This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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Here are the mechanisms for the reactions you requested: 1. Friedel-Crafts Alkylation This reaction involves the alkylation of an aromatic ring with an alkyl halide using a strong Lewis acid catalyst, typically AlCl_3. Step 1: Formation of the carbocation electrophile. The alkyl halide reacts with the Lewis acid to form a carbocation. R-Cl + AlCl_3 R^+ + AlCl_4^- (Note: Carbocation rearrangements can occur if a more stable carbocation can be formed.) Step 2: Electrophilic attack on the benzene ring. The carbocation attacks the electron-rich benzene ring, forming a sigma complex (arenium ion), which temporarily disrupts the aromaticity. C_6H_6 + R^+ H \\ R C_6H_5^+ (The positive charge is delocalized over the ring carbons ortho and para to the point of attack.) Step 3: Deprotonation and restoration of aromaticity. A base (e.g., AlCl_4^-) removes a proton from the sigma complex, restoring the aromaticity of the ring and regenerating the AlCl_3 catalyst. H \\ R C_6H_5^+ + AlCl_4^- C_6H_5R + HCl + AlCl_3 2. Friedel-Crafts Acylation This reaction involves the acylation of an aromatic ring with an acyl halide (or anhydride) using a strong Lewis acid catalyst, typically AlCl_3. Step 1: Formation of the acylium ion electrophile. The acyl halide reacts with the Lewis acid to form a resonance-stabilized acylium ion. R-CO-Cl + AlCl_3 R-C^+ + AlCl_4^- Step 2: Electrophilic attack on the benzene ring. The acylium ion attacks the electron-rich benzene ring, forming a sigma complex (arenium ion). C_6H_6 + R-C^+ H \\ CO-R C_6H_5^+ Step 3: Deprotonation and restoration of aromaticity. A base (e.g., AlCl_4^-) removes a proton from the sigma complex, restoring the aromaticity of the ring. H \\ CO-R C_6H_5^+ + AlCl_4^- C_6H_5CO-R · AlCl_3 + HCl (The ketone product forms a complex with AlCl_3.) Step 4: Hydrolysis. Aqueous workup is required to release the ketone product from its complex with AlCl_3. C_6H_5CO-R · AlCl_3 + H_2O C_6H_5CO-R + Al(OH)_3 + HCl 3. Nitration of Benzene This reaction involves the introduction of a nitro group (-NO_2) onto an aromatic ring using a mixture of concentrated nitric acid and sulfuric acid. Step 1: Formation of the nitronium ion electrophile. Nitric acid is protonated by sulfuric acid, and then water is eliminated to form the nitronium ion (NO_2^+), which is the active electrophile. HNO_3 + H_2SO_4 H_2NO_3^+ + HSO_4^- H_2NO_3^+ NO_2^+ + H_2O H_2O + H_2SO_4 H_3O^+ + HSO_4^- Overall: HNO_3 + 2H_2SO_4 NO_2^+ + H_3O^+ + 2HSO_4^- Step 2: Electrophilic attack on the benzene ring. The nitronium ion attacks the electron-rich benzene ring, forming a sigma complex (arenium ion). C_6H_6 + NO_2^+ H \\ NO_2 C_6H_5^+ Step 3: Deprotonation and restoration of aromaticity. A base (e.g., HSO_4^- or H_2O) removes a proton from the sigma complex, restoring the aromaticity of the ring. H \\ NO_2 C_6H_5^+ + HSO_4^- C_6H_5NO_2 + H_2SO_4 3 done, 2 left today. You're making progress.