In the following sequence of reactions,the co | vedclass

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(C) Step $1$: Reaction of isopropylamine with $HNO_2$ gives propan-$2$-ol $(A)$.$CH_3-CH(NH_2)-CH_3 + HNO_2 \rightarrow CH_3-CH(OH)-CH_3 + N_2 + H_2O$

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$2$-methylpropan-$2$-ol

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(C) Step $1$: Reaction of isopropylamine with $HNO_2$ gives propan-$2$-ol $(A)$.$CH_3-CH(NH_2)-CH_3 + HNO_2 \rightarrow CH_3-CH(OH)-CH_3 + N_2 + H_2O$Step $2$: Oxidation of propan-$2$-ol $(A)$ gives acetone $(B)$.$CH_3-CH(OH)-CH_3 \xrightarrow{[O]} CH_3-CO-CH_3$Step $3$: Reaction of acetone $(B)$ with methylmagnesium iodide $(CH_3MgI)$ followed by hydrolysis gives $2$-methylpropan-$2$-ol $(C)$.$CH_3-CO-CH_3 + CH_3MgI$ $\rightarrow (CH_3)_3C-OMgI$ $\xrightarrow{H_2O/H^+} (CH_3)_3C-OH$Thus,the final product $C$ is $2$-methylpropan-$2$-ol.

In the following sequence of reactions,the compound $C$ formed will be:
$CH_3-CH(NH_2)-CH_3$ $\xrightarrow{HNO_2} A$ $\xrightarrow{\text{Oxidation}} B$ $\xrightarrow[(ii) H^+/H_2O]{(i) CH_3MgI} C$

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In the following sequence of reactions,the compound $C$ formed will be:$CH_3-CH(NH_2)-CH_3$ $\xrightarrow{HNO_2} A$ $\xrightarrow{\text{Oxidation}} B$ $\xrightarrow[(ii) H^+/H_2O]{(i) CH_3MgI} C$