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
1 answer:
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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<em>The mass of a gold bar = 1.424 x 10⁴ gram</em>
<em></em>
<h3><em>Further Explanation</em></h3>Density is a quantity derived from the mass and volume
density is the ratio of mass per unit volume
With the same mass, the volume of objects that have a high density will be smaller than type
The unit of density can be expressed in g / cm3 or kg / m3
Density formula:
ρ = density
m = mass
v = volume
A common example is the water density of 1 gr / cm3
The mass itself is often equated with weight, even though it is different. Mass is the amount of matter in the matter while weight is related to the gravitational force
<em>Known variable</em>
volume gold bar 7.379 × 10⁻⁴ m³
a density of 19.3 g/cm³
<em>Asked</em>
the mass of a gold bar
<em>Answer</em>
volume is known to be 7.379 × 10⁻⁴ m³, then we change it first to units of cm³ adjusting to units of density
7.379 × 10⁻⁴ m³ = 7.379 × 10² cm³
then:
mass = volume x density
mass = 7.379 × 10² cm³ x 19.3 g/cm³
mass = 1.424 x 10⁴ gram
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Keywords: density, mass, volume, a gold bar
