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junior college 1 | H2 Maths
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May i know why after the 1st SN2 reaction, why don't the Iodide ion attack the nitrogen ion? Which part of my concept is wrong?
Since Iodide is negative means its an nucleophile and Nitrogen is positive means is an electrophile. Usually nucleophile attack electrophile so why can't Iodide iom attack nitrogen in this case?
Date Posted: 2 years ago
Views: 232
Do you mean, why isn't the I- attacking the N+ and then the N-H or C-N bond breaks, giving out H- or H3C- respectively?
Just in general why is that so when I- is nucleophile and nitrogen is electrophile but I- doesn't attack N+
You'll need to consider if that is feasible.
If the I- attacks the N+, either the C-N bond or N-H bond will be broken.
What you end up with is still a N with a formal charge (+) and now you have either a free hydride (H-) or a carbanion (-CH3)
There is charge separation (you have separated + and - charges) and these two species are highly unstable.
It would be much more feasible for the I- to act as a base and abstract a H , breaking the N-H bond. This is much faster.
The N now is neutral and you have HI.
If the I- attacks the N+, either the C-N bond or N-H bond will be broken.
What you end up with is still a N with a formal charge (+) and now you have either a free hydride (H-) or a carbanion (-CH3)
There is charge separation (you have separated + and - charges) and these two species are highly unstable.
It would be much more feasible for the I- to act as a base and abstract a H , breaking the N-H bond. This is much faster.
The N now is neutral and you have HI.
Oh yes that right, didn't think of that. Thank you for the explanation!
By the way, that N+ is not actually considered to be an electrophile.
N is more electronegative than C and H (N 3.04, C 2.55 and H 2.2) , so the electrons in the bond pair of C-H and N-C are actually closer to the N than to the H/C.
Whereas when you had the N: attacking the CH3I,
The I has a higher electronegativity than C (2.66 vs 2.55) so the C is electrophilic in this sense)
You may ask why shouldn't H leave as a H- instead.
That would be due to the stability of the anion formed.
The negative charge can be dispersed over the larger I- ion more effectively than the very small H- ion (hydride)
That's why HI is a very strong acid as the conjugate base I- is very stable.
If it were unstable, we wouldn't even have much dissociation and any conjugate base formed would quickly attack the H+ ions to form back the acid.
For further reading on this reaction :
https://www.masterorganicchemistry.com/2017/05/26/alkylation-of-amines-is-generally-a-crap-reaction/
N is more electronegative than C and H (N 3.04, C 2.55 and H 2.2) , so the electrons in the bond pair of C-H and N-C are actually closer to the N than to the H/C.
Whereas when you had the N: attacking the CH3I,
The I has a higher electronegativity than C (2.66 vs 2.55) so the C is electrophilic in this sense)
You may ask why shouldn't H leave as a H- instead.
That would be due to the stability of the anion formed.
The negative charge can be dispersed over the larger I- ion more effectively than the very small H- ion (hydride)
That's why HI is a very strong acid as the conjugate base I- is very stable.
If it were unstable, we wouldn't even have much dissociation and any conjugate base formed would quickly attack the H+ ions to form back the acid.
For further reading on this reaction :
https://www.masterorganicchemistry.com/2017/05/26/alkylation-of-amines-is-generally-a-crap-reaction/
Thank you!
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