Advertisements
Advertisements
प्रश्न
Illustrate the following reactions giving suitable example in each case
Acetylation of amines
Advertisements
उत्तर
Acetylation of amines: Acetylation is the process of introducing an acetyl group into a molecule. Aliphatic and aromatic primary and secondary amines undergo the acetylation reaction by nucleophilic substitution when treated with acid chlorides, anhydrides or esters. This reaction involves the replacement of the hydrogen atom of the −NH2 or > NH group by the acetyl group, which, in turn, leads to the production of amides. To shift the equilibrium to the right-hand side, the HCl formed during the reaction is removed as soon as it is formed. This reaction is carried out in the presence of a base (such as pyridine), which is stronger than the amine.
For example, the reaction of benzenamine with acetyl chloride in the presence of pyridine yields N-phenylethanamide.
\[{C_6 H_5 - {NH}_2}_{Benzenamine} + {{CH}_3 - CO - Cl}_{Acetyl chloride} \to^{Pyridine} {C_6 H_5 - NH - CO - {CH}_3}_{{N - Phenylethanamide}_\left( Soluble in alkali \right)} + HCl\]
APPEARS IN
संबंधित प्रश्न
What is the action of the following reagents on aniline?
Bromine water
Account for the following:
Although the amino group is o, p-directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline.
What is the action of acetic anhydride on diethylamine?
What is the action of the following reagents on aniline?
Hot and conc. sulphuric acid
A solution contains 1 g mol. each of p-toluene diazonium chloride and p-nitrophenyl diazonium chloride. To this 1 g mol. of alkaline solution of phenol is added. Predict the major product. Explain your answer.
Assertion: N, N-Diethylbenzene sulphonamide is insoluble in alkali.
Reason: Sulphonyl group attached to nitrogen atom is strong electron-withdrawing group.
Give reasons for the following observation:
Aniline is acetylated before nitration reaction.
Give reasons for the following observation:
Aniline does not react with methyl chloride in the presence of anhydrous AlCl3 catalyst.
How can the activating effect of the −NH2 group in aniline be controlled?
Aniline does not give Friedel-Crafts reaction. Give a reason.
