Hello and welcome to this video tutorial on how to name out Keynes using iUPackNomenclature. I'm going to show you how to name some basic out Keynes that will build up to more complex structures. First off, we'll look at this structure. Earlier we had a handle on how to name out Keynes structures where all the carbons in the molecule were completely saturated with hydrogen. In out Keynes structures there is at least one portion of the molecule that contains a double bond. In out Keynes the double bond gives each carbon a trigonal planar arrangement. So it looks like a flat triangle. The alkane carbon and the two atoms attached to it are all on the same plane. With the new functional group added we need to set a priority list for organic functional groups. Alcane and their related groups, alkyls, halogens, succuloccules have the same level of priority. With their multiple bonds, alkenes and alkynes are higher in priority than this batch of other functional groups, with alkenes higher than alkynes. Examining our molecule, we need to find the longest continuous carbon chain that includes the highest priority group. For this structure this is the alkyne functionality. So in numbering the longest chain we make sure to give the alkyne group the lowest possible carbon number. And that lowest carbon number is where the alkyne double bond is first encountered. The alkyne in this three carbon structure is first encountered at carbon one. This carbon number is placed in front of the parent name. Based on the link that the carbon chain the parent, if it were an alkane, the name would be propane. To change the name, to show an alkane exists instead. We drop the A in the A and E suffix and replace it with E to make an E ending. Ordinarily we would need to place the carbon number for the location of the alkene double bond in front of the parent name. But since an alkene cannot begin any other carbon in a short three carbon chain, in this case we drop using the one. Instead it is implied. So the name of this compound would be propane. Let's try a slightly different compound. Again, examining our molecule we need to find the longest continuous carbon chain that includes the highest priority group. In this structure our functional groups are halogen, bromine, and an alkyne double bond. The higher priority group here is the alkyne. The lowest carbon number in which the alkyne is encountered in the parent chain is carbon number one. This is a three carbon chain. For an alkyne this would be propane. But since this has an alkyne group we drop the A and replace it with E to make an E suffix. So propane. No need to add the location of the alkyne double bond since it is implied to be on carbon one. The only substituent on this parent chain is the bromine which is located on carbon number two. So the complete name of this compound is to bromopropane. Check out the next set of alkyne compounds to name in part two of this video tutorial. Thanks for stopping by.