The reaction scheme from the literature is that 1mmol trimethylamine react with 2mmole 2-chloro-1-methyl pyridium iodide (Mukaiyama reagent) with 1.5mmol NaNO2 in wet SiO2 (50%, w/w) in a solvent of n-hexane (3 mL) to produce "?" mmoles of N-Nitrosodimethylamine (NDMA)
I see that the reaction is very efficient being 75% of the theoretical yield, but I can't seem to figure out how many grams of NDMA they produced because they didn't notice the authors listing the product compound with the Iodine ion, though they probably extracted that on the silica beads with filtration like they did with the Mukaiyama reagent. I think that you might be able to figure it out if you look at table 1 entry 10 in the cited research paper by Roya Azadi & Khatereh Kolivand.
Source: Azadi, Roya, and Khatereh Kolivand. "Mukaiyama reagent/NaNO 2/wet SiO 2 as a mild and heterogeneous reagent system for the N-nitrosation of amines and preparation of azides from hydrazines." Tetrahedron Letters 56.41 (2015): 5613-5615.
http://www.sciencedirect.com/science/article/pii/S0040403915300113
Thanks so much!
Hi Adron,
I hope this helps.
The amount of NMe3 used is:
0.0010 mol of NMe3 x (59.11 g/mol NMe3 = 0.05911 g of NMe3
Knowing that the mol to mol ratio between NMe3 and NMe2NO is 1 to 1 is important.Therefore the theoretical yield can be calculated as:
0.05911 g NMe3 x (mol NMe3/59.11 g) x (1 mol NMe2NO/1 mol NMe3) x (74.08 g/mol NMe2NO) = 0.07408 g of NMe2NO
Knowing that the yield was 75%, we find that:
0.07408 g of NMe2NO x 0.75 = 0.05556 g or 55.56 mg of NMe2NO was obtained
NMe3 is the limiting reagent. Therefore, 1 mol of NMe3 leads to (at max) 1 mol of NMe2NO... This is general chemistry, not an "assumption".
See this for step by step directions:
http://www.csun.edu/~hcchm001/LIMITREG.pdf
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