Alkyl halide nomenclature and classification

- [Lecturer] You often see two different ways to name alkele halides. And so we'll start with the common way first. So think about alkele halides. First you wanna think about an alkele group, and this alkele group is an ethyl group, there are two carbons on it. So we write in here ethyl. And then since it's alkele halides, you wanna think about the halogen you have and end in i. So this is chlorine so it's gonna end in i, so chloride. So ethyl chloride would be the name for this compound. Now let's name the same molecule using IUPAC nomenclature. In this case it's gonna be named as a halo alkane. So for a two carbon alkane that would be ethane. So I write in here ethane. And of course our halogen is chlorine, so this would be chloro. So chloroethane is the name of this molecule. If I had fluorine instead of chlorine, it would be fluoroethane. So let me write in here fluoro, notice the spelling on that. If I had a bromine instead of the chlorine, it would be bromoethane. And finally if I had an iodine instead of the chlorine, it would be iodoethane. So let me write in here ioto. Let's name this compound using our common system. So again think about the alkyl group that is present. So we saw in earlier videos this alkyl group is isopropyl. So I write in here isopropyl. And again we have chlorine attached to that. So it would be isopropyl chloride using the common system. If I'm naming this using the IUPAC system, I look for my longest carbon chain, so that'd be one, two and three. I know that is propane so I write in here propane. And we have a chlorine attached to carbon two. So that would be 2-chloro, 2-chloropropane. Let's look at how to classify alkyl halides. We find the carbon that's directly bonded to our halogen and we see how many alkyl groups are attached to that carbon. There's only one alkyl group, this methyl group here, attached to this carbon so that's called primary. So ethyl chloride is an example of a primary alkyl halide. If you look at isopropyl chloride down here. This is the carbon that's bonded to our halogen and that carbon is bonded to two alkyl groups. So that's said to be a secondary alkyl halide. And let me draw in an example of another one here really fast. So for this compound the carbon that is bonded to our halogen is bonded to three alkyl groups. So three methyl groups here. So that's called a tertiary alkyl halide. And the name of this compound is tert-butyl chloride. So that's the common name for it. And that's the one that you see used most of the time. For larger molecules it's usually easier to use the IUPAC system. So let's name this compound using the IUPAC system. It's just like naming alkanes. First you wanna find you longest carbon chain and name it. So that would be one, two, three, four, five, six and seven. A seven carbon alkane we know is heptane. So I'm gonna write in here heptane. Next let's think about how to number this carbon chain. Do I wanna number it from the left or do I want to number it from the right? So let's try numbering it from the left first. So one, two, three, four, five, six, and seven. If I'm numbering it from the left, I have a bromine at two, a methyl at four, and another methyl at five, so two, four and five. Let's try numbering this compound from the right. So numbering our carbon chain from the right would mean this is carbon one, two, three, four, five, six and seven. This will give me a methyl group at three, a methyl group at four and a bromine at six, so three, four, and six. Our goal is to give the lowest number possible to our first substituent. So on the left that would be a two for the lowest number and on the right that would be a three. Obviously two is lower than three so we're gonna stick with the numbering system on the left. So we've already identified our substituents. Alright we have bromine and we have two methyl groups, that would be dimethyl. And you also wanna think about alphabetical order. So the bromine's gonna go first. So just for spacing purposes I'm gonna put in dimethyl here, so dimethyl and write that first. And that's at carbon four and five. So the four and five just identify where those two methyl groups are. And then we have a bromine at carbon two, so 2-bromo. So our full IUPAC name is 2-bromo-4,5-dimethylheptane. Alright let's name this compound. So we approach it the same way. One, two, three, four, five, six and seven. So again heptane would be our parent name here. Now numbering from the left or from the right? One, two, three, four, five, six and seven. This gives us a methyl group at two, another methyl group at five and a bromine at five. So two, five, five. For the one on the right, again same compound, I'm just gonna number this one starting from the right to see if this gives me a lower number or not. So this would give me a bromine at three, a methyl at three, and another methyl group at six. So three, three, six. Obviously the one on the left would win again because two is lower than three. Our goal is to give the lowest number possible to our first substituent. So if we're numbering it from the left, let's see we have a methyl group at two, and a methyl group at five. So it's like the previous situation, we have two methyl groups so dimethyl. So let me write that in here, so dimethyl. This time we have a methyl group at two and five. So I'll write in here two and five. And then a bromine at five, so it'd be 5-bromo. So our full name would be 5-bromo-2,5-dimethylheptane. Now let's name this compound. So it's the same approach so let's count up our carbons here in our longest carbon chain. One, two, three, four, and five. So a five carbon alkane is pentane so I write in here pentane. Next I think about how do I number my carbon chain to give the lowest number possible to my first substituent? If I number it from the left, I give the bromine here a one. So that's the best way to number it. So that'd be one, two, three, four, and five. So I have a methyl group at four, a chlorine at three, and a bromine at one. And we wanna put these in alphabetical order. So it's gonna be the bromine first, then the chlorine and then the methyl. So I'll go ahead and put the methyl in here. So this is at carbon four so 4-methyl. Next I have my chlorine at three so that would be 3-chloro. And finally my bromine at one, so 1-bromo. So the full name is 1-bromo-3-chloro-4-methylpentane. Now let's look at this compound. So we have one, two, three, four, five and six. So six carbons in our chain, a six carbon alkane is hexane. So I write in here hexane. Now I have two substituents, I have a bromine and I have a methyl group. So let's number from the left and see what happens. One, two, three, four, five and six. That gives me a bromine at two and a methyl at five. Now let's number this from the right. So this would be one, two, three, four, five, six. This gives me a methyl at two and a bromine at five. So just looking at numbers, we can't decide who wins here. So we have two versus two which is a tie, five versus five so that's a tie. So the way to break the tie is to think about the alphabet for your substituents. So we have bromine versus methyl so b versus m. Obviously b comes before m in the alphabet so the bromine's gonna win. We're gonna give the bromine the lower number. And that of course is the example on the left where the bromine is coming off of carbon two. So we're gonna choose the system on the left or the way of numbering the carbon chain from the left. Which means we have a methyl group at five. So 5-methylhexane. And then bromine at two, so 2-bromo. So 2-bromo-5-methylhexane would be the name. So what do we do if we have some stereochemistry in our compound? Well first let's ignore the stereochemistry and let's just name this how we've been naming the other molecules. We find our longest carbon chain. So that'd be one, two, three, and four so that's butane. So I write in here butane. We want to number our chain to give the lowest number possible to our substituents. So that's one, two, three, and four. So our substituent is the bromine coming off of carbon two. So 2-bromobutane would be the name. But now we have to worry about our stereochemistry. We know that we have one chiral center. So here is our chiral center. We know there's a hydrogen going away from us in space, that there's a bromine coming out at us. And we need to assign priority to those four groups that are bonded to our chiral center. So I showed you how to do this in an earlier video, so I won't go into too much detail. But let's assign priority really quickly. We know that bromine with the highest atomic number is gonna get the highest priority. So this group gets a number one. The ethyl group would get the second highest priority. The methyl group would get the third highest. And finally this hydrogen going away from us is the lowest priority group so this gets a number four. So we have one, two, and three going around clockwise. And we know that is R. So to complete the name, in parentheses here, I put R, and (R)-2-bromobutane is the IUPAC name.