Distinguish Tests Between Nitroalkanes And Alkyl Nitrities

Distinguish Tests Between Nitroalkanes And Alkyl Nitrities

1. Nitroalkane on reduction with produce amines while alkyl nitrites produce alcohols and

2. Nitroalkanes do not get hydrolysed in basic conditions while nitrites produce alcohols

Illustration 3. Nitrobenzene, but not benzene, is used as a solvent for Friedel – Craft alkylation of bromobenzene. Why?

Solution: If during alkylation of bromobenzene benzene is used as a solvent, alkylation of benzene will take place because benzene is more reactive in electrophilic substitution reactions than bromobenzene. On the other hand, nitrobenzene does not under goes Friedel – Craft reaction and can be used as solvent for alkylation of bromobenzene.

Illustration 4. An organic compound (A) C2H5NO2 gave on hydrolysis an alcohol (B) of molecular mass 46 and an acid. On reduction it gave the same alcohol (B) and NH3 gas. What is (A)?

Solution: (i) Molecular formula of A suggests that it is either.

(ii) Since (A) undergoes hydrolysis to give (B) and an acid, it is ethyl nitrite C₂H₅O⎯N =O.

Reaction:

Molecular mass of (B) is 46

Illustration 5. Convert benzene to m-nitro aniline.

Solution:

AMINES

Amines are derivatives of They are represented by general formula

General methods of preparation:

1. Alkylation of with alkyl halides or alcohols:

This reaction is called Hofmann’s Method. Exhaustive methylation is as follows:

Aryl halides show low reactivity for this reaction.

2. By the action of ammonia on alcohol:

This method yields a mixture of amines and salts which are separated from each other by means of Hinsberg method, Hofmann method and fractional of distillation. However, amines can be prepared in good yield by using excess of ammonia.

Illustration 6. Which of the following compounds exists as non resolvable racemic mixture?

(A)		(B)
(C)		(D)

Solution: 3° amine contain all three alkyl groups different give non resolvable mixture.

Hence (D) is correct

Illustration 7. The statement not correct for piperidine is

(A) it is heterocyclic compound (B) it is homocyclic compound

(C) it is more basic than pyridine (D) it is more basic than aniline

Solution: Piperidine is heterocyclic compound.

Hence (A), (C) and (D) are correct and (B) is wrong.

Illustration 8. N-Ethyl-N-methyl propanamine does not show optical activity why?

Solution:

The energy difference between I and II is very small, these are rapidly interverted into each other hence are non resolvable. This is called nitrogen inversion or umbrella effect.

Methods yielding only  AMINES:

1. By reduction of nitroalkanes:

2. Mendius reduction of alkyl cyanides:

Ethylamine

Reduction of alkyl isocyanides with amines eg.

3. By reduction of amides and oximes:

4. By Hofmann bromamide reaction:

5. By Gabriel phthalimide reaction:

Potassium phthalimide formed after reaction of phthalimide with KOH, on heating with alkyl halide gives N – alkyl phthalimide. This on hydrolysis with 20% hydrochloric acid under pressure gives 1° amine.

Illustration 9.

The product B is

(A)		(B)	PhCH2NH2
(C)		(D)

Solution: Gabrial pthalamide reaction.

Hence (B) is correct.

Illustration 10. Which of the following amines can not be prepared by Gabriel Phthalimide. synthesis?

1. CH₃CH₂NH₂
2. (C2H₅)₂NH
3. CH₃CH₂CH₂NH₂

Solution: (C₂H₅)₂NH is a secondary amine and hence can not be prepared by this method since this method can be used to get only primary amines.

6. By action of chloramine on Grignard’s reagent:

7. By decarboxylation of amino acids:

8. By Wurtz method:

9. Schmidt reaction:

Methods giving 2°  amines only:

1. By reduction of alkyl isocyanide with sodium and ethanol

2. By heating an alcoholic solution of 1°  amine with alkyl halide

3. By hydrolysis of p – nitroso dialkyl aniline with boiling alkali