Give the systematic name for each of the following compounds: a. b. c. d.

Name: Problem 27, Chapter 3: Organic Chemistry
Uploaded: 2021-07-19T21:18:13-08:00
Duration: 9 min 58 s
Channel: Allison Krajewski
Description: Give the systematic name for each of the following compounds: a. b. c. d.

Give the systematic name for each of the following...

Key Concepts

Substituent Naming

Substituent naming involves identifying and naming groups attached to the parent chain. These groups are treated as prefixes in the overall name of the compound. Correctly naming substituents includes using appropriate prefixes and their corresponding locants to specify their positions on the main carbon chain.

Locant Assignment

Locants are numerical designations used to indicate the position of substituents and functional groups on the parent chain. The assignment of locants follows the rule of lowest possible numbers to ensure clarity and minimize ambiguity in the compound’s structure.

Functional Group Prioritization

In systematic naming, functional groups are assigned priorities based on their reactivity and naming conventions. The highest priority functional group often dictates the suffix of the compound, while others are treated as substituents with corresponding prefixes. Correct prioritization is essential for ensuring that the molecule is named accurately according to IUPAC rules.

IUPAC Nomenclature

IUPAC nomenclature is a systematic method developed by the International Union of Pure and Applied Chemistry to name chemical compounds. It provides a set of rules to clearly describe the structure of a molecule by indicating the types, positions, and connectivity of atoms and functional groups, ensuring that each compound has a unique and universally accepted name.

Parent Chain Identification

This concept refers to selecting the longest continuous chain of carbon atoms in an organic molecule, which serves as the main framework for the compound’s name. The parent chain’s length and structure determine the root name and are fundamental in the application of numbering and naming rules.

Transcript

00:01 Okay, so for this problem, you're given a series of alkenes, and you want to provide suitable iupac or systematic naming to them.

00:15 So anytime you have a structure that you want to name, the first thing you want to do is identify the principal functional group.

00:23 And so in this case, your main functional group is going to be your alken.

00:28 So you're going to give your lowest priority of numbering to your alkyne.

00:36 So that's the first thing.

00:39 We identify our alkeen.

00:41 I actually not going to write this.

00:44 But from there, you want to identify your longest carbon chain that includes your alkyne.

00:51 So this first one is pretty easy.

00:54 So let's just start numbering from the left.

00:55 We can number one, two, three, four, five.

01:00 Six, seven, eight, nine.

01:04 So we have nine carbons.

01:07 So we know we have, we need to use the prefix non for nine.

01:15 From there, we need to give our alken the lowest possible number.

01:19 So we can number from left to right.

01:21 We could also potentially number from right to left.

01:25 So if we do that, we have one, two, three, four, five, six, seven, eight, nine.

01:34 And from there, we want to make sure that our alkenes has the lowest possible number.

01:38 So if we start numbering from the left, it is on four.

01:41 If we start numbering from the right, it's on five.

01:44 So we want it on the lower number, which is four.

01:48 So we want to use this numbering scheme and not that numbering scheme.

01:53 So our alken is on, our alken is on carbon four.

02:07 The last thing is, in this case, we want to, identify our substituents.

02:12 So we have two bromine groups.

02:14 So we have two bromines on carbon three and on carbon eight.

02:22 So if we put everything together, we have three eight dibromo.

02:32 And now we have our double bond on carbon four.

02:34 So we can call this four non -eem.

02:40 Sometimes you might see it written in this way.

02:43 So 3 -8 -di -bromo non -4 -e.

02:50 I think the second way is a little bit newer, but your textbook usually uses this first way, but just for your reference, i'll write it in the first way for the remaining problems, but both of these are correct.

03:09 Okay, so let's look at the next one.

03:11 So for right now, we're going to ignore e and z or e.

03:16 And and we're just going to do the naming for the structure.

03:22 So here we have our alken, we want to give this alken our lowest possible number.

03:27 So if we start numbering from the left, so we have one, two, three, because we have a longer chain on our ethyl versus our methyl group.

03:38 This would be carbon four.

03:40 And now we have two carbons up here versus four carbons down here.

03:45 So we name the chain five, six, six, seven, and eight.

03:53 I'll turn it.

03:53 So our, i guess, our longest chain is eight carbons, so this is going to be oct.

04:01 We could also potentially number from the right, but if we number from the right, we have one, two, three, four, five.

04:09 So our double bond is going to be on carbon five instead of on carbon three.

04:14 So we don't want to use that, but we do want to use what's shown in red.

04:17 So our double bond is going to be on carbon three.

04:24 And now we can look at our substituents.

04:27 So we have a methyl group on carbon three...

Please give Ace some feedback