Which of the following compounds is not solub | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) For a nitroalkane to be soluble in $NaOH$,it must possess at least one $\alpha$-hydrogen atom. The $\alpha$-hydrogen is acidic due to the electron

Vedclass · Accepted Answer

$(CH_3)_3C-NO_2$

Vedclass · Answer

(C) For a nitroalkane to be soluble in $NaOH$,it must possess at least one $\alpha$-hydrogen atom. The $\alpha$-hydrogen is acidic due to the electron-withdrawing effect of the $-NO_2$ group,allowing it to be removed by a base like $NaOH$ to form a water-soluble salt. $1.$ $CH_3-CH_2-NO_2$ (Nitroethane) has two $\alpha$-hydrogens. $2.$ $(CH_3)_2CH-NO_2$ ($2$-Nitropropane) has one $\alpha$-hydrogen. $3.$ $(CH_3)_3C-NO_2$ ($2$-Methyl$-2-$nitropropane) has no $\alpha$-hydrogen. $4.$ $Ph-CH_2-NO_2$ (Phenylnitromethane) has two $\alpha$-hydrogens. Since $(CH_3)_3C-NO_2$ lacks an $\alpha$-hydrogen,it cannot form a salt with $NaOH$ and is therefore insoluble.

Which of the following compounds is not soluble in $NaOH$?

Similar Questions

Which one of the following compounds undergoes bromination of its aromatic ring (electrophilic aromatic substitution) at the fastest rate?

Assertion : Acetamide reacts with $Br_2$ in presence of methanolic $CH_3ONa$ to form methyl $N$-methylcarbamate.
Reason : Methyl isocyanate is formed as an intermediate which reacts with methanol to form methyl $N$-methylcarbamate.

Choose the correct order of basic strength.

Compound $III$ is
$\xrightarrow[{ii)\,CuCl/HCl}]{{i)\,HNO_2\,(0 - 5\,^oC)}}I$ $\xrightarrow{{Sn/HCl}}II$ $\xrightarrow[{ii)\,H_3PO_2 + H_2O}]{{i)HNO_2\,(0 - 5\,^oC)\,}}III$

Reaction of propanamide with $Br_2 / KOH \ (aq)$ produces :

Vedclass Test Series

Exam Paper Generator

Online Exam Module