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MrRa [10]
3 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]3 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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The enthalpy of reaction for the combustion of ethane 2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O calculated from the average bond energies of the compounds is -2860 kJ/mol.

The reaction is:

2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O  (1)  

The enthalpy of reaction (1) is given by:

\Delta H = \Delta H_{r} - \Delta H_{p}   (2)

Where:

r: is for reactants

p: is for products

The bonds of the compounds of reaction (1) are:

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Hence, the enthalpy of reaction (1) is (eq 2):

\Delta H = \Delta H_{r} - \Delta H_{p}

\Delta H = 2*\Delta H_{CH_{3}CH_{3}} + 7\Delta H_{O_{2}} - (4*\Delta H_{CO_{2}} + 6*\Delta H_{H_{2}O})      

\Delta H = 2*(6*\Delta H_{C-H} + \Delta H_{C-C}) + 7\Delta H_{O=O} - (4*2*\Delta H_{C=O} + 6*2*\Delta H_{H-O})  

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Read more here:

brainly.com/question/11753370?referrer=searchResults  

I hope it helps you!        

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So the reaction is:


<span>CH3CH2CH2CH2C ||| CCH3+2Br2 ---->


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.
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</span>
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