[mistermister101]

Hey I'm new to these forums. Wish I had found out about them earlier.

Anyhow I'm stuck on these last 4 problems for my orgo homework. Ive tried them numerous times but to no avail.

Here they are:

http://imgur.com/QrBuX

I need to figure out the structures for the bolded letters. Thanks. Ill be here for the next 3 hours before work, and am open to working through the problems together. After that ill be back in 6 hours or so.

[OC pro]

They are not so easy ones...took 10 minutes for me to get to the solution  .

Any ideas by yourself? The hints do help a lot.

[mistermister101]

Ok for G I tried first moving that double bond up, then the double bonded oxygen to an O- to form an enolate. Then I followed the hints and did a Michael reaction and an aldol reaction. This ended up forming the compound with the aldehyde group added to the same carbon the methyl is on with a double bond on the 1st and 2nd carbons on that chain. This answer proved to be wrong.

Next i tried forming an enolate by taking the hydrogen to the right of the top ketone. Then I did the same reaction adding the aldehyde compound to the carbon to the right of the original ketone. This answer was also wrong.

I can't see the formation of an enolate on the bottom chain ketone. I guess I just dont know which enolate is the most stable.

For K I don't exactly understand Mannich reactions at all. I have never done one in which the aldehyde isn't just a single carbon, so the reaction confused me. I added the Nitrogen group to the end of one end of the aldehyde and basically did one Mannich reaction but I don't see how another can occur.

For I, J and H I am thoroughly confused haha...

I guess for H the concentrated Sulfuric acid donates a proton to the OH and it leaves, the heat then allows the formation of a double bond, possibly in the middle of the compound. O3 then divides the molecule making 2 ketones possibly? NaoH/heat then removes one of them? No idea to be honest...

For I it looks like we have a Grignard Reagent with a protecting group on it... I'm not exactly sure where this is attatching. I guess the MgBr leaves and u have a negative on the end of it and the rest of the molecule attacks a + carbon, or maybe the double bond attacks it.

Overall your right, I find these very difficult. Textbook hasnt helped much either as the examples are very simple.