[Dev]

The method i was taught is to check whether the 2 structures are non superimposable mirror images.
OR to designate R/S to every Carbon and if all of them don't match, then they are enantiomers.

But first method fails here since i can't find a single mirror image. Valencies are switched.
Second method is just way too long.

Any help?
I feel like I have a hard time visualizing these structures.
Any help with that would be great.

In the picture, you have to find the option out of the four given in the bottom which is an enantiomer of the structure given in the top.

[fledarmus]

Ah, they're being nasty.

Best bet - build models. Then you can rotate the carbons to see if they overlap.

If you absolutely have to work without models (for example, a test where no models are presented) then the very poor second choice is to assign R and S to each chiral center.

The third choice is to draw the terminal carbons in wedge/dashed wedge configuration and practice rotating it to see what Fischer projections are possible for different rotations.

[Dev]

Hey!

I think that the question requires us to use the third choice. (So i can't use models, and assigning RS to each C will take too long)

This question was in a test I took. The answer given was (a)

Is there a method to switch valencies in Fischer projection without changing the sterochemistry of the structure?

Also, how do you change a wedge-dash structure into a fischer projection?

[Dev]

K, I have this kind of a theory, and I just need one of the super-brains on this forum to read it and confirm it's authenticity.   

As we know, for a single tetrahedron Carbon, switching one valency with the other makes the new found structure the enantiomer of the original structure.
Which can also be explained in this way: switching one valency with any other valency on the same (sp3) Carbon changes R into S and vice versa.

so if i just check how many valencies have been switched, (because to make it into its enantiomer, each carbon should have odd number of valency switches) i can answer this question in a shorter time.

[fledarmus]

By definition on a Fischer projection, for each atom on the backbone, the bonds to the left and right come out of the paper (wedge bonds) and the bonds to the top and bottom go back into the paper (dashed bonds). See the attached illustration.

Valency is the charge that an atom has - I don't see how that relates to R and S structures. What you CAN say is that switching the sides of two groups connected to the same carbon from left to right will change the carbon from an R to an S, or vice versa. Changes on the end carbons are a little more difficult to visualize, and will be easier to see using the description above and rotating the carbon in your mind to see where the groups end up.