[discodermolide]

I think then that its the second option for the intra molecular reaction....

Your product in the latest picture has a pentavalent carbon atom!

[Twickel]

:@:@. damn it. I am not thinking,to be honest I do not know what the answer is, can someone explain it to me?

[discodermolide]

:@:@. damn it. I am not thinking,to be honest I do not know what the answer is, can someone explain it to me?

Sorry, I'm going to bow out here. All these pictures have totally confused me so that I do not know what you require to know.

[Twickel]

Can I just re draw the ONE question?

[orgopete]

I think I see some correct thinking occurring. That should render Q1 to a correct solution. However, this step makes no sense.

I think if you were to simply attempt an aldol cyclization from each of the possible enolates should answer the question.

[Twickel]

I didnt know the correct positions of the two carbonyls relative to each other, when I first drew it out, they were not alpha to one another  but Beta, so I just made that up to get the hydrogens to be alpha to the carbonylss

[Twickel]

I think I see some correct thinking occurring. That should render Q1 to a correct solution. However, this step makes no sense.

I think if you were to simply attempt an aldol cyclization from each of the possible enolates should answer the question.

The problem is my text book, only give one example.... I do not know how to do this reaction. Can anyoneplease just show me the correct mechanism I will give it one more shot, realised another mistake ive been making. It has to be B? is it possible to have mixed intra molecular aldol condensations?

[Twickel]

Is it B,  I worked with retro arrows, so the new pi bond alpha and beta to the carbony carbon is broken, I stick an Oh group on the Beta carbon, break that bond and I get something that looks like B,

[Dan]

Go one step at a time. For Q1, take each of the 4 substrates in turn, draw an aldol condensation mechanism and see what you get as the product. You cannot simply pick one because it "looks right", go through the intramolecular condensations for each starting material - the appropriate starting material is the one that gives you the product by a reasonable mechanism. Also - check the structures again! You posted the 4 options twice now, but compound C has changed from a dialdehyde to a ketoaldehyde, and D has changed from a ketoaldehyde to a diketone (see below) - which one is it?

Take your time, and check your work before posting it up. Slow down and don't cut corners - it is clear you have the concepts required to solve these problems, I just get the impression you are just rushing too much.

For Q3, what kind of reagent is diethyl carbonate - acid, base, nuclophile, electrophile?

[Twickel]

I think it is a weak acid, because it can lose a proton and the conjugate base can be stabilised through resonance.

In regards to Q1, I can not do it.... can someone please walk me through it. I will post the mechanism for A and from that it is wrong....

[Twickel]

Look I can not do it I have tried so many times, it is not working, I need help, someone please show me the correct mechanism, for some reason I get 6 membered rings.
But I think its the aldehyde that has to be enolised, because the carbonyl thats first enolised is not the one that gets protonated to form OH

[discodermolide]

Look I can not do it I have tried so many times, it is not working, I need help, someone please show me the correct mechanism, for some reason I get 6 membered rings.
But I think its the aldehyde that has to be enolised, because the carbonyl thats first enolised is not the one that gets protonated to form OH

If I understand correctly your product has an aldehyde attached to a 5 membered ring with a double bond in it, correct? As in my picture?

[Twickel]

Yes that is the product, I need to know where that came from

[discodermolide]

Yes that is the product, I need to know where that came from

OK then your starting material must have two CH2CHO groups in it. Like the picture.
Perhaps you can work it out from there?

[Twickel]

Thank you. The problem I have is forming the actual ring, I form the enolate ion on the bottom aldehyde. I then count the carbons number 1 being the carbon bonded to the negatively charged oxygem, number 6 being the carbon of the other aldehyde, that gives me 6. How do I correctly number the carbons when forming rings?