Name the following alkyl halides: (FIGURE CANNOT COPY) |...

Name: Problem 17, Chapter 10: Organic Chemistry
Uploaded: 2020-06-26T00:23:08-08:00
Duration: 13 min 6 s
Channel: Tom Rutherford
Description: Name the following alkyl halides: (FIGURE CANNOT COPY)

Key Concepts

IUPAC Nomenclature

IUPAC nomenclature is the systematic method for naming organic compounds. It involves a set of standardized rules that allow chemists to generate unique and unambiguous names based on the structure of the compound. This system ensures consistency in naming, whether the compound is simple or complex, and provides information about the molecular structure including the carbon chain length and the type and position of substituents.

Alkyl Halides

Alkyl halides, also known as haloalkanes, are organic compounds in which one or more hydrogen atoms in an alkane have been replaced by a halogen atom. They are important in organic chemistry due to their reactivity and use as intermediates in various chemical reactions. Their nomenclature follows IUPAC rules where the halogen substituent is identified and its position on the carbon chain is indicated.

Longest Carbon Chain Identification

Identifying the longest continuous carbon chain is a crucial step in naming organic compounds. This chain determines the parent name of the molecule, and all substituents, including halogen atoms, are numbered accordingly to give the lowest possible locants. Correct determination of the longest chain is essential for an accurate systematic name.

Halogen Substituent Identification

Halogen substituents are treated as prefixes in the IUPAC naming system. Each halogen (such as fluorine, chlorine, bromine, or iodine) is recognized by its specific prefix and its position on the carbon chain is identified by a numerical locant. This identification is key to ensuring that the name reflects the structure of the alkyl halide.

Locant Numbering

Locant numbering involves assigning numbers to the carbon atoms in the longest chain such that substituents receive the lowest possible numbers. This process is critical for accurately indicating the positions of halogen atoms and any other substituents in the molecule, which helps avoid ambiguity and maintain consistency in the naming process.

Transcript

00:01 Okay, this problem is asking us to name these different compounds.

00:03 In the textbooks, it gives you a different form.

00:06 I'm going to draw them in the skeletal structure just because i think it's easier to represent the name in that way.

00:10 Okay, so let's start.

00:12 In this first compound, i have my bromines, i have another roaming, and then i have the remainder of my compound.

00:22 Okay, so to begin, i need to identify the longest carbon chain.

00:26 I can either go in this direction, or i can go in this direction.

00:30 Let's decide which direction we should go in.

00:32 So i can either start this way, one, two.

00:35 I see at position two, i have a methyl group that is a substituent.

00:39 So let's note that.

00:40 Let's test the other way.

00:41 I have one, two.

00:42 I also have a methyl group.

00:44 So because i have a methyl group on both sides, whether i go in the right direction or the left direction, i run into a methyl group, so i need to keep going.

00:53 Three.

00:54 At three, i have a bromo.

00:56 In three, this direction, i don't have anything.

00:58 So i want to put my substituent on the lowest numbered carbon.

01:02 So that's why i'm going to go in this direction.

01:04 Okay, i'm going to erase this.

01:06 Let's continue.

01:07 So 1, 2, 3, 4, 5, 6, 7.

01:11 Okay, i can also name this number 7, but because i already named it this one number 7, i'm just going to stick with that.

01:18 So i'm going to have a methyl group here eventually.

01:20 Okay.

01:22 So because it is a 7 numbered carbon, that means that i'm going to have hept.

01:29 Hept represents 7 and because it is an alkan as opposed to an alken or alkyne, i'm going to end in aen.

01:36 So heptane is my parent compound and now i need to determine where the substituents lie in that.

01:43 Okay, so i have a bromo and i also have another bromo.

01:48 So that means that i have at position 3 and position 4, position 3 and in position 4, i have 2 bromos.

01:55 So die, represented two, bromo.

02:00 Okay, that covers my bromines, and now i have my methyl groups.

02:05 So my methyl groups are at number two and number six.

02:09 So i need to draw that out.

02:10 Number two, and number six, die, representing two, methyl.

02:17 Okay, i don't have any more substituents, so i just need to put these all together alphabetically.

02:23 So three, four, die bromo, dash, two, six, dimethyl.

02:40 And because there are no numbers and letters, or letters between numbers, i just go straight ahead into my heptane.

02:50 So if you notice, between letters and numbers, i have dashes, and between numbers, i have commas.

02:56 Between letters, i don't have anything.

02:58 Okay, and that is just standard.

03:00 Okay, so the final answer is 3 -4 -di -bromo, 2 -6 dimethyl hepatine.

03:05 Okay, moving on.

03:10 I'm going to draw this in the skeletal structure as well.

03:14 In the textbook, they don't give you any different between the cis and trans.

03:18 So in my final answer, i'm going to have the trans isomer because that's the way i drew it.

03:23 Ifr to draw it this way, then it'll be cis.

03:30 But for now, i'm just going to do the trans, just so we have some answer.

03:34 Okay.

03:35 So i need to determine which way i should go in, because this is one continuous carbon chain.

03:41 I need to know if i should go this way or i should go this way.

03:44 Because this is an alken, the alkenes takes precedence over the iodide substitute.

03:50 So that's why i should go this way because the alken is at 1, 2, rather than 1, 2, 3, 3, 4.

03:58 Okay, so that's why i'm going in the right direction.

04:03 Okay, let's clean up.

04:04 Okay, 1, 2, 3, 4, 5, 6.

04:08 This is a 6 number of carbon, so that's why i'm going to end up with a hex, and because it is an alken, it's going to be hexine.

04:19 Okay, so the hex represents six, the e 'en represents the alken.

04:24 Okay, and i also have my iodide at my number five, so that's why i'm going to have five ioto.

04:34 So the hexene, the ein, the alken, is at position number two.

04:39 So i'm going to draw two, hexene.

04:41 If i were to draw three hexene, then my alkyne would be in this position right here.

04:46 I don't want that, i want to keep it how it is.

04:48 So two hexene.

04:50 So now let's put it all together.

04:52 It's going to be 5, ioto, 2, hexene.

04:59 Again, between numbers and letters, i have dashes, and if i were to have letters and letters, i want to have anything, no dashes.

05:08 Okay, moving on, i have this compound.

05:21 I have a bromine, i have the chlorine, and then i have the remainder of my compound.

05:27 Okay, so let's figure out which way to number this and what the numbered carbon is.

05:32 So i can either go this way or this way.

05:35 Let's figure out which one...

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