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Chapter 15: Problem 42

Design a synthesis of benzene starting from inorganic compounds and organic compounds containing only hydrogen and no more than four carbon atoms.

Short Answer

Expert verified
Start from methane, convert to acetylene, and then trimerize acetylene to form benzene.

Step by step solution

01

Identify the building blocks

Benzene is a hydrocarbon with the molecular formula C6H6. First, identify suitable starting materials that contain carbon and hydrogen. Methane (CH4) and acetylene (C2H2) are good candidates as they fit the criteria of having no more than four carbon atoms each.
02

Generate acetylene

Methane can be converted to acetylene. First, produce acetylene by partial oxidation of methane using oxygen (O2): CH4 + O2 → C2H2 + H2O.
03

Formation of Benzene from Acetylene

Couple three molecules of acetylene to form benzene. The reaction involved is: 3 C2H2 → C6H6 This trimerization utilizes a catalyst such as iron or nickel under high temperature.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

benzene synthesis
Benzene is a simple aromatic hydrocarbon with a molecular formula of C6H6. It is widely used in the chemical industry as a starting material for various compounds. Synthesizing benzene from simpler organic molecules like methane and acetylene involves a couple of key reactions:

acetylene trimerization
Trimerization of acetylene is a key reaction in organic synthesis, where three acetylene (C2H2) molecules join together to form a single benzene (C6H6) molecule. This reaction is a type of cyclization that creates an aromatic ring:

methane partial oxidation
Methane partial oxidation to produce acetylene is a significant step in the synthesis of various organic compounds, including benzene. Partial oxidation is a controlled process where methane is partially converted to produce specific products instead of complete combustion to carbon dioxide and water:

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Most popular questions from this chapter

At low temperature, only a single product (A) is formed from the treatment of \(p\)-xylene with isopropyl chloride and \(\mathrm{AlCl}_{3}\), which is surely no surprise, as there is only one position open on the ring. However, at higher temperature some sort of rearrangement occurs, as two products are formed in roughly equal amounts. First, write a mechanism for formation of both products. Hint: Note that \(\mathrm{HCl}\) is a product of the first reaction to give A. Second, explain why product A rearranges to product \(\mathbf{B}\). What is the thermodynamic driving force for this reaction?

Explain why cyanobenzene (benzonitrile, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CN}\) ) substitutes preferentially in the meta position.

Provide syntheses for the following compounds, as free of other isomers as possible. You do not need to write mechanisms, and you may start from benzene, inorganic reagents of your choice, and organic reagents containing no more than three carbons.

When 1 -fluorobutane is dissolved in superacid at low temperature, it is possible to observe a carbocation. However, the \({ }^{13} \mathrm{C}\) NMR spectrum of that carbocation shows only two signals. (a) How many signals did you expect to see in the spectrum? (b) What carbocation do you think is observed? Although part (a) might be easy for you, part (b) actually poses a vexing problem. Can you write a mechanism for formation of the carbocation that shows only two signals in its \({ }^{13}\) C NMR spectrum? What problems appear as you try to write the mechanism?

In the presence of very strong acid, phenol is brominated in the meta position. Why do you suppose this is the case? When phenol is brominated in slightly basic conditions, tribromophenol is the major product. Why would basic conditions favor polybromination?
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