Chemical Forums
Chemistry Forums for Students => Organic Chemistry Forum => Topic started by: saN on October 26, 2007, 02:20:14 PM
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I have this problem to work out.
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My Answers
a) The product will not be optically active because there is no chiral center. Even though this mechanism is an SN1 rxn (no room for inversion), there are no four different substituents.
b) If C-1 is not chiral, how will I be able to assign R or S? If an R or S assignment can occur, there will not be a racemic mixture as the methyl group on not on C-1 will create partial blocking from the underside causing the compound to be a majority of one compound.
My question is, is this a SN1 rxn and do my analysis seem correct? I don't feel too confident in my answers.
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Just a question: shouln't occur here an elimination instead of a substitution? We have a tertiary halide and a strong base, so elimination should be favoured. However, any substitution that may occur will produce chiral molecules.
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We haven't covered eliminations rxn yet.
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As far as substitution goes, you are correct in stating that this would be a SN1. The rate limiting step would be the breaking of the C-Br bond and the formation of a carbocation which would be stabilized by hyperconjugation. The carbocation is available for attack from either the top or the bottom of the plane so you will have a mixture of R and S enantiomers at C-1. The mixture will depend on sterics and coan only be determined experimentally. The starting compound is optically active so unless elimination occurs forming an unsaturated bond from C-1 to C-2 your product will be chiral. Draw Fisher projections to determine optical activity, don't be confused by the cyclic system.
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The starting compound is optically active, but what about the product? In order for the product to optically active, there must be some chirality within the molecule. The C-1 carbon of the product would be attached to a hydroxyl, methyl, and the rest of the chain. Will that second methyl group have an effect on its chirality?
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Draw the Fisher projection including C-2 (with the methyl), hydrogen, hydroxyl and C-5 (methylene-CH2). The ring is not symmetric because of the methyl group on C-2. Both C-1 and C-2 will be chiral centers after a SN1 substitution on C-1.
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However, an SN1 reaction should produce a racemic or nearly racemic mixture so that the bulk solution will show no optical rotation, even though it is composed of optically active molecules.
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However, an SN1 reaction should produce a racemic or nearly racemic mixture so that the bulk solution will show no optical rotation, even though it is composed of optically active molecules.
Right, it is important to realize that the products are chiral and individually they rotate polarized light in opposite directions of the same degree but as a racemic solution they cancel each other (which is probably the purpose of this question as stated by Yggdrasil).
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Is this the proper Fisher Projection?
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Yes. So you should be able to assign R or S to C-2 and the product will be a mixture, probably racemic of R and S C-1.
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I got that at C-2 it is R. I am having trouble assigning priorities for C-1. I know that the oxygen has is #1 and the methyl is #4. There is a methyl attached on C-3, but just another CH2 on C-5. Will this methyl make the C-3 priority #2? because of the substituent; therefore, making it a R at C-1?
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The C-1 will be R if the hydroxyl group attaches from the top, but if from the bottom, C-1 will become S. That means the molecule can be R,R or R,S. So, if it is R,R, the product will be optically active as it will rotate the light dextrorotatory, but if the product is R,S, it overall molecule will not rotate light as C-1 and C-2 will cancel each other.
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New Answer:
a) The product will be optically active if the hydroxyl group attaches from the top of the reaction site making C-1 and C-2 both R. The product will not be optically active if the hydroxyl group attaches from the bottom of the reaction site making C-1 S and C-2 R. The rotation of light will cancel each other out.
b) C-1 will have a higher probability of being R than S because if the hydroxyl group attaches at the top, there is one methyl group that may block. If the hydroxyl group attaches at the bottom making it S, there is two methyl groups that may block. The methyl group on C-1 has an equal amount of blocking for a top and bottom attachment.
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Lets establish a system for stereochemistry as (C-1,C-2). Your possibilities are (R,R) and (S,R) as you have determined. Assume after reaction the hydroxyl group has attacked equally from the top and the bottom so you have 50% (R,R) and 50% (S,R). If you isolated (R,R) it would be optically active. If you isolated (S,R) it would also be optically active. It is the fact that (R,R) and (S,R) are in racemic (1:1) ratio that the R configuration rotation at C-1 will cancel the S configuration rotation at C-1. There is no theoretical way to determine the optical relationship between C-1 and C-2.
R stands for rectus and S stands for sinister. They have to so with the orientation of the stereocenter based on Cahn-Ingold-Prelog Priority Rules
http://en.wikipedia.org/wiki/Cahn_Ingold_Prelog_priority_rules
Determining R and S will not tell you which way polarized light will be rotated, but you should know that in a racemic mixture, like we have at C-1, the rotation of R will cancel S. R has the opposite rotational degree of S.
I suggest you find a similar reaction example with only one chiral center so that you can understand the optical rotation relationship that exists between a racemic mixture. Think about this same reaction but pretend as if C-2 contained 2 hydrogen (therefore not a chiral center).
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So, are you saying that the methyl group on C-2 will not have any blocking affects on how the hydroxyl group attacks the top or bottom? If so, than C-1 will have an equal probability of being R or S. As for optically active, both configurations will rotate light at each carbon, but overall as a molecule, the R,R will rotate light and R,S will not.
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The methyl group may have some blocking effects but for the simplicity and purpose of this problem you should assume the product is a racemic mixture of (R,R) and (S,R)
Lets say C-2 R rotates light +55, C-1 R rotates light -22 and C-1 S rotates light +22 (opposite R)
C-1 + C-2
So (R,R) rotates light -22 + 55 = 33 (isolated)
and (S,R) rotates light 22 + 55 = 77 (isolated)
but if you have 50% of each (racemic mixture) then
0.50 (33) (R,R) + 0.50 (77) (S,R) = 55
So essentially the chiral center at C-I is cancelled if the rotation of the mixture determined.
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Racemic mixtures to not bend polarized light. The thing that I am still confused on is if (R,R) is optically active, and (S,R) is not, how does (S,R) affect (R,R) to not bend light? Doesn't the molecule (S,R) cancel light bending in the molecule itself and has zero affect on how (R,R) is optically active? Then, the overall products will be optically active.
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It is only a racemic mixture at C-1, having equal probability of being either configurations; therefore, the product will be optically active due to C-2 being R.
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It is only a racemic mixture at C-1, having equal probability of being either configurations; therefore, the product will be optically active due to C-2 being R.
Correct.
(R,R) is optically active when isolated.
(S,R) is optically active when isolated.
If there is only one chiral stereocenter then a racemic mixture will not rotate polarized light.
The rotation at C-2 is retained and is completely independent of C-1.
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(S,R) is optically active when isolated.
How is this optically active when isolated? Do you mean that the C-1 and C-2 will rotate polarized light looking at step by step causing the net rotation to be zero or that C-1 and C-2 have different substituents causing the degree of rotation to partially cancel and have a net rotation not equal to zero?
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Each chiral stereocenter has a specific rotation of light that is independent of the other chiral centers within the molecule. The rotation is additive which is a net rotation.
Isolated means by itself.
Lets say C-2 R rotates light +55, C-1 R rotates light -22 and C-1 S rotates light +22 (opposite R)
C-1 + C-2
So (R,R) rotates light -22 + 55 = 33 (isolated)
and (S,R) rotates light 22 + 55 = 77 (isolated)
but if you have 50% of each (racemic mixture) then
0.50 (33) (R,R) + 0.50 (77) (S,R) = 55
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I think I understand now. I thought optically active dealt with the net rotation. Since each chiral stereocenter rotation of light is independent of other chiral centers, is it safe to say that any molecule with a chiral stereocenter will be optically active and that optically active does not deal with net rotation?
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The only time C-1 and C-2 will cancel each other is when the compound is meso. Optical activity does deal with net rotation.
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So, the (S,R) molecule will be optically active because it is not a meso compound, and there are chiral stereocenters that rotate polarized light; therefore, the net rotation of polarized light is not equal to zero. You also say that only meso compounds cancel each other out. Well, since this molecule is (S,R), wouldn't there be a degree of partial cancellation, but not completely as a meso compound would do?
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Well, since this molecule is (S,R), wouldn't there be a degree of partial cancellation, but not completely as a meso compound would do?
Yes, there could be, If C-2 R is (+) and C-1 S is (-).
But it could also be additive if C-2 R is (+) and C-1 S is (+)
It all depends on the specific rotation at each center. Reread the example I provided earlier.
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How would one measure the specific rotation of a chiral carbon? Is it only through experiments?
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Using a polarimeter. I think that was your point of confusion. Specific rotation of a chiral compound can not be determined theoretically. But you can know that any R configuration has the opposite rotation as any S configuration.
http://en.wikipedia.org/wiki/Polarimetry
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I will be posting up my final answers to this question in a bit. My other question is, why is this rxn not SN2? Is it because inversion at C-1 is nearly impossible due to the bayer strain and structure of the cyclic ring? The leaving group is also attached to a 3o carbon atom.
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a) The product will be optically active if the hydroxyl group attacks from the top making the reaction site C-1 and C-2 both R configurations. If the hydroxyl group attacks from the bottom, the C-1 will become S configuration and C-2 will become R configuration. The net rotation of light will not be zero because the degree of rotation of light from C-1 and C-2 are not opposite and equal. The rotation of light will cancel each other out only when the molecule is a meso structure. This molecule is not meso; therefore, either configuration will be optically active. The net rotation of light may be zero if C-1 and C-2 bend polarized light to the same degree but in opposite directions. In this case, the solution will not be optically active even though it contains optically active molecules.
b) At C-1, the reaction will proceed as SN1 because it is attached to a 3° carbon and the inversion of atoms from an SN2 reaction is nearly impossible due to the bayer strain and structure of the cyclic ring. After the carbocation becomes planar, the nucleophile can attack the top or the bottom. The majority of the molecules will be slightly more R than S due to the blocking effects. If the nucleophile attacks the top, it only has one methyl group to block. If the nucleophile attacks the bottom, it has two methyl groups to block.
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If the methyl groups had no blocking effects on the nucleophile, than C-1 will have a equal probability of being R or S configuration making it a racemic mixture.
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I will be posting up my final answers to this question in a bit. My other question is, why is this rxn not SN2? Is it because inversion at C-1 is nearly impossible due to the bayer strain and structure of the cyclic ring? The leaving group is also attached to a 3o carbon atom.
Thats why its Sn1, the carbon is sterically hindered because tertiary plus the carbocation is stabilized due to electron repulsion and sigma delocalization.
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The C-1 will be R if the hydroxyl group attaches from the top, but if from the bottom, C-1 will become S. That means the molecule can be R,R or R,S. So, if it is R,R, the product will be optically active as it will rotate the light dextrorotatory, but if the product is R,S, it overall molecule will not rotate light as C-1 and C-2 will cancel each other.
Always remember that different stereogenic centers will not necessarily rotate the light to the same extent, neither does the designation 'R' or 'S' tell us anything about the direction of rotation.
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Quote from earlier in the thread:
'It is the fact that (R,R) and (S,R) are in racemic (1:1) ratio that the R configuration rotation at C-1 will cancel the S configuration rotation at C-1.'
A 50:50 mixture of (R,R) and (S,R) is not a racemaic mixture, since RR is not the enantiomer of SR. These are mixtures of diasteromers. Even if you assume equal amounts of each diastereomers, there is no reason to believe their optical rotations would cancel.
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Thanks for everyones *delete me*!