The conversion of propan-1-ol to n-butylamine | vedclass

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Question

(A) The conversion of propan$-1-$ol $(CH_3CH_2CH_2OH)$ to $n-$butylamine $(CH_3CH_2CH_2CH_2NH_2)$ requires increasing the carbon chain length by one c

Vedclass · Accepted Answer

$(i) SOCl_2, (ii) KCN, (iii) H_2 / Ni 	ext{ or } Na(Hg) / C_2H_5OH$

Vedclass · Answer

(A) The conversion of propan$-1-$ol $(CH_3CH_2CH_2OH)$ to $n-$butylamine $(CH_3CH_2CH_2CH_2NH_2)$ requires increasing the carbon chain length by one carbon atom.Step $1$: Propan$-1-$ol reacts with $SOCl_2$ to form $1-$chloropropane $(CH_3CH_2CH_2Cl)$.Step $2$: $1-$chloropropane reacts with $KCN$ (nucleophilic substitution) to form butanenitrile $(CH_3CH_2CH_2CN)$.Step $3$: Butanenitrile is reduced using $H_2/Ni$ or $Na(Hg)/C_2H_5OH$ (Mendius reduction) to form $n-$butylamine $(CH_3CH_2CH_2CH_2NH_2)$.

The conversion of propan$-1-$ol to $n-$butylamine involves the sequential addition of reagents. The correct sequential order of reagents is:

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