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Eddi Din [679]
2 years ago
12

Be sure to answer all parts. ethyl butanoate, ch3ch2ch2co2ch2ch3, is one of the many organic compounds isolated from mangoes. wh

ich hydrogen is most readily removed when ethyl butanoate is treated with base? propose a reason for your choice, and using the data listed below, estimate its pka. h5ch02p59q

Chemistry
1 answer:
GrogVix [38]2 years ago
4 0

Ethyl Butanoate when treated with a base looses a proton which is more acidic in nature. In this case ethyl butanoate acts as a lowery bronsted acid. It donated the more acidic proton to lowery bronsted base.

Among the protons attached to different carbon atoms the hydrogen atoms next to carbonyl functional group (labelled as red in attached picture) are more acidic in nature and are readily donated on treatment with strong base. These hydrogen atoms are also called alpha hydrogen name after their position.

Acidity of Alpha Hydrogens:

The driving force behind the acidity of alpha hydrogens is the formation of enolates. The enolate formed is resonance stabilized. This stability is the main reason for the said acidity. The pKa value of said protons is approximately 20-25. Hence, the enolate formed is infact the conjugate base and can act as neucleophile.

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1 year ago
According to the law of conservation of mass in a chemical reaction the total starting mass of all the reactants equals the tota
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2.92 A 50.0-g silver object and a 50.0-g gold object are both added
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Answer:

82.9 mL  

Explanation:

1. Volume of silver

\begin{array}{rcl}\text{Density}&=& \dfrac{\text{Mass}}{\text{Volume}}\\\\\rho&=& \dfrac{m}{V}\\\\V &=& \dfrac{m}{\rho}\\\\& = & \dfrac{\text{50.0 g}}{\text{10.49 g$\cdot$mL}^{-1}}\\\\& = & \text{4.766 mL}\\\end{array}\\\text{The volume of the silver is $\large \boxed{\textbf{4.766 mL}}$}

2. Volume of gold

\begin{array}{rcl}V& = & \dfrac{\text{50.0 g}}{\text{19.30 g$\cdot$mL}^{-1}}\\\\& = & \text{2.591 mL}\\\end{array}\\\text{The volume of the gold is $\large \boxed{\textbf{2.591 mL}}$}

3. Total volume of silver and gold

V = 4.766 mL + 2.591 mL = 7.36 mL

4 New reading of water level

V = 75.5 mL + 7.36 mL = 82.9 mL

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