Assertion: Acetamide reacts with Br₂ in presence of methanolic CH₃ONa to form methyl N-methylcarbamate. Reason: Methyl isocyanate is formed as an intermediate which reacts with methanol.

Which compound gives a product without nitrogen with HNO₂?

Positive carbylamine test is given by: I. N,N-dimethylaniline II. 2,4-dimethylaniline III. N-methyl-o-methylaniline IV. p-methylbenzylamine

Nitration of aniline in ice-cold condition gives:

Identify compound A:

Positive carbylamine test is shown by:

Source of nitrogen in Gabriel synthesis is:

Which gives amine on heating with amide?

Which gives primary amine on reduction?

Final product after diazotization and Balz–Schiemann reaction:

Two amines: one more basic and one less basic than NH₃:

Identify compound A:

Incorrect reaction among the following:

Product D formed from aniline reactions is:

Reaction RCHO + NH₂NH₂ → RCH=N–NH₂ is:

Benzamide + POCl₃ gives:

m-Nitroaniline formation occurs because:

Maximum CH₃I reacting with methylamine:

Amine not liberating N₂ with HNO₂:

Phenyl isocyanide is formed from:

Reaction product C is:

In the Sandmeyer reaction, an aromatic diazonium salt reacts with CuCl/HCl to form:

Benzene diazonium chloride on heating with water gives:

Pungent smell due to formation of:

Secondary amine is:

Cannot be prepared by Gabriel synthesis:

Incorrect statement:

Identify compound A:

In the carbylamine reaction, aniline reacts with chloroform and alcoholic KOH to form:

Basic character in chlorobenzene is: