Sodium Amide (NaNH2) Strong Base Reagent

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Sodium Amide ($NaNH_2$)

The ultra-strong base for eliminations and deprotonating alkynes.

By chemca Team • Updated Oct 2026

Sodium Amide ($NaNH_2$) is an exceptionally strong base and a powerful nucleophile. Because the amide ion ($NH_2^-$) is the conjugate base of ammonia ($NH_3$), which is a very weak acid ($pK_a \approx 38$), $NaNH_2$ is significantly stronger than commonly used bases like Sodium Hydroxide ($NaOH$) or Potassium Hydroxide ($KOH$).

⚠️ Common Confusion Alert: Do not confuse $NaNH_2$ (Sodium Amide, a strong base) with $Na$ in Liquid $NH_3$ (Sodium dissolved in Ammonia, Birch Reduction, which is a reducing agent that provides solvated electrons). They do completely different things!

1. Synthesis of Alkynes (Double Dehydrohalogenation)

Elimination of Vicinal or Geminal Dihalides

Conditions: Strong heating with $NaNH_2$ in an inert solvent like mineral oil, or liquid ammonia.

Action: Converts dihalides to alkynes by removing two molecules of $HX$ via the $E2$ mechanism.

Why use $NaNH_2$ instead of alcoholic KOH?

Alcoholic $KOH$ is strong enough to remove the first $HX$ molecule to form a vinyl halide. However, vinyl halides are highly unreactive towards further elimination because the halogen's lone pairs are in resonance with the double bond. Removing the second $HX$ requires a much harsher, stronger base: $NaNH_2$.

$$ \underset{\text{Vicinal Dihalide}}{R-CH(Br)-CH_2(Br)} \xrightarrow[\text{1st Elimination}]{\text{Alc. KOH, } \Delta} \underset{\text{Vinyl Halide (stable)}}{R-C(Br)=CH_2} \xrightarrow[\text{2nd Elimination}]{NaNH_2, \Delta} \underset{\text{Alkyne}}{R-C\equiv CH} $$

2. Deprotonation of Terminal Alkynes

Formation of Sodium Alkynides

Context: Terminal alkynes ($R-C\equiv C-H$) have an acidic hydrogen because the carbon is $sp$ hybridized (50% s-character, making it highly electronegative). However, they are still weak acids ($pK_a \approx 25$). Standard bases like $NaOH$ ($pK_a$ of water $\approx 15.7$) cannot deprotonate them.

Action: $NaNH_2$ cleanly deprotonates terminal alkynes, releasing Ammonia gas ($NH_3$) and forming a powerful nucleophile: the Alkynide ion.

$$ R-C\equiv C-H + Na^+NH_2^- \longrightarrow \underset{\text{Sodium Alkynide}}{R-C\equiv C^-Na^+} + NH_3 \uparrow $$

Synthetic Application (Chain Elongation): The resulting alkynide ion is an excellent nucleophile. It can attack primary alkyl halides ($R'-X$) via an $S_N2$ mechanism to create larger, internal alkynes.

$$ R-C\equiv C^- + R'-CH_2-X \longrightarrow R-C\equiv C-CH_2-R' + X^- $$

3. The Chichibabin Reaction

Amination of Pyridine

Conditions: $NaNH_2$ in liquid ammonia or inert solvent, Heat ($\sim 100^\circ C$).

Action: $NaNH_2$ acts as a strong nucleophile and attacks electron-deficient aromatic heterocycles like Pyridine at the 2-position (ortho to Nitrogen) in a Nucleophilic Aromatic Substitution ($S_NAr$) reaction.

$$ \text{Pyridine} + NaNH_2 \xrightarrow{\Delta} \underset{\text{2-Aminopyridine}}{\text{2-Amino-Pyridine}} + H_2 \uparrow $$

• This reaction uniquely expels a hydride ion ($H^-$), which ultimately leaves as Hydrogen gas ($H_2$).
• It provides a direct way to introduce an amino group ($-NH_2$) into a pyridine ring.

4. Reagent Comparison

Property / Reaction	$NaOH$ / $KOH$ (Aq)	$KOH$ (Alcoholic)	$NaNH_2$ (Sodium Amide)
Base Strength	Moderate Base	Strong Base	Ultra-Strong Base
Role in Alkyl Halides	Nucleophile ($S_N2$ $\rightarrow$ Alcohols)	Base ($E2$ $\rightarrow$ Alkenes)	Very Strong Base ($E2$ $\rightarrow$ Alkynes)
Reaction with Terminal Alkynes	No Reaction	No Reaction	Deprotonates (Forms Alkynide)
Vinyl Halide to Alkyne?	No	Very Slow / Requires extreme heat	Yes, readily occurs

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