What crossed-aldol product results when butanal is heated in the presence of excess benzaldehyde and sodium hydroxide? draw the
molecule on the canvas by choosing buttons from the tools (for bonds), atoms, and advanced template toolbars. the single bond is active by defa?
1 answer:
Answer:
2-(hydroxy(phenyl)methyl)butanal is formed as cross aldol product.
Explanation:
The mild acidic proton present at alpha position in 1-Butanal is abstracted by strong base NaOH. The enolate formed is the conjugate base of 1-Butanal and cacts as a nucleophile. The enolate is then added across carbonyl group of Benzaldeyde and yield 2-(hydroxy(phenyl)methyl)butanal. The reaction is as follow,
2-(hydroxy(phenyl)methyl)butanal is formed as cross aldol product.
Explanation:
The mild acidic proton present at alpha position in 1-Butanal is abstracted by strong base NaOH. The enolate formed is the conjugate base of 1-Butanal and cacts as a nucleophile. The enolate is then added across carbonyl group of Benzaldeyde and yield 2-(hydroxy(phenyl)methyl)butanal. The reaction is as follow,
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Answer:
124.91mL
Explanation:
Given parameters:
P₁ = 1.08atm
V₁ = 250mL
T₁ = 24°C
P₂ = 2.25atm
T₂ = 37.2°C
V₂ = ?
Solution:
To solve this problem, we are going to apply the combined gas law;
P, V and T represents pressure, volume and temperature
1 and 2 delineates initial and final states
Convert the temperature to kelvin;
T₁ = 24°C, T₁ = 24 + 273 = 297K
T₂ = 37.2°C , T₂ = 37.2 + 273 = 310.2K
Input the variables and solve for V₂
V₂ = 124.91mL
Answer:
1.98 M
Explanation:
Given data
- Initial volume (V₁): 93.2 mL
- Initial concentration (C₁): 2.03 M
- Volume of water added: 3.92 L
Step 1: Convert V₁ to liters
We will use the relationship 1 L = 1000 mL.
Step 2: Calculate the final volume (V₂)
The final volume is the sum of the initial volume and the volume of water.
Step 3: Calculate the final concentration (C₂)
We will use the dilution rule.