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Hi, I'm stuck with the following chem problem:So I can't paste it but will describe.

I have (3S,4S)-3-bromo-3,4-dimethylhexane but I don't know based on on the given picture why that is. I cannot post the picture but here the description.

I start out with two carbons, then use solid lines to position Bromine and hydrogen opposite each other along the two carbons. So carbon 1 on the sketch has Bromine facing up and carbon 2 facing down.

Then I add the other two groups on each carbon.

Carbon 1 has ethyl group on nonsolid wedge (facing away from the picture) and methyl group with solid wedge (facing towards me)

Carbon 2 has ethyl group on solid wedge (facing towards me) and methyl group on nonsolid wedge (facing away).

The dilemma: I get how carbon 1, which really is carbon 3 in the whole parent chain, is S configuration. But how I don't get how carbon 2, which really is carbon 4 in the whole parent chain, is S configuration. How are you prioritizing the groups around carbon 4 which I refered to as carbon 2 to start with?

Please help!


I also want to add that similarly I don't get how (3S, 4R)-3-bromo-3,4-dimethylhexane's configuration is solved through the same type of drawing except the carbon 2 (which is really carbon 4) has its ethyl and methyl groups switched. So now the drawing has ethyl in the nonsolid wedge and methyl in the solid wedge.

I think my problem is priority and maybe where I position the eye to solve for the configurations. Right now I'm basing the configurations of the drawing by having my eye face the page. I'm not moving to the sides or anything to determine this. I know that if Hydrogen is not on the nonsolid wedge then the actual configuration is opposite of the configuration I get.

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2 Answers

I will draw the Fisher projection for (3S,4S)-3-bromo-3,4-dimethylhexane below:

Fischer projection:





where the ethyl and methyl groups point towards you and the Br and H groups point away from you.  In the structures above, the priorities for C-3 are:

Priority 1 = Br

Priority 2 = CHMeEt

Priority 3 = ethyl

Priority 4 = methyl

For C-4 the priorities are:

Priority 1 = CMe(Et)(Br)
Priority 2 = Ethyl

Priority 3 = methyl

Priority 4 = H

This IS a hard problem to approach without a picture. I will take a stab at it. I will not put in any hydrogens unless needed for clarity. Assume there are as many hydrogens present as needed to give every carbon 4 bonds.

The base molecule is hexane, which has 6 carbons all single bonded:




3-bromo means the third carbon has a bromine attached:




C - C - C - C - C - C

(In case that doesn't line up in your browser, the Br is above the third carbon and single bonded to it.)

3,4-dimethyl means that carbon 3 and carbon 4 both have a methyl group attached. A methyl is one carbon long.



C - C - C - C - C - C

          |    |

          C   C

That is the basic structure of 3-bromo-3,4-dimethylhexane without consideration of the stereochemistry.

3S, 4S means that around each of the chiral carbons (carbon 3 and carbon 4), the substituents have an S ordering. We order by atomic number. Take a look at carbon 3:

Carbon 3 has an ethyl group, a bromine, a methyl group, and a branched 4 carbon (methly propyl) group. The highest priority is the bromine because its atomic number is higher than the attached atoms in the other groups. All three of the other groups have carbons attached to carbon 3. Consider the next atom out from the carbon. Both the ethyl and the methyl propyl groups have a second carbon, while the methyl group only has hydrogens. So the methyl is lowest priority. The ethyl and methylpropyl groups are tied at the second carbon, so go another step out for them. That is when the ethyl runs out of carbons while the methypropyl still has carbons. The priority is:

Bromo > methylpropyl > ethyl > methyl. Place the lowest priority group, the away from you behind the chiral carbon. I will use Δ and Υ to indicate going into the plane of the page:



C - C - C - C - C - C

          Δ     |

          C    C

Looking at the highest 3 priority substituents of only carbon 3, we see it goes from Br to methylpropyl to ethyl. You would be looking down to hide the methyl behind carbon three. The circle is clockwise so the stereochemistry is R, the way I've drawn it. We want S. To convert from R to S, exchange any 1 single substituent for its neighbor:

   C-C   H


Br - C - C - C - C


      C   C

Now consider carbon 4 as I've drawn it.

The group priorities are bromo, methyl propyl group > ethyl > methyl > hydrogen.  The hydrogen is above carbon 4 and would be coming out of the page. I've used ♦ to mean a darkened wedge indicating coming out of the page. To hide the hydrogen behind carbon 4, you would be looking up from the bottom. The arrow connecting the groups looks like it is clockwise viewing it from the top, but from the bottom it would be counterclockwise, or S. Thus we have drawn 3S, 4S 3-bromo-3,4-dimethylhexane.

Please feel free to ask any clarifying questions. I hope the drawings make sense.