well welcome back everyone this is Mrs Hansen again carrying on with our chapter 2 lessons we're up to section four which is actually quite a short section carbon atoms with formal charges and they were highlighted a little bit in chapter 1 as well remember that um typically speaking an atom of carbon in its most stable form will have four bonds so that's simply because carbon lives in group 4 a we represent those with four veence electrons it needs to have an octet and so it will do so by creating four bonds we said those could be four single bonds it could be two single and a double it could be one single and a triple but the requirement there of course is four bonds so that means that carbon has no formal charge if it has four bonds but what happens if it does have a formal charge and that's kind of where it's getting in this section it says carbo cations have only three bonds and therefore a formal charge of plus one so for example here's a carbon and I see that it has only three bonds and what it has here is an actual empty orbital that empty orbital is changing its molecular geometry from sp3 which is four bonds to an SP2 hybridize which makes it a very flat trigonal planer molecular geometry since it's having a third uh you know the sp3 orbital is completely empty it's missing electrons so here's an example of a carbo cat ion I see that it only has two bonds therefore I know that it has a third Bond as a hydrogen we just don't see it and here at the terminal position I can see that it has a formal charge of plus one meaning it has just three bonds meaning it must have two hydrogens even though they're not drawn we know they're there now a carbo an ion a negative ion has three bonds and an extra electron that's how it uh receives a negative formal charge remember formal charge just to kind of help us remember formal charge is defined as the group number it lives in So that's its veence electrons just based on these uh that's not spelled right veence electrons just based on where it lives on the periodic table minus the assigned electrons so I drew a circle around this carbon and inside that circle is 1 2 3 four five assigned electrons lives in group four but there's five electrons assigned making it a -1 so both carbo cat ions and carb an ions have three bonds but the difference between them is the nature of that fourth orbital that sp3 orbital it's empty in a carbo cat ion and it's occupied by a lone pair in an anion so this again would be sp3 hybridized when we look at Bond line structures we're going to practice identifying lone pairs of electrons and sometimes we see the the electrons are drawn and often times we see that they're not drawn so the dots are optional but what must always be included as the formal charge so for example if I see in this chain I see an oxygen written with a NE one I now know that this has a formal charge of-1 so therefore since oxygen lives in group six what number of electrons must be on that to give it a negative one overall and we can see that that would be seven electrons 6 - 7 is a -1 so assigned electrons would be the one in the bond and two sets or U three sets of lone pair electrons giving it a negative one charge so even though they're not there I know they're there because I can see the negative one oxygen has really three possible bonding patterns you need to know these and remember that this is is the same three patterns for any second row atom with five six or seven veence electrons five veence electrons is nitrogen six veence electrons is oxygen seven veence electrons would be any hogen an oxygen that has seven electrons will always have a negative a negative formal charge that's what we just practiced so I know that there's three sets of dots one Bond and three lone pairs is a negative charge if oxygen who has no formal charge must have two bonds and two lone pairs that's just based on its electron dot configuration we're going to have no formal charge and if oxygen has three bonds and one lone pair now you're going it's going to have a plus one formal charge very similar to what the hydro ionium polyatomic ion is that h3o+ so again we're looking at negative 1 with seven electrons no charge if it has two sets of paired electrons and two bonds and a plus one charge if it has two bonds and only one lone pair of electrons so let's practice some of that skill all right draw all the lone pairs of the electrons on this particular structure and again I'm just drawing these from your homework problems to help you with practice here's an oxygen and it has a plus formal charge oxygen lives in group 6A to have a formal charge of plus one it means it only has five assigned electrons how many does it have so far it has one two three assigned electrons half of a bonded pair it needs to have five so I'm going to add one set of lone pair electrons to achieve the positive charge three bonds remember this chart three bonds and one lone pair well that's exactly what we have here three bonds and one loone pair to give it a plus one charge let's try these here's oxygen with no formal charge there's no plus or minus indicated so we know therefore to know formal charge it needs two bonds which it has and two sets of electron pairs here is an oxygen with two bonds it needs two sets of Lone pairs here is an oxygen with two bonds it needs two sets of Lone pairs so both of these homework examples gave us uh no charges all right how about these examples from your homework I pulled one that had a negative and one that had a positive and this one up here with no formal charge so that means that I can finish that off two bonds and two sets of dots means no formal charge one Bond and three sets of dots means it's a minus one let's remember why there are six ve valence electrons in the oxygen based on the fact it lives in group 6A there are 1 2 3 4 5 6 7even assigned electrons giving it a formal charge of minus1 so we needed three sets of dots to complete that and up here two sets of dots with the no formal charge and letter H it has a plus formal charge a plus one oxygen lives in group 6A how many dots does it need to end up with a plus one charge overall and that answer is five it has one two three already half of each bonded pair meaning it needs a set of dots one set of dots only to leave it with a plus one charge how about that for nitrogen the formal charge on nitrogen can be calculated very much the similar what we were just doing remember that nitrogen lives in group five 5 a so it needs three Bonds in one loone pair to not have any formal charge so three Bonds in one loone pair no formal charge so if it has a NE -1 formal charge we can see that it has an extra set of dots two bonds two lone pairs if it has a positive that means it has four bonds and no loone pairs let's practice this one draw the loan pair of electrons all right well let's start with this one here I'm going to draw a circle around this nitrogen and I see no formal charge nitrogen lives in group 5 a so the the formal charge is calculated uh is zero if nitrogen lives in group 5 a it has five assigned electrons we know that it needs five so that it has zero overall for its formal charge well it had three from the bond so I know it needs one set of Dots here nitrogen has 1 2 3 four assigned electrons already it lives in group five it has four assigned electrons the formal charge as it's written is A+ one so that doesn't get any dots at all so that falls under this guy if it's a plus one it has no dots on it that's how it's left with the the positive formal charge here are some more from your homework notice here this nitrogen has no formal charge nitrogen lives in group five it must be assigned five electrons to end up with a zero formal charge so far it has three assigned electrons so therefore I know it needs a set of dots one set of dots here's a nitrogen right now it has 1 2 3 four assigned electrons it lives in group five there's four assigned electrons leaving me a plus one formal charge this receives no further dots and here nitrogen is assigned one two electrons so far so here nitrogen lives in group five I want the overall to end with a minus one that's its formal charge so I can see that it needs four more that's a plus one Linda 5 - 6 is a NE 1 so I can see it needs six more electrons 1 2 3 4 5 6 7 8 what is going on in my head this nitrogen it lives in group 5 a it needs to have an overall charge of minus one it has two assigned electrons already right so I already have the two there so it means I need to subtract three more and so what will that look like two sets of dots nitrogen with two single bonds and two sets of dots will end up with a formal charge overall of a minus one just getting it correct 1 2 3 4 5 six assigned electrons giving me a minus one overall oh I worked through that one thank you in our sixth section we turn our attention to a dashed wedged approach to drawing molecules in a three-dimensional way so knowing that it's very difficult and this this truly is the heart of chemistry is trying to Envision molecules in a three-dimensional space when they lie flat on a piece of paper that we've drawn on so we're going to practice using a dashed line and solid wedges to show groups that point back into the paper and those that point out of the paper for instance the carbon backbone of this particular structure all lies in the same plane and I know that because they're all solid lines if I see a wedge attaching to an atom it means that it's coming out of the plane it's coming out right at you in a three-dimensional shape so it's coming out of the paper and if you see a dashed line like this here this represents that it's behind the plane so again the carbon backbone all lies within the same plane this o group is behind the paper and this o group is coming out of the paper here I'm noticing that this is a wedge this is kind of coming out this bicyclic structure is coming out of the paper at me as is this bromine and the methyl group The methyl group will actually be uh behind the paper yeah that's not what I wanted to do there we go there are other ways that we can see some three-dimensional structures and again we'll practice these and upcoming chapters for sure we need to recognize what a fisser projection is often times carbohydrates and biochemistry are represented by Fisher projections these are all chyro carbons we'll learn about that in chapter 5 and it's showing me that in the same plane these carbons are in the same plane in a backbone so all of these in the up and down and these are actually coming out as wedges they're kind of coming out to hug you and then up above here are the the dashes so again we'll practice those but recognize a complex carbohydrates for example um acyclic compounds can be used in Fisher projections very quick way to show three-dimensional structures and we'll practice drawing especially in chapter five when we get to chirality there's a Hayworth projection which are true for cyclic structures for instance if I take this carbon and actually attach it to this carbon to create a cyclic structure we could turn it into a Hayworth projection and we see a few of these already these are bicyclic compounds where there's two rings that are actually fused together so really just the shape of this compound kind of helps us to interact biologically especially for enzyme substrate chemistry um and gives us an accurate depiction and how to interpret 3D Bond Bond line structures at this point in our journey would just need to recognize them it's not a point of Mastery to draw them just yet but right now I want you to recognize what does the dash and the wedge represent what is a Fisher projection what is a Hayworth projection and how would we show a bicyclic structure such as what we have here so just keep in mind if we have an sp3 carbon two of the bonds lie in the same plane one of them must come out as a wedge and one of them must come out as a dash four bonds two in the same plane one's a dash one's a wedge and if you have a longer carbon chain let's put four carbons in a row you always want to make sure that the carbon backbone is in the same plane so that's when we draw these zigzags now if this carbon has two bonds that are already in the plane the third Bond must be a wedge the fourth Bond will come out as a dash that's coming out and behind so again one of these must be a wedge the other must be a dash and even if the terminal carbon has three hydrogens one will be in the same plane one will come out as a dash and wedge let's begin a talk really about through the end of this chapter about resonance structures so this is going to show us a little bit of a limitation to these Bond line structures which are truly generally efficient for most structures they're really the preferred way to draw a structure a pair of bonding electrons is always represented as a line drawn between two atoms so here is a bond drawn between two atoms it's a single bond between two carbon atoms here is a bond drawn between two carbon atoms and here is a d double bond drawn between two carbon atoms so we can typically see a pair of bonding electrons being represented by a line drawn between two atoms which implies really that the electrons are confined to that particular space which is true for this particular structure but it's not always true and that's why we're leading to this idea of resonance Pi bonds and or formal charges are often more spread out than a bond line structure can imply and that's what I mean by a drawback so let's consider for example an Al carbocation now what does this word mean Al means I have a double bond oh I just got a text I read it a carbon to carbon double bond and then next door to that is an alic position position the alyc carbocation for instance means that the carbon not in the bond but one directly next to the bond would have an empty p orbital this would be an SP2 hybridized carbon because it's has a formal charge of + one in this case the pond is actually distributed evenly throughout this entire structure structure and we know this is true by examining Bond lengths you remember a little while ago I said the longest Bond length is a single bomb a little bit shorter than that based on electron density pulling the atoms together would be a double bomb and even shorter net would be a triple bond that's exaggerated but we can get the idea that the more electron density between the atoms the closer those atoms are together they're being pulled closer so I might suspect that this carbon to carbon might appear to where the double bond here would be shorter in length and this Bond here would be longer in length if that is indeed true we would expect a difference in length between the position of carbon 1 and two which would be shorter than the bond length between Carbon 2 and three but what we have found is this is not true when there's an empty orbital in the alyc position of the double bond when I say alyc position the two carbons in the double bomb one position next to it is the alic position and but what we find is equal Bond lengths and that's critical in understanding the term resonance it's not a single it's not a double but it's a hybrid of both and so often times you would see this hybrid structure with an even distribution of the electron density of that Pi bond between carbon 1 2 and three so Carbon 2 has an equal distribution of bond length to the first as well as the third and and kind of a spread out Pi bond in equal directions from Carbon 2 that's what the term resonance means means a delocalized pi bond by delocalized I mean it's not spending its entire time between carbon 1 and two it's not spending its entire time between Carbon 2 and three but it's evenly distributed in all directions so that's what we're going to refer to as a delocalized pi Bond so if all the carbon atoms which they did in that alyc position have an hybridized p orbital then they all overlap side by side so you see this position here a moment ago we have this represented as a double bond and a positive charge but there's no difference in its actual electron orbital since they all have the ability to overlap they are said to be delocalized electrons that they will distribute themselves equally through all of three of those P orbitals now if you see an SP2 hybridized carbon you know that it will exhibit resonance so how can we use a bond line structure to accurately depict the equal distribution of the pi Bond and that's what you'll start to see when we represent resonance structures a resonance structure is an kind of um an opportunity to show where the electron distribution is at any one point in time notice this resonance arrow is a double-headed arrow that term that little arrow there is critically important to represent resonance saying these are two equal structures they are both valid ways to draw that particular structure of the alyc carbo cat ion when you compare a set of arrows that point forward and back keep in mind that's something that we learned represents chemical equilibrium that's not what we're representing here it's not to say that this compound turns into this compound which then turns back it's not going back and forth they are equally present equal means both are valid ways to draw the structure so the P electrons actually exists on both sides of that middle carbon and so you represent both structures in terms of resonance two valid structures to represent the compound and if you think about what we referred to as a hybrid compound that's where we would represent equal Bond lengths with dashed lines to say it is delocalized from Carbon 2 the pi bond is equal distributed back to carbon 1 and out to carbon 3 an equal distribution of the pi Bond now to show that hybrid structure we represent that with compound a where the double bond is between carbon 1 and two and then we represent that in in uh compound B by representing it between between Carbon 2 and three keep in mind I'm not just flipping over the structure I'm actually moving the pi Bond so so because neither of these contributes contributors exist it's not like it's not saying this one flips to this one and then this one flips back it doesn't mean that at all it's it's the average or hybrid is much more appropriate so think of this analogy a nectarine is a hybrid it's formed by mixing a peach and a plum I didn't know that a peach and a plum so a nectarine really doesn't switch back and forth between being a peach and a plum it's just always a nectarine the same is true for a resonance structure this compound doesn't really exist like this and then flip to this and then flip back but it exists as a delocalized pi bond in which we have an equal distribution of charge so the bondline structure using resonance structures uses a double-headed Arrow to say the structure is not truly the first it's not truly the second but really represents the hybrid picture of both these delocalized electrons which spread out therefore have a lower energy and lower energy means much more stable so anytime a molecule possesses resonance it is a much more stable ized molecule and electrons exist in orbitals that Grant a that span this greater distance giving electrons more freedom to move around therefore of course it's going to minimize repulsions so electrons spend time close to multiple nuclei all at once and maximize attractions so delocalization is another term for saying resonant structure and this charge is spread evenly over more than one atom and resulting in these charges that create a full stabilized structure resonance equal stability which is a good thing now how do we start moving these electrons to create one resonant structure to the next and that's the job of a curved arrow and boy friends you're going to be using curved arrows from now until forever in this course throughout the entire course we will use curved Arrow to show electron flow electron movement through a chemical reaction or a resonance structure and the faster and sooner you learn to master the skill the easier the course becomes all of the reaction mechanisms every type of reagent we use will constantly involve placing arrows and arrows track electrons and that's the heart of organic chemistry is tracking electron flow so the arrow starts where the electron are currently located and the arrow ends where the electrons will end up start where the electrons are and point to where they're going now that sounds easy enough but sometimes it just takes a little practice to get the hang of what's allowed and what's not allowed in drawing arrows and I mentioned we'll explore more and more of these curved arrows in every single chapter coming up so there's specific rules rules that we might as well introduce you to for using curved arrows to describe resonance now this is not reaction mechanism later in the chapters but this is just delocalization which is another term for resonance so here are some rules for using curved arrows when we show resonance structures which is delocalizing electrons giving them an equal distribution over more than one carbon to carbon Bond first of all rule never show a single which is a sigma Bond as being delocalized in other words the electrons that are being shared in a single Bond will never be delocalized if you were to delocalize there I got it if you were to put an arrow here and move it here these electrons will actually cause a bond to break that's not what resonance means we're not breaking bonds we're moving piie bonds so Sigma bonds are never involved in resonance structures only Pi bonds so never break single bonds that's not what resonance is all about rule two never exceed the octet rule for any row2 element they simply will never break the octet rule they have no D sublevels to expand their octet into as the third period does now remember when we wrote out our electron configurations the first period 1 S2 the second period 2 S2 2 P6 third period 3 S2 notice that it's the third period before you ever get to a d orbital so here the second period does not have D orbitals to expand the octed in the second period is limited to just eight electrons so don't break the octet rule for boron carbon nitrogen oxygen Florine so keep in mind examples of bad arrows if I have an Arrow moving a pair of electrons I would end up with a double bond which would put two many bonds on carbon bad Arrow here is a set of electrons being made into a bond too many bonds on nitrogen bad Arrow here's a set of electrons being made into a bond too many 2 4 6 8 2 4 6 8 10 too many electrons bad arrow and again if I took a pair of electrons and made them into a double bond this carbon which has two hydrogens on it has too many electrons that's a bad Arrow never break the octet rule for row two they may certainly have less than eight electrons and they would do so by developing formal charges but they will not expand their octet so let's practice what that means is this a good Arrow or a bad Arrow so ask yourself these two questions does the tail of the curved Arrow start at a pie Bond all right so let's make sure we understand does the tail start at a pie Bond and the answer is yes does the head of the curved Arrow violate the octet rule well the answer to that eyes in looking at what we have this is a positive charge so that means this carbon has one two three bonds the fourth bond is empty so even though it's not there there is a hydrogen there so let's move these electrons and see what would happen if I moved those electrons the double bond then places here and I still have that hydrogen does that carbon break the octet rule the answer is 1 2 3 4 that's perfect that's the octet rule at its best so because this plus sign meant that there was an empty orbital that it's a carbocation SP2 hybridized this is a very valid Arrow this is a good Arrow I did not break the octet rule and I originated at a pi Bond remember that to originate you have to be at a pi Bond or a pair of electrons a lone pair not a single bond is this a good arrow and the answer was yes the tail originated at a p Bond yay and the positive carbocation had an empty orbital so there was indeed room for that fourth Bond and we did not violate the octet rule how about another example from your homework is this a good Arrow well let's see right now nitrogen has 1 2 3 four assigned electrons that's why we see it having a positive formal charge draw the result and see what would happen this nitrogen now would have a double bond car carbon hydrogen hydrogen did we break any rule well rule one says did the tail originated a pi Bond so rule one is good rule two says you cannot break the octet rule let's count on nitrogen there is one a lone pair here so this is a shared pair 1 2 3 4 5 6 7 8 9 10 bad five Bonds on the nitrogen yep we found that is this good or bad well let's draw it out and see what we have this car carbon has four bonds so there's no hydrogen there but if I move a set of electrons that's going to create a double bond the oxygen then has just one set because we've moved a pair in does that break any rules first of all rule one says the tail has to start at a pi Bond or a set of electrons so rule one is good the tail is starting at a pair of electrons not a single Bond that's the critical thing there rule one says never started a single Bond and we're good with that rule two says are you breaking the octet rule look at here here's the carbon with 1 2 3 four five bonds so rule two says bad this carbon ended up with five bonds that is not a valid Arrow perfect how about one more let's see here's this carbon the direction it's moving to it already has three bonds and I don't see a formal charge there so I know it has a fourth Bond even though it's not drawn remember that the hydrogen is really there so let's draw the result of this shift the pi Bond would then move here there is a hydrogen on that carbon it didn't move so is this good or bad well rule one says don't start it at a single Bond well that's good we started at a pi Bond how about rule two never break the octet rule well what happened here here's carbon with one 2 three four five bonds so that's bad carbon has too many bonds now visually keep in mind you have to remember to consider the hydrogens that's the trick and often as a beginning student you might forget that but this fourth bond is present I don't have to show it it's a line structure but I know it's there so again bad Bond bad bad Arrow oh we got another one doing all your homework for you is this a good Arrow rule one says I can never ever ever start at a single Bond so right off the bat I broke rule one I cannot start the arrow at a single Bond bad Arrow if we have formal charges in resonance structures we need to consider those and moving our curved arrows around so so as we're deriving resonance structures you'll often have one or maybe even several formal charges to contend with and so remember we are required to indicate the formal charge all valid Lewis structures require formal charges to be drawn so as we begin practicing you know what structure um results as a term of you know the curved arrows we want to make sure that we incorporate um the formal charges needed so for instance here we have a structure with two curved arrows moving electrons and these are good arrows so I just want to draw what's there so the other side this is what we're trying to draw so I noticed that the oxygen had two sets of Lone pairs but I'm moving one in to be a double bond so I get a double bond there one of these comes out this particular double bond then moves to be a set of Lone pairs all right so that's the result of the moving of electrons this guy forms the double bond that's here and this guy here is going to put a pair of electrons on that carbon now this carbon had 1 2 3 four bonds no formal charge this carbon had 1 two three bonds and so therefore there is a hydrogen there so keep that in mind as you're deciding um formal charge you know we have to make sure that we take into consideration hydrogens let me say that again this carbon already has four bonds right 1 2 3 4 so no hydrogens there this carbon has one two three bonds but no formal charge meaning that there is a fourth Bond it's just not drawn so we know that there is indeed a hydrogen so now let's calculate formal charge so that our structure is completely correct draw a circle around the oxygen oxygen lives in group 6A so there's six assign six veence electrons it has 1 2 3 4 five assigned electrons around it meaning that its formal charge is A+ one we have to indicate that next to the symbol here's the carbon draw a circle around the carbon carbon lives in group four how many assigned electrons well half of every bonded pair so 1 2 3 four five it has five assigned electrons meaning it has a formal charge of negative so I have to show those in my resonance structures or it'll be wrong so this is a section just reminding us that make sure when you're drawing resonance structures and you're moving electrons to consider formal charge as you develop your structure we can draw the other resonance structures by following the instructions provided by the curved arrows which is exactly what we just did in the previous slide and then we added on the formal charge which we decided oxygen was a plus and the carbon here was a negative because we knew there was a hydrogen attached and that's going to be your practice and you look at some of these skill builders in this section we can look at a couple together you want to draw resonance structures you're just following the arrows and then drawing any kind of formal charge from it so for instance here we have an oxygen and I just like to get my skeleton set the oxygen had three sets of Lone pairs thus the negative charge but one of them now is being moved to create a pi Bond so there's the first electron movement this Pi bond is now going to form a set of Lone pairs down here so that's the other structure I'm not done yet CU I have to consider formal charge I draw a circle around the oxygen I know oxygen lives in group six it has 1 2 3 four five six assigned electrons it has no formal charge perfect this carbon by the way had a hydrogen on it right you remembering to check that 1 2 three oh it needs that fourth Bond so 1 2 3 4 five assigned electrons gives me a formal charge of a negative so as I redraw that structure just to kind of clean it up you have an oxygen double bonded and you need to have a set of Lone electron with a negative charge as that resonance structure and that's how we drew that structure at the end notice we put them in a bracket we have a double-headed Arrow to say they are in resonance two equal structures and of course we understand that it's a hybrid of both it's not one or the other it's not flipping back and forth but it's both at the same time want to try some more here we have a structure and again what we would do is draw a double arrow and we put it in a bracket we have the pi Bond moving to this side all right so here's a carbon has one Bond already I know that there's two hydrogens there to give it a plus one formal charge it only has three of the four bonds this carbon has two hydrogens cuz the two bonds are in double bond there so it's two more remember remember the hydrogens don't move they're still there this carbon now over here has four bonds two of them to hydrogens the other two in a double this is now where the positive formal charge is ended up on the first carbon that's an alyc double bond right the carbon directly next to the double bond is a positive it's an alyc carbocation let's draw with the resonance structure here we see that we have an aromatic functional group this is a alcohol just to kind of keep practicing those and I noticed that one set of the electrons on the Alcohol excuse me went through to become a double bond so there's the double bond from moving here that's a double bond not coming out very neat this double bond shifted down this double bond became dots and this double bond was untouched I just think that looks messy that's a little better so that's our resulting uh resonance structures again it's not one or the other they're not flipping back or forth um what about formal charge here on any of these um oxygen has 1 2 3 4 five assigned electrons so it right now has a positive formal charge you see one down here as well this carbon 1 2 3 this had a hydrogen on it didn't it they have that fourth Bond so the hydrogen still there 1 2 3 four five it only needs four so this is a minus formal charge and that's what we've drawn sometimes you're asked to go the other way and look the two resonance structures and try drawing the arrows that would bring you from one to the other so as you approach these types of problems I recommend that you kind of think about where the dots are to begin with and kind of remember even though they're not there we know they're there to be a negative charge we know that it has to have six dots so really what I'm doing is bringing up a set of electrons to break the pi Bond and put put a lone pair up on the oxygen hi puppy that's going to leave the carbon with only three bonds and thus that's why you see it's positive formal charge and often times since these are going back and forth the arrow should be placed on both structures so to go back and forth Arrow number one breaks the piie bond makes a double bond Arrow number two puts the lone pair back as a double bond this is going either direction from that Arrow here's another example draw curved arrows to represent how one resonance structure converts to the other here you notice that a piie bond is just moving location here you're noticing that the pi Bond would be moving location so that only requires one Arrow doesn't it and that was that pretty easy and another one from your homework now remember even though these aren't there you have to Envision them right so what we're doing is moving a pi bond to become a set of lone pair doesn't matter where it's going to land on the oxygen and I needed a second Arrow because I noticed the the pi bond in the cyclic structure moved location as well resulting in formal charge on this carbon with only three bonds and on this oxy oxygen now with seven assigned electrons if you wanted to go the opposite way we just move a pair of electrons back and we move the double bond back perfect I'm going to pause the video here and came call this our lesson two I Know lesson one was quite long this will be lesson two there's one more video Lesson awaiting for you to finish out the structures of resonance come on back when ready