[saN]

I have this problem to work out.


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.

[Yarr]

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.

[saN]

We haven't covered eliminations rxn yet.

[agrobert]

As far as substitution goes, you are correct in stating that this would be a S_N1.  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.

[saN]

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?

[agrobert]

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 S_N1 substitution on C-1.

[Yggdrasil]

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.

[agrobert]

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).

[saN]

Is this the proper Fisher Projection?

[agrobert]

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.

[saN]

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?

[saN]

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.

[saN]

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.

[agrobert]

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). 

[saN]

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.