fischer projections - how to rotate?

[dedicate]

I'm trying to figure this problem out. It's in EK orgo, but unfortunately, there is like almost NOTHING in the book about fischer projections.

I'm trying to figure out how to manipulate these quickly, but I simply don't remember much about it from orgo.

Can you shed some light on how to quickly solve this problem?

Also, if EK organic generally considered good or bad for the MCAT? Chapter one seems awfully lacking thus far.

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[Timaeus]

Well, you know you want to identify the meso compound.

It looks like the answer is B.

Before I give an explanation, let me know if that's right so I don't look like I'm milking a he-goat 🙂

 

[grotto]

B is a meso compound, with a fischer projection you can rotate the groups in the second stereo center (in this case one place to the right or left) and you have can form a line of symmetry between the two groups. A meso compound is a molecule with multiple stereo centers that has symmetry, so it isn't chiral.

... i think🙂

 

[BloodySurgeon]

The most accurate way to do this is to assign R or S to each chiral carbon. Think of the fisher projection as a person with a bow tie or two and this will remind you that the sides are protruding towards you and the top and bottom are going away from you. You want the atom with the lowest proton number or priority which is Hydrogen to be facing away from you, therefore if it is not, you should have the opposite rotation. Answer A is C2-R, C3-R; Answer B is C2-S, C3-R; Answer C is C2-R, C3-R; Answer D is C2-R, C3-R.

Unless I got the away or towards rule wrong, the answer should be B. It is also a meso compound which both optically inactive and are R, S configuration.

 

[dedicate]

Thanks for the feedback.

The answer is B. When I saw the question, I figured I was definitely looking for a meso compound. But the problem was, I had no idea how to quickly manipulate the different projections to find which one was meso.

 

[Timaeus]

Cool.

Another easy way to do it, which I don't think has been mentioned yet, is that you can exchange the locations of two pairs of substituents, and still get the same compound.

So if you look at compound B, switch the location of the bottom CH3 and the H. However, since you have to do 2 switches (a pair, if you will) to get the same compound, do it again. This time yous switched the CH3 and the Br on the bottom, and then you will see the internal plane of symmetry.

 

[sagtig]

Whenever you are looking at compounds with two chiral centers, and you want to change the arrangement, as timaeus mentioned, you can switch two pairs on one of the chiral centers...

I also figured another way to solve this problem which might not be very helpful in terms of content, but useful in terms of overall test taking strategy is this... find something in the answer that makes it different from all the other choices... if you notice all the other three choices had two molecules on side of the fischer projections directly reflect each other, except... B... so there you go... if you can find something about one of the answer choices stand out from the rest, then that answer choice is correct