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Andrej [43]
3 years ago
7

Dehydration of 2-methyl-2-pentanol forms one major and one minor organic product. Draw the structures of the two organic product

s of this reaction.
Chemistry
1 answer:
Korvikt [17]3 years ago
7 0

Answer:

The major product is 2-methyl-2-pentene [ CH₃-CH₂-CH=C(CH₃)₂ ] and a minor product 2-methyl-1-pentene [ CH₃-CH₂-CH₂-C(CH₃)=CH₂ ].

Explanation:

Dehydration reaction is a reaction in which a molecule loses a water molecule in the presence of a dehydrating agent like sulfuric acid (H₂SO₄).

<u>Dehydration reaction of 2-methyl-2-pentanol</u> gives a major product 2-methyl-2-pentene and a minor product 2-methyl-1-pentene.

CH₃-CH₂-CH₂-C(CH₃)₂-OH (2-methyl-2-pentanol)→ CH₃-CH₂-CH=C(CH₃)₂ (2-methyl-2-pentene, major) + CH₃-CH₂-CH₂-C(CH₃)=CH₂ (2-methyl-1-pentene, minor)

<u>Since more substituted alkene is more stable than the less substituted alkene. So, the trisubstituted alkene, 2-methyl-2-pentene is more stable than the disubstituted alkene, 2-methyl-1-pentene.</u>

<u>Therefore, the trisubstituted alkene, 2-methyl-2-pentene is the major product and the disubstituted alkene, 2-methyl-1-pentene is the minor product.</u>

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Al2(co3)3
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3 years ago
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A gaseous compound is 30.4 % N and 69.6% OF. A 5.25 g sample of the gas occupies a volume of 1.00 L and exerts a pressure of 958
Harman [31]

Answer:

The molecular formula = N2O4

Explanation:

<u>Step 1</u>: Data given

A gaseous compound is 30.4 % N and 69.6%

Mass of the compound = 5.25 grams

Volume of the gas = 1.00 L

Pressure of the gas = 958 mmHg = 1.26 atm

Temperature of the gas = -4 °C = 273 -4°C = 269 Kelvin

Molar mass of N = 14 g/mol

Molar mass of O = 16 g/mol

<u>Step 2</u>: Calculate mass of N

Mass of Nitrogen = 5.25 grams * 0.304 = 1.596 grams

<u>Step 3:</u> Calculate mass of O

Mass of Oxygen = 5.25 grams * 0.696 = 3.654 grams

<u>Step 4:</u> Calculate number of moles N

Number of moles N = Mass of N/ Molar mass of N

Moles of N = 1.596 grams / 14g/mol

Moles of N = 0.114 moles N

<u>Step 5:</u> Calculate moles of O

Moles O = 3.654 grams / 16 g/mol

Moles 0 = 0.2884 moles

<u>Step 6:</u> Calculate empirical formule

We calculate the empirical formule by dividing number of moles by the smallest number of mol

N : 0.114 / 0.114 = 1

O: 0.2284 / 0.114 = 2

Empirical formule = NO2

<u>Step 7: </u>Calculate number of moles of 5.25 g sample via gas law:

p*V = nRT

⇒ with p = the pressure = 1.26 atm

⇒ with v = 1.00 L

⇒ with n = the number of moles = TO BE DETERMINED

⇒ with R = the gas constant = 0.08206 L*atm/ K*mol

⇒ with T = the temperature = 269 K

number of moles n = (p*V)/(R*T)

n = (1.26*1L)/(0.08206*269)

n = 0.057 mol  

<u>Step 8:</u> Calculate molar mass of the compound

This means 5.25 grams of the gas = 0.057 moles

So 1 mol of the compound has a molar mass of: 5.25 / 0.057 = 92.11 g/mol

<u>Step 9</u>: Calculate molar mass of the empirical formula NO2

N = 14 g/mol

O = 16 g/mol

NO2 = 14 + 16 + 16 = 46 g/mol

The empirical formule NO2 has a molar mass of 46 g/mol

<u>Step 10</u>: Calculate molecular formula

92.11 / 46 = 2

This means the empirical formula should be multiplied by 2

2*(NO2) = N2O4

The molecular formula = N2O4

8 0
3 years ago
How many moles are in 8.35 g of H2O?
AlexFokin [52]
It’s is 130 that is the answer!!!!
8 0
1 year ago
For a particular isomer of C 8 H 18 , C8H18, the combustion reaction produces 5104.1 kJ 5104.1 kJ of heat per mole of C 8 H 18 (
vlada-n [284]

Answer:

The standard enthalpy of formation of this isomer of octane is -220.1 kJ/mol

Explanation:

Step 1: Data given

The combustion reaction of octane produces 5104.1 kJ per mol octane

Step 2: The balanced equation

C8H18(g) + 12.5 O2 ⟶ 8CO2 (g) + 9 H2O (g)  ∆H°rxn = -5104.1 kJ/mol

Step 3:

∆H°rxn = ∆H°f of products minus the ∆H° of reactants

∆H°rxn = ∆H°f products - [∆H°f reactants]

-5104.1 kJ/mol = (8*∆H°fCO2 + 9*∆H°fH20) - (∆H°fC8H18 + 12.5∆H°fO2)

∆H°f C8H18 = ∆H°f 8CO2 + ∆H°f 9H2O+ 5104.1 kJ/mol

∆H°f C8H18 = 8 * (-393.5 kJ)/mol + 9 * (-241.8 kJ/mol)] + 5104.1 kJ /mol

∆H°f C8H18 = -220.1 kJ/mol

The standard enthalpy of formation of this isomer of octane is -220.1 kJ/mol

7 0
3 years ago
34g aluminum are combined with 39g chlorine gas which is the limiting reactant
vampirchik [111]

The limiting reactant is chlorine (Cl2).

<u>Explanation</u>:

Limiting reactant is the amount of product formed which gets limited by the reagent without continuing it.

        2 Al + 3 Cl2 ==> 2 AlCl3  represents the balanced equation.

Number of moles Al present = 34 g Al x 1 mole Al / 26.98 g

                                                = 1.260 g moles of Al

Number of moles Cl2 present = 39 g Cl2 x 1 mole Cl2 / 35.45 g

                                                  = 1.10 g moles of Cl2

Dividing each reactant by it's coefficient in the balanced equation obtains:

1.260 moles Al / 2 = 0.63 g moles of Al

1.11 moles Cl2 / 3 = 0.36 g moles of Cl2

The reactant which produces a lesser amount of product is called as limiting reactant.

Here the Limiting reactant is Cl2.

6 0
3 years ago
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