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Kinetic vs Thermodynamic product; HELP please

Hi,

In practical we did the reaction of iodolactonisation of 3-cyclohexene-carboxylic acid. So, C6H9CO2H + NaHCO3 + I2 + KI -----> C7H9IO2 (two possible products).
It can be controlled under kinetic or thermodynamic properties. In my case it forms a kinetic product (3,2,1)iodolactone (connecting carbons 2 and 3 on cycloxehane). All were mixed and stirred. Then, extracted with CHCl3 and further washed with Na2S2O3. Dried and evaporated.

Based on the lab expt., I don't know why I got really low yield (28%), though it should be high because under kinetic control it can't form the other possible product.

Please, please any suggestions on how to improve the yield and why is it low.
THANK YOU!
Reply 1
Anyone? :frown:
Reply 2
You could still have a lot of unreacted material?
Reply 3
Original post by Bradshaw
You could still have a lot of unreacted material?


Maybe... Do you think it is also possible that the equilibrium between starting material and Transition state caused backwards reaction? So reacting with reagents lead back to starting material, and hence low product yield?? I am confused and don't know if this is possible ...
Original post by Studentik
Hi,

In practical we did the reaction of iodolactonisation of 3-cyclohexene-carboxylic acid. So, C6H9CO2H + NaHCO3 + I2 + KI -----> C7H9IO2 (two possible products).
It can be controlled under kinetic or thermodynamic properties. In my case it forms a kinetic product (3,2,1)iodolactone (connecting carbons 2 and 3 on cycloxehane). All were mixed and stirred. Then, extracted with CHCl3 and further washed with Na2S2O3. Dried and evaporated.

Based on the lab expt., I don't know why I got really low yield (28%), though it should be high because under kinetic control it can't form the other possible product.

Please, please any suggestions on how to improve the yield and why is it low.
THANK YOU!


Can you give the actual name for your product...... The name you wrote doesn't give any hits.

I'm looking at Iodolactonisation reactions and what springs to mind is how you go about getting lactonisation in your starting material if the double bond does reside at that position (if it's not equilibrated to a different position).
Also, surely you would need to break open the ring to get a stable lactone product as otherwise you could only get a 4-membered lactone, hardly stable.....

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