Name the following alkynes according to the IUPAC rules of nom...

Question

Name the following alkynes according to the IUPAC rules of nomenclature.

(c) Structural formula of an alkyne with multiple bonds, illustrating R and S configuration for nomenclature practice.

Accepted Answer

Hello everyone. Let's solve this problem. It says provide the name for the given Alkin using Iupac nomenclature. So let's list out the steps that we're going to use. In order to get this name first, we want to identify and number the parent chain, which is the longest continuous carbon chain. And we want to make sure we include the double bonds and triple bonds in the parent chain. When we number the parent chain, we are giving numbering priority two alkenes and Alkins. Alkins and Alkins are equivalent to one another. So it doesn't matter which one, but we just give general priority to either of them. And when we name the parent chain, we have to remember to alter the ending. So for double bonds, we alter it with, in the suffix in and for triple bonds, we alter it with that suffix I OK. Once we have the parent chain down, we are going to identify and number the substituent groups. And remember that when we put the name together, we list the substitutes in alphabetical order, right, not numerical order. And then we need to check if there is any configuration, we need to specify in the name, for example, Ez for alkenes or R and S configuration for chiral centers. So if we have R and S configuration, we must follow the Khan Ol pre log nomenclature rules. And these rules revolve around the priority of the atoms. So first, we have to assign the priorities to each group or each atom that is attached to the carbon, right? Because for a chiro carbon, that means we have four different groups attached to that carbon. And we assign the priorities which is based on atomic mass. OK. If we have a tie, we look at the adjacent atoms and compare those. And if we have a double bond or a triple bond, we count those multiple times. So carbon double bonded carbon would count as a carbon bonded to two other carbons and same for a triple bond. But times three. And once we have those priorities, we determine whether we have R or S configuration based on the position of the lowest priority group. So if group number four is dashed, there is a different criteria than if group number four is wedge. And we want to see if we connect groups 12 and three, if they connect in a circle clockwise or counterclockwise. So if group number four is dash, counterclockwise would be s configuration clockwise is R configuration. If group number four is wedge, counterclockwise is R clockwise, is S OK. So let's follow all of these rules and name this outcome. So identifying the longest parent chain, longest carbon chain, including the double and triple bond. So over here, we would not start with this carbon because then our double bond would be a substituent group. We need to include that double bond. So we would have carbons,  aN:aN:000NaN 123456789. So the parent name would be na and then we will modify the ending A&E for the double bond and the triple bond. And we also want to add the Logan of the double bond and the triple bond. So non one in right, for that double bond starting on carbon, number one, and the triple bond starts on carbon number four. So we have four I so non one E is the name of our parent chain. So that is step number one. Now let's identify our substituent. We have a methyl group on carbon number two, an isopropyl group on carbon number three and an ethyl group on carbon number six. So I'm going to start putting this name together and remember we do this in alphabetical order. So ethyl, isopropyl and methyl would be in that order. So we have six ethyl, three isopropyl and two methyl. OK. Now, do we have any special configuration we need to consider for this name? Well, our double bond would not have E or Z configuration because there are two hydrogens on the first carbon. So they are the same priority group, but we do have some chiral centers here. Carbon three is a chiral center and carbon six is a chiral center. So we need to determine the configuration for these two chiral carbons. So let's assign the priorities of the four groups attached to each of these carbon starting with carbon number three. And I'm going to draw the dashed hydrogen on carbon three because that isopropyl group is wedged. So that implies that our hydrogen is dashed hydrogen is going to be our lowest priority group. It always is, if it is present, then we have carbon atoms for each of the three groups, we have the carbon in the triple bond, the carbon in the double bond and the carbon that has two methyl groups in our isopropyl. So the carbon in the triple bond is going to be highest priority because we count that as that carbon having three other carbons bonded to it, then the carbon in our double bond will be second highest priority. And the carbon in our isopropyl group will be third priority, right? Because that alkene, we have the two carbons in the, in the double bond and then the carbon in the methyl. So that's also three carbons. And in our isopropyl group, that carbon only has two other carbons attached to it, right? And one hydrogen. So now group number four is dashed. So we're using this part of our table and connecting groups, 12 and three that is going counterclockwise. So carbon three will have s configuration OK. And let's do the same thing for carbon number six, that ethyl group is wedge. So that means we have a dashed hydrogen hydrogen is lowest priority as always, immediately that Alkin catches my eye. That will be our highest priority group. Then let's look at the other two groups, the ethyl group and then the 123 carbon chain on the other side. So in our carbon chain, we have two hydrogens and a carbon. In our ethyl group, we also have two hydrogens and a carbon. So then we go to the next adjacent atom. Carbon eight has two hydrogens and a carbon. And the end of our ethyl group has three hydrogens. So the ethyl group, since it has more hydrogens will be the next lowest priority and the rest of our carbon chain will be the second priority. All right, group number four is dashed here again. So when we connect groups 12 and three, that is going counterclockwise as well. So carbon six will also have s configuration. So we are going to list that out front of the name three S six S oops. So in the end, the name, the full name for this compound is three S six S six ethyl, three isopropyl, two methyl, non one E four I. That is it for this problem. I hope this was helpful and thanks for watching.

Name the following alkynes according to the IUPAC rules of nomenclature.
(c)

Key Concepts

Alkynes

IUPAC Nomenclature

Numbering and Substituents

Watch next