Draw structures corresponding to the following IUPAC names:
a. trans-2-Pentene
b. trans-3,4-Dimethyl-3-hexene
c. 2-Methyl-1,3-butadiene
d. trans-3-Heptene

Question

Draw structures corresponding to the following IUPAC names:

a. trans-2-Pentene

b. trans-3,4-Dimethyl-3-hexene

c. 2-Methyl-1,3-butadiene

d. trans-3-Heptene

Accepted Answer

Welcome back everybody. Our next problem says, draw the structures of the Iupac names below. We have two different names. So we'll look at each of them for number one, we have trans 45 dimethyl oct four. So we look at our route of octin. So we know we have eight carbons. So we'll draw that line structure. 12, 345678, always double check by counting 12345678 carbons. So that's correct. The double bond for that makes it an alkene is on carbon four. So from 4 to 51234 of our double bond from 4 to 5. And then we're told we have dimethyl two methyl groups at carbons four and five and they're in the trans arrangement. Well, as we look at our molecule, we see that the either branch of our chain. So there's essentially two propol groups on either side, 123123 of our double bond. And they're already in the trans arrangement because they are opposite each other. So if I highlight those two branches, I see that they're on opposite sides of the double bond. So we can fill in our metals on our structure as drawn because they will also end up in the transposition. So the two metals coming up off our double bond. So ch actually, we don't need to write ch three excuse me, this is a skeletal structure just our line to make our methyl group. And they are in the transposition uh on opposite sides of the double bond. So we have our transposition, our two methyl groups indicated our eight carbon chain and the one double bond. So that's our line structure for our first molecule. So we'll go ahead and highlight that. Now, let's go on to molecule number two, which says trans hex three E. So we have that hex prefix. So we know there are six carbons in our main chain. So 123456, double check the count 123456. And we see that our double bond is from carbon three to carbon four. Draw that in. And we have a prefixed trans showing that the two parts of our long chain should be on opposite sides of the double bond, which they already are. Hopefully, by both of these examples, you can see that when you draw your main chain in this zigzag style, that is sort of the standard starting point, you are automatically setting up your molecule to have a trans arrangement. So we don't have to do anything because we already drew it in a trans arrangement, there's no other substituents, it's trans because these two chains are opposite sides. And therefore the other things that would be in here are hydrogen atoms if we drew them on there. So that makes it pretty clear that we have the trans isomer in this case. So we'll just remove those hydrogens for now. And we have this line structure of our six carbon chain, no substituents, a double bond between carbons three and four. See you in the next video.

Draw structures corresponding to the following IUPAC names:
a. trans-2-Pentene
b. trans-3,4-Dimethyl-3-hexene
c. 2-Methyl-1,3-butadiene
d. trans-3-Heptene

Verified video answer for a similar problem:

Key Concepts

IUPAC Nomenclature

Stereochemistry

Alkenes and Their Structures

Draw structures corresponding to the following IUPAC names:a. trans-2-Penteneb. trans-3,4-Dimethyl-3-hexenec. 2-Methyl-1,3-butadiened. trans-3-Heptene