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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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Answer : The enthalpy of the given reaction will be, -1048.6 kJ

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

The main reaction is:

XCl_4(s)+2H_2O(l)\rightarrow XO_2(s)+4HCl(g)    \Delta H=?

The intermediate balanced chemical reactions are:

(1) H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(g)     \Delta H_1=-241.8kJ

(2) X(s)+2Cl_2(g)\rightarrow XCl_4(s)    \Delta H_2=+461.9kJ

(3) \frac{1}{2}H_2(g)+\frac{1}{2}Cl_2(g)\rightarrow HCl(g)    \Delta H_3=-92.3kJ

(4) X(s)+O_2(g)\rightarrow XO_2(s)    \Delta H_4=-789.1kJ

(5) H_2O(g)\rightarrow H_2O(l)    \Delta H_5=-44.0kJ

Now reversing reaction 2, multiplying reaction 3 by 4, reversing reaction 1 and multiplying by 2, reversing reaction 5 and multiplying by 2 and then adding all the equations, we get :

(1) 2H_2O(g)\rightarrow 2H_2(g)+O_2(g)     \Delta H_1=2\times 241.8kJ=483.6kJ

(2) XCl_4(s)\rightarrow X(s)+2Cl_2(g)    \Delta H_2=-461.9kJ

(3) 2H_2(g)+2Cl_2(g)\rightarrow 4HCl(g)    \Delta H_3=4\times -92.3kJ=-369.2kJ

(4) X(s)+O_2(g)\rightarrow XO_2(s)    \Delta H_4=-789.1kJ

(5) 2H_2O(l)\rightarrow 2H_2O(g)    \Delta H_5=2\times 44.0kJ=88.0kJ

The expression for enthalpy of main reaction will be:

\Delta H=\Delta H_1+\Delta H_2+\Delta H_3+\Delta H_4+\Delta H_5

\Delta H=(483.6)+(-461.9)+(-369.2)+(-789.1)+(88.0)

\Delta H=-1048.6kJ

Therefore, the enthalpy of the given reaction will be, -1048.6 kJ

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