Reagent Friday: Sodium Amide (NaNH2)

by James

in Organic Reagents

In a blatant plug for the Reagent Guide, each Friday  I profile a different reagent that is commonly encountered in Org 1/ Org 2. Version 1.2 just got released this week, with a host of corrections and a new page index. 

Note: there should also be another exciting announcement about the Reagent Guide coming up in the next little while or so… more details to come!

What’s small, ferocious, and can fit into tight spaces? (Photo credit: Wikipedia)

NaNH2 (Sodium amide) 

 What it’s used for: NaNH2 is a strong base and excellent nucleophile. It’s used for deprotonation of weak acids and also for elimination reactions.

Similar to: LDA (lithium diisopropylamide).

The NH2- anion is the conjugate base of ammonia (NH3). If you’ll recall, the weaker the acid, the stronger the conjugate base – and since NH3 has a pKa of 38, NH2 is a strong base indeed. (Note that although I’m talking about NaNH2 here, the bases LiNH2 and KNH2 essentially behave the same way.)

As a strong base, NaNH2 will deprotonate alkynes, alcohols, and a host of other functional groups with acidic protons such as esters and ketones.  It is also a very strong nucleophile.

As a base, it’s often used in situations where a strong, small base is required. Like a piranha, NaNH2 is small, fast, and has razor sharp teeth, and can find its way into tight, enclosed spaces.

One common application of NaNH2 is in the deprotonation of alkynes to give so-called “acetylide” ions. These ions are excellent nucleophiles and can go on to react with carbonyls in addition reactions.

A second application of NaNH2 is in the formation of alkynes from halogens. Treatment of either geminal dihalides (i.e. – two halogens on one carbon) or vicinal dihalides (halogens on adjacent carbons) with NaNH2 (2 equiv) will result in the formation of alkynes.

Since vicinal dihalides are easily made by the reaction of alkenes with halogens such as Br2 or I2, this is a useful way of converting alkenes to alkynes.

How it works.

Deprotonation of functional groups such as OH and even alkyne C-H should hopefully be straightforward, but the use of bases to make alkenes may require some explanation. This is what is known as an elimination reaction, in that the elements H and Br (in this example) are removed in order to form the alkene. Specifically, this is an example of an E2 reaction. 

Since the alkene still has a halide attached, this too can be removed to generate a second double bond (π bond). This is another example of the E2 in that the hydrogen has to be anti to the bromine that is eliminated, but is unusual in that it is an sp2 hydrogen that is affected here:

This arrangement is necessary in order for the reaction to occur in that the pair of electrons in the C–H bond that is breaking will simultaneously interact with the antibonding orbital of the C-Br bond, leading to formation of the new π bond and expulsion of the Br(-)

P.S. You can read about the chemistry of NaNH2 and more than 80 other reagents in undergraduate organic chemistry in the “Organic Chemistry Reagent Guide”, available here as a downloadable PDF.

EDIT/UPDATE. @chemistinjapan makes the following observation:

Screen shot 2013-05-23 at 5.06.32 PM

A note of caution on use of NaNH2 as a nucleophile. My trusty copy of March has the following to say:

“The conjugate base of ammonia  is sometimes used as a nucleophile, but in most cases offers no advantage over ammonia, since the latter is basic enough.”

Furthermore, since NaNH2 is a strong base, it has the significant disadvantage of promoting side reactions from elimination (this can occur when attempting an SN2 with NaNH2 as the nucleophile, for example). Therefore, it is generally wise to avoid using NaNH2 as a nucleophile in organic synthesis. Sodium azide (followed by reduction) is the usual substitute.

I suppose one could use NaNH2 as a nucleophile in a case like this one (below) but again it offers no significant advantage over NH3:

acyl halides amides

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{ 17 comments… read them below or add one }

Stephen Dale

The best short summary of NH2- in solution I’ve found.

Just wanted to let you know I’m refeerencing this page in my Honours paper.


Silas Cardwell

God bless you and Canada.



THANK YOU SO MUCH. Your website is probably the most useful orgo website on the damn internet. I was trying to figure out how to use this stupid reagent for hours because my book doesn’t explain “how” it works and every other website just uses a ton of complex terminology to explain a simple concept.






Why NaNH2 is called as Sodamide even though it does not contain amide group?



“Amide” can be confusing. It can refer to either C(O)NH2 or NH2(-) . The name “Sodamide” comes from shortening the name SODium AMIDE.



Is sodamide the same as a sodium metal and liquid ammonia reduction?



No, the sodium in a liquid ammoniua reduction is sodium metal (Na). Sodamide is NaNH2.


anupam chaubey

Could you please tell me why is HgSO4 used in organic chemstry??Is it an oxidising agent or reducing one?? Moreover what is its action upon alkenes, carbonyl compounds and acids??


Nimesh Khandelwal

sir I want to ask if we use nanh2 with alkyl halide then substitution because of strong nucleophile should occur then why there is elimination.



Thanks for this breakdown on this reagent. I was so confused on how to use something that’s a strong base AND a strong nucleophile, but your description of why NaNH2 is better served as a base really helped me out.



I recently came across a problem involving acetylene reacting with NaNH2, NH3 followed by CH3-I
Is it possible for the base to deprotonate both sides of acetylene since there are two acidic hydrogens present? Or do we just assume one mole and treat it as an SN2 reaction.




I’d assume that it deprotonates just one side of acetylene (1 mol of base) and reacts with CH3I to make 1-propyne.

Hope that answer gets to you in time! James




I find it odd that NaNH2 can both react with an alkene to form an alkyne but ALSO react with an alkyne to form an acetylide anion. Since this is the case, if I reacted an alkene with excess NaNH2, wouldn’t the excess NaNH2 react with the alkyne I just created to form an acetylide anion? This seems like it would be an issue in cases of synthesis problems if I am not mistaken. Any insight would be greatly appreciated!

Many thanks,



YES – you are absolutely correct. When forming a terminal alkyl through elimination, in practice one must use 3 equiv of NaNH2 because if only 2 equivalents are used the reaction doesn’t go to completion.

Many instructors/schools are inconsistent on this point, and just mention “2 equiv” but you are absolutely correct.



I had been searching the reagent for hours. Found it finally. The best post.:)



Honestly, I really wish I had found this website earlier instead of 3 days before my final. These are by far the clearest and most helpful explanations I’ve ever found. Thank you!!


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