Give the systematic (IUPAC) names of the following alkenes.

Question

(a)

(b)

Accepted Answer

Hey everyone, let's do this problem. It says write an appropriate pack name for each of the Elkins given below. And we have three Calkins here. So let's review the steps for how to do this. The first step which I guess could be optional. But I recommend it. Especially if you have substitution groups would be to draw out the compound again. It could be helpful if you have substitution groups which you'll see in the parentheses here because you might have to attach them to the parent chain using a double bond or it will help clear up the numbering of the parent chain. So I highly recommend taking a little bit extra time just to quickly draw a little rough sketch out. Then you're going to identify the longest carbon chain which is our parent chain. Right? And when we have al canes we have to make sure we include the double bond. Okay, include the double bond. So even if there is a longer carbon chain available, you have to do the longest one that includes the double bond. Okay then we will name our parent chain using the prefix from the number of carbons and the suffix in which is for our elkin. And if we have three or more carbons in that parent chain name, we're going to include the number carbon of the double bond. And of course we want to make sure we have the lowest number possible. Always trying to start our numbering close to that, the elk in the double bond. And remember that if we have multiple double bonds like a dying for two A. Trying for three tetra in for four double bonds etcetera. So you want to use those if we have multiple bonds and we can't forget about our substitue in groups. So after you have your parent chain all sorted out, you want to write out the name and location of the substitue ints. And we add that to the front of the name right in front of the parent chain. And I think that's about it. So let's do this for our compounds that we have. So we don't have any substitute prints here. Just a good old al keen. So it's pretty easy to number and we see that there's two double bonds on each side. This compound is symmetrical right? Ch two double bonds ch bonded to ch double bonds. Ch two. So it doesn't matter which way we're going to number. Let's start from the left. We have four carbons so that will be meth meth probe butte And we have two double bonds. So the prefix is actually going to be buta because we're adding a dying. So we have our double bonds on carbon one in carbon three. Including those numbers because we have more than three carbons dying. Okay so butA 13 dying Now. Any substitute Wint's. No so that is our name. You can also write it as 1, 3 Buta dying. Okay. Where you include the numbers out front. Does that make sense? Cool. Okay. Next up We have CH2. Oops I like to write it connecting the carbon. Ch two C. H C. H. And then we have two ch three substitution groups. But notice even though they look like substitue in groups in the condensed formula. This is why I like to draw it out because we see that we can extend our carbon chain not just to this third carbon but to one of the methyl groups. Right? So we can start numbering close to our alkaline 123 four carbons. I'll do it down there so it looks cohesive. So four carbons is what prefects Beauty again and are all keen is on carbon number one and we have our ending in So but one in and we have a substitution group coming off of that third carbon. Right? We have a methyl group. So we write three metal out front. So we have three methyl Beauty one in or three metal. Bring the number for the double bond out front. One. Beauty. Okay two down one to go. You guys are probably getting the hang of this by now. If not. That's okay that's why we have one more left to really solidify its CH three CH two C ch two and ch three. Okay well the problem said these were Elkins but where's my double bond. Did they lie to me? No notice how this carbon here only has three sigma bonds and so does the ch two group. So we need to have a double bond here in order for their octet to be satisfied. Tricky. Tricky. So again, that's why I recommend drawing these out really quick. So you don't miss that sort of stuff. And instead of this these four carbons being our parent chain, we're actually going to start from here that CH two because it has the all keen. Right? And remember in our parent chain we need to have the double bond. So we number 12 and then go to the left because it's a longer chain. 34. So again, four carbons leads us to butte Beauty mean? But we have the one in there again and our missile group similar to the previous compound except it is on carbon number two. So two metal but one in or two metal one Beauty already. That's it for this problem. I hope this video is helpful. And thanks for watching

Give the systematic (IUPAC) names of the following alkenes.
(a)
(b)

Key Concepts

IUPAC Nomenclature

Alkenes

Substituents and Chain Length

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