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2 years ago

Addition of water to an alkyne gives a keto‑enol tautomer product. Draw an enol that is in equilibrium with the given ketone

1 answer:

7 0

Addition of water to an alkyne gives a keto‑enol tautomer product and that is the product changed into 2-pentanone, then the alkyne need to had been 1-pentyne. 2-pentyne might have given a combination of 2- and 3-pentanone.

<h3>What is the keto-enol means in tautomer?</h3>

They carries a carbonyl bond even as enol implies the presence of a double bond and a hydroxyl group. The keto-enol tautomerization equilibrium is depending on stabilization elements of each the keto tautomer and the enol tautomer.

The enol that could provide 2-pentanone might had been pent-1- en - 2 -ol. Because an equilibrium favors the ketone so greatly, equilibrium isn't an excellent description.
If the ketone have been handled with bromine, little response might be visible because the enol content material might be too low.
If a catalyst have been delivered, NaOH for example, then formation of the enolate of pent-1-en - 2 - ol might shape and react with bromine.
This might finally provide a bromoform product. Under acidic conditions, the enol might desire formation of the greater substituted enol constant with alkene stability.

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3 years ago

10. Which one of the following statements about sulfuric acid is correct? A. Sulfuric acid has little effect on metals.

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3 years ago

Consider the following reaction at equilibrium. 2CO2 (g) 2CO (g) + O2 (g) H° = -514 kJ Le Châtelier's principle predicts that th

tiny-mole [99]

Answer:

C. at low temperature and low pressure.

Explanation:

Le Châtelier's principle states that when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.

For the reaction:

2CO₂(g) ⇄ 2CO(g) + O₂(g), ΔH = -514 kJ.

Effect of pressure:

When there is an increase in pressure, the equilibrium will shift towards the side with fewer moles of gas of the reaction. And when there is a decrease in pressure, the equilibrium will shift towards the side with more moles of gas of the reaction.

The reactants side (left) has 2.0 moles of gases and the products side (right) has 3.0 moles of gases.

So, decreasing the pressure will shift the reaction to the side with higher no. of moles of gas (right side, products), so the equilibrium partial pressure of CO (g) can be maximized at low pressure.

Effect of temperature:

The reaction is exothermic because the sign of ΔH is (negative).

So, we can write the reaction as:

2CO₂(g) ⇄ 2CO(g) + O₂(g) + heat.

Decreasing the temperature will decrease the concentration of the products side, so the reaction will be shifted to the right side to suppress the decrease in the temperature, so the equilibrium partial pressure of CO (g) can be maximized at low temperature.

So, the right choice is:

C. at low temperature and low pressure.

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3 years ago

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3 years ago