Comments on: Reagent Friday: Sodium Amide (NaNH2)

By: Jessica Reel

Jessica Reel — Fri, 01 Dec 2023 21:31:35 +0000

Hi again, thanks for the super valuable resource. I have some questions about the possible SN2 competition when doing the double-dehalogenation of a dihalide to form an alkyne. I would assume that -NH2 is a very good nucleophile; however, I also know that acid-base reactions tend to be faster (and since this reaction must be done at low temperature due to the liquid NH3, it would be under kinetic control, right?) So maybe that is one way that we prevent the SN2..? But then again, I’ve also been told that heat promotes the elimination product… so that feels like conflicting factors to me.

A related thought: would there be less SN2 competition with a geminal dihalide instead of a vicinal dihalide (due to sterics)? Therefore is it better to try to do this double-dehalogenation starting from a gem-dihalide? Thank you!

By: Manvendra Singh

Manvendra Singh — Wed, 18 Jan 2023 10:43:16 +0000

how to determine the assay of NaNH2

By: suparna bhattacharjee

suparna bhattacharjee — Fri, 10 Jun 2022 06:41:10 +0000

Hello James,
Thanks for the brilliant work
My query is regarding Example 3
If the mechanism is dehydrohalogenation, for each Br that eliminates, there are two beta C from which H+ can be eliminated. Then why But-1,3-diene is not a possible product?
please reply

By: Kevin

Kevin — Fri, 24 Sep 2021 04:20:26 +0000

Thank you for the part about the acetylide formation! Also I liked the piranha comparison xD.

By: Spider Holland

Spider Holland — Thu, 04 Jun 2020 05:41:41 +0000

Thanks for solving the doubt.

By: James Ashenhurst

James Ashenhurst — Wed, 03 Jun 2020 15:36:23 +0000

That is a great question. Unless you’re in an advanced class, I think the example you have should read NaNH2/NH3 since it’s a textbook example of elimination to give the acetylide.
That’s my short answer. Longer answer below.
——

If I was writing a scheme where NaNH2 was used as a base in ammonia, I would write NaNH2 / NH3
If I was writing a scheme where sodium metal was used as a reductant I would write Na/NH3.

*As written* it suggests a reducing agent. Say 1-bromo-1-phenylethene was added to a solution of sodium in liquid ammonia. What would it do?

I think what it would do is reduce the C-Br bond to give an anion, which would then be protonated by NH3, giving styrene as the product.

Of course it would depend on how many equivalents of sodium you had, since if there were more, the aromatic ring in styrene would reduce further. However since no alcohol is indicated, it wouldn’t do a full Birch reduction and then there would be … a mess, to make a long story short.

In practice, to get NaNH2, it’s not enough to just add Na to liquid ammonia (Na/NH3) It helps to add a small amount of oxidant (like Fe(III) ) to get the process going that results in formation of NaNH2 (and formation of hydrogen gas)

By: Spider Holland

Spider Holland — Wed, 03 Jun 2020 15:09:39 +0000

“NaNH2 is usually made by dissolving sodium metal in liquid ammonia, so this is the most common reagent/solvent combination.”

If so, how will we solve 1-bromo-1-phenylethene + Na/liq.NH3 =?
Will Na/liq.NH3 work as a reducing agent and form 1-bromo-1-(cyclobuta-1,4-diene)-ethane?
Or will the so-prepared NaNH2 work as a strong base and form phenyl acetylide?

TLDR: If given reagent is Na/liq.NH3, how to know if to take it as the reducing agent Na/liq.NH3 or as the strong base NaNH2.

By: James Ashenhurst

James Ashenhurst — Thu, 28 May 2020 14:45:09 +0000

In reply to Sristy Choudhary.

It occurs through a benzyne intermediate. See here: https://www.masterorganicchemistry.com/reaction-guide/nucleophilic-aromatic-substitution-via-arynes/

By: Sristy Choudhary

Sristy Choudhary — Thu, 28 May 2020 08:50:56 +0000

How does NaNH2 react with Substituted Benzene compounds ?
For eg –
https://hi-static.z-dn.net/files/d97/7ba9f763a7008724465f4814cf7ef723.jpg

https://imgflip.com/i/433uis

In the above NH2 (-) is added at meta position i.e. which has the MINimum electron density a/c to +R effect of OCH3.
In the 2nd link NH2(-) is once again added to meta position which has MAX electron density due to -R effect of NO2.
I am not able to figure out what exactly is happening here.Could anyone help me out ?

By: James Ashenhurst

James Ashenhurst — Sun, 26 Apr 2020 03:06:18 +0000

In reply to Shashwat Saxena. NaNH2 is usually made by dissolving sodium metal in liquid ammonia, so this is the most common reagent/solvent combination. It is also possible to buy sodium amide (NaNH2) and use it in other solvents.