[agrobert]

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.

[saN]

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.

or

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.

[agrobert]

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. 

 

[saN]

(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?

[agrobert]

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

[saN]

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?

[agrobert]

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.

[saN]

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?

[agrobert]

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.

[saN]

How would one measure the specific rotation of a chiral carbon? Is it only through experiments?

[agrobert]

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

[saN]

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 3^o carbon atom.

[saN]

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. 

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.

[ultrashogun]

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 3^o 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.

[ultrashogun]

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.