Aniline does not undergo Friedel-Crafts reaction. Why?

The amines are basic due to the presence of lone pair of electrons on the nitrogen atom. The reagent AlCl 3 is a Lewis acid, and aniline acts as a Lewis base, so aniline reacts with AlCl 3 and forms a salt. In this reaction, the nitrogen atom acquires a positive charge, therefore it acts as a deactivating group towards electrophilic substitution reaction. Because Friedel-Crafts reaction is an electrophilic substitution reaction. Thus, aniline does not undergo a Friedel-Craft reaction. C 6 H 5 - NH 2 Aniline + AlCl 3 → C 6 H 5 NH 2 + AlCl 3 -

Give the structures of A, B and C in the following reactions: CH 3 Br → KCN A → LiAlH 4 B → 273 K HNO 3 C

Methyl bromide on reaction with potassium cyanide gives cyanomethane which on reduction gives ethanamine. The resulting product on reaction with cold nitrous acid gives ethanol CH 3 Br → KCN CH 3 CN Cyanomethane A → LiAlH 4 CH 3 CH 2 NH 2 Ethanamine B → 0 ∘ C HNO 2 CH 3 CH 2 OH Ethanol C

An aromatic compound ‘A’ on treatment with aqueous ammonia and heating forms compound ‘B’ which on heating with Br, and KOH forms a compound ‘C’ of molecular formula C 6 H 7 N . Write the structures and IUPAC names of compounds A, B and C.

Hoffmann developed a method for the preparation of primary amines by treating an amide with bromine in an aqueous or ethanolic solution of sodium hydroxide. In this degradation reaction, migration of an alkyl or aryl group takes place from carbonyl carbon of the amide to the nitrogen atom. The amine so formed contains one carbon less than that present in the amide. So, here C 6 H 7 N is formed by heating the compound amide(B) on Br and KOH. So, C is an amine and the only amine having molecular formula aniline C 6 H 7 N and the molecular formula of amide is benzamide C 6 H 5 CONH 2 . Therefore, benzamide is formed by heating compound ‘A’ with aqueous ammonia. Therefore, compound ‘A’ must be benzoic acid. Here is the reaction between the following compound:

How do you convert the following? Aniline to phenyl isocyanide.

Aliphatic and aromatic primary amines on heating with chloroform and ethanolic potassium hydroxide form isocyanides or carbylamines which are foul smelling substances. Secondary and tertiary amines do not show this reaction. This reaction is known as carbylamine reaction or isocyanide test and is used as a test for primary amines.

How do you convert the following? Nitromethane to dimethylamine

Nitromethane is converted to aminomethane by reducing it with tin along with concentrated hydrochloric acid, and then it is converted to methyl isocyanide in the presence of chloroform and potassium hydroxide.The methyl isocyanide so formed is then treated with sodium metal in presence of ethanol when it is reduced to form dimethylamine, as per the given reaction below:

Arrange the following in ‘decreasing order of the pK b values : C 2 H 5 NH 2 , C 6 H 5 NHCH 3 , C 2 H 5 2 NH and C 6 H 5 NH 2 .

The decreasing order of basic strength of the above amines and ammonia follows the following order: C 6 H 5 NH 2 < C 6 H 5 - NHCH 3 < C 2 H 5 NH 2 < C 2 H 5 2 NH . However, pK b is inversely proportional to the basic character. Therefore, decreasing order of the pK b values is C 6 H 5 NH 2 > C 6 H 5 NHCH 3 > C 2 H 5 NH 2 > C 2 H 5 2 NH .

Write the diazotisation reaction of aniline.

The process of conversion of a primary aromatic amino compound into a diazonium salt, is known as diazotisation. The diazotisation of aniline is carried out by adding an aqueous solution of sodium nitrite in excess of HCl at a temperature below 5 ° C . The benzene diazonium chloride is formed as a result.

Give a chemical test for primary amines.

A chemical test that is most commonly used for the identification of primary, secondary and tertiary amines is called the Hinsberg test. Getting a primary amine to react with a Hinsberg reagent usually leads to an amide being formed, called N-ethylbenzenesulphonyl amide, whereas the reagent used is called benzene sulfonyl chloride. This is strongly acidic as the hydrogen that is attached as a part of the nitrogen compound. Thus, this solution is also soluble in alkali.

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Wlhat happens when aniline reacts with bromine water at room temperature?

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