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(C) Sodium bicarbonate $(NaHCO_3)$ is a weak base. Only acids stronger than carbonic acid $(H_2CO_3)$ can react with it to evolve $CO_2$ gas and becom

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$o-$Nitrophenol

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(C) Sodium bicarbonate $(NaHCO_3)$ is a weak base. Only acids stronger than carbonic acid $(H_2CO_3)$ can react with it to evolve $CO_2$ gas and become soluble.$2, 4, 6-$Trinitrophenol (picric acid),benzoic acid,and benzene sulphonic acid are stronger acids than carbonic acid and are soluble in $NaHCO_3$.$o-$Nitrophenol is a weaker acid than carbonic acid. Additionally,it forms intramolecular hydrogen bonding between the $-OH$ group and the $-NO_2$ group,which makes the acidic proton less available for reaction with the base. Therefore,it is not soluble in sodium bicarbonate.

Which one of the following compounds will not be soluble in sodium bicarbonate?

Similar Questions

What is $Z$ in the following sequence of reactions?
$p-chloronitrobenzene$ $\xrightarrow{NaOH, 443 \ K} X$ $\xrightarrow{(i) Sn + HCl (ii) NaNO_2 / HCl, 0-5^{\circ}C} Y$ $\xrightarrow{H_3O^{+}, 10^{\circ}C} Z$

In the reaction shown,the electrophile involved is:

What is the major product formed when phenol is treated with chloroform and aqueous alkali?

Assertion $(A)$: $pK_a$ of phenol is $4.19$ and that of benzoic acid is $10$.
Reason $(R)$: Phenoxide ion is stabilised by non-equivalent resonance structures,whereas benzoate ion is stabilised by two equivalent resonance structures.
The correct option among the following is:

The reaction of phenol with chloroform and sodium hydroxide to give $o$-hydroxybenzaldehyde involves the formation of which intermediate?

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