I am trying to form N-nitrophenyl carbazole my new calculation I will do it with
ID: 1000223 • Letter: I
Question
I am trying to form N-nitrophenyl carbazole my new calculation I will do it with 0.5 g of Carbazole here was my first trying I did it with 0.05g of carbazole as u see I get impure product if we compaired toHNMR info on original experiment (first photo) my question is what should I pay attention to it if I start with 0.5 g to get pure product ? otherwise, in general to get pure compound what we have to care about it ? where the chemicals are new not old !!ps; I did it as my prof told me for 6 hr instead of 12 hr his opinion we can Chang the time no problem ..
Thanks a lot for all great person here :)
22.3. Compound Ic A 100-ml two-necked flask equipped with a reflux condenser and a stirrer was charged with carbazole (2.56 g, 15.3 mmol), K2CO3 (10.56 g, 76.52 mmol), p-nitrofluorobenzene (6.5 ml, 61.4 mmol) and DMF (80 ml). The reaction mixture was boiled for 12 h, cooled, and poured into water (500 ml). A yellow precipitate was filtered off, dried, and crystallized from benzene. Yield, 3.79 g (86%): mp. 208-210 °c (lit. m.p. 209-211 c[91 H NMR (400 MHz, CDCl): 6 7.35 (t, 2H), 743-7.751 (m, 4H), 7.77 (d, 2H), 8.14 (d, 2H), 8.46 (d, 2H) ppm. Found (%): C, 74.96: H, 43.27; N, 9.85. For C18H12N2O2: Anal. Caled (%): C, 74.99; H, 4.21; N, 9.72.
Explanation / Answer
According to your H-NMR spectra and its information, your product would be there. So, whatever procedure you followed to synthesize the compound is acceptable. Here, in your product impurity is in comparable amount. I think it perhaps unreacted p-nitroflourobenzene. I could not understand, why you have taken this much of excess reagents. It may interrupt to get pure compound. You will conduct one experiment with 1.1 equiv of p-nitroflourobenzene and 1.5 equiv of base then see what happend?, I believe that it will give certainely pure product. Otherwise, go to coloumn chromatography to purify the compound.
If any doubts, fell free to ask me
Best wishes
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