Write structural formulas for the following compounds (include...

Question

Write structural formulas for the following compounds (includes both old- and new-style names).

g. 5,5-dibromo-4-phenylcyclooct-1-yne

h. (E)-6-ethyloct-2-en-4-yne

i.1,4-heptadiyne

Accepted Answer

Hey everyone. Let's do this problem and says draw the appropriate structural formulas for the following old and new style name of compounds. We have three compounds five bromo four fennel cyclo oct E six methyl two in four ein and 14 hex a die eine. So first I like to break down the name of these structures, starting with the root or right to left. Sometimes there are some functional group names at the end, so I like to start with the root name and then go from there to functional groups and then substitution groups. Then we will draw the parent chain, we will add functional groups, then add substitue in groups and then we will add any hydrogen two carbons that still don't have four bonds. Right? Because they need 44 bonds to fill their octet Phil carbons octane. Okay, so let's do this. Starting with five bromo four fennel cyclo oct 19. Okay, so breaking down the name, we're just sort of translating what each part of this name means structurally. So starting with the route we have cyclo oct which means we have a ring with eight carbons. Then one line means we have a triple bond on carbon carbon number one. On carbon number four we have a fennel group which is a benzene ring. And on carbon # five We have a bruning atom. So let's draw our parent chain which would be our eight member ring. So 12345678. And we can number this ring. I'll do it counterclockwise adding our functional group. We have our triple bond on carbon one. Which means it's going to go between carbons one and two. Not one and eight. Okay then we have our fennel group on carbon number four. So our six member ring 1, 2, 3, 456 with alternating double bonds. And then our bro mean group on carbon number five. All right, so now we just have to add hydrogen to the carbons to fill their octet. So our carbons with the triple bond don't need any more hydrogen. Any hydrogen that all carbons with no substitution groups in the ring need to hydrogen two hydrogen ions To give them four bonds. Carbons with one substitution group. Like the bro mean need one hydrogen. Same with the carbon with the fennel ring and the carbons on our fennel ring each get one hydrogen because they all have a double bond. So they only need one hydrogen. Except that carbon that's connecting the fennel ring to the eight members ring. Right, It already has four bonds. So that is it for that structure. Okay, Let's go to the next one. We have E six metal oCT two in 4 line. So starting with the root we have OCD which is eight carbons. That doesn't look like an eight. So we have our eight carbon chain and two E. Means we have a double bond on carbon to for I means we have a triple bond on carbon number four. Methyl means we have a ch three group on carbon number six. And what does this E mean? Well that relates to the double bond because we can either have E. Or Z. Configuration and E. Means that the two largest groups are on opposite sides of the double bond. So remember this would be E. This would be easy on the same side. Okay, so we want our double bond to look like this. So we'll have a little dent in our straight chain. So let's draw our h carbon chain. So we're going to draw 123. That's for the E configuration four, 4567, 8. Okay, now we have our double bond, let's number our carbons first, our double bond on carbon number two. So carbon 2 to 3 triple bond on carbon four. So 4 to 5. So again, we see that E configuration here because one group is going down this way, the other larger group is going up this way. If we had a line here, they are on either side of the line, in the hydrogen would go this way. So we'll draw those in just a second. But I just wanted to demonstrate again how this is E configuration. So adding our substitution groups, we have the methyl on carbon # six. And now let's add all the hydrogen. So our methyl group gets three hydrogen End carbons get three hydrogen. So carbon is one and carbons with a single bond get to hydrogen ins This carbon with a method will get one carbons with a triple bond. Do not need any hydrogen. And these carbons with the double bond like we said get one hydrogen. So they are on opposite side of the double bond as well. So, that is it for compound number two. All right. One more. We have 14 hex a. Die ein, 14 exa die fine. So that hex a We know is going to be a six carbon chain die. I means we'll have to triple bonds and those will be on carbons one and four. So let's draw that out. So six carbon chain 123456. We can number them adding our triple bonds on carbons one. So between one and two and carbon four between four and five and adding our hydrogen, we add one to our end carbon with a triple bond, 3 to our other end carbon and no hydrogen on the other carbons touching the triple bonds. This carbon has two single bonds. So it needs to hydrogen and that is it for our hydrogen. And that is it for this problem. I hope this video was helpful and thank you for watching

Write structural formulas for the following compounds (includes both old- and new-style names).
g. 5,5-dibromo-4-phenylcyclooct-1-yne
h. (E)-6-ethyloct-2-en-4-yne
i.1,4-heptadiyne

Key Concepts

IUPAC Nomenclature

Functional Groups

Stereochemistry

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