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MrRa [10]
4 years ago
7

Ethers are almost always used as solvents for Grignard reactions, all of the reasons why they work so well are not fully underst

ood. Some help comes from characterizing the ether solvents. State whether they would normally be characterized as polar or nonpolar and also whether they would normally be characterized as protic or aprotic

Chemistry
1 answer:
vfiekz [6]4 years ago
5 0

Answer:

Ether is used as a solvent because it is aprotic and can solvate the magnesium ion.

Explanation:

Solubility in Water

Because ethers are polar, they are more soluble in water than alkanes of a similar molecular weight. The slight solubility of ethers in water results from hydrogen bonds between the hydrogen atoms of water molecules and the lone pair electrons of the oxygen atom of ether molecules.

Ethers as Solvents

Ethers such as diethyl ether dissolve a wide range of polar and nonpolar organic compounds. Nonpolar compounds are generally more soluble in diethyl ether than alcohols because ethers do not have a hydrogen bonding network that must be broken up to dissolve the solute. Because diethyl ether has a moderate dipole moment, polar substances dissolve readily in it.

Ethers are aprotic. Thus, basic substances, such as Grignard reagents, can be prepared in diethyl ether or tetrahydrofuran. These ethers solvate the magnesium ion, which is coordinated to the lone pair electrons of diethyl ether or THF. Figure attached, shows the solvation of a Grignard reagent with dietheyl ether.

The lone pair electrons of an ether also stabilize electron deficient species such as BF3 and borane (BH3). For example, the borane-THF complex is used in the hydroboration of alkenes (Section 1

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Answer:

(4) 230 kPa

Explanation:

The temperature is constant, so the only variables are pressure and volume.

We can use Boyle’s Law.

p_{1}V_{1} = p_{2}V_{2}

Divide both sides of the equation by V_{2}.

p_{2} = p_{1} \times \frac{V_{1}}{ V_{2}}

p_{1} = \text{145 kPa}; V_{1} = \text{125 cm}^{3}}

p_{2} = ?; V_{2} = \text{80. cm}^{3}

p_{2} = \text{145 kPa} \times \frac{\text{125 cm}^{3}}{\text{80. cm}^{3}} = \textbf{230 kPa}

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3 years ago
How can a montain range cause a large dessert?
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Explanation:

An extensive mountain range can cause a large desert area.

Mountains are usually geologic features that have heights that are greater than 600m.

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Answer:

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Answer:

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In an IR spectrum the carbonyl group is associated with the C=O stretch which occurs as a strong peak around around 1700 cm-1. For alcohol the -corresponding O-H stretching frequency occurs as a strong broad peak between 3200-3600 cm-1.

Therefore, in the case of estradiol the presence a strong broad peak in the 3200-3600 cm-1 and the absence of the peak at around 1700 cm-1. would suggest that the transformation is complete.

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