[rita87]

Hi ,
can anyone help me with the mechanism (with curly arrows)for the following reactions. I know there are many different steps and that's a long question, but i really dont know how it works,couldn't find any usefull sources.
Thank you guys

P.S. this can be found in"Two-directional synthesis. Part 1: A short formal synthesis of (±)-histrionicotoxin and (±)-histrionicotoxin 235A "Tetrahedron Letters ,Volume 41, Issue 47, 18 November 2000, Pages 9163-9165

[azmanam]

cool sequence. 

3->5, 6->7 are both in the acetal mechanism (the lewis acid changes, but the mechanism doesn't). 

5->6 is a variation of the aldol.

7->8 starts with a variation of imine formation, then joins 8->9 as a variation of a Michael addition.

Lots of variations of common reactions... but fortunately the mechanisms are all the same as the common reactions.

[rita87]

3->5 :I know the mechanism for a normal hydrolysis(HCl in this case) of acetal,but i dont know how it works for the ring.I also don't know what THF does in this mechanism.

Thanks

[azmanam]

THF is solvent only.  Tetrahydrofuran.

Cyclic acetals operate exactly the same as 'typical-looking' acetals.  Whether the alcohols come from two molecules of methanol or one molecule of ethylene glycol... the mechanism - forward and backward - is exactly the same.

[rita87]

so is this mechanism correct( I skipped 2-3 steps)
Thanks
 

[azmanam]

step 1 is good.  not sure how you got to step 2, but a) you switched from ethylene glycol to methanol, and (with the carbon chain and proton also on R) carbon has 5 bonds.  intermediate 4 is back on track, do you know how to get from int 5->pdt?

[rita87]

I am not sure if the ring breaks into OMe-R-OMe or O-R-OCH2CH3 .
yes, you are right, C should have 4 bonds it should be R=O-Me.

[azmanam]

products will be ethylene glycol (2 eqiv) and your dialdehyde.

I've redrawn the steps in your mech with what the structures should look like.  Can you push the arrows to get through the mech?

[rita87]

Thanks mazman , yes i know how to draw the arrows.
the only thing i still dont get is that according to the normal mechanism(not cyclic),there should be ROH loss in step 2 . is that different in the cyclic form?

[azmanam]

no different.  the R in ROH is just a longer, more complex, carbon chain that's still attached to the molecule.  The cyclic acetal is a common protecting group for aldehydes and ketones.  It makes the carbonyl carbon no longer electrophilic.

[rita87]

do you know why ROH stays attached to the molecule, and not lost as in the non cyclic form?

[azmanam]

It doesn't.  That's the whole point.  There's no difference in the mechanism.  The only thing that changes is the carbon chain on the oxygen units.  One decomposes to give 2 mono alcohols.  One decomposes to give one diol.  See spiroketals for another variation for which the mechanism also doesn't change.  Those decompose to give one ketodiol.

[rita87]

thank you sooo much azmanam, you are always a gr8 help. thanks

[rita87]

for 5-> 6 i am using peterson olefination, but i am very confused with different stereoisomers , please can you explain how it works in this case.
I know that depending on acid or base used,elimination will be syn or anti, but dont know if it also effects stereoisomers formed etc.
I am very confused
thanks

[azmanam]

Start here:

http://www.organic-chemistry.org/namedreactions/peterson-olefination.shtm