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Oliga [24]
3 years ago
8

Chapter 15 of your textbook discusses conjugation and various reactions of 1,3-dienes. In CHE 321, we discussed various ways to

make alkenes, but we did not discuss methodology for making dienes. Fortunately, we can apply some of our standard CHE 321 reactions to this important problem. Choose the major product (compound A) of the following reaction sequence.
Chemistry
1 answer:
xxMikexx [17]3 years ago
5 0

Answer:

Dienes are alkenes that contain two carbon-carbon double bonds, so they have the same properties as these hydrocarbons.

In the attached file are the two reactions of dienes production.

Explanation:

Two ways to obtain dienes are as follows:

-Reaction of oxidative dehydrogenation of an alkane, is an exothermic process and occurs at lower temperatures, diene and water are formed, generating greater conversion at lower temperature levels.

-Dehydration of primary alcohols. The treatment of alcohols with acid at elevated temperatures produces dienes due to water loss. For example, heating ethanol in the presence of sulfuric acid produces ethene by the loss of a water molecule.

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2. Reaction will shift to right. <span>

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3 0
3 years ago
Read 2 more answers
How many moles of water are produced from 15 moles of oxygen?
Grace [21]

Answer:

30 moles

Explanation:

Water is H2O, meaning there is 2 Hydrogen atoms and 1 Oxygen atom. Oxygen is O2, because it is a diatomic molecule. (Hydrogen is also a diatomic molecule, so H2)

The equation, balanced, would have to be: 2H2 + O2 -----> 2H2O

I multiply 15 moles O2 by the molar ratio of (hydrogen/oxygen)

15 mol. O2 *  (2 mol. H2/1 mol O2) = 30 moles of water

8 0
3 years ago
For the following reaction, KcKc = 255 at 1000 KK.
bonufazy [111]

Answer :

The equilibrium concentration of CO is, 0.016 M

The equilibrium concentration of Cl₂ is, 0.034 M

The equilibrium concentration of COCl₂ is, 0.139 M

Explanation :

The given chemical reaction is:

                           CO(g)+Cl_2(g)\rightleftharpoons COCl_2(g)

Initial conc.      0.1550      0.173           0

At eqm.          (0.1550-x)  (0.173-x)         x

As we are given:

K_c=255

The expression for equilibrium constant is:

K_c=\frac{[COCl_2]}{[CO][Cl_2]}

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

255=\frac{(x)}{(0.1550-x)\times (0.173-x)}

x = 0.139 and x = 0.193

We are neglecting value of x = 0.193 because equilibrium concentration can not be more than initial concentration.

Thus, we are taking value of x = 0.139

The equilibrium concentration of CO = (0.1550-x) = (0.1550-0.139) = 0.016 M

The equilibrium concentration of Cl₂ = (0.173-x) = (0.173-0.139) = 0.034 M

The equilibrium concentration of COCl₂ = x = 0.139 M

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Leno4ka [110]

Answer:

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Answer:

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