A compound X with formula C_4H_{11}N reacts w | vedclass

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(C) $1$. The reaction of an aliphatic primary amine with $HNO_2$ gives an alcohol with the evolution of $N_2$ gas.$2$. The compound $Y$ reacts with $I

Vedclass · Accepted Answer

$CH_3-CH(NH_2)-CH_2-CH_3$

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(C) $1$. The reaction of an aliphatic primary amine with $HNO_2$ gives an alcohol with the evolution of $N_2$ gas.$2$. The compound $Y$ reacts with $I_2 / NaOH$ (iodoform test) to give a sodium salt of a carboxylic acid,which indicates that $Y$ must be a secondary alcohol containing the $CH_3-CH(OH)-$ group.$3$. Among the given options,$CH_3-CH(NH_2)-CH_2-CH_3$ (butan$-2-$amine) reacts with $HNO_2$ to form butan$-2-$ol $(CH_3-CH(OH)-CH_2-CH_3)$.$4$. Butan$-2-$ol contains the $CH_3-CH(OH)-$ group,which gives a positive iodoform test with $I_2 / NaOH$ to produce sodium propionate and iodoform $(CHI_3)$.

$A$ compound $X$ with formula $C_4H_{11}N$ reacts with $HNO_2$ to liberate nitrogen and produce compound $Y$. The compound $Y$ on treatment with $I_2 / NaOH$ produces a sodium salt of a carboxylic acid. The compound $X$ is

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