1. ASSIGN R OR S TO ALL STEREOCENTERS 2. CIRCLE MESO...

1. ASSIGN R OR S TO ALL STEREOCENTERS 2. CIRCLE MESO COMPOUNDS 3. FOR THE FOLLOWING THREE PROBLEMS IDENTIFY RELATIONSHIP DIFFERENT COMPOUNDS CONSTITUTIONAL ISOMERS IDENTICAL COMPOUNDS DIASTEREOMERS ENANTIOMERS 1) $H_2C=C(H)-C(CH_3)H_2$ $H_3C-C(H)=C(H)CH_3$ 2) $\text{NH}_2$ $H$ $OH$ $H$ $HO_2C$ $CH_3$ $HO$ $H$ $H_3C$ $HO_2C$ $\text{NH}_2$ 3) $HO\cdots C(CO_2H)(H)C(CO_2H)OH$ $H$ $CO_2H$ $H$ $CO_2H$ $OH$ $OH$

Transcript

00:01 Hello students, hope you are doing great.

00:03 In this question, we are asked to draw all the stereoisomers of 2 ,3 -butanediol.

00:08 So the structure of 2 ,3 -butanediol is like this.

00:11 Here we have a chain of 4 carbon, here we have oh and here we have oh.

00:17 So this is the structure of butanediol and here it should be noted that we have two stereocenters.

00:22 Considering the stereochemistry of these two stereocenters, we have three stereoisomers for this given molecule which are like this.

00:31 Here we have oh in dash, oh in wedge.

00:39 Name of this molecule is 2r3s2 ,3 -butanediol.

00:50 Next molecule is, next stereoisomer for the same molecule is this where both the oh are present in wedge and the name of this molecule is 2r3r2 ,3 -butanediol and the third stereoisomer for the given molecule is this where both the oh group are present below the plane that is in dash and the name of this molecule is 2s3s2 ,3 -butanediol.

01:36 Now here in the first, we are asked which molecule is the meso compound.

01:40 So here it should be noted that these two stereocenters are having their opposite stereo, opposite configuration.

01:49 Therefore, if we cut this molecule from here, we have the equal half of each other.

01:53 Therefore, this is the meso compound and b and c are enantiomer of each other, a and b and a and c.

02:13 These two pairs are diastereomers of each other.

02:20 Next, we have to draw all the stereoisomers of 2 ,4 -pentanediol.

02:28 The structure of 2 ,4 -pentanediol is like this.

02:30 We have chain of five carbon like this.

02:33 On the second carbon, we have oh group.

02:35 On the fourth carbon, we have oh group and these two are the stereocenters into the given molecule...

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