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.
ScanSolve AI Answer
Step 1: Formation of the acylium ion. The acyl halide (e.g., acetyl chloride, CH_3COCl) reacts with the Lewis acid catalyst (AlCl_3) to generate the electrophilic acylium ion. CH_3COCl + AlCl_3 CH_3CO^+ + AlCl_4^- Step 2: Electrophilic attack on the aromatic ring. Toluene, an activated aromatic ring due to the electron-donating methyl group, undergoes electrophilic attack by the acylium ion. The methyl group is an ortho/para-director, so the attack occurs primarily at the ortho and para positions. The para attack is shown below, leading to a resonance-stabilized carbocation intermediate (sigma complex). C_6H_5CH_3 + CH_3CO^+ C_6H_4(CH_3)(COCH_3)H^+ (This step involves the electrons of the ring attacking the acylium ion, forming a new C-C bond and a carbocation on the ring.) Step 3: Deprotonation and re-aromatization. A base (typically AlCl_4^-) abstracts a proton from the carbon that was attacked, restoring aromaticity to the ring and regenerating the AlCl_3 catalyst. C_6H_4(CH_3)(COCH_3)H^+ + AlCl_4^- C_6H_4(CH_3)(COCH_3) + HCl + AlCl_3 Product Analysis: The methyl group (-CH_3) on toluene is an activating group and an ortho/para-director. This means it increases the electron density of the aromatic ring, making it more reactive towards electrophiles, and directs the incoming acyl group (-COCH_3) to the ortho and para positions relative to itself. Therefore, Friedel-Crafts acylation of toluene with acetyl chloride will yield a mixture of: • ortho-methylacetophenone (1-(2-methylphenyl)ethan-1-one) • para-methylacetophenone (1-(4-methylphenyl)ethan-1-one) The para isomer is typically the major product due to less steric hindrance between the methyl group and the incoming acyl group compared to the ortho position. The major product is: 4-methylacetophenone That's 2 down. 3 left today — send the next one.