Kaplan Question-ochem
Started by Jo07
First think of the Br, D, OH as being able to rotate about the chrial carbon.
Rotate Br, D, OH so that Br is in the back. (Mentally carry D toward the left.)
That way, D is on the left (towards you), Br is in the back (away from you), and OH is to the right (also towards you). Draw out the stick version.
(Remember: horizontal lines are towards you....if you have a hard time remember....think of them as bow ties, or that they look like hands which are out to huge you.)
Next, do the same for the bottom part. The way it's drawn OH is already in the back. So simply translate into Fisher. F to the right, H to the left, OH on the bottom.
So combine and draw Fisher.
The way I decribed, it looks like...
Br on top
D to left
OH to right
H to left
F to right
OH on bottom
Now, one important thing to remember about Fischer projections is that you can switch around the elements on the chiral carbon. Things that are next to each other can be switched. Example, if we switch the Br and the D, the Fisher we get is an enanitomer. But, if we carry out another switch, then we get a different representation of the original Fisher. For example, if you carry out a switch between the Br and top OH. This Fisher would be the same as the original Fisher.
So the key is: 1 switch = Enantiomers, 2 switches: Identical.
For us to get the Fisher to look like the one showen next to A...we'd have to carry out 1 switch between the F on the right and the bottom OH. Thus, the Fisher drawn in A is an enantiomer of the true Fisher, thus it is not the correct answer.
Rotate Br, D, OH so that Br is in the back. (Mentally carry D toward the left.)
That way, D is on the left (towards you), Br is in the back (away from you), and OH is to the right (also towards you). Draw out the stick version.
(Remember: horizontal lines are towards you....if you have a hard time remember....think of them as bow ties, or that they look like hands which are out to huge you.)
Next, do the same for the bottom part. The way it's drawn OH is already in the back. So simply translate into Fisher. F to the right, H to the left, OH on the bottom.
So combine and draw Fisher.
The way I decribed, it looks like...
Br on top
D to left
OH to right
H to left
F to right
OH on bottom
Now, one important thing to remember about Fischer projections is that you can switch around the elements on the chiral carbon. Things that are next to each other can be switched. Example, if we switch the Br and the D, the Fisher we get is an enanitomer. But, if we carry out another switch, then we get a different representation of the original Fisher. For example, if you carry out a switch between the Br and top OH. This Fisher would be the same as the original Fisher.
So the key is: 1 switch = Enantiomers, 2 switches: Identical.
For us to get the Fisher to look like the one showen next to A...we'd have to carry out 1 switch between the F on the right and the bottom OH. Thus, the Fisher drawn in A is an enantiomer of the true Fisher, thus it is not the correct answer.
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Another example:
D. WHY IT IS THE ANSWER.
1) Mentally rotate and take D to the back (D to the back because it is dashed.)
2) Draw Fisher.
RESULT: Br to the left, D on top/back/away from us, OH to the right
3) Bottom part does not need to be rotated since OH (dashed) is already in the back.
4) Draw Fisher.
RESULT: H to right, F to left, OH on bottom.
5) Compare Fisher we drew w/ answer given.
We drew: D on top, Br to left, OH to right, F to left, H to right, OH bottom.
SWITCHES: (One way of doing it.)
Br & D
D & OH
F & OH
OH & H
Gives us the same answer. We made 4 switches (even number) = identical.
D. WHY IT IS THE ANSWER.
1) Mentally rotate and take D to the back (D to the back because it is dashed.)
2) Draw Fisher.
RESULT: Br to the left, D on top/back/away from us, OH to the right
3) Bottom part does not need to be rotated since OH (dashed) is already in the back.
4) Draw Fisher.
RESULT: H to right, F to left, OH on bottom.
5) Compare Fisher we drew w/ answer given.
We drew: D on top, Br to left, OH to right, F to left, H to right, OH bottom.
SWITCHES: (One way of doing it.)
Br & D
D & OH
F & OH
OH & H
Gives us the same answer. We made 4 switches (even number) = identical.
Sorry I'm lost.
This is the fischer that is correct...
Br
OH--------- D
H-----------OH
F
I don't understand how that coincides with "D on top, Br to left, OH to right, F to left, H to right, OH bottom."
"The way I decribed, it looks like...
Br on top
D to left
OH to right
H to left
F to right
OH on bottom"
I don't know, but I just don't see it 😕
This is the fischer that is correct...
Br
OH--------- D
H-----------OH
F
I don't understand how that coincides with "D on top, Br to left, OH to right, F to left, H to right, OH bottom."
"The way I decribed, it looks like...
Br on top
D to left
OH to right
H to left
F to right
OH on bottom"
I don't know, but I just don't see it 😕
Sorry...I didn't clearly label.
The first post is for why A is not the answer.
The second post is for why D is the answer.
They are seperate. Is that the confusion?
If that's not the issue, then in your question, which letter are you referring to A or D?
The first post is for why A is not the answer.
The second post is for why D is the answer.
They are seperate. Is that the confusion?
If that's not the issue, then in your question, which letter are you referring to A or D?
Ah okay I see... one thing- you say that you don't move OH to the back because it is already dashed... well isn't D already dashed and in the back as well?
So as long as I do an even amount of switches, it will be the same compound? 2 or 4 vs 1 or 3 being enantiomers?
Thank you for taking the time to explain this
So as long as I do an even amount of switches, it will be the same compound? 2 or 4 vs 1 or 3 being enantiomers?
Thank you for taking the time to explain this
All the answers choices require you to rotate the top part because they have an element w/ a dashed lines that is not drawn in the back...it's drawn to the right or towards us. (But dashed means away from us, so we rotate to get it away from us, then draw the Fischer.)
For instance:
A) OH would be to the back if we drew the Fischer the way it is drawn...but OH does not have a dashed line. So that would not be correct.
However, the bottom portions are all drawn correctly, the dashed element is drawn towards the back. In all the examples, OH is dashed and drawn to the back and either F or H is wedged and sticking out towards us.
Yes, if you remember that odd = enantiomers and even = same...
You'll be able to use that always to see if two given Fisher projections are identical or enantiomers. 🙂
Glad you took the time to get this. I remember 2 Fischer questions on the O. Chem section.
If you need more help, let me know. 🙂
For instance:
A) OH would be to the back if we drew the Fischer the way it is drawn...but OH does not have a dashed line. So that would not be correct.
However, the bottom portions are all drawn correctly, the dashed element is drawn towards the back. In all the examples, OH is dashed and drawn to the back and either F or H is wedged and sticking out towards us.
Yes, if you remember that odd = enantiomers and even = same...
You'll be able to use that always to see if two given Fisher projections are identical or enantiomers. 🙂
Glad you took the time to get this. I remember 2 Fischer questions on the O. Chem section.
If you need more help, let me know. 🙂
Okay I went through all of them and draw quick fischer projections...
A wouldn't be right because it's:
Br
D OH
H F
OH
and to make it match you would have to switch the OH and F making one switch--enantiomer. Since in the original Br on top and OH on bottom are dashed lines they will be the vertical lines in the fischer.
I did that for all of them and only D has 4 switches making it Identical.
Am I going about this the right way? Or is this totally screwed up?
A wouldn't be right because it's:
Br
D OH
H F
OH
and to make it match you would have to switch the OH and F making one switch--enantiomer. Since in the original Br on top and OH on bottom are dashed lines they will be the vertical lines in the fischer.
I did that for all of them and only D has 4 switches making it Identical.
Am I going about this the right way? Or is this totally screwed up?
Thank you z123, you explained it very well. I had problem with this question as well. Another question. Do you think if I get a question like this I should go all over them from a-e or there is better quicker way?
