Consider the nitration by electrophilic aromatic substitution of methyl benzoate

Question

Consider the nitration by electrophilic aromatic substitution of methyl benzoate to methyl m-nitrobenzoate.

A reaction was performed in which 2.65 mL of methyl benzoate was reacted with a mixture of concentrated nitric and sulfuric acids to make 3.19 g of methyl m-nitrobenzoate. Calculate the theoretical yield and percent yield for this reaction.

Consider the nitration by electrophilic aromatic substitution of methyl benzoate to methyl m-nitrobenzoate. A reaction was performed in which 2.65 mL of methyl benzoate was reacted with a mixture of concentrated nitric and sulfuric acids to make 3.19 g of methyl m-nitrobenzoate. Calculate the theoretical yield and percent yield for this reaction.

Explanation / Answer

Mass of methyl benzoate = Volume x density = 2.65 mL x 1.09 g/mL = 2.89 g

Given that in the chemical equation that,

1 mole of methyl benzoate gives 1 mole of methyl m-nitro benzoate

Moles of methyl benzoate = 2.89 g / 136.15 g/mol = 0.0212 mol

So, it means that 0.0212 mol of methyl benzoate gives 0.0212 mol of methyl m-nitrobenzoate

So, 0.0212 mol of methyl m-nitrobenzoate is to be converted to grams.

mass = 0.0212 mol x 181.14 g/mol = 3.84 g

So, theoretical yield = 3.84 g

Actual yield = 3.19 g

Percent yield = (3.19 / 3.84 ) x 100 = 83.1 %

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