Amines and nucelophiles and their synethesis

Amine - derivative of amino acid which 1 or more of the hydrogens in the molecule has been replaced with ALKYL or ARYL groups

1' Amine                               2' Amine                          3' Amine

• Alkyl - methyl amine, ethyl amine
• Aryl - NH2 directly to the benzene ring 

Physical properties

• Higher boiling point due to hydrogen bonding
• As the carbon chain gets longer--> increased van der waals forces--> higher bp

Solubility

• lower mass compunds soluble in water due to the H bonding with solvent
• Solvent decreases as the molecules gets heavier
• Soluble in Organic solvents

Amines can accept a H+-----> BRONSTED LOWRY BASES

RNH2 + H+ -------> RNH3+ PROTON ACCEPTOR

INDUCTIVE- gives electron density (pushes it onto the N atom)

ELECTRON WITHDRAWING - Takes electron density (pulls electron density away from N atom)

Strength - greater the electron density on N the better it can pick up an proton.

BAD base

• Electron withdrawing substituents (benzene rings)---> decreases basicity as the electron density on N is decreased and the lone pair is less effective

GOOD base (INDUCTIVE EFFECT)

• Electron releasing substituents (CH3 groups) increases basicity as the electron density is increased and the lone pair is more effective.

NUCLEOPHILIC SUBSTITUTION - Due to their lone pair

NH3      +    Rx --------> [RHN3]+   +   X-

[RNH3]  +     NH3 -------> RN2       +  [NH4]

Reagent - Haloalkane

Product - substitued amines

Ammonia is a nucleophile, when ammonia reacts with a haloalkane a primary amine is produced

Amines - nucleophiles- attack haloalkanes to produce 2' amines----. this is a nucleophile too and can react further to produce 3' amine -----> ionice Quatenary salt 4' amine

• C2H5NH2 + C2H5Br ----->  HBr + (C2H5)2 NH   DIETHYLAMINE 2'amine
• (C2H5)2 NH + C2H5Br --------> HBr + (C2H5)3 N      TRIETHYLAMINE 3'amine
• (C2H5)3 N   + C2H5Br --------> (C2H5)4 NA+Br-  TETRAETHYLAMMONIUM BROMIDE 4' amine

Creating Amines this way is not useful when a specific product is needed

Synthetic route produces 1',2',3',4' which is then separated by fractional distillation

BUT a larger excess of ammonium------> greater % of 1' amines produced.

Primary amines can be prepared from nitriles and haloalkane in a 2 step process

Step1                                                                                                                                                                                           Haloalkanes react with the cyanide ion in aqeous ethanol a NUCLEOPHILIC REACTION occurs

• RBr + CN-  ------------> R - C=N  + Br-
• 

Step2                                                                                                                                                                                                        Nitrile produc t can be reduced using catalytic hydrogenation (nickel and hydrogen)

• R - C =N + 2H2 -------------> R-CH2NH2

Purer product arises as 1' amine can only be formed

Increasing the chain length ---------> Stronger base

Adding Benzene rings ----------------> weaker base

Step2                                                                                                                                                                                                        Nitrile produc t can be reduced using catalytic hydrogenation (nickel and hydrogen)

• R - C =N + 2H2 -------------> R-CH2NH2

Purer product arises as 1' amine can only be formed

Increasing the chain length ---------> Stronger base

Adding Benzene rings ----------------> weaker base

Water - Amine dissolves in water produces weak alkaline solutions

CH3NH2(g)    +     H2O(l)     <---------->   CH3NH3+ (aq)    +    OH-(aq)

Acid and amines react with acids ----------> produces salts

C6H5NH2   +   H+    +    Cl-    ---------------> C6H5NH3  +  Cl-

                                                                         Soluble       Phenylammonium chloride

- phenyl- amine is insoluble but it will dissolve in excess HCl as it forms the soluble ionic salt

C6H5NH3+     +      Cl-     +     NaOH --------> C6H5NH2 + NaCl  + H20

                                                                     Insoluble amine and it forms a layer

If a strong base like NaOH is added it removes the H+ from the salt and makes the insoluble amine

Step 1

Benzene is reacted with a mixture of conc HNO2  +   H2SO4 -------> NITROBENZENE

Step 2

Nitrobenzene is reduced to Phenylamine using tin and Hcl as a reducing agent. They remove the oxygen atoms of the NO2 groups replacing them with hyderogen atoms