[a confused chiral girl]

Hi, I need help on this synthesis....

I don't know how to determine which one below could be a product of malonic ester synthesis. I know that the alpha carbon would be substituted...but I don't know how to see it on these diagrams. I think #1 doesn't look like it....but how about the other 3?

 

[a confused chiral girl]

I feel uncertain about this question, which is asking for the product that would be formed in this reaction. I know that after going through 2 of the seqences, we end up with a carboxylic acid that then gets treated with methanol.  I am thinking that the prodcut would either be #1 or #4, but don't know if we still have carboxylic acid like #4..

thanks!! 

[kremar]

well, in the first problem all the possible answers have the correct synthon for a malonic ester synthesis (-CH2-COOH)
all we can do now is look at the carbon this group is attached to and imagine the corresponding halide.
in #1 & #3 we end up with tertiary halides, which we know wont react with the enolate and probably will suffer base-catalyzed elimination. in #2 we have a sp2 carbon, which we know do not undergo Sn2 reactions. #4 would imply an allilic halide, which would react without problems, and i'd choose it as the correct product.

in the second problem, since you think its either #1 or #4 that means you got the first 2 steps right.
the third step is a decarboxilation. try to work all of its sub-steps and see what happens to the ester groups. re-read your ester hydrolisis and transesterification reactions, and you should have no problems finishing the rest.

ps: as a side note:
if you add your diethyl malonate to an ethanol/sodium ethoxide solution, and then add the MeI, you are gonna have a considerable amount of the dialkylation product. it goes like this: once the enolate reacts with a MeI molecule, the moloalkylated product is more acidic than the starting enolate. thus, the enolate will act as a base and deprotonate any monoalkylated product it comes into contact, and this will lead to low yields of the desired product. the way to go would be to first prepare the enolate with equimolar amounts of base, and then add the enolate to an excess of MeI.
in the second alkylation this care is not needed since the product doesnt have any more alpha hydrogens left.