Organic reaction question

Question

I have 2-methylcyclohex-2-enone and the series of steps are:

1.mCPBA

2.CH3CH2ONa

3. CH3I

4. NaCN/H

5. H3O / heat

i dont see what i should do on step 3. I know step 1 and 2...ill have an ether and an alcohol...the alcohol on the more substituted position...but where does the CH3I add to? after that step i understand that NaCN adds to the carbonyl and then hydrolyzes

Answer 1

I don't know what chemistry you are studying nor what answer your professor wants you to give, but the mCPBA will probably give a Baeyer-Villiger rearrangement. See Syn. Communication, Vol 19, p. 829 (1989) for an example. If so, then ethoxide will give an ester enolate. This will undergo a Claisen condensation to give 2-acetyl-2-methylcyclopentanone for step 3. Now, steps 4 and 5 become rather complicated so I doubt this is the intended reaction. So, I would go with Dr. J's answer as what was intended.

Epoxidation reactions, though concerted, react most rapidly with electron rich double bonds. Here, the double bond is very electron deficient due to its conjugation with the ketone. An epoxide is usually made with a peroxide and base to effect a conjugate addition and intramolecular substitution to give the epoxide. Hydrogen peroxide and base or sodium hypochlorite are good reagents to effect the conjugate addition and result in an epoxide.

If I am wrong, please post references.  

Answer 2

In step 3:  The -OH group in the 2-position, which was created by ring opening of the epoxide group in step 2, reacts with CH3I by an SN2 reaction, converting the -OH group to a -OCH3 group.  Looks like you have the rest.  The final molecule should be 3-ethoxy-1-hydroxy-2-methoxycyclohexane -1-carboxylic acid.

If you don't convert the -OH group to a -OCH3 group, the CN- in step 4 is a strong enough base to cause H2O elimination.