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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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user100 [1]

Answer:

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3 0

3 years ago

All organic compounds contain ______, hydrogen, and usually oxygen. nitrogen. phosphorus. sulfur carbon.

Murrr4er [49]

Carbon is the answer. all hydrocarbures have to contain carbon.

4 0

3 years ago

If the value of Kc for the reaction is 2.50, what are the equilibrium concentrations if the reaction mixture was initially 0.500

ratelena [41]

Answer: The equilibrium concentration of sulfur dioxide, nitrogen dioxide, sulfur trioxide, nitrogen monoxide is 0.196 M, 0.196 M, 0.309 M and 0.309 M respectively.

Explanation:

We are given:

Initial concentration of sulfur dioxide = 0.500 M

Initial concentration of nitrogen dioxide = 0.500 M

Initial concentration of sulfur trioxide = 0.00500 M

Initial concentration of nitrogen monoxide = 0.00500 M

The chemical reaction follows:

$SO_2+NO_2\rightleftharpoons SO_3+NO$

Initial: 0.500 0.500 0.005 0.005

At eqllm: 0.500-x 0.500-x 0.005+x 0.005+x

The expression of equilibrium constant for the above reaction follows:

$K_c=\frac{[SO_3][NO]}{[SO_2][NO_2]}$

We are given:

$K_c=2.50$

Putting values in above equation, we get:

$2.50=\frac{(0.005+x)\times (0.005+x)}{(0.500-x)\times (0.500-x)}\\\\x=0.304,1.37$

Neglecting the value of x = 1.37, because change cannot be greater than the initial concentration

So, equilibrium concentration of sulfur dioxide = $(0.500-x)=(0.500-0.304)=0.196M$

Equilibrium concentration of nitrogen dioxide = $(0.500-x)=(0.500-0.304)=0.196M$

Equilibrium concentration of sulfur trioxide = $(0.00500+x)=(0.00500+0.304)=0.309M$

Equilibrium concentration of nitrogen monoxide = $(0.00500+x)=(0.00500+0.304)=0.309M$

Hence, the equilibrium concentration of sulfur dioxide, nitrogen dioxide, sulfur trioxide, nitrogen monoxide is 0.196 M, 0.196 M, 0.309 M and 0.309 M respectively.

4 0

3 years ago

What has a species reached when it’s birth rate is equal to its death rate?

EleoNora [17]

This particular case in a species is called

B) Carrying capacity

... or when a population reaches its equilibrium.

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CHEERS!!

8 0

3 years ago

45.6 mL of 0.0600 M Ba(OH)2 is added to 25.0 mL of HCl of unknown concentration. What is the concentration of HCl?