An organic compound containing C, H, N and O. On analysis gave , C=40.57% , H=8.51% ,N=23.65% and the rest is oxygen.When treated with bromine and caustic potash, it gave a colourless gas(G) like ammmonia which produced white fumes with HCl gas.This gas when treated with HNO2 gave an alcohol and nitrogen. The molecular weight of the substance was found to be 59.The structural formula of the compound is.
An organic compound containing C, H, N and O. On analysis gave , C=40.57% , H=8.51% ,N=23.65% and the rest is oxygen.When treated with bromine and caustic potash, it gave a colourless gas(G) like ammmonia which produced white fumes with HCl gas.This gas when treated with HNO2 gave an alcohol and nitrogen. The molecular weight of the substance was found to be 59.The structural formula of the compound is.
Using the above provided information, we calculate the mass of carbon, hydrogen, oxygen and nitrogen present in the given compound.
Carbon=23.94 ~ 2 atoms of carbon
Hydrogen= 5.02~ 5 atoms of hydrogen
Oxygen= 16.09~ 1 atom of oxygen
Nitrogen= 13.96~ 1 atom of nitrogen
Therefore, the molecular formula is C2H5NO.
Now, in the second part of the question, it is given that the gas liberated when treated with HNO2 gives an alcohol and N2 gas. Since the compound has N, the above reaction is used to test for aliphatic primary amines. Aliphatic primary amines are the only amines that give out N2ÂÂÂ gas.
Thus, R-NH2 + HNO2→ R-OH + N2+ H2O
From the first part of the reaction, we find that amides when treated with Br2 and KOH give amines.
Thus, R-CO-NH2 + Br2 + KOH → R-NH2 + K2CO3+ KBr+ H2O
Using the formula calculated (C2H5NO),
Amide= CH3-CO-NH2
∴Amine= CH3NH2
And Alcohol= CH3-OH
Using the above provided information, we calculate the mass of carbon, hydrogen, oxygen and nitrogen present in the given compound.
Carbon=23.94 ~ 2 atoms of carbon
Hydrogen= 5.02~ 5 atoms of hydrogen
Oxygen= 16.09~ 1 atom of oxygen
Nitrogen= 13.96~ 1 atom of nitrogen
Therefore, the molecular formula is C2H5NO.
Now, in the second part of the question, it is given that the gas liberated when treated with HNO2 gives an alcohol and N2 gas. Since the compound has N, the above reaction is used to test for aliphatic primary amines. Aliphatic primary amines are the only amines that give out N2ÂÂÂ gas.
Thus, R-NH2 + HNO2→ R-OH + N2+ H2O
From the first part of the reaction, we find that amides when treated with Br2 and KOH give amines.
Thus, R-CO-NH2 + Br2 + KOH → R-NH2 + K2CO3+ KBr+ H2O
Using the formula calculated (C2H5NO),
Amide= CH3-CO-NH2
∴Amine= CH3NH2
And Alcohol= CH3-OH
