객관식
Which two stereoisomers are formed as products when 4-tert-butylcyclohexanone is reduced with a hydride reagent such as or ?
116
views
3. Acids and Bases
Organic Chemistry Reactions
객관식
Which of the following aromatic rings is alkylated most rapidly by chloroethane and in a Friedel-Crafts alkylation reaction?
A
Nitrobenzene ()
B
Benzene ()
C
Anisole ()
D
Chlorobenzene ()
검증된 단계별 안내1
Identify the nature of the substituents on each aromatic ring and classify them as either electron-donating or electron-withdrawing groups. This affects the ring's reactivity in electrophilic aromatic substitution reactions like Friedel-Crafts alkylation.
Recall that Friedel-Crafts alkylation proceeds faster on rings that are activated by electron-donating groups because these groups increase the electron density on the aromatic ring, making it more nucleophilic and more reactive toward the electrophile generated by chloroethane and AlCl\_3.
Analyze each substituent: Nitro group (NO\_2) is a strong electron-withdrawing group, which deactivates the ring and slows down alkylation; the methoxy group (OCH\_3) in anisole is a strong electron-donating group, activating the ring; chlorine (Cl) is electron-withdrawing by induction but donates electrons by resonance, making chlorobenzene moderately deactivated; benzene has no substituents and serves as the baseline reactivity.
Rank the rings in order of increasing reactivity toward Friedel-Crafts alkylation based on the substituent effects: nitrobenzene (least reactive), chlorobenzene, benzene, and anisole (most reactive).
Conclude that anisole will be alkylated most rapidly because its methoxy substituent strongly activates the aromatic ring toward electrophilic substitution by increasing electron density and stabilizing the intermediate sigma complex.
관련 영상
관련 실천
