The acidity of alcohol,phenol,and carboxylic acid differs. Demonstrate their reactivity with $Na$,$NaOH$,and $NaHCO_3$ using the table below.
Compound Reaction with $Na$ metal Reaction with $NaOH$ Reaction with $NaHCO_3$
Alcohol $H_2$ is liberated No reaction No reaction
Phenol $H_2$ is liberated Soluble salt formed No reaction
Carboxylic acid $H_2$ is liberated Soluble salt formed $CO_2$ gas is liberated

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) The acidity of these compounds follows the order: $Carboxylic \ acid > Phenol > Alcohol$.
$1$. $Na$ metal: All three compounds contain an acidic hydrogen atom ($-OH$ group),so they all react with $Na$ to liberate $H_2$ gas.
$2$. $NaOH$: Carboxylic acids and phenols are acidic enough to react with the strong base $NaOH$ to form water-soluble salts. Alcohols are too weakly acidic to react with $NaOH$.
$3$. $NaHCO_3$: Only carboxylic acids are strong enough to react with the weak base $NaHCO_3$ to liberate $CO_2$ gas. Phenols and alcohols are not acidic enough to decompose $NaHCO_3$.

Explore More

Similar Questions

Arrange the following compounds in increasing order of their $pK_{a}$ values:
$(a)$ $p$-nitrobenzoic acid
$(b)$ $p$-methoxybenzoic acid
$(c)$ $p$-nitrophenol
$(d)$ benzoic acid

Which of the following acids,upon heating,loses a water molecule to form an $\alpha, \beta$-unsaturated acid?

Difficult
View Solution

Ammonium acetate reacts with acetic acid at $110\,^{\circ}C$ to form:

Which of the following substances will not undergo the Hell-Volhard-Zelinsky reaction?

Derive an equation for $pK_{a}$ for carboxylic acid compounds. Also,give the relation between $pK_{a}$ and acidic strength.

Difficult
View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo