Draw all stereoisomers formed in each reaction. a. b. c. |...

Name: Problem 24, Chapter 10: Organic Chemistry
Uploaded: 2022-09-02T09:41:36-08:00
Duration: 7 min 23 s
Channel: Timothy Gailey
Description: Draw all stereoisomers formed in each reaction. a. b. c.

Key Concepts

Meso Compounds

Meso compounds are a special type of stereoisomer that contain chiral centers but are overall achiral due to an internal plane or center of symmetry. Recognizing when meso compounds are possible is important in correctly depicting the full set of stereoisomers, as they reduce the count of unique chiral products.

Enantiomers

Enantiomers are a pair of stereoisomers that are non-superimposable mirror images of each other. They are often formed when a reaction creates a chiral center. Mastery of the concept of enantiomers is required to correctly identify which stereoisomers are mirror images in the set of products.

Diastereomers

Diastereomers are stereoisomers that are not mirror images of each other. They can arise when a molecule has multiple chiral centers, leading to stereoisomers that have different physical and chemical properties. Understanding diastereomerism helps in cataloging all unique stereochemical outcomes of a reaction.

Stereoisomerism

This concept refers to the existence of molecules with the same molecular formula and connectivity of atoms but with a different three-dimensional arrangement of atoms in space. In the context of the question, understanding stereoisomerism is crucial for drawing all possible spatial arrangements of molecules resulting from a reaction.

Chirality

Chirality is the property of a molecule that makes it non-superimposable on its mirror image. Molecules with one or more chiral centers can exist as different stereoisomers (enantiomers), which are mirror images of each other. Recognizing chiral centers is essential when predicting and drawing the resulting stereoisomers.

Transcript

00:01 Draw all stereo isomer formed in each reaction.

00:03 So look at a.

00:05 We have an alken, a methyl group being attacked essentially by clcl.

00:11 So we'll have a contribution here, one electron.

00:16 It'll give a bond back, and it will donate to the cl.

00:19 So we'll end up with a front side bonded cl group that is positive, and this methyl group is now in the back.

00:33 Now this step, we could, and that's messy looking, this step could actually be reversed in that we could have the cl group go to the back, but we're going to keep it here for convention's sake, and we'll talk about it once we're done.

00:48 We now also have this negative cl group, who has too many electrons.

00:53 So we're going to have a donation, and the cl that's positive, it's going to want to get rid of a bond because he's positive.

00:58 So we need to donate the bond from the most stable carbon.

01:05 And notice that the bottom carbon is going to be a little more stable as a carboconon than the top one would be.

01:10 So we will donate the bond, an electron, to this cl.

01:16 At the same time, this negative cl will attack that carbon because there's an opening there.

01:26 So we now have a front -sided cl bond top that was.

01:32 The positive one and we have our negative cell which is coming from the back side who will be in the back upon into a front facing methyl group now we're supposed to draw each sterisomer that's possible so if we were to consider if we had done this the other way where we start with a cl in the back that our top seal would now be in the back because that one stayed the same orientation it does mean that everything else, though, would flip.

02:04 So this methyl group would now be in the back.

02:08 And our other sale group would be in the front.

02:11 So there we go.

02:12 That's the two possible versions.

02:13 It just attends if we can eventually decide to attach cl to the front or if we attach our cl group to the back.

02:20 Looking at b.

02:22 Really similar, we have the center al -keen, which is going to interact with our brbr, so we'll have a donation, interaction backwards, br will get the electrons.

02:38 I'm just going to put circles for brevity's sake here.

02:44 So we now have, and this bond is going to be kind of wacky too.

02:50 Okay, so we now have a front side bond with br and another front side bond, that same br, which means that these both are back -end bonds.

03:10 It's kind of strange.

03:12 We also, and this br is also positive, and we do have a negative br.

03:20 There's too many electrons who will attempt to either go top or bottom.

03:24 Now, you'll notice this is going to be a mirror situation.

03:27 No matter which where we go, it's going to be the same.

03:29 So let's just go ahead and assume he's going to go to the closest one...