Draw the products of each reaction, including stereoisomers....

Key Concepts

Reaction Mechanisms

This concept involves understanding the step-by-step process by which reactants are transformed into products. It requires knowledge of electron flow, intermediates, and how various conditions or reagents influence the movement of electrons in molecules, leading to the formation of specific products.

Structural Representation in Organic Chemistry

This key concept focuses on the skills used to accurately draw the structures of organic molecules. It includes understanding how to depict bonds, functional groups, and three-dimensional orientation of atoms, which is especially important when representing stereochemical information.

Stereoisomerism

Stereoisomerism addresses the existence of molecules that have the same molecular formula and connectivity but differ in the spatial arrangement of their atoms. Recognizing and drawing stereoisomers, such as enantiomers and diastereomers, is crucial in understanding the properties and reactivities of chemical compounds.

Predicting Reaction Products

This concept involves the ability to deduce the outcome of chemical reactions based on the nature of the reactants, reagents, and conditions. It includes understanding regioselectivity and stereoselectivity, which help in predicting and drawing all major products, including any stereoisomeric forms.

Transcript

00:01 This is the answer to chapter 10, problem number 57 from the smith organic chemistry textbook.

00:10 And this problem says, draw the products of each reaction, including stereo isomers.

00:17 Okay, and so that is what i am going to do.

00:20 So for a here, we are hydrating an alkyne.

00:27 And so the alcohol is going to go to the more substituted carbon and the hydrogen is going to go to the less substituted carbon.

00:45 And so what that is going to mean is that our products are going to look like this.

00:59 Sorry, i was thinking for a moment about the other side of the ring, and i will show you what i mean.

01:07 So if the alcohol adds to the front, the methyl group will be pushed to the back like that.

01:14 So there's the alcohol.

01:16 And then we do need to draw stereochemistry for this carbon, but it doesn't really matter what stereochemistry we give it, because it's going to be cyst to the alcohol in one of these, entranced to the alcohol in the other.

01:35 And i will show you what i mean.

01:57 But we should give it the same stereochemistry in both.

02:19 Okay.

02:20 And so those are going to be our product here.

02:23 All right.

02:24 And then so for b, we're going to get the markovnikov edition of this iodine and hydrogen.

02:32 And so our products are, again, going to be a mixture.

02:39 So we can have the iodine add from the front like this, or we can have the iodine add from the back.

02:50 And then it will be dashed like that.

02:53 Okay.

02:54 I'm going to move these up a little bit so that i have some room for c.

02:58 Okay.

03:00 So then c is going to be the addition of two chlorines across this double.

03:05 Bond.

03:07 And remember the stereochemistry of this reaction, the chlorines always add anti.

03:14 So there are going to be two products here, but in both products, the chlorines are only going to be anti.

03:23 So if this chlorine is on a wedge, that is going to mean that this chlorine has to be on a dash and then the opposite will be true as well.

03:51 So if this chlorine is on a dash, puts this methyl group on a wedge, and then this methyl group on a dash so that this chlorine can be on the wedge.

04:04 Okay.

04:05 So moving to the next page, this is a hydroboration oxidation reaction, and this is always a sin addition of the elements of water, but again, we have no facial control.

04:24 So this is a sin addition, but we're going to get, again, two products.

04:36 So that methyl group is going to be unchanged regardless.

04:41 And then if our elements of water come from the front, they will both be on wedges, and if our elements of water instead come from the back, they'll both be on dashes, which would make this methyl group wedged, and this methyl group will always be wedged.

05:18 Okay, so then looking at e, we have the addition of hbr across this double bond, and so again, it's always going to be a markovnikov product.

05:34 So the bromine is always going to go to the tertiary carbon rather than the secondary carbon, but we're not going to have any control over which face it adds from...

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