Optimization continued

kilomentor | 10 February, 2011 14:50

Results B

Experiment #	Yield	Rank Yield	Rank	(100-Rank)/(N+1)
1	30	15	1	7.1
2	63	23	2	14.3
3	65	26	3	21.4
4	26	30	4	28.6
5	37	37	5	35.7
6	15	42	6	42.9
7	50	50	7	50.0
8	23	63	8	57.1
9	75	65	9	64.3
10	94	75	10	71.5
11	42	76	11	78.5
12	91	91	12	86.7
13	78	94	13	92.8

Results C

Experiment #	Yield	Rank Yield	Rank	(100-Rank)/(N+1)
1	0	0	1	7.1
2	45	0	2	14.3
3	0	11	3	21.4
4	11	15	4	28.6
5	15	23	5	35.7
6	32	24	6	42.9
7	53	32	7	50.0
8	33	33	8	57.1
9	44	42	9	64.3
10	24	44	10	71.5
11	60	45	11	78.5
12	42	53	12	86.7
13	23	60	13	92.8

Following the teaching of Hendrix in the article noted above if one makes 13 random observations and then ranks them there is a 7.2% chance (100-92.8) of finding a yield greater than the 13^th measurement (which is the highest yield obtained) Here that is 88% within the reaction space defined for these five reagent/condition combinations taken together. So long as the data is reproducible (and the 88% example should be repeated immediately) we know immediately that at least an 88% yield can be obtained using the conditions of experiment #13. Experiment #13 in our example was (1.5 mol. equivalents) of trityl fluoroborate, the solvent acetonitrile and the temperature 80 C with a time up to 48 hours.Using different numbers, suppose the results were as shown below designated There is an 7.2% chance (100-92.8)of finding a yield greater than 94% within the reaction space defined for these five reagent/condition combinations taken together. So long as the data is reproducible (and the 94% example should be repeated immediately) we know immediately that at least a 94% yield can be obtained using the conditions of experiment #10. Experiment #10 was 2 mol % of the chromium diol complex; 3.0 equiv. peracetic acid; +10 C; methylene chloride/CCl4 ratio 3:1 These results are actually all more optimistic than are very likely to be obtained. Using a different set of random numbers suppose the results were as shown below designated Results C.From these results it can be concluded that there is only a 7% probability to find a yield of greater than 60% in this complex reaction space. The only promising lead is that only a single experiment was conducted with discrete variables ofr Condition D that is Ni(II) bromide(2.0 mol. equivalents)/benzoyl peroxide (3 mol.equivalents),solvent acetonitrile and the temperature 80 C and the time up to 48 hours. If the reaction space was now restricted to these discrete conditions the probability of a high yield might be improved. This is an update of an original publication as of February 24, 2007. The author has not actually used the proposed method in process optimization. It is a theoretical proposal unlike other advice of Kilomentor, which is based on practical application and experience. I expect it will work and prove to be the most efficient method to optimize a process step. Comments and suggestions from experienced process chemists are welcome.

Comments

Remainder of article

kilomentor | 11/02/2011, 04:36

If you press general above you can find the remainder of this article.

Re: Optimization continued

Not Process Chemist | 04/03/2011, 03:55

What is the theory based of probability formula "(100-Rank)/(N+1)"? Is 13th experiments can reflect to the probability of 93%, and whether choosing optimistic reaction conditions by chemist is subject to "random". IMO It's really hard to deal with discrete variables in reactions, too many parameters in "one" reaction (addition sequences, quench time, quench temp, catalyst or reagent purity), needless to say different reagents. it may be more suitable in some robust reactions(coupling or recrystallization?) or reactants(without labile group)for optimization method.

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