[Bibinou]

Hello,

In order to prepare me well to my organic chemistry exam, I tried to answer to old question exam. I realized that I have some difficulties to answer them. In the cases I have to put one or more reagent(s) or a product of the reaction. I have prepare these question in the enclosed document.

Reaction 1 : What is the majoritary product and the minoritary product?
Reaction 2 : Is my reaction correct
Reaction 3 : I don't know what to put as reagent
Reaction 4 : Is my reaction correct
Reaction 5 : Is my reaction correct
Reaction 6 : Is my reaction correct
Reaction 7 : Is the reagent correct, have you other possibilities
Reaction 8 : I don't know what to put as reagent
Reaction 9 : I don't know what to put as reagent
Reaction 10 : Is my reaction correct

I thank you very much for your prompt answer. This is very important for me to understand these chemical reactions.

[DavidW]

1. The hydride can attack from either face of the carbonyl, and the stereocenter next to it will influence that.  You may want to review the section on either Cram's rule or the Felkin-Ahn rule, depending on which one you were taught.

2. Your product is incorrect.

3. Are you sure the product is not missing a carbon?  It looks like a setup for the Reformatsky reaction.

4.  What you have drawn so far is correrct.

5. That is correct.

6.  The first product you drew is incorrect.  You are missing the carbon of DCC.  And that last reagent should be allyl alcohol in order to get that final product you drew.

7.  It's correct, except CeCl3 is not necessary.  You only need CeCl3 when you run the Luche reduction of an
 enone.

8. Hint: an organometallic reagent.  You may want to review the reactivity of Weinreb's amide.

9.  That's an intramolecular Aldol reaction, followed by elimination.

10.  Incorrect, you are missing two carbons.

Hope this helps.

[Albert]

Reaction 9 : I don't know what to put as reagent

To state the obvious...

9.  That's an intramolecular Aldol reaction, followed by elimination.

...you just need a base.

[Bibinou]

Thank you very much David! Of course it helps 

Concerning :

Reaction 1 : I was taught both models, but have problems in seing Felkin-Ahn in 3D. So I wrote 2 compouds in Cram representation

Reaction 2 : I changes the representation of the product. Is it correct now?

Reaction 3 : Indeed, I forgot one carbon! That's of course a reformatsky reaction

Reaction 4 : I wrote the last part, but it doesn't seem correct...

Reaction 5 : OK

Reaction 6 : I added a carbon to my DCC and changes the last product given that I have to use CH2-CH-CH2-OH

Reaction 7 : OK

Reaction 8 : OK of course I didn't remark the Winereb amide

Reaction 9 : OK

Reaction 10 : I added 2 carbones more.

Thank you very much for your help

[DavidW]

Thank you very much David! Of course it helps 

Concerning :

Reaction 2 : I changes the representation of the product. Is it correct now?
Reaction 4 : I wrote the last part, but it doesn't seem correct...

You're welcome.  Regarding reaction 2, you should have two methoxy groups on the same carbon instead of one.  Think about the mechanism:  the carbonyl group is first protonated, and methanol attacks.  Following loss of proton, you now have a hemiacetal, i.e., a carbon attached to -OH and -OMe.  The -OH is protonated again, and can leave as water (-H2O), leaving behind a carbocation.  This carbocation doesn't just "disappear" to give the product you drew, but instead can be attacked by another MeOH.  What you can end up having is two methoxy groups on on carbon, or a dimethyl acetal.

For Reaction 4:  the reagent in step 3, tetrabutyl ammonium fluoride (or TBAF), is a reagent that attacks silicon, and what you ended up having is the free enolate, and the byproduct is (Me3SiF).  This is the same kind of enolate that you can generate simply by treating the starting cyclohexanone with LDA, but of course the countercation is different.  The enolate can then attack MeI (step 4) and the product you get will be 2,5,5-trimethylcyclohexanone.   Oh by the way, the first product you drew should have three methyl groups attached to silicon, not just one.

[Bibinou]

And concerning reaction 1 and 6, is it allright now?