawing groups of fluorine make the o h hydroxyl group more acidic when it's near more exciting atoms how about this one find the most acidic proton they're not color-coded but i'm see only two protons that would be considered acidic protons here and i realize that there are two hydrogens here you know those could be part of the game but as soon as i go a r i o there's four protons there that i can consider the two that are attached to the carbon i take off the plate right away because here i have two attached to oxygen oxygen is more electronegative so quickly i can say all right now i'm really deciding between these two hydrogens because oxygen is more electronegative i can cancel out even worrying about the two hydrogens that are attached to the carbon in the center so let's do ariel a atom both of these that i have left highlighted in red are attached to oxygen okay so that's not going to make my decision i go to r which is resonance both of these will exhibit resonance the oxide here has a nearby pi bond that can help to delocalize so this will be a d localized electron pair that's the same thing over here these are carboxylic acids c-o-o-h this will also have a delocalized resonance structure so that's really not not a factor because it's the same for both so that leads us to i the only next letter i is going to have to make our decision the inductive effect which one is closer which oxygen is actually closer to the highly electronegative chlorines that would help to stabilize the next door electric charge well here i can see that there's the carbonyl carbon and then directly here so that's one two carbons away are the chlorines if i count here here's one two three carbons away before i get to the the chlorines the greater effect will occur when it's closer so therefore what i highlighted here would be more acidic it is closer to the electron withdrawing groups on those on carbon two it's only two carbons away instead of three carbons away the closer it is the more the inductive effect has the ability to stabilize the negative charge how about this one or this one which is most acidic ario they're both top attached to oxygen so this hydrogen and this hydrogen are attached to an oxygen so that's not a factor resonance this hydrogen is attached to yeah it's right next door to the pi bond these are both carboxylic acids cool they're exactly the same functional group so not a factor yes they both have the ability to be stabilized by resonance not a factor that leaves us to go down to i reo are any going to be you know have a greater influence in terms of the inductive effect notice we have the same distance away however this only has one chlorine where this has two chlorines two groups of electron withdrawing groups two groups of chlorine tell me that it has the twice the ability to stabilize the electric charge as compared to just one group of chlorine chlorine's not really a group an atom would be a better word wouldn't it caught myself two atoms of chlorine versus one atom of chlorine the more the better at stabilizing negative charge so which is more acidic obviously the one with two chlorines to help stabilize the negative charge over a greater surface area easy enough the letter o in ariel stands for the type of orbital so think about the ability of you know the the ability of stabilization of an electric charge and the distance it is to the nucleus to do so the more stable the charge the closer it is to the nucleus let's kind of just think that through what i'm trying to say when i think of orbitals i'm just thinking about the hybridization on the atom that's trying to stabilize the negative charge hybridization we know that if it's sp3 hybridized it's a tetrahedral molecular geometry tetrahedral sp2 would be trigonal planar it's attached to three atoms sp means that it has two electron domains and that it's linear when you think about tetrahedral trigonal planar and linear think about it this way when we have a straight chain carbon and all the carbons are attached to just hydrogens kind of you know all single bonds this is what we refer to as an alkane all single bonds these all have sp3 hybridization alrighty four bonds tetrahedral molecular geometry all single bonds compare that to an alkene so i'm coming along here and now all of a sudden instead of a all singles i'm going to have a double bond meaning that we've changed the molecular geometry just a little bit saying that it's now trigonal planar and this here now is what we refer to as sp2 hybridized in a double bond and i know this could continue up and on down but an sp2 at the terminal end is now trigonal planar this is what we refer to as an alkene and finally when we had a triple bond this is what we refer to as sp in terms of linear molecular geometry and this is what we refer to as alkynes we have an alkane alkene alkyne now the shorter the atomic orbital the closer it is to the nucleus let's bring this into a better picture yet here's the single bonds see how long single bonds are when we have a double bond they're shorter the electron density is pulling the atoms closer together making a shorter bond and when we have a triple bond the bonds are very close together and exaggerating but we have long single bonds and we have very very short tight triple bonds and the double lies in the middle doesn't it that the electron density is increasing with the triple bond between the two carbon atoms that tells me that this hydrogen on the terminal end of an alkyne triple bond is very easily removed because the negative charge that would form on this negative charge here is very easily stabilized because it's very very close to the nucleus of that carbon because of that triple bond pulling electron density so close to itself it's closest to the nucleus and that creates a more stable configuration compare that to a single bomb and if i were to draw the single bonds and draw you know those long electrons a shared pair of electrons when i draw a lone pair of electrons on this carbon the electrons are much longer or farther away from the nucleus right because this is a very long distance not really but a longer distance than a triple bomb therefore this is less stable because it's further away from the carbon's nucleus the shorter the bond the tighter everything is the electron density is being pulled really close to everything so when this proton is abstracted and this bond collapses these electrons get pulled all kinds of tight right there when this bond collapses these are very loosely held electrons therefore this is a less stable configuration so all of that to kind of talk through orbital when you compare a triple bond which we called an alkyne the bond distance here between these two carbons is very short that makes this hydrogen very easy to remove i should maybe say very easy i'll say easier easier to lose this proton because the negative charge is going to be stabilized and so just think about that negative ion that forms as much more stable with this electron rich area it's closest to the nucleus and that's where positive charges are the nucleus and therefore that creates stability compare over here this is a carbon that's involved in a double bond so remember this would be sp hybridized this is sp2 hybridized meaning that a double bond in that particular e the alkene part of this molecule when it forms its anion is not as readily stabilized as it was over with the triple bond because there's a longer distance between the pi bonds here that means that these electrons are a little further out from the nucleus and a little less stabilized than if it had electron rich triple bonds so the blue hydrogen is more acidic than the red hydrogen and the same is true if you consider sp hybridization as well so the most acidic comes from the triple bond the next would come from a double bond and the least acidic would come from a single bond what type of orbital also affects the stability of a negative charge the more s character in the orbital the more stable the negative charge what did i just say s p3 right sp3 that's four domains that's what we would say is a a tetrahedral four domains and that of course would be an alkane notice that the s is just one fourth of all types of orbitals one out of four orbitals so 25 percent are an s character what if i had sp2 here's an s p2 i could say that's p3 but i should have written maybe spp sppp is sp3 but now look this would be three domains it's trigonal planar this would represent the alkene family which is a double bond there is one out of three one s out of three total orbitals one third is 33 percent exhibiting s character and of course the alkyne is simply an sp orbital right sp is an alkyne there's only two orbitals in a linear sp right this is the linear alkyne 180 degrees so 50 of the orbitals in a linear molecule are the s quality so as the s character in the orbital increases the more stable the negative charge an alkyne is 50 s configuration the other 50 comes from a p it's the most stable conjugate base of the acids when we compare the acidity of the compounds the least acidic comes from the orbitals sp3 all single bonds in next we would have sp2 and the most acidic would come from sp a linear molecule the terminal end of an alkyne is much more acidic than the terminal end of an alkene which is more acidic than the terminal hydrogen of an alkane all with the ability to stabilize the negative charge based on the distance from the nucleus and the quality of the s character remember we said this is 25 s this would be 33.333 percent s character and this is 50 s character of the two orbitals half of them are s of the three orbitals a third of them are s of the fourth orbit you know sp3 of the four orbitals a quarter of them are s increasing the s contribution to the orbitals increases the acidity which one is more acidic the red or the blue well i know right it's going to be easy here this carbon has a double bond it is sp2 hybridized that means it has three domains it's attached to three other atoms sp2 hybridized this carbon is attached to four other atoms it's sp3 hybridized it has four domains it's attached to four other atoms the one that exhibits the higher quality of s we'll have the more we'll have this the proton that's more acidic double bond a double bond hydrogen is more acidic than a single bond proton these are that i mean that's really going to be this structure when we say orbital the highest comes from the triple bond this is most acidic the double bond these two hydrogens would be in second place and the least acidic would be the hydrogens that are coming from sp3 hybridized carbons so the most acidic to the least acidic right so sp is the most acidic second place sp2 last place sp3 structure two here's a hydrogen coming off of the terminal end of that carbon this is an sp orbital therefore it is going to be more acidic than any of these sp3 hydrogens at the terminal end of those carbons so all of those hydrogens that are there but we don't see them but we know they're there all of those are equivalent hydrogens all of them exhibit sp3 in the last terminal hydrogen of an sp this would be the most acidic proton in that carbon structure notice here we have nine equivalent hydrogens on this side which are identical to the nine equivalent hydrogens here those are all sp3 hybridized finally here we have one two there's a hydrogen here and a hydrogen here notice that these are what we call equivalent hydrogens they are both sp2 so there's nine hydrogens that are equivalent in sp3 there are two hydrogens that are equivalent in sp2 so of course sp2 would be the more acidic proton but it's exactly the same acidity as this particular proton would have they're the same okay hey we've been working hard on a r i o right when assessing the acidity of protons we generally use ario aereo as the importance of stabilizing the effect and i want you to go right down the order look at the type of atom if no decision can be made go to residence and if no condecion can be made look to see if there's an induction effect and then finally consider the orbitals what is the molecular geometry is it linear trigonal planar or tetrahedral based on those four types of things we are able to figure out which structure is going to be the better acid and that answer lies in which is better at stabilizing the negative charge now in the little disclaimer down here it says this works most of the time but it's not always a hundred percent reliable as with any set of rules there's always rule breakers these exceptions to the rule i'm going to tell you the vast majority the vast majority of these examples we will fall under follow the aerial and i'm going to show you one exception one exception that you should kind of just remember so for instance here we have ethanol and we're going to consider polypropylene here's ethanol it's an alcohol functional group here's the h here is the hydrogen on propylene this particular carbon right here is attached really to one two other three you know there's three equivalent hydrogens there keep in mind and its pka is 43. when we look at the conjugate base we remove the proton and let the bond collapse and we form a formal charge of oxygen and a formal charge on carbon clearly we would predict according to ariel that the oxygen because it's more electronegative would be better at stabilizing the charge and that therefore this is the more acidic proton and we'd be absolutely correct the type of atom is consistent with this fact even though this here this propylene if i go down the list a r i o the first consideration is the atom we were able to make a judgment right off the first criteria and say wait wait wait oxygen more electronegative than carbon stop the bus right there and select the more electronegative element if i chose to keep going and went to the r next even though i made a decision let's just say i'm oblivious to saying oh stop right there i'm going to go down to r you know this guy is resident stabilized right there is a resonance structure where the bond can form from these adopts and this bond goes to form dots there's a resonance structure so i might be fooled into thinking golly you know since that has resonance he might be more stable than the other structure which has no resonance but that's not the way the game works i want you to go down in order a adam if no decision can be made then go to our residence and then if no decision can be made go to i and finally the last consideration is oh the orbitals as soon as we were able to decide based on the atom stop going down the list ario is a guideline it's a priority guideline and it sometimes fails and i say that just so rarely it does that i don't want you to be alarmed and thinking that oh why should i you know kind of understand ariel and and use it if it's going to let me down frequently and the answer is well it really won't i promise promise it really won't there's just one common exception to aereo and if i show it to you you'll go okay i'll remember that this decision here's an acetylene this is the acidic proton of an sp hybridized carbon it has a pka of 25. over here we have an acid ammonia with a pka of 38. which one of those acids would be more stable so here we have just look at the conjugate base that forms c triple bond c puts a negative charge on carbon here the ammonia nh3 nh3 well this proton again there's no negative charge there nitrogen has no formal charge in that particular structure the proton that was removed was attached to a carbon the proton that's removed to force its conjugate base is attached to a nitrogen so its conjugate base would place a negative charge on nitrogen so this is the acid this is its conjugate base so if we have a negative charge on nitrogen and we have a negative charge on carbon we would consider using reo and rightly so that the nitrogen being the more electronegative element should be the stronger acid but it's not if we had to judge the conjugate base stability we would have concluded that the negative charge on nitrogen would be more stable and therefore i would conclude that the ammonia must have been the stronger acid but published pka values verify that that's not true it's just one exception when i look at the terminal end of an alkyne it's a pretty strong acid more so than even the more electronegative element nitrogen so just kind of remember this guy even though it's attached to a carbon it's kind of a strong acid just kind of look for that and you'll be able to use and and also know you'll have a pka chart so when in doubt just look up the pka values so there's always exception any rule this is just one exception i want to show you to where you know based on structure i would predict the stronger acid to be the nitrogen because it's more electronegative should be able to stabilize the charge better but the actual answer is not because this triple bond does an incredibly nice job of stabilizing a charge just because it's so close to the protons and the electron density here between the atoms is very rich it can actually stabilize the negative charge a little bit better when in doubt check your pka chart i'm going to just keep practicing a little bit if you're hanging with me i'm doing some homework problems it says let's use ariel to predict the more acidic proton well here we have a r i o let's check the atom first the acidic proton here is on an oxygen so we would have oxygen stabilizing the negative charge hydrogen here is also attached to an oxygen so it would have oxygen stabilizing the negative charge it's the same atom they're both oxygen so this is no decision how about r resonance this red proton will be a delocalized resonance structure because it has the carbon that's involved in a pi bond so you can see when i have this conjugate pair form and this oxygen has its negative charge i can see that i can distribute this negative charge and make this electron you know the extra negative charge ultimately end up on that oxygen atom so i have this drawn too far that's why it looked bad may i repair that quickly here is the oxygen this is the negative charge so that's where the double bond would form that's why it looked upsetting to me as i just had that drawn wrong and so the resonance structure would end up looking like and i know this goes on it goes on up there is a single bond a negative charge and a double bond here and so that's resonance stabilized so the red one is yes it is resonance stabilized and how about this blue does it exhibit resonance the actual answer is no because the carbon nearby does not have a pi bond involved and so there's no resonance if i went on and saw inductive effect the hydrogen exhibits inductive effect because i can see all types of chlorine here the orbitals are really no factor all right no factor because they're all i mean i don't see double triple bonds of carbons so keep in mind you're going in order as soon as i found that the left molecule exhibits resonance but the right molecule did not i don't have to go any further if i did go further i notice that the right blue proton does have influence by the inductive effect whereas the right the left side the red proton has no inductive effect so if i'm trying to decide between resonance and induction go right down the acronym a r i o residence beats induction and therefore the more acidic proton is the one in red how about here which one is more acidic well if you consider removing both protons and just look to see what conjugate base forms we have the first conjugate base that looks like this and the second conjugate base that would look like this right so a r i o the atom make that your first decision the red proton is attached to an oxygen right the blue proton is attached to a carbon a is a very first decision can we make a decision right now a attached to oxygen or attached to carbon well the answer is absolutely oxygen i can automatically make my decision right here that's the more acidic proton and yes i can see resonance involved there is no induction there is no orbitals to consider but both have resonance a beats r how about another well ario this red hydrogen is attached to an oxygen this blue hydrogen is attached to a nitrogen those are different atoms so right off the bat a is the atom oxygen is more electronegative than nitrogen so i'm going to pick this one just based on the first criteria as being more acidic both exhibit resonance there is no inductive effect this is a non-consideration so again i'm just going between a and r both exhibit resonance they're both you know there is a pi bond in the adjacent carbon so that is a delocalized set of electrons but that's the same for both and if i'm comparing which atom can stabilize an electric charge you want to pick the more electronegative element and that would be the oxygen how about another oh my goodness we're doing lots all your homework problems worked out for you if you're still with me wake up on you how about the red or the blue here well look at this the a r i o they're both attached to the same atom the atom that they're attached to are carbon so that's not really a good decision maker because it's the same atom so i'm going to go down to the next resonance when i look at drawing this negative charge on carbon it's adjacent to a carbon in a pi bond so yes this can exhibit resonance and if i were to put a set of electrons here oh my goodness look look left and look right and i have two places that can exhibit resonance so this has a greater resonance i'll do this greater number of resident structures because i can delocalize that bond across both carbonyl carbons remember the carbonyl functional group is a c double bond o this set of dots on this carb and ion has access to two carbonyl groups so that's going to say oh my gosh all kinds of resonance structures can be drawn the delocalized bond just kind of feeling that through would end up having a structure that would allow us to delocalize all over this particular surface area whereas the blue hydrogen only has access to delocalize across this particular surface area so greater number of resonance structures allows me to predict that the red will be more acidic the red hydrogen will be resident stabilized by two carbonyl atoms the two oxygen groups greater surface area to destabilize more resonance structure cool you know we can also use these structures this area system to help predict the equilibrium position for any acid or base let's see how long i've been talking it's been an hour and 20. i'm going to stop the video here and pause and then let you take a little break and come back you