- ✓$(CH_3)_2NH$
- B$CH_3NH_2$
- C$(CH_3)_3N$
- D$C_6H_5NH_2$
$\left(\mathrm{CH}_{3}\right)_{2} \bar {N}H>\mathrm{CH}_{3} \bar {N}H_{2}>$$\left(CH_{3}\right)_{3} \bar {N}>\mathrm{C}_{6}\ mathrm{H}_{5} \bar {N} \mathrm{H}_{2}$
As we know $\mathrm{pk}_{\mathrm{b}}=-\log \mathrm{K}_{\mathrm{b}}$
So $ (CH_3) _{2}$ $NH$ will have smallest $pK_{b}$ value. In the case of phenylamine, $N$ is attached to $s p^{2}$ hybridised carbon, hence it has highest $p K_{b}$ and least basic strength.
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Gas $(A)$ burns with a blue flame and is oxidised to gas $(B)$
Gas $(A) + Cl_2 \to (D)$
$A, B$ and $D$ are
$\begin{array}{*{20}{c}}
{\,\,O} \\
{\,\,||} \\
{C{H_3} - C - OH}
\end{array}$ $\xrightarrow[\Delta ]{{N{H_3}}}A\,\xrightarrow[\Delta ]{{{P_2}{O_5}'}}B$
$X$ is