[xshadow]

I have to explain the diastereoselectivity pf this product ,where COOEt and OH are in anti , in the zig-zag chain rappresentation



I've tried to use Houk model but I don't know where the electrophilie should "attack"...from upside or from downside? I thought that the Electrophilie  woul have attacked from the opposite side of alcoholic group ...but if I suppose this I'll get  OH/COOEt in syn(in the product are anti)


Thanks

[AWK]

See: Clayden - Stereoselective enolate alkylation

[xshadow]

See: Clayden - Stereoselective enolate alkylation

Hi!
I have Clayden text...but in that chapter it makes only one example with an alkyl group  as tue bulky group

Here I have an -OH group but If I consider it as the bulky group the electrophile (the allylic halide) should attack from the opposite side

But I get the "syn" (-OH and -COOEt) product ,not the anti!!

The only way to explain the stereochemistry  of the product is to consider the methyl group as the bulky group instead of the alcoholic -OH...

Thanks

[AWK]

You did not have OH group, it is O^-. Try protection of OH group.

[xshadow]

You did not have OH group, it is O^-. Try protection of OH group.

Sorry AWK but the exercise test says :
1.  2 equivalent LDA
2.  Br-CH2-CH=CH2
(Don't use protector group)

So it could be  -O^-( ok) or OLi ...

But in order to have COOEt in anti with the alcoholic OH group in the final product  the electrophile should attack from the same side of O-

If the electrophile  attacks from the OPPOSITE side of O-  (like I did ) I got  -COOEt and OH in anti...not in syn
Correct??

So it should attack from the same side of the oxygen?
Perhaps the OLi is less bulky then the -CH3 downside?

Thanks

[xshadow]

In order to explain better



I've applied Houk model on the enolate

BUT I get the product diastereselectivity (OH- and COOEt in anti) only when I suppose that the electrophile attacks from the SAME side of oxygen (the "alcoholic one" obv)

But why?
From downside I have instead a  methyl group.
If the electrophile attack upside it means that the CH3 is bulky...more than the O-/OLi ??!

This is my doubt!
Thanks

[AWK]

I don't know your procedure, I don't know if it's plausible. I have only given a possible alternative.

[xshadow]

I don't know your procedure, I don't know if it's plausible. I have only given a possible alternative.

And what should it happen with protection of alcoholic group ?

If I use a bulky group to protect the oxygen the electrophile should attack from the opposite site...correct?

[xshadow]

I  found the clayden solution but I think is wrong!!!



In the last step  when he rotates the COOEt should be away from the observer!!

He draws the other enantiomer!! (it can be checked by seeing the configuration of that chiral centre)...he changes its configuration through a simple molecule rotation?! It can't be!!?! Is wrong

Any thoughts?
Thnaks

[AWK]

I checked the Dreiding model - you are right.
But I was right to point out the additional interactions.

[xshadow]

I checked the Dreiding model - you are right.
But I was right to point out the additional interactions.

yeah
Now we know that the stereochemistry of the last molecule is wrong

The -COOEt should be away from the observer... the correct product written with this explanation is not the same of the one written in the  starting question .
SO what is wrong??

a) the product written in the question that I have to explain
b) the explanation in order to explain this (because this explanation gives the -COOEt away from us , not near us as indicated at the start)
?

According to you,I mean
Thanks

[AWK]

Fráter–Seebach alkylation

[xshadow]

Fráter–Seebach alkylation

Thanks!! Opposite from alcoholic oxygen
Now I know  also the name of this procedure!