[keleline]

I'm doing a lab report for my ochem class and I'm not sure I fully understand how to obtain the products by writing the mechanism.

I'm using Piperonaldehyde + acetophenone ----> [ aldol ] ---> -H2O ----> 3,4-methylenedioxychalcone

My question is how do I draw the final structure 3,4-methylenedioxychalcone in its cis form? And why is the trans form the only one that is obtained?

And, if I were using H-NMR, how could I determine if I have the trans isomer rather than the cis isomer? I know it has to do with coupling, but could you explain it to me?

Thank you for your help

[gfunk]

I love Aldol.  Anyhow, out of your two starting materials, which one has the only enolizable protons?  That compound will then perform a nucleophilic attact on the most electrophilic center of the other molecule (where is it?).  That's the Aldol addition.

From there, there's a formal removal of water -- the condensation.  This leaves behind a double bond.

Do you know what the product looks like?  As for cis vs. trans, draw out the transition state and observe.

As for determination by 1H-NMR, yes, you'd measure the coupling constants of the splitting pattern (in Hz).  Cis protons will be from 6 to 12 Hz (typically 10 Hz), and trans protons will be from 12 to 18 Hz (typically 17 Hz).

[keleline]

I figured out how the reaction goes between Piperonal and acetophenone looks.  I still need help figuring out how to draw the cis isomer and I know how the trans isomer looks since that is the only one that is produced.

I have attached a picture of the equation of aldol condensation. Can someone please show me how the cis isomer of the product would look?

Thank you

[Squirmy]

Check out this Wiki page:

http://en.wikipedia.org/wiki/Geometric_isomerism

The pics of cis-2-butene and trans-2-butene should help

[keleline]

  I think my brain was momentarily fried. I got it, thank you for the visual!!