Give a correct name for each compound.
(e)

Question

Give a correct name for each compound.

(e)

(f)

Accepted Answer

Hello everyone. Let's do this problem. It says write the correct Iupac names for the following structures. And we are given the structure of two chemical compounds. So let's review the IU pack naming steps, step one A is going to be identify and number the parent chain and the parent chain is the longest continuous carbon chain. And we want to make sure that our substituent get the lowest numbers possible. OK. Then we want to modify the root name. So if we have say an alcohol group or an A double bond or a triple bond, we need to modify the root name from A&E to in this case, I noticed we have several double bonds. So we will change that root name from A&E to ene the ending. OK. Then step two, we are going to list the substituent with their locus or the number of the carbon that they are attached to. And we are going to list these in front of the root name in alphabetical order as always. Then step three, if necessary, we will add the configuration. So if we have, let's say Carl carbons, we will add R or S configurations. If we have double bonds that are substituted in a certain way, we will have E or Z configuration. OK. And lastly, step four, we are going to combine all of these aspects of the name. So first the configuration, then the substituent, then the root name. OK. So let's apply these steps to our two compounds. So compound A, the longest continuous carbon chain is going to be our six membered ring, right? We have a five membered ring and a six membered ring. So the six membered ring will be our route. Ok? And since it is six carbons, that will be the name cyclohexane. But notice that we have double bonds, two double bonds. So we want to include the ending dye and the locus of those two double bonds. So the name will be Cyclo Xa 13 dying. Now, how did I know it was 13 dying? I forgot to number the parent chain. We want to number our ring and we are going to start where that other ring is attached, right? Our substituent group because like we said, we want our substituent groups to get the lowest number possible and we want our alkenes to also get the lowest number possible. So we will start numbering our ring there, which means the double bonds start on carbons, one and three. So that's psycho hea 13, dying. And our substituent group is a five membered ring, all single bonds. So that name is cyclo penal, right, an alkyl group. And that is our only substituent. We don't have any chiral carbons or EZ configuration to worry about. And that substituent group is on carbon number one. So the final name for this compound is one cyclo Penil, cyclo hea 13, dying. OK. Moving on to compound B our longest continuous carbon chain is going to be 1234567 carbons long. So that is heptane, but we have some double bonds here. Let's number our chain first. So which end will we start at? Our bromine is closer to the right end, right? And our double bond is also located there on carbon two if we number from that way. But on the left hand side, if we start numbering from there, our alkene will start on carbon one alk take higher priority than the bromine groups and the bromo methyl group. So we will start numbering from the left hand side, 1234567. So that our double bonds get the lowest number possible and those double bonds are on carbons, 13 and five. So we will have Hep 135 trie OK. That is our root name. Now, our substituent, we have a bromine group on carbon six and a bromo methyl group on carbon three. So let's list those in alphabetical order. So we will have six Romo three, bromo, methyl, right? Since bromo is shorter than bromo methyl that is technically comes first in alphabetical order and then HEP 135. But do we have any configuration here? Yes, we have some easy configuration we need to consider because look at the double bonds on carbons three and five at alkene on carbon three, carbon three has two groups. So that side is di substituted and carbon four is monos substituted right within our group. So we have four different groups on these two carbons combined. So we can therefore prioritize them and determine if they are on the same side of that double bond or the opposite side of the double bond. So we list the priorities by atomic mass. And if we look at on carbon three, the bromo methyl group, that will be highest priority over the a little group. And at carbon four, that carbon group is going to be higher priority over the hydrogen that is not drawn. So the higher priority groups are on the same side. So that is going to be Z configuration. So we have three Z. Now at carbon number five or al, number five, I should say we have the same situation and that carbon five has a hydrogen and a carbon group so that hydrogen automatically gets second priority. The carbon group gets higher priority. And the bromine on the on carbon six will get higher priority than the methyl group because bromine has a higher atomic mass than carbon. So again, our higher priority groups are on the same side. So that is Z configuration for carbon five as well. So the full name is three Z five, Z six bromo, three bromo, methyl Hep 135 trie. So that is it for this problem? I hope this video is helpful and thanks for watching.

Give a correct name for each compound.
(e)
(f)

Key Concepts

IUPAC Nomenclature

Functional Groups

Locants and Substituents

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