Attack of a second pyrrolidine on the product is disfavoured compared to the first. Compare the starting material and the product: What effect do you think the pyrrolidinyl group has on the electronics of the ring system compared to bromide?
Attack of a second pyrrolidine on the product is disfavoured compared to the first. Compare the starting material and the product: What effect do you think the pyrrolidinyl group has on the electronics of the ring system compared to bromide?
As a general rule of thumb a pyridine ring nitrogen may be thought to exert approximately the same activating effect as a C-NO2 susbtituent toward nucleophilic aromatic substitution.
Thus, for reaction with NaOCH3/CH3OH performed at 50 degrees C: 2-Chloropyridine undergoes nucleophilic aromatic substitution of Cl by OCH3 about 2.3 x 10exp8 faster than chlorobenzene.
Similarly, for reaction with NaOCH3/CH3OH performed at 85 degrees C: 4-Nitrochlorobenzene undergoes nucleophilic aromatic substitution of Cl by OCH3 about 7 x 10exp10 faster than chlorobenzene.
Admittedly, the foregoing might not be a fair comparison for the following reasons:
(i) The C-Cl bond is more distant in 4-nitrochlorobenzene than in 2-chloropyridine, so activation via the inductive effect of substituents (C-NO2 group vs. ring nitrogen) is attenuated in the former compound.
(ii) Different mechanisms may be operative! Thus, nucleophilic aromatic substitution in activated aromatic rings (i. e., those that bear ring nitrogen and/or electron-withdrawing substituents) proceeds with formation of a Meisenheimer complex, whereas reaction of (unactivated) chlorobenzene with methoxide ion, even at 50 degrees C, most likely proceeds via base-promoted abstraction of an ortho-H followed by formation of a benzyne intermediate.
As a general rule of thumb a pyridine ring nitrogen may be thought to exert approximately the same activating effect as a C-NO2 susbtituent toward nucleophilic aromatic substitution.
Thus, for reaction with NaOCH3/CH3OH performed at 50 degrees C: 2-Chloropyridine undergoes nucleophilic aromatic substitution of Cl by OCH3 about 2.3 x 10exp8 faster than chlorobenzene.
Similarly, for reaction with NaOCH3/CH3OH performed at 85 degrees C: 4-Nitrochlorobenzene undergoes nucleophilic aromatic substitution of Cl by OCH3 about 7 x 10exp10 faster than chlorobenzene.
Admittedly, the foregoing might not be a fair comparison for the following reasons:
(i) The C-Cl bond is more distant in 4-nitrochlorobenzene than in 2-chloropyridine, so activation via the inductive effect of substituents (C-NO2 group vs. ring nitrogen) is attenuated in the former compound.
(ii) Different mechanisms may be operative! Thus, nucleophilic aromatic substitution in activated aromatic rings (i. e., those that bear ring nitrogen and/or electron-withdrawing substituents) proceeds with formation of a Meisenheimer complex, whereas reaction of (unactivated) chlorobenzene with methoxide ion, even at 50 degrees C, most likely proceeds via base-promoted abstraction of an ortho-H followed by formation of a benzyne intermediate.
A: So the mechanism I've drawn is an SN1 reaction with bromine leaving first and the pyrrolidine then attacking the carbon. I've then said that the reason this works when X = N is because it has a lone pair and can push electron density onto the bromine and it can leave.
OK, draw the resonance form of that cation with the +ve charge on N. Can you see a problem?
Quote
A: The only thing I can think of is that the pyrrolidine is deactivating
Your answer is really just a rephrasing of the question, not an answer. The reactivity will become easier to explain once you get the mechanism correct, so concentrate on that first.
Start by identifying the polarisation of the pyridine ring - put δ+ and/or δ- around the ring and think about what could happen in the presence of a nucleophile.
A: So the mechanism I've drawn is an SN1 reaction with bromine leaving first and the pyrrolidine then attacking the carbon. I've then said that the reason this works when X = N is because it has a lone pair and can push electron density onto the bromine and it can leave.
OK, draw the resonance form of that cation with the +ve charge on N. Can you see a problem?
Quote
A: The only thing I can think of is that the pyrrolidine is deactivating
Your answer is really just a rephrasing of the question, not an answer. The reactivity will become easier to explain once you get the mechanism correct, so concentrate on that first.
Start by identifying the polarisation of the pyridine ring - put δ+ and/or δ- around the ring and think about what could happen in the presence of a nucleophile.
I've considered the effects of the pyrrolidinyl group on the molecule, If we consider -I inductive effects and take the pyrrolidinyl group as position 1, this would make the ortho and para positions more electrophilic, i don't see how this would effect the C-Br in the meta position.
If we consider the lone pair on the nitrogen on pyrrodinyl group and push electron density into the ring, the ortho and para positions would become less electrophilic. Again i don't see how this would effect the C-Br in the meta position.
Any help would be appreciated.
Edit: nvm i realise im an idiot, i was getting mixed up with directing effects.
Consider the relative stabilities of the Meisenheimer complexes formed in the first and second substitutions.
I've considered the effects of the pyrrolidinyl group on the molecule, If we consider -I inductive effects and take the pyrrolidinyl group as position 1, this would make the ortho and para positions more electrophilic, i don't see how this would effect the C-Br in the meta position.
If we consider the lone pair on the nitrogen on pyrrodinyl group and push electron density into the ring, the ortho and para positions would become less electrophilic. Again i don't see how this would effect the C-Br in the meta position.
Any help would be appreciated.
Edit: nvm i realise im an idiot, i was getting mixed up with directing effects.
Consider the relative stabilities of the Meisenheimer complexes formed in the first and second substitutions.
I'm going to paste a screenshot of the question from my thread in order to save time from having to write it all out.
http://gyazo.com/e9e09d60b4aae6443f2e0ffc33dde559
I was wondering if part ii) was correct and for part iii) i can't come to a reasonable conclusion to account for why further substitution doesn't occur.
I've considered the effects of the pyrrolidinyl group on the molecule, If we consider -I inductive effects and take the pyrrolidinyl group as position 1, this would make the ortho and para positions more electrophilic, i don't see how this would effect the C-Br in the meta position.
If we consider the lone pair on the nitrogen on pyrrodinyl group and push electron density into the ring, the ortho and para positions would become less electrophilic. Again i don't see how this would effect the C-Br in the meta position.
Any help would be appreciated.
Edit: nvm i realise im an idiot, i was getting mixed up with directing effects.
I'm going to paste a screenshot of the question from my thread in order to save time from having to write it all out.
http://gyazo.com/e9e09d60b4aae6443f2e0ffc33dde559
I was wondering if part ii) was correct and for part iii) i can't come to a reasonable conclusion to account for why further substitution doesn't occur.
I've considered the effects of the pyrrolidinyl group on the molecule, If we consider -I inductive effects and take the pyrrolidinyl group as position 1, this would make the ortho and para positions more electrophilic, i don't see how this would effect the C-Br in the meta position.
If we consider the lone pair on the nitrogen on pyrrodinyl group and push electron density into the ring, the ortho and para positions would become less electrophilic. Again i don't see how this would effect the C-Br in the meta position.
Any help would be appreciated.
Edit: nvm i realise im an idiot, i was getting mixed up with directing effects.
OK I've redone the mechanism - and this reaction only occurs for pyridine because the intermediate anion is stabilised by the electronegative nitrogen. But I'm still unsure of why only the monosubstituted product is formed.
OK I've redone the mechanism - and this reaction only occurs for pyridine because the intermediate anion is stabilised by the electronegative nitrogen. But I'm still unsure of why only the monosubstituted product is formed.
Attack of a second pyrrolidine on the product is disfavoured compared to the first. Compare the starting material and the product: What effect do you think the pyrrolidinyl group has on the electronics of the ring system compared to bromide?
Attack of a second pyrrolidine on the product is disfavoured compared to the first. Compare the starting material and the product: What effect do you think the pyrrolidinyl group has on the electronics of the ring system compared to bromide?
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Thus, for reaction with NaOCH3/CH3OH performed at 50 degrees C:
2-Chloropyridine undergoes nucleophilic aromatic substitution of Cl by OCH3 about 2.3 x 10exp8 faster than chlorobenzene.
Similarly, for reaction with NaOCH3/CH3OH performed at 85 degrees C:
4-Nitrochlorobenzene undergoes nucleophilic aromatic substitution of Cl by OCH3 about 7 x 10exp10 faster than chlorobenzene.
Admittedly, the foregoing might not be a fair comparison for the following reasons:
(i) The C-Cl bond is more distant in 4-nitrochlorobenzene than in 2-chloropyridine, so activation via the inductive effect of substituents (C-NO2 group vs. ring nitrogen) is attenuated in the former compound.
(ii) Different mechanisms may be operative! Thus, nucleophilic aromatic substitution in activated aromatic rings (i. e., those that bear ring nitrogen and/or electron-withdrawing substituents) proceeds with formation of a Meisenheimer complex, whereas reaction of (unactivated) chlorobenzene with methoxide ion, even at 50 degrees C, most likely proceeds via base-promoted abstraction of an ortho-H followed by formation of a benzyne intermediate.
Ah, well. Hope this helps.
Thus, for reaction with NaOCH3/CH3OH performed at 50 degrees C:
2-Chloropyridine undergoes nucleophilic aromatic substitution of Cl by OCH3 about 2.3 x 10exp8 faster than chlorobenzene.
Similarly, for reaction with NaOCH3/CH3OH performed at 85 degrees C:
4-Nitrochlorobenzene undergoes nucleophilic aromatic substitution of Cl by OCH3 about 7 x 10exp10 faster than chlorobenzene.
Admittedly, the foregoing might not be a fair comparison for the following reasons:
(i) The C-Cl bond is more distant in 4-nitrochlorobenzene than in 2-chloropyridine, so activation via the inductive effect of substituents (C-NO2 group vs. ring nitrogen) is attenuated in the former compound.
(ii) Different mechanisms may be operative! Thus, nucleophilic aromatic substitution in activated aromatic rings (i. e., those that bear ring nitrogen and/or electron-withdrawing substituents) proceeds with formation of a Meisenheimer complex, whereas reaction of (unactivated) chlorobenzene with methoxide ion, even at 50 degrees C, most likely proceeds via base-promoted abstraction of an ortho-H followed by formation of a benzyne intermediate.
Ah, well. Hope this helps.
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OK, draw the resonance form of that cation with the +ve charge on N. Can you see a problem?
Quote
Your answer is really just a rephrasing of the question, not an answer. The reactivity will become easier to explain once you get the mechanism correct, so concentrate on that first.
Start by identifying the polarisation of the pyridine ring - put δ+ and/or δ- around the ring and think about what could happen in the presence of a nucleophile.
OK, draw the resonance form of that cation with the +ve charge on N. Can you see a problem?
Quote
Your answer is really just a rephrasing of the question, not an answer. The reactivity will become easier to explain once you get the mechanism correct, so concentrate on that first.
Start by identifying the polarisation of the pyridine ring - put δ+ and/or δ- around the ring and think about what could happen in the presence of a nucleophile.
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VOTE
Consider the relative stabilities of the Meisenheimer complexes formed in the first and second substitutions.
Consider the relative stabilities of the Meisenheimer complexes formed in the first and second substitutions.
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http://gyazo.com/e9e09d60b4aae6443f2e0ffc33dde559
I was wondering if part ii) was correct and for part iii) i can't come to a reasonable conclusion to account for why further substitution doesn't occur.
I've considered the effects of the pyrrolidinyl group on the molecule, If we consider -I inductive effects and take the pyrrolidinyl group as position 1, this would make the ortho and para positions more electrophilic, i don't see how this would effect the C-Br in the meta position.
If we consider the lone pair on the nitrogen on pyrrodinyl group and push electron density into the ring, the ortho and para positions would become less electrophilic. Again i don't see how this would effect the C-Br in the meta position.
Any help would be appreciated.
Edit: nvm i realise im an idiot, i was getting mixed up with directing effects.
http://gyazo.com/e9e09d60b4aae6443f2e0ffc33dde559
I was wondering if part ii) was correct and for part iii) i can't come to a reasonable conclusion to account for why further substitution doesn't occur.
I've considered the effects of the pyrrolidinyl group on the molecule, If we consider -I inductive effects and take the pyrrolidinyl group as position 1, this would make the ortho and para positions more electrophilic, i don't see how this would effect the C-Br in the meta position.
If we consider the lone pair on the nitrogen on pyrrodinyl group and push electron density into the ring, the ortho and para positions would become less electrophilic. Again i don't see how this would effect the C-Br in the meta position.
Any help would be appreciated.
Edit: nvm i realise im an idiot, i was getting mixed up with directing effects.
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