Question 25

[McatCrusher]

Hi,
Apparantly, i have a bit of problem with #25. I did study the org 2.2 but it really did not help. The thing is that i know that 1 & 3 are non-superimposable mirror images. But i dont get it why they are non-superimposable??? I mean if you rotate around carbon two and three, they become pretty much superimposable, dont they?
In org 2.2, i suffer from the same problem. there are four molecules and i think if you rotate them enough they are all suoperimposable. Now i know i am wrong somewhere but i dont know where??? i would appreciate if you could help me here.

[admin]

I know that this might sound silly, but the best thing for you to do is to draw the 2 molecules (I and III) on white paper, with the molecules more or less the same size.

Now hold the 2 pieces of paper up to the light, with the molecules facing you, try to maneuver the molecules so that they are absolutely identical. If you try it once, the meaning of non-superimposable will be more clear.

[jellywing_2058]

Compounds I and III do not have an axe of symmetry.

The compounds are drawn as Fisher projections, meaning that carbons 2 and 3 are in the same plane as the page, the CO2H groups are going into the page and the horizontal side chains are coming out of the page

(OR carbons 2 and 3 are in the same plane as the page, the CO2H groups are out of the page(front) and the side groups are going into the page (behind)).

Build them using a molecular kit for clarification

[admin]

bump

[jfalcone1428]

Shouldn't the explanation say that Molecules I and III are enantiomers? The explanation reads "Molecules I and II are enantiomers of each other!" The answer to this problem is B (compound II). A meso compound has no enantiomer.

Thanks

[mcat_premed3832]

You are correct. There was a typo in the Explanation which has been updated. Good pick up!