Question

The synthesis of triphenylmethanol via the grignard reaction is given by the following equation (below):

A student decides to perform the above reaction. She starts with 200 mg of bromobenzene (M.W. = 157) and uses 45 mg of Mg (M.W. = 24.3) and 0.36 g of benzophenone (M.W. = 182).

a. Calculate the limiting reagent

b. Calculate the theoretical yield of triphenylmethanol (M.W. = 260.3)

c. She obtains 0.172 grams of of the product. What is her actual yield?

d. Explain how the student could use IR to determine if she had obtained the desired product.

(By the way, here is the structure meant to be placed above) -

Answer 1

Concept and reason

The reagent that is used in lesser amounts than required is the limiting reagent. The theoretical yield is calculated from the limiting reagent.

Fundamentals

Balanced chemical reaction is a reaction in which the number of atoms of each kind of element is the same on both sides of the reaction.

The reagent that is lesser quantity than required is the limiting reagent.

Infrared (IR) spectroscopy helps to distinguish between two or more compounds. It is because the amount of light absorbed is related to the functional groups and the structure.

a.

The given reaction is as follows:

Mass of bromobenzene = $200{\rm{ mg}}$

Convert mg to g as follows:

$\begin{array}{l}\\200{\rm{ mg}} = 200{\rm{ mg}} \times \frac{{1{\rm{ g}}}}{{1000{\rm{ mg}}}}\\\\{\rm{ = 0}}{\rm{.2 g}}\\\end{array}$

Molar mass of bromobenzene = $157{\rm{ g/mol}}$

Calculate the moles of bromobenzene using the formula ${\rm{Moles}} = \frac{{{\rm{Mass}}}}{{{\rm{Molar mass}}}}$ as follows:

$\begin{array}{l}\\{\rm{Moles}} = \frac{{{\rm{0}}{\rm{.2 g}}}}{{{\rm{157 g}}}}\\\\{\rm{ = 0}}{\rm{.0013 mol}}\\\end{array}$

Mass of magnesium = $45{\rm{ mg}}$

Convert mg to g as follows:

$\begin{array}{l}\\45{\rm{ mg}} = 45{\rm{ mg}} \times \frac{{1{\rm{ g}}}}{{1000{\rm{ mg}}}}\\\\{\rm{ = 0}}{\rm{.045 g}}\\\end{array}$

Molar mass of magnesium = $24.3{\rm{ g/mol}}$

Calculate the moles of magnesium as follows:

$\begin{array}{l}\\{\rm{Moles}} = \frac{{{\rm{0}}{\rm{.045g}}}}{{{\rm{24}}{\rm{.3 g}}}}\\\\{\rm{ = 0}}{\rm{.0019 mol}}\\\end{array}$

Mass of benzophenone = $0.36{\rm{ g}}$

Molar mass of benzophenone = $182{\rm{ g/mol}}$

Calculate the moles of benzophenone as follows:

$\begin{array}{l}\\{\rm{Moles}} = \frac{{{\rm{0}}{\rm{.36 g}}}}{{{\rm{182 g}}}}\\\\{\rm{ = 0}}{\rm{.0019 mol}}\\\end{array}$

From the reaction, $1{\rm{ mol}}$ bromobenzene requires $1{\rm{ mol}}$ magnesium and $1{\rm{ mol}}$ benzophenone.

$0.0019{\rm{ mol}}$ magnesium and $0.0019{\rm{ mol}}$ benzophenone requires $0.0019{\rm{ mol}}$ bromobenzene.

But only $0.0013{\rm{ mol}}$ bromobenzene is present.

So, bromobenzene is the limiting reagent.

b)

The limiting reagent is bromobenzene.

From the reaction, $1{\rm{ mol}}$ bromobenzene produces $1{\rm{ mol}}$ triphenylmethanol.

$0.0013{\rm{ mol}}$ bromobenzene produces

$\frac{{\left( {0.0013{\rm{ mol bromobenzene}}} \right)\left( {1{\rm{ mol triphenylmethanol}}} \right)}}{{\left( {1{\rm{ mol bromobenzene}}} \right)}} = 0.0013{\rm{ mol triphenylmethanol}}$

Molar mass of triphenylmethanol = $260.3{\rm{ g/mol}}$

Convert moles to mass as follows:

$\begin{array}{l}\\{\rm{Moles}} = \frac{{{\rm{Mass}}}}{{{\rm{Molar mass}}}}\\\\0.0013{\rm{ mol}} = \frac{{{\rm{Mass}}}}{{260.3{\rm{ g/mol}}}}\\\\{\rm{Mass}} = \left( {0.0013{\rm{ mol}}} \right)\left( {260.3{\rm{ g/mol}}} \right)\\\\{\rm{ = 0}}{\rm{.34 g}}\\\end{array}$

c)

Mass of the product obtained = $0.172{\rm{ g}}$

Theoretical yield = $0.34{\rm{ g}}$

Substitute the values as calculate the theoretical yield as follows:

$\begin{array}{l}\\{\rm{Actual yield}} = \frac{{0.172{\rm{ g}}}}{{0.34{\rm{ g}}}} \times 100\\\\{\rm{ = 50}}{\rm{.6 \% }}\\\end{array}$

d.

The structure of the product is as follows:

A strong broad stretch peak at approximately $3500{\rm{ c}}{{\rm{m}}^{ - 1}}$ corresponds to the $- {\rm{OH}}$ stretching.

The absorption peaks at $3030 - 2930{\rm{ c}}{{\rm{m}}^{ - 1}}$ corresponds to $s{p^2}{\rm{ }} = {\rm{C}} - {\rm{H}}$ stretching.

The absorption peaks at $1600 - 1490{\rm{ c}}{{\rm{m}}^{ - 1}}$ , indicates the presence of benzene ring.

Ans: Part a

Therefore, bromobenzene is the limiting reagent.

Part b

Therefore, the theoretical yield of triphenylmethanol is $0.34{\rm{ g}}$ .

Part c

Therefore, the actual yield of triphenylmethanol is $50.6\%$ .

Part d

Therefore, the presence of peaks at approximately $3500{\rm{ c}}{{\rm{m}}^{ - 1}}$ , $3030 - 2930{\rm{ c}}{{\rm{m}}^{ - 1}}$ , and $1600 - 1490{\rm{ c}}{{\rm{m}}^{ - 1}}$ indicated that the obtained compound is the desired product.