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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Step 1: Mechanism for 2.1 - Diazonium Coupling Reaction This is an electrophilic aromatic substitution reaction where the benzenediazonium ion acts as an electrophile and N,N-dimethylaniline acts as a highly activated nucleophile. The dimethylamino group is a strong activating and ortho/para-directing group. The para position is attacked due to less steric hindrance. &The lone pair on the nitrogen of N,N-dimethylaniline activates the ring. \\ &The electrons from the para position attack the terminal nitrogen of the benzenediazonium ion. \\ &This forms a -complex (arenium ion). \\ &A proton is removed from the para position to restore aromaticity. \\ &The final product is p-(dimethylamino)azobenzene. \\ [width=0.9]2.1_mechanism.png Step 2: Mechanism for 2.2a - Benzyne Reaction The reaction conditions (NaNH_2 in NH_3) are characteristic of a benzyne (elimination-addition) mechanism for aryl halides. This mechanism requires an ortho hydrogen atom relative to the leaving group (bromine) for the initial deprotonation step. Upon examining the given reactant, 2-bromo-1,3-dimethylbenzene, the bromine atom is at position 2. The ortho positions (positions 1 and 3) are both substituted with methyl groups (CH_3), meaning there are no hydrogen atoms available at these positions for abstraction by the strong base (NH_2^-). Therefore, the benzyne mechanism is not possible for the specific reactant shown. The reaction as depicted (direct substitution of bromine by an amino group at the same position) cannot proceed via the expected benzyne pathway under these conditions. Step 3: Mechanism for 2.2b - Nucleophilic Aromatic Substitution (SNAr) This is a nucleophilic aromatic substitution reaction. The chlorine atom is replaced by a methoxy group. The presence of the strong electron-withdrawing nitro (NO_2) group para to the leaving group (Cl) and the carboxylic acid (COOH) group ortho to the leaving group activates the ring towards SNAr. &The methoxide ion (MeO^-) acts as a nucleophile and attacks the carbon bearing the chlorine atom. \\ &This forms a Meisenheimer complex (an anionic -complex), where the negative charge is delocalized \\ &into the electron-withdrawing nitro group. \\ &The chloride ion (Cl^-) is then expelled, restoring aromaticity and forming the final product. \\ [width=0.9]2.2b_mechanism.png Step 4: Mechanism for 2.2c - Diazotization and Hydrolysis This reaction proceeds in two main parts: Part 1: Diazotization (formation of the arenediazonium salt) Nitrous acid (HNO_2) is generated in situ from NaNO_2 and HCl. HNO_2 is then protonated and loses water to form the electrophilic nitrosonium ion (NO^+). The nitrogen of 3-methylaniline attacks the nitrosonium ion, followed by a series of proton transfers and tautomerization, leading to the formation of the 3-methylbenzenediazonium ion. Part 2: Hydrolysis (replacement of the diazonium group with a hydroxyl group) When the arenediazonium ion is warmed in the presence of water, the diazonium group (N_2^+) departs as stable nitrogen gas (N_2), forming an aryl cation. Water then acts as a nucleophile, attacking the aryl cation. Subsequent deprotonation yields the 3-methylphenol product. &Part 1: Diazotization \\ &1.\ Formation of nitrosonium ion: \\ & NaNO_2 + HCl HNO_2 + NaCl \\ & HNO_2 + H^+ H_2NO_2^+ \\ & H_2NO_2^+ NO^+ + H_2O \\ &2.\ Attack of aniline on nitrosonium ion: \\ [width=0.9]2.2c_diazotization_part1.png &3.\ Proton transfers and tautomerization to form diazonium ion: \\ [width=0.9]2.2c_diazotization_part2.png &Part 2: Hydrolysis \\ &1.\ Loss of nitrogen gas to form aryl cation: \\ [width=0.9]2.2c_hydrolysis_part1.png &2.\ Nucleophilic attack by water and deprotonation: \\ [width=0.9]2.2c_hydrolysis_part2.png What's next? Send 'em! 📸