Ochem question

[airstrikee]

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Which nitrogen is likely to be protonated first? I know the answer, but I would like to know the reasoning behind it. Also it would be a great if someone could even put the nitrogen's in order of protonation and explain why one is above the other.

 

[Teleologist]

A, because that's the only sp3 hybridized nitrogen. Every other nitrogen is roughly sp2 hybridized.

 

[K1N6]

In my opinion, I think it'd be the most stable that get's protonated first- the one that can dissipate the charge most or the one that has resonance structure. So i'd have to think B- edit: drew it out, and I think B makes more sense :/ If B gets protonated, then it can have a plethora of resonance structures and I think the most stable structure will proceed the forward reaction most efficiently.

 

[popopopop]

D? If not, I need to revisit the first thread below lol.

http://forums.studentdoctor.net/thr...dary-or-tertiary-amine.1097898/#post-15675685
http://www2.chemistry.msu.edu/faculty/reusch/VirtTxtJml/amine1.htm

 

[Hadi7183]

You are right. The answer is D. Secondary amines are more reactive than primary amines, so A and C can be eliminated. B is involved in an aromatic ring and will not react easily.

 

[airstrikee]

We did this question in class and my professor actually said the answer was A.
We were doing iclicker questions in class and afterwards she showed the pKa values and the right answer was A.
I'm still not sure why?

A - pKa of 10
B - pKa of 5
C - pKa of 4.5
D - pKa of 0

 

[Hadi7183]

I see. Well, my reasoning could be wrong, but I am not convinced your teacher's answer is right either. For example, I don't see how an amine can have such small (acidic) pKa values. Hopefully other members can shed some light on this.

 

[popopopop]

Ahh, my more learneth friend next to me picked A also because he said the D's electrons can participate in conjugation with B (so it's resonance stabilized). I should have counted my electrons.

 

[airstrikee]

Ahh that makes sense now. All of the nitrogen lone pairs for B, C, and D are weaker bases due to delocalization and or resonance. Thank you all!

Also in terms of pKa, why would D have a pKa of 0 and why would B have a pKa of 5

 

[Teleologist]

I see. Well, my reasoning could be wrong, but I am not convinced your teacher's answer is right either. For example, I don't see how an amine can have such small (acidic) pKa values. Hopefully other members can shed some light on this.

Well, amines aren't especially acidic.

 

[Teleologist]

In my opinion, I think it'd be the most stable that get's protonated first- the one that can dissipate the charge most or the one that has resonance structure. So i'd have to think B- edit: drew it out, and I think B makes more sense :/ If B gets protonated, then it can have a plethora of resonance structures and I think the most stable structure will proceed the forward reaction most efficiently.

You mean the most stable end product. You are referring to the thermodynamic product. We don't care about thermodynamics here. We're looking for which nitrogen gets protonated "first," and that expressly implies that we are looking for the kinetic product. The first nitrogen to be protonated is likely 1 - the one with its lone pairs not tied up in resonance (delocalization).

 

[K1N6]

You mean the most stable end product. You are referring to the thermodynamic product. We don't care about thermodynamics here. We're looking for which nitrogen gets protonated "first," and that expressly implies that we are looking for the kinetic product. The first nitrogen to be protonated is likely 1 - the one with its lone pairs not tied up in resonance (delocalization).

Ah, I see. I guess I need to study my OChem a bit more before test day! Thank you.