Hi I would to start this reaction with 1.00 g of 4-nitrophenyl carbazole to redu

Question

Hi I would to start this reaction with 1.00 g of 4-nitrophenyl carbazole to reduced to corresponding amine. My question : i need somone help me to recalculation all chemicals using in this expt plz !!
Ps; there is mistake in the article they wrote III but actually it is IC in the scheme! the wrong just the letter not the compound name ok ..

Asap plz help me & i am always rate !!

2.2.5. Compound IIc A 50-ml three-necked flask equipped with a stirrer, a reflux condenser, and a dropping funnel was charged with N-(4 nitrophenyl)carbazole (III) (2.595 g, 9 mmol), 10% Pd/C (0.09 g), and ethanol (27 ml), and the mixture was heated to boiling. Then, hydrazine monohydrate (2.3 ml) was slowly added dropwise, and the reaction mixture was boiled for 10 h. The mixture was cooled to room temperature, and the residual catalyst was filtered off. The filtrate was evaporated in a rotary evaporator to obtain yellow crystals. Yield, 2.07 g (89%), m.p. 91-93°C (lit. m.p. 91-93°C [10]). IR (KBr): v 3432, 3351 cm-1 (N-H). 'H NMR (400 MHz, CDCl3): 8.21 (d, 2H), 7.745-7.35 (m, 8H), 6.90 (d, 2H), 3.92 (s, 2H, NH2) ppm Found (%): C, 83.30; H, 5.51; N, 10.75.

Explanation / Answer

4-nitrophenyl carbazole (2.595 g, 9 mmol) is taken in the procedure. If 1g is required to start with, that means you have to start with 3.47 mmol.

Procedure says: 10% Pd/C 0.09 g ----> you need 0.0347 g = 347 mg

Procedure says:Ethanol 27 mL ----> you need 10.41 mL

Procedure says:Hydrazine monohydrate 2.3 mL ---->you need 0.89 mL

Procedure says: Yield 2.07 g (89%)---> if you are expecting the same yield, you will be looking at 0.798 g = 798 mg (100% yield being 897 mg)

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