Question

Reduction of Nitroalkanes yields
(A) Acid
(B) Alcohol
(C) Amine
(D) Diazo Compounds

Answer 1

Hint: When there is said to be reduction of any compound then it is compulsory that hydrogen must be added to that compound or oxygen is removed from that compound because when reduction process takes place any of the process is definitely going to be done or if no hydrogen and oxygen process takes place then it must be changes in their oxidation state of that compound. So in the above question we are going to do reduction process in the given Nitroalkanes.

Complete Step by Step Solution:
As per we talked about when any reduction is take place than hydrogen must be added to that compound from playing this to the following reaction below we get,
$C{H_3} - C{H_2} - N{O_2} + 6[H]\xrightarrow{{Sn/HCl}}C{H_3} - C{H_2} - N{H_2} + 2{H_2}O$.

Now we get data from the above equation that, $C{H_3} - C{H_2} - N{O_2}$ which is the Nitro alkane which is reacted with hydrogen to take place the reduction. As a result we get from the reactants in the product that $C{H_3} - C{H_2} - N{H_2}$ Amine is formed as in the product with water. Therefore, as we reduce Nitroalkanes yields we get the product/answer as Amine.
Hence, the correct option is (C).

Note: When we see the whole reaction deeply as this reaction is a reduction reaction then we see that here in the Nitro alkane [ $C{H_3} - C{H_2} - N{O_2}$ ] the number of hydrogen atom in this compound is $5$ whereas in the product we get Ethylamine [ $C{H_3} - C{H_2} - N{H_2}$ ] the number of hydrogen atom present in this compound is $7$ . As we see an increase in hydrogen atom from which we can also say that this is a reduction reaction.