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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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This are the reactants:

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NaI

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The second stage of photosynthesis produces _____ and _____.
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g aqueous barium hydroxide (ba(oh)2) and nitric acid (hno3) participate in a complete neutralization reaction. in the molecular
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You want to slow down the following reaction:
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Methanol, ethanol, and n−propanol are three common alcohols. When 1.00 g of each of these alcohols is burned in air, heat is lib
KengaRu [80]

Answer:

<u>For methanol:</u> Heat of combustion = -22.6 kJ / 0.0312 moles = -724.3590 kJ/mol (negative sign signifies release of heat)

<u>For ethanol: </u>Heat of combustion = -29.7 kJ / 0.0217 moles = -1368.6636 kJ/mol (negative sign signifies release of heat)

<u>For propanol: </u>Heat of combustion = -33.4 kJ / 0.0166 moles = -2012.0482 kJ/mol (negative sign signifies release of heat)

Explanation:

Given:

Mass of Methanol = 1.0 g

Mass of ethanol = 1.00 g

Mass of n-propanol = 1.00 g

<u>For methanol:</u>

2 CH₃OH + 3 O₂ ----> 2 CO₂ + 4 H₂O, ∆H₀ = -22.6 kJ/g  (negative sign signifies release of heat)

1 g of methanol on combustion gives 22.6 kJ of energy

Calculation of moles of methanol:

moles=\frac{Mass(m)}{Molar\ mass (M)}

Molar mass of methanol = 32.04 g/mol

Thus moles of methanol = 1 g/ (32.04 g/mol) = 0.0312 moles

Hence energy in kJ/mol:

<u>Heat of combustion = -22.6 kJ / 0.0312 moles = -724.3590 kJ/mol (negative sign signifies release of heat)</u>

<u></u>

<u>For ethanol:</u>

C₂H₅OH + 3 O₂ ----> 2 CO₂ + 3 H₂O, ∆H₀ = -29.7 kJ/g  (negative sign signifies release of heat)

1 g of ethanol on combustion gives 29.7 kJ of energy

Calculation of moles of ethanol:

moles=\frac{Mass(m)}{Molar\ mass (M)}

Molar mass of ethanol = 46.07 g/mol

Thus moles of ethanol = 1 g/ (46.07 g/mol) = 0.0217 moles

Hence energy in kJ/mol:

<u>Heat of combustion = -29.7 kJ / 0.0217 moles = -1368.6636 kJ/mol (negative sign signifies release of heat)</u>

<u></u>

<u>For propanol:</u>

2 C₃H₇OH + 9 O₂ ----> 6 CO₂ + 8 H₂O, ∆H₀ = -33.4 kJ/g , (negative sign signifies release of heat)

1 g of methanol on combustion gives 33.4 kJ of energy

Calculation of moles of methanol:

moles=\frac{Mass(m)}{Molar\ mass (M)}

Molar mass of methanol = 60.09 g/mol

Thus moles of methanol = 1 g/ (60.09 g/mol) = 0.0166 moles

Hence energy in kJ/mol:

<u>Heat of combustion = -33.4 kJ / 0.0166 moles = -2012.0482 kJ/mol (negative sign signifies release of heat)</u>

5 0
3 years ago
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