Home A-Level AQA Chemistry Aromatic Chemistry Paracetamol is a medicine commonly used to relieve mild pain
Paracetamol is a medicine commonly used to relieve mild pain - AQA - A-Level Chemistry - Question 8 - 2019 - Paper 2 Question 8
View full question Paracetamol is a medicine commonly used to relieve mild pain.
Traditionally, paracetamol has been made industrially in a three-step synthesis from phenol.
Step 1
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View marking scheme Worked Solution & Example Answer:Paracetamol is a medicine commonly used to relieve mild pain - AQA - A-Level Chemistry - Question 8 - 2019 - Paper 2
Name the mechanism of the reaction in Step 1. Only available for registered users.
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The mechanism of the reaction in Step 1 is electrophilic substitution.
Complete the equation for the reaction in Step 2. Only available for registered users.
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The equation for the reaction in Step 2 is:
C 6 H 4 ( O H ) + H N O 2 → C 6 H 4 ( N O 2 ) ( O H ) + H 2 O C_6H_4(OH) + HNO_2 \rightarrow C_6H_4(NO_2)(OH) + H_2O C 6 ​ H 4 ​ ( O H ) + H N O 2 ​ → C 6 ​ H 4 ​ ( N O 2 ​ ) ( O H ) + H 2 ​ O
Complete the mechanism for the reaction of 4-aminophenol with ethanoyl chloride. Only available for registered users.
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The mechanism for the reaction of 4-aminophenol with ethanoyl chloride is as follows:
The lone pair on the nitrogen of 4-aminophenol attacks the carbonyl carbon of ethanoyl chloride, forming a tetrahedral intermediate.
The intermediate collapses, releasing HCl and forming paracetamol.
The overall equation can be illustrated as follows:
C 6 H 4 ( N H 2 ) ( O H ) + C H 3 C O C l → C 6 H 4 ( C O N H 2 ) ( O H ) + H C l C_6H_4(NH_2)(OH) + CH_3COCl \rightarrow C_6H_4(CONH_2)(OH) + HCl C 6 ​ H 4 ​ ( N H 2 ​ ) ( O H ) + C H 3 ​ COCl → C 6 ​ H 4 ​ ( CON H 2 ​ ) ( O H ) + H Cl
Give one reason why ethanoyl chloride is not used in the industrial synthesis. Only available for registered users.
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One reason why ethanoyl chloride is not used in industrial synthesis is that it is corrosive and can release harmful fumes during the reaction.
Draw the structure of one of these other aromatic products. Only available for registered users.
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One other aromatic product that can be formed is 4-nitrophenol, represented structurally as:
OH
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C6H4-NO2
Complete the equation for this second step. Only available for registered users.
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The equation for the reaction of hydroquinone with ammonium ethanoate is:
C 6 H 4 ( O H ) + N H 4 C 2 H 3 O 2 → C 6 H 4 ( C O N H 2 ) + H 2 O C_6H_4(OH) + NH_4C_2H_3O_2 \rightarrow C_6H_4(CONH_2) + H_2O C 6 ​ H 4 ​ ( O H ) + N H 4 ​ C 2 ​ H 3 ​ O 2 ​ → C 6 ​ H 4 ​ ( CON H 2 ​ ) + H 2 ​ O
Calculate the mass, in kg, of hydroquinone (M_r = 110.0) needed to produce 250 kg of paracetamol. Only available for registered users.
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To calculate the mass of hydroquinone needed:
Calculate the moles of paracetamol:
e x t M o l e s o f p a r a c e t a m o l = 250 kg 0.15 kg/mol = 1666.67 moles ext{Moles of paracetamol} = \frac{250 \text{ kg}}{0.15 \text{ kg/mol}} = 1666.67 \text{ moles} e x t M o l eso f p a r a ce t am o l = 0.15 kg/mol 250 kg ​ = 1666.67 moles
You'll need the same amount of hydroquinone (1:1 molar ratio in the reaction). So, moles of hydroquinone = 1666.67 moles.
Calculate the mass of hydroquinone:
e x t M a s s o f h y d r o q u i n o n e = moles × M r = 1666.67 moles × 110.0 g/mol = 183333.7 g = 183.34 kg ext{Mass of hydroquinone} = \text{moles} \times M_r = 1666.67 \text{ moles} \times 110.0 \text{ g/mol} = 183333.7 \text{ g} = 183.34 \text{ kg} e x t M a sso f h y d ro q u in o n e = moles × M r ​ = 1666.67 moles × 110.0 g/mol = 183333.7 g = 183.34 kg
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