in this video we're going to focus on curve error notation also known as electron pushing arrows now whenever you see a full Arrow this represents the flow of two electrons or an electron pair if you see a half Arrow it represents the flow of one electron typically you'll see that when dealing with radical reactions so let's go over a common acid-base reaction so here we have acetic acid and we're going to react it with hydroxide and let's put the lone pairs on each oxygen atom so what's going to be the products in this reaction and how can we show the mechanism using curved error notation we know that acetic acid is a weak acid hydroxide is a strong base so hydroxide is going to abstract a proton from the acid acids are proton donors bases they're proton acceptors so we know we're going to get water as a product and we're going to get the conjugate base of acetic acid which is acetate how can we show arrows that will help us to get the products that we see here on the right now when drawing the arrows you want to draw the arrow from the nucleophile to the electrophile the nucleophile is typically the base generally speaking it is the molecule that is electron rich the electrophile is typically the acid and it's usually electron deficient or electron poor so when drawing the arrows electrons will flow from a region of high electron density to a region of low electron density in other words it usually goes from a nucleophile to the electrophile so we could say that the electron density around an oxygen is pretty high because it has a negative charge this hydrogen is partially positive and this oxygen is partially negative whenever you have an o h Bond because oxygen is more electronegative than hydrogen it's going to Bear the negative partial charge so this hydroxide ion is attracted to The partially positive hydrogen so we're going to draw the arrow from the oxygen to the hydrogen now think about what's Happening Here this Arrow tells us that the oxygen is going to use two of its electrons or one lone pair to form a bond between itself and hydrogen so those two electrons are right here in this Sigma Bond they were used to connect the hydrogen and the oxygen Bond now this Bond broke right here the o h Bond where did those electrons go did they go to the hydrogen or to the oxygen looking at our products we can see that here the oxygen had two lone pairs now it has three so we can tell that the electrons in that Bond went back to the oxygen atom and so we're going to show the flow of those electrons with this Arrow so it's going to point towards the oxygen atom and so really that's how you can use the curve error notation remember it represents the flow of electrons where the electrons are going how they're being used either break a bond or form a bond and that's what we have here so those are the only two arrows that we need to show the progression of this reaction let's look at another example so here we have br2 and we're going to irradiate it with radiation let's say UV radiation and something's going to happen here when you irradiate a halogen like bromine iodine or even chlorine the bond that connects these two atoms can break when that Bond breaks where will the electrons go will they go to the bromine on the left or the bromine on the right so once this absorbs UV radiation that Bond will break now both of the Vermin atoms are equal they're equal in electronegativity so their ability to pull electrons to themselves are exactly equal therefore when that Bond breaks one electron is going to go to the bromine on the left and the other it's going to go to the bromine on the right because they both have an equal pull on those two electrons so in this case we'd use half arrows because there are two electrons in that Bond one is going to the left one is going to the right so remember a full Arrow represents the flow of two electrons but a half Arrow represents the flow of one electron so we're going to get two separated bromine atoms with an odd number of electrons whenever you have an atom with an odd number of electrons that's a radical so this type of reaction you'll see when going over radical reactions this type of bond cleavage is known as a homolytic Bon cleavage the prefix homo means the same each atom gets the same amount of electrons analytic means think of to split apart think of the word lysis you're splitting something down in the middle so we're Breaking the Bond evenly and those two electrons are being shared equally or being taken equally among those two atoms now let's look at a heterolytic bond cleavage so let's say the carbon bromine bond is about to break where will the electrons go will they go to the carbon atom or to the bromine atom carbon has an electronegativity value of 2.5 on the 4.0 pollen scale and bromine the electronegativity is 2.8 so because bromine is more electronegative than carbon when that carbon bromine Bond breaks bromine is going to have a stronger pull on those electrons so the full arrow is going to go towards bromine and what we're going to get is a carbocation with an MTP orbital and we're going to get a bromide ion with eight electrons or rather eight valence electrons so that is an example of a heterolytic Bond cleavage the prefix hetero means different so carbon didn't get any electrons when that Bond broke bromine got both of the electrons now let's look at another situation let's say if we want to break the carbon hydrogen bond how should we draw the arrows how should we show it using curve arrow notation now carbon is more electronegative than hydrogen in a previous example carbon was less electronegative so it had the partial positive charge bromine was more electronegative so it had the partial negative charge now for the carbon hydrogen bond the situation is different carbon is more electronegative so it's going to Bear the partial negative charge hydrogen is more positive it's less electronegative more electropositive so it's going to have the partial positive charge when that Bond breaks the electrons are going to go towards the more electronegative carbon atom it has a stronger pull on the electrons in that Bond so in this situation we're going to get a carb anion rather than a carbocation so that's another example of a heterolytic bond cleavage now for those of you who are studying for the first exam of organic chemistry I have a practice test that can help you with that and it's posted in the links in the description section below now for those of you who want access to the free version which is a little bit shorter if you go to YouTube and type in in the search box organic chemistry exam one the free version should show up with the full version it's going to be in the description section below for those of you who are interested but now let's get back to this video now let's consider the reaction between hydroxide and methyl bromide what do you think is going to happen in this reaction feel free to show the arrows that will help us determine what the products of this reaction will be hydroxide in this reaction is going to behave as a nucleophile when hydroxide abstracts a proton it's behaving as a base when it attacks a carbon it's behaving as a nucleophile the bromine is partially negative carbon is partially positive with respect to bromine so hydroxide is going to be attracted to the nucleus of the carbon atom and that makes it nucleophilic it makes the carbon atom electrophilic so we're going to draw the arrow from a region of high electron density to a region of low electron density so we could say hydroxide has a lot of electrons it has a negative charge the carbon with the partial positive charge that's a region of low electron density so the arrow is going to flow from hydroxide to carbon and what's going to happen is we're going to take two electrons from hydroxide and use that to form a bond between the oxygen and the carbon atom so here is hydroxide now it's attached to carbon and we still have the other hydrogen atoms attached to it now carbon can only form four bonds so evidently this bond has to break and that's what's going to happen when this Bond breaks we know the electrons are going to go towards the more electronegative bromine atom and so we're going to get bromide as one of the products so that's how we can represent this reaction using curve arrow notation now here we have two four pentadion and this particular dichetone it's a beta that Ketone has an acidic Alpha hydrogen the hydrogen that's on a carbon which is adjacent to a carbonyl group that hydrogen is relatively acidic and when it's between two carbonyl groups the pka for this hydrogen is nine the pka for a hydrogen that's next to one carbonyl group that's around it could be anywhere between 17 and 20. so like for aldehydes it's 17. for ketones it's like close to 19 but somewhere in this ballpark range so this is the most acidic hydrogen that's on this compound so hydroxide rather than being a nucleophile attacking a carbon it can act as a base and go for the hydrogen and that's what's going to happen in this example now we know that carbon is more negative than hydrogen so carbon is partially negative with respect to this hydrogen and that hydrogen is partially positive so hydroxide is going to be attracted to it so we're going to draw the arrow from a region of high electron density that is from the nucleophile to a region of low electron density that is the electrophile so it's going to go from hydroxide to the hydrogen so what's going to happen is we're going to use a lone pair to form a bond between oxygen and hydrogen and so we're going to get water as one product we have three lone pairs on oxygen we use a one to form a bond so now we're left with two now what happens when the carbon hydrogen bond breaks if we take away the hydrogen that bond has to bring where will the electrons go because carbon is more electronegative than hydrogen we know that when the CH Bond breaks the electrons will go towards the carbon atom and so we're going to get a carbanion now the reason why this hydrogen is so relatively acidic for a CH hydrogen is because this carbon of a negative charge it's part of a resonance system the negative charge can delocalize into the carbonyl group and it can be placed on an oxygen atom and it's much the situation is more stable if we put the negative charge on the oxygen atom rather than the carbon atom because oxygen is more electronegative it can better handle the negative charge so let's draw the resonance structure now this carbon that's part of the carbonyl system it's partially positive because it's attached to an electronegative oxygen atom which is partially negative so when drawing the arrows we're going to draw it from a region of high electron density to a region of low electron density so the arrow is going to flow from the negative charge towards the carbonyl carbon so what's going to happen is it's going to be used to form a pi Bond and then this Pi bond is going to break those electrons are going to go on the oxygen atom and so this is one resonance structure that we can draw so now we have a pi Bond here and we have three lone pairs on the oxygen now keep in mind we can also move the negative charge here and put the negative charge on the auction on the right so but these that resonance structure with this one highlighted in blue they're equivalent so we can draw a total of three resonance structures for that particular ion go ahead and try these two use curve arrow notation to draw the resonance structure of each example by the way for those of you who want more examples on resonance structures I'm going to post the full video in the comments I mean in the description section below this video but if you want the free version if you go to the YouTube search bar and type in resonance structures organic chemistry tutor it should come up so let's start with the first example how should we show the arrows to draw the resonance structure for this allylic carbocation now it really helps to identify the region of high electron density with the region of low electron density the region of low electron density is going to be that empty p orbital on the carbon with the plus charge so that is the electrophilic sensor the nucleophilic center has to do with the double bond that's a region of high electron density because we got two Pi electrons that can move into that empty p orbital and that's what's going to happen so we're going to take these two electrons and move it here as we do so we now have a pi Bond between those two carbon atoms but this carbon lost the bond so now it's going to be electron deficient which means it carries the positive charge notice that these two resonance structures are equivalent if we want to draw the resonance hybrid it's going to look something like this because the double bond is between these three carbon atoms we're going to have a dashed line between those three carbon atoms now the plus charge is shared among those two carbon atoms so we can put a partially positive charge on those two carbon atoms because they're the same we can quantify and say it's going to be half of a charge on each of those two carbon atoms but this is a good generic answer when dealing with these types of questions now let's move on to the next example how would you use curve arrow notation to show the progression of to show the formation of the resonance structure for that ion now which one has higher electron density would you say the double bond or the lone pair with the negative charge the double bond is neutral the lone pair has a negative charge so it's more nucleophilic it's more negative so in this case the double bond is going to be the electrophile here the double Bond's neutral the plus charge is positive so neutral is more nucleophilic than a carbon with a plus charge so this time the arrow is going to flow towards the double bond rather than away from it so it's going to flow from a region of high electron density towards a region of low electron density so the first Arrow what we're going to do here is we're going to take this lone pair use it to form a pi Bond and then we're going to break this Pi Bond and move those two electrons to form a lone pair on the other carbon atom so we're going to have a pi Bond here and here's the negative charge so those two structures are also equivalent and if we want to draw the resonance hybrid it's going to look very similar to the situation above the double bond can move anywhere between these three carbon atoms so we're going to use dashed lines to represent that and the negative charge is shared among those two carbon atoms so that's basically it for this video so now you know how to use curve arrow notation to not only predict products but also to show the progression of a chemical reaction and you know how to use it to draw resonance structures as well now for those of you who want to access my extended organic chemistry videos you can access it at my patreon membership program at patreon.com maths science tutor or in my YouTube membership program which you can join in the video below now if we click on this organic chemistry posts it'll show you all my full length organic chemistry videos so I have this video on PKA values for those of you who are studying acids and bases for those of you who are just studying organic chemistry this basic introduction video will really help you get started and I have some other videos as well hybridization resonance structures the free version of this video is about 20 minutes long on YouTube but the full length version is about an hour long acids and bases functional groups new projections the worksheet contains all of the problems in the full length video some of you have asked for worksheets you prefer to work through the problems that way instead of you know watching a long video so I have it for not all of my videos but some of them so we have chair confirmations now for my organic chemistry exam one video you may have to type it in in the search box to get it or if you look at the description section of the referring video it'll take you directly to that link but here is the worksheet now if you want to watch a seven hour video you can do it that way but if you if you prefer to print out the worksheet and work through the problems that way you could do that as well by the way post a comment in below this video I want to know what your thoughts are do you prefer to watch The Seven Hour video or do you prefer to study for the exam by getting a printout of the worksheet and working through the problems while you're in school let me know in the comment section below now I have other videos stereochemistry specific rotation sm1 sn2 reactions there's a practice test on that 77 practice problems so here's the video with the practice test but I haven't done the worksheet yet we got alkene reactions alkyne reactions alcohol reactions radical reactions and then my organic chemistry one final exam review video so the video is actually finished but the worksheet is coming soon so let me know for those of you who are actually interested in getting these worksheets do you prefer to watch the six hour final exam review video or do you prefer to have a printout of the 100 practice problems and work through at school let me know in the comment section below