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Oksanka [162]
3 years ago
15

Write a mechanism for the conversion of the aldol addition product, 3-hydroxy-3-(4-nitrophenyl)-1-(2-pyridyl)-1-propanone, to th

e aldol condensation product, (E)-3-(4-nitrophenyl)-1-(2-pyridyl)-1-propenone. Be as complete as possible and show electron flow for all steps.
Chemistry
1 answer:
LenaWriter [7]3 years ago
8 0

Solution :

"Aldol" stands for the abbreviation, aldehyde and alcohol. When a ketone or an aldehyde's enolate reacts with the carbonyl of a  molecule at the alpha carbon, under the acidic or basic conditions so as to obtained the ketone or β-hydroxy aldehyde, is known as an aldol reaction.

For the conversion of the aldol addition product of a 3-hydroxy-3-(4-nitrophenyl)-1-(2-pyridyl)-1-propanone to an aldol condensation product of (E)-3-(4-nitrophenyl)-1-(2-pyridyl)-1-propenone, the mechanism is given in the diagram a below :

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3 years ago
40 g of CaCO3 is how many moles of CaCO3?<br> 10 moles<br> 0.4 moles<br> 40 moles<br> 100 moles
AleksandrR [38]

Answer:

0.4 moles

Explanation:

To convert between moles and grams you need the molar mass of the compound. The molar mass of of CaCO3 is 100.09g/mol. You use that as the unit converter.

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This rounds to 0.4 moles CaCO3

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3 years ago
Consider a 0.10 M aqueous benzoic acid, CeHeCOOH. The K benzoic acid. 6.5 x 10 for A) Write a balanced equation that shows the r
7nadin3 [17]

Answer:

a) C6H5COOH + H2O ↔ H3O+  +  C6H5COO-

b) [ H3O+ ] = 2.517 E-3 M

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Explanation:

a) balanced equation:

C6H5COOH + H2O ↔ H3O+  +  C6H5COO-

⇒ Ka = ( [ H3O+ ] * [ C6H5COO- ] ) / [ C6H5COOH ] = 6.5 E-5

mass balance:

0.10 m = [ C6H5COO- ] + [ C6H5COOH ].....(1)

charge balance:

[ H3O+ ] = [ C6H5COO- ] + [ OH- ] .......[ OH- ] : comes from water, it's not significant

⇒ [ H3O+ ] = [ C6H5COO- ] .........(2)

b) (2) in (1):

⇒ 0.10 M = [ H3O+ ] + [ C6H5COOH ]

⇒ [ C6H5COOH ] = 0.10 - [ H3O+ ]

⇒ Ka = [ H3O+ ]² / ( 0.1 - [ H3O+ ] ) = 6.5 E-5

⇒ [ H3O+ ]² + 6.5 E-5 [ H3O+ ] - 6.5 E-6 = 0

⇒ [ H3O+ ] = 2.517 E-3 M

c) pH = - log [ H3O+ ]

⇒ pH = - Log ( 2.517 E-3 )

⇒ pH = 2.599

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3 years ago
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