Please Help Me with Organic Chemistry?
http://i110.photobucket.com/albums/n114/...
I am confused on where to start. Does the MCPBA turn the double bond into an epoxide?
Answer:
Yes. Peroxy acids react with olefins to yield epoxides. Since the double bond is planar, the epoxide can form on either side of the plane, giving a mixture of isomers (which may actually be the same compound, as in the epoxidation of 2-butene).
Epoxides are reactive to nucleophiles. In basic media, the ring opening occurs via a Sn2-like mechanism, in wich the nucleophile (in this case HO-) attacks the less-substituted carbon, with the alcoxide as a leaving group. Even though a stronger base (alkoxide anion) is formed from a weaker one (hydroxide anion), this reaction proceedes becouse of the ring strain present in the epoxide 3-member heterocycle.
Finally, you're left with an alkoxide, wich is a good nucleophile. Remember that iodide is a good leaving group, and methyl halides are very reactive to nucleophiles. So, the addition of an aklylating agent (such as MeI) will result in the O-alkylation product.
NOTE: On your reaction the second step is done in water. Since water is a protic solvent, the alkoxide will protonate to give the vicinal diol. This has two problems: 1) alkylation can now take place at either one of the oxygen atoms. the regiochemistry can be predicted knowing that primary alcohols are more reactive than secondary alcohols. 2) without a base catalyst, the reaction will ocurr with much lower yields (since an alcohol is a worse nucleophile than an alkoxide). Unless that is the desired effect, a polar non-protic solvent should be used instead (DMF, DMSO, etc.)
Here are the structures, step by step:
http://img247.imageshack.us/img247/8211/...
I believe you are right. The double bond in the alkene will open up and form the epoxide (oxygen) ring in a lot of electron pushing :).
I could type a long bit out of my textbook but yours probably has it too, "mCPBA epoxidation".
mCPBA:
http://upload.wikimedia.org/wikipedia/co...
Here's the mechanism, the phenol-chloride in mCPBA is the "R" group.
http://www.chem.ucalgary.ca/courses/351/...
----------
The OH-/H2O MeI is probably for the next step in the reaction after the epoxidation. Remember this must be in a solvent that is non-polar, which is a requirement for this reaction to take place, which means that the reaction cannot involve the formation of ions or any species with significant separation of unlike charges.
FYI, the only other criteria for this is that it is stereo selective. A cis-alkene --> cis epoxide and same for trans.
----------
Have fun with O Chem!
Here is the answer:
MCPBA epoxidizes the double bond. The OH then attacks the epoxide and opens it up giving a diol (the secondary alcohol will be a mixture of the R and S isomer). Depending on the number of equivalents of CH3I the hydroxyl groups are then methylated with CH3I to give the diethers if there are at least 2 equivalents. If there is only one equivalent of CH3I then only one of the hydroxyl groups of the diol will be methylated and that one will be the primary one not the secondary. I hope this helps. Good luck.
The answers post by the user, for information only, FunQA.com does not guarantee the right.
More Questions and Answers: