Aromatic, Nucleophilic Substitution Flashcards
(21 cards)
What effect do deactivating groups have on nucleophilic aromatic substitution (NAS) reactivity?
Deactivating groups (electron-withdrawing groups) increase the reactivity of a benzene ring towards nucleophilic aromatic substitution. They stabilize the negative charge formed during the intermediate stage.
What is the trend in reactivity toward nucleophilic substitution from example (a) to (d)?
The reactivity increases from (a) to (d):
a) No EWG: very high temp & pressure (350°C, 150 bar)
b) 1 NO₂ group: lower temp needed (160°C)
c) 2 NO₂ groups: even lower temp (130°C)
d) 3 NO₂ groups: can proceed just by warming in H₂O
Which groups are ortho/para directors and generally activating?
Groups like –NH₂, –OH, –OR, –NHCOR, –R (alkyl), and halogens (X) are ortho/para directors. Most are activating, but halogens are deactivating despite being ortho/para directing.
Which groups are meta directors and generally deactivating?
Meta directors include –NO₂, –CN, –COOH, –CHO, –SO₃H, –CO₂R, –COR, and –NR₃⁺. These are all electron-withdrawing and deactivate the ring toward electrophilic substitution but activate it toward nucleophilic substitution.
Why are ortho and para positions activated in aniline?
The NH₂ group is electron-donating, which increases electron density at the ortho and para positions, making them favorable for electrophilic attack (but not for nucleophiles).
Why are ortho/para positions unfavorable for nucleophilic substitution in aniline?
Due to the high electron density (negative charge), nucleophiles are repelled. Hence, electron-donating groups deactivate the ring toward nucleophilic substitution, and meta positions are favored instead.
Why is the nitro group considered a meta director?
The nitro group is strongly electron-withdrawing. It pulls electron density away from the ring, making the ortho and para positions relatively positive and thus attractive to nucleophiles.
How do resonance structures of nitrobenzene explain its meta-directing nature?
Resonance delocalization of the positive charge appears at the ortho and para positions, making them unfavorable for electrophilic substitution but favorable for nucleophilic attack.
What are the two key steps in the carbanion mechanism of nucleophilic aromatic substitution?
Addition of the nucleophile to the ring, forming a carbanion intermediate.
Loss of the leaving group (e.g., Cl⁻).
How is the carbanion intermediate stabilized?
It is stabilized by resonance, especially if electron-withdrawing groups are at the ortho or para positions relative to the leaving group.
Why is the carbanion intermediate stabilized by electron-withdrawing groups?
Electron-withdrawing groups (e.g., NO₂) pull electron density away, allowing the negative charge on the carbanion to delocalize via resonance. This added resonance stability makes nucleophilic substitution easier.
What is the key feature of a stable carbanion in nucleophilic aromatic substitution?
A stable carbanion has its negative charge positioned adjacent to an electron-withdrawing group (like NO₂), which stabilizes the intermediate through resonance.
When does the benzyne mechanism occur in nucleophilic aromatic substitution?
It occurs when there are no electron-withdrawing groups on the benzene ring. The reaction proceeds via a highly reactive benzyne intermediate.
What is the general two-step process for the benzyne mechanism?
Elimination of HCl to form the benzyne intermediate (slow).
Nucleophilic attack on benzyne (fast), forming the substituted product.
Why is the benzyne mechanism less favorable compared to the carbanion mechanism?
Because forming benzyne requires harsh conditions and lacks stabilization from electron-withdrawing groups, making it less efficient and more difficult to achieve.
Why is benzyne considered highly reactive?
The third π bond in benzyne forms from side-to-side overlap of two sp² orbitals (instead of normal π overlap), resulting in a weak and strained bond — this makes benzyne very reactive.
Is benzyne similar to an alkyne?
No, despite having a triple bond, benzyne is not a true alkyne because its extra bond is much weaker and results from poor orbital overlap.
What kind of benzyl halides undergo nucleophilic substitution easily?
Benzyl halides that contain electron-withdrawing groups (especially at ortho and/or para positions) undergo substitution reactions easily due to the stabilization of the intermediate.
Why are these reactions important in synthesis?
They help synthesize compounds with two meta-directing groups that are ortho to each other, like –CN and –NO₂, which can direct future substitutions predictably.
Why does 1-chloro-2,4-dinitrobenzene react faster with NaOCH₃ than 1-chloro-3,5-dinitrobenzene?
In 1-chloro-2,4-dinitrobenzene, both NO₂ groups are ortho and para to the leaving group, stabilizing the carbanion intermediate via resonance. In 1-chloro-3,5-dinitrobenzene, the NO₂ groups are meta, offering no such stabilization.
Which positions on the ring provide stabilization for the intermediate in NAS?
Ortho and para positions relative to the leaving group provide resonance stabilization if occupied by electron-withdrawing groups.
