0ne mole of hydrazine (N2H4) loses 10 moles of electrons in a reaction to form a new compound X. Assuming that all the nitrogen atoms in hydrazine appear in the new compound, what is the oxidation state of nitrogen in X? (Note - There is no change in the oxidation state of hydrogen in the reaction).
-1
-3
+3
+5
C.
+3
Given, 1 mol of H2N - NH2 (hydrazine), it loses 10 moles of electrons to form a new compound that contains both the N-atoms with same oxidation number means:
N2H4 → 10e- + X (product)
Since, oxidation number of both the N-atoms are same. New total oxidation number of new compound X = 4 - 10 + x = 0
x = +6
Therefore, each N-atom has oxidation state = +3
Which one of the following is used as a test for aliphatic primary amines?
Tollen's test
Fehling's test
lsocyanide test
Azo dye test
When methanamine is treated with benzoyl chloride, the major product is
N-phenylethanamide
N-methylbenzamide
benzanilide
acetophenone
Phenyl isocyanide is prepared from aniline by
Carbylamine reaction
Rosenmund's reaction
Koble's reaction
Reimer-Tiemann reaction
Gabriel's phthalimide synthesis can be used to prepare
ethanamine
N-methylmethanamine
benzene amine
N, N-dimethylmethanamine
Select the compound which on treatment with nitrous acid liberates nitrogen.
Nitroethane
Triethylamine
Diethylamine
Ethylamine
Positive carbylamine test is shown by
N, N-dimethylaniline
triethylamine
N-methylaniline
p-methylbenzylamine
Ethanoic acid on heating with ammonia forms compound A which on treatment with bromine and sodium hydroxide gives compound B. Compound B on treatment with NaNO3/dil.HCl gives compound C. The compounds A, B and C respectively are
ethanamide, methanamine, methanol
propanamide, ethanamine, ethanol
N-ethylpropanamide, methaneisonitrile, methanamine
ethanamine, bromoethane, ethanediazoniumchloride
n-butylamine (l), diethylamine (II) and N, N-dimethylethylamine (III) have the same molar mass. The increasing order of their boiling point is
III < II < I
I < II < III
II < III < I
II < I < III