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tatiyna
3 years ago
8

Use a sheet of paper to answer the following question. Take a picture of your answers and attach to this assignment. Draw the st

ructure of 2,2,4,4-tetramethyl-3-pentanone and explain why it does not form an enol tautomer Explain why cyclohexa-2,4-dien-1-one (also called 2,4-cyclohexadienone) exists exclusively in the enol form

Chemistry
1 answer:
pentagon [3]3 years ago
8 0

Answer:

Explanation:

Structure of the 2,2,4,4-tetramethyl-3-pentanone is give in the attachment

In 2,2,4,4-tetramethyl-3-pentanone, no alpha hydrogen is present, therefore, enol form is not possible and hence, exist only in keto form.

Explanation for existence of cyclohexa-2,4-diene-1-one only in enol form:

keto form of cyclohexa-2,4-diene-1-one not aromatic and hence less stable.

Whereas enol form it is aromatic which makes it highly stable. that's why cyclohexa-2,4-diene-1-one exists only in enol form.

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<em>ANSWER:</em>

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2 years ago
Consider the reaction between iodine gas and chlorine gas to form iodine monochloride. A reaction mixture at 298.15k initially c
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Step 1 : Write balanced chemical equation

The balanced chemical equation for the reaction between iodine gas and chlorine gas is given below.

I_{2}  (g) + Cl_{2}  (g) ----->  2 ICl (g)

Step 2 : Set up ICE table

We will set up an ICE table for the above reaction

Following points are considered while drawing ICE table

- The initial concentration of product is assumed as 0

- The change in concentration (C) is assumed as x. Change (x) is negative for reactants and positive for products

- The coefficients in balanced equation are considered while writing C values

Check attached file for ICE table

Step 3 : Set up equilibrium constant equation

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Step 4 : Solving for x

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Let us plug in the equilibrium values (E) for I₂, Cl₂ and ICl from ICE table

81.9 = \frac{(2x)^{2}}{(0.437-x) (0.269-x)}

81.9 = \frac{(2x)^{2}}{x^{2} -0.706x + 0.118}

(2x)^{2} = 81.9 [ x^{2} -0.706x + 0.118]

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Therefore the change in concentration of the gases during the reaction is 0.254 M

Hence, x = 0.254 M

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[ICl]_{eq} = 2 ( 0.254) = 0.508 M

The concentration of ICl when the reaction reaches equilibrium is 0.508 M

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