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Chemistry Forums for Students => Organic Chemistry Forum => Organic Chemistry Forum for Graduate Students and Professionals => Topic started by: Kruslin on September 03, 2019, 07:16:07 PM

Title: Secondary isotope effect - what to do?
Post by: Kruslin on September 03, 2019, 07:16:07 PM
We are trying to make a deuterated analog of a natural product for some biochemical investigations. In the penultimate step, a previously formed nucleophile (alkoxide) attacks the triple bond of the substrate, and a product with a double bond is formed (scheme attached).

This reaction proceeds smoothly with a non-deuterated nucleophile, and the yield is satisfactory. However, when done with an alkoxide which bears a deuterium at α position relative to the hydroxy group, the yield decreases for approx. 60%... After some consulting and research, our guess is that the secondary isotope effect is responsible for that.

Unfortunately due to metabolomics/biochemical reasons, we can't avoid the deuterium being at that position...

How might we "fight" against the secondary isotope effect and increase the yield?
Title: Re: Secondary isotope effect - what to do?
Post by: Vidya on September 05, 2019, 03:02:15 AM
You can carry out reaction without deuterating it is first step and can do it once the product is formed.In the product alpha hydrogen is still most acidic and can be easily exchanged with Deuterium.
Title: Re: Secondary isotope effect - what to do?
Post by: rolnor on September 05, 2019, 05:10:15 AM
Vidya, how do you know this hydrogen is acidic?
Title: Re: Secondary isotope effect - what to do?
Post by: Vidya on September 06, 2019, 01:00:36 AM
Vidya, how do you know this hydrogen is acidic?
It is closest to most EN atom of the molecule and its conjugate base is resonance stabilized.
Title: Re: Secondary isotope effect - what to do?
Post by: rolnor on September 06, 2019, 04:04:39 AM
This type of hydrogens are not acidic if the R groups are not electronwithdrawing. The vinyl ether can be deprotonated with strong base like LDA I think.
I dont think a large isotope-effect is possible with this system, the nucleophilic power of the deuterated alkoxide should be similar as the non-deuterated. I would repeat the reaction and be very carefull with pure, dry reagents.
Title: Re: Secondary isotope effect - what to do?
Post by: hollytara on September 06, 2019, 10:05:23 AM
Are the deuterated and non-deuterated reactions performed at the same scale? 

Reactions done at small scale often have lower isolated yields.  Let's assume that 250 mg of your product gets lost due to absorption on drying agent, filter paper, and other mechanical manipulations.  If your scale is 10 g, then your limiting yield is 9.75 g (97.5%).  But at 1 g, your yield is limited to 75%. 

The differences you see between deuterated and nondeuterated may come down to that. 

You usually see the 2° KIE when there is a change in geometry ("hybridization" if you think that way) of the atom with the  D attached.  This doesn't happen in your reaction - the D on the alkoxide is tetrahedral all the way through.  If there was a KIE, it would slow the reaction down by a factor of about 1.2 - unless you are stopping the reaction too quickly, that won't change the yield. 

If your reaction is under equilibrium control, there can be Equilibrium Isotope Effects.  But these are unusual, and the shift in Keq to drop a yield from 90% to 60% would be unexpected. 

If you are getting enough product to move forward with your research I wouldn't over analyze it.