How to draw all the stereoisomers possible for the following molecules

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How to draw all the stereoisomers possible for the following molecules

Chemistry
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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these I mean
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Step #1 Identify chiral carbons. A chiral carbon is a carbon that has four different groups attached to it
|dw:1443409031254:dw|
|dw:1443409143801:dw| In our figure, you can clearly see that OH group a wedge means that the group is coming outwards, AT you, while a dashed line means the group is away not facing at you. On this carbon you can clearly see that this carbon is chiral with the OH group right? it has a CH3 group, an OH group, and a C carbon double bond. now what do we do now? for each stereo center there are two possible configurations, R and S. to assign these we must look at the atoms individually, and rank them according to their atomic weight, from a priority from 1-4. we must make sure that the group numbered 4 usually but not always is H is in the back. usually you just count the atoms, and then flip. so the heaviest elements get the highest priority 1. Oxygen bonded to the carbon is the highest so that gets labeled as 1. now we move over and see the other groups . the hydrogen bonded to the chiral carbon we label that 4. now we have two other groups a methyl group and a carbon attached to a double bond. for this we must look for the first point of difference. the first point of difference here is the double bond: notice a double bond means two carbon carbon bonds. so we have a for the CH3 1 carbon 3 hydrogens while for the double bond there are two carbons two hydrogens. the double bond gets designation 2 while the methyl group gets 3. @aaronq i could use your expertise to possibly identify any errors. |dw:1443409584603:dw|
but, what if say we have another configuration for this
|dw:1443409680858:dw|
the dashed line means that the OH is now in the back which means it would look like this: |dw:1443409779626:dw| for this to assign priority you would have to do the same procedure and if you get clockwise it's S if it's counter clockwise it's R.
generally speaking, you would identify the chiral carbons and count them up, then the number of stereoisomers becomes if I remember correctly, \[2^{n} \] where n is the number of chiral carbons. if you have 2 chiral carbons you have 4 stereoisomers,
my last comment should help you in setting up / solving your problem
Thank you!!
@c17H19NO3 this is a long problem there's a lot of different things you have to take into account later on but the first one and the most important is that chiral centers have four different groups attached to them.

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