For each of the compounds described by the following names,

Question

1. draw a three-dimensional representation.

2. star (*) each chiral center.

3. draw any planes of symmetry.

4. draw any enantiomer.

5. draw any diastereomers.

6. label each structure you have drawn as chiral or achiral.

a. (S)-2-chlorobutane

b. (R)-1,1,2-trimethylcyclohexane

Accepted Answer

Hello everyone. Today, we have the following problem for R two, Romo butane and S 133 trimet Cyclin perform the following first draw a 3d representation using wedged and dashed lines. So that's what we will do for our first structure R two bromo butane, this is a four carbon chain. We will label that one as structure one. And then we have structure two. So structure one, we have a four carbon structure. What they bro mean on carbon two. So we have four carbons on carbon two. We have a bromine but this has a configuration of R. So what we wanna do is to ensure that this is our configuration, we will just place this bromine on a wedge, which we can switch to a dash accordingly. And we will identify the four different groups that are bound to that specific carbon. And we will sign priority based on atomic number. So the bromine will get a one, the hydrogen will get a four, the two, we go to the carbon that is bound to a carbon and two hydrogens. And the three, we go to the carbon that is bound to all hydrogens. We trace from 12 and three to give us a configuration of R. So we have drawn the correct structure. No, we do have to draw the representation using wedged and dashed lines. So we will actually just redraw this structure one as the following and the bromine will be on a wedge and it will mark the asymmetric carbon with an asterisk as part two says moving on to structure two. If we wanted to draw that structure out, that is S 133 meyl cyclopentolate structure, which is a five carbon molecule and a ring structure. And then we have three metals, we have one methyl on carbon one. If we wanted to label that carbon as carbon one and then on carbon three, we have two metals. Now we have that s configuration and that s configuration will actually be on our carbon one. And so what we will do is we will place that methyl group on carbon one on a dash and we will change accordingly. So we identify the four different groups bound to that carbon and I identify if it's asymmetric and it is. So the hydrogen will get a four, the group that will get a two will be the carbon to the right of the asymmetric carbon as both that carbon and the adjacent carbon, which you get a three are both bound to one carbon of two hydrogens, the next carbon or carbon three in that clockwise direction is bound to three carbons, which beats out the tie of the group that gets a three as the next carbon, which would be carbon four is bound to a carbon in two hydrogens. And then the four will go to the methyl as that is bound to three hydrogens. We trace or that will get a one excuse me, but she won. So we trace from 12 and three de me will get a three, which gives us our, however, the lowest part of the group or the hydrogen is on a wedge or to the front and we need it towards the back. So we will actually reverse this direction and making it counterclockwise or s so we have the correct configuration. So we should draw out structure two in a neater format as the following. And we will mark that carbon with an asterisk. So for part three show any planes of symmetry. If we look at our first molecule, there are no planes of symmetry because it's a cro molecule. If we look at our second structure, once again, there is no internal plan of symmetry or aligns with symmetry because there is a chiral molecule, we then move on to part four to draw any an anti and ditherer. So for our first molecule to determine if there are any isomers, we have to use the formula two race to the power of N where N is equal to the number of chiral centers. And So for molecule one, we will have two razor, the power of one giving us two isomers. And we will also have two isomers for a second molecule as that also has one chiral center. So what we can do is we can change the orientation of that bond that carbon bromine bond such that instead of the bromine being on a wedge, it can be on a dash. And then for our second structure, we can also change the orientation such that the metal on a carbon one is on a wedge and not a dash. And then last but not least, we have to identify each structure is a chiral or chiral. Well, for our first structure, both of these molecules are chiral as they both contain a carbon that has four different groups bound to it. And then if we look at our second structure, these both, both of these isomers are chiral as they both contain an asymmetric carbon. And with that, we have solved the problem overall, I hope this helped hand. So next time.

Key Concepts

Chirality

Stereoisomers

Planes of Symmetry

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