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TEA [102]
2 years ago
9

Give the product for the reaction of 1-butene with methanol in the presence of acid. The mechanism is the same as the mechanism

for the addition of water to alkenes.
(A) 1-butanol
(B) 1-methoxybutane
(C) 2-ethoxybutane
(D) 1-ethoxybutane
(E) 2-methoxybutane
Chemistry
1 answer:
Nuetrik [128]2 years ago
8 0

Answer:

The final result is <u>2-methoxybutane. </u>

<u />

Explanation:

1-butene has a carbon-carbon double bound between C1 and C2.

When it will react with methanol, in the presence of an acid, the result will be an ether.

C4H8 + CH3OH → C5H12O

An acid-catalyzed ether synthesis from alkenes is limited by carbocation stability.

In the first step, the double bound will disappear. The C2 atom will be a C+ atom, this becaus it has only 3 bounds and not 4.

This C+ -atom will atract the O- atom to form an ether. The CH3 of methanol will bind on the C3 atom, this is the most stable position.

2-methoxybutane will be formed. It has a structural formula of C5H12O

1-methoxybutane will not be formed, because it's less stable.

1-butanol will be formed when water is added to 1-butene. The mechanism has the same principle but not the same product.

1-ethoxybutane and 2-ethoxybutane have a structural formula of C6H14O, this will not be the final result.

The final result is <u>2-methoxybutane. </u>

<u />

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2 years ago
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3 0
2 years ago
A solution is prepared by condensing 4.00 l of a gas, measured at 27°c and 748 mmhg pressure, into 58.0 g of benzene. calculate
fgiga [73]
First, we are using the ideal gas law to get n the number of moles:

PV = nRT

when P is the pressure = 748 mmHg/760 = 0.984 atm

V is the volume = 4 L

R is ideal gas constant = 0.0821

T is the temperature in Kelvin = 300 K

∴ n =  0.984atm*4L/0.0821*300

       = 0.1598 moles

when the concentration = moles * (1000g / mass)

                                         = 0.1598 * (1000g / 58 g )

                                         = 2.755 M

when the freezing point = 5.5 °C

and Kf = - 5.12 °C/m

∴ the freezing point for the solution = 5.5 °C + (Kf*m)

                                                            = 5.5 °C - (5.12°C/m * 2.755m)

                                                            = -8.6 °C

8 0
2 years ago
How much volume (in cm3) is gained by a person who gains 11.1 lbs of pure fat?
kirill [66]

The density of human fat is 0.918 g/cm^{3}. The mass of the pure fat is 11.1 lbs.

First convert mass from lb to g as follows:

1 lb=453.592 g

Thus,

11.1 lb\rightarrow 11.1\times 453.592 g=5034.875 g

Density of a substance is defined as mass per unit volume, thus volume can be calculated as:

V=\frac{m}{d}

Putting the values,

V=\frac{5034.875 g}{0.918 g/cm^{3}}=5484.61 cm^{3}

Therefore, volume gained by person will be 5484.61 cm^{3}.

6 0
2 years ago
150.0 grams of AsF3 were reacted with 180.0 g of CCl4 to produce AsCl3 and CCl2F2. The theoretical yield of CCl2F2 produced, in
Kamila [148]
The chemical reaction would be written as 

2 AsF3<span> + 3 CCl4 = 2 AsCl3 + 3 CCl2F2
</span>
We use the given amounts of the reactants to first find the limiting reactant. Then use the amount of the limiting reactant to proceed to further calculations.

150 g AsF3 ( 1 mol / 131.92 g) = 1.14 mol AsF3
180 g CCl4 (1 mol / 153.82 g) = 1.17 mol CCl4

 Therefore, the limiting reactant would be CCl4 since it would be consumed completely. The theoretical yield would be:

1.17 mol CCl4 ( 3 mol CCl2F2 / 3 mol CCl4 ) = 1.17 mol CCl2F2

7 0
2 years ago
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