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ryzh [129]
3 years ago
12

Two steps in a synthesis of the analgesic ibuprofen include a carbonyl condensation reaction, followed by an alkylation reaction

. identify intermediates a and b in the synthesis of ibuprofen.

Chemistry
2 answers:
svetlana [45]3 years ago
5 0

The intermediate a is \boxed{{\text{diethyl 2 - }}\left( {{\text{4 - isobutylphenyl}}} \right){\text{malonate}}}.

The intermediate b is \boxed{{\text{diethyl 2 - }}\left( {{\text{4 - isobutylphenyl}}} \right){\text{ - 2 - methylmalonate}}}.

Further explanation:

Carbonyl condensation reaction is the reaction between two carbonyl-containing reactants, one of which must possess an alpha hydrogen atom. The reaction involves the removal of an alpha hydrogen atom by a base. The enolate anion formed from this removal attacks the carbonyl carbon of the second molecule.The result is the formation of a new carbon-carbon bond between the carbonyl carbon atom of one molecule and the alpha-carbon atom of another carbonyl molecule.

Alkylation is defined as the transfer of an alkyl group from one molecule to another. The alkyl group can be transferred in the form of alkyl carbocation, a free radical, a carbanion or a carbene. An alkyl group can be added to benzene molecule by an electrophile aromatic substitution reaction called the Friedel‐Crafts alkylation reaction.

Ibuprofen is the derivative of propionic acid in which one hydrogen atom at position 2 is substituted by a 4-(2-methylpropyl) phenyl group.

Synthesis of ibuprofen form carbonyl condensation and alkylation reaction is as follows:

1. Ibuprofen is synthesized by reacting ethyl 2-(4-isobutylphenyl) acetate with base and diethyl carbonate. In this, the base reacts with the acidic proton to form enolate anion.The enolate anion reacts with diethyl carbonate to generate an intermediate diethyl 2-(4-isobutylphenyl) malonate. This step is the carbonyl condensation reaction.

2. The intermediate 2-(4-isobutylphenyl) malonate reacts with base again and forms enolate. The enolate with treatment with methyl iodide produces an intermediate diethyl 2-(4-isobutylphenyl)-2-methylmalonatediethyl. This step is an alkylation reaction.

3. The second intermediate diethyl 2-(4-isobutylphenyl)-2-methylmalonate on hydrolysis give ibuprofen.

Hence, the intermediate a is diethyl 2-(4-isobutylphenyl) malonate and the intermediate b is diethyl 2-(4-isobutylphenyl)-2-methylmalonate. The synthesis of ibuprofen is shown in the image attached.

Learn more:

1. Calculate number of solutes brainly.com/question/8054051.

2. How many moles of Cl are there in 8 moles of CCl4 brainly.com/question/2094744

Answer details:

Grade: Senior school

Subject: Chemistry

Chapter: Chemical reactions

Keywords: Carbonyl condensation, Alkylation, Friedel‐Crafts alkylation reaction, Ibuprofen, ethyl2-(4-isobutylphenyl) acetate, diethyl 2-(4-isobutylphenyl) malonate and diethyl 2-(4-isobutylphenyl)-2-methylmalonate.

Tresset [83]3 years ago
4 0
Ibuprofen is synthesized by reacting ethyl 2-(4-isobutylphenyl)acetate with base, the base abstracts the acidic proton and enolate is formed which on reaction with diethyl carbonate generates diethyl 2-(4-isobutylphenyl)malonate (A). diethyl 2-(4-isobutylphenyl)malonate on treatment with Base again looses the acidic proton and forms enolate. The enolate with treatment with Methyl Iodide yields diethyl 2-(4-isobutylphenyl)-2-methylmalonate (B). diethyl 2-(4-isobutylphenyl)-2-methylmalonate on hydrolysis give Ibuprofen.

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Answer : The correct answer for the Theoretical Yield is 48.93 g of product .

Theoretical yield : It is amount of product produced by limiting reagent . It is smallest product yield of product formed .

Following are the steps to find theoretical yield .

Step 1) : Write a balanced reaction between Al and Cl₂ .

2 Al + 3 Cl₂→ 2 AlCl₃

Step 2: To find amount of product (AlCl₃) formed by Al .

Following are the sub steps to calculate amount of AlCl₃ formed :

a) To calculate mole of Al :

Given : Mass of Al = 35.5 g

Mole can be calculate by following formula :

Mole = \frac{given mass (g)}{atomic mass \frac{g}{mol}}

Mole = \frac{35.5 g }{26.9 \frac{g}{mol}}

Mole = 1.32 mol

b) To find mole ratio of AlCl₃ : Al

Mole ratio is calculated from balanced reaction .

Mole of Al in balanced reaction = 2

Mole of AlCl₃ in balanced reaction = 2.

Hence mole ratio of AlC; l₃ : Al = 2:2

c) To find mole of AlCl₃ formed :

Mole of AlCl_3 = Mole of Al * Mole ratio

Mole of AlCl_3 = 1.32 mol of Al * \frac{2}{2}

Mole of AlCl₃ = 1.32 mol

d) To find mass of AlCl₃

Molar mass of AlCl₃ = 133.34 \frac{g}{mol}

Mass of AlCl3 can be calculated using mole formula as:

1.32 mol of AlCl_3 = \frac{ mass (g)}{133.34 \frac{g}{mol}}

Multiplying both side by 133.34 \frac{g}{mol}

1.32 mole  * 133.34\frac{g}{mol} = \frac{mass (g)}{133.34\frac{g}{mol}} *133.34  \frac{g}{mol}

Mass of AlCl₃ = 176.00 g

Hence mass of AlCl₃ produced by Al is 176.00 g

Step 3) To find mass of product (AlCl₃) formed by Cl₂ :

Same steps will be followed to calculate mass of AlCl₃

a) Find mole of Cl₂

Mole of Cl_2 = \frac{39.0 g}{70.9\frac{g}{mol}}

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b) Mole ratio of Cl₂ : AlCl₃

Mole of Cl₂ in balanced reaction = 3

Mole of AlCl₃ in balanced reaction = 2

Hence mole ratio of AlCl₃ : Cl₂ = 2 : 3

c) To find mole of AlCl₃

Mole of AlCl_3 = mole of Cl_2 * mole ratio

Mole of AlCl_3 = 0.55  mole  * \frac{2}{3}

Mole of AlCl3 = 0.367 mol

d) To find mass of AlCl₃ :

0.367 mol of AlCl_3 = \frac{mass (g) }{133.34 \frac{g}{mol}}

Multiplying both side by

133.34 \frac{g}{mol}

0.367 mol of AlCl_3 * 133.34 \frac{g}{mol}  = \frac{mass(g)}{133.34\frac{g}{mol}}   * 133.34 \frac{g}{mol}

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Step 4) To identify limiting reagent and theoretical yield :

Limiting reagent is the reactant which is totally consumed when the reaction is complete . It is identified as the reactant which produces least yield or theoretical yield of product .

The product AlCl₃ formed by Al = 176.00 g

The product AlCl₃ formed by Cl₂ = 48.93 g

Since Cl₂ is producing less amount of product hence it is limiting reagent and 48.93 g will be considered as Theoretical yield .

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