Question

What is the action of lithium aluminum hydride in the presence of ether on the following compounds? 
a. Nitroethane 
b. 2-Methyl-1-nitropropane

Answer 1

a. Nitroethane is reduced to ethylamine by the action of lithium aluminium hydride in the presence of ether.
\[\ce{\underset{\text{Nitroethane}}{CH_3CH_2NO_2} ->[LiAlH_4][Ether]  \underset{\text{Ethylamine}}{CH_3CH_2NH_2}}\]

b. 2 - Methyl - 1 - nitropropane is reduced to 2 - methylpropan - 1 - amine by the action of lithium aluminium hydride in the presence of ether.
\[\begin{array}{cc}
\ce{CH3}\phantom{.........................}\ce{CH3}\phantom{......}\\
|\phantom{.............................}|\phantom{......}\\
\ce{\underset{\text{2-Methyl-1-nitropropane}}{CH3-CH-CH2-NO2} ->[LiAlH4][Ether] \underset{\text{2-Methylprpan-1-amine}}{CH3-CH-CH2-NH2}}\\
\end{array}\]