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Author Topic: Ethyne with Sodium amide in ammonia  (Read 1715 times)

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Nescafe

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Ethyne with Sodium amide in ammonia
« on: September 06, 2012, 04:14:49 AM »

Hi,

What is the role of sodium amide in the reaction?

The pka of ethyne is 25, that of ammonia is 33, would'nt ammonia be able to deprotonate ethyne without the need of sodium amide?

Nescafe.
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fledarmus

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Re: Ethyne with Sodium amide in ammonia
« Reply #1 on: September 06, 2012, 05:22:36 AM »

Write the reactions for which those are the measured pKa's...

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Nescafe

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Re: Ethyne with Sodium amide in ammonia
« Reply #2 on: September 06, 2012, 02:26:49 PM »

NH3 + B-   ::equil:: -NH2 + BH

H≡H + B-  ::equil:: H≡ - + BH

I do not have a clear understanding of this stuff and I tend to get confused. I do understand that in order to deprotonate an acidic proton, the pka of the conjugate acid of the base used must be greater than the pka of the acidic proton for the equilibrium shift to the right.

I think I kind of get what you are trying to make me learn here. NH3 is not a strong enough base to deprotonate the ethyne as the conjugate acid of ammonia, +NH4 has a pka of ~4.5 which is actually more acidic than the ethyne proton. Thus why we need the sodium amide. and ammonia is just there as a solvent or like water for sodium hydroxide?

So should I be writing it as


NH3 + -NH2   ::equil:: -NH2 + +NH3?

an endless supply of -NH2?

I just got confused about something else. FMOC deprotection, that carbon in the FMOC which gets deprotonated has a pka of 22.6, how is it that piperidine which has a pka of 11ish is able to deprotonate it? Would not the equilibrium be more to the left in such a case in the following equation?

R2N-FMOC + piperidine  :lequil: R2N--FMOC(-H) + +Piperidine-(+H)

Is it because the acetate is a better nucleophile and picks up the piperidine proton before -FMOC(-H) has a chance?

Bleh =(

Nescafe.


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Nescafe

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Re: Ethyne with Sodium amide in ammonia
« Reply #3 on: September 07, 2012, 07:09:28 PM »

Anybody?
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fledarmus

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Re: Ethyne with Sodium amide in ammonia
« Reply #4 on: September 10, 2012, 04:56:59 AM »

NH3 + B-   ::equil:: -NH2 + BH

H≡H + B-  ::equil:: H≡ - + BH



I think I kind of get what you are trying to make me learn here. NH3 is not a strong enough base to deprotonate the ethyne as the conjugate acid of ammonia, +NH4 has a pka of ~4.5 which is actually more acidic than the ethyne proton. Thus why we need the sodium amide. and ammonia is just there as a solvent or like water for sodium hydroxide?

Exactly right.

Quote

So should I be writing it as


NH3 + -NH2   ::equil:: -NH2 + +NH3?

an endless supply of -NH2?

Well, first, you got an extra positive charge on the right side, but this isn't an endless supply of anything. All you are demonstrating is that in a solution of amide in ammonia, you can't determine which nitrogen is bonded to which hydrogen - the protons can move from ammonia molecule to amide throughout the solution. The short answer is, no, there is usually no point in writing this reaction. Although the reaction is occurring, and occurring rapidly, it isn't doing anything - the starting material is the same as the product. It is the same thing in water - if you add sodium hydroxide to water, you get hydroxide ions, and protons shuffle rapidly between the water molecules and the hydroxide ions.



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Nescafe

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Re: Ethyne with Sodium amide in ammonia
« Reply #5 on: September 11, 2012, 05:02:31 PM »

Right-on.

I don't know what I would do without you  >:( 

Nescafe.
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saptajung555

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Re: Ethyne with Sodium amide in ammonia
« Reply #6 on: September 11, 2012, 11:21:28 PM »

 ;D ;D ;D
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