Dashed lines to Fischer conversion?

[steelrfan8]

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This is #71 from Kaplan BB practice exam in the back... I can't figure out how to convert the dashed lines into the Fischer projection. I tried the R and S ranking for each chiral center to compare to the Fischer projection and that doesn't work so I must be be doing something wrong.... can anyone tell me how I have to look at the dashed lines in order to convert to Fischer?

More specifically, when I assign R,S config to the chiral centers, I get the top chiral configuration to match up with the correct Fischer chiral config given in the back for the answers.... the bottom ones I keep getting the opposite config of what the answer is supposed to be so I must be looking at the bottom incorrectly. So basically, if someone could explain how to assign the correct R,S config and also explain how I can look at the given dashed projection and covert it into Fischer without assigning configs it would be great.

I attached the picture of the problems.

Thanks a lot

 

[americanpierg]

This is #71 from Kaplan BB practice exam in the back... I can't figure out how to convert the dashed lines into the Fischer projection. I tried the R and S ranking for each chiral center to compare to the Fischer projection and that doesn't work so I must be be doing something wrong.... can anyone tell me how I have to look at the dashed lines in order to convert to Fischer?

More specifically, when I assign R,S config to the chiral centers, I get the top chiral configuration to match up with the correct Fischer chiral config given in the back for the answers.... the bottom ones I keep getting the opposite config of what the answer is supposed to be so I must be looking at the bottom incorrectly. So basically, if someone could explain how to assign the correct R,S config and also explain how I can look at the given dashed projection and covert it into Fischer without assigning configs it would be great.

I attached the picture of the problems.

Thanks a lot

I don't think you can assign R,S config because of the "D." I have a really hard time turning these into Fischer projections in my head, so i always make sure to check my work afterwards by converting them into newman projections, then converting newman into fischer. takes longer than normal... but once i find the right one i answer it, mark it, and come check the others later if i have time.

i try to find one to check by first doing it quickly in my head. Just to make it quick, you know the answer is D. on the bottom left Carbon, OH is pointing away, H is pointing towards, and F is in the plane. So you know you'll have OH-F-H going that way clockwise at the bottom of the Fischer.

At the top right carbon, you have D facing away from you. That puts BR on its left and OH on its right. So, it goes D-OH-Br in clockwise order.

Draw these, do some bond rotations, and you end up with the answer.

Checking it with newman projection-
Front molecule- F top, H bottom right, OH bottom left
Back molecule- OH bottom, Br top right, D top left

convert newman to fischer, you get same fischer projection

this is how i do it... hope that helps

 

[steelrfan8]

I don't think you can assign R,S config because of the "D." I have a really hard time turning these into Fischer projections in my head, so i always make sure to check my work afterwards by converting them into newman projections, then converting newman into fischer. takes longer than normal... but once i find the right one i answer it, mark it, and come check the others later if i have time.

i try to find one to check by first doing it quickly in my head. Just to make it quick, you know the answer is D. on the bottom left Carbon, OH is pointing away, H is pointing towards, and F is in the plane. So you know you'll have OH-F-H going that way clockwise at the bottom of the Fischer.

At the top right carbon, you have D facing away from you. That puts BR on its left and OH on its right. So, it goes D-OH-Br in clockwise order.

Draw these, do some bond rotations, and you end up with the answer.

Checking it with newman projection-
Front molecule- F top, H bottom right, OH bottom left
Back molecule- OH bottom, Br top right, D top left

convert newman to fischer, you get same fischer projection

this is how i do it... hope that helps

I think this was the step I was missing... converting to Newman (since I know how to do Newman to Fischer).

Quick question though, when you're looking at the dashed line projection, before converting to Newman, you have to look at it so that the dashed line atom is going AWAY from you (similar to R,S config rules) and then you figure out whether the 2 remaining atoms are going to be on the left/right? Or do you have to look straight into the front Carbon for example and then picture how the other 3 atoms are arranged?

Thanks a lot for the help BTW

 

[americanpierg]

I think this was the step I was missing... converting to Newman (since I know how to do Newman to Fischer).

Quick question though, when you're looking at the dashed line projection, before converting to Newman, you have to look at it so that the dashed line atom is going AWAY from you (similar to R,S config rules) and then you figure out whether the 2 remaining atoms are going to be on the left/right? Or do you have to look straight into the front Carbon for example and then picture how the other 3 atoms are arranged?

Thanks a lot for the help BTW

When you convert to Newman, you just draw it as it is. Imagine you're looking at the C-C bond from the bottom left side. This one is convenient becuase F and OH are top and bottom, but I don't know how it would be done if the OH was top too at the back, but anyways. Just draw the Newman how you see it. F and OH are in the plane, up and down, respectively. Draw the others on how theyre facing towards and away from you. Draw the fischer, keep Carbon with F still, rotate C with Br until OH is at the bottom. Then flip entire configuration 180 degrees. Hard to explain, but yea.

Just remember that when you go from Newman to Fischer, the back two carbon facing outside of the plane are switched.

NOTE: I just realized that the back carbon is always reverse in configuration than what you think you would normally get, which may be why you are always getting one chirality center wrong in your application or R/S. I'm pretty sure you're supposed to switch the configuration for the back carbon, since you reverse it when you change from Newman to Fischer too. Not sure why, but thats how Newman to Fischer works so I'm assuming this is no different. The substituents attached to the carbon closest to you (bottom left in this case) stays as it is (F-H-OH clockwise) while the back carbon is reversed (D-Br-OH clockwise becomes D-OH-Br clockwise)

 

[steelrfan8]

I'm starting to think that either I'm totally missing something important or the problem in the book is wrong. I just watched this video and it explains exactly what I been having trouble with and after reading what you explained I think I understand how to make the conversion.

http://www.youtube.com/watch?v=KKOY1sdYEes

Sticking with example D, as we've been talking about, for the Fischer projection it looks like that the atom connected to the solid line is the atom that goes on the top or bottom.... so in D, we should have OH on top and F on the bottom in the Fischer projection. Then we rotate the bottom to align the F on the solid line with the OH on the solid line above, and end up with OH on the left and H on the right. The top, since there is no rotation, ends up with D on the left and Br on the right. So for Fischer it should look at follows: (this follows logic from the video I posted)

.....OH
D ----- Br
OH ----- H
......F

The Fischer answer given in the book thats supposed to correspond with figure in D looks like:

......Br
OH ----- D
H ----- OH
......F

Why is the Br thats supposed is pointing forward in diagram D on top instead of the OH? Am I incorrect in assuming that the solid lines are the standard for the atoms that go on top and bottom?

Also, lets say we convert D into Newman as you said

Front molecule- F top, H bottom right, OH bottom left
Back molecule- OH bottom, Br top right, D top left

I draw this out, convert into Fischer which gives us (this is how Destroyer shows to convert Newman into Fischer)

.....F
OH----H
Br----D
....OH

Without going into details because our answers differ, we are consistant at keeping the F and OH (which are on solid lines in D) on the vertical axis.... the answer given in Kaplan puts Br and F on the vertical axis and this is where I get totally lost. The vid I posted makes perfect sense to me but doesn't go correspond with what Kaplan has.

Thanks for taking the time answering, I knwo this is super long but its driving me a bit crazy since its the only thing that doesn't seem to click for orgo and I know it'll probably be on the real exam in one form or another.

 

[porterhouse]

This may help a bit.

http://forums.studentdoctor.net/showthread.php?t=664727

 

[steelrfan8]

This may help a bit.

http://forums.studentdoctor.net/showthread.php?t=664727

Thats a good explanation.... I'm now almost 100% convinced that the problem in Kaplan BB is completly wrong. The explanation given in the youtube vid I posted along with what you posted contradicts the information given in Kaplan.

Main thing is that on a sawhorse projection, the atoms on the SOLID lines end up on the top or bottom in both Newman and Fischer projections. After those are arranged, its somewhat simple to figure out the config of the 2 remaining atoms.

I'm kinda pissed off because I now wasted almost an entire day of studying on this single problem... but as of now, I'm going to have to conclude that Kaplan is way off on this.

 

[americanpierg]

I'm starting to think that either I'm totally missing something important or the problem in the book is wrong. I just watched this video and it explains exactly what I been having trouble with and after reading what you explained I think I understand how to make the conversion.

http://www.youtube.com/watch?v=KKOY1sdYEes

Sticking with example D, as we've been talking about, for the Fischer projection it looks like that the atom connected to the solid line is the atom that goes on the top or bottom.... so in D, we should have OH on top and F on the bottom in the Fischer projection. Then we rotate the bottom to align the F on the solid line with the OH on the solid line above, and end up with OH on the left and H on the right. The top, since there is no rotation, ends up with D on the left and Br on the right. So for Fischer it should look at follows: (this follows logic from the video I posted)

.....OH
D ----- Br
OH ----- H
......F

The Fischer answer given in the book thats supposed to correspond with figure in D looks like:

......Br
OH ----- D
H ----- OH
......F

Why is the Br thats supposed is pointing forward in diagram D on top instead of the OH? Am I incorrect in assuming that the solid lines are the standard for the atoms that go on top and bottom?

Also, lets say we convert D into Newman as you said



I draw this out, convert into Fischer which gives us (this is how Destroyer shows to convert Newman into Fischer)

.....F
OH----H
Br----D
....OH

Without going into details because our answers differ, we are consistant at keeping the F and OH (which are on solid lines in D) on the vertical axis.... the answer given in Kaplan puts Br and F on the vertical axis and this is where I get totally lost. The vid I posted makes perfect sense to me but doesn't go correspond with what Kaplan has.

Thanks for taking the time answering, I knwo this is super long but its driving me a bit crazy since its the only thing that doesn't seem to click for orgo and I know it'll probably be on the real exam in one form or another.

It doesn't matter whats on the vertical axis or not, because you can always rotate the molecule to get whatever molecule you want to be there. To rotate, you have to keep one molecule the same and rotate the other three, in this case, keep the carbon the same and rotate the other three around the carbon until you get whichever one you want to be on the vertical axis.

You also put the B and D backwards. Notice in destroyer, the arrangement in the back carbons after going from newman to fischer are inverted. I don't know the reasoning why, but they say so and it works for every question, so I stuck with it and it indeed works for everything. If you switch B and D, then what you just came up with is the answer Kaplan gives. All you have to do is rotate the molecules into the correct order.

 

[steelrfan8]

It doesn't matter whats on the vertical axis or not, because you can always rotate the molecule to get whatever molecule you want to be there. To rotate, you have to keep one molecule the same and rotate the other three, in this case, keep the carbon the same and rotate the other three around the carbon until you get whichever one you want to be on the vertical axis.

You also put the B and D backwards. Notice in destroyer, the arrangement in the back carbons after going from newman to fischer are inverted. I don't know the reasoning why, but they say so and it works for every question, so I stuck with it and it indeed works for everything. If you switch B and D, then what you just came up with is the answer Kaplan gives. All you have to do is rotate the molecules into the correct order.

If you draw out the Newman projection in the same way you posted in your first reply:

Front molecule- F top, H bottom right, OH bottom left
Back molecule- OH bottom, Br top right, D top left

and covert that into Fischer, you invert the Br and D on the back molecule so in the Fischer diagram the Br ends on up the left and D ends up on the right on the bottom chiral carbon (like it says in Destroyer)....

I also think we may be having a bit of a miscommunication; I believe you're explaining how to go from Newman to Fischer which I understand... the problem that I'm having is in the Kaplan sawhorse projections, I can't seem to translate the sawhorse into either Newman or Fischer projection in that specific problem I posted.

In your 1st post, you attempted to translate the sawhorse projection I posted in "D" into a Newman projection.... when I translated your Newman projection into a Fischer projection, it doesn't correspond with the "correct" Fischer diagram answer given in Kaplan (unless I'm missing an additional step with the Fischer projection I get).

When you say this:

It doesn't matter whats on the vertical axis or not, because you can always rotate the molecule to get whatever molecule you want to be there. To rotate, you have to keep one molecule the same and rotate the other three, in this case, keep the carbon the same and rotate the other three around the carbon until you get whichever one you want to be on the vertical axis.

Are you talking about rotation on a Fischer projection where 1 group is held steady to allow 180 degree rotation? So if we once again translate your Newman projection into Fischer, you're saying to rotate that Fischer projection while holding 1 group steady for it to correspond to the answer given in the book?

 

[americanpierg]

Are you talking about rotation on a Fischer projection where 1 group is held steady to allow 180 degree rotation? So if we once again translate your Newman projection into Fischer, you're saying to rotate that Fischer projection while holding 1 group steady for it to correspond to the answer given in the book?

I think what you're confusing here is how to minipulate the fischer projection.

You can turn the entire molecule 180 degrees, or if you hold one group still, you can rotate the other three any way you want, 90, and 270 as well as the normal 180, and it doesn't make a difference.

And I misunderstood your previous post, but this is what you said you got.

Also, lets say we convert D into Newman as you said



I draw this out, convert into Fischer which gives us (this is how Destroyer shows to convert Newman into Fischer)

.....F
OH----H
Br----D
....OH

And this is the correct answer. Just rotate the bottom 3 so that Br is on the bottom, then rotate the entire molecule 180 and you have what Kaplan has.