in this lesson we're going to talk about resonance structures how can we draw the resonance structure for this allylic carbocation the first thing you should do is you should start the arrow from the double bond and it's going to flow towards the positive charge electrons will flow from a region of high negative charge towards a region of low negative charge or from negative to positive and so the arrow is always going to go from the nucleophilic region of the molecule to the electrophilic region of the molecule or ion in this case so the resonance structure for this allelic cation looks like this the double bond is going to move to the right side and it's going to trade places with the positive charge since this carbon lost pi electrons it now has the positive charge this carbon lost the pi bond but it regained it on the right side so it doesn't have the positive charge now which of these structures is the major resonance contributor is it the one on the left or the one on the right both of these are primary allylic carbocations so they're equally stable so there is no major or minor resonance contributor in this example the resonance hybrid is actually a blend of these two resonance structures so the pi bond is shared among the three carbon atoms and so is the positive charge it's actually shared between these two carbon atoms so they both carry a partial positive charge or you could say half of a positive charge if you divide it by two go ahead and try this problem draw as many resonance structures as you can and identify which one is the major resonance contributor or which resonance structure is the most stable so the pi bond is going to move towards the carbocation and the positive charge is going to jump two places or two carbon atoms towards the pi bond so the positive charge will now be here and so that's one resonance structure that we can draw and we can draw another one so the plus charge is going to jump two carbons to the left so it's now on that carbon so in our first example we have a primary carbocation or allylic carbocation and in the second example it's tertiary and in the third one it's secondary so therefore this one is the major resonance contributor because a tertiary allylic carbocation is more stable than a primary or secondary allylic carbocation so recall that tertiary carbocations are more stable than secondary ones which are more stable than primary ones so if we compare let's say a tertiary carbocation with a primary carbocation it's important to understand that the methyl groups are electron donating groups relative to hydrogen as a result they can donate electron density to the carbocation by means or through the sigma bond which is known as the inductive effect or through the overlap of atomic orbitals which is known as hyperconjugation and so a carbocation with more methyl groups or more carbon atoms attached to it is going to be more stable than one that has less now let's move on to our next example in this example we have a benzylic carbocation draw as many resonance structures as possible and identify which structure is the major resonance contributor so we can move the double bond towards the positive charge and just like before the positive charge is going to jump two carbons toward that double bond and so we have this resonance structure and then we can move this double bond here giving us our third resonance structure and then we can move this double bond in that location so this is a primary carbocation or technically benzylic carbocation this one is secondary this is secondary and this is secondary so this one is going to be the minor resonance contributor because it's the least stable now these are equally stable so they form part of the major resonance contributor now going back to the last problem there's one more resonance structure that we can draw out of the common ones there's a lot of uncommon resonance structures that we can draw so just keep that in mind i prefer to focus on the common ones so we can take this double bond move it to the plus charge and we'll get this answer it's similar to the original answer but notice that the double bonds in the benzene ring have been shifted now let's move on to our next example draw the resonance structure of the benzene molecule all we can do here is simply rotate the pi bonds and so this is the resonance structure for benzene and that's all we can do here now what about for the carboxylate ion what resonance structure can we draw go ahead and try that problem so what we can do is we can take a lone pair from the negatively charged oxygen use it to create a pi bond and then break this pi bond and so that's going to give us this particular structure now these two structures are identical and so they're equally stable there is no major or minor resonance structure in this example now let's move on to the next example so let's say if we have a sulfur atom instead of an oxygen atom identify the major resonance contributor in this case so we could follow the same pattern in order to draw the resonance structure so the question is is it better to put a negative charge on the oxygen atom or on the sulfur atom now granted oxygen is more electronegative than sulfur so that fact favors oxygen however sulfur is bigger than oxygen and let's say if we have a small atom versus a large atom if we were to make each atom an ion by giving it a negative charge which charge is more stable a big ion is more stable than a small iron assuming they have the same charge because this has more volume where it can stabilize that negative charge so this negative charge is you could think of it as being diluted over a larger surface area this one here it's more concentrated and so larger ions can are more stable than smaller ions the bigger the atom the better it can stabilize a negative charge so therefore it's better to put the negative charge on a sulfur atom then on an oxygen atom so this is going to be the major resonance contributor on the left this one here is the minor resin is contributor here's another example let's say we have an alcohol but adjacent to the o h group there's a positive charge draw the resonance structure and identify the major resonance contributor so we know the electrons will flow from a region of high negative charge towards the region of low negative charge so the arrow is going to point towards like from the lone pair but towards the positive charge and so we can use that lone pair to create a pi bond and so now the oxygen has one lone pair as opposed to two because this carbon atom it gained a pi bond the positive charge on that carbon atom has been neutralized the oxygen lost the lone pair so now it has a positive formal charge so is it better to put a positive charge on an oxygen atom or on a carbon atom now oxygen is more electronegative than carbon and so as a result it's better to put the positive charge on the carbon atom however it turns out that the structure on the left is the major residence contributor and the one on the i mean this is the right side not the left side the one on the left is the minor resonance contributor so why is this the major resonance contributor if we have a positive charge on an electronegative atom well we know it's better to put a positive charge on a more electropositive atom like carbon the answer has to do with the octet rule so if you look at the structure on the left the carbon atom doesn't obey the oxide rule it doesn't have eight electrons it has three bonds no lone pairs so it has six electrons if we look on the right side this carbon atom has four bonds so it has eight electrons it satisfies the octet rule the oxygen also satisfies the octet rule it has three bonds that's six electrons plus a lone pair so it has eight electrons on the left side the oxygen satisfies the octet rule has one bond plus the oh bond that's two that's four electrons plus two lone pairs so it has eight so the reason why this is the minor resonance contributor is because carbon does not obey the octet rule when it has a positive charge and so that's why you'll see that when dealing with resonance structures it's better to put the positive charge on the oxygen atom as opposed to the carbon atom to satisfy the octet rule so electronegativity is not a priority in this example you