A J то گای ه +0 Also calculate the amount of starting materials chlorobenzaldehyde and p-chloroacetophenone required to prepare 400 mg of the given chalcone product 1, 3-bis(4-chlorophenyl)prop-2-en-1-one molar mass ok 1,3-bis(4-Chlorophenyl) prop-2-en-1-one = 277.1591m01 number of moles= 0.400/277.15 = 0.00144 moles 2 x 0.00 144=0.00288 moves arams of acetophenone = 0.00144 X 120.16 = 0.1739 0.1739x2=0.3469 grams of benzaldehyde = 0.00144X106.12=0.1539 0.1539x2 = 0.3069 Starting materials: 0.3469 Ox acetophenone, 0.3069 of benzaldehyde 3
A J то گای ه +0 Also calculate the amount of starting materials chlorobenzaldehyde and p-chloroacetophenone required to prepare 400 mg of the given chalcone product 1, 3-bis(4-chlorophenyl)prop-2-en-1-one molar mass ok 1,3-bis(4-Chlorophenyl) prop-2-en-1-one = 277.1591m01 number of moles= 0.400/277.15 = 0.00144 moles 2 x 0.00 144=0.00288 moves arams of acetophenone = 0.00144 X 120.16 = 0.1739 0.1739x2=0.3469 grams of benzaldehyde = 0.00144X106.12=0.1539 0.1539x2 = 0.3069 Starting materials: 0.3469 Ox acetophenone, 0.3069 of benzaldehyde 3
Chapter9: Alkynes: An Introduction To Organic Synthesis
Section9.SE: Something Extra
Problem 32AP
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Transcribed Image Text:A
J
то
گای ه
+0
Also calculate the amount of starting materials chlorobenzaldehyde and p-chloroacetophenone
required to prepare 400 mg of the given chalcone product 1, 3-bis(4-chlorophenyl)prop-2-en-1-one
molar mass ok 1,3-bis(4-Chlorophenyl) prop-2-en-1-one = 277.1591m01
number of moles= 0.400/277.15 = 0.00144 moles
2 x 0.00 144=0.00288 moves
arams of acetophenone = 0.00144 X 120.16 = 0.1739
0.1739x2=0.3469
grams of benzaldehyde = 0.00144X106.12=0.1539
0.1539x2 = 0.3069
Starting materials:
0.3469 Ox acetophenone,
0.3069 of benzaldehyde
3
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