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Helen [10]
3 years ago
5

1-hexanol was prepared by reacting an alkene with either hydroboration-oxidation or oxymercuration-reduction. Draw the structure

of the alkene that was used to prepare the alcohol in highest yield.

Chemistry
1 answer:
slava [35]3 years ago
8 0

Answer:

hex-1-ene with a hydroboration-oxidation reaction

Explanation:

In this case, we can start with the structure of 1-hexanol. In this alcohol, the <u>"OH" is on the first carbon</u>. Thus, the alkene that this alcohol can produce must have a double bond between carbons 1 and 2, that is, <u>hex-1-ene</u>.

Now, it is important to know which of the reactions we can use, <em>hydroboration-oxidation, </em>or <em>oxymercuration-reduction</em>. To decide which reaction to using, we must look again at the alcohol we want to produce. In this alcohol, the "OH" is attached to a primary carbon. Therefore, the addition of the "OH" must be made on the <u>least substituted carbon</u>, that is, carbon number 1.

Because of this, an <u>anti-Markovnikov reaction</u> (a reaction in which the "OH" is added on the less substituted carbon) must be used. The reaction, which serves us then is: <u>hydroboration-oxidation</u> since this reaction is anti-markovnikov.

See figure 1

I hope it helps!

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Explanation : Given,

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Gibbs–Helmholtz equation is :

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As per question the reaction is spontaneous that means the value of \Delta G is negative or we can say that the value of \Delta G is less than zero.

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The above expression will be:

0>\Delta H-T\Delta S

T\Delta S>\Delta H

T>\frac{\Delta H}{\Delta S}

Now put all the given values in this expression, we get :

T>\frac{178500J/mole}{161.0J/mole.K}

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The mole of a substance is related to it's mass and molar mass according to the following equation:
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How to determine the mole of C₂H₅OH
From the question given above, the following data were obtained:
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5 0
2 years ago
The partial pressure of CO2 gas in a bottle of carbonated water is 4.60 atm at 25 ºC. How much CO2 gas (in g) will be released f
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If the partial pressure of CO₂ in a bottle of carbonated water decreases from 4.60 atm to 1.28 atm, the mass of CO₂ released is 0.265 g.

The partial pressure of CO₂ gas in a bottle of carbonated water is 4.60 atm at 25 ºC. We can calculate the concentration of CO₂ using Henry's law.

C = k \times P = \frac{1.65 \times 10^{-3} M }{atm}  \times 4.60 atm = 7.59 \times 10^{-3} M

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Now, we will repeat the same procedure for a partial pressure of 1.28 atm.

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The mass of CO₂ released will be equal to the difference in the masses at the different pressures.

m = 0.367 g - 0.102 g = 0.265 g

If the partial pressure of CO₂ in a bottle of carbonated water decreases from 4.60 atm to 1.28 atm, the mass of CO₂ released is 0.265 g.

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<em>The partial pressure of CO₂ gas in a bottle of carbonated water is 4.60 atm at 25 ºC. How much CO₂ gas (in g) will be released from 1.1 L of the carbonated water when the partial pressure of CO2 is lowered to 1.28 atm? At 25 ºC, the Henry’s law constant for CO₂ dissolved in water is 1.65 x 10⁻³ M/atm, and the density of water is 1.0 g/cm³.</em>

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