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Genrish500 [490]
1 year ago
7

Mixed aldol condensations involving benzaldehyde yield typically a single product. draw the structure of the partner that was us

ed to synthesize this molecule:
Chemistry
1 answer:
Andrej [43]1 year ago
6 0

The partner that is used along with benzaldehyde is acetophenone to yield  1,3-Diphenylprop-2-en-1-one.

<h3>What is Adol Condensation?</h3>

In adol condensation, aldehydes or ketones having α - H atoms undergo condensation in the presence of a dilute alkali to form β - hydroxy aldehydes or ketones.

<h3>What is Mixed Adol Condensation?</h3>

Under mixed or crossed adol condensation,

  • two different aldehyde molecules or
  • 2 different ketone molecules or
  • one aldehyde and one ketone molecule

undergoes condensation in the presence of dilute alkali to form a mixture of β - hydroxy aldehyde or ketone.

The following reaction takes place:

Disclaimer: The question was given incomplete on the portal. Here is the complete question.

Question: Mixed aldol condensations involving benzaldehyde yield typically a single product. draw the structure of the partner that was used to synthesize this molecule:

Learn more about adol condensation:

brainly.com/question/27178362

#SPJ4

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Evaluate the carbon dioxide molecule. Explain how to determine if double or triple bonds exist in the molecule.
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It contain double Bond.

Explanation:

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If the sun is next to the earth on the left, and the moon is next to the earth on the right, what type of tide is it?
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2 years ago
Which of the following reactions is an example of a single replacement reaction?
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5 0
2 years ago
If you combine 230.0 mL 230.0 mL of water at 25.00 ∘ C 25.00 ∘C and 120.0 mL 120.0 mL of water at 95.00 ∘ C, 95.00 ∘C, what is t
Thepotemich [5.8K]

<u>Answer:</u> The final temperature of the mixture is  49°C

<u>Explanation:</u>

To calculate the mass of water, we use the equation:

\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}

  • <u>For cold water:</u>

Density of cold water = 1 g/mL

Volume of cold water = 230.0 mL

Putting values in above equation, we get:

1g/mL=\frac{\text{Mass of water}}{230.0mL}\\\\\text{Mass of water}=(1g/mL\times 230.0mL)=230g

  • <u>For hot water:</u>

Density of hot water = 1 g/mL

Volume of hot water = 120.0 mL

Putting values in above equation, we get:

1g/mL=\frac{\text{Mass of water}}{120.0mL}\\\\\text{Mass of water}=(1g/mL\times 120.0mL)=120g

When hot water is mixed with cold water, the amount of heat released by hot water will be equal to the amount of heat absorbed by cold water.

Heat_{\text{absorbed}}=Heat_{\text{released}}

The equation used to calculate heat released or absorbed follows:

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)]      ......(1)

where,

q = heat absorbed or released

m_1 = mass of hot water = 120 g

m_2 = mass of cold water = 230 g

T_{final} = final temperature = ?°C

T_1 = initial temperature of hot water = 95°C

T_2 = initial temperature of cold water = 25°C

c = specific heat of water = 4.186 J/g°C

Putting values in equation 1, we get:

120\times 4.186\times (T_{final}-95)=-[230\times 4.186\times (T_{final}-25)]

T_{final}=49^oC

Hence, the final temperature of the mixture is  49°C

4 0
3 years ago
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