2017 Examples Example 3: p (compound of formula VI reparation of 7-4-1 mesyloxyb

Question

2017 Examples Example 3: p (compound of formula VI reparation of 7-4-1 mesyloxybutoxy)-1H-quinolin-2-one o H3C trethylamine o MW 311.35 (CH2) 40H MW 233.26 MW 114.55 ydroxybutoxy-1H-quinolin-2-one (50 tetrahydrofuran (500 ml), and trimethylamine (35 g) were mixed and the solution was cooled to 20 oc. Methane and sulfonyl chloride (32 g) was added to the reaction mixture at 20-25 oC over one hour the reaction mass was maintained for two hours. After the reaction was complete, water and stirred for one hour at room temperature (500 mL) was charged to the reaction mass at 50 °C to obtain T-4-mesyloxybutoxyy1H- was washed with water (100 mL) and dried butoxy) 1H-quinolin-2-one were quinolin-2-one (55 g) moles of 7-(4-hydroxy 1) (5 points) How many used? 2) (5 points) How many moles of methane sulfonyl chloride were used?

Explanation / Answer

1)

one mole of 7-(4-hydroxybutoxy)-1H-quinolin-2-one = 233.11 g

for 50 g of  7-(4-hydroxybutoxy)-1H-quinolin-2-one = 50/233 = 0.21 mole

2)

one mole of methane sulfonyl chloride = 114.55

for 32 g of methane sulfonyl choride = 32/114 = 0.28 mole

3)

Density of methane sulfonyl chloride = 1.48 g/cm3

Density = mass / volume = 32/1.48 = 21 mL

4)

when 233 g of  7-(4-hydroxybutoxy)-1H-quinolin-2-one is used 311 g of product will give 100 % yield

for 50 g of  7-(4-hydroxybutoxy)-1H-quinolin-2-one = 311 x 233/50 = 66 g

5)

actual yield = 55/66 x 100 = 83 %

6)

one mole prouduct = 311 g

for 66 g = 66/311 = 0.21 mole

7)

triethyl amine is added to neutalize the hydrochloric acid which formed as by-product

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