Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

How is nitroethane converted into
i) Ethylamine ii) N-ethyl hydroxylamine iii) Acetic acid ?

seo-qna
SearchIcon
Answer
VerifiedVerified
460.5k+ views
Hint: Recall the various preparation methods of amines from nitro compounds. Nitroethane is a nitro compound and nitro compounds can be converted into amines by passing hydrogen gas in the presence of Ni/Pd/Pt. This is called reduction reaction of nitro compounds.

Complete step by step answer:
The required conversion reactions are as follows:
(i) Conversion of nitroethane into ethylamine
Nitroethane can be converted into corresponding ethylamine by doing the reduction reaction of nitroethane. By passing hydrogen gas in the presence of finely divided nickel, platinum or palladium, nitroethane can be reduced to ethylamine.
$C{H_3} - C{H_2} - N{O_2}\xrightarrow[\Delta ]{{{H_2}/Ni}}C{H_3} - C{H_2} - N{H_2}$

(ii) Conversion of nitroethane into N-ethyl hydroxylamine
When nitroethane is reduced with zinc metal in the presence of ammonium chloride, it forms N-ethyl hydroxylamine. This reaction is carried out in the neutral medium. Nitro compounds are basic, so we need to add acid in the reaction to make the reaction medium neutral.
$C{H_3} - C{H_2} - N{O_2} + 4[H]\xrightarrow[{N{H_4}Cl}]{{Zn}}C{H_3} - C{H_2} - NH - OH + {H_2}O$

(iii) Conversion of nitroethane into acetic acid
To carry out this conversion, first we need to do reduction of nitroethane to form ethylamine. Then, ethylamine is converted into ethanenitrile. After this, ethanenitrile is converted into acetic acid by reflux hydrolysis with hydrochloric acid. Therefore, this conversion is carried out in three steps as follows.
$C{H_3} - C{H_2} - N{O_2}\xrightarrow[\Delta ]{{{H_2}/Ni}}C{H_3} - C{H_2} - N{H_2}\xrightarrow{{NaN{H_2}}}C{H_3} - C \equiv N\xrightarrow[{{H_2}O}]{{HCl}}C{H_3} - COOH$

Note: Nitro Compounds can also be converted into corresponding nitroalkanes by their reduction with metals in acidic medium. This reaction can be written as:
$C{H_3} - C{H_2} - N{O_2}\xrightarrow[{or,Fe/HCl}]{{Sn/HCl}}C{H_3} - C{H_2} - N{H_2}$

Usually, reduction with iron scrap and hydrochloric acid is preferred because $FeC{l_2}$ is formed which gets hydrolysed to release hydrochloric acid during the process of reaction. Thus, only a small amount of hydrochloric acid is required to initiate the reaction.