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Chapter 18: 18-31P (page 677)

What steps are needed to convert benzene into p-isobutyl acetophenone, a synthetic intermediate used in the synthesis of the anti-inflammatory agent ibuprofen.

Short Answer

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Steps in the synthesis of Ibuprofen

Step by step solution

01

Clemmensen reduction

Clemmensen reduction is the reduction of carbonyl groups in aldehydes and ketones to methylene group.

The reaction is done with zinc amalgam and hydrochloric acid.

02

Steps in converting benzene into p-isobutyl acetophenone

Benzene is converted into p-isobutyl acetophenone by Friedel-craft's acylation followed by Clemmensen reduction (Zn (Hg) + HCl).

The reduced product is again subjected to Friedel-craft's acylation to get p-isobutyl acetophenone.

03

Steps in converting p-isobutyl acetophenone to ibuprofen

  • Sulfur ylide reacts with carbonyl compound and forms epoxide.
  • Epoxide ring opening is followed by hydride transfer to form aldehyde.
  • Aldehyde undergoes oxidation to form ibuprofen.

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Most popular questions from this chapter

Label each compound as more or less reactive than benzene in electrophilic aromatic substitution.

Question: Explain the following observation. Ethyl 3-phenylpropanoate (C6H5CH2CH2CO2CH2CH3) reacts with electrophiles to afford ortho- and para-disubstituted arenes, but ethyl 3-phenylprop-2-enoate (C6H5=CHCO2CH2CH3) reacts with electrophiles to afford meta-disubstituted arenes.

a. Draw resonance structures for the carbocation formed after loss of a leaving group from phytyl diphosphate.

b. Draw the two-step mechanism for Friedel鈥揅rafts alkylation of 1,2-dihydroxynaphthoic acid with this carbocation to form X.

Question: What is the major product of electrophilic addition of HBr to the following alkene? Explain your choice.

Benzyl bromide\({\bf{(}}{{\bf{C}}_{\bf{6}}}{{\bf{H}}_{\bf{5}}}{\bf{C}}{{\bf{H}}_{\bf{2}}}{\bf{Br)}}\) reacts rapidly with \({\bf{C}}{{\bf{H}}_{\bf{3}}}{\bf{OH}}\)to afford benzyl methyl ether\({\bf{(}}{{\bf{C}}_{\bf{6}}}{{\bf{H}}_{\bf{5}}}{\bf{C}}{{\bf{H}}_{\bf{2}}}{\bf{OC}}{{\bf{H}}_{\bf{3}}}{\bf{)}}\). Draw a stepwise mechanism for the reaction, and explain why this 1掳 alkyl halide reacts rapidly with a weak nucleophile under conditions that favor an\({{\bf{S}}_{\bf{N}}}{\bf{1}}\) mechanism. Would you expect the para-substituted benzylic halides \({\bf{C}}{{\bf{H}}_{\bf{3}}}{\bf{O}}{{\bf{C}}_{\bf{6}}}{{\bf{H}}_{\bf{4}}}{\bf{C}}{{\bf{H}}_{\bf{2}}}{\bf{Br}}\) and \({{\bf{O}}_{\bf{2}}}{\bf{N}}{{\bf{C}}_{\bf{6}}}{{\bf{H}}_{\bf{4}}}{\bf{C}}{{\bf{H}}_{\bf{2}}}{\bf{Br}}\) to each be more or less reactive than benzyl bromide in this reaction?Explain your reasoning.
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