[Telamond]

So I have two questions regarding Aldol reactions:
1) Are all aldol reactions followed (automatically) by an aldol condensation?
2) Also, I have a hard time visualizing the Zimmerman-Traxler transition state.

So what I know about Zimmerman-Traxler is that it's a method to tell if the product from an aldol reaction is either syn or anti.
Here's how I would proceed in how to deduce if the product is syn or anti:
1) Draw a 6-ring chair conformation.
2) Exchange the carbons for oxygen, ligand and double bond.
3) Put the R group from the enolate equatorial.
This is where I have a hard time knowing the next step. How to draw the substituents on carbon after the double bond.

I've got a guess that I'd like to be confirmed. (I've even colour coded it!) The picture is from Wikipedia.
On the E-alkene the R and the Me are on the same side. Therefore they should be pointing towards the same direction (axiell + equatorial).
On the Z-alkene, they are placed across from each other, therefore they should be pointing as far away from each other (axial + axial).

4) After placing out the substituents on that carbon, the next step would be to place out the substituents on the aldehyde. The largest substituent is always equatorial.

The transition state is finished.

The product I have a hard time drawing out as well.
I just would like a confirmation on this as well,
It seems like the wiki picture they have chosen to have the -OH group always facing down. The difference is where the substituent is directed, syn or anti, thus the whole meaning of the Zimmerman-Traxler.

To know where the substituent is pointing, you check where the hydrogens are pointing (marked on the picture). If they're on the same side (equatorial + axial) then they are syn products. If they're on the opposite sides (axial + axial) then they are anti products.

Are these correct assumptions? :S
If something needs clarifications, please ask. I'll be hitting the refresh button every 2 minutes.

[azmanam]

Your drawing of the T.S. is fine.

Here's what I did a lot in my studies.  this is from  my notes for Felkin-Ahn control, but the thought experiment is the same.  When you exchange dashed lines for real bonds and replace the enolate with the ketone, you can still drawn a 6-membered ring with the L.A. still coordinated, only now it's the chair structure of the real product.

The interesting stereochemistry is in the newly formed bond, i.e. the C-C bond that is currently dashed in your color coded T.S.  you need to look at that bond as a different representation of the sawhorse orientation.  For my F.A. selectivity, the initial product was a Newman projection which I rotated into a sawhorse.  From there (or from a Newman projection), you can rotate the various C-C single bond (but I would only recommend rotating one bond at a time so as not to confuse yourself) until you have the "in the plane" bonds as the vertical bonds in the sawhorse.  Once they're the vertical bonds, the wedges and dashes should be apparent from the sawhorse. 

In my F.A. example, the cyclohexyl group eventually ended Vertical And To The Left, so when I drew it flat, it was in the plane to the left.  The Incoming nucleophile was Vertical And To The Right, so when flat, it was in the plane and to the right.  In rotating the sawhorse to a flat, the methyl and the alcohol both go to the back.

Try a strategy like that for drawing the flat stereochem for the aldol adducts. 

[azmanam]

no. not all aldol additions are followed by condensation.  Aldols under basic conditions rarely are.  Typically you need acid and potentially a bit of heat.

[Telamond]

Thanks for the reply azmanam!
 
I had no idea the Felkin-Ahn models could be used for aldol reactions. When the professor has been going through it, they've mainly applied it to two situations
1) Grignard addition to carbonyl compounds
2) Reduction of carbonyl compounds.

I'll check up on what you did there. Need some time processing the info (I'm slow  )

[azmanam]

I had no idea the Felkin-Ahn models could be used for aldol reactions.

nonono.  the F.A. was to show you how you can look at a stereocenter-forming reaction in 2-d on paper and elucidate the 3-d orientation of the substitutents in the product.

The theory is a unifying theory of predicting stereochemistry.  it's not that the F.A. gives you the same product as the aldol, it's that looking at a flat TS, you can convert it to a sawhorse which helps you see the 3-d orientation better.

sorry for confusing you.

[azmanam]

I'll check up on what you did there. Need some time processing the info

yes, practice is key.  keep practicing over and over and eventually you'll find your own system/shortcut and you'll be able to "see" the product stereochem faster.

[Telamond]

Oh! Sorry for misunderstanding! Now I understand what you were meaning a lot more.  

[coolakul]

To visualize the zimmerman traxler principle for syn or anti selectivity check out this page on PharmaXChange.info:
http://pharmaxchange.info/ they have explained how to derive the zimmerman traxler transition state pretty well there and have shown the six membered transition state. They have even shown how to rotate the bonds so as to achieve the answer in a easy and simplified manner.

[orgopete]

So I have two questions regarding Aldol reactions:
1) Are all aldol reactions followed (automatically) by an aldol condensation?
2) Also, I have a hard time visualizing the Zimmerman-Traxler transition state.
…
This is where I have a hard time knowing the next step. How to draw the substituents on carbon after the double bond.
…
The product I have a hard time drawing out as well.
I just would like a confirmation on this as well,
It seems like the wiki picture they have chosen to have the -OH group always facing down. The difference is where the substituent is directed, syn or anti, thus the whole meaning of the Zimmerman-Traxler.

To know where the substituent is pointing, you check where the hydrogens are pointing (marked on the picture). If they're on the same side (equatorial + axial) then they are syn products. If they're on the opposite sides (axial + axial) then they are anti products.

This is how I was understanding the question. "Given the transition state illustrated, how does one arrive at the product?" That is, the drawing did two steps at once. First they formed the bonds and then they changed the orientation of the product. This is how I would go about this. I would draw out each step. I would not try to combine steps or operations. When I make errors, it is frequently because I have attempted to perform too many mental steps. The second thing that may help, make models of the transition state and product, if you have models. Then you can rotate the bonds to give the indicated products. 

I am going to disagree with azmanam on the condensation. I understand that it would be difficult for someone learning organic chemistry to make this prediction. However, this has been the advice I give. If heat or high concentrations of base are present, then expect the dehydration to also occur.

By comparison, if a kinetic or thermodynamic enolate is prepared at -78°C and condensed with a carbonyl compound, followed by an immediate work up, then you should expect the simple aldol addition product.

We did a fairly standard mixed or crossed aldol condensation of acetone and benzaldehyde catalyzed by 3M NaOH. The dehydration accompanied the reaction. We also experienced some variability in the success of the reaction. What I discovered is that if the NaOH has aged and presumably begun to form carbonates, the reaction fails. Either starting with fresh 3M NaOH or simply adding a pellet or two of NaOH will catalyze the reaction. I think if you did a standard aldol condensation of an aldehyde, it would dehydrate.