NaOH (Conc./Dil.) & Ba(OH)2 Reagents | chemca

chemca.in

Home Articles Quizzes About

Reagents

NaOH and Ba(OH)₂

The crucial role of concentration: Dilute vs. Concentrated Base and Baryta Water.

By chemca Team • Updated Jan 2026

Sodium Hydroxide ($NaOH$) and Barium Hydroxide ($Ba(OH)_2$) are strong bases. The reactivity of NaOH changes dramatically with concentration, distinguishing between reactions requiring $\alpha$-hydrogens (Aldol) and those without them (Cannizzaro).

1. Dilute NaOH (10-20%)

Aldol Condensation

Dilute alkali removes an acidic $\alpha$-proton from carbonyl compounds to form an enolate ion, which undergoes nucleophilic addition.

Requisite: Aldehydes/Ketones MUST have at least one $\alpha$-Hydrogen.

$$ 2CH_3CHO \xrightarrow{\text{dil. } NaOH} \underset{\text{Aldol (3-Hydroxybutanal)}}{CH_3-CH(OH)-CH_2-CHO} \xrightarrow{\Delta, -H_2O} CH_3-CH=CH-CHO $$

Other Uses:

• Saponification: Hydrolysis of esters to form alcohols and sodium carboxylates.
• Hydrolysis: Conversion of alkyl halides ($R-X$) to alcohols ($R-OH$).

2. Concentrated NaOH (50%)

Cannizzaro Reaction

In the absence of acidic $\alpha$-hydrogens, concentrated alkali causes self-oxidation and reduction (Disproportionation).

Requisite: Aldehydes with NO $\alpha$-Hydrogen (e.g., HCHO, PhCHO, $(CH_3)_3CCHO$).

$$ 2HCHO \xrightarrow{\text{Conc. } NaOH} \underset{\text{Methanol}}{CH_3OH} + \underset{\text{Sodium Formate}}{HCOONa} $$

$$ 2C_6H_5CHO \xrightarrow{\text{Conc. } NaOH} C_6H_5CH_2OH + C_6H_5COONa $$

3. Reaction with Non-Metals (Disproportionation)

Phosphorus, Sulphur, and Halogens

A. White Phosphorus ($P_4$):
Heated with conc. NaOH in inert atmosphere to form Phosphine ($PH_3$).

$$ P_4 + 3NaOH + 3H_2O \rightarrow \underset{\text{Phosphine}}{PH_3 \uparrow} + \underset{\text{Sod. Hypophosphite}}{3NaH_2PO_2} $$

B. Halogens ($Cl_2, Br_2, I_2$):
Cold & Dilute: Forms Hypohalite ($NaOX$).

$$ Cl_2 + 2NaOH \rightarrow NaCl + NaOCl + H_2O $$

Hot & Concentrated: Forms Halate ($NaXO_3$).

$$ 3Cl_2 + 6NaOH \rightarrow 5NaCl + NaClO_3 + 3H_2O $$

C. Sulphur ($S_8$):
Forms Sulphide ($S^{2-}$) and Thiosulphate ($S_2O_3^{2-}$).

4. Barium Hydroxide ($Ba(OH)_2$) - Baryta Water

Specific Applications

A. Preparation of Diacetone Alcohol: $Ba(OH)_2$ is used for the Aldol condensation of Acetone. Being a slightly milder base than NaOH, it prevents the subsequent dehydration of the aldol product (ketol), allowing isolation of Diacetone Alcohol.

$$ 2CH_3COCH_3 \xrightarrow{Ba(OH)_2} \underset{\text{Diacetone Alcohol}}{(CH_3)_2C(OH)CH_2COCH_3} $$

B. Detection of $CO_2$: Baryta water turns milky with $CO_2$ due to formation of $BaCO_3$.

$$ Ba(OH)_2 + CO_2 \rightarrow \underset{\text{White ppt}}{BaCO_3 \downarrow} + H_2O $$

Knowledge Check

Test your understanding of Base Reactions

© 2026 chemca.in. All rights reserved.

Optimized for learning Organic Chemistry.

Quiz Revision Notes Downloads Class Notes Video Lectures