Outline a synthesis of cumene from cheap, readily available hydrocarbons. Then synthesize phenol from cumene.

Short Answer

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To synthesize cumene from cheap, readily available hydrocarbons, we start with benzene and propylene and perform the Friedel-Crafts alkylation reaction in the presence of a Lewis acid catalyst, aluminum chloride (AlCl3). This forms cumene. Then, we synthesize phenol from cumene by first converting cumene to cumene hydroperoxide through oxidation using molecular oxygen, and then performing an acid-catalyzed cleavage using sulfuric acid, yielding phenol and acetone.

Step by step solution

01

Formation of Cumene from Benzene and Propylene

In the first step, we perform the Friedel-Crafts alkylation reaction to obtain cumene by combining benzene and propylene in the presence of a Lewis acid catalyst such as aluminum chloride (AlCl3). Benzene (\(C_6H_6\)) + Propylene (\(C_3H_6\)) + AlCl3 → Cumene (\(C_9H_{12}\)) + HCl + AlCl3 This reaction represents the synthesis of cumene by Friedel-Crafts alkylation, in which the alkyl group of propylene is attached to the benzene ring. The Lewis acid catalyst, AlCl3, helps generate carbocation from propylene that attacks the benzene ring. The chloride ion (Cl-) reacts with the hydrogen in the benzene's ring and forms HCl.
02

Oxidation of Cumene to Cumene Hydroperoxide

To obtain phenol from cumene, we need to convert cumene into cumene hydroperoxide. The oxidation of cumene is performed by bubbling air (O2) through the solution of cumene. Cumene (\(C_9H_{12}\)) + O2 → Cumene Hydroperoxide (\(C_9H_{12}O_2\)) During this reaction, molecular oxygen is added across the benzylic C-H bond of cumene, and cumene hydroperoxide is formed. This step is a radical reaction and requires a radical initiator like benzoyl peroxide.
03

Acid-Catalyzed Cleavage of Cumene Hydroperoxide

Now, we perform the acid-catalyzed cleavage of cumene hydroperoxide to obtain phenol and acetone. The reaction takes place in the presence of a strong acid, such as sulfuric acid (H2SO4). Cumene Hydroperoxide (\(C_9H_{12}O_2\)) + H2SO4 → Phenol (\(C_6H_5OH\)) + Acetone (\(CH_3COCH_3\)) + H2SO4 Under acidic conditions, cumene hydroperoxide rearranges to form phenol and acetone products. This step is an important industrial process since both phenol and acetone are valuable chemicals in the chemical industry. In summary, we have successfully outlined the synthesis of cumene from cheap, readily available hydrocarbons (benzene and propylene) through Friedel-Crafts alkylation. We then synthesized phenol from cumene by forming cumene hydroperoxide through oxidation, followed by an acid-catalyzed cleavage to yield phenol and acetone products.

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Most popular questions from this chapter

(a) Why is phenol \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\) a stronger acid than an alcohol? (b) Would para-acetylphenol be a stronger or weaker acid than phenol? (c) Would 2, 4 -dinitrophenol be a stronger or weaker acid than \(p\) -nitrophenol? Draw contributing structures to support the answers.

Ortho-nitrophenol L and para-nitrophenol L \(\mathrm{N}_{2}\) have different acidities, melting points, and boiling points. In all cases, the ortho- nitrophenol has lower values in these areas. Why?

Give structures of all compounds below: (a) \(p\) -nitrophenol \(+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}+\mathrm{NaOH}(\mathrm{aq}) \rightarrow \mathrm{A}\left(\mathrm{C}_{8} \mathrm{H}_{9} \mathrm{O}_{3} \mathrm{~N}\right)\) \(\mathrm{A}+\mathrm{Sn}+\mathrm{HCL} \rightarrow \mathrm{B}\left(\mathrm{C}_{8} \mathrm{~B}_{11} \mathrm{ON}\right)\) \(\mathrm{B}+\mathrm{NaNO}_{2}+\mathrm{HCL}\), then phenol \(\rightarrow \mathrm{C}\left(\mathrm{C}_{14} \mathrm{H}_{14} \mathrm{O}_{2} \mathrm{~N}_{2}\right)\) \(\mathrm{C}+\) ethyl sulfate \(+\mathrm{NaOH}\) (aq) \(\rightarrow \mathrm{D}\left(\mathrm{C}_{16} \mathrm{H}_{18} \mathrm{O}_{2} \mathrm{~N}_{2}\right)\) \(\mathrm{D}+\mathrm{SnCL}_{2} \rightarrow \mathrm{E}\left(\mathrm{C}_{8} \mathrm{H}_{11} \mathrm{ON}\right)\) \(E+\) acetyl chloride phenacetin \(\left(\mathrm{C}_{10} \mathrm{H}_{13} 2 \mathrm{~N}\right)\), an analgesic ("pain-killer") and antypyretic ("fever- killer") (b) \(\beta-(-0-\) hy droxyphenyl \()\) ethyl alcohol \(+\mathrm{HBr} \rightarrow \mathrm{F}\left(\mathrm{C}_{8} \mathrm{H}_{9} \mathrm{OBr}\right)\) \(\mathrm{F}+\mathrm{KOH} \rightarrow\) coumarane \(\left(\mathrm{C}_{8} \mathrm{H}_{8} \mathrm{O}\right)\), insoluble in \(\mathrm{NaOH}\).

How much difference in physical properties would you expect for o-and \(p\) -cyanophenol isomers? Explain.

Write structures for each of the following names: (a) m-cresol (b) 3 -hydroxybenzenesulfonamide (c) 3 -chloro-1, 2 -benzoquinone (d) o-methoxyphenol (e) \(\mathrm{p}(\mathrm{t}-\) tolyl \()\) azophenol (f) benzyl pheny 1 ether (g) \(3-\) (o-hydroxy phenyl pentanoic acid) (h) 2 -methoxy-1, 4 -naphthoquinone

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