Give IUPAC names for the following compounds:

Name: Problem 4, Chapter 7: Organic Chemistry
Uploaded: 2020-04-13T08:48:03-08:00
Duration: 10 min 31 s
Channel: Ricajoy Montero
Description: Give IUPAC names for the following compounds:

Key Concepts

Stereochemistry

Stereochemistry deals with the three-dimensional arrangement of atoms in a molecule. When applicable in IUPAC naming, descriptors such as R/S or cis/trans are used to convey the spatial orientation of substituents, which is essential for a complete description of the molecule's structure.

Substituent Naming and Positioning

Substituent naming involves identifying groups attached to the parent chain and assigning them position numbers to indicate their location. These substituents are listed in the compound name in alphabetical order, which is key for creating a clear and organized nomenclature.

Parent Chain Identification

Identifying the parent chain, which is typically the longest continuous chain of carbon atoms that includes the principal functional group, is a critical step in organic nomenclature. The parent chain forms the basis for the molecule’s name and numbering system.

Organic Compound Nomenclature Rules

This set of rules involves selecting the longest carbon chain as the parent structure, identifying and naming substituents, and applying appropriate prefixes and suffixes. It is essential for providing a clear and systematic approach to naming complex organic molecules.

IUPAC Nomenclature

IUPAC nomenclature is a systematic method of naming chemical compounds as established by the International Union of Pure and Applied Chemistry. It provides a uniform standard for naming compounds, ensuring that each name uniquely identifies a specific chemical structure.

Functional Groups

Functional groups are specific groups of atoms within molecules that dictate the chemical reactivity and properties of the compounds. Recognizing and correctly naming these groups is vital because they influence the suffixes and overall structure of the IUPAC name.

Transcript

00:01 Problem 58, we're still on aiupac naming.

00:04 So let's go and start with part a.

00:07 As usual, let's go and find the longest chain we could find.

00:11 And i think for this one, it's pretty obvious.

00:15 All right.

00:16 And here we have a double bond, right? and the rule is whenever we have a double bond or a triple bond, we count from the side where we're closer to that bond.

00:30 Okay, so what i mean by that is this double bond right here looks like it's closer to the right side.

00:39 And i mean you don't really have to estimate.

00:41 You could check, but i think it's pretty clear to see that there's only two carbons on the right side of the bond and there's three on the left.

00:50 So it's closer to the right side.

00:54 So we count starting on the right side.

00:58 So let's go and count.

00:59 We got one, two, two.

01:01 Three, four, five on our main chain, okay? so let me give a space for part a down here.

01:11 So we got five, meaning we have a pentine.

01:15 So pent for five and pentine because we have a double bond.

01:23 And that double bond starts on carbon two, right? it starts on carbon two.

01:30 So this would actually be two pentene, okay? just kind of ran out of space there.

01:36 But, okay, so we have two pentene.

01:39 Next is we check our branch points.

01:42 And we see here on carbon four that we have a methyl group branching out.

01:47 So we have a four methyl, four methyl, okay.

01:53 So now that we have all the information we need, we could combine all of it and come up with our answer.

02:01 Okay so we have only one one group that we need to take care of so we got four methyl four methyl two pantine so that's your bear answer for part a let's go to part b now okay so part b all right so again as usual let's find the longest chain in this looks like our longest chain right here right we got two carbons up here compared to just one that we go along a straight line okay and again we count from where our double bond is closer and for this one it's pretty obvious that it's on the right side okay in fact it's it's at the end of the right side here okay so we're counting from the right side here one two to three three, four, five, okay? so just like part a, we got pantene.

03:22 But this time, our double bond starts on carbon one, right? so we have one, pantene, because there's five carbons, and pantene, e and e, because we have a double bond.

03:41 Okay? so next is we take care of our branch points, and for this one, we have a actually only have one we have a methyl group connected to carbon 3 okay so we have i mean we use a different color pen here okay we have three methyl three methyl okay so i think that's all the information we need and we could proceed to combining all these information so we have three methyl we name our branching group first so 3 -methyl, 1, bentene.

04:31 And that's our answer for part b.

04:36 So part c now, we have a benzene ring.

04:41 And connected to our benzene ring are two chlorine atoms.

04:48 So we got to count one on the carbon where our functional group is or our atom is.

04:55 So this would be one.

04:56 And we got to count two on the closest branching group as well.

05:01 So this has to be two.

05:04 Okay? because if we count two here, then this chlorine would be carbon six, and we don't want that.

05:12 So this chlorine would be on carbon two...