[farcus]

http://img102.imageshack.us/img102/497/helpvp7.jpg

here is the problem, which one of those four is like the other? I know the answer is D but after reading the explanation I still don't get it.

[Dan]

I will walk you through the first one, if you repeat this process by yourself for the others you will see that only D is correct.

OK, so we have Fischer projection 1 (see bottom of this post). In Fischer projections the vertical line of the longest carbon chain, in this case from Br to F, can be visualised as an arc coming towards you.

This is shown in structure 2, the C-C bond is lying in the plane of the page, the halogens are pointing down behind the page and the H/D/OH substituents are pointing at you.

With this in mind we can now draw a sawhorse type (eclipsed) structure 3.

By rotating about the C-C bond we can get the staggered structure 4, in the style of the question, which allows easy comparison. We can see that 4 is not the eclipsed structure suggested in the question, so A is incorrect.

Try this with all of them to confirm only D is correct.

[Dan]

Now, that is a fairly long winded way of working out the answer, but it is the most fundamental and reliable. There is a shortcut, if what follows here confuses you, stick to the above.

I will call the carbon centre with the Br substituent C1, and the centre with the F C2.

If we compare the staggered structure in the original question part A with the Fischer projection (1), we can see that C1 matches up ok, and it is C2 that is different. The C1 centres in part A are assigned as S (according to the CIP rules).

Look at the staggered structures in A, B, and C. The configuration at C1 is S in all cases. We worked out in part A that in the Fischer projection C1 has the S configuration if the OH at C1 points to the right.

We know A is incorrect (because the structures differ at C2) and in B and C both staggered structures have S configuration at C1, however the OH's at C1 in the fischer projections for B and C point left, and so must be R. So we quickly spot that in B and C the structures differ at least at C1, and B and C must be incorrect.

By process of elimination D is the answer. This saves alot of drawing.

[farcus]

Umm I thought in fisher the horizontal lines are the ones coming towards you while vertical is ones going back.

[farcus]

also i'm lost in the rotation of the C-C bond, so when I make one bond wedged does this mean the wedge one before will takes its place or what?

I must be doing something wrong because when I rotate the C-C bond, i get answer for second one also, here is what I did with crappy ms paint drawing.


I noticed that on the 3 -> 4 projections you didn't change the dashed line of Br, i think that is what I am doing wrong? But on the forth choice and many others the halogen aren't always in the back(dashed) so what is going on? I guess i'm confused as to how to rotate them on the paper.

[Dan]

Umm I thought in fisher the horizontal lines are the ones coming towards you while vertical is ones going back.

Yes, that's right. The horizontals point at you. The verticl is an arc coming towards you - so like the arc of an archers bow if he was pointing it right at you, here the tips of the bow are the halogens.

I must be doing something wrong because when I rotate the C-C bond, i get answer for second one also, here is what I did with crappy ms paint drawing.

Your rotation there is fine, but you have gone from one eclipsed conformation to another - rotate to a staggered formation (180 degrees, so the OH and F are trans), and you'll see that B is incorrect.

I noticed that on the 3 -> 4 projections you didn't change the dashed line of Br, i think that is what I am doing wrong? But on the forth choice and many others the halogen aren't always in the back(dashed) so what is going on? I guess i'm confused as to how to rotate them on the paper.

In 3 to 4 there is only a rotation of 60 degrees, so the Br is still pointing back from that perspective.

I think it's worth getting yourself a model set (try ebay or amazon). These can be pricey, but if you're on a budget you can use cocktail sticks (toothpicks) and different coloured balls of plasticine