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1 year ago

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25. 22 the reaction of phenylmagnesium bromide with methyl benzoate followed by acidification produces triphenylcarbinol. what a

re the sources of phenyl groups in triphenylcarbinol?

1 answer:

ExtremeBDS [4]1 year ago

3 0

Triphenylcarbinol( or triphenylmethanol) gets its phenyl groups from phenylmagnesium bromide.

Synthesis of triphenylcarbinol:

A typical laboratory experiment for demonstrating the Grignard reaction is the synthesis of triphenylmethanol using methyl benzoate, benzophenone, and phenylmagnesium bromide. Diethyl carbonate is a possible starting material.

Characteristics:

An organic substance is a triphenylmethanol. It is a white, crystalline substance that dissolves well in alcohol, diethyl ether, and benzene but not in water or petroleum ether. Due to the creation of a persistent "trityl" carbocation, it generates a bright yellow hue in very acidic solutions. Important dyes include several triphenylmethanol derivatives.
Triphenylmethanol has a core tetrahedral carbon atom, three phenyls (Ph) rings, and an alcohol group attached to it. With lengths of around 1.47Å for all three C-Ph bonds and 1.42Å for the C-O link, these connections are characteristic of sp3-sp2 carbon-carbon bonds.
Three neighboring phenyl groups give off specific qualities that are reflected in the alcohol's reactivity.

Learn more about triphenylmethanol here:

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<u>Answer:</u> The new pressure for oxygen gas is 6 atm.

<u>Explanation:</u>

To calculate the new pressure of the gas, we use the equation given by Boyle's Law.

This law states that the pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.

Mathematically,

$P\propto \frac{1}{V}$

or,

$P_1V_1=P_2V_2$

where,

$P_1\text{ and }V_1$ are the initial pressure and volume of the gas.

$P_2\text{ and }V_2$ are the final pressure and volume of the gas.

We are given:

$P_1=8atm\\V_1=36L\\P_2=?atm\\V_2=48L$

Putting values in above equation, we get:

$8atm\times 36L=P_2\times 48L\\\\P_2=6atm$

Hence, the new pressure for oxygen gas is 6 atm.

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Answer:

The enthalpy change for the given reaction is 424 kJ.

Explanation:

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We have :

Enthalpy changes of formation of following s:

$\Delta H_{f,Cu(NO_3)_2}=-302.9 kJ/mol$

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$\Delta H_{f,NO_2}= 33.2 kJ/mol$

$\Delta H_{f,O_2}= 0 kJ/mol$ (standard state)

$\Delta H_{rxn}=\sum [\Delta H_f(product)]-\sum [\Delta H_f(reactant)]$

The equation for the enthalpy change of the given reaction is:

$\Delta H_{rxn}$ =

$=(2 mol\times \Delta H_{f,CuO}+4\times \Delta H_{f,NO_2}+1 mol\times \Delta H_{f,O_2})-(2mol\times \Delta H_{f,Cu(NO_3)_2})$

$\Delta H_{rxn}$ =

$(2mol\times (-157.3 kJ/mol)+4\times 33.2 kJ/mol=1 mol\times 0 kJ/mol)-(2 mol\times (-302.9 kJ/mol)$

$\Delta H_{rxn}=424 kJ$

The enthalpy change for the given reaction is 424 kJ.

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Answer:

$V_2=18 \ L \ H_2$

General Formulas and Concepts:

<u>Chemistry - Gas Laws</u>

STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K
Charles' Law: $\frac{V_1}{T_1} =\frac{V_2}{T_2}$

Explanation:

<u>Step 1: Define</u>

Initial Volume: 5.0 L H₂ gas

Initial Temp: 273 K

Final Temp: 985 K

Final Volume: ?

<u>Step 2: Solve for new volume</u>

Substitute: $\frac{5 \ L \ H_2}{273 \ K} =\frac{x \ L \ H_2}{985 \ K}$
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Multiply: $4925 \ L \ H_2 \cdot K = 273x \ L \ H_2 \cdot K$
Isolate <em>x</em>: $18.0403 \ L \ H_2 = x$
Rewrite: $x=18.0403 \ L \ H_2$

<u>Step 3: Check</u>

<em>We are given 2 sig figs as the smallest. Follow sig fig rules and round.</em>

<em /> $18.0403 \ L \ H_2 \approx 18 \ L \ H_2$ <em />

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