Draw structures corresponding to the following names:
f. 3-Methoxy-4-methyl-1-pentene

Question

Draw structures corresponding to the following names:

f. 3-Methoxy-4-methyl-1-pentene

Accepted Answer

Hi, everybody. Here's our next problem give the structure of the following compound five oxy, 24 dimethyl hept tune. And we have four answer choices labeled A through D and they're all different structures. It's written as a sort of semi condensed structure, not a proper condensed structure because the branches are drawn on here as branches, not in parentheses. So first of all, let's come up with a structure and then match it to one of our answer choices. So we look for this, our main carbon chain and we look to the end of the word which says we have a Hept to in. So it means we have seven carbons with that Hept refix. So we'll draw them on here. So we'll write in seven CS 23, 456 seven. And as we said, we have this as sort of a semi condensed structure. So we have all of the hydrogens indicated on there, we have our seven carbons, we know we have a double bond and it's written as two E. So the double bond goes from carbon two to carbon three. So we indicate that in there. Now, let's look at our substituent groups. We'll deal with the simplest ones here, we have two methyl groups on carbons two and four. So we find carbon two and write CH three as a branch on there and then carbon four at the other methyl group. And then finally, the first part of our name is five oxy. This means it's an ether with one R group being the entire carbon chain there and the other R group being an apple group, that two carbon group that's on carbon five. So over two carbon, number five, since it's in uh since it's an ether, our branch goes first to an oxygen atom and then it's an ethyl branch. So after the och two ch three, so we have our main carbons and our substituents, all we have to do is add the hydrogens. So we add three hydros hydrogens to each terminal carbon and then go along too. Carbon number two, it has four bonds because it has a branch and a double bond. So it doesn't get any hydrogens. Carbon three has three bonds. So it gets one hydrogen, carbon four has three bonds since it has a methyl branch. So it gets one hydrogen carbon, 53 bonds. So it gets one hydrogen carbon six, no substituents. So it gets two hydrogens. So we have our structure written out, let's compare it to our answer choices. If we look at choice. A, we see that we have our eth oxy group on carbon two along with the double bond. So that's not correct ethyl group is in the wrong place. In choice B, the epoxy group is also on carbon too. So that choice B can't be our correct answer. When we look at choice C, we see we have a methyl group on carbon two. So that's correct double bond from 2 to 3 methyl group on four and oxy group on five. So num letter C is correct, we can double check all of our carbons, all all of our hydrogens, I mean, and they are all correct. So choice C will be our correct answer to be thorough. We'll just take a look at choice D and we see with choice D, our oxy group is also in the wrong place. It's added on the final carbon instead of a branch. So choice D is incorrect. So there we have it. The structure of the compound that's named for us is our choice C option with methyl branches on carbons, two and four. A double bond between carbons two and three and an oxy oxy group on carbon five. See you in the next video.

Draw structures corresponding to the following names:
f. 3-Methoxy-4-methyl-1-pentene

Verified video answer for a similar problem:

Key Concepts

IUPAC Nomenclature

Functional Groups

Structural Isomerism

Draw structures corresponding to the following names:f. 3-Methoxy-4-methyl-1-pentene

Verified video answer for a similar problem:

Key Concepts

IUPAC Nomenclature

Functional Groups

Structural Isomerism

Draw structures corresponding to the following names:
f. 3-Methoxy-4-methyl-1-pentene