What is the compound formed when sodium nitrite solution is added to butylamine and concentrated hydrochloric acid respectively ?
Both of them formed NO2 as i know, but what is the other compound formed?
or what was remained in the solution ?
It depends on the particular butylamine (isomer...is the amine primary or secondary)..see the last paragraph of
http://chemistry2.csudh.edu/rpendarvis/aminrxn.html
ok , thanks a lot!!!
To determine the compounds formed when sodium nitrite solution is added to butylamine and concentrated hydrochloric acid, we need to consider the reaction that takes place.
1. Reaction with sodium nitrite solution:
Sodium nitrite (NaNO2) reacts with butylamine (C4H9NH2) to form a specific compound. The reaction involves the replacement of the hydrogen in butylamine with a nitrite group (-NO2).
The balanced chemical equation for this reaction can be represented as follows:
C4H9NH2 + NaNO2 → C4H9N2NO + NaOH
The compound formed is called N-nitroso-N-butylamine (C4H9N2NO) and sodium hydroxide (NaOH) is also produced as a byproduct.
2. Reaction with concentrated hydrochloric acid:
N-nitroso-N-butylamine (C4H9N2NO) can react with concentrated hydrochloric acid (HCl) to form a new compound. In this reaction, the nitroso group (-NO) is replaced by a chloride ion (-Cl).
The balanced chemical equation for this reaction can be represented as follows:
C4H9N2NO + 2HCl → C4H9NCl2 + H2O + NO
The compound formed is called N,N'-dichlorobutylamine (C4H9NCl2) in addition to water (H2O) and nitric oxide (NO) as byproducts.
Therefore, the compounds formed when sodium nitrite solution is added to butylamine and concentrated hydrochloric acid are N-nitroso-N-butylamine (C4H9N2NO) and N,N'-dichlorobutylamine (C4H9NCl2) respectively.