Draw the structure that corresponds to the compound names show...

Question

Draw the structure that corresponds to the compound names shown.

(b) (5R,6E)-5-bromooct-6-en-1-yne

Accepted Answer

Hello everyone. Let's do this problem. It says what is the line angle formula that corresponds to the following molecule? SZ four Chloro no two E seven. I let's review the IU pack nomenclature rules so that we can convert this name into our structure. So the first step is to identify and number the parent chain. So in this case, going from the name to the structure, we will look at the end of the name for the parent chain name and remember that the locat goes before any modifications. And we want to look for double bond and triple bond modification. So for a double bond, the name will be modified to have an ending. And for a triple bond, the name is modified to have an ending. And again, the location of those double and triple bonds are right before these modified names. And that is it for the parent chain. The last thing is that the double bond and triple bond get numbering priority as well. So when we draw the structure, we want to assign the numbering properly and the numbering priority goes to the alkenes and the Alkins. OK. Then after we have the parent chain identified and drawn, we will look for the substituent, we usually identify and number the substituent. In this case, going from the name to the structure, we will look at the Logan that is before the modification of the name and that is the location of the substituent on the compound. And lastly, we have to apply any configurations to the structure. And in this case, I see SNZ. So we are dealing with EZ configuration which is for double bonds and R or S configuration, which is for chiral centers. So let's review how to identify those configurations. So for RS configuration, we use the coning pre rules. So first, we assign the priority of the four different groups on that chiro carbon. And we do so by atomic mass, if there is a tie, we look at the adjacent atoms and if there is a double bond or a triple bond, we count that atom as bonded multiple times then to assign R or S we will look at the orientation of the lowest priority group. So is group number four dashed or is it which? And when you are going from the name to the structure, you just need to draw it one way, assign the priorities and see if the way you drew it matches. If it doesn't, then you just switch the wedge and dash orientation. So if group number four is dash, we connect groups 12 and three. If they go in a counterclockwise position that is S configuration clockwise is R configuration. And the opposite is true. For when group number four is wedge, for easy configuration, we assign the priorities to the groups attached across that double bond. And if the high priority groups on each side of that double bond are on the same side of the fence, then that would be EZ configuration. And if the high priority group is on the opposite side, that would be E configuration. OK. So let's apply all these rules to draw our structure. So our parent chain, we look at the end of our compound non two E seven, I, so non tells us we have a nine carbon chain two. E tells us we have a double bond between carbons, two and three and 79. Tells us we have a triple bond between carbon seven and 81234567. OK. I'm going to have to redraw my structure because triple bonds are 180 degrees, right? Those atoms. So we don't have the bend 89. OK. So there is our parent chain. Now adding our substituent, we only have four Chloro. So chlorine on carbon four and let's add that wedge, let me number my chain. OK. So let's look at the configuration now. So did I draw the double bond configuration correctly? Well, what are the groups, the priorities of the groups across that double bond on carbon two, we have a methyl group in a hydrogen. On carbon three, we also have a hydrogen and then a carbon chain. So the highest priority groups are going to be those carbon groups in the lower priority groups will be the hydrogens. So if we draw a line for our fence, notice that the higher priority groups are not on the same side of the fence. So this is incorrect. So we have to draw our double bond like so, ok, so now our carbon groups are on the same side of that fence, right? Ok. And what about the s configuration? We have a croc carbon here. It's going to be carbon number four. So we have four different groups. Let's draw that hydrogen. So far I drew the chlorine as wedge, the hydrogen as dash. Let's see if that is the correct orientation here. So chlorine is highest priority. Hydrogen is lowest priority. Then we have the carbon that's in the double bond that carbon is attached to carbon carbon and hydrogen and the other carbon has one carbon and two hydrogens. So the carbon with more carbons and less hydrogens will be higher priority. So the carbons to the left carbon number three will be higher priority. Number two and carbon number five group will be third priority. So in this case, group, number four is dashed. We connect groups 12 and three that is going counterclockwise. So that is four S configuration or S configuration I just said four because that's carbon number four. So that is the correct way to draw. So let's add that chlorine as wedged configuration on carbon number four. Let me redraw this. It's a little wonky 123456. And then there in our wedge chlorine. So this is the correct line angle formula for SZ four Chloro non two E 79. That's it for this problem. I hope this was helpful and thanks for watching.

Draw the structure that corresponds to the compound names shown.
(b) (5R,6E)-5-bromooct-6-en-1-yne

Key Concepts

Stereochemistry

Nomenclature of Organic Compounds

Functional Groups

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