Anyone good at organic chemistry. I have a few problems i am stuck on

[jdpaul88]

Any hints or pointers will be appreciated. I think for the first one, one of the products might be a cyclic ring with an oxygen and two methyls coming off

 

[Giog]

Damn! I took organic chemistry 1 and 2 (with labs) in the summer and I couldn't even BEGIN to help you. I certainly did not encounter such challenging problems...Good luck is all I can say!

 

[PlurredChemistry]

You are correct for the first one.

The acid will protonate the alcohol group and then the formed H2O group will leave, forming a secondary carbocation.

Whenever you have a secondary carbocation you need to think of rearrangement, so see how the chain could reform to make a more stable tertiary carbocation.

Then the double bond acts as an nucleophile and attacks the new carbon center (either rearranged or unrearranged).

//

That was fun, haven't had to think about this stuff in a LONG time. Gotta run to a party, might come back later to check out the others.

 

[Roger&Me]

Number 3 could be accomplished like this: start with the condensation of benzaldehyde to yield benzoin, which is reduced by clemmenson reduction to stilbene, which is then reduced by monodeuterated LAH (or red-Al) to yield the product (you can reduce stilbene with LAH and the like because of the inductive effect of the two phenyls).

You might want to check out this paper for more info.

 

[atara]

The fourth might start from 3-bromopropionaldehyde. The enol ester is formed with Ac2O / H2SO4, a Barbier reaction with acrolein gives the right side of the molecule, hydrolysis of the ester gives back an important carbonyl group, the addition of 1-butyne gives a very suggestive hydroxyalkyne, and of course reduction gives the target compound in the appropriate stereochemistry.

You can cheat, sort of, since 3-bromopropionaldehyde enol formate technically only has four carbons -- and that saves you a step!